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NPTEL Syllabus
Flight dynamics I - Airplane
performance - Web course
COURSE OUTLINE
FLIGHT DYNAMICS - I - AIRPLANE PERFORMANCE
1. Introduction.
Definition and subdivisions of flight dynamics.
Forces and moments acting on vehicles in flight. .
Equations of motion and simplification for performanceanalysis.
2. Earth's atmosphere and International StandardAtmosphere.
3. Drag polar.
Various types of drags.
Methods of estimating drag polar.
Drag polar of vehicles from low speed to hypersonicspeeds.
High lift devices.
4. Review of the variations of thrust or power output andSFC with altitude and velocity for various air breathingengines.
5. Performance analysis.
Steady level flight - Maximum speed, minimum speedand their variations with altitude.
Steady climb - Maximum rate of climb, angle of climb andtheir variations with altitude; absolute ceiling and serviceceiling.
Range and endurance - Breguet formulae; range in
constant velocity flight; effect of wind on the range.
Accelerated level flight.
NPTELhttp://nptel.iitm.ac.in
AerospaceEngineering
Pre-requisites:
The student is expected to
have undergone courses on:
1. Vectors.
2. Rigid body dynamics.
3. Aerodynamics
4. Aircraft engines.
Additional Reading:
1. Miele, A. "Flightmechanics Vol I"Addison Wesley(1962).
2. Hale, F.J., "Introduction toaircraft performance,selection and design",John Wiley (1984).
3. Anderson, Jr. J.D"Introduction to flight"
if h di i G ill
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Manoeuvres - flight in vertical plane (loop); turn(minimumradius of turn and maximum rate of turn and theirvariations with the altitude).
V - n diagram.
Flight limitations.
Estimations of take-off distance and landing distance.
6. Examples of estimation of the drag polar andperformance of a piston engined and a jet enginedairplane.
COURSE DETAIL
A Web course shall contain 40 or more 1 hour lectureequivalents.
S.No Topics No.ofHours
1 Chapter 1 : Introduction 3
2 Chapter 2 : Earth’s atmosphere 2
3 Chapter 3 : Drag polar 7
4 Chapter 4 : Engine characteristics 4
5 Chapter 5 :Performance analysis I –Steady level flight
4
6 Chapter 6 : Performance analysis II –Steady climb,descent and glide
3
7 Chapter 7 : Performance analysis III –Range and endurance
3
8 Chapter 8 : Performance analysis IV–Accelerated level flight and climb
1
4. Roskam, J. "Methods forestimating drag polarsof subsonic airplanes"published by author1973.
Coordinators:
Prof. E.G. Tulapurkara Department of AerospaceEngineeringIIT Madras
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9 Chapter 9 : Performance analysis V –Manoeuvres
4
10 Chapter 10 : Performance analysis VI –Take-off and landing
3
11 Performance analysis of a piston-engined airplane
3
12 Performance analysis of a subsonic jettransport
3
Total 40
References:
1. Houghton and Carruthers, "Aerodynamics forengineering students", Edward Arnold (1982).
2. McCormick B.W, "Aerodynamics, aeronautics and flight
mechanics", John Wiley (1995).
3. Anderson, Jr. J.D "Aircraft performance and design"McGraw Hill International edition (1999).
4. Eshelby , M.E."Aircraft performancetheory and practice" ,Butterworth-Heinemann, Oxford,U.K., (2001).
5. Pamadi, B., "Performance, stability, dynamics and controlof an airplane", AIAA (2004).
6. Phillips, W.F. "Mechanics of flight" 2nd Edition, John Wiley(2010).
A joint venture by IISc and IITs, funded by MHRD, Govt of India http://nptel.iitm.ac.in
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 1Introduction
(Lectures 1, 2 and 3)
Keywords: Definition and importance of flight dynamics; forces acting on an
airplane; degrees of freedom for a rigid airplane; subdivisions of flight dynamics;
simplified treatment of performance analysis; course outline.
Topics1.1 Opening remarks
1.1.1 Definition and importance of the subject
1.1.2 Recapitulation of the names of the major components of the airplane1.1.3 Approach in flight dynamics
1.1.4 Forces acting on an airplane in flight
1.1.5 Body axes system for an airplane
1.1.6 Special features of flight dynamics
1.2 A note on gravitational force
1.2.1 Flat earth and spherical earth models
1.3 Frames of reference
1.3.1 Frame of reference attached to earth
1.4 Equilibrium of airplane
1.5 Number of equations of motion for airplane in flight
1.5.1 Degrees of freedom
1.5.2 Degrees of freedom for a rigid airplane
1.6 Subdivisions of flight dynamics
1.6.1 Performance analysis
1.6.2 Stability and control analysis
1.7 Additional definitions
1.7.1 Attitude of the airplane
1.7.2 Flight path
1.7.3 Angle of attack and side slip
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
1.8 Simplified treatment of performance analysis
1.9 Course outline
1.10 Background expected
References
Exercises
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
Chapter 1
Lecture 1
Introduction – 1
Topics
1.1 Opening remarks
1.1.1 Definition and importance of the subject
1.1.2 Recapitulation of the names of the major components of the airplane
1.1.3 Approach in flight dynamics
1.1.4 Forces acting on an airplane in flight
1.1.5 Body axes system for an airplane
1.1.6 Special features of flight dynamics
1.2 A note on gravitational force
1.2.1 Flat earth and spherical earth models
1.3 Frames of reference
1.3.1 Frame of reference attached to earth
1.1 Opening remarks
At the beginning of the study of any subject, it is helpful to know its definition,
scope and special features. It is also useful to know the benefits of the study of
the subject, background expected, approach, which also indicates the limitations,
and the way the subject is being developed. In this chapter these aspects are
dealt with.
1.1.1 Definition and importance of the subject
The normal operation of a civil transport airplane involves take-off, climb to
cruise altitude, cruising, descent, loiter and landing (Fig.1.1). In addition, the
airplane may also carry out glide (which is descent with power off), turning
motion in horizontal and vertical planes and other motions involving
accelerations.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 4
Fig.1.1 Typical flight path of a passenger airplane
Apart from the motion during controlled operations, an airplane may alsobe subjected to disturbances which may cause changes in its flight path and
produce rotations about its axes.
The study of these motions of the airplane – either intended by the pilot or
those following a disturbance – forms the subject of Flight dynamics.
Flight dynamics: It is a branch of dynamics dealing with the motion of an object
moving in the earth’s atmosphere.
The study of flight dynamics will enable us to (a) obtain the performance of the
airplane which is described by items like maximum speed, minimum speed,
maximum rate of climb, distance covered with a given amount of fuel, radius of
turn, take-off distance, landing distance etc., (b) estimate the loads on the
airplane, (c) estimate the power required or thrust required for desired
performance, (d) determine the stability of the airplane i.e. whether the airplane
returns to steady flight conditions after being disturbed and (e) examine the
control of the airplane.
Flight dynamics is a basic subject for an aerospace engineer and its
knowledge is essential for proper design of an airplane.
Some basic ideas regarding this subject are presented in this chapter. The topics
covered herein are listed in the beginning of this chapter.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
In this course, attention is focused on the motion of the airplane. Helicopters,
rockets and missiles are not covered.
1.1.2 Recapitulation of the names of the major components of the airplane
At this stage it may be helpful to recapitulate the names of the major
components of the airplane. Figures 1.2a, b and c show the three-view drawings
of three different airplanes.
Fig.1.2a Major components of a piston engined airplane
(Based on drawing of HANSA-3 supplied by
National Aerospace Laboratories, Bangalore, India)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
Fig.1.2b Major components of an airplane with turboprop engine
(Based on drawing of SARAS airplane supplied by
National Aerospace Laboratories, Bangalore, India)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
Fig.1.2c Major components of an airplane with jet engine
(Note: The airplane shown has many features, all of which may not be there in a
single airplane).
1.1.3 Approach
The approach used in flight mechanics is to apply Newton’s laws to the
motion of objects in flight. Let us recall these laws:
Newton’s first law states that every object at rest or in uniform motion
continues to be in that state unless acted upon by an external force.
The second law states that the force acting on a body is equal to the time
rate of change of its linear momentum.
The third law states that to every action, there is an equal and opposite
reaction.
Newton’s second law can be written as:
F = ma ; a = dV / dt ; V = dr / dt (1.1)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 8
Where F = sum of all forces acting on the body, m = mass, a = acceleration,
V = velocity, r = the position vector of the object and t = time
(Note: quantities in bold are vectors).
Acceleration is the rate of change of velocity and velocity is the rate of
change of position vector.
To prescribe the position vector, requires a co-ordinate system with
reference to which the position vector/displacement is measured.
1.1.4 Forces acting on an airplane
During the analysis of its motion the airplane will be considered as a rigid
body. The forces acting on an object in flight are:
– Gravitational force
– Aerodynamic forces and
– Propulsive force.
The gravitational force is the weight (W) of the airplane.
The aerodynamic forces and moments arise due to the motion of the
airplane relative to air. Figure 1.3 shows the aerodynamic forces viz. the drag
(D), the lift (L) and the side force (Y).
The propulsive force is the thrust(T) produced by the engine or the engine-
propeller combination.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
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Fig.1.3 Forces on an airplane
1.1.5 Body axes system of an airplane
To formulate and solve a problem in dynamics requires a system of axes.
To define such a system it is noted that an airplane is nearly symmetric, in
geometry and mass distribution, about a plane which is called the ‘Plane of
symmetry’ (Fig.1.4a). This plane is used for defining the body axes system.
Figure 1.4b shows a system of axes (OXbYbZb) fixed on the airplane which
moves with the airplane and hence is called ‘Body axes system’. The origin ‘O’ of
the body axes system is the center of gravity (c.g.) of the body which, by
assumption of symmetry, lies in the plane of symmetry. The axis OXb is taken
positive in the forward direction. The axis OZb is perpendicular to OXb in the
plane of symmetry, positive downwards. The axis OYb is perpendicular to the
plane of symmetry such that OXbYbZb is a right handed system.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
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Fig.1.4a Plane of symmetry and body axis system
Fig.1.4b The forces and moments acting on an airplane and the components of
linear and angular velocities with reference to the body axes system
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
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Figure 1.4b also shows the forces and moments acting on the airplane
and the components of linear and angular velocities. The quantity V is the
velocity vector. The quantities X, Y, Z are the components of the resultant
aerodynamic force, along OXb, OYb and OZb axes respectively. L’, M, N are the
rolling moment, pitching moment and yawing moment respectively about OXb,
OYb and OZb axes; the rolling moment is denoted by L’ to distinguish it from lift
(L). u,v,w are respectively the components, along OXb, OYb and OZb, of the
velocity vector (V). The angular velocity components are indicated by p, q, and r.
1.1.6 Special features of Flight Dynamics
The features that make flight dynamics a separate subject are:
i)During its motion an airplane in flight, can move along three axes and can
rotate about three axes. This is more complicated than the motions of machinery
and mechanisms which are restrained by kinematic constraints, or those of land
based or water based vehicles which are confined to move on a surface.
ii)The special nature of the forces, like aerodynamic forces, acting on the
airplane(Fig.1.3). The magnitude and direction of these forces change with the
orientation of the airplane, relative to its flight path.
iii)The system of aerodynamic controls used in flight (aileron, elevator, rudder).
1.2 A note on gravitational force
In the case of an airplane, the gravitational force is mainly due to the
attraction of the earth. The magnitude of the gravitational force is the weight of
the airplane (in Newtons).
W = mg; where W is the gravitational force, m is the mass of the airplane and g
is the acceleration due to gravity.
The line of action of the gravitational force is along the line joining the
centre of gravity (c.g.) of the airplane and the center of the earth. It is directed
towards the center of earth.
The magnitude of the acceleration due to gravity (g) decreases with
increase in altitude (h). It can be calculated based on its value at sea level (g o),
and using the following formula.
(g / g0) = [R / (R + h)]2 (1.2)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 12
where R is the radius of the earth,
R = 6400 km (approx.) and g0 = 9.81ms-2
However, for typical airplane flights (h < 20 km), g is generally taken to be
constant.
1.2.1 Flat earth and spherical earth models
In flight mechanics, there are two ways of dealing with the gravitational
force, namely the flat earth model and the spherical earth model.
In the flat earth model, the gravitational acceleration is taken to act
vertically downwards (Fig 1.5).
When the distance over which the flight takes place is small, the flat earth
model is adequate. Reference 1.1, chapter 4 may be referred to for details.
Fig.1.5 Flat earth model
In the spherical earth model, the gravitational force is taken to act along
the line joining the center of earth and the c.g. of the airplane. It is directed
towards the center of the earth (Fig.1.6).
The spherical earth model is used for accurate analysis of flights involving
very long distances.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
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Fig.1.6. Spherical earth model
Remarks:
In this course the flat earth model is used. This is adequate for the
following reasons.
i) The distances involved in flights with acceleration are small and the
gravitational force can be considered in the vertical direction by proper choice of
axes.
ii) In unaccelerated flights like level flight, the forces at the chosen instant of timeare considered and the distance covered etc. are obtained by integration. This
procedure is accurate as long as it is understood that the altitude means height
of the airplane above the surface of the earth and the distance is measured on a
sphere of radius equal to the sum of the radius of earth plus the altitude of
airplane.
iii) As mentioned in section 1.1.4, the forces acting on the airplane are the
gravitational force, the aerodynamic forces and the propulsive force. The first one
has been discussed in this section.The discussion on aerodynamic forces will be
covered in chapter 3 and that on propulsive force in chapter 4.
1.3 Frame of reference
A frame of reference (coordinate system) in which Newton’s laws of
motion are valid is known as a Newtonian frame of reference.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
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Since Newton’s laws deal with acceleration, a frame of reference moving
with uniform velocity with respect to a Newtonian frame is also a Newtonian
frame or inertial frame.
However, if the reference frame is rotating with an angular velocity (ω),
then, additional accelerations like centripetal acceleration {ω x (ω x r )} and
Coriolis acceleration (V x ω) will come into picture.
Reference 1.2,chapter 13 may be referred to for further details on non-Newtonian
reference frame.
1.3.1 Frame of reference attached to earth
In flight dynamics, a co-ordinate system attached to the earth is taken to
approximate a Newtonian frame (Fig.1.7).
The effects of the rotation of earth around itself and around the sun on this
approximation can be estimated as follows.
It is noted that the earth rotates around itself once per day. Hence
ω = 2 / (3600x24) = 7.27x10-5 s-1;
Since r roughly equals 6400 km; the maximum centripetal acceleration (ω2r)
equals 0.034 ms-2
.
The earth also goes around the sun and completes one orbit in approximately
365 days. Hence in this case,
ω = 2 / (365 x 3600 x 24) = 1.99x10-7s-1;
Further, in this case, the radius would be roughly the mean distance between the
sun and the earth which is 1.5x1011m. Consequently, ω2 r = 0.006 ms-2.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
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Fig.1.7 Earth fixed and body fixed co-ordinate systems
Thus, it is observed that the centripetal accelerations due to rotation of earth
about itself and around the sun are small as compared to the acceleration due to
gravity.
These rotational motions would also bring about Coriolis acceleration
(V x ω). However, its magnitude, which depends on the flight velocity, would be
much smaller than the acceleration due to gravity in flights up to Mach number of
3. Hence, the influence can be neglected.
Thus, taking a reference frame attached to the surface of the earth as a
Newtonian frame is adequate for the analysis of airplane flight. Figure 1.7 shows
such a coordinate system.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 1
Lecture 2
Introduction – 2
Topics
1.4 Equilibrium of airplane
1.5 Number of equations of motion fo r airplane in flight
1.5.1 Degrees of freedom
1.5.2 Degrees of freedom for a rigid airplane
1.6 Subdivisions of f light dynamics
1.6.1 Performance analysis
1.6.2 Stability and control analysis
1.7 Additional definitions
1.7.1 Attitude of the airplane
1.7.2 Flight path
1.7.3 Angle of attack and side slip
1.4 Equilibrium of airplaneThe above three types of forces (aerodynamic, propulsive and
gravitational) and the moments due to them govern the motion of an airplane in
flight.
If the sums of all these forces and moments are zero, then the airplane is
said to be in equilibrium and will move along a straight line with constant velocity
(see Newton's first law). If any of the forces is unbalanced, then the airplane will
have a linear acceleration in the direction of the unbalanced force. If any of the
moments is unbalanced, then the airplane will have an angular acceleration
about the axis of the unbalanced moment.
The relationship between the unbalanced forces and the linear
accelerations and those between unbalanced moments and angular
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
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accelerations are provided by Newton’s second law of motion. These
relationships are called equations of motion.
1.5 Number of equations of motion for an airplane in flight
To derive the equations of motion, the acceleration of a particle on the
body needs to be known. The acceleration is the rate of change of velocity and
the velocity is the rate of change of position vector with respect to the chosen
frame of reference.
1.5.1 Degrees of freedom
The minimum number of coordinates required to prescribe the motion is
called the number of degrees of freedom. The number of equations governing
the motion equals the degrees of freedom. As an example, it may be recalled
that the motion of a particle moving in a plane is prescribed by the x- and y-
coordinates of the particle at various instants of time and this motion is described
by two equations.
Similarly, the position of any point on a rigid pendulum is describe by just
one coordinate namely the angular position (θ) of the pendulum (Fig.1.8). In this
case only one equation is sufficient to describe the motion. In yet another
example, if a particle is constrained to move on a sphere, then its position is
completely prescribed by the longitude and the latitude. Hence, this motion has
only two degrees of freedom.
From the discussion in this subsection it is clear that the coordinates needed to
prescribe the motion could be lengths and/or angles.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
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Fig.1.8 Motion of a single degree of freedom system
1.5.2 Degrees of freedom for a rig id airp lane
To describe its motion, the airplane is treated as a rigid body. It may be
recalled that in a rigid body the distance between any two points is fixed. Thus
the distance r in Fig.1.9 does not change during the motion. To decide the
minimum number of coordinates needed to prescribe the position of a point on a
rigid body which is translating and rotating, one may proceed as follows.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
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Fig.1.9 Position of a point on a rigid airplane
A rigid body with N particles may appear to have 3N degrees of freedom,
but the constraint of rigidity reduces this number. To arrive at the minimum
number of coordinates, let us approach the problem in a different way. Following
Ref.1.3, it can be stated that to fix the location of a point on a rigid body one does
not need to prescribe its distance from all the points, but only needs to prescribe
its distance from three points which do not lie on the same line (points 1, 2 and 3
in Fig.1.10a). Thus, if the positions of these three points are prescribed with
respect to a reference frame, then the position of any point on the body is known.
This may indicate nine degrees of freedom. This number is reduced to six
because the distances s12, s23 and s13 in Fig.1.10a are constants.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
Fig.1.10a Position of a point with respect to three reference points
Another way of looking at the problem is to consider that the three
coordinates of point 1 with respect to the reference frame are prescribed. Now
the point 2 is constrained, because of rigid body assumption, to move on a
sphere centered on point 1 and needs only two coordinates to prescribe its
motion. Once the points 1 and 2 are determined, the point 3 is constrained, again
due to rigid body assumption, to move on a circle about the axis joining points 1
and 2. Hence, only one independent coordinate is needed to prescribe the
position of point 3. Thus, the number of independent coordinates is six (3+2+1).
Or a rigid airplane has six degrees of freedom.
In dynamics the six degrees of freedom associated with a rigid body,
consist of the three coordinates of the origin of the body with respect to the
chosen frame of reference and the three angles which describe the angular
position of a coordinate system fixed on the body (OXbYbZb) with respect to the
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
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fixed frame of reference (EXeYeZe) as shown in Fig.1.10b. These angles are
known as Eulerian angles. These are discussed in ch.7 of flight dynamics- II. See
also Ch.4 of Ref.1.3.
Fig.1.10b Coordinates of a point (P) on a rigid body
Remarks:
i) The derivation of the equations of motions in a general case with six degrees of
freedom (see chapter 7 of Flight dynamics-II or Ref.1.4 chapter 10, pt.3 or
Ref.1.5, chapter 10) is rather involved and would be out of place here.
ii) Here, various cases are considered separately and the equations of motion
are written down in each case.
1.6 Subdivisions of flight dynamics
The subject of flight dynamics is generally divided into two main branches viz.
(i) Performance analysis and (ii) Stability and control
1.6.1 Performance Analysis
In performance analysis, only the equilibrium of forces is generally
considered. It is assumed that by proper deflections of the controls, the moments
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
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can be made zero and that the changes in aerodynamic forces due to deflection
of controls are small. The motions considered in performance analysis are steady
and accelerations, when involved, do not change rapidly with time.
The following motions are considered in performance analysis
- Unaccelerated flights,
• Steady level flight
• Climb, glide and descent
- Accelerated flights,
• Accelerated level flight and climb
• Loop, turn, and other motions along curved paths which are
called manoeuvres
• Take-off and landing.
1.6.2 Stabilit y and control analyses
Roughly speaking, the stability analysis is concerned with the motion of
the airplane, from the equilibrium position, following a disturbance. Stability
analysis tells us whether an airplane, after being disturbed, will return to its
original flight path or not.
Control analysis deals with the forces that the deflection of the controls
must produce to bring to zero the three moments (rolling, pitching and yawing)and achieve a desired flight condition. It also deals with design of control
surfaces and the forces on control wheel/stick /pedals. Stability and control are
linked together and are generally studied under a common heading.
Flight dynamics - I deals with performance analysis. By carrying out this
analysis one can obtain various performance characteristics such as maximum
level speed, minimum level speed, rate of climb, angle of climb, distance covered
with a given amount of fuel called ‘Range’, time elapsed during flight called
‘Endurance’, minimum radius of turn, maximum rate of turn, take-off distance,
landing distance etc. The effect of flight conditions namely the weight, altitude
and flight velocity of the airplane can also be examined. This study would also
help in solving design problems of deciding the power required, thrust required,
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fuel required etc. for given design specifications like maximum speed, maximum
rate of climb, range, endurance etc.
Remark:
Alternatively, the performance analysis can be considered as the analysis
of the motion of flight vehicle considered as a point mass, moving under the
influence of applied forces (aerodynamic, propulsive and gravitational forces).
The stability analysis similarly can be considered as motion of a vehicle of finite
size, under the influence of applied forces and moments.
1.7 Additional definitions
1.7.1 Attitude
As mentioned in section 1.5.2 the instantaneous position of the airplane,
with respect to the earth fixed axes system (EXeYeZe), is given by the
coordinates of the c.g. at that instant of time. The attitude of the airplane is
described by the angular orientation of the OXbY
bZ
b system with respect to
OXeYeZe system or the Euler angles. Reference 1.4, chapter 10 may be referred
to for details. Let us consider simpler cases. When an airplane climbs along a
straight line its attitude is given by the angle ‘γ’ between the axis OXb and the
horizontal (Fig.1.11a). When an airplane executes a turn, the projection of OXb
axis, in the horizontal plane, makes an angle Ψ with reference to a fixed
horizontal axis (Fig.1.11b). When an airplane is banked the axis OYb makes an
angle with respect to the horizontal (Fig.1.11c) and the axis OZb makes an
angle with respect to the vertical.
Fig.1.11a Airplane in a climb
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Fig.1.11b Airplane in a turn - view from top
Fig.1.11c Angle of bank ( )
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1.7.2 Flight path
In the subsequent sections, the flight path, also called the trajectory,
means the path or the line along which the c.g. of the airplane moves. The
tangent to this curve at a point gives the direction of flight velocity at that point on
the flight path. The relative wind is in a direction opposite to that of the flight
velocity.
1.7.3. Angle of attack and side slip
While discussing the forces acting on an airfoil, the chord of the airfoil is
taken as the reference line and the angle between the chord line and the relative
wind is the angle of attack (α). The aerodynamic forces viz. lift (L) and drag (D) ,
produced by the airfoil, depend on the angle of attack (α) and are respectively
perpendicular and parallel to relative wind direction (Fig.1.11 d).
Fig.1.11d Angle of attack and forces on a airfoil
In the case of an airplane the flight path, as mentioned earlier, is the line along
which c.g. of the airplane moves. The tangent to the flight path is the direction of
flight velocity (V). The relative wind is in a direction opposite to the flight velocity.
If the flight path is confined to the plane of symmetry, then the angle of attack
would be the angle between the relative wind direction and the fuselage
reference line (FRL) or OXb axis (see Fig.1.11e). However, in a general case the
velocity vector (V) will have components both along and perpendicular to the
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plane of symmetry. The component perpendicular to the plane of symmetry is
denoted by ‘v’. The projection of the velocity vector in the plane of symmetry
would have components u and w along OXb and OZb axes (Fig.1.11f). With this
background the angle of sideslip and the angle of attack are defined as follows.
Fig.1.11e Flight path in the plane of symmetry
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Fig.1.11f Velocity components in a general case and definition of angle of attack
and sideslip
The angle of sideslip (β) is the angle between the velocity vector (V) and the
plane of symmetry i.e.
β = sin-1 (v/ |V|); where |V| is the magnitude of V.
The angle of attack (α) is the angle between the projection of velocity vector (V)
in the Xb - Zb plane and the OXb axis or
-1 -1 -1
2 2 2 2
w w wα = tan = sin = sin
u | | -v u +wV
Remarks:
i) It is easy to show that, if V denotes magnitude of velocity (V), then
u = V cos α cos β, v = V sin β; w = V sin α cos β.
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ii) By definition, the drag (D) is parallel to the relative wind direction. The lift force
lies in the plane of symmetry of the airplane and is perpendicular to the direction
of flight velocity.
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Chapter 1
Lecture 3
Introduction – 3
Topics
1.8 Simplif ied treatment of performance analysis
1.9 Course out line
1.10 Background expected
1.8 Simplified treatment in performance analysisIn a steady flight, there is no acceleration along the flight path and in a
level flight; the altitude of the flight remains constant. A steady, straight and level
flight generally means a flight along a straight line at a constant velocity and
constant altitude.
Sometimes, this flight is also referred to as unaccelerated level flight. To illustrate
the simplified treatment in performance analysis, the case of unaccelerated level
flight is considered below.
The forces acting on an airplane in unaccelerated level flight are shown in the
Fig.1.12.
They are: Lift (L), Thrust (T), Drag (D) and Weight (W) of the airplane.
It may be noted that the point of action of the thrust and it’s direction depend on
the engine location. However, the direction of the thrust can be taken parallel to
the airplane reference axis.
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Fig.1.12 Forces acting in steady level flight
The lift and drag, being perpendicular to the relative wind, are in the
vertical and horizontal directions respectively, in this flight. The weight acts at the
c.g. in a vertically downward direction.In an unaccelerated level flight, the components of acceleration in the
horizontal and vertical directions are zero.
Hence, the sums of the components of all the forces in these directions
are zero. Resolving the forces along and perpendicular to the flight path (see
Fig.1.12.), gives the following equations of force equilibrium.
T cos α – D = 0 (1.3)
T sin α + L – W = 0 (1.4)
Apart from these equations, equilibrium demands that the moment about
the y-axis to be zero, i.e.,
Mcg = 0
Unless the moment condition is satisfied, the airplane will begin to rotate
about the c.g.
Let us now examine how the moment is balanced in an airplane. The
contributions to Mcg come from all the components of the airplane. As regards the
wing, the point where the resultant vector of the lift and drag intersects the plane
of symmetry is known as the centre of pressure. This resultant force produces a
moment about the c.g. However, the location of the center of pressure depends
on the lift coefficient and hence the moment contribution of wing changes with
the angle of attack as the lift coefficient depends on the angle of attack. For
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convenience, the lift and the drag are transferred to the aerodynamic center
along with a moment (Mac). Recall, that moment coefficient about the a.c. (Cmac)
is, by definition, constant with change in angle of attack.
Similarly, the moment contributions of the fuselage and the horizontal tail
change with the angle of attack. The engine thrust also produces a moment
about the c.g. which depends on the thrust required.
Hence, the sum of the moments about the c.g. contributed by the wing,
fuselage, horizontal tail and engine changes with the angle of attack. By
appropriate choice of the horizontal tail setting (i.e. incidence of horizontal tail
with respect to fuselage central line), one may be able to make the sum of these
moments to be zero in a certain flight condition, which is generally the cruise
flight condition. Under other flight conditions, generation of corrective
aerodynamic moment is facilitated by suitable deflection of elevator (See
Fig.1.2a, b and c for location of elevator). By deflecting the elevator, the lift on the
horizontal tail surface can be varied and the moment produced by the horizontal
tail balances the moments produced by all other components.
The above points are illustrated with the help of an example.
Example 1.1
A jet aircraft weighing 60,000 N has it’s line of thrust 0.15 m below the line
of drag. When flying at a certain speed, the thrust required is 6000 N and the
center of pressure of the wing lift is 0.45 m aft of the airplane c.g. What is the lift
on the wing and the load on the tail plane whose center of pressure is 7.5 m
behind the c.g.? Assume unaccelerated level flight and the angle of attack to be
small during the flight.
Solution:
The various forces and dimensions are presented in Fig.1.13. The lift on
the wing is LW and the lift on the tail is LT. Since the angle of attack (α) is small, it
may be considered that cos α = 1 and sin α = 0. Thus, the force equilibrium (Eqs.
1.3 and 1.4), yields :
T – D = 0
LW + LT – W = 0
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i.e. D = T = 6000 N and LT + LW = 60000 N
From Fig. 1.13., the moment equilibrium about the c.g. gives:
Mcg = T (zd + 0.15) – D.zd – 0.45.LW – 7.5.LT = 0 where zd is the distance of drag
below the c.g; not shown in figure as it is of no significance in the present
context.
Fig.1.13 Forces acting on an airplane in steady level flight
Solving these equations, gives :
LW = 63702.13 N and LT = -3702.13 N
Following observations can be made.
A) The lift on the wing is about 63.7 kN. The lift on the tail is only 3.7 kN and is in
the downward direction.
B) The contribution of tail to the total lift is thus small, in this case, about 6% and
negative. This negative contribution necessitates the wing lift to be more than the
weight of the airplane. This increase in the lift results in additional drag called trim
drag.
C) The distance zd is of no significance in this problem as the drag and thrust
form a couple whose moment is equal to the thrust multiplied by the distance
between them.
D) Generally, the angle of attack (α) is small. Hence, sin α is small and cos α is
nearly equal to unity. Thus, the equations of force equilibrium reduce to
T – D = 0 and L – W = 0.
E) It is assumed that the pitching moment equilibrium i.e. Σ Mcg = 0 is achieved
by appropriate deflection of the elevator. The changes in the lift and drag due to
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
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elevator deflections are generally small and in performance analysis, as stated
earlier, these changes are ignored and the simplified picture as shown in Fig.1.14
is considered adequate.
Fig.1.14. Simplified picture of the forces acting on an airplane in level flight.
1.9 Course outl ine
Let us consider the background material required to carry-out the
performance analysis. It is known that :
L = (1/2) ρ V2 S CL
D = (1/2) ρ V2 S CD
where CL and CD are the lift and drag coefficients; S is the area of the wing.
The quantities CL and CD depend on α , Mach number (M = V / a) and Reynolds
number (Re = ρ V l /µ); where l is the reference length. Thus
CD = f (CL, M, Re) (1.6)
The relation between CL and CD at given M and Re is known as the drag
polar of the airplane.
This has to be known for carrying the performance
analysis. The density of air (ρ) depends on the flight altitude. Further the Mach
number depends on the speed of sound, which in turn depends on the ambient
air temperature. Thus, performance analysis requires the knowledge of the
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-1
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variations of pressure, temperature, density, viscosity etc. with altitude in earth’s
atmosphere.
The evaluation of performance also requires the knowledge of the engine
characteristics such as, variations of thrust (or power) and fuel consumption with
the flight speed and altitude.
Keeping these aspects in view, following will be the contents of this course.
Earth’s atmosphere (chapter 2)
Drag polar (chapter 3)
Engine characteristics (chapter 4)
Performance analysis. ( chapters 5 to 10)
These topics will be taken up in the subsequent chapters.
The Appendices ‘A’ and ‘B’ present the performance analyses of piston-engined
and jet airplane respectively.
1.10 Back ground expected
The student is expected to have undergone courses on (a) Vectors (b)
Rigid body dynamics (c) Aerodynamics and (d) Aircraft engines.
Remark: References 1.5 to 1.14 are some of the books dealing with airplane
performance. They can be consulted for additional information.
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Chapter 1
References
1.1 Miele, A. “Flight mechanics Vol I” Addison Wesley (1962).
1.2 Shames, I.H. and Krishna Mohana Rao, G. “Engineering mechanics – statics
and dynamics”, 4th Edition, Dorling Kindersley (India), licensees of Pearson
Education (2006).
1.3 Goldstein H. “Classical mechanics “Second edition Addison Wesley (1980).
1.4 Davies, M. (Editor) “The standard handbook for aeronautical and
astronautical engineers” McGraw Hill (2003).
1.5 Perkins, C.D. and Hage, R. E. “Airplance performance, stability and
control” John Wiley (1963).
1.6 Dommasch, D.O. Sherby, S.S. and Connolly, T.F. “Airplane
aerodynamics” Pitman (1967).
1.7 Houghton E.L. and Carruthers N.B. “Aerodynamics for engineering
students”, Edward Arnold (1982).
1.8 Hale, F.J. “Introduction to aircraft performance, selection and design”,
John Wiley (1984).1.9 McCormick B.W. “Aerodynamics, aeronautics and flight mechanics”, John
Wiley (1995).
1.10 Anderson, Jr. J.D. “Aircraft performance and design” McGraw Hill
International edition (1999).
1.11 Eshelby, M.E. ”Aircraft performance-theory and practice”, Butterworth-
Heinemann, Oxford, U.K., (2001).
1.12 Pamadi, B. “Performance, stability, dynamics and control of an
airplane”, AIAA (2004).
1.13 Anderson, Jr. J.D. “Introduction to flight” Fifth edition, McGraw-Hill,
(2005).
1.14 Phillips, W.F. “Mechanics of flight” 2nd Edition John Wiley (2010).
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1.15 Jackson, P. (Editor) “Jane’s all the world’s aircraft” Published annually
by Jane’s information group Ltd., Surrey, U.K..
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Chapter 1
Exercises
1. Sketch the three views of an airplane and show it’s axes systems.
2. Define, with neat sketches, the following terms.
(a) flight path
(b) flight velocity
(c) body axes system
(d) angle of attack
(e) angle of slide slip and
(f) bank angle.
3.“Jane’s All the World Aircraft” (Ref.1.15) is a book published annually andcontains details of airplanes currently in production in various countries. Refer to
this book and study the three view drawings, geometrical details and
performance parameters of different types of airplanes.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
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Chapter 2Earth’s atmosphere (Lectures 4 and 5)
Keywords: Earth’s atmosphere; International standard atmosphere;
geopotential altitude; stability of atmosphere.
Topics
2.1 Introduction
2.2 Earth’s atmosphere
2.2.1 The troposphere
2.2.2 The stratosphere
2.2.3 The mesosphere2.2.4 The ionosphere or thermosphere
2.2.5 The exosphere
2.3 International standard atmosphere (ISA)
2.3.1 Need for ISA and agency prescribing it.
2.3.2 Features of ISA
2.4 Variations of properties with altitude in ISA
2.4.1 Variations of pressure and density with altitude
2.4.2 Variations with altitude of pressure ratio, density ratio speed of
sound, coefficient of viscosity and kinematic viscosity.
2.5 Geopotential altitude
2.6 General remarks
2.6.1 Atmospheric properties in cases other than ISA
2.6.2 Stability of atmosphere
References
Exercises
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Chapter 2
Lecture 4
Earth’s atmosphere – 1
Topics
2.1 Introduction
2.2 Earth’s atmosphere
2.2.1 The troposphere
2.2.2 The stratosphere
2.2.3 The mesosphere
2.2.4 The ionosphere or thermosphere
2.2.5 The exosphere
2.3 International standard atmosphere (ISA)
2.3.1 Need for ISA and agency prescribing it.
2.3.2 Features of ISA
2.1 Introduction
Airplanes fly in the earth’s atmosphere and therefore, it is necessary toknow the properties of this atmosphere.
This chapter, deals with the average characteristics of the earth’s
atmosphere in various regions and the International Standard Atmosphere (ISA)
which is used for calculation of airplane performance.
2.2 Earth’s atmosphere
The earth’s atmosphere is a gaseous blanket around the earth which is
divided into the five regions based on certain intrinsic features (see Fig.2.1).
These five regions are: (i) Troposphere, (ii) Stratosphere, (iii) Mesosphere,
(iv) Ionosphere or Thermosphere and (v) Exosphere. There is no sharp
distinction between these regions and each region gradually merges with the
neighbouring regions.
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Fig.2.1 Typical variations of temperature and pressure in the earth’s atmosphere
2.2.1 The troposphere
This is the region closest to the earth’s surface. It is characterized by
turbulent conditions of air. The temperature decreases linearly at an approximate
rate of 6.5 K / km. The highest point of the troposphere is called tropopause. The
height of the tropopause varies from about 9 km at the poles to about 16 km at
the equator.
2.2.2 The stratosphere
This extends from the tropopause to about 50 km. High velocity winds
may be encountered in this region, but they are not gusty. Temperature remains
constant up to about 25 km and then increases. The highest point of the
stratosphere is called the stratopause.
2.2.3 The mesosphere
The mesosphere extends from the stratopause to about 80 km. The
temperature decreases to about -900C in this region. In the mesosphere, the
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pressure and density of air are very low, but the air still retains its composition as
at sea level. The highest point of the mesosphere is called the mesopause.
2.2.4 The ionosphere or thermosphere
This region extends from the mesopause to about 1000 km. It is
characterized by the presence of ions and free electrons. The temperature
increases to about 00C at 110 km, to about 10000C at 150 km and peak of about
17800C at 700 km (Ref.2.1). Some electrical phenomena like the aurora borealis
occur in this region.
2.2.5 The exosphere
This is the outer fringe of the earth’s atmosphere. Very few molecules are
found in this region. The region gradually merges into the interplanetary space.
2.3 International Standard Atmosphere (ISA)
2.3.1 Need for ISA and agency prescr ibing it
The properties of earth’s atmosphere like pressure, temperature and
density vary not only with height above the earth’s surface but also with the
location on earth, from day to day and even during the day. As mentioned in
section 1.9, the performance of an airplane is dependent on the physical
properties of the earth’s atmosphere. Hence, for the purpose of comparing
(a) the performance of different airplanes and (b) the performance of the same
airplane measured in flight tests on different days, a set of values for atmospheric
properties have been agreed upon, which represent average conditions
prevailing for most of the year, in Europe and North America. Though the agreed
values do not represent the actual conditions anywhere at any given time, they
are useful as a reference. This set of values called the International Standard
Atmosphere (ISA) is prescribed by ICAO (International Civil Aviation
Organization). It is defined by the pressure and temperature at mean sea level,
and the variation of temperature with altitude up to 32 km (Ref.1.11, chapter 2).
With these values being prescribed, it is possible to find the required physical
characteristics (pressure, temperature, density etc) at any chosen altitude.
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Remark:
The actual performance of an airplane is measured in flight tests under
prevailing conditions of temperature, pressure and density. Methods are
available to deduce, from the flight test data, the performance of the airplane
under ISA conditions. When this procedure is applied to various airplanes and
performance presented under ISA conditions, then comparison among different
airplanes is possible.
2.3.2 Features of ISA
The main features of the ISA are the standard sea level values and the
variation of temperature with altitude. The air is assumed as dry perfect gas.
The standard sea level conditions are as follows:
Temperature (T0) = 288.15 K = 150C
Pressure (p0) = 101325 N/m2 = 760 mm of Hg
Rate of change of temperature:
= - 6.5 K/km upto 11 km
= 0 K/km from 11 to 20 km
= 1 K/km from 20 to 32 km
The region of ISA from 0 to 11 km is referred to as troposphere. That
between 11 to 20 km is the lower stratosphere and between 20 to 32 km is the
middle stratosphere (Ref.1.11, chapter 2).
Note: Using the values of T0 and p0 , and the equation of state, p = ρRT, gives the
sea level density (ρ0) as 1.225 kg/m3.
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Chapter 2
Lecture 5
Earth’s atmosphere – 2
Topics
2.4 Variations of properties with altitude in ISA
2.4.1 Variations of pressure and density with altitude
2.4.2 Variations with altitude of pressure ratio, density ratio speed of
sound, coefficient of viscosity and kinematic viscosity.
2.5 Geopotential altitude
2.6 General remarks
2.6.1 Atmospheric properties in cases other than ISA
2.6.2 Stability of atmosphere
Atmospher ic properties o f ISA (Table 2.1)
2.4 Variations of properties with altitude in ISA
For calculation of the variations of pressure, temperature and density with
altitude, the following equations are used.The equation of state p = ρ R T (2.1)
The hydrostatic equation dp/dh = - ρ g (2.2)
Remark:
The hydrostatic equation can be easily derived by considering the balance of
forces on a small fluid element.
Consider a cylindrical fluid element of area A and height Δh as shown in Fig.2.2.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
Fig.2.2 Equilibrium of a fluid element.
The forces acting in the vertical direction on the element are the pressure forces
and the weight of the element.
For vertical equilibrium of the element,
pA – {p + (dp /dh) Δh} A – ρ g A Δh = 0
Simplifying, dp /dh = - ρ g
2.4.1 Variations of pressure and density with altitude
Substituting for ρ from the Eq.(2.1) in Eq.(2.2) gives:
dp / dh = -(p/RT) g
Or (dp/p) = -g dh/RT (2.3)
Equation (2.3) is solved separately in troposphere and stratosphere, taking into
account the temperature variations in each region. For example, in the
troposphere, the variation of temperature with altitude is given by the equation
T = T0 – λ h (2.4)
where T0 is the sea level temperature, T is the temperature at the altitude h and λ is the temperature lapse rate in the troposphere.
Substituting from Eq.(2.4) in Eq.(2.3) gives:
(dp /p) = - gdh /R (T0 – λ h) (2.5)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
Taking ‘g’ as constant, Eq.(2.5) can be integrated between two altitudes h1 and
h2. Taking h1 as sea level and h2 as the desired altitude (h), the integration gives
the following equation, the intermediate steps are left as an exercise.
(p/p0) = (T/T0)(g/λR) (2.6)
where T is the temperature at the desired altitude (h) given by Eq.(2.4).
Equation (2.6) gives the variation of pressure with altitude.
The variation of density with altitude can be obtained using Eq.(2.6) and
the equation of state. The resulting variation of density with temperature in the
troposphere is given by:
(ρ/ρ0) = (T/T0)(g/λR)-1 (2.7)
Thus, both the pressure and density variations are obtained once the
temperature variation is known.
As per the ISA, R = 287.05287 m2sec-2 K and g = 9.80665 m/s2.
Using these and λ = 0.0065 K/m in the troposphere yields (g/Rλ) as 5.25588.
Thus, in the troposphere, the pressure and density variations are :
(p/p0) = (T/T0)5.25588 (2.8)
(ρ/ρ0) = (T/T0)4.25588 (2.9)
Note: T= 288.15 - 0.0065 h; h in m and T in K.
In order to obtain the variations of properties in the lower stratosphere (11
to 20 km altitude), the previous analysis needs to be carried-out afresh with λ = 0
i.e., ‘T’ having a constant value equal to the temperature at 11 km (T = 216.65 K).
From this analysis the pressure and density variations in the lower stratosphere
are obtained as :
(p / p11) = (ρ / ρ11) = exp { -g (h - 11000) / RT11 } (2.10)
where p11, ρ11 and T11 are the pressure, density and temperature respectively at
11 km altitude.
In the middle stratosphere (20 to 32 km altitude), it can be shown that (note in
this case λ = -0.001 K / m):
(p / p20) = (T / T20)- 34.1632 (2.11)
(ρ / ρ20) = (T/ T20)- 35.1632 (2.12)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 4
where p20, ρ20 and T20 are pressure, density and temperature respectively at
20 km altitude.
Thus, the pressure and density variations have been worked out in the
troposphere and the stratosphere of ISA. Table 2.1 presents these values.
Remark:
Using Eqs.(2.1) and (2.2) the variations of pressure and density can be worked
out for other variations of temperature with height (see exercise 2.1).
2.4.2. Variations with altitude of pressure ratio, densi ty ratio , speed of
sound, coefficient of viscosity and kinematic viscosity
The ratio (p/p0) is called pressure ratio and is denoted by δ. Its value in ISA can
be obtained by using Eqs.(2.8),(2.10) and (2.11). Table 2.1 includes these
values.
The ratio (ρ / ρ0) is called density ratio and is denoted by σ. Its values in ISA can
be obtained using Eqs.(2.9),(2.10) and (2.12). Table 2.1 includes these values.
The speed of sound in air, denoted by ‘a’, depends only on the temperature and
is given by:
a = (γ RT)0.5 (2.13)
where γ is the ratio of specific heats; for air γ = 1.4. The values of ‘a’ in ISA can
be obtained by using appropriate values of temperature. Table 2.1 includes thesevalues.
The kinematic viscosity ( ) is given by:
= μ / ρ where μ is the coefficient of viscosity.
The coefficient of viscosity of air (μ) depends only on temperature. Its variation
with temperature is given by the following Sutherland formula.
3/2-6 T
μ = 1.458X10 [ ]T+110.4
, where T is in Kelvin and μ is in kg m-1 s-1 (2.14)
Table 2.1 includes the variation of kinematic viscosity with altitude.
Example 2.1
Calculate the temperature (T), pressure (p), density (ρ ), pressure ratio
(δ ) , density ratio (σ ), speed of sound (a) , coefficient of viscosity (μ ) and
kinematic viscosity ( ) in ISA at altitudes of 8 km, 16 km and 24 km.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
Solution:
It may be noted that the three altitudes specified in this example, viz.
8 km, 16 km and 24 km, lie in troposphere, lower stratosphere and middle
stratosphere regions of ISA respectively.
(a) h = 8 km
Let the quantities at 8 km altitude be denoted by the suffix ‘8’.
In troposphere: 0T = T -λh
where, T0 = 288.15 K, λ = 0.0065 K /m
Hence, 8T = 288.15 - 0.0065 8000 = 236.15K
From Eq.(2.8)
5.25588 5.2558888 0
0
p = δ = T/T = 236.15/288.15 = 0.35134p
Or 2
8p = 0.35134 × 101325 = 35599.5 N/m
3
8 8 8
35599.5ρ = p / RT = = 0.52516 kg/m
287.05287×236.15
8 8 0σ = ρ /ρ = 0.52516/1.225 = 0.42870
a8 = (γ RT8)0.5
0.5= 1.4×287.05287×236.15 = 308.06 m/s
From Eq.(2.14):
1.5 1.5-6 -6 -5 -1 -18
8
8
T 236.15μ = 1.458×10 = 1.458×10 = 1.5268×10 kg m s
T +110.4 236.15+110.4
-5 -5 2
8 8 8= μ /ρ = 1.5268×10 /0.52516 = 2.9072×10 m /s
Remarks:
(i) The values calculated above and those in Table 2.1 may differ from each
other in the last significant digit. This is due to the round-off errors in the
calculations.
(ii) Consider an airplane flying at 8 km altitude at a flight speed of 220 m/s.
The Mach number of this flight would be: 220/308.06 = 0.714
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
(iii) Further if the reference chord of the wing (cref ) of this airplane be 3.9 m,
the Reynolds number in this flight, based on cref , would be:
6ref e -5
Vc 220×3.9R = = = 29.51×10
2.9072×10
(iv) For calculation of values at 16 km altitude, the values of temperature,
pressure and density are needed at the tropopause viz. at h=11 km.
Now 11T = 288.15-0.0065×11000 = 216.65 K
5.25588 2
11p = 101325 216.65/288.15 = 22632 N/m
3
11ρ = 22632/ 287.05287×216.65 = 0.36392 kg/m
(b) h = 16 km
In lower stratosphere Eq.(2.10) gives :
11
11 11
p ρ= = exp -g h-11000 /RT
p ρ
Consequently,
16 16
11 11
p ρ= = exp -9.80665 16000-11000 / 287.05287×216.65 = 0.45455
p ρ
Or 2
16
p = 22632×0.45455 = 10287 N/m
3
16ρ = 0.36392×0.45455 = 0.16541kg/m
16δ = 10287 /101325 = 0.10153
16σ = 0.16541/1.225 = 0.13503
0.5
16a = 1.4×287.05287×216.65 = 295.07m/s
1.5-6 -5 -1 -1
16
216.65μ = 1.458×10 = 1.4216×10 kg m s
216.65+110.4
-5 -5 2
16 = 1.4216×10 /0.16541= 8.594×10 m /s
Remark :
To calculate the required values at 24 km altitude, the values of T and p are
needed at h = 20 km. These values are :
T20 = 216.65
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
20
11
p= exp -9.80665 20000-11000 / 287.05287×216.65 = 0.24191
p
Or 2
20p = 22632 0.24191= 5474.9 N/m
(c) h = 24 km
24T = 216.65+0.001 24000-20000 = 220.65K
From Eq.(2.11):
-34.163224
24 20
20
p= T /T
p
Or -34.1632 2
24p = 5474.9 220.65/216.65 = 2930.5N/m
24ρ = 2930.5/ 287.05287×220.65 = 0.04627
Hence, 24δ = 2930.5/101325 = 0.02892
and 24σ = 0.04627/1.225 = 0.03777
0.5
24a = 1.4×287.05287×220.65 = 297.78 m/s
1.5-6 -5 -1 -1
24
220.65μ = 1.458×10 = 1.4435×10 kg m s
220.65+110.4
-5 -4 2
24 = 1.4435×10 /0.04627 = 3.12×10 m /s
Answers:
h (km) 8 16 24
T (K) 236.15 216.65 220.65
p (N/m2) 35599.5 10287.0 2930.5
0δ = p/p 0.35134 0.10153 0.02892
3ρ kg/m 0.52516 0.16541 0.04627
0σ = ρ/ρ 0.42870 0.13503 0.03777
a (m/s) 308.06 295.07 297.78
-1 -1μ kg m s 1.5268 x 10-5 1.4216 x 10-5 1.4435 x 10-5
2m /s 2.9072 x 10-5
8.594 x 10-5
3.12 x 10-4
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 8
2.5 Geopotential alti tude
The variations of pressure, temperature and density in the atmosphere
were obtained by using the hydrostatic equation (Eq.2.2). In this equation ‘g’ is
assumed to be constant. However, it is known that ‘g’ decreases with altitude.
Equation (1.1) gives the variation as:
0G
Rg = g ( )
R+h
where ‘R ’ is the radius of earth and ‘hG’ is the geometric altitude above earth’s
surface.
Thus, the values of p and ρ obtained by assuming g =0
g are at an
altitude slightly different from the geometrical altitude (hG). This altitude is called
geopotential altitude, which for convenience is denoted by ‘h’. Following Ref.1,
the geopotential altitude can be defined as the height above earth’s surface in
units, proportional to the potential energy of unit mass (geopotential), relative to
sea level. It can be shown that the geopotential altitude (h) is given, in terms of
geometric altitude (hG), by the following relation. Reference 1.13, chapter 3 may
be referred to for derivation.
G
Rh = h
R-h
It may be remarked that the actual difference between h and hG is small
for altitudes involved in flight dynamics; for h of 20 km, hG would be 20.0627 km.
Hence, the difference is ignored in performance analysis.
2.6 General remarks:
2.6.1 Atmospheric properties in cases other than ISA
It will be evident from chapters 4 to 10 that the engine characteristics and
the airplane performance depend on atmospheric characteristics. Noting that ISAonly represents average atmospheric conditions, other atmospheric models have
been proposed as guidelines for extreme conditions in arctic and tropical regions.
Figure 2.3 shows the temperature variations with altitude in arctic and tropical
atmospheres along with ISA. It is seen that the arctic minimum atmosphere has
the following features. (a) The sea level temperature is -500C (b) The
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
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temperature increases at the rate of 10 K per km up to 1500 m altitude. (c) The
temperature remains constant at -350C up to 3000 m altitude. (d) Then the
temperature decreases at the rate of 4.72 K per km up to 15.5 km altitude (e)
The tropopause in this case is at 15.5 km and the temperature there is -94 0c.
The features of the tropical maximum atmosphere are as follows.
(a) Sea level temperature is 450 C.
(b) The temperature decreases at the rate of 6.5 K per km up to 11.54 km
and then remains constant at -300 C.
Fig.2.3 Temperature variations in arctic minimum, ISA and tropical maximum
atmospheres (Reproduced from Ref.1.7, Chapter 3 with permission of author)
Note:
(a) The local temperature varies with latitude but the sea level pressure (p0)
depends on the weight of air above and is taken same at all the places i.e.
101325 N/m2. Knowing p0 and T0, and the temperature lapse rates, the pressure,
temperature and density in tropospheres of arctic minimum and tropical
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
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maximum can be obtained using Eqs. (2.4), (2.6) and (2.7). (see also exercise
2.1).
(b) Some airlines/ air forces may prescribe intermediate values of sea level
temperature e.g. ISA +150C or ISA +200C. The variations of pressure,
temperature and density with altitude in these cases can also be worked out from
the aforesaid equations.
2.6.2 Stabilit y of atmosphere
It is generally assumed that the air mass is stationary. However, some
packets of air mass may acquire motion due to local changes. For example, due
to absorption of solar radiation by the earth’s surface, an air mass adjacent to the
surface may become lighter and buoyancy may cause it to rise. If the
atmosphere is stable, a rising packet of air must come back to its original
position. On the other hand, if the air packet remains in the disturbed position,
then the atmosphere is neutrally stable. If the rising packet continues to move up
then the atmosphere is unstable.
Reference 1.7, chapter 3 analyses the problem of atmospheric stability
and concludes that if the temperature lapse rate is less than 9.75 K per km, then
the atmosphere is stable. It is seen that the three atmospheres, representing
different conditions, shown in Fig.2.3 are stable.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
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Altit-
ude
(m)
Tempe-
rature
(K)
Pressure
(N/m2)
δ
(p/po)
Density
(kg/m3)
σ
(ρ/ρo)
speed
of
sound
(m/s)
Kinematic
viscosity
(m2/s)
0 288.15 101325.0 1.00000 1.22500 1.00000 340.29 1.4607E-005
200 286.85 98945.3 0.97651 1.20165 0.98094 339.53 1.4839E-005
400 285.55 96611.0 0.95348 1.17864 0.96216 338.76 1.5075E-005
600 284.25 94321.6 0.93088 1.15598 0.94365 337.98 1.5316E-005
800 282.95 92076.3 0.90872 1.13364 0.92542 337.21 1.5562E-005
1000 281.65 89874.4 0.88699 1.11164 0.90746 336.43 1.5813E-005
1200 280.35 87715.4 0.86568 1.08997 0.88977 335.66 1.6069E-005
1400 279.05 85598.6 0.84479 1.06862 0.87234 334.88 1.6331E-005
1600 277.75 83523.3 0.82431 1.04759 0.85518 334.10 1.6598E-005
1800 276.45 81489.0 0.80423 1.02688 0.83827 333.31 1.6870E-005
2000 275.15 79494.9 0.78455 1.00649 0.82162 332.53 1.7148E-005
2200 273.85 77540.6 0.76527 0.98640 0.80523 331.74 1.7432E-005
2400 272.55 75625.4 0.74636 0.96663 0.78908 330.95 1.7723E-005
2600 271.25 73748.6 0.72784 0.94716 0.77319 330.16 1.8019E-0052800 269.95 71909.7 0.70969 0.92799 0.75754 329.37 1.8321E-005
3000 268.65 70108.2 0.69191 0.90912 0.74214 328.58 1.8630E-005
3200 267.35 68343.3 0.67450 0.89054 0.72697 327.78 1.8946E-005
3400 266.05 66614.6 0.65744 0.87226 0.71205 326.98 1.9269E-005
3600 264.75 64921.5 0.64073 0.85426 0.69736 326.18 1.9598E-005
3800 263.45 63263.4 0.62436 0.83655 0.68290 325.38 1.9935E-005
4000 262.15 61639.8 0.60834 0.81912 0.66867 324.58 2.0279E-005
4200 260.85 60050.0 0.59265 0.80197 0.65467 323.77 2.0631E-005
4400 259.55 58493.7 0.57729 0.78510 0.64090 322.97 2.0990E-005
4600 258.25 56970.1 0.56225 0.76850 0.62735 322.16 2.1358E-005
4800 256.95 55478.9 0.54753 0.75217 0.61402 321.34 2.1734E-005
Table 2.1 Atmospheric properties in ISA (Cont..)
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5000 255.65 54019.4 0.53313 0.73611 0.60091 320.53 2.2118E-005
5200 254.35 52591.2 0.51903 0.72031 0.58801 319.71 2.2511E-005
5400 253.05 51193.7 0.50524 0.70477 0.57532 318.90 2.2913E-005
5600 251.75 49826.4 0.49175 0.68949 0.56285 318.08 2.3324E-0055800 250.45 48488.8 0.47855 0.67446 0.55058 317.25 2.3744E-005
6000 249.15 47180.5 0.46564 0.65969 0.53852 316.43 2.4174E-005
6200 247.85 45900.9 0.45301 0.64516 0.52666 315.60 2.4614E-005
6400 246.55 44649.5 0.44066 0.63088 0.51501 314.77 2.5064E-005
6600 245.25 43425.9 0.42858 0.61685 0.50355 313.94 2.5525E-005
6800 243.95 42229.6 0.41677 0.60305 0.49229 313.11 2.5997E-005
7000 242.65 41060.2 0.40523 0.58949 0.48122 312.27 2.6480E-0057200 241.35 39917.1 0.39395 0.57617 0.47034 311.44 2.6974E-005
7400 240.05 38799.9 0.38292 0.56308 0.45965 310.60 2.7480E-005
7600 238.75 37708.1 0.37215 0.55021 0.44915 309.75 2.7998E-005
7800 237.45 36641.4 0.36162 0.53757 0.43884 308.91 2.8529E-005
8000 236.15 35599.2 0.35134 0.52516 0.42870 308.06 2.9073E-005
8200 234.85 34581.2 0.34129 0.51296 0.41875 307.21 2.9629E-005
8400 233.55 33586.9 0.33148 0.50099 0.40897 306.36 3.0200E-005
8600 232.25 32615.8 0.32189 0.48923 0.39937 305.51 3.0784E-005
8800 230.95 31667.6 0.31254 0.47768 0.38994 304.65 3.1383E-005
9000 229.65 30741.9 0.30340 0.46634 0.38069 303.79 3.1997E-005
9200 228.35 29838.2 0.29448 0.45521 0.37160 302.93 3.2627E-005
9400 227.05 28956.1 0.28577 0.44428 0.36268 302.07 3.3272E-005
9600 225.75 28095.2 0.27728 0.43355 0.35392 301.20 3.3933E-005
9800 224.45 27255.2 0.26899 0.42303 0.34533 300.33 3.4611E-005
10000 223.15 26435.7 0.26090 0.41270 0.33690 299.46 3.5307E-005
10200 221.85 25636.2 0.25301 0.40256 0.32862 298.59 3.6020E-005
10400 220.55 24856.4 0.24531 0.39262 0.32050 297.71 3.6752E-005
10600 219.25 24096.0 0.23781 0.38286 0.31254 296.83 3.7503E-005
Table 2.1 Atmospheric properties in ISA (Cont..)
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10800 217.95 23354.4 0.23049 0.37329 0.30473 295.95 3.8274E-005
11000 216.65 22631.5 0.22336 0.36391 0.29707 295.07 3.9065E-005
11200 216.65 21929.4 0.21643 0.35262 0.28785 295.07 4.0316E-005
11400 216.65 21248.6 0.20971 0.34167 0.27892 295.07 4.1608E-00511600 216.65 20588.9 0.20320 0.33106 0.27026 295.07 4.2941E-005
11800 216.65 19949.7 0.19689 0.32079 0.26187 295.07 4.4317E-005
12000 216.65 19330.4 0.19078 0.31083 0.25374 295.07 4.5736E-005
12200 216.65 18730.2 0.18485 0.30118 0.24586 295.07 4.7202E-005
12400 216.65 18148.7 0.17911 0.29183 0.23823 295.07 4.8714E-005
12600 216.65 17585.3 0.17355 0.28277 0.23083 295.07 5.0275E-005
12800 216.65 17039.4 0.16817 0.27399 0.22366 295.07 5.1886E-00513000 216.65 16510.4 0.16294 0.26548 0.21672 295.07 5.3548E-005
13200 216.65 15997.8 0.15789 0.25724 0.20999 295.07 5.5264E-005
13400 216.65 15501.1 0.15298 0.24925 0.20347 295.07 5.7035E-005
13600 216.65 15019.9 0.14823 0.24152 0.19716 295.07 5.8862E-005
13800 216.65 14553.6 0.14363 0.23402 0.19104 295.07 6.0748E-005
14000 216.65 14101.8 0.13917 0.22675 0.18510 295.07 6.2694E-005
14200 216.65 13664.0 0.13485 0.21971 0.17936 295.07 6.4703E-005
14400 216.65 13239.8 0.13067 0.21289 0.17379 295.07 6.6776E-005
14600 216.65 12828.7 0.12661 0.20628 0.16839 295.07 6.8916E-005
14800 216.65 12430.5 0.12268 0.19988 0.16317 295.07 7.1124E-005
15000 216.65 12044.6 0.11887 0.19367 0.15810 295.07 7.3403E-005
15200 216.65 11670.6 0.11518 0.18766 0.15319 295.07 7.5754E-005
15400 216.65 11308.3 0.11160 0.18183 0.14844 295.07 7.8182E-005
15600 216.65 10957.2 0.10814 0.17619 0.14383 295.07 8.0687E-005
15800 216.65 10617.1 0.10478 0.17072 0.13936 295.07 8.3272E-005
16000 216.65 10287.5 0.10153 0.16542 0.13504 295.07 8.5940E-005
16200 216.65 9968.1 0.09838 0.16028 0.13084 295.07 8.8693E-005
16400 216.65 9658.6 0.09532 0.15531 0.12678 295.07 9.1535E-005
Table 2.1 Atmospheric properties in ISA (Cont..)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
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16600 216.65 9358.8 0.09236 0.15049 0.12285 295.07 9.4468E-005
16800 216.65 9068.2 0.08950 0.14581 0.11903 295.07 9.7495E-005
17000 216.65 8786.7 0.08672 0.14129 0.11534 295.07 1.0062E-004
17200 216.65 8513.9 0.08403 0.13690 0.11176 295.07 1.0384E-00417400 216.65 8249.6 0.08142 0.13265 0.10829 295.07 1.0717E-004
17600 216.65 7993.5 0.07889 0.12853 0.10492 295.07 1.1060E-004
17800 216.65 7745.3 0.07644 0.12454 0.10167 295.07 1.1415E-004
18000 216.65 7504.8 0.07407 0.12068 0.09851 295.07 1.1780E-004
18200 216.65 7271.9 0.07177 0.11693 0.09545 295.07 1.2158E-004
18400 216.65 7046.1 0.06954 0.11330 0.09249 295.07 1.2547E-004
18600 216.65 6827.3 0.06738 0.10978 0.08962 295.07 1.2949E-00418800 216.65 6615.4 0.06529 0.10637 0.08684 295.07 1.3364E-004
19000 216.65 6410.0 0.06326 0.10307 0.08414 295.07 1.3793E-004
19200 216.65 6211.0 0.06130 0.09987 0.08153 295.07 1.4234E-004
19400 216.65 6018.2 0.05939 0.09677 0.07900 295.07 1.4690E-004
19600 216.65 5831.3 0.05755 0.09377 0.07654 295.07 1.5161E-004
19800 216.65 5650.3 0.05576 0.09086 0.07417 295.07 1.5647E-004
20000 216.65 5474.9 0.05403 0.08803 0.07187 295.07 1.6148E-004
20200 216.85 5305.0 0.05236 0.08522 0.06957 295.21 1.6694E-004
20400 217.05 5140.5 0.05073 0.08251 0.06735 295.34 1.7257E-004
20600 217.25 4981.3 0.04916 0.07988 0.06521 295.48 1.7839E-004
20800 217.45 4827.1 0.04764 0.07733 0.06313 295.61 1.8440E-004
21000 217.65 4677.9 0.04617 0.07487 0.06112 295.75 1.9060E-004
21200 217.85 4533.3 0.04474 0.07249 0.05918 295.89 1.9701E-004
21400 218.05 4393.4 0.04336 0.07019 0.05730 296.02 2.0363E-004
21600 218.25 4257.9 0.04202 0.06796 0.05548 296.16 2.1046E-004
21800 218.45 4126.8 0.04073 0.06581 0.05372 296.29 2.1752E-004
22000 218.65 3999.7 0.03947 0.06373 0.05202 296.43 2.2480E-004
22200 218.85 3876.7 0.03826 0.06171 0.05038 296.56 2.3232E-004
Table 2.1 Atmospheric properties in ISA (Cont..)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
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22400 219.05 3757.6 0.03708 0.05976 0.04878 296.70 2.4009E-004
22600 219.25 3642.3 0.03595 0.05787 0.04724 296.83 2.4811E-004
22800 219.45 3530.5 0.03484 0.05605 0.04575 296.97 2.5639E-004
23000 219.65 3422.4 0.03378 0.05428 0.04431 297.11 2.6494E-00423200 219.85 3317.6 0.03274 0.05257 0.04291 297.24 2.7376E-004
23400 220.05 3216.1 0.03174 0.05091 0.04156 297.38 2.8287E-004
23600 220.25 3117.8 0.03077 0.04931 0.04026 297.51 2.9228E-004
23800 220.45 3022.6 0.02983 0.04776 0.03899 297.65 3.0198E-004
24000 220.65 2930.4 0.02892 0.04627 0.03777 297.78 3.1200E-004
24200 220.85 2841.1 0.02804 0.04482 0.03658 297.92 3.2235E-004
24400 221.05 2754.6 0.02719 0.04341 0.03544 298.05 3.3302E-00424600 221.25 2670.8 0.02636 0.04205 0.03433 298.19 3.4404E-004
24800 221.45 2589.6 0.02556 0.04074 0.03325 298.32 3.5542E-004
25000 221.65 2510.9 0.02478 0.03946 0.03222 298.45 3.6716E-004
25200 221.85 2434.7 0.02403 0.03823 0.03121 298.59 3.7927E-004
25400 222.05 2360.9 0.02330 0.03704 0.03024 298.72 3.9178E-004
25600 222.25 2289.4 0.02259 0.03589 0.02929 298.86 4.0468E-004
25800 222.45 2220.1 0.02191 0.03477 0.02838 298.99 4.1800E-004
26000 222.65 2153.0 0.02125 0.03369 0.02750 299.13 4.3174E-004
26200 222.85 2087.9 0.02061 0.03264 0.02664 299.26 4.4593E-004
26400 223.05 2024.9 0.01998 0.03163 0.02582 299.40 4.6056E-004
26600 223.25 1963.9 0.01938 0.03064 0.02502 299.53 4.7566E-004
26800 223.45 1904.7 0.01880 0.02969 0.02424 299.66 4.9124E-004
27000 223.65 1847.3 0.01823 0.02878 0.02349 299.80 5.0732E-004
27200 223.85 1791.8 0.01768 0.02788 0.02276 299.93 5.2391E-004
27400 224.05 1737.9 0.01715 0.02702 0.02206 300.07 5.4102E-004
27600 224.25 1685.8 0.01664 0.02619 0.02138 300.20 5.5868E-004
27800 224.45 1635.2 0.01614 0.02538 0.02072 300.33 5.7690E-004
28000 224.65 1586.2 0.01565 0.02460 0.02008 300.47 5.9569E-004
Table 2.1 Atmospheric properties in ISA (Cont…)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 16
28200 224.85 1538.7 0.01519 0.02384 0.01946 300.60 6.1508E-004
28400 225.05 1492.6 0.01473 0.02311 0.01886 300.74 6.3508E-004
28600 225.25 1448.0 0.01429 0.02239 0.01828 300.87 6.5572E-004
28800 225.45 1404.8 0.01386 0.02171 0.01772 301.00 6.7700E-00429000 225.65 1362.9 0.01345 0.02104 0.01718 301.14 6.9896E-004
29200 225.85 1322.2 0.01305 0.02040 0.01665 301.27 7.2161E-004
29400 226.05 1282.8 0.01266 0.01977 0.01614 301.40 7.4497E-004
29600 226.25 1244.7 0.01228 0.01916 0.01564 301.54 7.6906E-004
29800 226.45 1207.6 0.01192 0.01858 0.01517 301.67 7.9391E-004
30000 226.65 1171.8 0.01156 0.01801 0.01470 301.80 8.1954E-004
30200 226.85 1137.0 0.01122 0.01746 0.01425 301.94 8.4598E-00430400 227.05 1103.3 0.01089 0.01693 0.01382 302.07 8.7324E-004
30600 227.25 1070.6 0.01057 0.01641 0.01340 302.20 9.0136E-004
30800 227.45 1038.9 0.01025 0.01591 0.01299 302.33 9.3035E-004
31000 227.65 1008.1 0.00995 0.01543 0.01259 302.47 9.6026E-004
31200 227.85 978.3 0.00966 0.01496 0.01221 302.60 9.9109E-004
31400 228.05 949.5 0.00937 0.01450 0.01184 302.73 1.0229E-003
31600 228.25 921.4 0.00909 0.01406 0.01148 302.87 1.0557E-003
31800 228.45 894.3 0.00883 0.01364 0.01113 303.00 1.0895E-003
32000 228.65 867.9 0.00857 0.01322 0.01079 303.13 1.1243E-003
Table 2.1 Atmospheric properties in ISA
Note: Following values / expressions have been used while preparing ISA table.
2 -2
2R=287.05287m sec Kg= 9.80665m/s
Sutherland formula for viscosity:
3/2-6 T
μ = 1.458X10 [ ]T+110.4
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 17
In troposphere (h = 0 to 11000 m): T= 288.15 - 0.0065 h.
p = 101325 [1-0.000022588h] 5.25588
ρ = 1.225 [1-0.000022588h]
4.25588
.
In lower stratosphere (h = 11000 to 20000 km): T=216.65 K.
p = 22632 exp {-0.000157688 (h-11000)}
ρ = 0.36391 exp {-0.000157688 (h-11000)}
In middle stratosphere (h = 20000 to 32000 km):
T = 216.65 + 0.001h
p = 5474.9 [1+0.000004616(h-20000)]-34.1632
ρ = 0.08803 [1+0.000004616(h-20000)]-35.1632
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 2
Reference
2.1 Gunston, B, “The Cambridge aerospace dictionary” Cambridge University
Press (2004).
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 2
Exercises
2.1 On a certain day the pressure at sea level is 758 mm of mercury
(101059 N / m2) and the temperature is 25oC. The temperature is found to fall
linearly with height to -55oC at 12km and after that it remains constant upto
20 km. Calculate the pressure, density and kinematic viscosity at 8km and 16km
altitude.
(Hint : When the temperature variation is linear, Eqs. (2.6) and (2.7) can be used
to obtain the pressure and density at a chosen altitude by using appropriate
values of p0, T0, 0ρ and λ . As regards the constant temperature region, an
equation similar to Eq (2.10) can be used; note that, in this exercise, the
tropopause is at 12 km altitude)
[Answers:
p8 = 36,812 N/m2, 8 = 0.5238 kg/m3, 8 = 3.002 x 10-5 m2/sec,
p 16 = 10897 N/m2, 16 = 0.1740 kg/m3, 16 = 8.218 x 10-5 m2/sec]
Remark : Due to round off errors in calculations, the student may get the
numerical values which are slightly different from those given as answers. Values
within 0.5% of those given as answers can be regarded as correct.
2.2 If the altimeter in an airplane reads 5000m, on the day described in exercise
2.1, what is the altitude of airplane above mean sea level? What would be the
indicated altitude after landing on aerodrome at sea level?
(Hint: An altimeter is an instrument which senses the ambient pressure and
indicates height in ISA corresponding to that pressure. It does not read the
correct altitude when the atmospheric conditions differ from ISA.
To solve this exercise, obtain the pressure corresponding to 5000 m altitude inISA. Then find the altitude corresponding to this pressure in the atmospheric
conditions prevailing as in exercise 2.1. As regards the second part of this
exercise, the pressure at the sea level on that day is 101059 N/m2. When the
airplane lands at sea level, the altimeter would indicate altitude, in ISA,
corresponding to this pressure. In actual practice, the air traffic control would
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
inform the pilot about the local ambient pressure and the pilot would adjust zero
reading of his altimeter.)
[Answers: 5152 m, 22.3 m].
2.3 An altimeter calibrated according to ISA reads an altitude of 3,600 m. If the
ambient temperature is –60 C, calculate the ambient density.
[Answer: 0.847 kg/m3].
2.4 During a flight test for climb performance, the following readings were
observed at two altitudes:
Record Number 1 2
Indicate altitude (m) 1,300 1,600
Ambient temperature (0C) 16 14
The altimeter is calibrated according to ISA. Obtain the true difference of height
between the two indicated altitudes.
(Hint: Note that the ambient temperatures are different from those in ISA at 1300
and 1600 m altitudes. Hence the actual altitudes are different from the indicated
altitudes. To get the difference between these two altitudes ( Δh), obtain
pressures at 1300 and 1600 m heights in ISA. Let the difference in pressures be
Δp. Calculate density at the two altitudes using corresponding pressures and
temperature. Take average of the two densities (avg). Using Eq. (2.2) : Δh ≈ - Δp / {avg x g} )
[Answer: 311 m]
Remark:
The difference between the actual altitudes (311 m) and the indicated
altitudes (300 m) is small. Since altimeters of all the airplanes are calibrated
using ISA, the difference between indicated altitudes and actual altitudes of two
airplanes will be small. To take care of any uncertainty, the flight paths of two
airplanes are separated by several hundred meters. However, with the
availability of Global Positioning System (GPS) the separation between two
airplanes can be reduced.
2.5 A light airplane is flying at a speed of 220 kmph at an altitude of 3.2 km.
Assuming ISA conditions and the mean chord of the wing to be 1.5 m, obtain the
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
Reynolds number, based on wing mean chord, and the Mach number in this
flight.
[Answers: Re = 4.83 x 106, M = 0.186]
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 2
Table 2.1 Atmospheric properties in ISA
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
Altit-
ude
(m)
Tempe-
rature
(K)
Pressure
(N/m2)
δ
(p/po)
Density
(kg/m3)
σ
(ρ/ρo)
speed
of
sound
(m/s)
Kinematic
viscosity
(m2/s)
0 288.15 101325.0 1.00000 1.22500 1.00000 340.29 1.4607E-005
200 286.85 98945.3 0.97651 1.20165 0.98094 339.53 1.4839E-005
400 285.55 96611.0 0.95348 1.17864 0.96216 338.76 1.5075E-005
600 284.25 94321.6 0.93088 1.15598 0.94365 337.98 1.5316E-005
800 282.95 92076.3 0.90872 1.13364 0.92542 337.21 1.5562E-005
1000 281.65 89874.4 0.88699 1.11164 0.90746 336.43 1.5813E-005
1200 280.35 87715.4 0.86568 1.08997 0.88977 335.66 1.6069E-005
1400 279.05 85598.6 0.84479 1.06862 0.87234 334.88 1.6331E-005
1600 277.75 83523.3 0.82431 1.04759 0.85518 334.10 1.6598E-005
1800 276.45 81489.0 0.80423 1.02688 0.83827 333.31 1.6870E-005
2000 275.15 79494.9 0.78455 1.00649 0.82162 332.53 1.7148E-005
2200 273.85 77540.6 0.76527 0.98640 0.80523 331.74 1.7432E-005
2400 272.55 75625.4 0.74636 0.96663 0.78908 330.95 1.7723E-005
2600 271.25 73748.6 0.72784 0.94716 0.77319 330.16 1.8019E-005
2800 269.95 71909.7 0.70969 0.92799 0.75754 329.37 1.8321E-005
3000 268.65 70108.2 0.69191 0.90912 0.74214 328.58 1.8630E-005
3200 267.35 68343.3 0.67450 0.89054 0.72697 327.78 1.8946E-005
3400 266.05 66614.6 0.65744 0.87226 0.71205 326.98 1.9269E-005
3600 264.75 64921.5 0.64073 0.85426 0.69736 326.18 1.9598E-005
3800 263.45 63263.4 0.62436 0.83655 0.68290 325.38 1.9935E-0054000 262.15 61639.8 0.60834 0.81912 0.66867 324.58 2.0279E-005
4200 260.85 60050.0 0.59265 0.80197 0.65467 323.77 2.0631E-005
4400 259.55 58493.7 0.57729 0.78510 0.64090 322.97 2.0990E-005
4600 258.25 56970.1 0.56225 0.76850 0.62735 322.16 2.1358E-005
Table 2.1 Atmospheric properties in ISA (Cont..)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-2
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4800 256.95 55478.9 0.54753 0.75217 0.61402 321.34 2.1734E-005
5000 255.65 54019.4 0.53313 0.73611 0.60091 320.53 2.2118E-005
5200 254.35 52591.2 0.51903 0.72031 0.58801 319.71 2.2511E-005
5400 253.05 51193.7 0.50524 0.70477 0.57532 318.90 2.2913E-0055600 251.75 49826.4 0.49175 0.68949 0.56285 318.08 2.3324E-005
5800 250.45 48488.8 0.47855 0.67446 0.55058 317.25 2.3744E-005
6000 249.15 47180.5 0.46564 0.65969 0.53852 316.43 2.4174E-005
6200 247.85 45900.9 0.45301 0.64516 0.52666 315.60 2.4614E-005
6400 246.55 44649.5 0.44066 0.63088 0.51501 314.77 2.5064E-005
6600 245.25 43425.9 0.42858 0.61685 0.50355 313.94 2.5525E-005
6800 243.95 42229.6 0.41677 0.60305 0.49229 313.11 2.5997E-0057000 242.65 41060.2 0.40523 0.58949 0.48122 312.27 2.6480E-005
7200 241.35 39917.1 0.39395 0.57617 0.47034 311.44 2.6974E-005
7400 240.05 38799.9 0.38292 0.56308 0.45965 310.60 2.7480E-005
7600 238.75 37708.1 0.37215 0.55021 0.44915 309.75 2.7998E-005
7800 237.45 36641.4 0.36162 0.53757 0.43884 308.91 2.8529E-005
8000 236.15 35599.2 0.35134 0.52516 0.42870 308.06 2.9073E-005
8200 234.85 34581.2 0.34129 0.51296 0.41875 307.21 2.9629E-005
8400 233.55 33586.9 0.33148 0.50099 0.40897 306.36 3.0200E-005
8600 232.25 32615.8 0.32189 0.48923 0.39937 305.51 3.0784E-005
8800 230.95 31667.6 0.31254 0.47768 0.38994 304.65 3.1383E-005
9000 229.65 30741.9 0.30340 0.46634 0.38069 303.79 3.1997E-005
9200 228.35 29838.2 0.29448 0.45521 0.37160 302.93 3.2627E-005
9400 227.05 28956.1 0.28577 0.44428 0.36268 302.07 3.3272E-005
9600 225.75 28095.2 0.27728 0.43355 0.35392 301.20 3.3933E-005
9800 224.45 27255.2 0.26899 0.42303 0.34533 300.33 3.4611E-005
10000 223.15 26435.7 0.26090 0.41270 0.33690 299.46 3.5307E-005
10200 221.85 25636.2 0.25301 0.40256 0.32862 298.59 3.6020E-005
10400 220.55 24856.4 0.24531 0.39262 0.32050 297.71 3.6752E-005
Table 2.1 Atmospheric properties in ISA (Cont..)
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10600 219.25 24096.0 0.23781 0.38286 0.31254 296.83 3.7503E-005
10800 217.95 23354.4 0.23049 0.37329 0.30473 295.95 3.8274E-005
11000 216.65 22631.5 0.22336 0.36391 0.29707 295.07 3.9065E-005
11200 216.65 21929.4 0.21643 0.35262 0.28785 295.07 4.0316E-00511400 216.65 21248.6 0.20971 0.34167 0.27892 295.07 4.1608E-005
11600 216.65 20588.9 0.20320 0.33106 0.27026 295.07 4.2941E-005
11800 216.65 19949.7 0.19689 0.32079 0.26187 295.07 4.4317E-005
12000 216.65 19330.4 0.19078 0.31083 0.25374 295.07 4.5736E-005
12200 216.65 18730.2 0.18485 0.30118 0.24586 295.07 4.7202E-005
12400 216.65 18148.7 0.17911 0.29183 0.23823 295.07 4.8714E-005
12600 216.65 17585.3 0.17355 0.28277 0.23083 295.07 5.0275E-00512800 216.65 17039.4 0.16817 0.27399 0.22366 295.07 5.1886E-005
13000 216.65 16510.4 0.16294 0.26548 0.21672 295.07 5.3548E-005
13200 216.65 15997.8 0.15789 0.25724 0.20999 295.07 5.5264E-005
13400 216.65 15501.1 0.15298 0.24925 0.20347 295.07 5.7035E-005
13600 216.65 15019.9 0.14823 0.24152 0.19716 295.07 5.8862E-005
13800 216.65 14553.6 0.14363 0.23402 0.19104 295.07 6.0748E-005
14000 216.65 14101.8 0.13917 0.22675 0.18510 295.07 6.2694E-005
14200 216.65 13664.0 0.13485 0.21971 0.17936 295.07 6.4703E-005
14400 216.65 13239.8 0.13067 0.21289 0.17379 295.07 6.6776E-005
14600 216.65 12828.7 0.12661 0.20628 0.16839 295.07 6.8916E-005
14800 216.65 12430.5 0.12268 0.19988 0.16317 295.07 7.1124E-005
15000 216.65 12044.6 0.11887 0.19367 0.15810 295.07 7.3403E-005
15200 216.65 11670.6 0.11518 0.18766 0.15319 295.07 7.5754E-005
15400 216.65 11308.3 0.11160 0.18183 0.14844 295.07 7.8182E-005
15600 216.65 10957.2 0.10814 0.17619 0.14383 295.07 8.0687E-005
15800 216.65 10617.1 0.10478 0.17072 0.13936 295.07 8.3272E-005
16000 216.65 10287.5 0.10153 0.16542 0.13504 295.07 8.5940E-005
16200 216.65 9968.1 0.09838 0.16028 0.13084 295.07 8.8693E-005
Table 2.1 Atmospheric properties in ISA (Cont..)
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16400 216.65 9658.6 0.09532 0.15531 0.12678 295.07 9.1535E-005
16600 216.65 9358.8 0.09236 0.15049 0.12285 295.07 9.4468E-005
16800 216.65 9068.2 0.08950 0.14581 0.11903 295.07 9.7495E-005
17000 216.65 8786.7 0.08672 0.14129 0.11534 295.07 1.0062E-00417200 216.65 8513.9 0.08403 0.13690 0.11176 295.07 1.0384E-004
17400 216.65 8249.6 0.08142 0.13265 0.10829 295.07 1.0717E-004
17600 216.65 7993.5 0.07889 0.12853 0.10492 295.07 1.1060E-004
17800 216.65 7745.3 0.07644 0.12454 0.10167 295.07 1.1415E-004
18000 216.65 7504.8 0.07407 0.12068 0.09851 295.07 1.1780E-004
18200 216.65 7271.9 0.07177 0.11693 0.09545 295.07 1.2158E-004
18400 216.65 7046.1 0.06954 0.11330 0.09249 295.07 1.2547E-00418600 216.65 6827.3 0.06738 0.10978 0.08962 295.07 1.2949E-004
18800 216.65 6615.4 0.06529 0.10637 0.08684 295.07 1.3364E-004
19000 216.65 6410.0 0.06326 0.10307 0.08414 295.07 1.3793E-004
19200 216.65 6211.0 0.06130 0.09987 0.08153 295.07 1.4234E-004
19400 216.65 6018.2 0.05939 0.09677 0.07900 295.07 1.4690E-004
19600 216.65 5831.3 0.05755 0.09377 0.07654 295.07 1.5161E-004
19800 216.65 5650.3 0.05576 0.09086 0.07417 295.07 1.5647E-004
20000 216.65 5474.9 0.05403 0.08803 0.07187 295.07 1.6148E-004
20200 216.85 5305.0 0.05236 0.08522 0.06957 295.21 1.6694E-004
20400 217.05 5140.5 0.05073 0.08251 0.06735 295.34 1.7257E-004
20600 217.25 4981.3 0.04916 0.07988 0.06521 295.48 1.7839E-004
20800 217.45 4827.1 0.04764 0.07733 0.06313 295.61 1.8440E-004
21000 217.65 4677.9 0.04617 0.07487 0.06112 295.75 1.9060E-004
21200 217.85 4533.3 0.04474 0.07249 0.05918 295.89 1.9701E-004
21400 218.05 4393.4 0.04336 0.07019 0.05730 296.02 2.0363E-004
21600 218.25 4257.9 0.04202 0.06796 0.05548 296.16 2.1046E-004
21800 218.45 4126.8 0.04073 0.06581 0.05372 296.29 2.1752E-004
22000 218.65 3999.7 0.03947 0.06373 0.05202 296.43 2.2480E-004
Table 2.1 Atmospheric properties in ISA (Cont..)
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22200 218.85 3876.7 0.03826 0.06171 0.05038 296.56 2.3232E-004
22400 219.05 3757.6 0.03708 0.05976 0.04878 296.70 2.4009E-004
22600 219.25 3642.3 0.03595 0.05787 0.04724 296.83 2.4811E-004
22800 219.45 3530.5 0.03484 0.05605 0.04575 296.97 2.5639E-00423000 219.65 3422.4 0.03378 0.05428 0.04431 297.11 2.6494E-004
23200 219.85 3317.6 0.03274 0.05257 0.04291 297.24 2.7376E-004
23400 220.05 3216.1 0.03174 0.05091 0.04156 297.38 2.8287E-004
23600 220.25 3117.8 0.03077 0.04931 0.04026 297.51 2.9228E-004
23800 220.45 3022.6 0.02983 0.04776 0.03899 297.65 3.0198E-004
24000 220.65 2930.4 0.02892 0.04627 0.03777 297.78 3.1200E-004
24200 220.85 2841.1 0.02804 0.04482 0.03658 297.92 3.2235E-00424400 221.05 2754.6 0.02719 0.04341 0.03544 298.05 3.3302E-004
24600 221.25 2670.8 0.02636 0.04205 0.03433 298.19 3.4404E-004
24800 221.45 2589.6 0.02556 0.04074 0.03325 298.32 3.5542E-004
25000 221.65 2510.9 0.02478 0.03946 0.03222 298.45 3.6716E-004
25200 221.85 2434.7 0.02403 0.03823 0.03121 298.59 3.7927E-004
25400 222.05 2360.9 0.02330 0.03704 0.03024 298.72 3.9178E-004
25600 222.25 2289.4 0.02259 0.03589 0.02929 298.86 4.0468E-004
25800 222.45 2220.1 0.02191 0.03477 0.02838 298.99 4.1800E-004
26000 222.65 2153.0 0.02125 0.03369 0.02750 299.13 4.3174E-004
26200 222.85 2087.9 0.02061 0.03264 0.02664 299.26 4.4593E-004
26400 223.05 2024.9 0.01998 0.03163 0.02582 299.40 4.6056E-004
26600 223.25 1963.9 0.01938 0.03064 0.02502 299.53 4.7566E-004
26800 223.45 1904.7 0.01880 0.02969 0.02424 299.66 4.9124E-004
27000 223.65 1847.3 0.01823 0.02878 0.02349 299.80 5.0732E-004
27200 223.85 1791.8 0.01768 0.02788 0.02276 299.93 5.2391E-004
27400 224.05 1737.9 0.01715 0.02702 0.02206 300.07 5.4102E-004
27600 224.25 1685.8 0.01664 0.02619 0.02138 300.20 5.5868E-004
27800 224.45 1635.2 0.01614 0.02538 0.02072 300.33 5.7690E-004
Table 2.1 Atmospheric properties in ISA (Cont..)
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28000 224.65 1586.2 0.01565 0.02460 0.02008 300.47 5.9569E-004
28200 224.85 1538.7 0.01519 0.02384 0.01946 300.60 6.1508E-004
28400 225.05 1492.6 0.01473 0.02311 0.01886 300.74 6.3508E-004
28600 225.25 1448.0 0.01429 0.02239 0.01828 300.87 6.5572E-00428800 225.45 1404.8 0.01386 0.02171 0.01772 301.00 6.7700E-004
29000 225.65 1362.9 0.01345 0.02104 0.01718 301.14 6.9896E-004
29200 225.85 1322.2 0.01305 0.02040 0.01665 301.27 7.2161E-004
29400 226.05 1282.8 0.01266 0.01977 0.01614 301.40 7.4497E-004
29600 226.25 1244.7 0.01228 0.01916 0.01564 301.54 7.6906E-004
29800 226.45 1207.6 0.01192 0.01858 0.01517 301.67 7.9391E-004
30000 226.65 1171.8 0.01156 0.01801 0.01470 301.80 8.1954E-00430200 226.85 1137.0 0.01122 0.01746 0.01425 301.94 8.4598E-004
30400 227.05 1103.3 0.01089 0.01693 0.01382 302.07 8.7324E-004
30600 227.25 1070.6 0.01057 0.01641 0.01340 302.20 9.0136E-004
30800 227.45 1038.9 0.01025 0.01591 0.01299 302.33 9.3035E-004
31000 227.65 1008.1 0.00995 0.01543 0.01259 302.47 9.6026E-004
31200 227.85 978.3 0.00966 0.01496 0.01221 302.60 9.9109E-004
31400 228.05 949.5 0.00937 0.01450 0.01184 302.73 1.0229E-003
31600 228.25 921.4 0.00909 0.01406 0.01148 302.87 1.0557E-003
31800 228.45 894.3 0.00883 0.01364 0.01113 303.00 1.0895E-003
32000 228.65 867.9 0.00857 0.01322 0.01079 303.13 1.1243E-003
Table 2.1 Atmospheric properties in ISA
Note: Following values / expressions have been used while preparing ISA table.2 -2
2
R=287.05287m sec K
g= 9.80665m/s
Sutherland formula for viscosity:
3/2-6 T
μ = 1.458X10 [ ]T+110.4
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In troposphere (h = 0 to 11000 m): T= 288.15 - 0.0065 h.
p = 101325 [1-0.000022588h] 5.25588
ρ = 1.225 [1-0.000022588h]4.25588 .
In lower stratosphere (h = 11000 to 20000 km): T=216.65 K.
p = 22632 exp {-0.000157688 (h-11000)}
ρ = 0.36391 exp {-0.000157688 (h-11000)}
In middle stratosphere (h = 20000 to 32000 km):
T = 216.65 + 0.001h
p = 5474.9 [1+0.000004616(h-20000)]-34.1632
ρ = 0.08803 [1+0.000004616(h-20000)]-35.1632
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 3
Drag polar
(Lectures 6 to 12)
Keywords: Various types of drags; streamlined body and bluff body; boundary
layers; airfoil characteristics and designations; drags of airplane components;
drag polars at subsonic, transonic, supersonic and hypersonic speeds; high lift
devices
Topics
3.1. Introduct ion- Need and definition of drag polar
3.1.1 Contributions to airplane drag3.1.2 Interference drag
3.1.3 Contributions to airplane lift
3.1.4 Contributions to airplane pitching moment
3.1.5 Drag coefficient, lift coefficient and pitching moment coefficient of the
airplane
3.1.6 Categorization of airplane components
3.2 Estimation of drag polar at low subsonic speeds
3.2.1 Angle of attack of airplane, wing incidence and tail incidence
3.2.2 Skin friction drag, pressure drag and profile drag of an airfoil
3.2.3 Summary of lift coefficient, drag coefficient, pitching moment
coefficient, centre of pressure and aerodynamic centre of an airfoil
3.2.4 Examples of pressure coefficient distributions
3.2.5 Introduction to boundary layer theory
3.2.6 Boundary layer over a flat plate – height of boundary layer,
displacement thickness and skin friction drag
3.2.7 Boundary layer separation, adverse pressure gradient and
favourable pressure gradient
3.2.8 Boundary layer transition
3.2.9 Turbulent boundary layer over a flat plate
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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3.2.10 General remarks on boundary layers
3.2.11 Presentation of aerodynamic characteristics of airfoils
3.2.12 Geometric characteristics of airfoils
3.2.13 Airfoil nomenclature\designation
3.2.14 Induced drag of wing
3.2.15 Drag coefficient of fuselage
3.2.16 Drag coefficients of other components
3.2.17 Parabolic drag polar, parasite drag, induced drag and Oswald
efficiency factor
3.2.18 Parasite drag area and equivalent skin friction coefficient
3.2.19 A note on estimation of minimum drag coefficients of wings and
bodies
3.2.20 Typical values of CDO, A, e and subsonic drag polar
3.2.21 Winglets and their effect on induced drag
3.3 Drag polar at high subsonic, transonic and supersonic speeds
3.3.1 Some aspects of supersonic flow - shock wave, expansion fan
and bow shock
3.3.2 Drag at supersonic speeds
3.3.3 Transonic flow regime - critical Mach number and drag
divergence Mach number of airfoils, wings and fuselage.
3.3.4 Parabolic drag polar at high speeds
3.3.5 Guidelines for variations of CDo and K for subsonic jet transport
airplanes
3.3.6 Variations of CDo and K for a fighter airplane
3.3.7 Area ruling
3.4 Drag polar at hypersonic speeds
3.5 Lift to drag ratio
3.6 Other types of drags
3.6.1 Cooling drag
3.6.2 Base drag
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
3.6.3 External stores drag
3.6.4 Leakage drag
3.6.5 Trim drag
3.7 High li ft devices
3.7.1 Need for increasing maximum lift coefficient (CLmax)
3.7.2 Factors limiting maximum lift coefficient
3.7.3 Ways to increase maximum lift coefficient viz. increase in camber,
boundary layer control and increase in area
3.7.4 Guidelines for values of maximum lift coefficients of wings with
various high lift devices
References
Exercises
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 4
Chapter 3
Lecture 6
Drag polar – 1
Topics
3.1. Introduct ion- Need and definition of drag polar
3.1.1 Contributions to airplane drag
3.1.2 Interference drag
3.1.3 Contributions to airplane lift
3.1.4 Contributions to airplane pitching moment
3.1.5 Drag coefficient, lift coefficient and pitching moment coefficient of the
airplane
3.1.6 Categorization of airplane components
3.2 Estimation of drag polar at low subsonic speeds
3.2.1 Angle of attack of airplane, wing incidence and tail incidence
3.2.2 Skin friction drag, pressure drag and profile drag of an airfoil
3.1 Introduction- Need and definition of drag polar
As mentioned in section 1.9, to obtain the performance of an airplane
requires the value of the drag coefficient of the airplane (CD) when the lift
coefficient (CL) and Mach number (M) are given. The relationship between the
drag coefficient and the lift coefficient is called ‘Drag polar’. It may be pointed out
that aerodynamics generally deals with the drag, lift and pitching moment of
individual components like wing, fuselage etc. Whereas, for the estimation of the
airplane performance the knowledge of the drag, lift and pitching moment of the
entire airplane is required.
Equation (1.6) indicates that the drag coefficient is a function of lift
coefficient (CL), Mach number (M) and Reynolds number (Re). However, for a
given airplane a single drag polar can be used for flights upto critical Mach
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number (Ref.1.4 section 10.14); see sections 3.3.3 to 3.3.5 for details of critical
Mach number. For airplanes flying at transonic and supersonic speeds, the drag
polar depends on Mach number. Hence, the usual practice is to obtain the drag
polar of subsonic airplanes at a suitable flight speed (generally the cruising
speed) and for a high speed airplane, the drag polars are obtained at suitable
values of Mach numbers spread over the range of operating Mach numbers.
In this chapter the estimation of the drag polar at subsonic, transonic and
supersonic speeds is discussed. The topic of drag polar at hypersonic speed is
also touched up on.
3.1.1 Contributions to airplane drag
The usual method to estimate the drag of an airplane is to add the drags
of the major components of the airplane and then apply correction for the
interference effects.
The major components of the airplane which contribute to drag are wing,
fuselage, horizontal tail, vertical tail, nacelle(s) and landing gear.
Thus,
D = Dwing
+ Dfuse
+ Dht
+ Dvt + D
nac + D
lg+ D
etc + D
int (3.1)
where Dwing
, Dfuse
, Dht, D
vt , D
nac and D
lg denote drag due to wing, fuselage,
horizontal tail, vertical tail , nacelle(s) and landing gear respectively.
Detc
includes the drag of items like external fuel tanks, bombs, struts etc.
Dint
is the drag due to interference which is described in the next section.
3.1.2 Intereference drag
While presenting the data on the drag of wing or fuselage or any other
component of the airplane, the data generally refers to the drag of that
component when it is alone in the airstream and free from the influence of any
other component. Whereas, in an airplane, the wing, the fuselage and the tails
are present in close proximity of each other and the flow past one component is
influenced by the others. As a result, the drag of the airplane as a combination of
different components is different from the sum of the drags of individual
components. To appreciate this, let us consider the case examined in Ref.3.1.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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Flow past a rather thick airfoil section, shown in Fig.3.1a, is examined at a
Reynolds number of 420,000. The maximum thickness and the chord of the
airfoil are denoted respectively by ‘d’ and ‘c’. The thickness ratio (d/c) for the
airfoil in Fig.3.1a is 33.3%.
The drag coefficient is defined as :
d 212
DC =
ρV cb ; b = span of the airfoil model
The drag coefficient (Cd) is found to be 0.0247.
Subsequently, another identical airfoil is placed side by side with a spacing(s)
as shown in Fig.3.1b. The tests were carried out for different values of s/d. It is
found that for large values of s/d, say s/d > 5, the flows past the two sections do
not interfere and the total drag coefficient of the combination is equal to the sum
of the drags of each airfoil namely (Cd)combination = 0.0494.However, at closer
spacings the results presented in table 3.1 are obtained.
s/d 1.16 1.4 1.8 2.0 2.6 4 5
(Cd)combination 0.1727 0.1194 0.0824 0.0761 0.0627 0.0527 0.0494
Cdint 0.2233 0.070 0.033 0.0267 0.0133 0.0033 0.0
Table 3.1 Interference drag coefficient for different spacings between two airfoils
Note:
(Cd)combination = (Cd)airfoil1 + (Cd)airfoil2 + Cdint
It is evident that Cdint depends on the relative positions and could be very large.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
(a) Single airfoil
(b) Configuration with airfoils placed side by side as seen in plan view
(c) Circular cylinder with splitter plate at rear
Fig.3.1 Interference effects
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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Remarks:
(i)The drag coefficient of the individual airfoil in this example is large as the airfoil
is thick and Reynolds number is rather low. Airfoils used on airplanes would have
thickness ratio (t/c) of 12 to 18% and the values of Cd, for Reynolds number of
6 x 106, would be around 0.006.
(ii) Ways to reduce interference drag
A large number of studies have been carried out on interference drag and
it is found that Dint
can be brought down to 5 to 10% of the sum of the drags of all
components, by giving proper fillets at the junctions of wing and fuselage and
tails and fuselage ( Fig.3.2 ).
Fig.3.2 Reduction of interference drag using fillets
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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(iii) Favorable interference effect
The interference effects need not always increase the drag. As an
example the drag of a circular cylinder with a splitter plate (Fig.3.1c) is lower than
the drag of a cylinder without it at certain Reynolds numbers (Ref.3.2). In an
another example, the birds flying in formation flight experience lower drag than
when flying individually.
(iv) Chapter VIII of Ref.3.3 can be consulted for additional information on
interference drag.
3.1.3 Contributions to airplane lift
The main contribution to the lift comes from wing-fuselage combination
and a small contribution from the horizontal tail i.e.
L = Lwing + fuselage + Lht (3.2)
For airplanes with wings having aspect ratio greater than six, the lift due to the
wing-fuselage combination is roughly equal to the lift produced by the gross wing
area. The gross wing area (S) is the planform area of the wing, extended into the
fuselage, up to the plane of the symmetry.
3.1.4 Contributions to airplane pitching moment
The pitching moment of the airplane is taken about its center of gravity
and denoted by Mcg
.
Main contributions to Mcg are from wing, fuselage, nacelle(s) and
horizontal tail i.e.
Mcg
= Mwing
+ Mfuselage
+ Mnac
+ Mht (3.3)
3.1.5 Drag coefficient, lift coefficient and pitching moment coefficient of the
airplane
To obtain the non-dimensional quantities namely drag coefficient (CD), lift
coefficient (CL) and pitching moment coefficient (Cmcg) of the airplane, the
reference quantities are the free stream dynamic pressure (½ ρ2V
), the gross
wing area (S) and the mean aerodynamic chord of the wing ( c ). Consequently,
cgD L mcg2 2 21 1 1
2 2 2
MD LC = ; C = ; C =
ρV S ρV S ρV Sc
(3.4)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 10
However, the drag coefficient and lift coefficient of the individual
components are based on their own reference areas as given below.
(a) For wing, horizontal tail and vertical tail the reference area is their planform
area.
(b)For fuselage, nacelle, fuel tanks, bombs and such other bodies the reference
area is either the wetted area or the frontal area. The wetted area is the area of
the surface of the body in contact with the fluid. The frontal area is the maximum
cross-sectional area of the body.
(c) For other components like landing gear the reference area is given along with
the definition of CD.
Remarks:
(i)The reference area, on which the CD and C
L of an individual component is
based, is also called proper area and denoted by S; the drag coefficient based
on S is denoted by C
D
.
(ii)The reference areas for different components are different for the following
reasons. The aim of using non-dimensional quantities like CD is to be able to
predict the characteristics of many similar shapes by carrying out computations
or tests on a few models. For this to be effective, the phenomena causing thedrag must be taken into account while specifying the reference qualities. In this
context the drag of streamline shapes like wing and slender bodies is mainly due
to the skin friction and depends on the wetted area. Whereas, the drag of bluff
bodies like the fuselage of a piston-engined airplane, is mainly the pressure drag
and depends on the frontal area. It may be added that for wings, the usual
practice is to take the reference area as the planform area because it (planform
area) is proportional to the wetted area.
(iii)At this stage the reader is advised to the revise the background on
aerodynamics (see for examples References 1.9, 1.10 and 1.12).
(iv) Following the above remarks, the total drag of the airplane can be expressed
as:
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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2 2 2 2Dwing fuse Dfuse nac Dnac ht Dht
2 2 2vt Dvt lg Dlg etc Detc int
1 1 1 1D = ρV SC + ρV S C + ρV S C + ρV S C
2 2 2 2
1 1 1+ ρV S C + ρV S C + ρV S C +D (3.5)
2 2 2
It may be recalled that Setc and CDetc refer to areas and drag coefficients of other
items like external fuel tanks, bombs, struts etc..
Or D 212
DC =
ρV S
lgfuse ht vt nac etcDwing Dfuse Dht Dvt Dnac Dlg Detc Dint
SS S S S S= C + C + C + C + C + C + C + C (3.6)
S S S S S S
The data on drag, lift and pitching moment, compiled from various sources, isavailable in references 1.9, 1.10, 1.12 and 3.3 to 3.9.
3.1.6 Categorization of airplane components
During the discussion in the previous section it was mentioned that (a) for
wing, horizontal tail and vertical tail, the planform area is taken as the reference
area, (b) for fuselage, the wetted area or the frontal area is taken as the
reference area. The reason for these specifications lies in the fact that in
aerodynamics the airplane components are categorised as (a) wing type
surfaces, (b) bodies and (c) others. This categorisation, described below, is
based on common geometrical features of certain airplane components.
Figure 3.3 shows the geometric parameters of a wing. It is observed that the
span (b) of the wing is much larger than the chord (c) of the wing section (or the
airfoil) and in turn the chord is much larger than the thickness (t) of the airfoil. For
wings of subsonic airplanes the ratio (b/c) is between 5 to 12 and the ratio (t/c)
for the commonly used profiles is 0.10 to 0.18 or 0.1t/c and 0.1c/b . This
separation of sizes ( or scales in more technical terms) enables the simplification
that the flow past a wing can be analysed as a study of flow past an airfoil and
then applying correction for the effect of finite wing span. It may be recalled that
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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in aerodynamics an airfoil is treated as a wing of infinite span or a two-
dimensional problem.
Fig.3.3 Geometric parameters of a wing
Hence, subsections 3.2.2 to 3.2.13 deal with various aspects of flow past airfoils
which are relevant to the estimation of drag polar. The subsequent subsection
deals with the induced drag which is the result of finite span. It may be added
that in aerodynamics, the quantity finite aspect ratio (A) is employed instead of
the finite span. The aspect ratio is defined as :
A = b2/S; b = wing span, S = wing planform area
Remarks :
(i)When the aspect ratio is less than about 5, which is characteristic of wings of
high speed airplanes, the flow past the wing has to be treated as three-
dimensional.
(ii) Horizontal tail, vertical tail and streamlined struts, seen on some low speed
airplanes, come under the category of wing type surfaces.
Figure 3.4a shows the fuselage of a jet airplane. Here the length (l f ) is much
larger than the height (h) and width (w), but ‘h’ and ‘w’ are generally not very
different in their dimensions. Hence, the flow past a fuselage cannot be
considered as two-dimensional. However, for jet airplanes, l f /h is around 6 to 10
and the analysis of flow past fuselage can be simplified by assuming the fuselage
to be a slender/streamlined body.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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Fig.3.4 Fuselage parameters
Figure 3.4 b shows the fuselage of a low speed airplane. Here l f /h is rather low
and the fuselage is treated as a bluff body.
Precise definitions of the streamlined body and bluff body are given in the
subsequent sections.
Remarks:
(i) As regards the analysis of flow is concerned, the fuselage, nacelle, external
fuel tanks, bombs, and antenna masts have common geometric features and are
categorised as “bodies”.
(ii) Components of airplane like landing gear, which do not fall under the above
two categories, are designated as ‘others’.
3.2. Estimation of drag polar at low subsonic speeds
As mentioned in the previous section, the drag polar of an airplane can be
obtained by summing-up the drags of individual components and then adding 5
to 10% for the interference drag. As the drag coefficient depends on the angle of
attack, this exercise has to be carried-out at different angles of attack. The
definition of the angle of attack of the airplane and brief descriptions of the drag
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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coefficients of the airplane components are presented before discussing the drag
polar.
3.2.1 Angles of attack of the airplane, wing incidence and tail inc idence
For defining the angle of attack of an airplane, the fuselage reference
line(FRL) is taken as the airplane reference line (Figs.1.9 and 3.5).The angle
between the free stream velocity and FRL is the angle of attack of the airplane.
However, the angles of attack of the wing and tail are not the same as that of the
fuselage.
The wing is fixed on the fuselage such that it makes an angle, iw, to the
fuselage reference line (Fig.3.5). This angle is called wing incidence. The angle
iw is generally chosen such that during the cruising flight the wing can produce
enough lift when fuselage is at zero angle of attack. This is done because the
fuselage produces least drag when it is at zero angle of attack and that is ideal
during the cruising flight. In other words, during cruise the wing produces the lift
required to balance the weight whereas the fuselage, being at zero angle of
attack, produces least drag.
The horizontal tail is set on fuselage at an angle i t (Fig.3.5). This angle is
called tail incidence. It is generally chosen in a manner that during cruise the lift
required from the tail, to make the airplane pitching moment zero, is produced by
the tail without elevator deflection. This is because, the drag, at low angles of
attack, is least when the required lift is produced without elevator deflection.
Remark :
The angles iw and it are measured clockwise from FRL. The angle iw is
positive but the angle it is generally negative.
Fig.3.5 Wing incidence (iw) and tail incidence (it)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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3.2.2 Skin f riction drag, pressure drag and, profi le drag of an airfoi l
The drag coefficient of a wing consist of the (i) the profile drag due to
airfoil (Cd) and (ii) the induced drag due to the finite aspect ratio of the wing (C Di).
The symbols dC and Cl with lower case suffices refer to the drag coefficient and
lift coefficient of the airfoil. The profile drag of the airfoil consists of the skin
friction drag and the pressure drag. It may be added that an element on the
surface of an airfoil, kept in a flow, experiences shear stress tangential to the
surface and pressure (p) normal to it (Fig.3.6). The shear stress multiplied by the
area of the element gives the tangential force. The component of this tangential
force in the free stream direction when integrated over the profile gives the skin
friction drag. Similarly, the pressure distribution results in normal force on the
element whose component in the free stream direction, integrated over the profile
Fig.3.6 Shear stress ( ) and pressure(p) on an airfoil
gives the pressure drag. The pressure drag is also called form drag. The sum of
the skin friction drag and the pressure drag is called ‘Profile drag’. The profile
drag depends on the airfoil shape, Reynolds number, angle of attack and surface
roughness.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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Chapter 3
Lecture 7
Drag polar – 2
Topics
3.2.3 Summary of lift coefficient, drag coefficient, pitching moment
coefficient, centre of pressure and aerodynamic centre of an airfoil
3.2.4 Examples of pressure coefficient distributions
3.2.5 Introduction to boundary layer theory
3.2.6 Boundary layer over a flat plate – height of boundary layer,
displacement thickness and skin friction drag
3.2.3 Summary of the lift coefficient, drag coefficient, pressure coefficient,
pitching moment coefficient, centre of pressure and aerodynamic centre of
an airfoil
In order to understand the dependence of pressure drag and skin friction
drag on various factors, it is appropriate, at this stage, to present brief
discussions on (I) generation of lift, drag and pitching moment from the
distributions of pressure (p) and shear stress ( ) and (II) outline of boundary
layer theory. These and the related topics are covered in this subsection and in
the subsections 3.2.4 to 3.2.10. In subsections 3.2.11 to 3.2.13 the airfoil
characteristics and their nomenclature are dealt with. Subsequently, the
estimation of the drags of wing, fuselage and the entire airplane at subsonic
speeds are discussed(sections 3.2.14 to 3.2.21).
Figure 3.7 shows an airfoil at an angle of attack ( α )kept in a stream of
velocityV . The resultant aerodynamic force (R) is produced due to the
distributions of the shear stress( ) and the pressure (p). The distributions also
produce a pitching moment (M). By definition, the component of R perpendicular
to the free stream direction is called lift (L) and the component along the free
stream direction is called drag (D). The resulant aerodynamic force (R) can also
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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be resolved along and perpendicular to the chord of the airfoil. These
components can be denoted by C and N respectively(Fig.3.7). From the
subsequent discussion in this section, it will be evident that it is more convenient
to evaluate N and C from the distributions of shear stress ( ) and pressure (p)
and then evaluate L and D.
Fig.3.7 Aerodynamic forces and moment on an airfoil
From Fig.3.7 it can be deduced that :
L = Ncosα - Csinα (3.7)
D = Nsinα+Ccosα (3.8)
Figure 3.8 shows elements of length and dsu and dsl at points Pu and P
l
on the upper and lower surfaces of the airfoil respectively. The cartesian
coordinates of points Pu and Pl are (xu,yu) and (x
l, y
l) respectively. Whereas su
and sl are respectively the distances along the airfoil surface, of the points Pu
and Pl measured from the stagnation point (Fig.3.8).
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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Fig.3.8 Pressure and shear stress at typical points on upper
and lower surfaces of an airfoil
To obtain the forces at points Pu and Pl , the local values of p and are
multiplied by the local area. Since the flow past an airfoil is treated as two-
dimensional, the span of the airfoil can be taken as unity without loss of
generality. Hence, the local area is (ds x 1) and the quantities, L, D, N and C, on
the airfoil, are the forces per unit span. Keeping these in mind, the local
contributions, dNu and dCu, to N and C respectively, from the element at point Pu are obtained as:
u u u u u u udN = -p ds cosθ - ds sinθ (3.9)
u u u u u u udC = -p ds sinθ + ds cosθ (3.10)
Note that the suffix ‘u’ denotes quantities at point Pu and the positive direction of
the angle uθ is as shown in Fig.3.8 .
Expressions similar to Eqs.(3.9) and (3.10) can be written down for the
contributions to N and C from element at point Pl .
Integrating over the entire airfoil yields :
u u u u u
uppersurface lowersurface
N = - p cosθ + sinθ ds + p cosθ - sinθ dsl l l l l
(3.11)
u u u u u
uppersurface lowersurface
C = -p sinθ + cosθ ds + p sinθ + cosθ dsl l l l l
(3.12)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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Proceeding in a similar manner, it can be shown that M le, the pitching
moment about leading edge of the airfoil, per unit span, is :
le u u u u u u u u u u u
uppersurface
M = p cosθ + sinθ x - p sinθ - cosθ y ds
lower surface
+ -p cosθ + sinθ x + p sinθ + cosθ y dsl l l l l l l l l l l
(3.13)
Note: Once N and C are known, the lift per unit span (L) and drag per unit
span (D) of the airfoil can be obtained using Eqs.(3.7) and (3.8).
It is convenient to work in terms of lift coefficient (Cl) and drag coefficient
(Cd). The definitions of these may be recalled as :
l
2
LC =
1ρV c
2
(3.14)
and
d2
DC =
1ρV c
2
(3.15)
It may be pointed out, that integration of a constant pressure, say p ,
around the body would not give any resultant force i.e.
p ds = 0 (3.16)
Hence, instead of ‘p’ the quantity p-p can be used in Eqs.(3.11), (3.12)
and (3.13). At this stage the following quantities are also defined.
pressure coefficient : p2
p-pC =
1ρV
2
(3.17)
skin friction drag coefficient :
f 2
c = 1ρV
2
(3.18)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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:
n2
c2
lemle
2 2
NNormalforcecoefficient C =
1ρV c
2
C
Chordwiseor axial forcecoefficient: C = (3.19)1ρV c
2
MPitchingmoment coefficient: C =
1ρV c
2
It may be noted that dx = ds cos θ and dy = -ds sin θ, where “ds” is an
elemental length around a point P on the surface and θ is the angle between the
normal to the element and the vertical (Fig.3.8). Note that θ is measured positive
in the clockwise sense. It can be shown that :
n p pu fu f
0 upper surface lower surface
c
c fu f pu p
0 upper surface lowersurface
c1
C = C -C dx + c dy + c dyc
(3.20)1
C = c +c dx + C dy - C dyc
l l
l l
Following section 10.2 of Ref.1.4, the expressions for Cn, Cc and Cmle can be
rewritten as:
2
1
1
1
un p pu fu f
0 0
uc pu p fu f
0 0
umle pu p fu f
0 0
upu fu u p2
0 0
c c
c c
c c
c c
dy dyC = C -C dx+ c +c dx
c dx dx
dy dy1C = C -C dx + c -c dx
c dx dx
dy dyC = C -C xdx - C +c x dx
c dx dx
dy d+ C +c y dx+ -C
c dx
l
l l
l
l l
l
l l
l
f
(3.21)
y+c y dx
dxl
l l
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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Remarks:
(i) From Cn and Cc the lift coefficient (Cl) and drag coefficient (Cd) are obtained
as :
n cC = C cosα - C sinα
l
(3.22 )
d n cC = C sinα + C cosα (3.23)
(ii) Centre of pressure : The point on the airfoil chord through which the
resultant aerodynamic force passes is the centre of pressure. The aerodynamic
moment about this point is zero. It may be noted that the location of centre of
pressure depends on the angle of attack or the lift coefficient.
(iii) Aerodynamic cent re: As the location of the centre of pressure depends on
lift coefficient (Cl ) the pitching moment coefficient about leading edge (Cmle) alsochanges with C
l. However, it is found that there is a point on the airfoil chord
about which the pitching moment coefficient is independent of the lift coefficient.
This point is called ‘Aerodynamic centre‘. For incompressible flow this point is
close to the quarter chord point of the airfoil.
(iv) If the distributions of Cp and cf are obtained by analytical or computational
methods, then the pressure drag coefficient (Cdp) and the skin friction drag
coefficient(Cdf ) can be evaluated.In experimental work the pressure distribution on an airfoil at different angles of
attack can be easily measured. However, measurement of shear stress on
an airfoil surface is difficult.The profile drag coefficient (Cd) of airfoil, which is the
sum of pressure drag coefficient and skin friction drag coefficient, is measured in
experiments by ‘Wake survey technique’ which is described in Chapter 9, section
‘f’ of Ref.3.10. In this technique, the momentum loss due to the presence of the
airfoil is calculated and equated to the drag (refer section 7.5.1 of Ref.3.11 for
derivation).
3.2.4 Examples of pressure coefficient d istributions
Though the expression for lift coefficient (Cl) involves both the pressure
coefficient (Cp) and the skin friction drag coefficient (cf ), the contribution of the
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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former i.e. Cp is predominant to decide Cl. On the other hand, the pressure drag
coefficient (Cdp) is determined by the distribution of Cp and the skin friction drag
coefficient (Cdf ) is decided by the distribution of shear stress .
In this subsection the distributions of CP in typical cases and their implications for
Cl and Cdp are discussed.
The distribution of the pressure coefficient is generally plotted on the outer side
of the surface of the body (Fig.3.9a). The length of the arrow indicates the
magnitude of Cp. As regards the sign convention, an arrow pointing towards the
surface indicates that Cp is positive or local pressure is more than the free stream
pressure p . An arrow pointing away from the surface indicates that Cp is
negative i.e. the local pressure is lower than p .
(a) Ideal fluid flow (b) Real fluid flow
Fig.3.9 Distribution of Cp around a circular cylinder
Figure 3.9 shows distributions of Cp in ideal fluid flow and real fluid flow past a
circular cylinder. It may be recalled that an ideal fluid is inviscid and
incompressible whereas a real fluid is viscous and compressible. From thedistribution of Cp in ideal fluid flow (Fig.3.9a) it is seen that the distribution is
symmetric about X-axis and Y-axis. It is evident that in this case, the net forces in
vertical and horizontal directions are zero. This results in Cl = 0, Cdp = 0. These
results are available in books on fluid mechanics and aerodynamics. In the real
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fluid flow case, shown in Fig.3.9b, it is seen that the flow separates from the body
(see description on boundary layer separation in section 3.2.7) and the pressure
coefficient behind the cylinder is negative and nearly constant. However, the
distribution is still symmetric about horizontal axis. Thus in this case Cl = 0 but
dpC > 0 .
The distributions of Cp over symmetrical and unsymmetrical foils at Cl = 0
and Cl > 0 are shown in Figs.3.10 a to d. Note also the locations of centre
pressure and the production of pitching moment for the unsymmetrical airfoil.
Flow visualization pictures at three angles of attack( α ) are shown in Figs.3.36 a,
b and c. An attached flow is seen at low angle of attack. Some separated flow is
seen at moderate angle of attack and large separated flow region is seen near α
close to the stalling angle ( stallα ). It may be pointed out that theoretical calculation
of skin friction drag using boundary layer theory can be done, when flow is
attached. This topic is discussed in the next subsection.
(a)Distribution of pressure coefficient on symmetrical airfoil at Cl= 0 and α= 0
Note : Lu = Ll
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(b) Distribution of pressure coefficient on symmetrical airfoil at Cl > 0 and α> 0
(c) Distribution of pressure coefficient on cambered airfoil at Cl = 0, α< 0 ;
Note: Lu and Llform a couple; centre of pressure is at infinity, Cmac < 0,
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(d) Distribution of pressure coefficient on cambered airfoil at Cl > 0, α> 0
Note : Cmac same as in Fig.(c)
Fig.3.10 Distributions of pressure coefficient on symmetrical and unsymmetric
airfoils at Cl = 0 and C
l> 0
3.2.5 Introduction to boundary layer theory
Under conditions of normal temperature and pressure a fluid satisfies the
‘No slip condition’ i.e. on the surface of a solid body the relative velocity between
the fluid and the solid wall is zero. Thus, when the body is at rest the velocity of
the fluid layer on the body is zero. In this and the subsequent subsections, the
body is considered to be at rest and the fluid moving past it. Though the velocity
is zero at the surface, a velocity of the order of free stream velocity is reached in
a very thin layer called ‘Boundary layer’. The velocity gradient normal to the
surface
Uy
is very high in the boundary layer. Hence even if the coefficient of
viscosity μ is small, the shear stress,
Uμ
y,in the boundary layer may be
large or comparable to other stresses like pressure. Outside the boundary layer
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the gradient / U y is very small and viscous stress can be ignored and flow
treated as inviscid. It may be recalled from text books on fluid mechanics, that in
an inviscid flow the Bernoulli’s equation is valid.
Features of the boundary layer over the surface of a streamlined body are shownin Fig.3.11a. On the surface of a bluff body the boundary layer develops upto a
certain extent and then separates (Fig.3.11b). The definitions of the streamlined
body and bluff body are presented at the end of this subsection.
(a) boundary layer over a streamlined body
(b) Boundary layer over a bluff body
Fig.3.11 Boundary layer over different shapes (not to scale)
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The features of the flow are as follows.
1.Near the leading edge (or the nose) of the body the flow is brought to rest.This
point is called the ‘Stagnation point’. A laminar boundary layer develops on the
surface starting from that point. It may be recalled, from topics on fluid
mechanics, that in a steady laminar flow the fluid particles move downstream in
smooth and regular trajectories; the streamlines are invariant and the fluid
properties like velocity, pressure and temperature at a point remain the same
with time. In an unsteady laminar flow the fluid properties at a point may vary but
are known functions of time. In a turbulent flow, on the other hand, the fluid
properties at a point are random functions of time. However, the motion is
organized in such a way that statistical averages can be taken. In a laminar
boundary layer the parameter which mainly influence its development is the
Reynolds number x eR = ρU x/μ ; x being distance along the surface, from the
stagnation point.
2.Depending on the Reynolds number (RX), the pressure gradient and other
parameters, the boundary layer may separate or become turbulent after
undergoing transition. The turbulent boundary layer may continue till the trailing
edge of the body (Fig.3.11a) or may separate from the surface of the body (point
‘S’ in Fig 3.11b). It may be added that the static pressure across the boundary
layer at a station ‘x’, is nearly constant with ‘y’. Hence the pressure gradient
referred here is the gradient (dp/dx) in the flow outside the boundary layer.
3.Nature of boundary layer decides the drag and the heat transfer from the body.
If the boundary layer is separated, the pressure in the rear portion of the body
does not reach the freestream value resulting in a large pressure drag (Fig.3.9b).
Incidently a streamlined body is one in which the major portion of drag is skin
friction drag. For a bluff body the major portion of drag is pressure drag. A
circular cylinder is a bluff body. An airfoil at low angle of attack is a streamlined
shape. But, an airfoil at high angle of attack like stallα is a bluff body.
Remark:
General discussion on boundary layer is a specialised topic and the
interested reader may consult Ref.3.11 for more information. Here, the features
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of the laminar and turbulent boundary layers on a flat plate are briefly described.
While discussing separation, the boundary layer over a curved surface is
considered.
3.2.6 Laminar boundary layer over flat plate – height o f boundary layer,
displacement thickness and skin fri ction drag
The equations of motion governing the flow of a viscous fluid are called
‘Navier-Stokes (N-S) equations’. For derivation of these equations refer to
chapter 15 of Ref.3.12. Taking into account the thinness of the boundary layer,
Prandtl simplified the N-S equations in 1904. These equations are called
‘Boundary layer equations’ (Chapter 16 of Ref.3.12). Solution of these equations,
for laminar boundary layer over a flat plate with uniform external stream, was
obtained by Blasius in 1908. Subsequently many others obtained the solution.
The numerical solution by Howarth, presented in Ref.3.10, chapter 7, is given in
Table 3.2. In this table U is the local velocity, Ue is the external velocity (which in
this particular case is V ), and η is the non-dimensional distance from the wall
defined as :
eUη = y
x
(3.24)
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Table 3.2 Non-dimensional velocity profile in a laminar boundary layer over a
flat plate
Height o f boundary layer
It is seen from table 3.2 that the external velocity (Ue) is attained very
gradually. Hence the height at which U/Ue equals 0.99 is taken as the height of
the boundary layer and denoted by 0.99δ . From table 3.2, eU/U 0.99 is attained
at η = 5 . Noting the definition of η in Eq.(3.24) gives :
e
0.99
U5 = δ x
Or 0.99 ex
e x
δ U x5 5= = ; R =
x U x R
(3.25)
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Figure 3.12 shows a typical non-dimensional velocity profile in a laminar
boundary layer. While presenting such a profile, it is a common practice to plot
U/Ue on the abscissa and ( 0.99y/δ ) on the ordinate.
Fig.3.12 Non-dimensional velocity profile in laminar and turbulent boundary
layers on a flat plate
It is seen from Eq.(3.25) that 0.99δ grows in proportion to
1
2
x (see Fig.3.13). It
may be added that in this special case of laminar boundary layer on flat plate, the
velocity profiles are similar at various stations i.e. the non-dimensional profiles of
U/Ue vs (y/ 0.99δ ) are same at all stations.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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Fig.3.13 Schematic growth of boundary layer
Displacement thickness and skin fric tion drag coefficient
The presence of boundary layer causes displacement of fluid and skin
friction drag. The displacement thickness
1
δ is defined as :
1
e0
Uδ = 1- dy
U
(3.26)
The local skin friction coefficient ( f C or cf )is defined as :
wallf f wall wall
2y= 0
e
uC = c = ; = μ ;Note: is a function of 'x'.
1 yρU
2
(3.27)
If the length of the plate is L, then the skin friction drag per unit span of the
plate (Df ) is :
f wall
0
L
D = dx
Hence, skin friction drag coefficient Cdf is given by:
f df
2
DC =
1ρV L
2
(3.28)
From the boundary layer profile (table 3.2) it can be shown that for a flat
plate of length, L, the expressions for 1δ and Cdf are:
1
L
δ 1.721=
L R (3.29)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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df L
L
V L1.328C = ; R =
R
(3.30)
Remark :
Reference 1.11, chapter 6 may be consulted for additional boundary
layer parameters like momentum thickness ( 2δ ), shape parameter (H = 1δ / 2δ )
and energy thickness ( 3δ ) of a boundary layer.
Example 3.1
Consider a flat plate of length 500 mm kept in an air stream of velocity 15 m/s.
Obtain (a) the boundary layer thickness 0.99δ and the displacement thickness
1δ at the end of the plate (b) the skin drag coefficient. Assume -6 2= 15×10 m /s
and the boundary layer to be laminar.
Solution:
L = 0.5 m, V = 15 m/s , -6 2= 15×10 m /s
Hence, 5
L -6
0.5×15R = = 5×10
15×10
Consequently, from Eq.(3.25):
-30.99
5L
δ 5 5= = = 7.07×10
L R 5×10
Or -3 -3
0.99δ = 7.07×10 ×0.5 = 3.54×10 m = 3.54mm
From Eq.(3.29):
-31
5L
δ 1.721 1.721= = = 2.434×10
L R 5×10
Or -5 -3
1δ = 2.434×10 ×0.5 = 1.217×10 m = 1.217 mm
From Eq.(3.30):
df 5
L
1.328 1.328C = = = 0.00188
R 5×10
Remark:
0.99δ /L is found to be 7.07 x 10-3. Hence the assumption of the thinness of
boundary layer is confirmed by the results.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 3
Lecture 8
Drag polar – 3
Topics
3.2.7 Boundary layer separation, adverse pressure gradient and
favourable pressure gradient
3.2.8 Boundary layer transition
3.2.9 Turbulent boundary layer over a flat plate
3.2.10 General remarks on boundary layers
3.2.7 Boundary layer separation, adverse pressure gradient and favourable
pressure gradient
When the flow takes place around airfoils and curved surfaces, the velocity
outside the boundary layer is not constant. From Bernoulli’s equation it can be
deduced that when the velocity decreases the pressure increases and vice-
versa. When the velocity is decreasing i.e. dp/dx is positive, the pressure
gradient is called ‘Adverse pressure gradient’. When dp/dx is negative it is called
‘Favourable pressure gradient’.Figure 3.14 shows the development of a boundary layer in an external stream
with adverse pressure gradient (dp/dx > 0). Such a flow may occur on the upper
surface of an airfoil beyond the point of maximum thickness. Since the static
pressure at a station remains almost constant across the boundary layer, the
pressure inside the boundary layer at stations separated by distance Δx also
increases in the downstream direction.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
Fig.3.14 Flow in boundary layer before and after point of separation (not to
scale)
Figure 3.14 also shows a small element ABCD in the boundary layer. The
pressure on the face AD is p whereas that on the face BC is p+ dp/dx Δx .
Since dp/dx is positive in this case, the net effect causes a deceleration of the
flow, in addition to that due to viscosity. The effect is more pronounced near the
surface and the velocity profile changes as shown in Fig.3.14. Finally at point S
the slope of the velocity profile at the wall, wall
U/ y , becomes zero. Besides
the change in shape, the boundary layer also thickens rapidly in the presence of
adverse pressure gradient. Downstream of the point S, there is a reversal of the
flow direction in the region adjacent to the wall. A line can be drawn (indicated as
dotted line in Fig.3.14) in such a way that the mass flow above this line is the
same as that ahead of point S. Below the dotted line, there is a region of
recirculating flow and the value of the stream function ψ for the dotted line is
zero. However, ahead of the point S, the ψ = 0 line is the surface of the body.
Thus, after the point S, it is observed that between the main flow (i.e. region
above ψ = 0 line) and the body surface lies a region of recirculating flow. When
this happens the flow is said to be ‘Separated’ and S is referred to as the ‘Point
of separation’. Due to separation, the pressure recovery, which would have taken
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
place in an unseparated flow, does not take place and the pressure drag of the
body increases.
Remarks:
(i) If the adverse pressure gradient is very gradual then separation may
not take place (Refer to Ch. 8 of Ref.3.11 for dp/dx needed for
separation).
(ii) Separation does not take place when the pressure gradient is
favourable.
(iii) In the two-dimensional case shown in Fig.3.14 the gradient / U y is
zero at the point of separation. Hence c f is zero at this point. This
behaviour is used in computations to determine the location of the
separation point.
3.2.8 Boundary layer transition
In a laminar boundary layer, either the flow variables at a point have
constant values or their values show a definite variation with time. However, as
Reynolds number increases, it is found that the flow variables inside the
boundary layer show chaotic variation with time. Such a boundary layer is called
‘Turbulent boundary layer’. The change over from laminar to turbulent boundary
layer is called ‘Transition’ and takes place over a distance called ‘Transitionlength’.
Initiation of transition in a boundary layer, can be studied as an instability
phenomenon. In this study, the flow is perturbed by giving a small disturbance
and then examining whether the disturbance grows. Details of the analysis are
available in chapter 15 of Ref.3.11 and chapter 5 of Ref.3.13. The salient
features can be summarized as follows.
1. In a boundary layer on a flat plate with uniform external subsonic stream (Ue =
constant), the flow becomes sensitive to some disturbances as Rx exceeds
5 x 105 . This is called ‘Critical Reynolds number (Rcrit)’. For boundary layers in
other cases, Rcrit depends on Mach number, surface curvature, pressure gradient
in external stream and heat transfer from wall.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 4
2. After Rcrit is exceeded, some disturbances grow. These are called ‘Tollmien –
Schlichting (T-S) waves’.
3. The T-S waves lead to three-dimensional unstable waves and formation of
isolated large scale vortical structures called turbulent spots.
4. The turbulent spots grow and coalesce to form fully turbulent flow.
Remarks:
(i) As Rcrit is exceeded only some disturbances grow and hence in flows with very
low free stream turbulence level, Rcrit as high as 2.8 x 106 has been observed in
experiments. It may be recalled from fluid mechanics that the flow in pipe can
become turbulent when Reynolds number, based on pipe diameter (Red),
exceeds 2000. But laminar flow has been observed, in very smooth pipes, even
at Red = 40,000.
(ii) Transition process takes place over a length called transition length.
Reference 3.11, chapter 15 gives some guidelines for estimating this length.
Surface roughness reduces this length.
(iii) In flows with external pressure gradient, the transition is hastened by adverse
pressure gradient. It is generally assumed that transition does not take place in
favourable pressure gradient.
3.2.9 Turbulent boundary layer over a flat plate
When the flow is turbulent, one of its dominant features is that the velocity at
a point is a random function of time(Fig.3.15). When a quantity varies in a
random manner, one cannot say as to what the value would be at a chosen time,
though the values may lie within certain limits. In such a situation, the flow
features are described in terms of statistical averages. For example, the average
Uof a fluctuating quantity U is given by :
0
0
T +T
0
T -T
1U T = Udt2T (3.31)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
Fig.3.15 Typical turbulence signal
If the quantity U is independent of T0 then the phenomenon is called ‘Stationary
random phenomenon’. The discussion here is confined to this type of flow. In
such a case, the instantaneous value, U t , is expressed as :
U t = U+u t ; u = U-U .
By definition u = 0 . Hence, to distinguish different turbulent flows the root mean
square (r.m.s.) of the fluctuating quantity is used.
2
rmsu = 2u' =
T
2
-T
1u dt
2T (3.32)
T
r.m.s. value of u is 2u'
Another feature of turbulent flows is that even if the mean flow is only in one
direction, the fluctuations are in all three directions i.e. the instantaneous velocity
vector ( V ) at a point would beV = U+u i+v j+ w k; u ,v andw are the components of the
fluctuating velocity along x, y and z directions.
Because of the random fluctuations, the transfer of heat, mass and momentum is
many times faster in turbulent flows than in laminar flows. However, the part of
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
the kinetic energy of the mean motion which gets converted into the random
fluctuations is finally dissipated into heat and as such losses are higher when the
flow is turbulent.
Characteristics of turbulent boundary layer:
Analysis of turbulent boundary layer is more complicated than that of laminar
boundary layer. Reference 3.11, chapters 16 to 22 can be referred to for details.
A few results are presented below.
Velocity profile:
Velocity profile of a turbulent boundary layer on a flat plate with zero
pressure gradient is also shown in Fig.3.12. The profile can be approximated by
a power law like:
1/7 5 7
e
U= y/δ ; 5×10 < Re < 10
U (3.33)
This approximation is called ‘1/7th power law profile’.
Boundary layer thickness 0.99δ :
Though the velocity gradient U/ y near the wall is much higher for
turbulent boundary layer than for the laminar case, the gradient is lower away
from the wall and 0.99δ is much higher for a turbulent boundary layer. Reference
3.13, chapter 6, gives the following expression for 0.99δ .
0.99 turb
1/7
x
δ 0.16=
x R (3.34)
Skin friction:
The value of wall
U/ y is higher for turbulent boundary layer than for
laminar boundary layer (Fig.3.12). Hence, the skin friction drag for turbulentboundary layer is much higher than that for a laminar boundary layer. Reference
3.13, chapter 6 gives the following expression for df C .
df 1/7
L
0.031C =
R (3.35)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
Remarks:
(i) In Eqs.(3.34) and (3.35) it is assumed that the boundary layer is turbulent
from the leading edge. Corrections to these expressions can be applied by
taking the start of the transition region as the origin of the turbulent boundary
layer. However, at the values of RL obtained in actual airplanes the error in Cdf ,
by ignoring the laminar region is small.
(ii) In certain references following expressions are found for 0.99δ and Cdf.
1
5/ 0.37 /0.99 xδ x = R and1
50.072 /df LC = R .
However, Ref.3.13 chapter 6 shows that Eqs. (3.34) and (3.35) are more
accurate.
(iii) For the 1/7th
power law profile of the turbulent boundary layer (Eq.3.33),
it can be shown using Eqs.(3.26) and (3.33) that :
1
δδ =
8 for turbulent boundary layer (3.36)
Example 3.2
Consider a case with L = 0.5 m, V = 30m/s , -6 2= 15×10 m /s which gives
RL = 106. Obtain the values of 0.99δ , 1δ and Cdf in the following cases.
(i) Assume that the boundary layer is laminar throughout even at RL = 106
(ii) Assume that the boundary layer is turbulent from landing edge of plate.
Solution:
(i) Laminar flow
0.99
6L
δ 5 5= = = 0.005
L R 10 or 0.99δ = 2.5 mm
1
6L
δ 1.721 1.721= = = 0.001721
L R 10 or 1δ = 0.86 mm
df 6
L
1.328 1.328C = = = 0.001328
R 10
(ii) Turbulent flow
0.99
1/7
L
δ 0.16= = 0.02223
L R or 0.99
δ = 11.12 mm
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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1
δδ = 1.39 mm
8
df 1/7 67L
0.031 0.031C = = = 0.00431
R 10
Remark :
The comparison of the above results points out that the values of 0.99δ and 1δ are
larger when the boundary layer is turbulent than when it is laminar. The value of
Cdf in the former case is nearly three times of that in the later case.
3.2.10 General remarks on boundary layers
In this subsection the following four topics are briefly touched upon.
(i) Calculation of boundary layer, (ii) Separation of turbulent boundary layer,
(iii) Laminar flow airfoil and (iv) Effect of roughness on transition and skin friction
(i) Calculation of boundary layer
To calculate the boundary layer over an airfoil the first step is to obtain the
velocity distribution using potential flow theory. It may be recalled from
aerodynamics, that in potential flow analysis the velocity on the surface of the
body is not zero. It is assumed that this velocity distribution, given by potential
flow, would roughly be the distribution of velocity outside the boundary layer (Ue).
From this velocity distribution and using Bernoulli’s equation, the first estimate of
dp/dx is obtained. Based on this data the growth of laminar boundary layer and
the location of transition point are determined. After the transition, the growth of
turbulent boundary layer is calculated. After obtaining the boundary layers the
displacement thickness 1δ is added to the airfoil shape and calculations are
repeated till the displacement thickness assumed at the beginning of an iteration
is almost same as that obtained after calculation of the boundary layer.
Subsequently, the skin friction drag can be calculated. Section 18.4 of Ref.3.11
may be consulted for details. Presently, Computational Fluid Dynamics (CFD) is
used for these calculations.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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(ii) Separation of turbulent boundary layer
A turbulent boundary layer may also separate from the surface when it is
subjected to adverse pressure gradient. However, due to turbulent mixing the
value of w
U/ y for separation to take place is much higher than that in the
case of laminar boundary layer. Hence, a turbulent boundary layer has a higher
resistance to separation. This behaviour is used in bluff bodies to delay the
separation and reduce their pressure drag. For example, in the case of a circular
cylinder the laminar boundary layer separates at around 800 leaving a large
Fig.3.16 Schematic of flow past circular cylinder (a) Laminar
separation (b) Turbulent separation
separated region (Fig.3.16a). However, if the transition to turbulent flow takes
place before separation of laminar boundary layer, the separation is delayed. A
turbulent layer separates at around o108 , giving a smaller separated region
(Fig.3.16b). Since the drag of a bluff body is mainly pressure drag, the total drag
decreases significantly when the flow is turbulent before separation.
For example, the drag coefficient of a circular cylinder is around 1 when the
separation is laminar and it is 0.3 when the separation is turbulent (Refer chapter
1 of Ref.3.11)
(iii) Laminar flow airfoils
For a streamlined body, like an airfoil at low angle of attack, the drag is mainly
skin friction drag. Figure 3.17 and Eqs.(3.30) and (3.35) indicate that Cdf is much
higher when boundary layer is turbulent. Hence, to reduce the drag of the airfoil,
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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its shape is designed in such a way that the transition to turbulence is delayed
and the flow remains laminar over a longer portion of the airfoil.
These airfoils are called ‘Laminar flow or low drag airfoils’. Presently, efforts are
in progress to delay the transition by boundary layer control (see remark in
section 3.7.2).
Fig.3.17 Skin friction drag coefficient at various Reynolds numbers and
levels of roughness
(iv) Effect of roughness on transition
It was mentioned that the critical Reynolds number (Rcrit) depends on factors like
pressure gradient, Mach number, surface curvature and heat transfer. However,
the onset of transition may be delayed when disturbance like free stream
turbulence is low. However, if the surface is rough, this delay may not be
observed when roughness exceeds a certain value (see chapter 15 of Ref.3.11)
(v) Effect of roughness on sk in friction in turbulent boundary layerEquation (3.35) indicates that Cdf is proportional to -1/7
LR i.e. Cdf decreases with
RL. However, when the surface is rough it is observed that the decrease in Cdf
stops after a certain Reynolds number(Fig.3.17). This Reynolds number is called
‘Cut-off Reynolds number’ and is denoted by (Re)cut-off .
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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The roughness is quantified by the parameter ( l /k), where
l = characteristic length e.g. the length (L) in case of a flat plate and chord (c) in
case of an airfoil.
k = height of roughness referred to as equivalent sand roughness.
Following Ref.3.6, chapter 3, typical values of k are given in table 3.3.
Type of surface Equivalent sand roughness (m)
Natural sheet metal 4.06 x 10-6
Smooth paint 6.35 x 10-6
Standard camouflage paint 1.02 x 10-5
Mass production paint 3.048 x 10-5
Table 3.3 Equivalent sand roughness for typical surfaces
Based on Ref.3.11, chapter 18, Fig.3.17 shows typical plots of Cdf vs Re for
turbulent boundary layer with l /k as parameter. For example, when l /k = 105 , Cdf
remains almost constant at 0.0032 beyond Re = 7 x 106. Reference 3.6 section
3.1 may be seen for plot of (Re)cutoff vs ( l /k) with Mach number as parameter.
These plots are based on section 4.1.5, of Ref.3.5.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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Chapter 3
Lecture 9
Drag polar – 4
Topics
3.2.11 Presentation of aerodynamic characteristics of airfoils
3.2.12 Geometric characteristics of airfoils
3.2.13 Airfoil nomenclature\designation
3.2.14 Induced drag of wing
3.2.15 Drag coefficient of fuselage
3.2.16 Drag coefficients of other components
3.2.17 Parabolic drag polar, parasite drag, induced drag and Oswald
efficiency factor
3.2.11. Presentation of aerodynamic characteristics of airfoils
As mentioned in the beginning of subsection 3.2.3, the ways of presenting
the aerodynamic and geometric characteristics of the airfoils and the
nomenclature of the airfoils are discussed in this and the next two subsections.Figure 3.18 shows typical experimental characteristics of an aerofoil. The
features of the three plots in this figure can be briefly described as follows.
(I) Lift coefficient (Cl) vs angle of attack (α). This curve, shown in Fig.3.18a, has
four important features viz. (a) angle of zero lift ( 0α
l), (b) slope of the lift curve
denoted by dCl / dα or a
0 or αC
l, (c) maximum lift coefficient ( maxC
l) and (d)
angle of attack (αstall
) corresponding to maxCl
.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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Fig.3.18 Aerodynamic characteristics of an airfoil
(a) Clvs α (b) C
l vs C
d
(c) Cmc/4
vs α
(II) Drag coefficient (Cd) vs Cl. This curve, shown in Fig.3.18b, has two
important features viz. (a) minimum drag coefficient (Cdmin
) and (b) lift coefficient
( optCl
) corresponding to Cdmin
. In some airfoils, called laminar flow airfoils or low-
drag airfoils, the minimum drag coefficient extends over a range of lift coefficients
(Fig.3.18b). This feature is called ‘Drag bucket’. The extent of the drag bucket
and the lift coefficient at the middle of this region are also characteristic features
of the airfoil. It may be added that the camber decides optCl
and thickness ratio
decides the extent of the drag bucket.
(III) Pitching moment coefficient about quarter-chord Cmc/4 vs α . This curve is
shown in Fig.3.18c. Sometimes this curve is also plotted as Cmc/4 vs Cl. From
this curve, the location of the aerodynamic center (a.c.) and the moment about it
(Cmac
) can be worked out. It may be recalled that a.c. is the point on the chord
about which the moment coefficient is independent of C l .(IV) Stall pattern : Variation of the lift coefficient with angle of attack near the stall
is an indication of the stall pattern. A gradual pattern as shown in Fig.3.18a is a
desirable feature. Some airfoils display abrupt decrease in Cl after stall. This
behaviour is undesirable as pilot does not get adequate warning regarding
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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impending loss of lift. Airfoils with thickness ratio between 6 – 10% generally
display abrupt stall while those with t/c more than 14% display a gradual stall. It
may be added that the stall patterns on the wing and on the airfoil are directly
related only for high aspect ratio (A > 6) unswept wings. For low aspect ratio
highly swept wings three-dimensional effects may dominate.
3.2.12 Geometrical characteristics of airfoils
To describe the geometrical characteristics of airfoils, the procedure given in
chapter 6 of Ref.3.14 is followed. In this procedure, the camber line or the mean
line is the basic line for definition of the aerofoil shape (Fig.3.19a). The line
joining the extremities of the camber line is the chord. The leading and trailing
edges are defined as the forward and rearward extremities, respectively, of the
mean line. Various camber line shapes have been suggested and they
characterize various families of airfoils. The maximum camber as a fraction of the
chord length (ycmax/c) and its location as a fraction of chord (xycmax/c) are the
important parameters of the camber line.
Various thickness distributions have been suggested and they characterize
different families of airfoils Fig.3.19b. The maximum ordinate of the thickness
distribution as fraction of chord (ytmax/c) and its location as fraction of chord
(xytmax
/c) are the important parameters of the thickness distribution.
Airfoi l shape and ordinates:
The aerofoil shape (Fig.3.19c) is obtained by combining the camber line
and the thickness distribution in the following manner.
a) Draw the camber line shape and draw lines perpendicular to it at various
locations along the chord (Fig.3.19c).
b) Lay off the thickness distribution along the lines drawn perpendicular to the
mean line (Fig.3.19c).
c) The coordinates of the upper surface (xu, yu) and lower surface (xl, yl) of the
airfoil are given as :
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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u t
u c t
l t
l c t
x = x - y sinθ
y = y + y cosθ
x = x + y sinθ
y = y - y cosθ
3.37
where yc and yt are the ordinates, at location x, of the camber line and the
thickness distribution respectively; tan θ is the slope of the camber line atlocation x (see also Fig.3.19d).d) The leading edge radius is also prescribed for the aerofoil. The center of the
leading edge radius is located along the tangent to the mean line at the
leading edge (Fig.3.19c).
e) Depending on the thickness distribution, the trailing edge angle may be zero
or have a finite value. In some cases, thickness may be non-zero at the
trailing edge.
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Fig.3.19 Airfoil geometry
3.2.13 Airfoi l nomenclature/designation
Early airfoils were designed by trial and error. Royal Aircraft
Establishment (RAE), UK and Gottingen laboratory of German establishment
which is now called DLR(Deutsches Zentrum f ϋr Luft-und Raumfahrt – German
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
Centre for Aviation and Space Flight) were pioneers in airfoil design. Clark Y
airfoil shown in Fig.3.20a is an example of a 12% thick airfoil with almost flat
bottom surface which has been used on propeller blades.
Taking advantage of the developments in airfoil theory and boundary
layer theory, NACA (National Advisory Committee for Aeronautics) of USA
systematically designed and tested a large number of airfoils in 1930’s. These
are designated as NACA airfoils. In 1958 NACA was superseded by NASA
(National Aeronautic and Space Administration). This organization has
developed airfoils for special purposes. These are designated as NASA airfoils.
Though the large airplane companies like Boeing and Airbus, design their own
airfoils the NACA and NASA airfoils are generally employed by others. A brief
description of their nomenclature is presented below. The description of NACA
airfoils is based on chapter 6 of Ref.3.14.
NACA four-digi t series airfoi ls
Earliest NACA airfoils were designated as four-digit series. The thickness
distribution was based on successful RAE & Gottigen airfoils. It is given as :
2 3 4
t
ty = 0.2969 x - 0.1260 x - 0.3516 x +0.2843 x -0.1015 x
20 (3.38)
where t = maximum thickness as fraction of chord.
The leading radius is : r t = 1.1019 t2
Appendix I of Ref.3.14 contains ordinates for thickness ratios of 6%, 9%, 10%,
12%, 15%, 18%, 21% and 24%. The thickness distributions are denoted as
NACA 0006, NACA 0009,……..,NACA 0024. Figure 3.20b shows the shape of
NACA 0009 airfoil. It is a symmetrical airfoil by design. The maximum thickness
of all four-digit airfoils occurs at 30% of chord. In the designation of these airfoils
the first two digits indicate that the camber is zero and the last two digits indicate
the thickness ratio as percentage of chord.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
a) Clark Y – Airfoil with flat bottom surface, used on propeller blades
b) NACA 0009 – Symmetrical airfoil used on control surfaces
c) NACA 23012 – Airfoil with high Clmax , used on low speed airplanes
d) NACA 662 – 215 – Laminar flow or low drag airfoil
e) NASA GA(W) -1 or LS(1) - 0417 – Airfoil specially designed for generalaviation airplanes
f) NASA SC(2)-0714 – Supercritical airfoil with high critical Mach number,specially designed for high subsonic airplanes
Fig.3.20 Typical airfoils
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 8
The camber line for the four-digit series airfoils consists of two parabolic arcs
tangent at the point of maximum ordinate. The expressions for camber(yc) are :
2
c ycmax2
2
ycmax2
m
y = 2px - x ; x xp3.39
m= 1-2p +2px- x ; x > x
1-p
m = maximum ordinate of camber line as fraction of chord
p = chordwise position of maximum camber as fraction of chord
The camber lines obtained by using different values of m & p are denoted by two
digits, e.g. NACA 64 indicates a mean line of 6% camber with maximum camber
occuring at 40% of the chord. Appendix II of Ref.3.14 gives ordinates for NACA
61 to NACA 67 mean lines. The ordinates of other meanlines are obtained by
suitable scaling. For example, NACA 24 mean lines is obtained by multiplying the
ordinates of NACA 64 mean line by (2/6).
A cambered airfoil of four-digit series is obtained by combining meanline and
the thickness distribution as described in the previous subsection. For example,
NACA 2412 airfoil is obtained by combining NACA 24 meanline and NACA 0012
thickness distribution. This airfoil has (a) maximum camber of 2% occurring at
40% chord and (b) maximum thickness ratio of 12%.
Refer appendix III of Ref.3.14, for ordinates of the upper and lower surfaces of
several four-digit series airfoils. Appendix IV of the same reference presents the
low speed aerodynamic characteristics at M = 0.17 and various Reynolds
numbers. Chapter 7 of the same reference gives details of experimental
conditions and comments on the effects of parameters like camber, thickness
ratio, Reynolds number and roughness on aerodynamic characteristics of airfoils.
NACA five-digit series airfoi lsDuring certain tests it was observed that maxC
l of the airfoil could be
increased by shifting forward the location of the maximum camber. This finding
led to development of five-digit series airfoils. The new camber lines for the five-
digit series airfoils are designated by three digits. The same thickness distribution
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 9
was retained as that for NACA four-digit series airfoils. The camber line shape is
given as :
3 2 2
c 1
3
1
1
y = k x -3mx +m 3-m x , 0 < x m6(3.40)
1= k m 1- x ; m<x<1
6
The value of ‘m’ decides the location of the maximum camber and that of k 1 the
design lift coefficient( iCl or optC
l). A combination of m = 0.2025 and k1 = 15.957
gives iCl = 0.3 and maximum camber at 15% of chord. This meanline is
designated as NACA 230. The first digit ‘2’ indicates that iCl= 0.3 and the
subsequent two digits (30) indicate that the maximum camber occurs at 15% of
chord.
A typical five-digit cambered airfoil is NACA 23012. Its shape is shown in
Fig.3.20c. The digits signify :
First digit(2) indicates that iCl = 0.3.
Second & third digits (30) indicate that maximum camber occurs at 15% of chord.Last two digits (12) indicate that the maximum thickness ratio is 12%.
Remarks:
(i) Refer Appendices II, III and IV of Ref.3.14 for camber line shape,
ordinates and aerodynamic characteristics of five-digit series airfoils.
(ii) Modified four and five digit series airfoils were obtained when leading
edge radius and position of maximum thickness were altered. For
details Ref.3.14, chapter 6 may be consulted.
Six series airfoils
As a background to the development of these airfoils the following points may
be mentioned.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 10
(i) In 1931 T.Theodorsen presented ’Theory of wing sections of arbitrary
shape’ NACA TR 411 which enabled calculation flow past airfoils of
general shape .
(ii) Around the same time the studies of Tollmien and Schlichting on
boundary layer transition, indicated that the transition process, which
causes laminar boundary layer to become turbulent, depends
predominantly on the pressure gradient in the flow around the airfoil.
(iii) A turbulent boundary layer results in a higher skin friction drag
coefficient as compared to when the boundary layer is laminar. Hence,
maintaining a laminar boundary layer over a longer portion of the airfoil
would result in a lower drag coefficient.
(iv) Inverse methods, which could permit design of meanline shapes and
thickness distributions, for prescribed pressure distributions were also
available.
Taking advantage of these developments, new series of airfoils called low drag
airfoils or laminar flow airfoils were designed. These airfoils are designated as
1-series, 2-series,…….,7-series. Among these the six series airfoils are
commonly used airfoils. Refer Ref.3.14, chapter 6 for more details.
When the airfoil surface is smooth. These airfoils have a Cdmin
which is lower
than that for four-and five-digit series airfoils of the same thickness ratio. Further,
the minimum drag coefficient extends over a range of lift coefficient. This extent
is called drag bucket (see Fig.3.18b).
The thickness distributions for these airfoils are obtained by calculations which
give a desired pressure distribution. Analytical expressions for these distributions
are not available. Appendix I of Ref.3.14 gives symmetrical thickness
distributions for t/c between 6 to 21%.
The camber lines are designated as : a = 0, 0.1, 0.2 …., 0.9 and 1.0. For
example, the camber line shape with a = 0.4 gives a uniform pressure distribution
from x/c = 0 to 0.4 and then linearly decreasing to zero at x/c = 1.0. If the camber
line designation is not mentioned, ‘a’ equal to unity is implied.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 11
An airfoil with a designation as NACA 662-215 is shown in Fig.3.20d. It is
obtained by combining NACA 662 – 015 thickness distribution and a = 1.0 mean
line. The digits signify :
1st digit ‘6’ indicates that it is a 6 series airfoil
2nd digit ‘6’ denotes the chordwise position of the minimum pressure in tenths of
chord for the symmetrical airfoil at Cl= 0. i.e. the symmetrical section
(NACA 662 - 015) would have the minimum pressure at x/c = 0.6 when producing
zero lift.
The suffix ‘2’ indicates that the drag bucket extends ±0.2around optCl
.
The digit ‘2’ after the dash indicates that optCl
is 0.2. Thus in this case, drag
bucket extends for Cl = 0.0 to 0.4.
The last two digits ”15” indicate that the thickness ratio is 15%.
Since the value of ‘a’ is not explicitly mentioned, the camber line shape
corresponds to a = 1.0.
Remarks:
(i) Refer appendices I, II, III and IV of Ref.3.14 for details of thickness
distribution, camber distribution, ordinates and aerodynamic
characteristics of various six series airfoils.
(ii) The lift coefficient at the centre of the drag bucket ( optCl
) depends on
the camber. The extent of drag bucket depends on the thickness ratio
and the Reynolds number. The value given in the designation of the
airfoil is at Re = 9 x 106. The extent is about ±0.1 for t/c of 12%,±0.2
for t/c of 15% and ±0.3 for t/c of 18%. When the extent of the drag
bucket is less than ±0.1 , the subscript in the designation of the airfoil
is omitted, e.g. NACA 66-210
NASA airfoils
NASA has developed airfoil shapes for special applications. For
example GA(W) series airfoils were designed for general aviation airplanes. The
‘LS’ series of airfoils among these are for low speed airplanes. A typical airfoil of
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 12
this category is designated as LS(1)-0417. In this designation, the digit ‘1’ refers
to first series, the digits ‘04’ indicate optCl
of 0.4 and the digits ‘17’ indicate the
thickness ratio of 17%. Figure 3.20c shows the shape of this airfoil. For the
airfoils in this series, specifically designed for medium speed airplanes, theletters ‘LS’ are replaced by ‘MS’.
NASA NLF series airfoils are ‘Natural Laminar Flow’ airfoils.
NASA SC series airfoils are called ‘Supercritical airfoils’. These airfoils have a
higher critical Mach number. Figure 3.20f shows an airfoil of this category.
Chapter 3 of Ref.1.9 may be referred to for further details.
Remarks:
(i)Besides NACA & NASA airfoils, some researchers have designed airfoils for
specialized applications like (a) low Reynolds number airfoils for micro air
vehicles, (b) wind mills, (c) hydrofoils etc. These include those by Lissaman,
Liebeck, Eppler and Drela. Reference 3.9, chapter 4, and internet
(www.google.com) may be consulted for details.
(ii)The coordinates of NACA, NASA and many other airfoils are available on the
website entitled ‘UIUC airfoil data base’.
3.2.14 Induced drag of wing
In the beginning of section 3.2.2 it was mentioned that the drag of the
wing consists of (i) the profile drag coefficient due to airfoil (Cd) and (ii) the
induced drag coefficient (CDi) due to finite aspect ratio of the wing. Subsections
3.2.3 to 3.2.13 covered various aspects of profile drag. In this subsection the
induced drag of the wing is briefly discussed.
For details regarding the production of induced drag and derivation of the
expression for the induced drag coefficient, the books on aerodynamics can be
consulted e.g. Ref.3.12, chapter 5. Following is a brief description of the induced
drag.
Consider a wing kept at a positive angle of attack in an air stream. In this
configuration, the wing produces a positive lift. At the wing root, the average
pressure on the upper surface is lower than the free stream pressure p and
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 13
the average pressure on the lower surface is higher than p . Since the span of
the wing is finite it has the wing tips and at these tips there cannot be a pressure
discontinuity or the pressure at the wing tips would be the same on the upper
side and the lower side. The pressure at the wing tips is expected to be mean of
the pressures on the upper and lower sides at the root section. Because of the
difference of pressures between the root and the tip, the pressure on the upper
surface of the wing increases from root to the tip in the spanwise direction.
Similarly, the pressure on the lower surface of the wing, decreases from the root
to the tip in the spanwise direction. These pressure gradients on the upper and
lower surfaces would lead to cross flows on these surfaces. Thus, at a given
spanwise station, the airstreams from the upper and lower surfaces would meet,
at the trailing edge, at an angle. This would cause shedding of vortices from the
trailing edge. Viewed from the rear, the vortices would appear rotating clockwise
from the left wing and anticlockwise from the right wing. These vortices soon roll
up to form two large vortices springing from positions near the wing tips. As a
consequence of these vortices the air stream in the vicinity and behind the wing
acquires a downward velocity component called induced downwash.
This downwash tilts the aerodynamic force rearwards resulting in a component in
the free stream direction called induced drag. The induced drag coefficient (CDi)is given as :
2 2L L
Di
wing
C 1+δ CC =
A Ae
(3.41)
Where A is the wing aspect ratio (A = b2/S) and δ is a factor which depends on
wing aspect ratio, taper ratio, sweep and Mach number. The quantity ‘ewing ‘ is
called Oswald efficiency factor for wing.
It may be added that a wing with elliptic chord distribution has the minimum
induced drag i.e. δ = 0 in Eq.(3.41).
Reference 3.6 section 3.3, gives the following expression for ewing which is
based on Ref.3.5 section 4.1.5.2.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 14
LαW
wing
LαW
1.1 C /Ae =
CR + 1-R
A
(3.42)
where,
LαW2
12 2
2
2 2
2 AC =
tan Λ A β
2+ 1+ +4β
(3.43)
LαWC = slope of lift curve of wing in radians
A = aspect ratio of wing
R = a factor which depends on (a) Reynolds number based on leading
edge radius, (b) leading edge sweep ( LE ), (c) Mach number (M), (d) wing
aspect ratio (A) and (e) taper ratio (λ ).
2β = 1-M
1/2 = sweep of semi-chord line
= ratio of the slope of lift curve of the airfoil used on wing divided by 2 .
It is generally taken as unity.
Remarks:
(i)Example 3.3 illustrates the estimation of ewing for an unswept wing. Section 2.5
of Appendix ‘B’ illustrates the steps for estimating ewing of a jet airplane.
(ii) When a flap is deflected, there will be increments in lift coefficient and also in
profile drag coefficient and induced drag coefficient. Refer section 2.9 of
Appendix ‘A’.
(iii) The drags of horizontal and vertical tails, can be estimated by following a
procedure similar to that for the wing. However, contributions to induced drag
from the tail surfaces are generally neglected.
3.2.15 Drag coeff icient of fuselage
The drag coefficient of a fuselage (CDf
) consists of (a)the drag of the fuselage at
zero angle of attack (CDo
)f plus (b) the drag due to angle of attack. Following
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 15
Ref.3.7, section 19.3 it can be expressed as:
2
(3.44)Df Dof C = C 1+K15
where α is the angle of attack of fuselage in degrees.
For a streamlined body (CDo
)f is mainly skin friction drag and depends on
(i) Reynolds number, based on length of fuselage ( f l ), (ii) surface roughness
and (iii) fineness ratio (Af ). The fineness ratio is defined as:
Af = f l /de (3.44a)
‘de‘ is the equivalent diameter given by:
( /4)de2 = A
fmax
where Afmax
equals the area of the maximum cross-section of the fuselage.
When the fineness ratio of the fuselage is small, for example, in case of
general aviation airplanes, the fuselage may be treated as a bluff body. In such a
case the term CDof is mainly pressure drag and the drag coefficient is based on
the frontal area (Afmax
). However, the expression for (CD0)f given in Ref.3.6,
section 3.1.1 includes the effect of pressure drag and is also valid for general
aviation airplanes (refer section 2 of Appendix A).
The quantity ‘K’ in Eq.(3.44) has a value of 1 for a circular fuselage and 4 to 6 for
a rectangular fuselage. However, the general practice is to include the increase
in drag of fuselage, due to angle of attack, by adding a termfuselage
1
e
to
wing
1
e
.
Remark:
The drag coefficients of other bodies like engine nacelle, external fuel tanks and
bombs suspended from the wing, can also be estimated in a manner similar to
that of fuselage.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 16
3.2.16 Drag coeffic ients of other components
The drag coefficients of other components like landing gear are based on
areas specific to those components. They should be obtained from the sources
of drag data mentioned earlier. The change in drag of these components, with
angle of attack, is included by adding a termother
1
e
to (fuselage
1
e +
wing
1
e ) i.e.
for the entire airplane,wing fuselage other
1 1 1 1= + +
e e e e (3.44b)
Reference 3.6, section 3.2 recommendsother
1
e
as 0.05.
3.2.17 Parabolic drag polar, parasite drag, induced drag and Oswald
efficiency factor
It was mentioned earlier that the drag polar can be obtained by adding the
drag coefficients of individual components at corresponding angles of attack.
This procedure needs a large amount of detailed data about the airplane
geometry and drag coefficients. A typical drag polar obtained by such a
procedure or by experiments on a model of the airplane has the shape as shown
in Fig.3.21a. When this curve is replotted as CD
vs CL
2 (Fig.3.21b), it is found
that over a wide range of 2
LC the curve is a straight line and one could write:
CD = C
D0 + KC
L
2 (3.45)
CD0 is the intercept of this straight line and is called zero lift drag coefficient or
parasite drag coefficient (Fig.3.21b).
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 17
Fig.3.21a Typical drag polar of a piston – engined airplane
Fig.3.21b Drag polar replotted as CD vrs.
2
LC
The term 2
LKC is called induced drag coefficient or more appropriately lift
dependent drag coefficient. K is written as:
1K =
A e
(3.46)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 18
The quantity ‘e’ in Eq.(3.46) is called ‘Oswald efficiency factor’. It includes the
changes in drag due to angle of attack of the wing, the fuselage and other
components as expressed by Eq.(3.44b). It may be added that in the original
definition of Oswald efficiency factor, only the contribution of the wing was
included. However, the expression given by Eq.(3.44b) is commonly
employed(Ref.1.12, Chapter 2 and Ref.3.6, Chapter 2).
The drag polar expressed by Eq.(3.45) is called ‘Parabolic drag polar’.
Remarks:
i) A parabolic expression like Eq.(3.45) fits the drag polar because the major
contributions to the lift dependent drag are from the wing and the fuselage and
these contributions are proportional to the square of the angle of attack or CL.
ii) Rough estimate of CDo :
Based on the description in Ref.1.9, chapter 4 and Ref.3.7, chapter 14, the
parasite drag (Dparasite or D0) of an airplane can be approximately estimated as the
sum of the minimum drags of various components of the airplane plus the
correction for the effect of interference.
Modifying Eq.(3.1), the parasite drag can be expressed as:
Dparasite = D0 = (Dmin)wing + (Dmin)fuse + (Dmin)ht + (Dmin)vt + (Dmin)nac + (Dmin)lg +
(Dmin
)etc
+ Dint
(3.46a)
Modifying Eq.(3.5), the above equation can be rewritten as :
( )
2 2 20 Dmin fuse Dmin nac Dminwing fuse nac
2 2ht Dmin vt Dminht vt
2 2lg Dmin etc Dmin intlg etc
1 1 1D = ρV S C + ρV S C + ρV S C +
2 2 2
1 1ρV S C + ρV S C +
2 2
1 1ρV S C + ρV S C +D 3.46b
2 2
Dividing Eq.(3.46b) by
21
ρV S2 yields:
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 19
( )
fuse nacD0 Dmin Dmin Dminwing fuse nac
lght vtDmin Dmin Dminht vt lg
etcDmin Dintetc
S SSC = C + C + C +
S S S
SS SC + C + C +
S S S
S C +C 3.46cS
To simplify Eq.(3.46c) the minimum drag coefficient of each component is
denoted by CD
and the area on which it is based is called ‘Proper drag area’ and
denoted by S. Thus, when the contribution of fuselage to CDo
is implied, then
CD
refers to Dmin fuseC and S refers to fuseS . With these notations Eq.(3.46c)
simplifies to :
D0C = DintD1
C S + CS
(3.46d)
The product D0C S is called ‘Parasite drag area’.
Note :
1)See example 3.3 for estimation of D0C of a low speed airplane.
2) In Appendices A and B the parasite drag coefficient ( D0C ) is estimated using
the procedure given in Ref.3.6, chapter 3, which in turn is based on Ref.3.5,
section 4.5.3.1. In this procedure the contributions of the wing and fuselage to
D0C are estimated togather as wing – body combination and denoted by D0WBC
iii) The parabolic polar is an approximation. It is inaccurate near CL= 0 and
CL= C
Lmax (Fig.3.21b). It should not be used beyond CLmax
.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 3
Lecture 10
Drag polar – 5
Topics
3.2.18 Parasite drag area and equivalent skin friction coefficient
3.2.19 A note on estimation of minimum drag coefficients of wings and
bodies
3.2.20 Typical values of CDO, A, e and subsonic drag polar
3.2.21 Winglets and their effect on induced drag
3.3 Drag polar at high subsonic, transonic and supersonic speeds
3.3.1 Some aspects of supersonic flow - shock wave, expansion fan
and bow shock
3.3.2 Drag at supersonic speeds
3.3.3 Transonic flow regime - critical Mach number and drag
divergence Mach number of airfoils, wings and fuselage
3.2.18 Parasite drag area and equivalent skin friction coefficient As mentioned in remark (ii) of the previous subsection, the product C
Do x S is
called the parasite drag area. For a streamlined airplane the parasite drag is
mostly skin friction drag. Further, the skin friction drag depends on the wetted
area which is the area of surface in contact with the fluid. The wetted area of the
entire airplane is denoted by Swet
. In this background the term ‘Equivalent skin
friction coefficient (Cfe)’ is defined as:
CDo x S = Cfe x Swet
Hence, Cfe = CDo
xwet
S
S and wet
DO f e
SC = C
S (3.47)
Reference 3.9, Chapter 12 gives values of Cfe for different types of airplanes.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
Example 3. 3
A quick estimate of the drag polar of a subsonic airplane is presented in this
example which is based Ref.3.7, section 14.8. However, modifications have been
incorporated, keeping in view the present treatment of the drag polar.
An airplane has a wing of planform area 51.22 m2 and span 20 m. It has a
fuselage of frontal area 3.72 m2 and two nacelles having a total frontal area of
3.25 m2. The total planform area of horizontal and vertical tails is 18.6 m2. Obtain
a rough estimate of the drag polar in a flight at a speed of 430 kmph at sea level,
when the landing gear is in retracted position.
Solution:
Flight speed is 430 kmph = 119.5 m/s.
Average chord of wing( wingc ) = S / b = 51.22/20 = 2.566 m.
The value of kinematic viscosity ( ) at sea level = -614.6× 10 m2
Reynolds number (Re) based on average chord is:
6
-6
119.5× 2.566= 21×10
14.6× 10
It is assumed that NACA 23012 airfoil is used on the wing. From Ref.3.14,
Appendix IV, the minimum drag coefficient, (Cd)min, of this airfoil at Re = 9 x 106 is
0.006. However, the value of drag coefficient is required at Re = 621×10 .
Assuming the flow to be turbulent (Cd)min can be taken proportional to1
-7
eR (Eq.
3.35). Thus, Cdmin at wRe = 21 x 106 would roughly be equal to:
6 61 17 70.006 21 10 / 9 10
= 0.0053
As regards the fuselage and nacelle, the frontal areas are specified. Hence, they
are treated as a bluff bodies. The value of (CDmin)fuselage can be taken as 0.08
(Ref.3.4). The nacelle generally has a lower fineness ratio and (CDmin
)nac
can be
taken as 0.10.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
As regards the horizontal and vertical tails, the Reynolds number based on their
average chords (Retail) can be calculated if the areas and spans of these were
given. The following is suggested to obtain a rough estimate of Retail.
2
ht vtS S 18.6/2 9.3 m . Then
tail tail
wing wing
c S
c S
and tail
w
tail
wing
cRe 9.3= 0.426
Re c 51.22
Hence, Re tail 6 6 621×10 ×0.426 = 8.95×10 9×10
At this Reynolds number (Cdmin)tail can be assumed to be 0.006
The calculation of the parasite drag coefficient (CDo) is presented in Table E 3.3.
Part S (m2) CD
CD
S (m2)
Wing 51.22 0.0053 0.271
Fuselage 3.72 0.080 0.298
Nacelles 3.25 0.1 0.325
Tail surfaces 18.6 0.006 0.112
Total 1.006
Table E3.3 Rough estimate of CDo
Adding 10% for interference effects, the total parasite drag area (CD
S ) is:
1.006 + 0.1006 = 1.1066 m2.
Hence,- CD0 = 1.1066/51.22 = 0.0216
Wing aspect ratio is 202 / 51.22 = 7.8
To obtain Oswald efficiency factor for the airplane (e) , the quantities ewing,
efuselage and eother are obtained below.
Equations (3.42) and (3.43) give :
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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LαW
Wing
LαW
1.1 C /Ae =
CR + 1-R
A
2
22 4
LαW12 2
2
2 AC =
tan2Λ A β
1+ +β
Here
A = 7.8, M = V/a = 119.5/340 = 0.351
Hence, 2 2β = 1-M = 1-0.351 = 0.936
For the purpose of calculating ewing, the taper ratio (λ ), the quarter chord
sweep ( 14
) and the quantity , are taken as 0.4, 0 and 1 respectively.
Consequently, o
LEΛ = 3.14
Hence, Lα2 2
2 ×7.8C = = 5.121rad
7.8 ×0.9362+ +4
1
-1
From Ref.3.14, chapter 6, the leading edge radius, as a fraction of chord, for
NACA 23012 airfoil is :
1.109 t2 = 1.019 x 0.122 = 0.016
Rle = 0.016 x c = 0.016 x 2.566 = 0.041 m
Reynolds number, based leading edge radius ( eLERR ), is :
eLER
5-6
0.041×119.5R = = 3.35×10
14×10
Hence, 2
eLER LE LER cotΛ 1-M cosΛ 5 2= 3.35×10 ×18.22× 1-0.351 ×0.998
= 57.16 x 105
Further,LE
Aλ 7.8×0.4= = 3.13
cosΛ 0.998
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
Corresponding to the above values of ( 2
eLER LE LER cotΛ 1-M cosΛ ) and
(LE
Aλ
cosΛ), Fig 3.14 of Ref.3.6, gives R = 0.95.
Hence,
wing
1.1× 5.121/7.8e = = 0.925
0.95 × 5.121 /7.8 + 0.05×
To obtain efuselage , it is assumed that the fuselage has a round cross section.
In this case, Fig.2.5 of Ref 3.6 gives: fuselage
fuselage
1/ S /S
e
= 0.75 when A = 7.8.
Consequently,
fuselage
1 = 0.75×3.72/51.22 = 0.054e
others
1
e is recommended as 0.05(Ref.3.6, section 2.2)
Thus,wing fuselage other
1 1 1 1= + +
e e e e =
1+0.054+0.05 = 1.185
0.925
Or e = 0.844
Hence,
1 1
= = 0.0484 Ae × 7.8 × 0.844
Hence, a rough estimate of the drag polar is:
2LD
C = 0.0216+0.0484C
Answer: A rough estimate of the drag polar is : 2LD
C = 0.0216+0.0484C
Remark:
i) A detailed estimation of the drag polar of Piper Cherokee airplane is
presented in appendix A.
3.2.19 Note on estimation of minimum drag coeffic ients of wings and
bodies
Remark (ii) of section 3.2.17 mentions that the parasite drag coefficient of
an airplane (CD0
) is given by :
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
D0C = DintD
1C S + C
S
where the values of DC
represent the minimum drag coefficients of various
components of the airplane.In example 3.3 the minimum drag coefficients of wing, fuselage, nacelle,
horizontal tail and vertical tail were estimated using experimental data. However,
the minimum drag coefficients of shapes like the wing, the horizontal tail, the
vertical tail and the streamlined bodies can be estimated using the background
presented in subsections 3.2.5 to 3.2.10. The procedure for such estimation are
available in Ref.3.6 which in turn is based on Ref.3.5. The basis of this procedure
is that the minimum drag coefficient of a streamlined shape can be taken as the
skin friction drag coefficient of a flat plate of appropriate characteristic length,
roughness and area.
With these aspects in view, the procedure to estimate the minimum drag
coefficient of the wing can be summerised as follows. It is also illustrated in the
sections on drag polar in Appendices A & B.
(a) The reference length ( l ) is the mean aerodynamic chord of the exposed wing
i.e. the portion of wing outside the fuselage. This chord is denoted by ec . Obtain
roughness parameter ( l /k) with ec as ‘ l ’ and value of k from Table 3.3.
(b) The flow is assumed to be turbulent over the entire wing.
(c)Choose the flight condition. Generally this is the cruising speed (Vcr ) and the
cruising altitude (hcr ). Obtain the Reynolds number (Re) based on Vcr , kinematic
viscosity cr at hcr and the reference length as ec i.e. e cr
cr
c VRe =
.
Obtain (Re)cutoff corresponding to ( l /k) using Fig.3.2 of Ref.3.6. Obtain Cdf
corresponding to lower of Re and (Re)cut-off . Following Ref.3.6 this value is
denoted by Cfw in Appendices A & B.
(d) Apply correction to Cfw for type of airfoil and its thickness ratio. Multiply this
value by (Swet/Sref ), where Swet is the wetted area of the exposed wing and Sref is
the reference area of the wing. Refer to section 3.1 of Ref.3.6 for estimation of
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
Swet and correction for airfoil shape. When the shape of the airfoil changes along
the wingspan, a representative section is taken for estimation of Swet.
Similar procedure can be used to estimate the minimum drag coefficients of the
horizontal tail and vertical tail.
As regards estimation of the minimum drag coefficient of fuselage, the reference
length is taken as the length of fuselage ( f l ) and the roughness factor is taken as
( f l /k). Correction is applied for fineness ratio ( f l /de) of the fuselage. Where ‘de’ is
the equivalent diameter of the fuselage (see section 3.2.15). The wetted area in
this case is the wetted area of the fuselage.
Finally, correction is applied for wing-body interference effect (see Appendices A
& B for details).
Similar procedure can be used to estimate the minimum drag coefficients of
bodies like nacelle, external fuel tanks, bombs etc.
3.2.20 Typical values of CDO, A, e and subsonic drag polar.
Based on the data in Ref.3.9, chapter 4 , Ref.3.18 vol. VI , chapter 5 and
Ref.3.15 , chapter 6, the typical values of CD0
, A, e and the drag polar for
subsonic airplanes are given in Table 3.4.
Type of
airplane CD0
A e Typical polar
Low speed
(M <0.3)
0.025 to
0.04 6 to 8
0.75 to
0.85
0.025 + 0.06CL2
Medium speed
(M around 0.5)
0.02 to
0.024 10 to 12
0.75 to
0.85
0.022 + 0.04CL2
High subsonic
(M around 0.8,
Swept wing)
0.014 to
0.017 6 to 9
0.65 to
0.75
0.016 +0.045CL2
Table 3.4 Typical values of CD0
, A, e and subsonic drag polar
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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Remarks:
(i) Table 3.4 shows that CD0
for low speed airplanes is higher than other
airplanes. This is because these airplanes have exposed landing gear, bluff
fuselage (see Fig.1.2a) and struts when a high wing configuration is used. The
CD0
for high subsonic airplanes is low due to smooth surfaces, thin wings and
slender fuselage. It may be added that during the design process, the values of
airfoil thickness ratio, aspect ratio and angle of sweep for the wing are obtained
from considerations of optimum performance.
(ii) The low speed airplanes have a value of K (=1/ Ae ) higher than the other
airplanes. One of the reasons for this is that these airplanes have only a
moderate aspect ratio (6 to 8) so that the wing-span is not large and the hanger-
space needed for parking the plane is not excessive.
(iii) See section 2 of Appendix A for estimation of the drag polar of a subsonic
airplane in cruise and take-off conditions.
3.2.21 Winglets and their effect on induced drag
According to Ref.2.1, a Winglet is an upturned wing tip or added axialliary airfoil
above and / or below the wing tips. Figure 1.2c shows one type of winglets at
wing tips. The winglets alter the spanwise distribution of lift and reduce the
induced drag. Reference 1.9, chapter 4 can be referred for a simplified analysis
of the effect of winglets. However, along with reduction in induced drag, the
winglets increase the weight of the wing and also the parasite drag. After trade-
off studies which take into account the favourable and unfavourable effects of the
winglets, the following approximate dimensions are arrived at for the winglets.
Root chord of about 0.65 ct, tip chord of about 0.2 ct and height of about ct ;
where ct is the tip chord of the wing. As regards the effect on induced drag,
Ref.3.15, chapter 5 suggest that the effect of winglets can be approximatelyaccounted for by increasing the wing span by an amount equal to half the height
of the winglet. The procedure is illustrated in example 3.4
Example 3.4
Consider a wing, with the following features. Area (S) = 111.63 m2,
Aspect ratio (A) = 9.3, span (b) = 32.22 m, root chord (cr ) = 5.59 m,
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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tip chord (ct) = 1.34 m
Further, the airplane has (a) parasite drag coefficient (CDO) = 0.0159 ; (b) Oswald
efficiency factor (e) = 0.8064 (c) lift coefficient during cruise (CLcr )=0.512.
Examine the benefits of fitting winglets to this wing.
Solution :
The drag polar of the existing airplane is :
CD = 2 2
L L
10.0159+ C = 0.0159+0.04244C
π×9.3×0.8064
When CL = 0.512, CD = 0.0159 + 0.04244 x 0.5122 = 0.02703
With winglets, the wing span effectively increases to :
te
c 1.34b = b+ = 32.22+ = 32.89 m
2 2
Hence, the effective aspect ratio (Ae) =2 2eb 32.89
= = 9.691S 111.63
Consequently, the drag polar approximately changes to :
2 2L L
10.0159+ C = 0.0159+0.0407C
×9.691×0.8064
At CL = 0.512, the CD of the wing with winglet is :
0.0159 + 0.04074 x 0.5122 = 0.02658
Reduction in drag coefficient is 0.02703 – 0.02658 = 0.00045 or 1.7%
Note :
(CL/CD)existing wing = 0.512/0.02703 = 18.94
(CL/CD)modified wing = 0.512/0.02658 = 19.26
3.3 Drag polar at high subsonic, transonic and supersonic speeds
At this stage, the reader is advised to revise background on compressible
aerodynamics and gas dynamics. References.1.9 & 1.10 may be consulted.
Before discussing the drag polar at high subsonic, transonic and supersonic
speeds, the relevant features of supersonic and transonic flows are briefly
recapitulated in the following subsections.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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3.3.1 Some aspects of supersonic flow – shock wave, expansion fan and
bow shock
When the free stream Mach number roughly exceeds a value of 0.3, the
changes in the fluid density, within the flow field, become significant and the flow
needs to be treated as compressible. In a compressible flow, the changes of
temperature in the flow field may be large and hence the speed of sound
(a = γRT ) may vary from point to point. When the free stream Mach number
exceeds unity, the flow is called supersonic. When a supersonic flow
decelerates, shock waves occur. The pressure, temperature, density and Mach
number change discontinuously across the shock. The shocks may be normal or
oblique. The Mach number behind a normal shock is subsonic; behind an oblique
shock it may be subsonic or supersonic. When supersonic flow encounters a
concave corner, as shown in Fig.3.22a, the flow changes the direction across a
shock. When such a flow encounters a convex corner, as shown in Fig.3.22b, the
flow expands across a series of Mach waves called expansion fan. A typical flow
past a diamond airfoil at supersonic Mach number is shown in Fig.3.23. If the
free stream Mach number is low supersonic (i.e. only slightly higher than unity)
and the angle θ, as shown in Fig.3.23, is high then instead of the attached shock
waves at the leading edge, a bow shock wave may occur ahead of the airfoil. A
blunt-nosed airfoil can be thought of an airfoil with large value of ‘θ’ at the leading
edge and will have a bow shock at the leading edge as shown in Fig.3.24.
Behind a bow shock there is a region of subsonic flow (Fig.3.24).
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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(a) (b)
Fig.3.22 Supersonic flow at corners(a) Concave corner (b) Convex corner
Fig.3.23 Supersonic flow past a diamond airfoil
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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Fig.3.24 Bow shock ahead of blunt-nosed airfoil
3.3.2 Drag at supersonic speeds
At supersonic Mach numbers also the drag of a wing can be expressed as
sum of the profile drag of the wing section plus the drag due to effect of finite
aspect ratio. The profile drag consists of pressure drag plus the skin friction drag.
The pressure drag results from the pressure distribution caused by the shock
waves and expansion waves (Fig.3.23) and hence is called ‘Wave drag’.It is denoted by Cdw. Figures 3.25a and b show the distributions of pressure
coefficients (Cp) on an airfoil at angles of attack ( α ) of 00 and 20.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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(b) oα = 2
Fig.3.25 Pressure distributions over a diamond airfoil (a) oα = 0 (b) oα = 2
From the distributions of pC at oα = 0 , on various faces of the diamond airfoil, it is
observed that the distributions are symmetric about the X-axis but not symmetric
about the Y-axis. This indicates Cl= 0 but Cdw > 0. From the distributions of Cp at
oα = 2 , it is seen that the distributions are unsymmetric about both X- and
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Y-axes. Thus in this case, Cl > 0 and Cdw > 0. It may be added that a leading
edge total angle of 10 would give a thickness ratio of 8.75%, which is rather high.
Supersonic airfoils would have (t/c) between 3 to 5%.
At supersonic speed the skin friction drag is only a small fraction of the wavedrag. The wave drag of a symmetrical airfoil (Cdw) can be expressed as (Ref.1.9,
chapter 5):
2 2
2dW
4C = [α +(t/c) ]
M -1
(3.48)
where α = angle of attack in radians and
t / c = thickness ratio of the airfoil
The wave drag of a finite wing at supersonic speeds can also be expressed as
KCL
2 (refer Ref.1.9, chapter 5 for details). However, in this case K depends on
the free stream Mach number (M∞), aspect ratio and leading edge sweep of the
wing (refer Ref.1.9 chapter 5 for details).
The estimation of the wave drag of a fuselage at supersonic speeds is more
involved than that of the wing. It is considered as flow past a combination of a
nose cone, a cyclindrical mid-body and a conical after body. It may be added that
the supersonic airplanes generally have low aspect ratio wings and the wavedrag of the wing-body is analysed as a combination. Reference 1.9, chapter 5
deals with some of these aspects. Reference 3.5 is generally used to estimate
the drag of wing-body-tail combination at desired values of Mach numbers.
3.3.3 Transonic flow regime, critical Mach number and drag divergence
Mach number of airfoi l , wing and fuselage
A transonic flow occurs when the free stream Mach number is around one.
The changes in the flow and hence in the drag occurring in this range of Mach
numbers can be better understood from the following statements.
I) In the subsonic flow past an airfoil the flow velocity is zero at the stagnation
point. Subsequently, the flow accelerates, it reaches a maximum value (Vmax) and
later attains the free stream velocity (V∞). The ratio (Vmax /V∞) is greater than
unity and depends on (a) the shape of airfoil (b) the thickness ratio (t/c) and
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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( c ) the angle of attack (α). As (Vmax / V∞ ) is greater than unity, the ratio of the
maximum Mach number on the airfoil ( M max) and free stream Mach number
(M∞) would also be more than unity. However, (Mmax/ M∞) would not be equal to
(Vmax /V∞) as the speed of sound varies from point to point in the flow.
II) Critical Mach number: As M∞ increases, Mmax also increases. The free stream
Mach number for which the maximum Mach number on the airfoil equals unity is
called the critical Mach number (Mcrit
).
III) The changes in flow patterns when the free stream Mach number changes
from subcritical (i.e. critM M ) to supersonic (M > 1 ) are highlighted below .
(A) When M is less than or equal to Mcrit then the flow is subsonic everywhere
i.e. in the free stream, on the airfoil and behind it (Fig.3.26a).
(B) When M∞ exceeds Mcrit
, a region of supersonic flow occurs which is
terminated by a shock wave. The changes in flow pattern are shown in
Figs.3.26b and c.
(C) As free stream Mach number increases further the region of supersonic flow
enlarges and this region occurs on both the upper and lower surfaces of the
airfoil (Figs.3.26c, d & e).
(D) At a free stream Mach number slightly higher than unity, a bow shock is seen
near the leading edge of the airfoil (Fig.3.26f).
(E) At a still higher Mach numbers, the bow shock approaches the leading edge
and if the leading edge is sharp, then the shock waves attach to the leading edge
as shown in Fig.3.23.
Fig.3.26 (a) Mach number subsonic everywhere
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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Fig.3.26 (b) M∞ only slightly higher than Mcrit ; shock waves are not discernible
Fig.3.26 (c) M∞ greater than Mcrit ; shock wave seen on the upper surface
Fig.3.26 (d) M∞ greater than Mcrit ; shock waves seen on both the upper and
lower surfaces
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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Fig.3.26 (e) M∞ greater than Mcrit; shock waves seen on both the upper
and lower surfaces at the trailing edge
Fig.3.26 (f) M∞ greater than unity; bow shock wave seen ahead of the airfoil;
shock waves also seen at the trailing edge on both upper and lower surfaces
Fig.3.26 Flow past airfoil in transonic range at α=20
(Adapted from Ref.3.16, chapter 9 with permission from author). The angle of
attack (α) being 20 is mentioned in Ref.3.17 chapter 4.
(IV) Transonic flow regime
When M∞ is less than Mcrit the flow every where i.e. in the free stream,
and on the body and behind it, is subsonic. It is seen that when Mcrit
< M∞ < 1,
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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the free stream Mach number is subsonic but there are regions of supersonic
flow on the airfoil (Figs.3.26c, d & e). Further, when M∞ is slightly more than
unity i.e. free stream is supersonic; there is bow shock ahead of the airfoil
resulting in subsonic flow near the leading edge (Fig.3.24). When the shock
waves are attached to the leading edge (Fig.3.23) the flow is supersonic every-
where i.e. in the free stream and on the airfoil and behind it.
Based on the above features, the flow can be classified into three regimes.
(a) Sub-critical regime - when the Mach number is subsonic in the free stream
as well as on the body (M∞ < Mcrit
).
(b) Transonic regime - when the regions of both subsonic and supersonic flow
are seen within the flow field.
(c) Supersonic regime - when the Mach number in the free stream as well as
on the body is supersonic.
The extent of the transonic regime is commonly stated as between 0.8 to
1.2. However, the actual extent of this regime is between Mcrit
and the Mach
number at which the flow becomes supersonic everywhere. The extent depends
on the shape of the airfoil and the angle of attack. In the transonic regime the lift
coefficient and drag coefficient undergo rapid changes with Mach number
(Figs.3.27a, b and c). It may be recalled that Cd and Cl refer to the drag
coefficient and lift coefficient of an airfoil respectively.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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Fig.3.27a Variation of lift coefficient (Cl) in transonic range for the airfoil in
Fig.3.26 (α=20). (Adapted from Ref. 3.16, chapter 9 with permission from author)
Note: The points A, B, C, D, E and F corresponds to those in Figs.3.26a, b, c, d,
e and f respectively.
Fig.3.27b Typical variations of drag coefficient (Cd) in transonic region for
airfoils of different thickness ratios (Adapted from Ref.3.17,
chapter 4 with, permission of McGraw-Hill book company)
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Figure 3.27a shows the variation of the lift coefficient ( Cl) with Mach number at a
constant value of angle of attack. It is seen that at sub critical Mach numbers, Cl
increases with Mach number. This is due to the effect of compressibility on
pressure distribution. However, as the critical Mach number is exceeded theformation of shocks changes the pressure distributions on the upper and lower
surfaces of the airfoil and the lift coefficient decreases (points C & D in
Fig.3.27a). This phenomenon of decrease in lift due to formation of shocks is
called ‘Shock stall’. For a chosen angle of attack the drag coefficient begins to
increase near Mcrit
and reaches a peak around M∞ = 1 (Fig.3.27b).
(V) Drag divergence Mach number (MD)
The critical Mach number (Mcrit) of an airfoil has been defined in statement (II) of
this subsection. It is the free stream Mach number (M∞) for which the maximum
Mach number on the airfoil equals one. The critical Mach number is a theoretical
concept. It is not possible to observe this (Mcrit
) in experiments as the changes in
flow, when M∞ just exceeds Mcrit
, are very gradual. Hence, a Mach number
called ‘Drag divergence Mach number (MD)’ is used in experimental work. The
basis is as follows.
When the change in Cd with Mach number is studied experimentally, the effects
of changes in flow, due to the appearance of shock waves, are noticed in the
form of a gradual increase in the drag coefficient. The Mach number at which the
increase in the drag coefficient is 0.002 over the value of Cd at sub-critical Mach
numbers is called ‘Drag divergence Mach number’ and is denoted by MD.
Figure 3.27c shows a typical variation of Cd with M and also indicates MD.
The following may be added. (a) For a chosen angle of attack the value of Cd
remains almost constant when the Mach number is sub-critical. (b) The drag
divergence Mach number of an airfoil depends on its shape, thickness ratio and
the angle of attack. (c) The increase in the drag coefficient in the transonic region
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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is due to the appearance of shock waves. Hence, this increment in Cd is called
wave drag.
Fig.3.27c Definition of drag divergence Mach number(The curve corresponds to t / c =0.12 in Fig.3.27b)
Remark:
Supercritical airfoil
For airplanes flying at high subsonic speeds the lift coefficient under
cruising condition (CLcr ) is around 0.5. At this value of lift coefficient, the older
NACA airfoils have drag divergence Mach number (MD) of around 0.68 for a
thickness ratio (t/c) of around 15%.
With the advancements in computational fluid dynamics (CFD) it was
possible, in 1970’s to compute transonic flow past airfoils. This enabled design of
improved airfoils, called supercritical airfoils, which have MD around 0.75 for t/c of
15% (Ref.3.18 part II, chapter 6). For comparison, the shapes of older airfoil
(NACA 662 – 215) and a supercritical airfoil are shown in Fig.3.20d and f. Note
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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the flat upper surface of the supercritical airfoil (refer Ref. 1.9 chapter 3 for
additional information).
(VI) Drag divergence Mach number of a wing
The drag divergence Mach number of an unswept wing depends on the
drag divergence Mach number of the airfoil used on the wing and its aspect ratio.
The drag divergence Mach number of the wing can be further increased by
incorporating sweep (Λ) in the wing. Figure 3.3 shows the geometrical
parameters of the wing including the sweep. The beneficial effects of sweep on
(a) increasing MD, (b) decreasing peak value of wave drag coefficient (C
Dpeak)
and (c) increasing the Mach number, at which CDpeak
occurs, are evident from
Fig.3.28.
Fig.3.28 Effect of wing sweep on variation of CD
with Mach number.
(Adapted from Ref.3.3, chapter 16 with permission)
(VII) Drag divergence Mach number of fuselage
It can be imagined that the flow past a fuselage will also show that the
maximum velocity (Vmax
) on the fuselage is higher than V∞. Consequently, the
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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fuselage will also have critical Mach number (Mcritf
) and drag divergence Mach
number. These Mach numbers depend on the fineness ratio of the fuselage. For
the slender fuselage, typical of high subsonic jet airplanes, Mcritf
could be around
0.9. When Mcritf is exceeded the drag of the fuselage will be a function of Mach
number in addition to the angle of attack.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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Chapter 3
Lecture 11
Drag polar – 6
Topics
3.3.4 Parabolic drag polar at high speeds
3.3.5 Guidelines for variations of CDo and K for subsonic jet transport
airplanes
3.3.6 Variations of CDo and K for a fighter airplane
3.3.7 Area ruling
3.4 Drag polar at hypersonic speeds3.5 Lift to drag ratio
3.6 Other types of drags
3.6.1 Cooling drag
3.6.2 Base drag
3.6.3 External stores drag
3.6.4 Leakage drag
3.6.5 Trim drag
3.3.4 Parabolic drag polar at high speeds
The foregoing sections indicate that the drag coefficients of major airplane
components change as the Mach number changes from subsonic to supersonic.
Consequently, the drag polar of an airplane, being the sum of the drag
coefficients of major components, will also undergo changes as Mach number
changes from subsonic to supersonic. However, it is observed that the
approximation of parabolic polar is still valid at transonic and supersonic speeds,
with CD0
and K becoming functions of Mach number i.e.:
CD = C
D0 (M) + K (M)CL
2 (3.49)
Detailed estimation of the drag polar of a subsonic jet airplane is presented in
section 2 of Appendix B.
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3.3.5 Guidelines for variations of CDo and K for subsonic jet transport
airplanes
Subsonic jet airplanes are generally designed in a manner that there is no
significant wave drag up to the cruise Mach number (Mcruise
). Further, the drag
polar of the airplane for Mach numbers upto Mcruise can be estimated, using the
methods for subsonic airplanes. Section 2 of Appendix B illustrates the
procedure for estimation of such a polar. However, to calculate the maximum
speed in level flight (Vmax
) or the maximum Mach number Mmax
, guidelines are
needed for the increase in CD0 and K beyond Mcruise. Such guidelines are
obtained in this subsection by using the data on drag polars of B727-100 airplane
at Mach numbers between 0.7 to 0.88.
Reference 3.18 part VI, chapter 5, gives drag polars of B727-100 at M = 0.7,
0.76, 0.82, 0.84, 0.86 and 0.88. Values of CD and C
L corresponding to various
Mach numbers were recorded and are shown in Fig.3.29 by symbols. Following
the parabolic approximation, these polars were fitted with Eq.(3.49) and CD0
and
K were obtained using least square technique. The fitted polars are shown as
curves in Fig.3.29. The values of CD0
and K are given in Table 3.5 and presented
in Figs.3.30 a & b.
Fig.3.29 Drag polars at different Mach numbers for B727-100
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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M CD0
K
0.7
0.76
0.820.84
0.86
0.88
0.01631
0.01634
0.016680.01695
0.01733
0.01792
0.04969
0.05257
0.061010.06807
0.08183
0.103
Table 3.5 Variations of CD0
and K with Mach number
Fig.3.30a Parameters of drag polar - CD0
for B727-100
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Fig.3.30b Parameters of drag polar - K for B727-100
It is seen that the drag polar and hence CD0
and K are almost constant up to
M = 0.76. The variations of CD0
and K between M = 0.76 and 0.86, when fitted
with polynomial curves, give the following equations (see also Figs.3.30 a & b).
CD0= 0.01634 -0.001( M-0.76)+0.11 (M-0.76)2 (3.50)
K= 0.05257+ (M-0.76)2 + 20.0 (M-0.76)3 (3.51)
Note: For M ≤ 0.76, CD0
= 0.01634, K = 0.05257
Based on these trends, the variations of CD0
and K beyond Mcruise but upto
Mcruise + 0.1 are expressed by the following two equations.
CD
= DOcr C - 0.001 ( M-Mcruise) + 0.11 (M-Mcruise)2 (3.50 a)
K = Kcr + (M-Mcruise)2 + 20.0 (M-Mcruise)3 (3.51 a)
where DOcr C and Kcr are the values of CD0
and K at cruise Mach number for the
airplane whose Vmax
or Mmax
is required to be calculated. It may be pointed out
that the value of 0.01634 in Eq.(3.50) has been replaced by DOcr C in Eq.(3.50a).
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This has been done to permit the use of Eq.(3.50a) for different types of
airplanes which may have their own values of DOcr C (see section 4.2 of Appendix
B). For the same reason the value of 0.05257 in Eq.(3.51) has been replaced by
Kcr
in Eq.(3.51a).
Section 4.2 of Appendix B illustrates the application of the guidelines given in this
subsection.
3.3.6 Variations of CD0
and K for a fighter airplane
Reference 1.10, chapter 2 has given drag polars of F-15 fighter airplane at
M = 0.8, 0.95, 1.2, 1.4 and 2.2.These are shown in Fig.3.31. These drag polars
were also fitted with Eq.(3.49) and CD0
and K were calculated. The variations of
CD0 and K are shown in Figs.3.32a & b. It is interesting to note that CD0 has a
peak and then decreases, whereas K increases monotonically with Mach
number. It may be recalled that the Mach number, at which CD0
has the peak
value, depends mainly on the sweep of the wing.
Fig.3.31 Drag polars at different Mach numbers for F15 (Reproduced from
Ref.1.10, chapter 2 with permission from McGraw-Hill book company)
Please note: The origins for polars corresponding to different Mach numbers are
shifted.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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Fig.3.32a Typical variations of CD0
with Mach number for a fighter airplane
Fig.3.32b Typical variations of K with Mach number for a fighter airplane
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3.3.7 Area rul ing
The plan view of supersonic airplanes indicates that the area of cross section
of fuselage is decreased in the region where wing is located. This is called area
ruling. A brief note on this topic is presented below.
It was observed that the transonic wave drag of an airplane is reduced when the
distribution of the area of cross section of the airplane, in planes perpendicular to
the flow direction, has a smooth variation. In this context, it may be added that
the area of cross section of the fuselage generally varies smoothly. However,
when the wing is encountered there is an abrupt change in the cross sectional
area. This abrupt change is alleviated by reduction in the area of cross section of
fuselage in the region where the wing is located. Such a fuselage shape is called
‘Coke-bottle shape’. Figure 3.33c illustrates such a modification of fuselage
shape.
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Fig.3.33 Design for low transonic wave drag
(a) Abrupt change in cross sectional area at wing fuselage junction
(b) Coke-bottle shape
Figure 3.34, based on data in Ref.1.9 , chapter 5, indicates the maximum wave
drag coefficient, in transonic range, for three configurations viz (i) a body of
revolution (ii) a wing-body combination without area ruling and (iii) a wing-body
combination with area ruling (Ref.1.9, chapter 5 may be referred to for further
details). Substantial decrease in wave drag coefficient is observed as a result of
area ruling. Figure 3.35. presents a practical application of this principle.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
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CDWave = 0.0035 0.008 0.0045
Fig.3.34 Maximum transonic wave drag coefficient of three different shapes
(a) body of revolution (b) wing-body combination without area ruling (c) wing-
body combination with area ruling
Fig.3.35 An example of area ruling - SAAB VIGGEN
(Adapted from http://upload.wikimedia.org)
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3.4 Drag polar at hypersonic speeds
When the free stream Mach number is more than five, the changes in
temperature and pressure behind the shock waves are large and the treatment of
the flow has to be different from that at lower Mach numbers. Hence, the flows
with Mach number greater than five are termed hypersonic flows. Reference 3.19
may be referred to for details. For the purpose of flight mechanics it may be
mentioned that the drag polar at hypersonic speeds is given by the following
modified expression (Ref.1.1, chapter 6).
CD= C
D0 (M) + K (M)CL
3/2 (3.52)
Note that the exponent of the CL term is 1.5 and not 2.0.
3.5 Lift to drag ratioThe ratio C
L/ C
D is called lift to drag ratio. It is an indicator of the aerodynamic
efficiency of the design of the airplane. For a parabolic drag polar CL/ C
D can be
worked out as follows.
CD= C
D0 +KC
L
2
Hence, CD / CL = (C
D0/ C
L) +KC
L (3.53)
Differentiating Eq.(3.53) with CL and equating to zero gives CLmd
which
corresponds to minimum of (CD / CL
) or maximum of (CL/ C
D).
CLmd
= (CD0 / K)1/2 (3.54)
CDmd
= CD0
+ K (CLmd
)2 = 2 CD0
(3.55)
(L/D)max = (CLmd / CDmd
) =D0
1
2 C K (3.56)
Note:
To show that CLmd
corresponds to minimum of (CD / CL
), take the second
derivative of the right hand side of Eq.(3.53) and verify that it is greater than zero.
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3.6 Other types of drag
Subsections 3.1.1, 3.2.2, 3.2.14, 3.2.17 and 3.3.2. dealt with the skin
friction drag, pressure drag (or form drag), profile drag , interference drag ,
parasite drag, induced drag, lift dependent drag and wave drag. Following
additional types of drags are mentioned briefly to conclude the discussion on this
topic.
3.6.1Cooling drag
The piston engines used in airplanes are air cooled engines. In such engines, a
part of free stream air passes over the cooling fins and accessories. This causes
some loss of momentum and results in a drag called cooling drag.
3.6.2 Base drag
If the rear end of a body terminates abruptly, the area at the rear is called a
base. An abrupt ending causes flow to separate and a low pressure region exists
over the base. This causes a pressure drag called base drag.
3.6.3 External stores drag
Presence of external fuel tank, bombs, missiles etc. causes additional parasite
drag which is called external stores drag. Antennas, lights etc. also cause
parasite drag which is called protuberance drag.
3.6.4 Leakage drag
Air leaking into and out of gaps and holes in the airplane surface causes
increase in parasite drag called leakage drag.
3.6.5 Trim drag
In example 1.1 it was shown that to balance the pitching moment about c.g.
(Mcg), the horizontal tail which is located behind the wing produces a lift (- LT) in
the downward direction. To compensate for this, the wing needs to produce a lift
(LW
) equal to the weight of the airplane plus the downward load on the tail i.e. LW
= W + LT. Hence, the induced drag of the wing, which depends on Lw, would be
more than that when the lift equals weight. This additional drag is called trim drag
as the action of making Mcg equal to zero is referred to as trimming the airplane.
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It may be added that a canard surface is located ahead of the wing and the lift on
it, to make Mcg equal to zero, is in upward direction. Consequently, the lift
produced by the wing is less than the weight of the airplane. SAAB Viggen
shown in Fig.3.35, is an example of an airplane with canard. Reference1.15 and
internet (www.google.com) may be consulted for details of this airplane.
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Chapter 3
Lecture 12
Drag polar – 7
Topics
3.7 High l ift devices
3.7.1 Need for increasing maximum lift coefficient (CLmax)
3.7.2 Factors limiting maximum lift coefficient
3.7.3 Ways to increase maximum lift coefficient viz. increase in camber,
boundary layer control and increase in area
3.7.4 Guidelines for values of maximum lift coefficients of wings with
various high lift devices
3.7 High l ift devices
3.7.1 Need for increasing maximum lift coeffic ient (CLmax)
An airplane, by definition, is a fixed wing aircraft. Its wings can produce lift only
when there is a relative velocity between the airplane and the air. The lift (L)
produced can be expressed as :2
L
1L= ρV SC
2 (3.57)
In order that an airplane is airborne, the lift produced by the airplane must be
atleast equal to the weight of the airplane i.e.
2
L
1L = W = ρ V S C
2 (3.58)
OrL
2WV =
ρSC
(3.59)
However, CL has a maximum value, called LmaxC , and a speed called ‘Stalling
speed (Vs)’ is defined as :
sLmax
2WV =
ρSC (3.59a)
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The speed at which the airplane takes-off ( T0V ) is actually higher than the
stalling speed.
It is easy to imagine that the take-off distance would be proportional 2T0V and in
turn to 2SV . From Eq.(3.59a) it is observed that to reduce the take-off distance (a)
the wing loading (W/S) should be low or (b) the CLmax
should be high. Generally,
the wing loading of the airplane is decided by considerations like minimum fuel
consumed during cruise. Hence, it is desirable that CLmax
should be as high as
possible to reduce the take-off and landing distances. The devices to increase
the CLmax
are called high lift devices.
3.7.2 Factors limiting maximum lif t coefficient
Consider an airfoil at low angle of attack (α). Figure 3.36a shows a flow
visualization picture of the flow field. Boundary layers are seen on the upper and
lower surfaces. As the pressure gradients on the upper and lower surfaces of the
airfoil are low at the angle of attack under consideration, the boundary layers on
these surfaces are attached. The lift coefficient is nearly zero. Now consider the
same airfoil at slightly higher angle of attack (Fig.3.36b). The velocity on the
upper surface is higher than that on the lower surface and consequently the
pressure is lower on the upper surface as compared to that on the lower surface.
The airfoil develops higher lift coefficient as compared to that in Fig.3.36a.
However the pressure gradient is also higher on the upper surface and the
boundary layer separates ahead of the trailing edge (Fig.3.36b). As the angle of
attack approaches about 15o the separation point approaches the leading edge
of the airfoil (Fig.3.36c). Subsequently, the lift coefficient begins to decrease
(Fig.3.36d) and the airfoil is said to be stalled. The value of α for which Cl equals
maxCl is called stalling angle (αstall). Based on the above observations, the stalling
should be delayed to increase maxCl
.
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Fig.3.36a Flow past an airfoil at low angle of attack. Note: The flow is from left to
right (Adapted from Ref.3.20, chapter 6 with permission of editor)
Fig.3.36b Flow past an airfoil at moderate angle of attack.
Note: The flow is from right to left
(Adapted from Ref. 3.21, part 3 section II B Fig.200 with permission from
McGraw-Hill book company)
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Fig.3.36c Flow past an airfoil at angle of attack near stall. Note: The flow is from
left to right (Adapted from Ref.3.12, chapter 6 with permission of editor)
Fig.3.36d Typical Cl vrs α curve
Remark:
Since stalling is due to separation of boundary layer, many methods have
been suggested for boundary layer control. In the suction method, the airfoil
surface is made porous and boundary layer is sucked (Fig.3.37a). In the blowing
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method, fluid is blown tangential to the surface and the low energy fluid in the
boundary layer is energized (Fig.3.37b). Blowing and suction require supply of
energy and are referred to as active methods of control. The energizing of the
boundary layer can be achieved in a passive manner by a leading edge slot
(Fig.3.37c) and a slotted flap which are described in section 3.7.3. Reference
3.11, chapter 11 may be referred for other methods of boundary layer control and
for further details.
a. Suction
b. Blowing
c. Blowing achieved in a passive manner
Fig.3.37 Boundary layer control with suction and blowing
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3.7.3. Ways to inc rease maximum li ft coefficient v iz. increase in camber,
boundary layer control and increase in area
Beside the boundary layer control, there are two other ways to increase
the maximum lift coefficient of an airfoil ( maxCl
) viz. increase of camber and
increase of wing area. These methods are briefly described below.
I) Increase in maximum lift coefficient due to change of camber
It may be recalled that when camber of an airfoil increases, the zero lift
angle (0
αl) decreases and the C
l vs α curve shifts to the left (Fig.3.38). It is
observed that αstall does not decrease significantly due to the increase of
camber and a higher maxCl
is realized (Fig.3.38). However, the camber of the
airfoil used on the wing is chosen from the consideration that the minimum dragcoefficient occurs near the lift coefficient corresponding to the lift coefficient
during cruise. One of the ways to achieve a temporary increase in the camber
during take-off and landing is to use flaps. Some configurations of flaps are
shown in Fig.3.39. In a plain flap the rear portion of the airfoil is hinged and is
deflected when maxCl
is required to be increased (Fig.3.39a). In a split flap only
the lower half of the airfoil is moved down (Fig.3.39b). To observe the change in
camber brought about by a flap deflection, draw a line in-between the upper andlower surfaces of the airfoil with flap deflected. This line is approximately the
camber line of the flapped airfoil. The line joining the ends of the camber line is
the new chord line. The difference between the ordinates of the camber line and
the chord line is a measure of the camber.
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Fig.3.38 Increase in Clmax
due to increase of camber
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Fig.3.39 Flaps, slot and slat
II) Increase in maximum lift coefficient due to boundary layer control
In a slotted flap (Fig.3.39c) the effects of camber change and the
boundary layer control (see remark at the end of section 3.7.2) are brought
together. In this case, the deflection of flap creates a gap between the main
surface and the flap (Fig.3.39c). As the pressure on the lower side of airfoil is
more than that on the upper side, the air from the lower side of the airfoil rushes
to the upper side and energizes the boundary layer on the upper surface. This
way, the separation is delayed and maxCl
increases (Fig.3.40). The slot is
referred to as a passive boundary layer control, as no blowing by external source
is involved in this device. After the success of single slotted flap, the double
slotted and triple slotted flaps were developed (Figs.3.39d and e).
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 9
Fig.3.40 Effects of camber change and boundary layer control on maxCl
III) Increase in Clmax due to change in wing area
Equation (3.57) shows that the lift can be increased when the wing area
(S) is increased. An increase in wing area can be achieved if the flap, in addition
to being deflected, also moves outwards and effectively increases the wing area.
This is achieved in a Fowler flap (Fig.3.39f). Thus a Fowler flap incorporates
three methods to increase maxCl
viz. change of camber, boundary layer control
and increase of wing area. It may be added that while defining the maxCl
, in case
of Fowler flap, the reference area is the original area of the wing and not that of
the extended wing.
A zap flap is a split flap where the lower portion also moves outwards as
the flap is deflected.
IV) Leading edge devices
High lift devices are also used near the leading edge of the wing. A slot
near the leading edge (Fig.3.39g) also permits passive way of energizing the
boundary layer. However, a permanent slot, in addition to increasing the lift, also
increases the drag and consequently has adverse effects during cruise. Hence, a
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 10
deployable leading edge device called ‘Slat’ as shown in Fig.3.39h is used. When
a slat is deployed it produces a slot and increases maxCl
by delaying separation.
On high subsonic speed airplanes, both leading edge and trailing edge
devices are used to increasemaxCl
(Fig.1.2c).
Remarks:
i) References 1.9, 1.10, 1.12 and 3.9 may be referred for other types of high
lift devices like Kruger flap, leading edge extension, blown flap etc.
ii) Reference1.10, chapter 1 may be referred for historical development of
flaps.
3.7.4 Guide lines for values of maximum lift coefficients of w ings with
various high lift devices
An estimate of the maximum lift coefficient of a wing is needed to calculate
the stalling speed of the airplane. It may be added that the maximum lift
coefficient of an airplane depends on (a) wing parameters (aspect ratio, taper
ratio and sweep) (b) airfoil shape, (c) type of high lift device(s), (d) Reynolds
number , (e) surface finish , (f) the ratio of the area of the flap to the area of wing
and (g) interference from nacelle and fuselage.
Table 3.6 presents the values of CLmax
which are based on (a) Ref.1.10,
chapter 5, (b) Ref.3.9 chapter 5 and (c) Ref.3.15 chapter 5. These values can be
used for initial estimate of CLmax
for subsonic airplanes with unswept wings of
aspect ratio greater than 5.
The quarter chord sweep( 1/4 ) has a predominant effect on CLmax
. This effect,
can be roughly accounted for by the following, cosine relationship:
(CLmax
)Λ = (C
Lmax)Λ=o
cos 1/4
For example, when the unswept wing without flap has CLmax
of 1.5, the same
wing with 30o sweep would have a CLmax
of 1.5 x cos 30o or 1.3. Similarly, an
unswept wing with Fowler flap has CLmax
of 2.5. The same wing with 30o sweep
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 11
would have CLmax
of 2.5 x cos 30o or 2.17. With addition of leading edge slat, this
can go upto 2.43.
Type of flap Guideline for CLmax
in landing configuration
No flap 1.5
Plain flap 1.8
Single slotted flap 2.2
Double slotted flap 2.7
Double slotted flap with slat 3.0
Triple slotted flap 3.1
Triple slotted flap with slat 3.4Fowler flap 2.5
Fowler flap with slat 2.8
Table 3.6 Guidelines for CLmax
of subsonic airplanes with unswept wings of
moderate aspect ratio
Figure 3.41 shows CLmax
for some passenger airplanes. The solid lines
correspond to the cosine relation given above.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 12
Fig.3.41 Maximum lift coefficient of passenger airplanes operating at high
subsonic Mach numbers
(Adapted from Ref.3.22, Chapter 8 with permission of authors)
Remarks:
i) The value of CLmax
shown in Table 3.6 can be used in landing configuration.
The flap setting during take-off is lower than that while landing. The maximum lift
coefficient during take-off can be taken approximately as 80% of that during
landing.
ii) The values given in Table 3.6 should not be used for supersonic airplanes
which have low aspect ratio wings and airfoil sections of small thickness ratio.
Reference 3.5, section 4.1.3.4 may be referred to for estimating CLmax
in these
cases.
iii) As the Mach number (M) increases beyond 0.5, the maxCl
of the airfoil section
decreases due to the phenomena of shock stall (see item IV in section 3.3.3).
Hence CLmax of the wing also decreases for M > 0.5. The following relationship
between CLmax
at M between 0.5 to 0.9, in terms of CLmax
at M = 0.5, can be
derived based on the plots of CLmax
vs M in Ref.3.23, chapter 9, and Ref.3.9
chapter 12.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 13
Lmax M
Lmax M=0.5
(C )= - 0.418M + 1.209 , 0.5 M 0.9
(C )
For example at M = 0.9, CLmax
would be about 0.833 of CLmax
at M = 0.5.
Note: The maximum lift coefficient (CLmax
) in transonic Mach number range is not
likely to be monotonic as seen in Fig.3.27a. At transonic and supersonic Mach
numbers, CLmax
must be estimated at each Mach number. Reference 3.5,
section 4.1.3.4 may be consulted for this estimation.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 3
References
3.1 Biermann, D. and Herrnitein Jr., W.H.“The interference between struts invarious combinations” NACA TR 468, (1934). Note: This report can be
downloaded from the site “NASA Technical Report Server(NTRS)”.
3.2 Apelt, C.J. and West, G.S. “The effects of wake splitter plates on bluff body
flow in the range of 104 < Re < 5 x 104 part-2” J.Fluid Mech. Vol.71, pp 145-160,
(1975).
3.3. Hoerner, S.F. “Fluid dynamic drag” Published by author (1965).
3.4. Royal Aeronautical Society data sheets – Now known as Engineering
Sciences Data Unit (ESDU).
3.5. Hoak, D.E. et al. “USAF stability and control DATCOM,” Air Force Wright
Aeronautical Laboratories Technical Report 83-3048, October 1960. (Revised
April 1978). Note: USAF Digital DATCOM can be accessed from net.
3.6 Roskam, J. “Methods for estimating drag polars of subsonic airplanes”
Roskam aviation and engineering (1973).
3.7. Wood K.D. “Aerospace vehicle design Vol.I” Johnson Publishing Co.,
Boulder, Colarado (1966).
3.8. Torenbeek, E. “Synthesis of subsonic airplane design” Delft University Press
(1982).
3.9. Raymer D.P.“Aircraft design: A conceptual approach” AIAA Educational
Series, Fourth Edition (2006).
3.10 Schlichting,H. “Boundary layer theory” McGraw-Hill (1968).
3.11 Schlichting, H. and Gersten , K. “Boundary layer theory” 8th Edition,
Spinger-Verlag, (2000).3.12 Anderson,Jr. J.D. “Fundamentals of aerodynamics” McGraw-Hill,
International Edition (1988).
3.13 White,F.M. “Viscous fluid flow” 2nd Edition, McGraw-Hill (1991).
3.14 Abbott, I.H and Von Doenhoff, A.E. “Theory of wing sections” Dover (1959).
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
3.15 Howe, D. “Aircraft conceptual design synthesis” Professional Engineering
Publishing Limited, London (2000).
3.16 Duncan, W.J., Thom, A.S., and Young, A.D. “Mechanics of fluids” E.L.B.S.
and Edward Arnold, (1975).
3.17 Schlichting, H.and Truckenbrodt, E.D. “Aerodynamics of the airplane”
translated by H.J. Ramm, McGraw Hill, (1979).
3.18 Roskam, J “Airplane design volume I-VIII” Roskam Aviation and
engineering (1990).
3.19 Anderson, Jr, J.D. “Hypersonic and high temperature gas dynamics”
McGraw Hill (1989).
3.20 Oertel, H. (Editor) “Prandtl’s essentials of fluid mechanics” Second edition
Springer-Verlag,(2004).
3.21 Kaufmann, W. “Fluid mechanics” McGraw Hill ,(1963).
3.22 Jenkinson L. R., Simpkin P. and Rhodes D. “Civil jet aircraft design” Arnold,
(1999).
3.23 Huenecke. K, “Combat aircraft design” Airlife Pub. Co. (1987).
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 3
Exercises
3.1 Following data relate to a light airplane.
W =11000 N
Wing: S = 15 m2
, CD0
= 0.007, A.R. = 6.5, taper ratio () = 1.0, e = 0.9.
Fuselage: Has a drag of 136N at V = 160 km/hr at sea level when the angle of
attack is zero.
Horizontal tail: CD0
= 0.006, St = 2.4 m2
Vertical tail: CD0
= 0.006, Sv = 2.1 m2
Other components: CD0 based on wing area = 0.003Estimate the drag polar of the airplane assuming the contribution of the fuselage
to the lift dependent drag as small.
[Answer: CD = 0.0193 + 0.0544 C
L
2 ]
[Remark : The CDO of wing in this exercise appears higher than CDO of tails. It is
likely that the airfoil section used on Wing may be thicker (say 15 to 18%) and
that on tail be thinner (say 9%).]
3.2 A drag polar is given as:
CD = C
D0 + KC
L
n
Show that:
CLmd
= 1/nD0C{ }K(n-1)
, CDmd
=n
n-1C
D0
(CL/C
D) max = (n-1)/n 1/n
1/n D0
1 1{ }
n C K(n-1)
n-1
Verify that when n = 2, the above expressions reduced to those given by
Eqs. (3.54),(3.55) and (3.56).
3.3 Based on data in Ref.1.1, chapter 6, the drag polar of a hypersonic glider is
given in the table below.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-3
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
CL 0 0.05 0.1 0.15 0.2
CD 0.028 0.0364 0.05 0.07 0.0907
Fit Eq.(3.52) to this data and obtain CD0
and K. Also obtain CLmd
, CDmd
and
(CL/C
D)max
. The expressions mentioned in exercise 3.2 can be used.
[Answers: CD0
= 0.0283, K = 0.703, CD = 0.0283+0.0703 C
L
3/2
CLmd
= 0.1865, CDmd
= 0.0849, (CL/C
D)max
= 2.197].
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 1
Chapter 4
Engine characteristics
(Lectures 13 to 16)
Keywords: Engines for airplane applications; piston engine; propeller
characteristics; turbo-prop, turbofan and turbojet engines; choice of engine for
different applications.
Topics
4.1 Introduction
4.1.1 Engines considered for airplane applications
4.2 Piston engine-propeller combination
4.2.1 Operating principle of a piston engine
4.2.2 Effect of flight speed on the output of a piston engine
4.2.3 Effect of altitude on the output of a piston engine
4.2.4 Specific fuel consumption (SFC)
4.2.5 The propeller
4.2.6 Propeller efficiency
4.2.7 Momentum theory of propeller
4.2.8 Parameters for describing propeller performance and typical
propeller characteristics
4.2.9 Selection of propeller diameter for chosen application
4.2.10 Procedure for obtaining propeller efficiency for given h,V, BHP
and N
4.2.11 Variations of THP and BSFC with flight velocity and altitude
4.2.12 Loss of propeller efficiency at high speeds
4.3 Gas turbine engines
4.3.1 Propulsive efficiency
4.3.2 Why turboprop, turbo fan and turbojet engines?
4.3.3 Characteristics of a typical turboprop engine
4.3.4 Characteristics of a typical turbofan engine
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 2
4.3.5 Characteristics of a typical turbojet engines
4.4 Deducing output and SFC of engines where these characteristics are
not available directly
4.5 A note on choice of engines for different range of flight speeds
References
Excercises
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 3
Chapter 4
Lecture 13
Engine characteristics – 1
Topics
4.1 Introduction
4.1.1 Engines considered for airplane applications
4.2 Piston engine-propeller combination
4.2.1 Operating principle of a piston engine
4.2.2 Effect of flight speed on the output of a piston engine
4.2.3 Effect of altitude on the output of a piston engine
4.2.4 Specific fuel consumption (SFC)
4.2.5 The propeller
4.2.6 Propeller efficiency
4.2.7 Momentum theory of propeller
4.1. Introduction
To evaluate the performance of an airplane we need to know theatmospheric characteristics, the drag polar and the engine characteristics like
variations of thrust (or power) output and specific fuel consumption with flight
speed and altitude. In this chapter the engine characteristics are briefly reviewed.
4.1.1 Engines considered for airplane applications
Following power plants are considered for airplane applications.
(a) Piston engine-propeller combination.
(b) Gas turbine engines - turboprop, turbofan and turbojet.
(c) Ramjets.
(d) Rockets.
(e) Combination power plants like ramrocket and turboramjet.
At present, piston engine-propeller combination and gas turbine engines are the
power plants used on airplanes. Ramjets offer simplicity of construction and have
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 4
been proposed for hypersonic airplanes. However, a ramjet cannot produce any
thrust when flight speed is zero. Hence, it is proposed to use a rocket or turbojet
engine to bring it (ramjet) to a flight speed corresponding to Mach number (M) of
2 or 3 and then the ramjet engine would take over. Consequently, the
combination power plants viz. ramrocket or turboramjet have been proposed.
Rockets have sometimes been used on airplanes as boosters to increase
the thrust for a short duration e.g. during take-off.
4.2 Piston engine-propeller combination
In this case the output of the engine viz. brake horse power (BHP) is
available at the engine shaft and is converted into thrust by the propeller.
4.2.1 Operating principle of a piston engine
A few relevant facts about the operation of piston engines, used on
airplanes, are mentioned here. In these engines a certain amount for fuel-air
mixture is taken in, it is compressed, then ignition, due to a spark, takes place
which is followed by the power stroke and the exhaust stroke (Fig.4.1).
Remarks:
i ) The piston engine in which the ignition is caused by a spark from the spark
plug is called a spark-ignition engine. There are other types of piston engines in
which the pressure and temperature at the end of the compression stroke are
high enough to cause spontaneous ignition. Such engines do not need a spark
plug and are known as compression-ignition engines.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 5
Fig.4.1 Four stroke cycle of a spark-ignition engine
ii) The volume of the air-fuel mixture taken in, is almost equal to the swept
volume i.e., product of the area of cross-section of the engine cylinder and the
length of the piston stroke. The mass of fuel taken in per power stroke is thus
approximately equal to:
(swept volume) X (density of air) / (air-fuel ratio).
4.2.2 Effect of flight speed on the output of a piston engineFor a given altitude and r.p.m. (N) the power output changes only slightly
with flight speed. This is because the piston engines are generally used at low
speeds (M < 0.3) and at these low Mach numbers, the increase in manifold
pressure due to the deceleration of air in the engine manifold is negligible. Hence
power output increases only slightly with flight speed. This increase is generally
ignored.
4.2.3 Effect of altitude on the output of a piston engine
To understand the effect of altitude on the output of the piston engine, the
following three facts need to be noted. (a) As stated at the end of the subsection
4.2.1, the mass of fuel taken in per stroke is equal to the product of swept volume
and density of air divided by air-fuel ratio. (b) For complete combustion of fuel,
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 6
the air-fuel ratio has a definite value (around 15, the stoichiometric ratio). (c) As
the flight altitude increases, the density of air decreases.
Thus, for a given engine r.p.m. and air-fuel ratio, the mass of air and
consequently, that of the fuel taken in decreases as the altitude increases. Since,
the power output of the engine depends on the mass of the fuel taken in, it
(power output) decreases with altitude. The change in power output (P) with
altitude is roughly given as (Ref.3.7,Appendix 1 A-5 and Ref.4.3, chapter 14):
(P / P0) =1.13σ – 0.13 (4.1)
where P0 is the power output at sea level under ISA conditions and σ is the
density ratio.
Remark:
(i) Reference 3.15, chapter 3, gives the following alternate relationship for
decrease of power output with altitude :
(P / P0) = σ1.1
(4.1a)
(ii) Figure 4.2 shows the performance for a typical piston engine. To prepare
such a performance chart, the engine manufacturer carries out certain tests, on
each new engine. During these tests the engine is run at a chosen RPM and
different loads are applied. The throttle setting is adjusted to get steady
conditions. The quantities like (a) engine RPM(N), (b) torque developed, (c)
manifold air pressure(MAP) and (d) the fuel consumed in a specific interval of
time, are measured.These tests are conducted at different RPM’s. From these
test data the power output and the fuel flow rate per hour are calculated. The
data are also corrected for any difference between the ambient conditions during
the test, and the sea level standard conditions. The left side of Fig.4.2 presents
the sea level performance of a Lycoming engine. The upper part of the figure
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 7
Fig.4.2 Typical piston engine performance (Lycoming 0-360-A)
(with permission from Lycoming aircraft engines )
shows the power output at different MAP’s with RPM as parameter. The lower
part of the figure shows the fuel flow rate in US gallons per hour.
To obtain the effect of altitude on the engine output, the power output is
measured at different RPM’s and MAP’s during flight tests at different altitudes.
Typical altitude performance of Lycoming engine is presented in the right side of
Fig.4.2.
From such a chart, the output of the engine and the fuel flow rate can beobtained for a chosen combination of altitude, RPM, MAP and ambient
temperature. The steps to obtain these are explained with the help of examples
4.1 and 4.2
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 8
It may be added that the units used in Fig.4.2, which is reproduced from
manufacturer’s catalogue, are in FPS system. However, SI units are used in this
and the subsequent chapters.
4.2.4 Specific fuel consumption (SFC)
In engine performance charts, the fuel consumption is presented as fuel
flow rate per hour. However, in engineering practice the fuel consumption is
expressed as specific fuel consumption (SFC). It is defined as :
Fuel flow rate in Newton per hour SFC =
BHP in kW (4.1b)
Remarks :
(i) The output of a piston engine or turboprop engine is available as power at the
engine shaft. It is called BHP and measured in HP when FPS system is used. In
SI units the output is measured in kW. On the other hand, the output of a
turbofan or a turbojet engine is available as thrust, which is measured in ‘lb’ in
FPS system and in Newton in SI units.
The specific fuel consumption of a jet engine is defined as:
SFCFuel flow rate in Newton per hour
=Thrust in Newton
(4.1c)
(ii) To distinguish the specific fuel consumption of a piston or a turboprop engine,
from that of a jet engine, the SFC defined by Eq.(4.1b), is denoted as BSFC i.e.
Fuel flow rate in Newton per hour BSFC =
BHP in kw with units of N/kW-hr (4.1d)
The specific fuel consumption of a turbofan or a turbojet engine is denoted by
TSFC i.e.
Fuel flow rate in Newton per hour TSFC =
Thrust in Newton with units of hr -1 (4.1e)
(iii) BSFC in metric units is also expressed as mg/W-s
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 9
Example 4.1
Obtain the power output and BSFC for the Lycoming engine when
operating at sea level at an RPM(N) of 2400 and MAP of 24 of mercury (Hg).
Solution :
From plots in the left side of Fig.4.2, for N = 2400 and MAP = 24 of Hg
the power output is 136 HP and the fuel flow rate is 10.7 US gallons/hr.
Taking 1 US gallon = 3.78 litre and density of petrol as 0.76 kg/m3 gives:
1 gallon per hour of petrol = 3.78 x 0.76 kg/hr
= 3.78 x 0.76 x 9.81 N/hr
= 28.18 N/hr of petrol
Hence, the fuel flow rate in the case under study is :
10.7 x 28.18 = 301.5 N/hr.
Noting that 1 lb/hr = 4.45 N/hr, The fuel flow rate in this case is 67.75 lbs/hr.
Further, 1 HP is 0.7457 kW. Hence, power output of 136 HP equals 101.4 kW.
Hence, BSFC in SI units is: 301.5/101.4 = 2.973 N/kW-hr
In FPS units it is: 67.75/136 = 0.498 lb/HP-hr
Answers:
For the given engine, the power output, fuel flow rate and BSFC at N = 2400 and
MAP = 24 of Hg under sea level standard conditions are :
(i)Power output = 101.4 kW = 136 HP, (ii) Fuel flow rate = 10.7 US gallons/hr
or 301.5 N/hr or 67.75 lb/hr of petrol (iii) BSFC = 2.973 N/kW-hr = 0.498 lb/HP-hr
Example 4.2
Obtain the power output and BSFC for the Lycoming engine when
operating at 8000 altitude, RPM (N) of 2200 and MAP of 20 of Hg.
Solution :
Reference 1.9 chapter 6, gives the following procedure to obtain the
output and fuel flow rate using left and right sides of Fig.4.2.
(i)At sea level for N = 2200 and MAP of 20 of Hg the output would be 97.5 BHP.
This is indicated by point ‘B’ in the left hand side of Fig.4.2. This side of the
diagram is also called sea level performance.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 10
(ii)Transfer this point to the right hand side of Fig.4.2 at sea level which is
indicated by point ‘C’. The right side of the diagram is also called altitude
performance.
(iii)Locate a point on the altitude curve corresponding to N = 2200 and MAP of
20 of Hg. This point is indicated by ‘A’.
(iv)Join points C and A by a dotted line. The value at 8000 on this line (the point
‘D’) is the output at h = 8000 corresponding to N = 2200 and MAP = 20 of Hg.
It is seen that the value is 107 HP.
(v)To get the fuel flow rate, mark a point ‘F’ on the sea level performance at 107
HP and N = 2200. The MAP at this point is observed to be 21.2 of Hg. The fuel
flow rate corresponding to N = 2200 and MAP of 21.2 of Hg, from the lower part
of figure in the left side is 8.25 gallons per hour. This point is indicated by ‘G’
Hence, at h = 8000 , N = 2200 and MAP of 20 . The output is 107 HP (79.79
kW) and the fuel flow rate is 8.25 gallons / hr (232.5 N/hr or 52.2 lbs/hr of petrol).
Consequently, BSFC =232.5
= 2.914 N/kW-hr 79.79
or in FPS units, BSFC =52.2
= 0.488 lb/HP-hr 107
Answers :
At h = 8000 , N = 2200 and MAP = 20 of Hg :
Output = 79.79 kW =107 HP and BSFC =2.914 N/kW-hr = 0.488 lb/BHP-hr
Note : Reference 1.9, chapter 6 may be referred to obtain the correction to the
output if the ambient temperature is different from that in ISA.
4.2.5 The propeller
The output of the engine is converted into thrust by the propeller. A typical
engine with a two bladed propeller is shown in Fig.4.3. Depending on the engine
power and the operating conditions, the propeller may have two to four blades.
Special propellers with five or six blades have also been used in practice when
required.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 11
Fig.4.3 Typical engine-propeller combination
(Source: www.flickr.com)
The propeller blade, as seen in Fig.4.3, is like a wing with significant amount of
twist. Refer Fig.3.3 for geometric parameters of a wing. The geometry of the
propeller is defined by the following features. (a) The variation of the chord,
shape and thickness of the airfoil section (also called blade element) over the
span of the blade. (b) The angle between the chord of the blade element and the
plane of rotation. This is also one of the definitions of the pitch angle (β).
The pitch angle (β) varies along the span of the blade for the following reason.
Since the propeller blade moves forward as it rotates, the blade element has a
forward velocity of V and a circumferential velocity of 2πrn ; where ‘r’ is the
radius of the blade element and ‘n’ is the revolutions per second of the propeller.
The blade element experiences a relative wind which is resultant of the forward
and circumferential velocities. As ‘r’ varies from the root to the tip, the blade
elements at various spanwise locations of the propeller are subjected to a
relative wind which varies significantly, in magnitude and direction, along the
span. Further, each blade element being an airfoil, must operate at a moderate
angle of attack. These two considerations require that the blade elements along
the span of the blade make different angles to the plane of rotation or have
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 12
different pitch angles (β). The pitch of the blade is generally the pitch of the blade
element at r/R = 0.75, where R is the radius of the blade.
For other definitions of pitch Ref.2.1 and chapter 6 of Ref.1.9, be
consulted. For details of the geometry of propellers refer chapter 6 of Ref.1.9,
chapter 16 of Ref.3.7 and Ref.4.1.
4.2.6 Propeller efficiency
Consider that an engine located in an airplane is developing certain output
indicated as BHP. The propeller attached to this engine produces a thrust T
when the airplane moves with a speed V . In this situation, the power output
called ‘Thrust horse power (THP)’, is (TV / 1000) in kW. The efficiency of the
propeller is therefore defined as:
ηρ = THP / BHP = T V /(1000 x BHP) ( 4.2 )
Note: T is in Newton, V is in m/s and THP and BHP are measured in kW.
The efficiency of a propeller can be estimated by analysing the flow through a
propeller. The momentum theory of propeller is briefly discussed in the next
subsection. Subsequent subsections deal with determination of propeller
efficiency from experimental results.
4.2.7 Momentum theory of propeller
As the name suggests, this theory is based on the idealization that the
thrust produced by the propeller is the result of the increase in momentum
imparted to the airstream passing through the propeller. It is assumed that the
propeller can be thought of as an actuator disc. This disc is an idealised device
which produces a sudden pressure rise in a stream of air passing through it. This
pressure rise integrated over the disc gives the thrust developed by the propeller.
Figure 4.4 shows the actuator disc and the flow through it. It is assumed that : (i)
the flow is incompressible and inviscid, (ii) the increase in pressure is constant
over the disc (iii) there is no discontinuity in flow velocity across the disc (iv) in
the flow behind the disc, called slipstream, there is no swirl.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 13
Fig.4.4 Flow through an actuator disc
(a) Stream tube (b) Pressure variation
In Fig.4.4 the actuator disc is located at plane AA. Far upstream, the velocity is
V and the pressure p is the atmospheric pressure. The velocity V equals theforward speed of the airplane on which the propeller is mounted. A stream tube
enclosing the disc is also shown in Fig.4.4. As the stream approaches the front
face of the disc the fluid velocity reaches a value V1 at the disc. As the flow is
assumed to be inviscid and incompressible, Bernoulli’s equation is valid till the
front face of the disc and the pressure decreases, to a value p1. At the disc,
energy is added in the form of increase in pressure by an amount Δp while the
velocity remains the same as V1 through the disc (Fig.4.4a). After the disc the
pressure gradually returns to the atmospheric value of p . Bernoulli’s equation is
again valid behind the disc and the fluid velocity increases to a value V j. The
changes in pressure and velocity are shown in Fig.4.4a.
Applying Bernoullis equation ahead and behind the disc gives :
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 14
Total head ahead of disc = H =1
2 2
1
1 1p + ρV = p + ρV
2 2 (4.3)
Total head behind the disc = H1 = 2 2
1 1 j
1 1p + Δp + ρV = p + ρV
2 2 (4.4)
Consequently, 2 21 1
2 2
2 2
1 j j1 Δp = H - H = p ρV p ρV = ρ V - V2
(4.5)
Since Δp is the change in pressure over the disc, the thrust acting on the disc is:
2 2
j
ρT = A Δp=A V - V
2 (4.6)
where, A = area of disc = 2d4
; d = diameter of the propeller
Alternatively, the thrust produced can also be obtained as the rate of change of
momentum of the stream i.e.
jT = m V - V (4.7)
where, m = rate of mass flow through the disc = 1ρ A V (4.8)
Hence, 1 jT = ρ A V V - V (4.9)
Equating Eqs.(4.6) and (4.9) yields :
2 2
1 j j
ρρ AV V - V = A V -V
2
Or j
1
V + VV =
2 (4.10)
Thus, the momentum theory shows that the velocity at the disc (V1) is the
average of V j & V . In other words, half of the increase in velocity takes place
ahead of the disc and the remaining half behind it.
The efficiency of the actuator disc can be obtained by considering the ratio of
power output to the power input.
The power output = work done = jT V = m V V V (4.11)
The power input is the energy imparted to the fluid stream. This is the energy of
the stream far behind the disc minus the energy of the stream far ahead of the
disc. i.e.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 15
Power input = 2 2
j
1 1mV - mV
2 2
(4.12)
Hence, propeller efficiency is:
2
2
1
j
p2 j j
j
mV V V 2Vpower output
η = = = =m Venergy input V VV V
2 V
(4.13)
Remarks:
(i)Equation (4.13) gives the propeller efficiency under ideal conditions and
represents an upper limit on efficiency obtainable. In practical situations, the
efficiency would be lower due to losses associated with (a) profile drag of blades,
(b) swirl in slip stream and (c) the pressure at the blade tips being the same
ahead and behind the disc.
(ii)For production of thrust, V j must be greater than V . But for high propeller
efficiency V j must be only slightly higher than V . Hence to get adequate amount
of thrust with high propeller efficiency a large mass of air should be given a small
velocity increment.
(iii)Propeller theories like blade element theory, and vortex theory take into
account effects of drag of blades, finite span of blade etc. For details of these
theories refer to chapter 6 of Ref.1.9.Example 4.3
A propeller of diameter 1.8 m is mounted on an airplane. When
moving at a speed of 200 kmph it produces a thrust of 2070 N under standard
sea level conditions. Calculate the velocity of slip stream far behind the propeller
and the ideal efficiency of the propeller.
Solution :
Diameter of propeller = d = 1.8 m
Free stream velocity = V = 200 kmph = 55.56 m/s
Slipstream velocity = V j
Consequently, Thrust = T =
jm V -V
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 16
j2
V +Vm = ρ d
4 2
Hence, j2
j
V +55.562070 = 1.225× ×1.8 V -55.56
4 2
Or 1328.1 = 2 2
jV -55.56
Or jV = 66.45 m/s
Ideal propeller efficiency = j
2 2= = 0.9107 = 91.07 %
V 66.451+1+
55.56V
Answers :
Velocity of slip stream far behind propeller = 66.45 m/s = 239.22 kmphIdeal propeller efficiency = 91.07 %
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 1
Chapter 4
Lecture 14
Engine characteristics – 2
Topics
4.2.8 Parameters for describing propeller performance and typical
propeller characteristics
4.2.9 Selection of propeller diameter for chosen application
4.2.8 Parameters for describing propeller performance and typical propeller
characteristics
As pointed out at the end of the previous subsection, the momentum
theory of propeller has limitations. Though the refined theories are helpful in
design of propeller blades, the propeller characteristics obtained from the wind
tunnel tests are used for estimation of airplane performance. These
characteristics are presented in terms of certain parameters. First these
parameters are defined and then typical characteristics of propellers are
presented. The procedures for (a) selection of the propeller diameter and (b)obtaining the propeller efficiency for given h, v, BHP and N, are given in the next
two subsections.
Following Ref.4.1 and Ref.3.7 chapter 6, the propeller performance is
expressed in terms of the following coefficients. It may be pointed out that FPS
units are used in these references whereas SI units are used here.
Advance ratio : J = V/nd (4.14)
Power coefficient: PC = P/ρn3d5; P in Watts (4.15)
Thrust coefficient: CT = T/ρn2d
4 (4.16)
Speed power coefficient: Cs = V (ρ/ Pn2)1/5 = 5P
J/ C (4.17)
Propeller efficiency: pη =TV / P; P in Watts
= J (CT / PC ) (4.18)
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 2
Torque coefficient:2 5Q
Q=ρn d
C (4.19)
Torque speed coefficient: 3
S QQ = J/ C = V ρd /Q (4.20)
where, P = Power in watts, T = thrust (N); V = flight velocity (m/s), n = rotationalspeed (rev/s),
d = diameter of propeller (m)
Q = Torque (Nm) = P/ 2 n
In FPS units:
T = thrust (lbs); P = power (ft lbs/s) = 550 BHP
V = velocity (ft / s), BHP = brake horse power
The performance of a propeller is indicated by thrust coefficient (CT), power
coefficient ( PC ) and efficiency ( pη ). These quantities depend on advance ratio
(J) and pitch angle β . Based on Ref.4.1, the experimental characteristics of a
two bladed propeller are presented in Figs.4.5a to d.
Figure 4.5a presents the variation of pη vs J with β as parameter. It is seen that
pη is zero when V is zero; J is also zero in this case by virtue of its
definition(Eq.4.14). Equation (4.2) also indicates that pη is zero when V is zero.
This is because even though the engine is working and producing thrust, no
useful work is done when V is zero. This is like a person pressing an immovable
wall. He spends muscular energy to push the wall but the output and hence the
efficiency is zero as the wall does not move and no useful work is done.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 3
Fig.4.5a Propeller efficiency ( pη ) vs advance ratio (J) with pitch angle (β) as
parameter.
For a chosen value of β , the efficiency ( pη ) increases as J increases. It reaches
a maximum for a certain value of J and then decreases (Fig.4.5a). The maximum
value of pη is seen to be around 80 to 85%. However, the value of J at which the
maximum of pη occurs, depends on the pitch angleβ . This indicates that for a
single pitch or fixed pitch propeller, the efficiency is high (80 to 85%) only over a
narrow range of flight speeds (Fig.4.5a). Keeping this behaviour in view, the
commercial airplanes use a variable pitch propeller. In such a propeller the entireblade is rotated through a chosen angle during the flight and the pitch of all blade
elements changes. Such propellers have high efficiency over a wide range of
speeds. However, propellers with variable pitch arrangements are expensive and
heavy. Hence, personal airplanes, where cost of the airplane is an important
consideration, employ a fixed pitch propeller. As a compromise, in some designs,
propellers with two or three pitch settings are employed.
Figure 4.5b presents the variation of power coefficient (P
C ) vs J with β and CT
as parameters. This chart is useful to obtain pη for given values of altitude,
velocity, RPM and BHP (see subsection 4.2.10).
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 4
Fig.4.5b Power coefficient ( PC ) vs advance ratio (J) with pitch angle (β) and
thrust coefficient (CT) as parameters.
Figure 4.5c presents the variations of CS vs J and CS vs pη with βas parameter.
This figure is designated as ‘Design chart’ and is used for selection of the
diameter of the propeller. A brief explanatory note on this topic is as follows.
Using defintions of J and PC , the parameter sC , defined below, is obtained. It is
observed that this parameter does not involve the diameter (d) of the propeller.
2 1/51/5P
sJ
C = = V (ρ / Pn )C
(4.21)
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 5
It is also observed that the parameter sC depends on V, ρ , P and N.
Consequently, this parameter can be evaluated when the power output (P),
engine RPM(N) and flight condition viz. V and h are specified.
The design problem involves obtaining the value of J which would give the
maximum value of pη for a specified value of sC . This is arrived at in the
following manner.
Fig.4.5c Design chart
Using the data in Figs.4.5b & a , the values of sC can be obtained for constant
values of J or β . For example, for β= 15o the values given in Table 4.1 are
obtained.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 6
J PC
From Fig.4.5b
sC
From Eq.(4.21)
pη
From Fig.4.5a0 0.04 0 0
0.2 0.04 0.381 0.43
0.4 0.037 0.773 0.69
0.6 0.025 1.255 0.805
0.8 0.005 3.685 0.35
Table 4.1 variation ofs
C with J for β = o15
Similar calculations at , , ,o o o o o oβ = 20 25 30 ,35 40 and 45 yield additional values.
From these values the curves for sC vs pη and sC vs J at different values of β
can be plotted. These are shown in the upper and lower parts of Fig.4.5c. Based
on these plots, the dotted line in the lower part of Fig.4.5c gives the values of J
and β which would give maximum pη . This line is designated as ‘Line of
maximum efficiency for sC ’. For example, corresponding to a value of sC = 1.4,
the dotted line gives J = 0.74 and β = o20 . The upper part of the Fig.4.5c gives
pη = 82% for the chosen value of sC = 1.4.
From the value of J, the propeller diameter is obtained as d = V/(nJ) ; note that
the values of V and n are already known. Subsection 4.2.9 gives additional
details and example 4.4 illustrates the procedure to select the propeller diameter.
Figure 4.5d presents the variation of thrust coefficient (CT) vs J with β as
parameter. It is observed that when J is zero, CT is not zero as the propeller
produces thrust, even when ‘V’ is zero. The curves in Fig.4.5d are useful to
estimate the thrust developed by the propeller especially during the take-off flight.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 7
Fig.4.5d Thrust coefficient (CT) vs advance ratio (J) with pitch angle β as
parameter.
Fig.4.5 Typical characteristics of a two bladed propeller
(Adapted from Ref. 4.1)
Remark :
Reference 4.1 contains information on propellers with three and four
blades. Reference 3.7 chapter 16 contains information on six bladed propellers.
Additional information can be obtained from Ref.4.2 which is cited in chapter 17of Ref.4.3.
4.2.9 Selection of propeller diameter for chosen application
A propeller is selected to give the best efficiency during a chosen flight
condition which is generally the cruising flight for transport airplanes. Some
companies may design their own propellers but it is an involved task. Hence, the
general practice is to use the standard propellers and the charts corresponding to
them. As a first step, the number of blades of the propeller is decided depending
on the amount of power to be absorbed by the propeller.
The designer of a new airplane generally chooses the diameter of the propeller
using the design chart (e.g. Fig.4.5c) appropriate to the propeller. Let us consider
a two bladed propeller. Following steps are used to select the diameter of a
propeller.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 8
(a) Choose a level flight condition i.e. altitude ( ch ) and speed ( cV ).
(b) Obtain lift coefficient (CL) in this flight using :
CL = W/ 2c0.5ρV S . Obtain the corresponding CD from the drag polar of the
airplane.
(c) Obtain THP required during the flight using : THP = 3
DC0.5ρV SC /1000
(d) Assume pη = 0.8 .
(e) Obtain BHP = THP/0.8. Then RPM (N) which will give this power output at the
chosen hc with low BSFC is known from the engine curves e.g. Fig.4.2.
Calculate n = N/60.
(f) Calculate 2
1/5
SC = V ρ /Pn .
(g) From the design chart like Fig.4.5c, obtain the value of J on the dotted line,
corresponding to the value of CS in step (f). Also obtain the value of β from the
same curve. Obtain the value of pη from the upper part of the design chart.
(h) Since V, n and J are known, obtain propeller diameter (d) using : d = V/n J
(i) If the value of pη obtained in step (g) is significantly different from the value of
0.8 assumed in step (d), then iterate by using the value ofp
η obtained in step (g).
Finally round-off the propeller diameter to nearby standard value.
Remark :
The choice of the parameters of the propeller like, diameter, pitch, blade
size are also influenced by factors like noise level of the propeller, ground
clearance, and natural frequency of the blade. Refer chapter 6 of Ref.1.9.
Example 4.4
Consider the case of Piper Cherokee airplane dealt with in Appendix A
and obtain the diameter of the propeller for this airplane. According to chapter 6
of Ref.1.9, the chosen speed and altitude for propeller design are 132 mph
(212.4 kmph or 59 m/s) and sea level standard conditions respectively. The
engine operates at 75% of the maximum power at an RPM of 2500.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 9
Solution :
From Appendix ‘A’ the following data are obtained on Piper Cherokee
airplane.
Weight of airplane = W = 10673.28 N
Drag polar : CD = 0.0349 + 0.0755 2LC
Wing area = S = 14.864 m2 .
At sea level ρ= 1.225 kg/m3
CL under chosen flight condition is =
1.2
10673.28= 0.3368
1× 225 × 59 ×14.864
2
CD = 0.0349 + 0.0755 x 0.33682 = 0.04346
Hence, thrust horse power required (THPr ) is :
THPr =
31× 1.225 × 59 × 14.864 × 0.04346
2 = 81.26 kW1000
As a first step, assume pη = 0.8.
Consequently, the required BHP is :
BHPr = 81.26/0.8 = 101.6 kW = 101600 W
Noting that at sea level the maximum power is 135 kW, the BHP r of 101.6 kW is
close to 75% of that value which is prescribed in the exercise.
N = 2500. Hence, n = revolutions per second = 2500/60 = 41.67
Consequently, CS = V 1
1/52 25ρ / pn = 59 1.225 /101600 × 41.67 = 1.38
The airplane has a two bladed propeller of standard design and hence Fig.4.5c is
applicable. From this figure, corresponding to CS of 1.38, the dotted line gives
J = 0.74 ,o
β = 20 , pη = 0.83.
Consequently, the first estimate of propeller diameter is :
V 59d = = = 1.91m
nJ 41.67×0.74
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 10
Since, the value of pη obtained is somewhat different from the value of 0.8
assumed earlier, the steps are repeated with pη = 0.83.
BHPr = 81.26/0.83 = 97.90 kW = 97900 W
CS = 59 (1.225/97960 x 41.472
)1/5
= 1.390
From Fig.4.5c corresponding to CS of 1.39, the dotted line gives:
J = 0.75 and oβ = 20 and pη = 0.83.
Consequently, the second estimate of propeller diameter is :
d =59
= 1.89 m41.67×0.75
Since the latest value of pη is same as the value with which the steps were
repeated, the propeller diameter is taken as 1.89 m.
Remark:
The value of the propeller diameter obtained above is very close to the value of
1.88 m in the actual airplane.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 1
Chapter 4
Lecture 15
Engine characteristics – 3
Topics
4.2.10 Procedure for obtaining propeller efficiency for given h,V, BHP
and N
4.2.11 Variations of THP and BSFC with flight velocity and altitude
4.2.12 Loss of propeller efficiency at high speeds
4.3 Gas turbine engines
4.3.1 Propulsive efficiency
4.3.2 Why turboprop, turbo fan and turbojet engines?
4.2.10 Procedure for obtaining THP for given h, V, BHP and N
For calculating the performance of the airplane, the thrust horse power (THP)
is needed at different values of engine RPM(N), break horse power (BHP), flight
speed (V) and flight altitude (h). In this context the following may be noted.
(a) The engine output (BHP) depends on the altitude, the RPM (N) and themanifold air pressure (MAP).
(b) The propeller absorbs the engine power and delivers THP; pTHP = η BHP
(c) The propeller efficiency depends, in general, on BHP, V, N and β.
(d) The three quantities viz. d, V and n can be combined as advance ratio
(J = V/nd).
(e) Once pη is known :
THP = pη x BHP and T = THP×1000/ V .
The steps required to obtain pη depend on the type of propeller viz. variable pitch
propeller, constant speed propeller and fixed pitch propeller. The steps in the
three cases are presented below.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 2
I) Variable pitch propeller
In this type of propeller the pitch of the propeller is changed during the flight
so that the maximum value of pη is obtained in various phases of flight. The steps
are as follows.
(a) Obtain the ambient density ρ for the chosen altitude.
(b) Obtain 3 5PC = P / ρ n d ; P is BHP in watts
(c) Obtain J = V/nd
(d) Calculate 1/5PSC = J/C
(e) From the design chart for the chosen propeller (e.g. Fig.4.5c for a two
bladed propeller), obtain β which will give maximum efficiency. Obtain
corresponding pη . Consequently,
THP = pη x BHP and T = THP x 1000 / V
(f) To get the thrust (T) at V = 0, obtain CT at J = 0 for a pitch setting which
would give high value of CT. In Fig.4.5d, it is seen that for a two bladed
propeller, CT at J = 0, is nearly maximum for oβ 30 . Having known CT, the
thrust(T) is given by :
T = 2T
4ρ n d C
II) Constant speed propeller
The variable pitch propellers were introduced in 1930’s. However, it was
noticed that as the pilot changed the pitch of the propeller, the engine torque
changed and consequently the engine RPM deviated from its optimum value.
This rendered, the performance of the engine-propeller combination,
somewhat suboptimal. To overcome this problem, the constant speed
propeller was introduced. In this case, a governor mechanism alters the fuel
flow rate so that the required THP is obtained even as rpm remains same.
The value of β is adjusted to give maximum possible pη .
The steps to obtain pη are the same as mentioned in the previous case.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
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III) Fixed pitch propeller
From Fig.4.5b it is observed that a fixed pitch propeller has a definite
value of CP for a chosen value of advance ratio (J). Consequently, the
propeller can absorb only a certain amount of power for a given value of J.
Thus when the flight speed changes, the power absorbed by the propeller
also changes. However, for the engine-propeller combination to be in
equilibrium i.e. run at a constant r.p.m, the power absorbed by the propeller
and that produced by the engine must be the same. This would render the
problem of determining power output as a trial and error procedure. However,
it is observed that the torque of a piston engine remains nearly constant over
a wide range of r.p.m’s. Using this fact, the torque coefficient (CQ) and torque
speed coefficient (Qs) are deduced in Ref.3.7, chapter 16, from the data on
CP & CT . Further a procedure is suggested therein to obtain pη at different
flight speeds.
Herein, the procedure suggested in the Appendix of Ref.4.1 is presented. It is
also illustrated with the help of example 4.5.
It is assumed that the propeller is designed for a certain speed, altitude, rpm
and power absorbed.Let V0 = design speed (m/s)
N0 = design rpm ; n0 = N0 / 60
BHP0 = BHP of the engine under design condition (kW)
d = diameter of propeller (m)
J0 = Advance ratios under design condition = V0 / n0 d
0β = design blade angle; this angle is fixed
0η = efficiency of propeller under design conditionThe steps, to obtain the THP at different flight speeds, are as follows.
1. Obtain from propeller charts, CT and CP corresponding to J0 and 0β .
These values are denoted by CTO and CPO.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
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2. Choose values of J from 0 to a suitable value at regular intervals. Obtain
from the relevant propeller charts, the values of CT and CP at these values
of J’s and the constant value of 0β .
3. Calculate J/J0, CT/CP and CP0/CP from values obtained in step 2.
4. Calculate:
T0 = 0 0 0η BHP ×1000/ V and (4.22)
0 0 PO TOK = T C /C (4.23)
5. The assumption of constant torque (Q0) gives that N and P are related.
Note: 0 0 0Q = P / 2πn
This yields:
PO
0 P
CN =N C
(4.24)
0
0 0
J NV = V × ×
J N (4.25)
and P0 T T0 0
T0 P P
C C CT = T = K
C C C (4.26)
Consequently, THP = TV/1000 and BHP = THP/ pη
The procedure is illustrated with the help of example 4.5.Example 4.5
Obtain the thrust and the thrust horse power at sea level for V upto 60 m/s
for the propeller engine combination of example 4.4
Solution:
From example 4.4 it is noted that the propeller is designed to absorb
97.9 kW at 2500 rpm at V = 59 m/s.The propeller diameter is 1.88 m and β = 20o.
Hence, V0 = 59 m/s, N0 = 2500, n0 = 41.67, 0β = 20o
BHP0 = 97.9 kW, 0η = 0.83
00
0
V 59J = = = 0.753
n d 41.67×1.88
From Fig.4.5d, CTO = 0.046
From Fig.4.5b, CPO = 0.041
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
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Hence, CTO/CPO = 0.046/0.041 = 1.122
0
97.9×1000×0.83T = = 1377.24N
59
PO
0TO
C 0.041K = T = 1377.24× = 1227.54
C 0.046
The remaining calculations are presented in Table E 4.5
J J/J0 CT
*
PC
$
T
P
C
C
CP0/CP
N/N0
£
V
€
T
(N)
€€
N
#
pη
**
THP
$$
BHP
££
0 0 0.104 0.066 1.576 0.621 0.788 0 1927 1971 0 0 -
0.1 0.133 0.104 0.065 1.589 0.629 0.793 6.21 1951 1983 0.17 12.15 71.23
0.2 0.266 0.104 0.065 1.606 0.636 0.792 12.49 1971 1993 0.33 24.61 74.60
0.3 0.398 0.102 0.062 1.631 0.657 0.811 19.05 2002 2027 0.49 38.14 77.83
0.4 0.531 0.093 0.060 1.545 0.683 0.827 25.91 1897 2067 0.62 49.15 79.28
0.5 0.664 0.082 0.058 1.420 0.712 0.844 33.05 1743 2109 0.70 57.61 82.29
0.6 0.797 0.070 0.059 1.306 0.765 0.875 41.12 1603 2187 0.77 65.91 85.60
0.7 0.930 0.055 0.046 1.185 0.884 0.900 51.55 1455 2350 0.81 75.00 92.60
0.8 1.062 0.040 0.036 1.099 1.126 1.061 66.50 1349 2653 0.83 89.71 108.1
*From Fig.4.5d ; $ From Fig.4.5b; £ From Eq.(4.24); € From Eq.(4.25);
€€ From Eq.(4.26); # N = (N/N0)x N0 ; ** From Fig.4.5a; $$ THP = TV/1000 ;
££ BHP = THP/ pη
Table E4.5 Thrust and power output of an engine-propeller combination with
fixed pitch propeller
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
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The results are shown in Figs.E4.5a and b
Fig.E 4.5 Variations of thrust (T) and thrust horse power (THP) with velocity(V)
(a) T vs V (b) THP vs V
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
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Answers :
The variations of T and THP with V are given in table below.
V(m/s) 0 6.21 12.49 19.05 25.91 33.05 41.12 51.55 66.50
T (N) 1927 1951 1971 2002 1897 1743 1603 1455 1349
THP(kW) 0 12.15 24.61 38.14 49.15 57.61 65.91 75.00 89.71
BHP(kW) - 71.23 74.10 77.83 79.28 82.29 85.60 92.60 108.1
N (RPM) 1971 1983 1993 2027 2067 2109 2187 2350 2653
4.2.11 Variations of THP and BSFC with flight velocity and altitude
As mentioned earlier,THP equals pη × BHP . Thus, the variations of THP with V
and h depends on variations of pη and BHP with V and h. In this context, the
following may be recalled.
(i)At a given altitude and RPM, the engine output (BHP) is almost constant with
flight velocity. (ii) BHP decreases with altitude as given by Eqs.(4.1) or (4.1a).
(iii) The propeller efficiency pη depends on BHP, h, V, n and β . For a variable
pitch propeller pη remains nearly constant over a wide range of flight speeds.
Thus for an airplane with variable pitch propeller, the THP vs V curve for a
chosen RPM and h remains flat over a wide range of flight speeds. A typical
variations of THP with V, at chosen ‘RPM(N)’ and with ‘h’ as parameter are
shown in Fig.4.6.
From the engine charts the fuel flow rate and BSFC are known at chosen
MAP & N. From these values the BSFC at the chosen MAP & N, can be
calculated using Eq.(4.1d) . See section 6 of Appendix A for typical calculations.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
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Fig.4.6 Schematic variation of THP with flight speed for an engine-propeller
combination with variable pitch propeller
4.2.12 Loss of propeller efficiency at high speeds
As noted earlier, the propeller blade is like a rotating wing with forward
motion. The resultant velocity at the propeller tip (VRtip
) would be the highest. It is
equal to:
VRtip
= { V2 + (2 π n R)2}1/2, where R is the radius of the propeller.
When the Mach number corresponding to VRtip
exceeds the critical Mach number
for the airfoil used on the propeller, the drag coefficient of the airfoil would
increase and the lift coefficient would decrease (see subsection 3.3.3).
Consequently, the efficiency of the propeller would decrease. This loss ofefficiency can be delayed to higher flight Mach numbers by use of advanced
propellers. These propellers have swept blades and are being used on turboprop
airplanes up to flight Mach number of 0.7. Figure 4.7a shows one such propeller
placed in a wind tunnel and Fig.4.7b shows another propeller mounted on
ATR 72 airplane.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
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Fig.4.7a Advanced propeller being tested in a Wind tunnel
(Adapted from Ref 4.4)
Fig.4.7b Advanced propeller mounted on ATR72 airplane
(Source : www.fspilotshop.com)
Fig.4.7 Features of an advanced propeller
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
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4.3 Gas Turbine Engines
A gas turbine engine consists of a diffuser to decelerate the air stream
entering the engine, a compressor, a combustion chamber, a turbine and a
nozzle (Fig.4.8a). In some turbojet engines, an afterburner is incorporated
between the exit of the turbine and the entry of the nozzle (Fig.4.8b).The hot
gases leaving the combustion chamber expand partly in the turbine and partly in
the nozzle. The need for three variants of gas turbine engines viz. turboprop,
turbofan & turbojet can be explained by considering their propulsive efficiencies.
Fig.4.8 Turbojet engine
(Source : http://www.aerospaceweb.org)
4.3.1 Propuls ive efficiency
Propulsive efficiency is the ratio of useful work done by the air stream and
the energy supplied to it.
In a gas turbine engine, the velocity of the air stream ( V ) is augmented
to V j,the velocity of the jet stream, thereby imparting kinetic energy at the rate of :
(m /2) [ V j 2 - V
2] (4.27)
where m is the mass flow rate.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
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The engine develops a thrust T and hence results in a useful work of T V .
Noting that:
T = m (V j - V ), (4.28)
the propulsive efficiency (ηpropulsive) is:m(V -V )(V ) 2 j
η = =propulsive m V2 2 j(V -V ) 1+ j2 V
(4.29)
4.3.2 Why turboprop, turbofan and turbojet engines?
The overall efficiency of a gas turbine engine is the product of items like
cycle efficiency, combustion efficiency, mechanical efficiency and propulsive
efficiency. The cycle efficiency depends on the engine cycle and in turn on the
maximum temperature / pressure in the engine. The combustion efficiency and
mechanical efficiency are generally of the order of 95%. Thus propulsive
efficiency finally decides the overall efficiency of a gas turbine engine as a
propulsive system.
Remark:
The action of a propeller is also similar to that of a jet engine i.e. it also enhances
velocity of the free stream from V to V j, In this case, V j is the velocity of the
stream far behind the propeller(see subsection 4.2.7). Hence, the propulsive
efficiency of a propeller which was called ideal efficiency of propeller, is also
given by Eq.(4.29), which is same as given by Eq.(4.13).
The variation of propulsive efficiency with flight speed provides the reason for
use of turboprop, turbofan and turbojet engines in airplanes operating at different
range of flight speeds.Consider the variation of propulsive efficiency with flight
speed. For this purpose, a subsonic jet engine with convergent nozzle is
considered. In this case, the Mach number at the exit, would be unity and thetemperature of the exhaust gases would be around 600 K. Under these
conditions, V j, the velocity of jet exhaust would be around 500 m/s. Using
Eq.(4.29), the values of propulsive efficiency obtained at different flight speeds
( V ) are given in the Table 4.2.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
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V (m/s) 100 125 166.7 250 333.3 400
V j / V 5 4 3 2 1.5 1.25
ηρ % 33.3 40.0 50.0 66.7 80.0 88.9
Table 4.2 Variation of propulsive efficiency with flight speed for V j = 500 m/s
Remarks:
i) Turboprop engine
It is seen from Table 4.2 that ηp will be low if a pure jet engine is used at
low speeds. An analysis of Eqs.(4.28 and 4.29) points out that for havingadequate thrust and high propulsive efficiency at low flight speeds, a small
increment in velocity should be given to a large mass of air. This is effectively
done by a propeller. Thus for airplanes with flight Mach number less than about
0.5, a turbo-prop engine is used (Fig.4.9). In this case, the turbine drives the
compressor and also the propeller through a gearbox (Fig.4.9). The gear box is
needed because the turbine r.p.m. would be around 15000-20000 whereas, the
propeller rotates at about 3000 r.p.m.
For practical reasons, the expansion of the gases coming out of the combustion
chamber is not allowed to take place completely in the turbine and a part of the
expansion is carried out in the nozzle. Hence, in a turboprop engine, about 80 to
90% of the total output is produced through the propeller and the rest 20 to 10%
as output from the jet coming out of the nozzle.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
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Fig.4.9 Turboprop engine
(Source: www.aircraftenginedesign.com)
ii) Turbofan engine
As the flight Mach number increases beyond 0.7, the propeller efficiency
decreases rapidly due to the formation of shock waves at the tip of the propeller
blade. Hence, for airplanes flying near Mach number of unity, a turbo-fan engine
is used (Fig.4.10).In this engine a major portion of the power output (about 60%)
is obtained as jet thrust and the rest as thrust from the fan. A fan has a smaller
diameter as compared to the propeller and it is generally placed inside a duct. A
ducted fan has a higher propulsive efficiency than a propeller.
It is observed in Fig.4.10 that all the air taken in by the fan does not go
through the turbine. Incidentally the part of the engine consisting of the
compressor, combustion chamber, turbine and nozzle is called ‘Gas generator’.
The ratio of the mass of the air that passes through the fan to the mass of air that
passes through the gas generator is called ‘Bypass ratio’.
Early turbofan engines had bypass ratio of 1:1. At present, it is around 6.5:1 and
is likely to increase in future.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
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Fig.4.10 Turbofan engine
(Source : http://www.aerospaceweb.org)
iii) Turbojet engine
At supersonic Mach numbers, up to three, a turbo-jet engine is used. In this
engine entire power output is through the jet thrust.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
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Chapter 4
Lecture 16
Engine characteristics – 4
Topics
4.3.3 Characteristics of a typical turboprop engine
4.3.4 Characteristics of a typical turbofan engine
4.3.5 Characteristics of a typical turbojet engines
4.4 Deducing output and SFC of engines where these characteristics are
not available directly
4.5 A note on choice of engines for different range of flight speeds
4.3.3 Characteristics of a turboprop engine
As noted earlier, in this engine, a major portion of the output is available at
the propeller shaft (SHP) and a small fraction through the jet thrust (T j). Hence,
the output is represented as:
THP = ηp SHP + (T j V /1000) (4.30)
where SHP = shaft horse power available at propeller shaft in kW,ηp = propeller
efficiency and T j = jet thrust
The total output of a turbo-prop engine, also called ‘Equivalent shaft horse
power (ESHP)’, is defined as :
ESHP = SHP + {T j V / (0.8x1000) } (4.31)
Note : (i) For the purpose of defining ESHP, the value of ηρ is taken as 0.8 in
Eq.(4.31). The ESHP and SHP are in kW.
(ii) Equation (4.31) would not be able to account for the contribution, to ESHP, of
the thrust produced when the flight velocity (V) is zero or the static condition. For
this case and when V < 100 knots (or 185 kmph), the convention is to define
ESHP as follows (Ref.4.3, chapter 14).
ESHP = SHP + (T j / 14.92) (4.31a)
where ESHP and SHP are in kW and T j is in N.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
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For example a turboprop engine developing SHP of 746 kW and jet thrust of
503 N, under sea level static condition, would have :
ESHP = 746 + (503/14.92) = 780 kW.
Characteristics of a typical turbo-prop engine are shown in Fig.4.11. It is
observed that the power output increases with flight speed. This increase is due
to two factors viz. (a) the mass flow through the engine ( ;i im = ρ A V Ai and Vi
being the area of intake, and the velocity at the intake) increases with flight
speed and (b) the pressure rise due to the deceleration of the flow in the inlet
diffuser also increases with flight Mach number.
Figure 4.11 also shows the influence of ambient temperature on power output. It
is observed that there is a significant fall in ESHP as the ambient temperature
rises.
From the curves regarding fuel flow rate in Fig.4.11, the BSFC can be obtained
at various speeds and altitudes as:
BSFC = (Fuel flow/hr) / ESHP
Remark:
Reference 3.9 Appendix E.3 gives performance curves for a large turboprop
engine with sea level static power of 6500 HP. It may be noted that the ‘Sea levelstatic power’ is the engine output at sea level at zero velocity. Reference 1.9,
chapter 6 gives characteristics of an engine of around 1700 HP.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
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Fig.4.11 Characteristics of PT6A-25 turboprop engine
(Adapted from Brochure of Pratt & Whitney Canada Corp. 1000, Marie-Victorin,
Longueuil Quebec J4G 1A1, Canada © Pratt & Whitney Canada Corp.
Reproduced with permission)
4.3.4 Characterisitcs of typical turbofan engine
In the early turbofan engines the thrust output used to remain fairly
constant with flight speed. In the modern turbofan engines the performance at
low speeds and low altitudes (up to about 5 km) has been improved so that the
ratio of the sea level static thrust and that (thrust) in high speed-high altitudeflight is much higher than the early turbofan engines. The ‘Sea level static thrust’
is the engine output at M=0 at sea level. Higher sea level static thrust helps in
reducing the distance required for take-off. Figure 4.12 shows the variations of
thrust with Mach number at different altitudes for an engine with bypass ratio of
4.9. The figure also shows the values of the specific fuel consumption (TSFC).
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
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Remark:
Chapter 9 of Ref.3.22 gives the performance, in terms of non-dimensional
parameters, for engines with bypass ratios of 3, 6.5, 8 and 13. The curves are
also presented for take-off rating, climb rating and cruise rating. It may be added
that the ‘Take-off rating’ is the engine output which can be availed for about 5
min. The engine can be run at ‘Climb rating’ for about half an hour and at ‘Cruise
rating’ for long periods.
Fig.4.12 Characteristics of Pratt and Whitney PW4056 turbofan engine -
maximum cruise thrust
(With permission from Pratt and Whitney, East Hartford)4.3.5 Characterisitcs of typical turbojet engine
The characteristics of a supersonic turbojet engine are shown in Figs.4.13a to d.
It is observed that at subsonic speeds the thrust is fairly constant, but it increases
considerably at supersonic speeds. This rise is due to increased ram pressure
in the intake, as a result of the deceleration of the supersonic flow. The Mach
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
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number at which the peak value of thrust occurs depends on the design of the
engine.
Fig.4.13a Characteristics of Pratt and Whitney
JT4A-3 turbojet engine (estimated thrust, TSFC, and airflow) under standard
atmospheric condition and 100% RAM recovery. h = sea level
(With permission from Pratt and Whitney, East Hartford)
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
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Fig.4.13b Characteristics of engine in Fig.4.13a, h = 15000 ft
(With permission from Pratt and Whitney, East Hartford)
Fig.4.13c Characteristics of engine in Fig.4.13a, h = 30000 ft
(With permission from Pratt and Whitney, East Hartford)
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
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Fig.4.13d Characteristics of engine in Fig.4.13a, h = 45000 ft
(With permission from Pratt and Whitney, East Hartford)
Remarks:
i)In Fig.4.13a to d the true airspeed is given in knots;one knot is equal to
1.852 kmph. Further, the speed of sound at h = 0, 15000’ , 30000’ and 45000’ is
respectively 661, 627, 589 and 574 knots.
ii) Bypass supersonic turbofan engines are also being considered for supersonic
flight. Reference 3.9, gives, in Appendix E, typical curves for an engine with sea
level static thrust of 30000 lb (133 kN). Similarly Ref.4.5, chapter 8 also presents
curves for an engine with 33000 lb (146.3 kN) sea level static thrust. Figures 4.13
a to d also indicate the values of specific fuel consumption (TSFC) and the air
flow rate.iii) Figure 4.8b shows an afterburner duct between the turbine exit and the entry
of the nozzle. The same figure also shows the fuel spray bars and the flame
holder. An afterburner is used to increase the thrust output for a short duration.
When the fuel is burnt in the afterburner, the temperature of the gases goes up
and the thrust increases when these gases subsequently expand in the nozzle.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
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However, the specific fuel consumption also goes up considerably and the
afterburner operation is resorted to only for a short duration like during take-off or
transonic acceleration.
4.4 Deducing output and SFC of engines where these characteristics are
not available directly
The detailed information about engine performance (i.e. variations with
altitude and flight velocity of the thrust (or power) and TSFC (or BSFC) is
generally available only in a limited number of cases. To get the performance of
an engine with other rating, scaling of the available data is carried out. For this
purpose, the values of thrust(or power) of the engine, whose characteristics are
known, are multiplied by a suitable factor which will bring the output of the
existing engine equal to the output of the desired engine. It is assumed that the
SFC values will be the same for the two engines. This kind of scaling is generally
applicable for outputs within ± 25% of the output of the known engine (Ref.4.5,
chapter 8).
4.5 A note on choice of engines for different range of flight speeds
The topic of choice of engine for different types of airplanes is generally covered
in airplane design. Here some salient points are mentioned to conclude the
discussion on engines.
The following five criteria are used to select a power plant for a specific
application.
1.Overall efficiency 0η : This quantity is the product of (a) thermodynamic
cycle efficiency tη (b) Combustion efficiency cη (c) mechanical efficiency
mη and (d) propulsive efficiency pη . The thermodynamic efficiency depends
on the thermodynamic cycle on which the engine operates. The details regarding
estimation of tη are available in books on thermodynamics. However, it is of the
order of 40 to 50%. The combustion efficiency and mechanical efficiency would
be around 95%. The propulsive efficiency of the propeller and gas turbine
engines have been described in subsections 4.2.7, 4.2.8 and 4.3.2. It has been
pointed out there that pη depends on flight speed or Mach number.
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 9
The specific fuel consumption (SFC) is an indication of the overall efficiency.
Based on Ref.3.9 chapter 3, it can be mentioned that the piston engine-propeller
combination would have lowest SFC for Mach number (M) upto about 0.3. The
turboprop engine would have lowest SFC in the range of Mach number from 0.3
to 0.6 which may extend to M 0.7 with the use of a transonic propeller. The
high bypass ratio turbofans have lowest SFC between for M 0.7 to 1.0 and the
low by-pass ratio ones between M 1 to1.6 . Turbojets are more suited for
1M .6 to about 3.5 and ramjets later upto M 8 . It may be recapitulated that a
ramjet engine requires another powerplant to bring it to Mach number of about
1.5.
2. Variation of thrust (or power) with flight speed and altitude:
The shaft horse powers of piston engine and turboprop engine do not change
significantly with flight speed. Consequently, the thrust outputs of these engines
decrease significantly with flight speed or Mach number. The output of a turbofan
engine decreases with Mach number, especially at low altitudes (Fig.4.12). The
thrust of a jet engine is fairly constant at subsonic speeds but increases
considerably at supersonic speeds (Fig.4.13 c & d). As regards the effect of flight
altitude Eq.(4.1a) shows that for a piston engine 1.1
slP/P = σ where σ is the
density ratio and the suffix ‘sl’ denotes a quantity at sea level.
For a turbo-prop engine (from Ref 1.10 chapter 3), 0.7
slP/P σ . From
Ref.3.15, chapter 3, (T/Tsl) for turbofan and turbojet engines is also roughly
proportional to 0.7σ
3. Weight of the engine:
The weight of the engine contributes to the gross weight of the airplane and
hence it should be as low as possible.This quantity is indicated by the ratio Wpp
/T
or Wpp/BHP, where WPP is the weight of the power plant. This ratio depends on
the type of engine and the engine rating; it (ratio) decreases as the rating
increases. Based on data in Ref.1.15, it can be mentioned that the weight per
unit BHP for a piston engine is around 9N/kW for an engine with a rating of
around 150 kW and about 6N/kW for a rating of around 500 kW. For a turboprop
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 10
engine WPP/ESHP is around 2.9 N/kW for rating of 500 kW, 2.3 N/kW for a rating
of 2500 kW and 1.4 N/kW for a rating of 7500 kW. For a turbofan engines the
ratio WPP/T could be around 0.25 N/N for a rating of around 100 kN and about
0.15 N/N for a rating of about 250 kN.
4. Frontal area:
The frontal area of an engine contributes to the parasite drag of the airplane.
Hence, a lower frontal area is a desirable feature of the engine. For a given
output the piston engine-propeller combination generally has the highest frontal
area. Turboprop, turbofan and turbojet follow in the decreasing order of the
frontal area.
5.Other considerations :
Gas turbine engines have mechanical simplicity as compared to a piston
engine. However, gas turbine engines are costlier than the piston engines as
some of the components of the gas turbine engines operate at higher
temperature and RPM. This requires special materials and fabrication
techniques.
Keeping these factors in view the different types of engine are used in
the speed range/application as given in Table 4.3
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 11
Type of engine Speed / Mach number
range
Application – airplanes
in the following
categories
Piston engine-propeller
combination Upto 300 kmph
General aviation, trainer,
agricultural and sports.
Turboprop 250 to 600 kmph;
upto 750 kmph with
advanced propeller
Short and medium range
transport/cargo, aerial
survey, feeder liner and
executive transport.
Turbofan M from 0.7 to 1.0
Medium and long range
transports, cargo,
maritime patrol, executive
transport, jet trainer.
Turbojet M from 1 to 3
Trainers, supersonic
transport, fighter,
interceptor, bomber.
Ramjet M from 2 to 8
Intended for hypersonic
transport.
Table 4.3 Speed range and applications of different types of engines
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 1
Chapter 4
References
4.1 Hartman, E.P. and Biermann, D. “The Aerodynamic characteristics of full-
scale propellers having 2, 3, and 4 blades of clark Y and R.A.F. 6 airfoil sections”
NACA TR 640, Nov.1937. This report can be downloaded from the site “NASA
Technical Report Server (NTRS)”.
4.2 “Generalized method for propeller performance estimation” Hamilton
Standard Division, Hamilton Standard Publication PDB6101A, United Aircraft
Corp., 1963.
4.3 Nicholai, L.M. and Carichner, G.E “Fundamentals of aircraft and airship
design – Vol I – Aircraft design” AIAA educational series (2010).
4.4 Mikkelson D.C. and Mitchell G.A. “High speed turboprop for executive aircraft
– potential and recent test results“ NASA TM 31482, Jan 1980. This report can
be downloaded from the site “NASA Technical Report Server (NTRS)“.
4.5 Jenkinson L.R., Marchman III J.F. “Aircraft design projects” Butterworth-
Heinemann (2003).
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Flight dynamics-I Prof. E.G.TulapurkaraChapter IV
Indian Institute of Technology, Madras 1
Chapter 4
Exercises
4.1) What are the different types of engines used on airplanes? State thespeed/Mach number range in which they are used.
4.2) Sketch a typical BHP vrs altitude curve for a piston engine. Why does the
power output of a piston engine decrease rapidly with altitude? Supercharger is
needed to delay this loss of power to higher altitudes. Look for information on
supercharger from books (e.g. Ref.1.9) and internet (www.google.com).
4.3) What are the essential differences between turboprop, turbofan and turbojet
engines? Derive an expression for the propulsive efficiency and justify the range
of flight Mach numbers in which these engines are used.
4.4) A propeller of 2 m diameter is mounted on an airplane flying at a speed of
216 kmph. If the velocity of air far behind the propeller be 81 m/s, calculate the
propulsive efficiency and the thrust developed by the propeller.
[Answers: ηp = 85.1%, T = 5695 N]
4.5) Neatly sketch the following:
(a) variation of propeller efficiency vs flight velocity with propeller pitch angle as
parameter.
(b) Variation of thrust vs. Mach number with altitude as parameter for a
turbofan engine.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 5
Lecture 19
Performance analysis I – Steady level flight – 3
Topics
5.8 Influence of level flight analysis on airplane design
5.9 Steady level flight performance with a given engine
5.10 Steady level flight performance with a given engine and parabol ic
polar
5.10.1 Airplane with jet engine
5.8 Influence of level flight analysis on airplane design
The significant manner in which the performance analysis helped in
evolution of the airplane configuration can be appreciated from the following
discussion.
(a) The low speed airplanes are powered by engines delivering BHP or ESHP. In
this case, the major portion of the power required is induced power, which
depends on the factor K in drag polar (Eq.5.10). This factor is given as 1 / A e
where A is the aspect ratio of the wing and ‘e’ is the Oswald’s efficiency factor
(see Eq.3.46). Hence the low speed airplanes and gliders have high aspect ratio
wings. It may be added that personal airplanes have aspect ratio between 6 to 8
as hanger space is also an important consideration. However, medium speed
commercial airplanes have aspect ratio between 10 to 12. Gliders have aspect
ratio as high as 16 to 20.
(b) For high subsonic airplanes most of the drag is parasite drag which depends
on DOC (see Eq.5.12). Hence, high speed airplanes have features like smooth
surfaces, thin wings, streamlined fuselage, smooth fairings at wing-fuselage joint
and retractable landing gear. These features reduce DOC . Manufacturing
techniques have also been improved to achieve smooth surface finish. High
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
speed airplanes also have high wing loading (W/S) to reduce the wing area.
Table 3.4 may be referred to for typical values of DOC , A and e of different types
of airplanes. The reciprocal of (CD / CL
) is (CL / C
D). It is called lift-drag ratio
(L / D). The maximum value of this ratio, (L / D)max, is an indication of the
aerodynamic efficiency of the airplane. (L / D)max
lies between 12 to 22 for a
subsonic airplanes and between 5 to 8 for supersonic airplanes.
(c) When the weight of an airplane increases the thrust required increases in
proportion to W and the power required increases in proportion to W3/2
(Eqs.5.3
and 5.4). Hence, airplane design bureaus have a group of engineers which
keeps a close watch on any increase in the weight of the airplane.
5.9 Steady level flight performance with a given engine
At the outset the following three points may be noted.
(I)In steady level flight the thrust must be equal to drag (Eq.5.1).
(II) The thrust is provided by the engine or the engine-propeller combination and
from chapter 4, it is noted that the thrust or power output varies with engine RPM,
flight speed and altitude.
(III) For airplanes with piston engine or turboprop engine, the output is the power
available at the engine shaft. Hence, to estimate the performance of such
airplanes the calculations are carried-out in terms of BHP or THP. For airplanes
with turbofan or turbojet engines, the output is in terms of thrust and to estimate
the performance of such airplanes the calculations are carried-out in terms of
thrust.
Typical variations, with altitude and flight speed, of the maximum thrust available
(Ta) and the maximum thrust horse power available (THP)a are shown in
Figs.5.5 and 5.6a respectively. The thrust required and power required curves
are also shown in same figures.
Consider the curves of Ta and Tr corresponding to sea level conditions. It is seen
that the power or thrust available is much more than the minimum power or thrust
required. Hence, flights over a wide range of speeds are possible by controlling
the engine output with the help of throttle and ensuring thrust equals drag.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
However, as the speed increases above the speed for minimum power or thrust
( Vmp or Vmd), the power or thrust required increases and at a certain speed the
power or thrust required is equal to the maximum available engine output (point
A in Figs.5.5 & 5.6a). This speed is called the ‘Maximum speed(Vmax)’. Similar
intersections between power available and power required curves or thrust
available and thrust required curves are seen at higher altitudes (points B, C and
D in Fig.5.5, point B in Fig.5.6a and point C in Fig.5.6b).
Similarly, when the flight speed decreases below Vmp or Vmd the power or
thrust required increases and there is a speed at which the power or thrust
required is equal to the available power or thrust - point D’ in Fig.5.5 and point C’
in Fig.5.6b. Figure 5.6b is drawn separately from Fig.5.6a to show the points Cand C’ clearly.
Thus, the minimum speed can be limited by available thrust or power output. It is
denoted by (Vmin
)e. However, in level flight the requirement of lift equal to weight
should also be satisfied(Eq.5.1). Hence, level flight is not possible below stalling
speed. Thus, two factors viz. the thrust or power available and the stalling, limit
the minimum flight speed of an airplane. Satisfying both these requirements, the
minimum speed of the airplane at an altitude will be the higher of the two speeds
viz. (Vmin
)e and V
S.
Typical variations of Vmax
, (Vmin
)e and V
S are shown for a jet engined airplane in
Fig.5.9. The details of the calculations are given in Appendix B. Similarly, typical
variations of these speeds in case of a piston engined airplane are shown in
Fig.5.10 with details of calculation given in Appendix A. The following
observations are made.
(i)For a jet airplane Vmax
may slightly increase initially with altitude and then
decrease. However, there is an altitude at which the thrust required curve is
tangential to the thrust available curve and flight is possible only at one speed.
This altitude is called ‘Ceiling’ and denoted by hmax
. Above hmax
the thrust
available is lower than the minimum thrust required and level flight is not possible
as the requirement of T = D cannot be satisfied.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 4
Fig.5.9 Vmax and Vmin for jet airplane
Fig.5.10 Vmax
, (Vmin
)e and Vs for airplane with engine-propeller combination
(ii)The minimum speed of a jet airplane is the stalling speed (Vs) at low altitudes.
However, near the ceiling, the minimum speed is that limited by the thrust
available i.e. (Vmin
)e.
0
1000
2000
3000
4000
5000
6000
0 10 20 30 40 50 60 70 80
Velocity (m/s)
Vs
Vmax
(Vmin)e
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
(iii)In the case of a piston engined airplane, the maximum speed seems to
decrease with altitude. In this case also there is a ceiling altitude beyond which
the power available is lower than the minimum power required and hence level
flight is not possible. The ceiling in this case, is lower than in the case of a jet
airplane because the power output of a piston engine decreases rapidly with
altitude. As regards the minimum speed, it is also limited by stalling at low
altitudes and by power available near the ceiling altitude.
5.10 Steady level flight wi th a given engine and parabolic polar
If the drag polar is parabolic and the engine output can be assumed to be
constant with speed, then Vmax
and (Vmin
)e from the engine output consideration,
can be calculated analytically. i.e. by solving an equation. It may be noted from
Figs.5.5 & 5.6 that the assumption of Ta or Pa as constant with V appears
reasonable near the speeds where Vmax occurs.
5.10.1 Airplane with jet engine:
The steps to calculate Vmax
and (Vmin
)e are as follows.
(1) Choose an altitude ‘h’. Let Ta be the thrust available in the range of speeds
where Vmax is likely to occur.
(2) Tr =Ta= W(CD / CL)
Hence,
DOT CCa D= = +KCLW C CL L
Or DOT2 aKC - C + C = 0LL W
(5.25)
Equation (5.25) is a quadratic in CL. Its solution gives two values of C
L at which
level flight with the given thrust is possible. Let these values of CL be denoted as
CL1
and CL2
. Then, the corresponding flight speeds, V1 and V
2, are given as:
1 12 22W 2W
V = and V =1 2ρSC ρSCL1 L2
(5.26)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
It may be pointed out that the same results can be obtained by using Eq.(5.12),
i.e.
1
2DO
22W2T = T = ρV S C + Ka r 2ρV S
Or AV4
– BV2
+ C = 0 (5.27)
where,1
2DO
22KW A = ρSC , B = T and C =a
ρS
For given value of thrust (Ta), Eq.(5.27) also gives two solutions for level flight
speeds V1 and V
2.
Let V1 be the higher among V
1 and V
2.Then, V
1 is the maximum speed and V2 is
the minimum speed, based on engine output i.e. (Vmin
)e. The higher of (V
min)e
and the stalling speed (Vs) will be the minimum speed at the chosen altitude.
The example 5.2 illustrates the procedure.
Remarks:
i) Calculate the Mach number corresponding to V1. If it is more than the critical
Mach number then DOC and K would need correction and revised calculation,
would be required.
ii) Obtain, from the engine charts, the thrust available at V1 . Let it be denoted by
Ta1. If the thrust available (Ta), assumed at the start of the calculation(step 1), is
significantly different from Ta1, then the calculations would have to be revised
with new value of Ta. However, it is expected that the calculations would
converge to the correct answer in a few iterations.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 5
Performance analysis I – Steady level fl ight
(Lectures 17 to 20)
Keywords: Steady level flight – equations of motion, minimum power required,
minimum thrust required, minimum speed, maximum speed; stalling speed;
equivalent airspeed.
Topics
5.1 Introduction
5.1.1 Subdivisions of performance analysis
5.1.2 Importance of performance analysis
5.1.3 Approach in performance analysis5.2 Equations of motion for s teady level flight
5.3 Stalling speed
5.4 Equivalent airspeed
5.4.1 Airspeed indicator
5.5 Thrust and power required in steady level flight – general case
5.6 Thrust and power required in s teady level flight when drag
polar is independent of Mach number
5.7 Thrust and power required in s teady level flight – consideration of
parabolic polar
5.8 Influence of level flight analysis on airplane design
5.9 Steady level flight performance with a given engine
5.10 Steady level flight performance with a given engine and parabol ic
polar
5.10.1 Airplane with jet engine
5.10.2 Parameters influencing Vmax of a jet airplane
5.10.3 Airplane with engine-propeller combination
5.11 Special feature of steady level flight at supersonic speeds
References
Exercises
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
Chapter 5
Lecture 17
Performance analysis I – Steady level flight – 1
Topics
5.1 Introduction
5.1.1 Subdivisions of performance analysis
5.1.2 Importance of performance analysis
5.1.3 Approach in performance analysis
5.2 Equations of motion for s teady level flight
5.3 Stalling speed5.4 Equivalent airspeed
5.4.1 Airspeed indicator
5.5 Thrust and power required in steady level flight – general case
5.1 Introduct ion:
During its normal operation an airplane takes –off, climbs to the cruising
altitude, cruises at almost constant altitude, descends and lands. It may also fly
along curved paths like turns, loops etc. The flights along curved paths are also
called manoeuvres. Analyses of various flights are the topics under the
performance analysis. A revision of section 1.6 would be helpful at this stage.
5.1.1 Subdivisions of performance analysis
Performance analysis covers the following aspects.
I) Unaccelerated flights:
(a) In a steady level flight an airplane moves with constant velocity at a constant
altitude. This analysis would give information on the maximum level speed and
minimum level speed at different altitudes.
(b) In a steady climb an airplane climbs at constant velocity. This analysis would
provide information on the maximum rate of climb, maximum angle of climb and
maximum attainable altitude (ceiling).
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
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(c) In a steady descent an airplane descends with constant velocity. A glide is a
descent with zero thrust. This analysis would give the minimum rate of sink and
time to descend from an altitude.
(d) Range is the horizontal distance covered, with respect to a given point on the
ground, with a given amount of fuel. Endurance is the time for which an airplane
can remain in air with a given amount of fuel.
II) Accelerated flights:
(a) In an accelerated level flight an airplane moves along a straight line at
constant altitude and undergoes change in flight speed. This analysis provides
information about the time required and distance covered during acceleration
over a specified velocity range.
(b) In an accelerated climb, an airplane climbs along a straight line accompanied
by a change in flight speed. This analysis gives information about the change in
the rate of climb in an accelerated flight as compared to that in a steady climb.
(c) Loop is a flight along a curved path in a vertical plane whereas a turn is a
flight along a curved path in a horizontal plane. This analysis would give
information about the maximum rate of turn and minimum radius of turn. These
items indicate the maneuverability of an airplane.
(d) During a take-off flight an airplane starts from rest and attains a specified
height above the ground.This analysis would give information about the take-off
distance required.
(e) During a landing operation the airplane descends from a specified height
above the airport, lands and comes to rest. This analysis would provide
information about the distance required for landing.
5.1.2 Impor tance of performance analysis
The performance analysis is important to asses the capabilities of an
airplane as indicated in the previous subsection. Moreover, from the point of
view of an airplane designer, this analysis would give the thrust or power
required, maximum lift coefficient required etc. to achieve a desired performance.
This analysis would also point out the new developments required, in airplane
aerodynamics and engine performance, to achieve better airplane performance.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
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5.1.3 Approach in performance analysis
As mentioned in subsection 1.1.3 the approach here is to apply the
Newton’s laws and arrive at the equations of motion. The analysis of these
equations would give the performance.
Remarks:
i) References 1.1, 1.5 to 1.13 may be referred to supplement the analysis
described in this and the subsequent five chapters.
ii) It would be helpful to recapitulate the following points.
(a) A ‘Flight path’ is the line along which the centre of gravity (c.g.) of the airplane
moves. Tangent to the flight path gives the direction of the ‘Flight velocity’ (see
Fig.5.1).
Fig.5.1 Flight path
(b) The external forces acting on a rigid airplane are:
(I) Aerodynamic forces (lift and drag)
(II) Gravitational force
(III) Propulsive force (thrust)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
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(c) The forces produced due to control deflection, needed to balance the
moments, are assumed to be small as compared to the other forces. With this
assumption all the forces acting on the airplane are located at the centre of
gravity (c.g.) of the airplane (Fig.5.2) and its motion is simplified to that of a point
mass moving under the influence of aerodynamic, propulsive and gravitational
forces.
Fig.5.2 Steady level flight
5.2 Equations of motion for s teady level flight
In this flight the c.g. of the airplane moves along a straight line at a
constant velocity and at a given altitude. The flight path, in this case, is a
horizontal line. The forces acting on the airplane are shown in Fig.5.2. ‘T’ is
Thrust, ‘D’ is Drag, ’L’ is lift and ‘W’ is the weight of the airplane.The equations of
motion are obtained by resolving, along and perpendicular to the flight direction,
the forces acting on the airplane. In the present case, the following equations areobtained.
T - D = m ax
L - W = m az
where, m is the mass of airplane and ax, and az, are the components of the
acceleration along and perpendicular to the flight path respectively.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
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As the flight is steady i.e. no acceleration along the tangent to the flight path,
implies that ax = 0. Further, the flight is straight and at constant altitude, hence,
az= 0.
Consequently, the equations of motion reduce to:
T – D = 0, L – W = 0 (5.1)
Noting that, L = (1/2)ρV2SCL and L = W in level flight, gives :
W = (1/2)ρV2SCL
Or V = (2W / ρSCL)1/2 (5.2)
Further, (1/2)ρV2S = W / CL
Noting that, D = (1/2)ρV2SCD and T = D in level flight, gives
the thrust required (Tr ) as :
Tr = D =(1/2) ρV2SCD
Substituting for (1/2) ρV2S as W / CL, yields:
Tr = W (CD/ CL) (5.3)
The power required (Pr ), in kiloWatts, is given by:
Pr = Tr V/1000 (5.3a)
where Tr is in Newton and V in m/s.
Substituting for V and Tr from Eqs. (5.2) and (5.3) in Eq.(5.3a) yields:
CW 2WDP = × ×r 1000 C ρS CL L
Or3 C1 2W DPr 3/21000 ρ S C
L
(5.4)
Remarks:
i) Equations (5.1) to (5.4) are the basic equations for steady level flight and would
be used in subsequent analysis of this flight.
ii) To fly in a steady level flight at chosen values of h and V, the pilot should
adjust the following settings.
(a) The angle of attack of the airplane to get the desired lift coefficient so that the
lift(L) equals the weight(W).
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
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(b) The throttle setting of the engine, so that thrust equals drag at the desired
angle of attack. He (pilot) will also have to adjust the elevator so that the airplane
is held in equilibrium and the pitching moment about c.g. is zero at the required
angle of attack. As noted earlier, the forces (lift and drag) produced due to the
elevator deflection are neglected.
5.3 Stalling speed:
Consider that an airplane which has weight (W) and wing area (S), is flying at
an altitude (h). From Eq.(5.2) it is observed that, the flight velocity (V) is
proportional to 1/CL
1/2
. Thus, the value of CL required would increase as the flight
speed decreases. Since CL cannot exceed C
Lmax, there is a flight speed below
which level flight is not possible. The flight speed at which CL equals C
Lmax is
called ‘Stalling speed’ and is denoted by Vs. Consequently ,
Vs= (2W / ρSCLmax
)1/2 (5.5)
It is evident from Eq.(5.5) that Vs increases with altitude since the density (ρ)
decreases with height.The variations of Vs with h for a typical piston engined
airplane and a typical jet airplane are presented in Figs.5.3a and b respectively.
Appendices A & B give the details of calculations.
Remark:
The maximum lift coefficient (CLmax
) depends on the flap deflection (δf ). Hence,
Vs will be different for the cases with (a) no flap (b) flap with take-off setting (c)
flap with setting for landing. Figure 5.3a presents the variations of stalling speed,
with altitude, for four cases viz. with no flap and with three different flap settings.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 8
Fig.5.3a Variations of stalling speed with altitude for a low speed airplane
Fig.5.3b Variations of stalling speed with altitude for a jet transport
5.4 Equivalent airspeed
Equivalent airspeed (Ve) is defined by the following equation.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 9
,
2 21 1ρV = ρ Vo e2 2
Noting σ = ρ/ρ V can be expressed as :o e
2W1/2V = Vσ =eρ SCo L
(5.6)
Remarks:
i) From Eq.(5.6) it is evident that for a given wing loading (W/S), the equivalent
airspeed in steady level flight is proportional to 1/CL1/2 and is independent of
altitude. Thus the stalling speed, for a given airplane configuration, when
expressed as equivalent airspeed is independent of altitude.
ii) To avoid confusion between equivalent airspeed (Ve ) and the actual speed of
the airplane relative to the free stream (V), the latter is generally referred to as
true airspeed.
5.4.1 Airspeed indicator
The equivalent airspeed is also significant from the point of view of
measurement of speed of the airplane using Pitot-static system. It may be
recalled from the topics studied in fluid machanics that a Pitot-static tube senses
the Pitot (or total) pressure (pt) and the static pressure ( sp ). The difference
between pt and sp is related to the velocity of the stream ( V ) by the following
equation.
1;
2
2 42
st
M Mp - p = ρV 1+ + +.... M = V /a,
4 40
a = speed of sound (5.6a)
Thus, at low speeds (M < 0.2),
1
2
2stp - p ρV
It may be pointed out that, in the case of an airplane, the air is stationary and the
airplane is moving. Hence, the quantity V in the above expressions, equals the
speed of the airplane(V). Hence, at low speeds:
2 2s et 0
1 1p -p ρV = ρ V
2 2 (5.6b)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 10
In an airplane the Pitot pressure is sensed by a Pitot tube mounted on the
airplane and static pressure is sensed by a hole located at a suitable point on the
airplane.These two pressures are supplied to the airspeed indicator mounted in
the cockpit of the airplane. The mechanism of the airspeed indicator in low speed
airplanes is such that it senses t sp -p and indicates eV . Note that 0ρ is a
constant value.
At subsonic speeds, when the compressibility effects become significant,
the airspeed indicator mechanism is calibrated to indicate ‘Calibrated airspeed
(Vcal)’, based on the following equation which is a simplified form of Eq.(5.6a).
1 11
2 4
2
2
2
calst 0 cal
0
Vp -p = ρ V
a
(5.6c)
where, a0 = speed of sound under sea level standard conditions.
For further details like construction of airspeed indicators and measurement of
airspeed at supersonic Mach numbers, Refs. 5.1, 5.2 and 5.3 may be consulted.
Information is also available on the internet. However, it may be added that the
static pressure sensed by the static pressure hole may be influenced by the flow
past the airplane. It may be slightly different from the free stream static pressure
and hence the speed indicated by the airspeed indicator may be slightly different
from eV or calV . The speed indicated by the airspeed indicator is called ‘Indicated
airspeed’ and denoted by Vi.
Remark:
On high speed airplanes the speed with respect to ground called ‘Ground
speed’ is deduced from the coordinates given by the global positioning system
(GPS). However, airspeed indicator based on Pitot static system is one of the
mandatory instruments on the airplane.
5.5 Thrust and power required in steady level flight – general case
From Eqs.(5.3) and (5.4) it is noted that :
3C C1 2WD DT = W and P =r r 3/2C 1000 ρSL CL
.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 11
The drag coefficient (CD)depends on the lift coefficient (C
L) and the Mach
number. The relationship between CD and C
L, the drag polar, is already known
from the estimation of the aerodynamic characteristics of the airplane.
Thus, when the drag polar, the weight of the airplane and the wing area areprescribed, the thrust required and the power required in steady level flight at
various speeds and altitudes can be calculated for any airplane using the above
equations. The steps are as follows.
i) Choose an altitude (h).
ii) Choose a flight velocity (V).
iii) For the chosen values of V and h, and given values of the weight of
airplane (W) and the wing area(S) calculate CL as :
L 2
2WC =
ρSV
where corresponds to the density at the chosen ‘h’.
iv) Calculate Mach number from M = V/a; ‘a’ is the speed of sound at the chosen
altitude.
v) For the values of CL and M, calculated in steps (iii) and (iv), obtain C
D from the
drag polar. It (drag polar) may be given in the form of Eqs.(3.45) or (3.49). The
drag polar can also be given in the form of a graph or a table.
vi) Knowing CD, The thrust required (Tr ) and power required (Pr ) can now be
calculated using Eqs.(5.3) and (5.4).
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 5
Lecture 18
Performance analysis I – Steady level flight – 2
Topics
5.6 Thrust and power required in s teady level flight when drag
polar is independent of Mach number
5.7 Thrust and power required in s teady level flight – consideration of
parabolic polar
5.6 Thrust and power required in steady level flight when drag polar is
independent of Mach number
When the Mach number is less than about 0.7, the drag polar is generally
independent of Mach number. In this case, CD / C
L and C
D / CL
3/2 can be
calculated for different values of CL. The curves shown in Figs.5.4a and b are
obtained by plotting CD / CL
and CD / CL
3/2 as functions of C
L. From these curves
it is observed that CD / CL is minimum at a certain value of CL. This CL is denoted
by CLmd
as the drag is minimum at this CL. The power required is minimum when
CD / CL
3/2 is minimum. The C
L at which this occurs is denoted by C
Lmp . Thus in
steady level flight:
Trmin = W (C
D / CL)min
(5.7)
3 C2W DP =rmin 3/2ρS CL min
(5.8)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
Fig.5.4a Variation of CD / CL
with CL
Fig.5.4b Variation of CD / CL
3/2
with CL
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
The speeds, at which the drag and the power required are minimum, are denoted
by Vmd and Vmp respectively. The expressions for Vmd and Vmp are:
2W 2WV = , V =md mpρSC ρSCLmd Lmp
(5.9)
Note:
i) CLmd
and CLmp
are not equal and the corresponding speeds are different. As the
density occurs in the denominator of Eq.(5.9), it implies that Vmd and Vmp
increase with altitude.
ii) Since for Mach number is lower than about 0.7, the drag polar is assumed to
be independent of Mach number, the values of CLmd, CLmp , (CD / CL)min and
(CD / C L
3/2)min
are also independent of Mach number. From Eqs.(5.7) and (5.8) it
is seen that Trmin
is independent of altitude whereas Prmin
increases with altitude
in proportion to 1/ σ1/2
.
iii) It is also observed in Fig.5.4a that a line drawn parallel to the X-axis cuts the
curve at two points A and B. This shows that for the same value of CD / CL
or the
thrust {Tr = W(CD / CL) }, an airplane can have steady level flight at two values of
lift coefficients viz. CLA
and CLB
. From Eq.(5.2) each value of CL corresponds to a
velocity. Hence for the same amount of thrust, in general, flight is possible at two
speeds (V A
and VB). These speeds are:
V A
= (2W / ρSCLA
)1/2
, VB = (2W / ρSC
LB)1/2 (5.9a)
Similarly, from Fig.5.4b it is observed that with the same power, in general, level
flight is possible at two values of lift coefficient viz. CLA
and CLB and
correspondingly at two flight speeds viz. V A
and VB
.
iii) Typical variations of thrust required with flight speed and altitude are shown in
Fig.5.5. Following interesting observations are made in this case where the drag
polar is independent of Mach number. From Eq.(5.7) the minimum drag depends
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 4
only on W and (CD / CL
)min
and hence is independent of altitude. However, the
speed corresponding to minimum drag (Vmd) increases with altitude (Eq.5.9).
Hence, the thrust required curves at various altitudes have the same minimum
thrust at all altitudes and the curves have a horizontal line, corresponding toT = T
rmin , as a common tangent (see Fig.5.5). This feature should be kept in
mind when thrust required curves for subsonic airplanes are plotted.
Fig.5.5 Thrust required and thrust available for subsonic jet airplane
iv) Typical variations of power required with flight speed and altitude are shown
in Fig.5.6a. Interesting observations are made in this case also. From Eq.(5.8)
the minimum power required (Prmin) depends on W3/2 , (CD/CL3/2)min and ρ
-1/2 .
From Eq.(5.9) it is observed that Vmp depends on ρ-1/2 . Noting that for airplanes
with piston engine or turboprop engine, the flight Mach number is less than 0.7,
the drag polar is independent of Mach number. However, due to dependence on
ρ-1/2 , the Prmin and Vmp increase with altitude (Fig.5.6a) . It may be added that
the slope of a line, joining a point on the Pr vs V curve and the origin, is P r / V or
Tr . However, as pointed out earlier, Tr has a minimum value (Trmin) which is
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
independent of altitude. Hence, all Pr vs. V curves have a common tangent
passing through the origin. Such a tangent is shown in Fig.5.6a. This feature
should be pointed out when Pr vs. V curves are plotted at different altitudes. Note
that the common tangent to Pr vs. V curves does not touch at Vmp but at Vmd.
Fig.5.6a Power required and power available curves
Fig.5.6b Power required and power available at an altitude near ceiling
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
5.7 Thrust and power required in steady level flight – consideration of
parabolic drag polar
The discussion in section 5.6, was with reference to a general drag polar
which may be given in tabular form or a plot. Consider the parabolic polar given
by :
CD = C
D0 + KCL
2 (5.10)
Since an equation is available for the drag polar, it is possible to obtain
mathematical expressions for the power required and thrust required. In this
section it is assumed that CDO and K are constant with Mach number.
Substituting for CD
in expression for thrust required gives:
Tr = D = (1/2)ρV2SCD
= (1/2) ρV2 S (C
D0+KC
L
2) (5.11)
Substituting for CL as W /{(1/2)ρV2S} in Eq.(5.11) yields:
2
1 2W2T = ρ V S C + Kr Do 12 2ρV S
2
Or 2 2 2= + / ( )r DOT V SC KW ρV S
1ρ 2
2 (5.12)
In Eq.(5.12) the first term (½) ρ V2 S CD0 is called ‘Parasite drag’. The second
term 2 K W2 / (ρV
2
S) is called ‘Induced drag’. Typical variations of the parasite
drag, induced drag and total drag are shown in Fig.5.7.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
Fig.5.7 Variation of drag with flight speed
It is observed that from Fig.5.7 that the parasite drag, being proportional to V2
,
increases rapidly with speed. The induced drag being proportional to 1/ V2
is high
at low speeds but decreases rapidly as speed increases. The total drag, which isthe sum of the induced drag and the parasite drag, is approximately equal to
induced drag at low speeds and approaches parasite drag at high speeds. It has
a minimum value at a speed (Vmd
) where the parasite drag and induced drag are
equal to each other (Fig.5.7). This can be verified by differentiating Eq.(5.12) with
respect to V and equating it to zero i.e.
2dT 2KW (-2)r = ρ V S C + = 0md D0 3dV ρS Vmd
Or
12
1/42W K
V =mdρS CD0
(5.13)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 8
Substituting Vmd in Eq.(5.12) gives minimum thrust required i.e.
Trmin = W (C
D0 K)1/2+W(CD0
K)1/2 = 2W(CD0
K)1/2 (5.14)
From Eq.(5.14) it is observed that when V equals Vmd , the parasite drag and
induced drag both are equal to W (CD0 K)1/2. This is also shown in Fig.5.7.
Expression for power required in the present case is given by :
T V 1 1 3rP = = ρV SCr D1000 1000 2
Substituting for CD from Eq.5.10 gives:
1 1 3 2
P = ρ V S [C + KC ]r D0 L100021 1 W3 2Or P = ρ V S [C + K ( ) ]r D0 2110002 ρV S
2
Or21 1 KW3P = ρV S C +r D0
2000 500 ρVS (5.15)
The first term in Eq.(5.15) is called ‘Parasite power’ and the second term is called
‘Induced power’. The variations with flight velocity (V) of induced power, parasite
power and the total power required are shown in Fig.5.8.
It is observed that the minimum power occurs at a speed, Vmp , at which the
induced power is three times the parasite power. This can be verified by
differentiating Eq.(5.15) with respect to V and equating it to zero. The verification
is left as an exercise to the student.
1/2
1/42W K
V =mpρS 3CDo
(5.16)
1/2
1/431 2W 256 3P = C Krmin Do1000 ρS 27
(5.17)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 9
Fig.5.8 Variation of power required with flight speed
Remarks:
i) The expressions given in Eqs.(5.13) and (5.14) can be obtained in the following
alternate way.
Tr = W (CD / CL
)
Hence, Trmin = W (C
D / CL)min (5.18)
But, for a parabolic polar
CC D0D = + K CLC CL L
(5.19)
The value of CL at which (C
D / CL) is minimum i.e. (C
Lmd) is given by :
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 10
Cd(C /C ) D0D L = 0 or - + K = 02dCL CLmd
This gives CLmd
as:
CLmd = (C
Do / K)1/2 (5.20)
The corresponding drag coefficient, CDmd
is
Dmd DODO
DOKC
C = C + = 2CK
(5.21)
Equation (5.21) shows that when Tr equals Tmin
, both parasite drag coefficient
and induced drag coefficient are equal to DOC . Hence under this condition, the
parasite drag and induced drag both are equal to (1/2)ρ V
2
S DOC .
Further,
12
12Dmd DO
LmdDO
DDO
L
C 2CC= = = 2 (C K)
C C C / Kmin
(5.22)
Hence, Trmin and Vmd are:
DO DO
2W1/2 1/2 1/4T = 2 W (C K) and V = ( ) (K / C )rmin mdρS
,
which are the same as Eqs.(5.14) & (5.13).
(ii) Exercise 5.4 gives expressions for Tr in terms of V/Vmd and Trmin.
(iii) Similarly, expressions given in Eqs.(5.16) and (5.17) can be obtained in the
following alternate manner.
1/23 C1 2W DP =r 3/21000 ρS C
L
Hence, Prmin occurs when CD/CL
3/2
is minimum. For a parabolic polar
1/2DOCCD = + K CL3/2 3/2C C
L L
Therefore,
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 11
12
DO
5/2L L
3/2d C / CD CL 3 1 K = - +
dC 2 2L C C
Equating the R.H.S. to zero, the value of CL at which the power required is
minimum ( LmpC ) is given as:
LmpC = (3 DOC / K)
1/2
(5.23)
Then the drag coefficient, corresponding toLmp
C is given by:
Dmp DO DODO3KC
C = C + = 4 CK
(5.24)
Equation (5.24) shows that when Pr equals P
rmin the parasite drag coefficient is
equal to DOC and the induced drag coefficient is equal to 3 DOC . Consequently,
the parasite power is (1/2) ρ V3 S DOC and induced power is 3 times of that.
Hence,
DODO
DO
1/44CC 256 3D = = C K
3/2 3/4 27C (3C / K)L min
(5.24a)
1/2 1/2 1/42W 2W KV = =mp
ρSC ρS 3 CLmp DO
1= V 0.76 Vmd md1/43
(5.24b)
The above expression for Vmp is the same as in Eq.(5.16).
Example 5.1
An airplane weighing 100,000 N is powered by an engine producing
20,000 N of thrust under sea level standard conditions. If the wing area be
25 m2 , calculate (a) stalling speeds at sea level and at 10 km altitude,
(b) (CD / C
L)min
, (CD / 3/2
LC )
min, T
rmin, P
rmin, Vmd and Vmp under sea level
conditions.
Assume CLmax = 1.5, C
D = 0.016 + 0.064 2
LC .
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 12
Solution:
The given data are : W = 100,000 N, T = 20,000 N, CD = 0.016 + 0.064 C
L
2
,
S = 25 m2
, CLmax
=1.5
a) SLmax
2WV = ,
ρSC
at s.l. ρ = 1.225 kg/m3
,
at 10 km = 0.413 kg / m3
Hence, at sea level,2 × 100000
V =S1.225 × 25 × 1.5
= 66 m/s = 237.6 kmph
At 10 km altitude, 2 × 100000V =S0.413 × 25× 1.5
= 113.6 m/s = 409.0 kmph.
b)CD0C = = 0.016/0.064 = 0.5Lmd
K
DmdC 2= = 0.032DOC
Hence, (CD / C
L)min
= 0.032/0.5 = 0.064 and
Trmin
= W (CD / C
L)min
= 100000 x 0.064 = 6400 N
C = 3C /K = 0.866Lmp DO
C = 4C = 0.064Dmp DO
3/2 3/2(C /C ) = 0.064/0.866 = 0.0794D minL
2W 2 ×100000V = = = 114.5 m/s = 412.2 kmphmd
ρ S C 1.225× 25 ×0.5Lmd
2 ×100000V = = 86.30 m/s = 310.7 kmphmp
1.225× 25 ×0.866
Note: Vmp = Vmd / 31/4
3 31 2W 1 2 × 1000003/2P = C / C = × 0.0794 = 641.5 kW.rmin D L1000 ρS 1000 1.225 × 25min
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 13
Answers :
a) SV at sea level = 237.6 kmph
SV at 10 km altitude = 409.0 kmph
b) (CD / CL)min = 0.064 ; 3/2(C / C ) = 0.0794D minL ; Trmin = 6400 N
At sea level : P = 641.5 kW;rmin V = 412.2 kmph;md V = 310.7 kmphmp
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 5
Lecture 20
Performance analysis I – Steady level flight – 4
Topics
5.10.2 Parameters influencing Vmax of a jet airplane
5.10.3 Airplane with engine-propeller combination
5.11 Special feature of steady level flight at supersonic speeds
5.10.2 Parameters influencing Vmax of a jet airplane
From Eq.(5.27), an analytical expression for Vmax can be deduced when it is
assumed that the thrust available (Ta) ,CDO and K remain constant with flight
speed. The derivation is as follows.
22
2DOa1 2W
T = ρV S C K2 ρV S
(5.27)
or AV4 – BV2 + C = 0
where, DO
1 A = ρSC
2
, B = Ta and22KW
C =
ρS
. (5.27a)
When Ta, CDO and K have constant values, Eq.(5.27a) gives :
122B ± B -4AC
V =2A
Consequently, Vmax being the larger of the two solutions, is :
2
max
12
B+ B -4ACV =2A
Substituting for A, B and C from Eq.(5.27a) yields :
2 2
max 2 2 2 2 2
1
2
DO DO
a a
DO
T T W KV = + - 4
ρSC ρ S C S ρ C
(5.27b)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
Multiplying and dividing some of the terms in Eq.(5.27b) by ‘W’ gives:
12
max
2 2 2
DO2 2 2 2
DO DO DO
a aT /W W/S T /W W/S KCWV = + - 4
ρC Sρ C ρ C
Simplifying yields :
Or
DO
max
DO
22
1/2
a aW
T /W W/S + T /W -4C KS
V =ρC
(5.27c)
Equation (5.27c) shows that Vmax depends on thrust to weight ratio ( aT /W), wing
loading (W/S), CDO , K and ρ .The maximum speed (Vmax) increases with increaseof ( aT /W) and (W/S) and decreases with increase of CDO and K. The term ‘ρ ’ in
the denominator of Eq.(5.27c) indicates that Vmax would be higher at higher
altitudes because ρ decreases with altitude. In section 4.5 it is pointed out that
the thrust output decreases as 0.7σ . Taking this into account, Eq.(5.27c) indicates
that Vmax would increase slightly upto a certain altitude as shown in Fig.5.9.
The trend of Vmax, decreasing after a certain altitude, observed in Fig.5.9, can be
explained as follows.
From atmospheric characteristics (Chapter 2), it is observed that, with the
increase of altitude the speed of sound decreases. Thus for a given Vmax the
Mach number corresponding to it would increase with altitude. When the Mach
number exceeds the critical Mach number, CDO & K would no longer be constant
but actually increase. This would result in lowering of Vmax as compared to that
obtained with constant values of CDO and K. In section 4.2 of Appendix ‘B’ the
values of Vmax at different altitudes are obtained by a graphical procedure which
takes into account the changes in CDO and K when Mach number is greater than
0.8.
5.10.3 Airplane with engine-propeller combination
The steps to calculate Vmax
and (Vmin
)e in this case, are as follows.
(1) Assume an altitude ‘h’. Let Pa be the THP available in kW at this altitude.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
(2) From Eq.(5.15) :
21 1 KW3
P = P = ρ V S +r a DO2000 500 ρVS
C
or A1V
4
– B1V + C1 = 0 (5.28)
where,2
, =1DO1 1 KW
A = ρSC , B = P C1 1 a2000 500 ρS
.
Equation (5.28) is not a quadratic. An iterative method of solving Eq.(5.28) is
given in example 5.3. Equation (5.28) has two solutions V1 and V
2. The higher of
these two gives Vmax
and the lower value gives (Vmin
)e. The minimum speed at
the chosen altitude is higher of (Vmin
)e and Vs (see example 5.3).
Remark:
Obtain power available at V1 calculated above and denote it by Pa1. If Pa
assumed at the beginning of the calculation in step (1), is significantly different
from Pa1, then the calculations would need to be revised with the new value of
Pa1. However, it is expected that the calculations would converge in a few
iterations.
Example 5.2For the airplane in example 5.1 obtain the maximum and minimum speed in
steady level flight at sea level.
Solution:
The given data are :
W = 100,000N, T = 20,000N, S = 25 m2, 2C = 0.016 + 0.064 CD L, CLmax = 1.5
In this case, T / W = 20000 / 100000 = 0.2 = CD / CL
0.0160.2 = + 0.064CL
CL
2Or 0.064 C - 0.2 C + 0.016 = 0LL
Solving the above equation gives: CL = 3.04 and 0.0822. The corresponding
speeds are :
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 4
2 ×100000V = = 281.8 m/smax
1.225× 25 × 0.0822
and2 ×100000
( V ) = = 45.4 m/smin e1.225× 25 × 3.04
Since VS , as calculated in example 5.1, is 66.0 m/s, the minimum speed is
decided by VS and equals 66.0 m/s.
The Mach number corresponding to Vmax
is :
281.8 / 340.29 = 0.828.
This value of Mach number is likely to be greater than Mcrit
. As a possible
assumption let us assume Mcruise
= 0.8 and obtainDO
ΔC and ΔK from Eqs.3.50a
and 3.51a. Consequently,
DO ΔC = - 0.001 (M - 0.8) + 0.11 (M - 0.8)2 and ΔK = (M - 0.8)2 + 20(M - 0.8)3
For M = 0.828,DO
ΔC = 0.000055 and ΔK = 0.00122
Hence, the drag polar at M = 0.828 is likely to be:
CD = (0.016 + 0.000055) + (0.064+0.00122)C
L
2
= 0.016055 + 0.06522 CL
2
Using this polar and revising the calculations, gives: Vmax = 281.3 m/s
This revised value of Vmax
is very close to the value of 281.8 m/s obtained earlier
and hence further revision is not needed.
(Answers: Vmax = 281.3 m/s =1012.7 kmph, Vmin = 66.00 m/s = 237.6 kmph)
Example 5.3
A piston-engined airplane has the following characteristics.
W = 11,000 N, S = 11.9 m2, CD = 0.032 + 0.055 C
L
2, CLmax
= 1.4.
Obtain the maximum and minimum speeds in level flight at an altitude of 3 kmassuming that the engine BHP is 103 kW and the propeller efficiency is 83%.
Solution :
W = 11,000 N, S = 11.9 m2, CD = 0.032 + 0.055 C
L
2
CLmax
= 1.4, ρ at 3km altitude = 0.909 kg/m3
,
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
Pa = η x BHP = 0.83 x 103 = 85.5 kW
From Eq.(5.15):
21 2K W3P = P = ρV S C +a r Do
2000 1000 ρSV
Or21 2 0.055 ×11000385.5 = × 0.909 × 11.9× 0.032×V + ×
2000 1000 0.909 × 11.9 × V
= 1.731 x 10-4 V3 +1230.5
V (5.28a)
Equation (5.28a) is not a quadratic. However, it can be solved for Vmax
and (Vmin
)e
by an iterative procedure.
Solution for Vmax:
When solving for V max
, by an iterative procedure, it is assumed that the first
approximation (Vmax1
) is obtained by retaining only the term containing the
highest power of V in Eq.(5.28a) i.e.
1st approximation: 85.5 = 1.731 x 10-4 V3
max1
This gives Vmax1
= 79.05 m/s
To obtain the 2nd approximation, substitute Vmax1
in the second term on RHS of
Eq.(5.28a). Note that this term was ignored in the first approximation.
3max2
1230.5-485.5 = 1.731 x 10 V +79.05
Or V max2
= 73.93 m/s
To obtain the 3rd approximation, substitute Vmax2
in the second term on RHS of
Eq.(5.28a), i.e.
3max31230.5
-485.5 = 1.731× 10 V + 73.93
Or Vmax3
= 73.54 m/s
To obtain the 4th approximation, substitute Vmax3
in the second term on RHS of
Eq.(5.28a), i.e.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
3max4
1230.5-485.5 = 1.731× 10 V +73.54
Or Vmax4
= 73.51 m/s
Since the 3rd and 4th approximations are close to each other, Vmax
is taken as
73.51 m/s.
Solution for (Vmin
)e:
When solving for (Vmin
)e, by an iterative procedure, it is assumed that the first
approximation 1min e
V , is obtained by retaining only the term containing the
lowest power of V in Eq.(5.28a) i.e.
1min e
1230.585.5 =
V
Or 1min e
V = 14.4 m/s.
To obtain the 2nd approximation, substitute (Vmin
)e1 in the first term on RHS of
Eq.(5.28a). Note that this term was ignored in the first approximation.
2min e
1230.5-4 385.5 = 1.731× 10 × 14.4 +(V )
Or 2min e(V ) = 14.48 m/s
Since the second approximation is very close to the first one,
(Vmin
)e is taken as 14.48 m/s
The stalling speed at 3 km altitude is :
2W 2 × 11000V = = = 38.2m/ss
ρSC 0.909×11.9 ×1.4Lmax
Since Vs is greater than (Vmin
)e, the minimum speed is 38.2 m/s.
Answers: At 3 km altitude:
Vmax = 73.51 m/s = 265.0 kmph , V
min = 38.20 m/s = 137.4 kmph
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
5.11 Special features of steady level flight at supersonic speeds
At transonic and supersonic speeds the variations of CD0
, K and Ta with
Mach number do not permit simple mathematical treatment of the performance
analysis. The thrust required (Tr ) increases rapidly as the Mach number
approaches unity (Fig.5.11).
Fig.5.11 Level flight performance at high speeds
The thrust available also increases but the increase is not as fast as that
of Tr and the thrust available and thrust required curves may intersect at many
points (points A, B, and C in Fig.5.11). It is interesting to note that if the airplane
can go past Mach number represented by point B in Fig.5.11, then it can fly up to
Mach number represented by point C with the same engine. To overcome the
rapid drag rise in transonic region (Fig.5.11), the afterburning operation of the
engine is resorted to. It may be mentioned that an afterburner duct is located
between the turbine and the nozzle (Fig.4.8b). When the afterburner is on,
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 8
additional fuel is burnt in the afterburner duct. This gives additional thrust.
However, the specific fuel consumption is very high with afterburner on and
hence this operation is resorted to only for a short duration.
The thrust with afterburner on is shown schematically by a dotted line in Fig.5.11.
It is observed that the thrust available is more than the thrust required and
airplane can accelerate beyond point B. When the Mach number is close to that
represented by point C, the afterburner can be shut down and the airplane runs
with normal engine operation.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 5
References
5.1 Pallett, E.H.J. “Aircraft instrument integrated systems” 3rd Edition, Longman
Science & Technology, (1992).
5.2 Illman, P.E. “The pilot’s handbook of aeronautical knowledge“ 3rd Edition,
Tab books division of McGraw Hill (1995).
5.3 Perkins, C.D. (Editor) “AGARD Flight test manual, Vol.I – Performance”
Pergamon Press (1959).
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-5
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 5
Exercises
5.1 Obtain the maximum speed and minimum speed in steady level flight at sea
level for the following airplane:
W = 36,000 N; S = 26.0 m2; CD = 0.032 + 0.043C
L
2 BHP = 503 kW; Propeller
efficiency = 82%; CLmax = 1.5
[ Answers: Vmax = 324.6 kmph; Vmin = 139.8 kmph]
5.2 A jet engined airplane has a weight of 64,000 N and wing area of 20 m2
. If
the engine output at 5 km altitude be 8000 N, calculate the maximum and
minimum speeds in level flight. Given that
CDO
= 0.017, A = 6.5, e = 0.80, CLmax
= 1.4.
[ Answers: Vmax
= 877 kmph,Vmin = 283.6 kmph]
5.3 An airplane stalls at M=0.2 at sea level. What will be the Mach number and
equivalent airspeed when it stalls at 5 km altitude? Compare the thrust required
to maintain level flight near stall at the two altitudes. Assume the weight of the
airplane to be same at the two altitudes.
[ Answers: M = 0.274, Ve
= 68.06 m/s, (Tr )s.l
= (Tr )5 km
as CL
is same]
5.4 (a) Show that the thrust required in steady level flight at a speed V for an
airplane with parabolic drag polar is given by:
V AW2T = D = AW ( ) +r 2Vmd (V / V )md
where, Vmd = speed for minimum drag, W = weight of airplane and A = (CD0
K)1/2
.
(b) Also show that if V = mVmd, then the thrust required (Tr ) in terms of the
minimum thrust required (Trmin) is given by :
2 -2r
rmin
T 1= m +m
T 2
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 6
Performance analysis II – Steady climb, descent and gl ide
(Lectures 21,22 and 23 )
Keywords: Steady climb – equations of motion, thrust and power required;
maximum rate of climb; maximum angle of climb; absolute ceiling; service ceiling;
glide – equations of motion, minimum angle of glide, minimum rate of sink;
hodographs for climb and glide.
Topics
6.1 Introduction
6.2 Equations of motion in steady climb
6.3 Thrust and power required for a prescribed rate of cl imb at a given
flight speed
6.4 Climb performance with a given engine
6.4.1 Iterative procedure to obtain rate of climb
6.5 Maximum rate of climb and maximum angle of climb
6.5.1 Parameters influencing (R/C)max of a jet airplane
6.5.2 Parameters influencing (R/C)max
of an airplane with engine-propeller
combination
6.6. Climb hydrograph
6.7. Absolute ceiling and service ceiling
6.8 Time to cl imb
6.9 Steady descent
6.10 Glide
6.10.1 Glide performance – minimum angle of glide, minimum rate of sink
and maximum range and endurance in glide.
6.11 Glide hodograph
Exercises
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
Chapter 6
Lecture 21
Performance analysis II – Steady climb, descent and gl ide – 1
Topics
6.1 Introduction
6.2 Equations of motion in steady climb
6.3 Thrust and power required for a prescribed rate of cl imb at a given
flight speed
6.4 Climb performance with a given engine
6.4.1 Iterative procedure to obtain rate of climb
6.1. Introduction
In this chapter the steady climb, descent and glide are dealt with. A glide
is a descent with thrust equal to zero. The approach in this chapter is as follows.
(a) Present the forces acting on the airplane in the chosen flight,
(b) Write down equations of motion using Newton’s second law,
(c) Derive expressions for performance items like rate climb, angle of climb.(d) Obtain variation of these with flight velocity and altitude.
6.2 Equations of motion in a steady climb
During a steady climb the center of gravity of the airplane moves at a
constant velocity along a straight line inclined to the horizontal at an angle γ
(Fig.6.1). The forces acting on the airplane are shown in Fig.6.1.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
Fig.6.1 Steady climb
Since the flight is steady, the acceleration is zero and the equations of motion in
climb can be obtained by resolving the forces along and perpendicular to the
flight path and equating their sum to zero i.e.T – D – W sin = 0 (6.1)
L – W cos = 0 (6.2)
Hence, sin = (T- D / W) (6.3)
From the velocity diagram in Fig.6.1, the vertical component of the flight velocity
(Vc) is given by:
Vc = V sin = (T- D / W) V (6.4)
The vertical component of the velocity (Vc) is called rate of climb and also
denoted by R/C. It is also the rate of change of height and denoted by (dh / dt).
Hence,
Vc = R/C = dh/dt =T-D
Vsin = VW
(6.5)
Rate of climb is generally quoted in m/min.
Remarks :
i) Multiplying Eq.(6.1) by flight velocity V, gives:
T V = D V + W V sin = D V + W Vc dh d
= DV+ mg = DV+ mghdt dt (6.6)
In Eq.(6.6) the terms ‘TV’, ‘DV’ and ‘ d
mghdt
’ represent respectively, the power
available, the energy dissipated in overcoming the drag and the rate of increase
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 4
of potential energy. Thus, when the airplane climbs, its potential energy
increases and a part of the engine output is utilized for this gain of potential
energy.
Two facts may be pointed out at this juncture. (a) Energy supply to the airplane
comes from the work done by the engine which is represented by the term‘TV’ in
Eq.(6.6). (b) The drag acts in a direction opposite to that of the flight direction.
Hence, energy has to be spent on overcoming the drag which is represented by
the term ‘DV’ in Eq.(6.6). This energy (DV) is ultimately lost in the form of heat
and is appropriately termed as ’Dissipated’. Continuous supply of energy is
needed to overcome the drag. Thus, a climb is possible only when the engine
output is more than the energy required for overcoming the drag.
It may be recalled from section 5.9 that in a level flight, at speeds equal to V max
and (Vmin)e , the power (or thrust ) available is equal to the power (or thrust)
required to overcome the drag (see points D and D’ in Fig.5.5 and points C and
C’ in Fig.5.6b). Hence, the rate of climb will be zero at these speeds. The climb is
possible only at flight speeds in between these two speeds viz. Vmax and (Vmin)e.
It is expected that there will be a speed at which the rate of climb is maximum.
This flight speed is denoted by V(R/C)max
and the maximum rate of climb is
denoted by (R/C)max. The flight speed at which the angle of climb () is maximum
is denoted by V max
.
ii) In a steady level flight, the lift is equal to weight but in a climb, the lift is less
than weight as cos is less than one, when is not zero. Note that when an
airplane climbs vertically, its attitude is as shown in Fig.6.2. It is observed that in
this flight, the resolution of forces along and perpendicular flight direction gives:
L = 0, T = D + W
These expressions are consistent with Eqs.(6.1) and (6.2) when = 90o is
substituted in them. Note that in this flight the thrust is more than the weight.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
Fig.6.2 Airplane in vertical climb
6.3 Thrust and power required for a prescribed rate of climb at a given
flight speed
Here it is assumed that the weight of the airplane (W), the wing area (S) and the
drag polar are given. The thrust required and power required for a chosen rate of
climb (Vc) at a given altitude (h) and flight speed (V) can be obtained, for a
general case, by following the steps given below. It may be pointed out that the
lift and drag in climb are different from those in level flight. Hence, the quantities
involved in the analysis of climb performance are, hereafter indicated by the
suffix ‘c’ i.e. lift in climb is denoted by Lc
i) Since Vc and V are prescribed, calculate the angle of climb γ from:
γ = sin-1 (Vc / V)
ii) From Eq.(6.2) the lift required in climb (Lc ) is :
Lc = W cos γ (6.7)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
iii) Calculate the lift coefficient in climb (CLc ) as:
CLc =L W cosc =
1 212 ρV SρV S 22
(6.8)
iv) Obtain the flight Mach number; M = V/a ; a = speed of sound at the chosen
altitude.
v) Corresponding to the values of CLc and M, obtain the drag coefficient in climb
(CDC
) from the drag polar. Hence, drag in climb (Dc) is given by:
Dc = (1/2 ρV2 S C
DC) (6.9)
vi) The thrust required in climb (Trc) is then given by:
Trc = W sin γ + Dc (6.10)
and the power required in climb (Prc) is :
Prc =T Vrc in kW1000
(6.11)
Example 6.1
An airplane weighing 180,000N has a wing area of 45 m2 and drag polar given
by CD = 0.017 + 0.05 2
LC . Obtain the thrust required and power required for a
rate of climb of 2,000 m/min at a speed of 540 kmph at 3 km altitude.
Solution:
The given data are:
W = 180,000 N, S = 45 m2, CD = 0.017 + 0.05 2
LC
Vc = 2,000 m/ min = 33.33 m/s, V = 540 kmph = 150 m/s.
at 3 km altitude = 0.909 kg/m3
sin = Vc / V = 33.33/150 = 0.2222 or = 12o-50’, cos = 0.975
Lc = W cos = 180000 x 0.975
Or CLc =2
2W cos
ρ V S
=
180000 × 0.975 × 2= 0.381
0.909 × 150×150 × 45
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
Hence, CDC
= 0.017 + 0.05 X0.3812 = 0.02426
Dc = (1/2 ρ V2 S) C
DC
= (1/2) X 0.909 X 150 X 150 X 45 X0.02426 = 11163 N
Hence, Trc = W sin + Dc = 180000 X 0.2222 + 11163 = 51160 N
Prc = TrcV/1000 = 51160 X 150/1000 = 7674 kW
Answers:
Thrust required in climb (Trc) = 51,160 N
Power required in climb (Prc) = 7,674 kW
6.4 Climb performance with a given engine
In this case, the engine output is prescribed at a certain altitude and flight
speed. This is in addition to the data on weight of the airplane (W), the wing area
(S) and the drag polar.The rate of climb (Vc) and the angle of climb(γ) are
required to be determined at the prescribed altitude and flight speed.
The solution to this problem is not straightforward as sin γ depends on
(T- Dc) and the drag in climb (Dc) depends on the lift in climb (Lc ), which in turn
depends on W cos γ. Hence, the solution is obtained in an iterative manner. This
is explained later in this section. However, if the drag polar is parabolic with
constant coefficients, an exact solution can be obtained using Eqs. (6.1) to (6.4).
The procedure is as follows.
From Eq.(6.4), sin = Vc / V.
Using Eq.(6.7), the lift during climb (Lc) = W cos = W (1-sin2 )1/2
122= W 1-(V /V)c
(6.12)
Hence, Lift coefficient during climb
122W 1-(V /V)cLcC = =Lc 1 212 ρV SρV S 22
(6.13)
By its definition, D = (1/2) ρ V2 SC
D.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 8
When the polar is parabolic, the drag in climb (Dc) can be expressed as :
Dc = (1/2) ρ V2 S (C
DO+K 2C
Lc) DO
22 V1 KW2 c= ρ V S C + 1-212 VρV S
2
(6.14)
From Eq.(6.10), the thrust required in climb (Trc) is given by :
Trc = W sin γ + Dc =W Vc + Dc
V
Substituting for Dc , yields :
Trc DO
22 V WV1 KW2 c c = ρ V S + 1- +212 V VρV S
2
C
(6.15)
or2V Vc c A - B + C =0
V V
(6.16)
where,2KW
A = , B = W21 ρV S
2
and21 2KW2C = T - ρV SC -DO 22 ρV S
(6.17)
Equation (6.16) is a quadratic in (Vc / V), and has two solutions. The solution
which is less than or equal to one, is the valid solution because Vc / V equals
sin γ and sin γ cannot be more than one. Once (Vc / V) is known, the angle of
climb and the rate of climb can be immediately calculated. This is illustrated in
example 6.2.
Example 6.2.
For the airplane in example 6.1, obtain the angle of climb and the rate of climb at
a flight speed of 400 kmph at sea level, taking the thrust available as 45,000 N.
Solution:
In this case, W = 1,80,000 N, S = 45 m2
, CD = 0.017 + 0.052
CL
V = 400 kmph = 111.1 m/s, T = 45,000 N
From Eq.(6.15), Trc DO
22 V WV1 KW2 c c = ρ V S + 1- +212 V VρV S
2
C
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 9
Substitution of various quantities yields:
2145000 = × 1.225 × 111.1 × 45 × 0.0172
22 V V0.05 × 180000 c c + × 1 - 180000 ×
2 21 V× 1.225 × 111.1 × 45 V2
Simplifying,2
V Vc c- 37.82V V
+ 7.24 = 0
Solving the above quadratic gives : ( Vc / V) = 37.62, 0.192.
Since sin cannot be larger than unity, the first value is not admissible.
Hence, Vc / V = sin = 0.192 or = 11o 4
Vc = 0.192 111.1 = 21.33 m/s = 1280 m/min.
Answers:
Angle of climb () = 11o 4 ; Rate of climb (Vc) = 1280 m/min
6.4.1 Iterative procedure to obtain rate of climb
When the drag polar is not given by a mathematical expression, an
iterative procedure is required to obtain the rate of climb for a given thrust ( aT ) or
thrust horse power (THPa). The need for an iterative solution can be explained as
follows.
From Eq.(6.10), sin =T - Da c
W (6.18)
To calculate sin , the drag in climb (Dc) should be known. The term Dc
depends on the lift in climb (Lc). In turn Lc is W cos , but cos is not known in
the beginning.
To start the iterative procedure, it is assumed that the lift during climb (Lc) is
approximately equal to W. Using this approximation, calculate the first estimate
of the lift coefficient (CL1
) as :
(CL1
) = W / (1/2)ρ V2 S
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 10
From CL1
and the flight Mach number obtain CD1
from the drag polar. Calculate
the first approximation of drag (D1) as:
D1= (1/2) ρ V
2 S C D1
Hence, the first approximation to the angle of climb (1) is given by:
sin 1
T - Da 1=W
(6.19)
In the next iteration, put L = W cos 1 and carry out the calculations and get a
second approximation to the angle of climb (2). The calculations are repeated till
the values of after consecutive iterations are almost the same. Once the angle
is known, Vc
is given as V sin .
It is observed that the convergence is fast and correct values of and Vc are
obtained within two or three iterations. This is due to the following two reasons.
(a)When is small (i.e. less than 10o), cos is almost equal to one, and the
approximation, L = W, is nearly valid. (b) When is large the lift dependent part
of the drag, which is affected by the assumption of L W , is much smaller than
Ta . Consequently, the value of given by Eq.(6.18) is close to the final value.
Example 6.3 illustrates the procedure.
Example 6.3
An airplane weighing 60,330 N has a wing area of 64 m2 and is equipped
with an engine-propeller combination which develops 500 kW of THP at 180
kmph under standard sea-level conditions. Calculate the rate of climb at this flight
speed. The drag polar is given in the table below.
CL 0.0 0.1 0.2 0.3 0.4 0.5 0.6
CD 0.022 0.0225 0.024 0.026 0.030 0.034 0.040
CL 0.7 0.8 0.9 1.0 1.2
CD 0.047 0.055 0.063 0.075 0.116
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 11
Solution:
The given data are: W = 60,330 N, S = 64 m2, V=180 kmph = 50 m/s,
THPa = 500 kW. Hence, T
a = (THP
ax 1000)/V = 500 x 1000 / 50 = 10,000 N
The values of and Vc are obtained by the iterative procedure explained in
section 6.4.1.
1st approximation: L W =1
2
2ρ V S CL1
Hence, CL1
60330 × 2= = 0.615
1.225× 50 × 50 × 64
CD1
: By interpolating between the values given in the above table, the value of
C D1
is 0.041, corresponding to CL1
of 0.615.
Hence, D1 = (1/2) 1.225 50 50 64 X0.041= 4030 N
From Eq.(6.19) : sin 1
T - Da 1=W
Or sin 1=
10000-4030= 0.099
60330
Or 1= 5o 41
Hence, Vc1
= 50 x 0.099 = 4.95 m/s
cos 1 = 0.995
2nd approximation:
L = W cos 1= 60330 0.995 = 60036 N
L2
60036 × 2C = = 0.612
1.225 × 50 × 50 × 64
From above table CD2
is 0.0408 corresponding to CL2
of 0.612.
Hence, D2 = (1/2) 1.225 50 50 64 0.0408 = 4010 N
sin 2=
10000-4010= 0.0993
60330
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 12
Hence, Vc2
= 50 x 0.0993 = 4.965 m/s
The two approximations, Vc1 and V
c2 are fairly close to each other. Hence, the
iteration process is stopped.
Vc = 4.965 m/s = 298 m/min.
Remark:
In the present example, is small (5041’) hence 2nd iteration itself gives
the correct value. For an interceptor airplane which has very high rate of climb
(about 15000 m/min) few more iterations may be needed.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 6
Lecture 22
Performance analysis II – Steady climb, descent and gl ide – 2
Topics
6.5 Maximum rate of climb and maximum angle of climb
6.5.1 Parameters influencing (R/C)max of a jet airplane
6.5.2 Parameters influencing (R/C)max of an airplane with engine-propeller
combination
6.5 Maximum rate of climb and maximum angle of climb
Using the procedure outlined above, the rate of climb and the angle of
climb can be calculated at various speeds and altitudes. Figures 6.3a to 6.3f
present typical climb performance of a jet transport. Figure 6.4a to 6.4d present
the climb performance of a piston engined airplane. Details of the calculations for
these two cases are presented in Appendices B and A respectively.
Fig.6.3a Climb performance of a jet transport - rate of climb
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
Fig.6.3b Climb performance of a jet transport - angle of climb
Fig.6.3c Climb performance of a jet transport - V(R/C) max
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
Fig.6.3d Climb performance of a jet transport-Vmax
Fig.6.3e Climb performance of a jet transport - variation of (R/C)max with altitude
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 4
Fig.6.3f Climb performance of a jet transport - variation of max with altitude
Fig.6.4a Climb performance of a piston engined airplane- rate of climb
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
Fig.6.4b Climb performance of a piston engined airplane- angle of climb
Fig.6.4c Climb performance of a piston engined airplane - V(R/C)max, and Vmax
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
Fig.6.4d Climb performance of a piston engined airplane - Variation of (R/C)max
with altitude
Remarks:
i) At V = Vmax
the available thrust or THP is equal to the thrust required or power
required in level flight. Hence climb is not possible at this speed. Similar is the
case at (Vmin)e limited by engine output (Figs.6.3a and 6.4a). For the same
reasons, at Vmax
and (Vmin)e the angle of climb () is also zero (Figs.6.3b and
6.4b). It may be recalled from subsection 5.9, that at low altitudes the minimum
speed is decided by stalling and hence the calculations regarding the rate of
climb and the angle of climb are restricted to flight speeds between Vmin and
Vmax.
ii) The speed at which R/C is maximum is denoted by V(R/C)max
, and the speed at
which γ is maximum is called Vmax . Figures 6.3c and d and Fig.6.4c show the
variation of these speeds with altitudes for a jet transport and a piston engined
airplane respectively. It may be noted that V(R/c)max
and Vmax
are different from
each other. For a jet airplane V(R/c)max
is higher than Vmax
at low altitudes . The
two velocities approach each other as the altitude increases. For a piston
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
engined airplane V(R/C)max
is lower than Vmax at low altitudes . The two velocities
approach each other as the altitude increases. These trends can be explained as
follows.
From Eqs.(6.3) and (6.4), it is observed that is proportional to the excess thrusti.e. (Ta - Dc) and the rate of climb is proportional to the excess power i.e.
(TaV – DcV). It may be recalled that for a piston engined airplane the power
available remains roughly constant with velocity and hence, the thrust available
(Ta = Pa / V) will decrease with velocity. On the other hand, for a jet airplane the
thrust available is roughly constant with velocity and consequently the power
available increases linearly with velocity (see exercise 6.3). The differences in
the variations of Ta and Pa with velocity, in the cases of jet engine and piston
engine, decide the aforesaid trends.
iii) As the excess power and the excess thrust decrease with altitude, (R/C)max
and max also decrease with altitude.
6.5.1 Parameters influencing (R/C)max of a jet airplane
In subsection 5.10.2, the parameters influencing Vmax were identified by
simplifying the analysis with certain assumptions. In this subsection theparameters influencing (R/C)max are identified in a similar manner. The limitations
of the simplified analysis are pointed out at the end of this subsection.
The case of a jet airplane is considered in this subsection.
From Eq.(6.5) it is noted that :
C
T-DV = R/C = V
W
Following simplifying assumptions are made to identify the parameters
influencing (R/C)max.
(a)Drag polar is parabolic with constant coefficients i.e. CDO and K are constants.
(b) Though the angle of climb is not small, for the purpose of estimating the
induced drag, the lift (L) is taken equal to weight. See comments at the end of
section 6.4.1 for justification of this approximation.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 8
(c) At a chosen altitude, the thrust available (Ta) is constant with flight speed.
With these assumptions, the expression for drag simplifies to that in the level
flight i.e.
2
2 2
2D DO1 1 2WD = ρV SC = ρV S C + K2 2 ρSV
Hence,
CaT -D
V = VW
Or
2
2DOa
c
1ρVT 2K W2V = V - C -
W W/S ρV S
Or -13
C DOaT 1 2K W
V = V - ρV W/S C -W 2 ρV S
(6.20)
To obtain the flight speed corresponding to (Vc)max , Eq.(6.20) is differentiated
with respect to V and equated to zero i.e.
-12
2DOacdV T 3 2K W
= - ρV W/S C + = 0dV W 2 ρV S
(6.21)
Simplifying Eq.(6.21) yields:
2(R/C)max (R/C)max
2
2
4
DO DO
a2 T /W W/S 4K W/SV - V - = 0
3ρC 3ρ C (6.22)
Noting from Eq.(3.56) that 2
maxDO
1L/D =
4C K, yields:
2
(R/C)max (R/C)max
max
2
22
4
2DO DO
a2 T /W W/S W/SV - V - = 0
3ρC 3ρ C L/D
Thus,
(R/C)max 2
2
max
12
DO
a
a
T /W W/S 3V = 1 ± 1+
3ρC LT /W
D
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 9
The negative sign in the above equation, would give an imaginary value
for R/C max
V and is ignored.
Hence,
R/C max 2
DOmax
12
2a
a
T /W W/S 3V = 1+ 1+
3ρC L/D T /W
1
2
DO
aT /W W/S Z= ,
3ρC
(6.23)
where,
22
max a
3Z = 1 + 1 +
L/D T /W (6.24)
Substituting V(R/C)max from Eq.(6.23) in Eq.(6.5) yields:
12
DODOmax
DO DO
a aa
a
2 W/S K 3ρCT /W W/S Z T /W W/S ZCT 1R/C = × - ρ -
3ρC W 2 3ρC W/S ρ T /W W/S Z
Or
DO
DO
12
max
a a
a a
T /W W/S Z T 6KCZR/C = - T /W -
3ρC W 6 T /W Z
=
DO
1/2 3/2
2 2
max
a
a
W/S Z T Z 31- -
3ρC W 6 2 T /W L/D Z
(6.25)
Remarks:
The following observations can be made based on Eqs.(6.23) to (6.25)
(i) In Eq.(6.24), the quantity Z appears to depend on (L/D)max and (a
T /W). In this
context it may be noted that for jet airplanes (a) the value of (L/D)max would be
around 20 and (b) the value of ( aT /W) would be around 0.25 at sea level and
around 0.06 at tropopause. With these values of (L/D)max and ( aT /W), Z would be
around 2.1 at sea level and 2.7 at tropopause.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 10
However, in Eqs.(6.23) and (6.25) the terms involving Z, appear as 1/2Z or Z/6.
Hence, the dependence of R/C max
V and maxR/C on Z does not appear to be of
primary importance. The important parameters however, are ( aT /W), (W/S),ρ
and CDO. It may be recalled from section 4.5 that for a turbofan engine, aT
decreases with altitude in proportion to 0.7σ ; σ being the density ratio.
(ii) From Eq.(6.25) it is observed that for given values of W/S and C DO ,
maxR/C decreases with altitude. Hence, suitable value of ( aT /W) is required to
achieve the specified rates of climb at different altitudes.
The same equation also indicates that the rate of climb increases when wing
loading increases and CDO decreases. However, the performance during cruise
and landing generally place a limit on the value of (W/S).
(iii) From Eq.(6.23) it is observed that the flight speed for maximum rate of
climb(V(R/Cmax), increases with ( aT /W), (W/S) and altitude. In this context it may
be pointed out that the Mach number corresponding to V(R/C)max, should always
be worked out and corrections to the values of CDO and K be applied when this
Mach number exceeds Mcruise. Without these corrections, the values of
maxR/C obtained may be unrealistic.
(iv) Figure 4.12 shows typical variations of thrust vs Mach number with altitude as
parameter. It is observed, that the thrust decreases significantly with Mach
number for altitudes equal to or less than 25000 (7620 m). Thus, the
assumption of thrust being constant with flight speed is not a good approximation
for h 25000 .
6.5.2 Parameters influencing (R/C)max of an airplane with engine-propeller
combination
In this subsection the simplified analysis is carried out for climb performance of
an airplane with engine-propeller combination.
From Eq.(6.5)
C
V T-D TV- DVV = R/C = =
W W
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 11
TV = 1000 p aη P ; aP = power available in kW
DV = Power required to overcome drag = 3D
1ρV SC
2
Following assumptions are made to simplify the analysis and obtain parameters
which influence (R/C)max are V(R/C)max in this case.
(a) Drag polar is parabolic with CDO and K as constants.
(b) L = W for estimation of induced drag.
(c) Power available is constant with flight speed (V).
Consequently,
2
3DO 2
1 2WDV = ρV S C +K
2 ρSV
(6.26)
Since Pa is assumed to be constant, the maximum rate of climb would be
obtained when DV is minimum. This occurs at the flight speed corresponding to
minimum power mpV .
Hence, in this case : R/C max mpV = V
The expression for mpV is given in Eq.(5.24b), consequently :
R/C max
114
2mp
DO
2W KV = V =ρS 3C
(6.27)
Substituting V(R/C)max from Eq.(6.27) in Eq.(6.5) gives:
R/C max
R/C max
R/C max
2
2
2DO
pmax
a V1000η P 1 2WR/C = - ρV SC + K
W W 2 ρSV
-1
R/C max
p a
DODO
DO
1000η P 2K W/S1 2 K W= - V ρ W/S C +
2W 2 ρ 3C S ρ K/ 3C W/Sρ
(6.28)
Noting that ,max
DO
1L/D =
2 C K and
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
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1/ 3 + 3= 1.155, yields
2:
R/C max
p a
max
max
1000η P 1.155R/C = - V
W L/D (6.29)
Substituting for V(R/C)max from Eq.(6.27) yields :
p 1/2a
maxDO max
1000η P 2 K 1.155R/C = - W/S
W ρ 3C L/D (6.30)
Remarks:
(i) From Eq.(6.27) it is observed that V(R/C)max increases with wing loading (W/S).
(ii) From Eq.(6.30) it is observed that (R/C)max increases as pη , Pa and (L/D)max
increase. However, the second term on the right hand side of this equation
indicates that (R/C)max decreases with increase of wing loading. This trend is
opposite to that in the case of jet airplanes. Thus, for a specified (R/C) max , the
wing loading for an airplane with engine-propeller combination should be rather
low, to decrease the power required.
(iii) The first term in Eq.(6.30) involves pη and Pa. From subsection 4.2.2 it is
noted that Pa is nearly constant with flight speed (V). However, the assumption
of pη being constant with V is roughly valid only when the airplane has a variable
pitch propeller. For a fixed pitch propeller pη varies significantly with V (Fig.4.5a)
and the assumption of Pa being constant with V is not appropriate in this case.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 6
Lecture 23
Performance analysis II – Steady climb, descent and gl ide – 3
Topics
6.6. Climb hydrograph
6.7. Absolute ceiling and service ceiling
6.8 Time to cl imb
6.9 Steady descent
6.10 Glide
6.10.1 Glide performance – minimum angle of glide, minimum rate of sink
and maximum range and endurance in glide.
6.11 Glide hodograph
6.6 Climb hodograph
From Fig.6.1 it is observed that in a climb, the vertical velocity is the rate of climb
(Vc) and the horizontal velocity is Vh. From the discussion in section 6.5 it is
observed that for a chosen altitude, the vertical velocity (Vc) and the horizontalvelocity (Vh) change with the flight speed (V). A plot of the values of Vc and Vh at
a particular altitude, in which Vc is plotted on y-axis and Vh is plotted on the x-axis
is called ‘Climb hodograph’. Figure 6.5 shows a hodograph, based on the sea
level climb performance (Fig.6.3) of a jet airplane.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
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Fig.6.5 Climb hodograph
In a hodograph the line, joining the origin to a point on the curve, has the length
proportional to the flight velocity (V) and the angle this line makes to the
horizontal axis (Vh - axis) is the angle of climb ( ). This becomes evident when it
is noted that Vc and Vh are the components of the flight velocity (V) (see
Fig.6.1).
A line from the origin which is tangent to the hodograph gives the value of max
and also the velocity corresponding to it (Fig.6.5). Actually, a climb hodograph
gives complete information about the climb performance at the chosen altitude
especially max
, Vmax
, (R/C)max
, (R/C)max
, V(R/C)max
, (R/C)max
and Vmax
. These
quantities are marked in Fig.6.5
6.7 Absolute ceiling and service ceiling
Figures 6.3e and 6.4d present the variations of maximum rate of climb,
(R/C)max
, with altitude. It is observed that (R/C)max
decreases as the altitude
increases. The altitude at which the (R/C)max is zero is called ‘Absolute ceiling’. It
is denoted by h max
. At this altitude level flight is possible only at one speed(see
sec. 5.9).
Near absolute ceiling, the rate of climb is very small and the time to climb
becomes very large.It is not possible to reach the absolute ceiling (see remark (ii)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
in section 6.8). Hence, for practical purposes an altitude at which the maximum
rate of climb is 100 ft /min(30.5 m/min) is used as ‘Service ceiling’.
To obtain the absolute ceiling and service ceiling the values of (R/C)max
at
different altitudes are plotted as shown in Figs.6.3e and 6.4d. Subsequently, the
(R/C)max
vs h curve is extrapolated till (R/C)max
= 0. The altitude at which (R/C)max
equals zero is the absolute ceiling. The altitude at which (R/C)max
equals 100
ft/min (30.5 m/min) is the service ceiling. From Fig.6.3e and Appendix ‘B’ the
absolute ceiling and service ceiling for the jet transport are 11.95 and 11.71 km
respectively. From Fig.6.4d and Appendix A the values of these ceilings for a
piston engined airplane are 5.20 and 4.61 km respectively.
6.8 Time to cl imb:
From the knowledge of the variation of rate of climb with altitude, the time
required (t) to climb from an altitude h1 to h2 can be calculated as follows.
dh dhV = or dt =c
dt Vc
Hence, t =2
1
hdh
Vch (6.31)
The rate of climb (Vc) in Eq.(6.31) depends on the speed and altitude at which
the climb takes place. The appropriate values of Vc can be taken from plots
similar to those given in Figs.6.3e or 6.4d.
Remarks:
i) It may be noted that in a climb which attempts to fly at (R/C) max at each altitude,
the flight velocity, V(R/C)max , increases with altitude (Figs.6.3c and 6.4c).
Consequently, such a flight is an accelerated climb and the values of Vc obtained
using steady climb analysis will need to be appropriately corrected for the
acceleration (see section 8.3.2 on accelerated climb).
ii) As an exercise the student should plot the height (h) on y-axis and the time to
climb (t) on x-axis. It is observed that this curve reaches the absolute ceiling
(hmax) in an asymptotic manner. In other words, the time taken to reach absolute
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
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ceiling would be infinite and not practically attainable. Hence, service ceiling is
used for practical purposes.
6.9 Steady descent
Figure 6.6 shows an airplane in a steady descent. In such a flight thrust is
less than drag. The equations of motion are as follows.
T + W sin - D = 0 (6.32)
L – W cos = 0 (6.33)
Fig.6.6 Descent or glide
Hence, sin =D - T
W (6.34)
Rate of descent ( Vd ) =D-T
VW
(6.35)
The rate of descent is also called rate of sink and denoted by (R/S).
6.10 Glide
In a glide the thrust is zero. This may happen in a powered airplane due
to failure of engine while in flight. In a class of airplanes called gliders there is no
engine and the thrust is always zero. With thrust equal to zero, the following
equations of motion for glide, are obtained from Eqs.(6.32) and (6.33).
W sin - D =0 (6.36)
L - W cos =0 (6.37)
Hence,
sin = D / W and (6.38)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
Vd = V sin =D V
W (6.39)
The angle of glide, , is generally small. Hence, L W and one can write,
sin = =
CD D D
W L CL (6.40)
1/2 1/22L 2W
Vρ SC ρSCL L
and
1/2
3/2
CDV DV 2W DV =dW L ρS C
L
(6.41)
Remarks:
i) Multiplying Eq.(6.31) by V gives:
W V sin -D V = 0
Or W Vd – D V = 0
Noting that Vd is the rate of descent and equals dh / dt,
dhW - DV = 0
dt (6.42)
From Fig.6.6 it is to be noted that ‘V’ is along the glide path and hence in the
downward direction. Consequently in Eq.(6.42) dh/dt is negative as the altitude
is decreasing. As a result, the potential energy of the glider decreases with time.
This loss of potential energy is utilized to provide for the energy required to
overcome the drag (the second term in Eq.6.42). Hence, for a glider to stay aloft,
it must be brought to a certain height and speed before it can carry out the glide.
This is done by launching the glider by a winch or by towing the glider by another
powered airplane.
6.10.1 Glide performance – minimum angle of glide and minimum rate of
sink and maximum range and endurance in glide
The performance in a glide is stated in terms of the following four quantities.
(a) Minimum angle of glide (γmin) (b) Minimum rate of climb ((R/S)min or Vdmin).
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
(c) The distance covered in glide from a certain height or ’range in glide (Rglide)’
(d) The time elapsed in descending from a given height or ‘Endurance in glide
(Eglide)’.
From Eq.(6.40) the minimum angle of glide (γmin), occurs when CD / CL is
minimum or at CL = CLmd . From Eq.(6.41) the minimum rate of sink (R/S)min or
Vdmin occurs when CD / 3/2CL
is minimum or at CL = CLmp . This can be
understood from the following alternate explanation. When a glider sinks, it is
expending energy to overcome the drag, which comes from the potential energy
initially imparted to it. Thus, the rate of sink would be minimum when the rate of
power consumption is minimum and this occurs when V equals Vmp .
Gliders with very low rate of sink (around 0.5 m/s) are called ‘Sail planes’. From
Eq.(6.41) it is observed that a low rate of sink is achieved by (a) low wing loading
(b) low CDOwith smooth surface finish and (c ) large aspect ratio (16 to 20) to
reduce K. Note from Eq.(5.24a ) that (CD / 3/2CL
)min depends on 1/4C
DO and K3/4.
If a glider is left at a height ‘h’ above the ground, then the horizontal distance
covered in descending to the ground is called ‘Range in glide’ and denoted by
Rglide. Assuming γ to be constant during the glide, the range in glide can beexpressed as:
L
D
Ch hR = = hglide
tan C (6.43)
Thus, the range in glide would be maximum when the flight is at LC
corresponding to CLmd or at V= Vmd.
The time to descend from a height h is called Eglide. Assuming Vd to be constant,
Eglide equals (h / Vd). The quantity Eglide would be maximum when the descent
takes place at CL = CLmp or V = Vmp.
It is evident from the above discussion that the flight speeds for min
and (R/S) min
are different .
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
6.11 Glide hodograph
In section 6.6, the climb hodograph was discussed. Similarly, a glide hodograph
is obtained when horizontal velocity (Vh) is plotted on the x-axis and the rate of
sink (Vd) is plotted on the y-axis. A typical diagram is shown in Fig.6.7. Such a
diagram gives complete information about glide performance at an altitude
especially, min
, Vmin , (R/S)min, V(R/S)min and (R/S)min .
Fig.6.7 Glide hodograph
Example 6.4
A glider weighing 4905 N has a wing area of 25 m2,DO
C = 0.012, A = 16
and e = 0.87. Determine (a) the minimum angle of glide, minimum rate of sink
and corresponding speeds under sea level standard conditions (b) the greatest
duration of flight and the greatest distance that can be covered when glided froma height of 300 m. Neglect the changes in density during glide.
Solution:
(a) D DO2L
2C 1LC = + = 0.012 + C A e 3.14 × 16 ×0.87
C
= 0.012 + 0.023 CL
2
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 8
12
DO
DO
L min
D min
L(R/S)min
D(R/S)min
C = C = (0.012/0.023) = 0.721Lmd
C = 2C = 0.024
C = C = 3 C = 1.25Lmp Lmd
C = 4C = 0.048
Hence,
= (C /C ) = 0.024 / 0.721min D L min =0.0332 radians or 1.9o
12
3/22W(R/S) = (C /C )min D minLρS
Noting that the density ()has been assumed to be constant and equal to that at
sea level i.e. = 1.225 kg/m3 , the above equation gives :
122 × 4905 0.048
(R/S) = = 0.615 m/smin 3/21.225 × 25 1.25
12
1 12 2
2 × 4905V = = 21.05 m/s.min
1.225 × 25 × 0.721
2W 2 × 4905
V = = = 16m/s(R/S)min ρ S C 1.225 × 25 ×1.25Lmp
(b) The greatest distance, in descending from 300 m to sea level, ( Rglide)max,
is (note is assumed constant during glide) :
( Rglide)max = 300 /0.0332 = 9040 m = 9.04 km.
Longest time taken in descending from 300 m to sea level (Eglide
)max
is (note R/S
is assumed constant during glide) :
(Eglide)max = 300/0.615 = 487 s = 8 min 7s.
Note:
The rate of sink, in a flight when the greatest distance is covered, is higher than
the minimum rate of sink. Hence, the time of flight will be shorter in this case than
in a flight for longest endurance in glide. From the above data, the student may
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 9
show that (R/S) in the flight corresponding to (Rglide)max is 0.7015 m/s and the
endurance in this flight is 427 s.
Remarks:
i) If the glide takes place from a sufficiently high altitude (as may happen for an
airplane having an engine failure in cruise), the rate of sink (R/S) cannot be taken
as constant during the descent. Equation (6.41) should be used to calculate the
rates of sink at various altitudes.
ii) The time elapsed during glide (Eglide), in a general case is given by:
2
1
hdh
E =glideVdh
; (6.44)
where Vd is the rate of descent corresponding to an altitude ‘h’.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 6
Exercises
6.1 An airplane powered by a turbojet engine weighs 180,000 N, has a wing area
of 50 m2 and the drag polar is CD = 0.016 + 0.048CL2. At sea level a rate of
climb of 1200 m/min is obtained at a speed of 150 m/s. Calculate the rate of
climb at the same speed when a rocket motor giving an additional thrust of
10,000 N is fitted to the airplane.
(Answer: 1702 m /min.)
6.2 A glider having a wing loading of 185 N / m2 has the following drag polar.
CL 0.0 0. 1 0.2 0.4 0.6 0.8 1.0 1.2 1.4
CD 0.0145 0.014 0.0155
0.0183
0.0231
0.0299
0.0385 0.0491 0.062
Obtain the minimum rate of sink, minimum angle of glide and corresponding
speeds at sea level.
(Hint: Obtain CD
/ CL and C
D / C
L
3/2 from the given data, plot them, obtain
(CD / CL
) min , (C
D / CL
3/2)min and proceed.)
(Answers: (R/S)min
= 0.647 m/s,min
= 2.13o, V(R/S)min
= 54.2 kmph,
V min
= 71.35 kmph)
6.3 Consider a subsonic jet airplane. Assume that (a) thrust available (Ta) is
roughly constant, (b) L ≈ W in climb or the drag in climb (D) is roughly equal to
the drag in level flight and (c) the drag polar is parabolic. With these
assumptions and from exercise 5.4 which gives:
DOV A W2D = A W( ) + , A = C K,
2Vmd (V/V )md
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-6
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
show that (V / Vmd
) for (R/C)max
i.e. (V / Vmd
)(R/C)max
is given by:
T T 2 2a a± ( ) +12AV W W=
V 6Amd (R/C)max
Further taking DOC = 0.016 and K = 0.05625 or A = 0.03 obtain the following
table.
Ta / W 0.2 0.15 0.1 0.06
V
Vmd (R/C)max
1.54 1.36 1.16 1.0
Note that, as the altitude increases, Ta / W decreases and as a consequence
V
Vmd (R/C)max
tends to 1. At absolute ceilingV
Vmd (R/C)max
= 1 but
(R/C)max is zero !.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 7
Performance analysis III – Range and endurance
(Lectures 24-26)
Keywords: Range; endurance; safe range; gross still air range; Breguet
formulae; cruising speed and altitude; cruise climb; effect of wind on range.
Topics
7.1 Introduction
7.2 Definitions of range and endurance
7.2.1 Safe range
7.2.2 Head wind, tail wind, gust and cross wind
7.2.3 Gross still air range (GSAR)
7.3 Rough estimates of range and endurance
7.4 Accurate estimates of range and endurance
7.4.1 Dependence of range and endurance on flight plan and remark on
optimum path
7.4.2 Breguet formulae for range and endurance of airplanes with engine-
propeller combination and jet engine
7.4.3 Discussion on Breguet formulae – desirable values of lift coefficient
and flight altitude
7.4.4 Important values of lift coefficient
7.4.5 Influence of the range performance analysis on airplane design
7.5 Range in constant velocity - constant altitude flight (Rh,v)
7.6 Cruising speed and cruising altitude
7.7 Cruise climb
7.8 Effect of w ind on range and endurance
References
Excercises
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
Chapter 7
Lecture 24
Performance analysis III – Range and endurance – 1
Topics
7.1 Introduction
7.2 Definitions of range and endurance
7.2.1 Safe range
7.2.2 Head wind, tail wind, gust and cross wind
7.2.3 Gross still air range (GSAR)
7.3 Rough estimates of range and endurance
7.4 Accurate estimates of range and endurance
7.4.1 Dependence of range and endurance on flight plan and remark on
optimum path
7.1 Introduction
Airplane is a means of transport designed to carry men and materials
safely over a specified distance. Hence, the fuel required for a trip or the distance
covered with a given amount of fuel are important items of performance analysis.
Similarly, airplanes used for training, patrol and reconnaissance would be
required to remain in air for a certain period of time. Thus, the fuel required to
remain in air for a certain length of time or the time for which an airplane can
remain in air with a given amount of fuel are also important aspects of
performance analysis. These two aspects viz. distance covered and the time for
which an airplane can remain in air are discussed under the topic of range and
endurance and are the subject matter of this chapter.
7.2 Definitions of range and endurance
Range (R) is the horizontal distance covered, with respect to a given point
on the ground, with a given amount of fuel. It is measured in km. Endurance (E)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
is the time for which an airplane can remain in air with a given amount of fuel. It
is measured in hours. The above definition of range is very general and terms
like safe range and gross still air range are commonly used. These terms include
details of the flight plan and are explained in the subsequent subsections.
7.2.1 Safe range
It is the maximum distance between two destinations over which an
airplane can carry out a safe, reliably regular service with a given amount of fuel.
This flight involves take-off, acceleration to the speed corresponding to desired
rate of climb, climb to the cruising altitude, cruise according to a chosen flight
plan, descent and landing. Allowance is also given for the extra fuel requirement
due to factors like (i) head winds (see next subsection) normally encountered en-
route (ii) possible navigational errors (iii) need to remain in air before permission
to land is granted at the destination and (iv) diversion to alternate airport in case
of landing being refused at the scheduled destination.
7.2.2 Head wind, tail wind, gust and cross wind
Generally the performance of an airplane is carried out assuming that the flight
takes place in still air. However the air mass may move in different directions.
Following three cases of air motion are especially important.
(a) Head wind and tail wind: In these two cases the direction of motion of air (Vw)
is parallel to the flight direction. If Vw is opposite to that of the flight direction, it is
called ‘Head wind’. When Vw is in the same direction as the flight direction, it is
called ‘Tail wind’ (Fig.7.1a). In the presence of wind, the velocitiy of the airplane
with respect to air (Va) and that with respect to ground (Vg) will be different. For
the head wind case, Vg = Va - Vw, and for the tail wind case, Vg = Va + Vw.
(b) Gust: When the velocity of the air mass is perpendicular to flight path and
along the vertical direction, it is called gust. Here the velocity of gust is denoted
by Vgu (Fig.7.1b). This type of air movement would result in a change of the
angle of attack of the airplane.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 4
(c ) Cross wind: When the velocity of the air mass is perpendicular to flight path
and parallel to the sideward direction, it is called ‘Cross wind’. Here it is denoted
by ‘v’ (Fig.7.1c).
Fig.7.1a Head wind and tail wind
Fig.7.1b Gust
Fig.7.1c Side wind
7.2.3 Gross still air range (G.S.A.R.)
The calculation of safe range depends on the route on which the flight
takes place and other practical aspects. It is not a suitable parameter for use
during the preliminary design phase of airplane design. For this purpose, gross
still air range (G.S.A.R.) is used. In this case, it is assumed that the airplane is
already at the cruising speed and cruising altitude with desired amount of fuel
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
and then it carries out a chosen flight plan in still air, till the fuel is exhausted.
The horizontal distance covered in this flight is called ‘Gross still air range’. In the
subsequent discussion the range will mean gross still air range.
Remark:
As a guideline G.S.A.R. is roughly equal to one and a half times the safe range.
7.3 Rough estimates of range and endurance
If the weight of the fuel available (Wf in N) and the average rate of fuel
consumption during the flight are known, then the rough estimates of range (R)
and endurance (E) are given as follows.
R = Wf x (km / N of fuel)average
(7.1)
E = Wf x (hrs / N of fuel) average
(7.2)
The estimation procedure is illustrated with the help of example 7.1.
Example 7.1
An airplane has a weight of 180,000 N at the beginning of the flight and
20% of this is the weight of the fuel. In a flight at a speed of 800 kmph the lift to
drag ratio (L/D) is 12 and the TSFC of the engine is 0.8. Obtain rough estimates
of the range and endurance.
Solution:
W1 = Weight at the start of the flight = 180,000N
Wf = Weight of the fuel = 0.2x180,000 = 36,000N
W2 = Weight of the airplane at the end of the flight
= 180,000 - 36,000= 144,000N.
Hence, the average weight of the airplane during the flight is :
Wa =
180000 + 144000= 162000N
2
Consequently, the average thrust (Tavg
) required during the flight is:
Tavg
= Wa / (L / D) = 162000/12 = 13500 N
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
The average fuel consumed per hour is:
Tavg x TSFC = 13500 x 0.8 = 10800 N
Since the average speed is 800 kmph, the distance covered in 1 hr is 800 km.
Noting that the fuel consumed in 1 hr is 10,800 N, gives:(km / N of fuel)
average = 800/10800.
Consequently,800
R = 36000 × = 2667km10800
and
the endurance E = 36000 x1
= 3.33 hrs.10800
7.4 Accurate estimates of range and endurance
For accurate estimates of range and endurance, the continuous variation
of the weight of the airplane during the flight and consequent changes in the
following quantities are considered.
(a) The thrust required (or power required),
(b) TSFC (or BSFC) and
(c) Flight velocity and lift coefficient.
It may be recalled from subsection 4.2.4, that is the specific fuel consumption
(SFC) of an engine delivering shaft horse power to a propeller is denoted by
BSFC and the SFC of a jet engine is denoted by TSFC. The units of BSFC and
TSFC are respectively N/kW-hr and N/N-hr (or hr -1).
The steps to accurately estimate the range and endurance are as follows.
Let W be the weight of the airplane at a given instant of time and Wfi be the
weight of the fuel consumed from the beginning of the flight up to the instant
under consideration.
Then, W = W1 - Wfi (7.3)
where, W1 = weight of the airplane at the start of the flight.
Let dR and dE be the distance covered in km and the time interval in hours
respectively, during which a small quantity of fuel dWf is consumed. Then,
dR = dWf x (km/N of fuel) (7.4)
and dE = dWf x (hrs/ N of fuel) (7.5)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
Following Ref.1.5, chapter 4, the Eqs.(7.4) and (7.5) are rewritten as:
dR = dWf km/hr
N of fuel /hr
(7.6)
and dE = dWf x { 1 / (N of fuel / hr)} (7.7)
It may be pointed out that (a) km/hr = 3.6 x V, where V is the flight speed in m/s.
(b) the fuel / hr in Newtons is equal to BSFC x BHP for an airplane with engine-
propeller combination and equal to TSFC x T for a jet airplane.
Hereafter, the airplane with engine-propeller combination is referred to as “E.P.C”
and the jet airplane as “J.A” Note that in the case of an engine-propeller
combination, the engine could be a piston engine or a turboprop engine and in
the case of a jet airplane the engine could be a turbofan or a turbojet engine.
Equations (7.6) and (7.7) can be rewritten as :
3.6VdR = dWf
BSFC × BHP For E.P.C. (7.8)
and3.6V
dR = dWf TSFC × T
For J.A. (7.8a)
dWf dE =BSFC × BHP
For E.P.C. (7.9)
anddWf dE =
TSFC × T For J.A. (7.9a)
Recall from section 5.2 that in a level flight,
C 1 2DT = D = W , L = W = ρV SCLC 2L
,
1122
L 0 L
2W 2WV = =
ρS C σ ρ S C
(7.10)
Using, ρ0 = 1.225 kg/m3 yields:
W 1/2V = 1.278 ( )σ SCL
. (7.11)
Substituting for T and V from Eqs.(7.10) and (7.11), the expression for BHP is:
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 8
12 3/2
D L1 T V 1 3/2BHP = = W / η (σ S) C / Cpη 1000 782.6p
(7.12)
where p is the propeller efficiency.
During the analysis of range, the rate of change of weight of the airplane is only
due to the consumption of fuel. Hence,
dWf = - dW
Substituting for V, BHP, T and dWf in Eqs. (7.8),(7.8a),(7.9) and (7.9a) gives:
-3600 η dWpdR =
BSFC × W (C /C )D L For E.P.C. (7.13)
and1/2 1/2
- 4.6 dWdR =
TSFC (σ S W) (C /C )D L
For J.A. (7.13a)
1/2782.6 η (σS) dWpd E = -
3/2 3/2BSFC × W C /CD L
For E.P.C. (7.14)
and
-dWdE =
TSFC × W C /CD L For J.A (7.14a)
Let W2 be the weight of the airplane at the end of the flight. Integrating
Eqs.(7.13), (7.13a), (7.14) and (7.14a), the range and endurance are given as:
2 2
1 1
W W3600 η dWp
R = dR = -BSFC × W (C /C )D LW W
For E.P.C. (7.15)
and2
1122
1
W- 4.6 dW
R =TSFC (σ S W) (C /C )DW L
For J.A (7.15a)
1/22 2
1 1
W W 782.8 η (σS) dWpE = dE = -
3/2 3/2BSFC × W (C /C )DW W L
For E.P.C. (7.16)
and2
1
W-dW
E =TSFC × W (C /C )D LW
For J.A. (7.16a)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 9
7.4.1 Dependence of range and endurance on flight plan and remark on
optimum path
The set of Eqs.(7.15),(7.15a),(7.16) and (7.16a) or (7.8), (7.8a), (7.9) and (7.9a)
when integrated, give the range and endurance. However, while doing this, itshould be noted that the weight of the aircraft decreases continuously as the fuel
is consumed. Further, the flight is treated as steady level flight and hence, T = D
and L= W must be satisfied at each instant of time. Consequently, the thrust and
power required and the flight speed may change continuously. Hence, it is
necessary to prescribe the flight plan i.e., the manner in which the velocity
changes with time during the flight. The following three types of flight plans can
be cited as examples.
(a) Level flight at a constant velocity. In this flight, the lift coefficient decreases
gradually as the weight of the airplane decreases (Eq.7.10). Simultaneously, the
thrust required also decreases continuously.
(b) Level flight with constant lift coefficient (or constant angle of attack) . In this
flight, in accordance with Eq.(7.10), the flight velocity and the thrust required
decrease continuously as the weight of the airplane decreases.
(c) Level Flight with constant thrust. In this case, the continous decrease in the
airplane weight during the flight, requires that the flight velocity and the lift
coefficient (CL) be adjusted so that at each instant of time, the thrust balances
the drag and the lift balances the weight.
As mentioned earlier, the airplanes are commercial means to transport men and
materials. Hence, maximization of range and endurance are important
requirements. However, the right hand sides of Eqs.(7.15),(7.15a),(7.16) and
(7.16a) involve integrals. The optimization of an integral is different from the
optimization of an expression. The latter is done by taking the derivative of the
expression and equating it to zero. Whereas, in the case of an integral, it is to be
noted that the value of the integral depends on how the integrand varies with the
independent variable. This variation, in mathematical terms, is called a path. For
example, as mentioned above, the range will depend on the flight plan viz.
constant angle of attack flight or constant velocity flight or constant thrust flight.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 10
The problem of optimization is to find out the path that will maximize the integral.
The branch of Mathematics which deals with optimization of integrals is called
‘Calculus of variation’. This topic is outside the scope of the present introductory
course. Interested reader may refer, chapter 20 of Ref.7.1.Remark:
It can be shown, using calculus of variation, that if the specific fuel consumption,
propeller efficiency and altitude are assumed constant, then the maximum range
is obtained in a flight with constant lift coefficient. With these assumptions
Eqs.(7.15),(7.15a),(7.16) and (7.16a) become easy to integrate. The expressions
for range and endurance, obtained with these assumptions, are called ‘Breguet
formulae’. These are derived in the next subsection. It may be pointed out that
Breguet was a French pioneer in aeronautical engineering.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 7
Lecture 25
Performance analysis III – Range and endurance – 2
Topics
7.4.2 Breguet formulae for range and endurance of airplanes with engine-
propeller combination and jet engine
7.4.3 Discussion on Breguet formulae – desirable values of lift coefficient
and flight altitude
7.4.4 Important values of lift coefficient
7.4.2 Breguet formulae for range and endurance of airplanes with engine-
propeller combination and jet engine
The derivations of these formulae are based on the assumptions that during
the flight:
(i) BSFC or TSFC is constant
(ii) p is constant for engine propeller combination (E.P.C).
(iii) altitude is constant
(iv) CL is constant and
(v) flight Mach number is below critical Mach number so that the drag polar is
independent of Mach number.
With, these assumptions, certain terms in Eqs.(7.15) and (7.15a) can be taken
outside the integral and the equations reduce to:
2
1
w3600 η dWp
R = -
WBSFC (C /C )D wL For E.P.C
and2
1
W-4.6 dW
R =1/2 1/2 1/2TSFC(σS) (C /C ) WD WL
For J.A.
Hence,
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
8289.3 η Wp 1R = log10
BSFC (C /C ) WD L 2
For E.P.C. (7.17)
and
1/2W W9.2 1/21 2R = ( ) 1-
1/2 σS W1TSFC(C /C )D L
For J.A. (7.17a)
Similarly, from the above assumptions, Eqs.(7.16) and (7.16a) reduce to :
1/2 2
1
W782.6 η (σS) dWp
E = -3/23/2 WBSFC × C / CD WL
For E.P.C
and2
1
W1 dW
E = -TSFC(C / C ) W
D L W
For J.A.
Hence,
1/2 1/21565.2 η σSp 1/2E = [ ] W - W2 13/2 W1BSFC × C /CD L
For E.P.C. (7.18)
and 10W2.303 1E = log
TSFC (C / C ) WD L 2
For J.A. (7.18a)
7.4.3 Discuss ion of Breguet formulae – desirable values of lift coefficient
and flight altitude
The following conclusions can be drawn from the above expressions for
range and endurance viz. Eqs.7.17, 7.17a, 7.18 and 7.18a.
(1) For range and endurance to be high, it is evident that p should be high and
the TSFC and BSFC should be low.
(2) Desirable values of lift coefficients for an airplane with engine-propeller
combination: The endurance is maximum (Eq.7.18) when the lift coefficient is
such that CD/C
L
3/2 is minimum, i.e., C
L = C
Lmp. This can be understood from the
fact, that with BSFC being assumed constant, the rate of fuel consumption per
hour would be minimum, in this case, when the power required is minimum.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
From Eq.(7.17), for range to be maximum, in this case, CD/C
L should be
minimum or CL = C
Lmd. This can be understood from the fact that the range, in
this case is proportional to V/THP or L / D. Hence, range is maximised when CL
corresponds to minimum drag.
(3) Desirable values of lift coefficients for a jet airplane: From Eq.(7.18a) it is
observed that the endurance is maximum when CD / CL
is minimum or CL= C
Lmd.
This can be understood from the fact that with TSFC being assumed constant,
the fuel flow rate per hour would be minimum when the thrust required is
minimum.
From Eq.(7.17a) the range, in this case is maximum when CD/C
L
1/2 is minimum.
This can be understood from the fact that the range, in this case, is proportional
to (V / T) or 1/2
L DC /C . The CL corresponding to (C
D/C
L
1/2
)min
is denoted here by
CLmrj
.
(4) Desirable values of flight altitude : Equation (7.17a), also shows that for a jet
airplane, the range would be high, when (a) the wing loading (W/S) is high and
(b) density ratio () is low or the altitude is high. Hence, the jet airplanes have
wing loading of the order of 4000 to 6000 N/m2, which is much higher than that
for the low speed airplanes which have a wing loading of 1000 to 2500 N/m2
. The
jet airplanes also cruise at high altitude (10 to 12 km) which is not much below
the ceiling altitude of 12 to 14 km for these airplanes.
From Eq.(7.18) it is observed that the endurance of an airplane with engine-
propeller combination is high when (a) the wing loading is low and (b) is high or
flight takes place near sea level.
It may be added that the final wing loading chosen for an airplane is a
compromise between requirements of cruise, climb, take-off and landing. The
take-off and landing distances increase in direct proportion to the wing loading
(subsections 10.4.5 and 10.5.3), and hence, a high wing loading is not desirable
from this point of view.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 4
Remarks:
i) If the drag polar is parabolic, an expression for CLmrj
can be derived as follows.
1/2 1/2
1/2
2
3/2
C = C + K CD DO L
CC DODHence, = + K CL
C CL L
d(C / C ) C 3D -3/2 1/2DOL = C + K C = 0Lmrj LmrjdC 2 2L
Or CLmrj = (C
DO / 3K)1/2
7.4.4 Important values of lift coefficientThe points on the drag polar at which C
L is equal to C
Lmax, C
Lmp, C
Lmd and C
Lmrj
are shown in Fig.7.2. The importance of these values of lift coefficient can be
reemphasized as follows.
(i) The maximum lift coefficient (CLmax
) decides the stalling speed which is one of
the criterion for the minimum speed of the airplane. It also affects the minimum
radius of turn (see subsection 9.3.3) and the take-off and landing distances (see
subsections 10.4.5 and 10.5.3)(ii) The lift coefficient corresponding to minimum power required (CLmp) influences
the performance of airplanes with engine-propeller combination. It decides the
flight speeds corresponding to maximum rate of climb, minimum rate of sink and
maximum endurance of these airplanes.
(iii) The lift coefficient corresponding to minimum thrust required ( LmdC ) is also
the value of CL at which (L/D) is maximum. From Fig.7.2 it is observed that the
slope of a line joining the origin to a point on the curve, is equal to (CL
/ CD
). At,
CL = LmdC this line, from the origin, is tangent to the drag polar and has the
maximum slope (Fig.7.2). The value of CLmd decides the flight speed for
maximum range of an airplane with engine-propeller combination and the
maximum endurance of a jet airplane.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
(iv) The lift coefficient corresponding to max
1/2
L DC /C or CLmrj
decides the flight
speed for maximum range of jet airplanes.
Fig.7.2 Important points on a drag polar
Example 7.2
An airplane having an engine-propeller combination weighs 88,290 N and
has a wing area of 45 m2
. Its drag polar is given by: CD = 0.022 + 0.059C
L
2.
Obtain the maximum range and endurance at sea level in a steady level flight at
a constant angle of attack from the following additional data.
Weight of fuel and oil = 15,450 N, BSFC = 2.67 N/kW-hr,
propeller efficiency (p) = 85%.
Note: Along with the fuel, the lubricating oil is also consumed and this fact is
taken into account in this example, by specifying the weight of the oil along with
the weight of fuel.
Solution:
W1 = 88290 N, W2 = 88290 – 15450 = 72840 N
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
Since BSFC, p and CL are constant, the maximum range and endurance occur
when CL has the values of C
Lmd and CLmp respectively.
C Lmd = (CDO/K)
1/2= (0.022/0.059)1/2 = 0.6106,
CDmd
= 2 CDO = 0.044
C Lmp = (3CDO/K)
1/2= (3 x 0.022/0.059)1/2 = 1.058, C
Dmp = 4 CDO = 0.088
Hence, (CD /C
L)min
= 0.044/0.6106 = 0.0721
and 3/2 3/2(C /C ) = 0.088/(1.058)D minL= 0.0808
From Eq. (7.17):
8289.3 η W 8289.3 × 0.85 88290p 1R = log = × log10 10BSFC × (C /C ) W 2.67 × 0.0721 72840D L min 2
= 3058 km.
Remark:
Since CL is constant during the flight, the flight velocity and the power required
change as the fuel is consumed. In the present case, the following results
illustrate the changes.
Velocity at the beginning of the flight:
V1= 1/2 1/22W /ρ S C = (2 × 88290/1.225 ×45 × 0.6106)1 Lmd = 72.41 m/s.
= 260.7kmph.
Velocity at the end of flight:
= 1/222W /ρ S CLmd
= (2 x 72840/1.225 x 45 x 0.6106)1/2 = 65.8 m/s = 236.8 kmph.
Power required in the beginning of the flight:
1 D L 11 1W × C /C ×VT × V 88290 × 0.044
= = = ×72.41= 460.7kW1000 1000 1000 × 0.6106
Power required at the end of flight:
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
2 22 2 D LW × C /C ×VT × V 72840 × 0.044
= = = × 65.8 = 345.5kW1000 1000 1000 × 0.6106
Maximum endurance:
From Eq.(7.18) the maximum endurance is :
1/2 1/2
1/2 1/2
η W1565.2 σSp 1E = -1max 3/2BSFC W W1 2C /CD L min
1565.2 ×0.85 1.00 ×45 88290= -1 = 14.06hrs
2.67 × 0.0808 88290 72840
In this flight CL equals 1.058. Proceeding in a manner similar to the remark
above, it can be shown that the speeds at the beginning and end of the flight forE
max are 197.8 kmph and 179.7 kmph respectively. The power outputs required
at the beginning and the end of this flight are 402.8 kW and 302.0 kW
respectively.
Example 7.3
A jet airplane has a weight of 922,140 N and wing area of 158 m2. The
weight of the fuel and oil together is 294,300 N. The drag polar is given by:
2C = 0.017 + 0.0663 CD L
Obtain the maximum range in constant CL flight at an altitude of 10 km assuming
the TSFC to be 0.95 hr -1.
Solution:
In a flight with constant C L the maximum range occurs when
1/2C = C = (C / 3K)L Lmrj D0
1/20.017
C = = 0.292Lmrj3 × 0.0663
CDmrj
= 0.017 + 0.0663 (0.292)2 = 0.02265
1/2 1/2( )C /C = 0.02265 / 0.292 = 0.04192Dmrj Lmrj
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 8
σ at 10 km altitude is: 0.3369
From Eq.(7.17a):
1/21/2
1/2
1/2 1/2
W W9.2 1 2R = 1 -maxσS W1TSFC × C /CD L
9.2 922140 922140-294300= × 1 - = 5317 km
0.95 × 0.04192 0.3369 × 158 922140
Remarks:
i)The flight velocity corresponding to a CL of 0.292 at an altitude of 10 km is equal
to: [2 x 922140 / (158 x 0.413 x 0.292)]
1/2
= 311.1 m/s.
The speed of sound at 10 km is 299.5 m/s. Thus the Mach number at this speed
would be 311.1/299.5 = 1.04. This value is definitely higher than the critical Mach
number of the airplane. Consequently, the prescribed drag polar is not valid. The
CD will actually be much higher and the range much lower.
As an alternative, let the critical Mach number be taken as 0.85 and the range be
calculated in a flight at constant CL which begins at this Mach number.
Consequently, V = 0.85 x 299.5 = 254.5 m/s.
Hence,CL = (2 x 922140/0.413 x 158 x 254.52) = 0.436
Consequently, CD = 0.017 + 0.0663 x (0.436)2 = 0.0296
and 1/2 1/2C /C = 0.0296/0.436 = 0.0448D L
The range in a constant CL flight with CL=0.436 would be:
1/2 1/2
4975
9.2 922140 922140-294300
= × 1- km0.95 × 0.0448 0.3369×158 922140
.
(ii) The data given in this example, roughly corresponds to that of Boeing 727,
the famous jetliner of 1970’s. The value of TSFC corresponds to engines of that
period. The value of K equal to 0.0663 includes the change in K, when Mach
number lies in the transonic range.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 7
Lecture 26
Performance analysis III – Range and endurance – 3
Topics
7.4.5 Influence of the range performance analysis on airplane design
7.5 Range in constant velocity - constant altitude flight (Rh,v)
7.6 Cruising speed and cruising altitude
7.7 Cruise climb
7.8 Effect of w ind on range and endurance
7.4.5 Influence of the range performance analysis on airp lane design
In section 5.8 it was pointed out that the analysis of level flight
performance led to improvements in design of airplanes. Similarly, the analysis of
range also helped in improvements in airplane design in the following way.
The high speed airplanes are jet airplanes and for these airplanes the range
(R) is proportional to :
1 11/2TSFC C /CD L
or C1 1L1/2TSFC CD CL
Noting that 1/ 1/2LC
is proportional to flight speed (V),
C1 LR VTSFC CD
Since, high speed airplanes fly in lower stratosphere, where speed of sound is
constant,
C1 LR MTSFC CD
(7.19)
The quantityC1 L M
TSFC CD can be referred to as figure of merit (FM) for the
following reasons.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
(a) A low value of TSFC in an indicator of high engine efficiency and (b) A high
value of (CL/CD) is an indicator of high aerodynamic efficiency.
The figure of merit provided guidelines when the supersonic airplane
Concorde was being designed in early 1960’s.The subsonic jets of that periodlike Boeing 707 would fly around M = 0.8, have (L / D)
max around 16 and TSFC
around 0.9. These values would give the FM of 0.8x16/0.9 or 14.2. If Concorde
were to compete with subsonic jets, it needed to have a similar value of FM. The
fighter airplanes of that period flying at Mach number of two had TSFC of 1.5 and
(L/D) max
of 5. This would give FM of (2 x 5) /1.5 = 6.66 which was far too low as
compared to that for subsonic airplanes. Hence the targets for Concorde, which
was being designed for a Mach number of 2.2, were fixed at (L/D) max of 7.5 andTSFC of 1.2. This would give FM of 2.2(1/1.2) x 7.5 = 13.75, which was
comparable to the FM of subsonic airplanes. However, to achieve a TSFC of 1.2
at M =2.2, a large amount of research was carried out and the Olympus engine
used on Concorde was developed jointly by Rolls-Royce of U.K. and SNECMA of
France. Similarly, to achieve an (L/D) max of 7.5 at M = 2.2 needed a large amount
of computational and experimental effort. A picture of Concorde, a technological
marvel, is shown in Fig.7.3.It may be added that for Concorde the Mach number was limited to 2.2 as
the designers had chosen to use aluminum as structural material. At M = 3 the
FM could be greater than that of subsonic airplanes but the aerodynamic heating
would cause surface temperatures of around 300oC at which the strength and
modulus of elasticity of aluminum will be significantly reduced.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
Fig.7.3 Concorde
(Source: www.airplane-pictures.net)
The B787 (Fig.7.4) being brought out by Boeing and called ‘Dream liner’ has
M = 0.85, (L/D) max
of 22 and TSFC of 0.54 hr -1. These values of (L/D) max and
TSFC indicate steady improvements in aerodynamics and engine performance
over the last five decades .
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 4
Fig.7.4 Boeing 787 Dream liner
(Source: www.lotz.com)
7.5 Range in constant velocity - constant altitude flight (Rh,v)
The assumption of constant CL during cruise gives the longest range(R).
However, it is more convenient for the pilot to fly the airplane at a constant speed
or Mach number. He just needs to keep an eye on the airspeed indicator or
Machmeter and adjust other parameters like the angle of attack and engine
setting.
To derive an expression for range in level flight at constant speed (Rh,v),
an airplane with jet engine is considered and it is assumed that TSFC is
constant. Equation (7.8a) is the basic equation for range of a jet airplane. When
V is constant, the equation takes the following form.
2
1
1
W3.6 V dW
R =h vTSFC Tr
W
(7.20)
Tr = thrust required
Assuming a parabolic polar,
Tr =2 21 2K W KW2 21ρ V S C + = q S C + ; q = ρ VDO DO2 22 qSρ V S
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
Note: The dynamics pressure (q), is constant in a constant velocity and constant
altitude flight.
Substituting for Tr in Eq.(7.20) gives:
2
1
1
w3.6 V -dWR =h v 2TSFC q S CD0 1+aWw
where, a =K
2 2q S CD0
1
3.6 V -1 -1Or R = [tan a W - tan a W ]h v 1 2qS C TSFC aD0
(7.20a)
where W2 = weight of the airplane at the end of the flight.
LetWf ζ =W1
, where Wf = weight of fuel
Hence, W2 = W
1(1-) ;
Further, let E1 = W1/D1 = initial lift to drag ratio,
2
11 D1 DO 2 2
KWD = qSC = qS C +
q S
2
1DO
KW= qSC +
qS
CL1
= CL at start of flight =
W121 ρV S
2
= 1W
qS
Emax
= 1
2 K CD0
Noting that ,
-1 -1 -1 1 21 2
1 2
θ -θtan θ - tan θ = tan
1 + θ θ
,
Equation (7.20a) can be rewritten as :
-1 1 2h,v
1 2
DO
DO
aW - aW3.6VR = tan
1+aW WKTSFC qSC
qS C
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
1 2-1
2
1DO2 2
DO
a W - W3.6V= tan
WKTSFC KC1 + 1- ζ
C q S
1
DO-1max
2
12 2
DO
WKζ
C qS7.2E V= tan
WKTSFC1+ 1-ζ
C q S
Multiplying the denominator and numerator of the terms in square brackets by
qSCDO gives :
DO 1-1maxh,v 2 2
1 1DO
KC W ζ7.2E VR = tanKW KWTSFC
qSC + - ζqS qS
DO 1-1max
2
11
KC W ζ7.2E V= tan
WTSFCD - K ζ
qS
Dividing the numerator and denominator of the term in square brackets by D1,
gives :
1DO
-1max 1h,v
1 1
1
WKC ζ
7.2E V DR = tan
TSFC W W1 - K ζ
qS D
Or
-1max 1h,V
max L1 1
7.2E V E ζR = tan
TSFC 2E 1-KC E ζ
(7.21)
For an airplane with an engine-propeller combination, the range at constant
speed and constant altitude (Rh,v) is given as:
2
1
1
W3600 η dWp
R =h vBSFC T
W (7.22)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
Assuming BSFC and ηp to be constant and the drag polar as parabolic i.e.
1
2
22KW2T = ρ V S C +DO 2ρSV and substituting in Eq.(7.22) gives:
1
7200 η E ζp -1 1R = E tanh v maxBSFC 2E (1-KC E ζ)max L1 1
(7.23)
Remarks:
i) Comparing the ranges in the constant velocity and constant CL flights, Ref.1.1,
chapter 9, shows that the maximum range in a constant velocity flight is only
slightly lower than that in a constant CL flight.
ii) In actual practice BSFC (or TSFC) and ηp may vary during the cruise. If
detailed information about their variations is available, then better estimates of
range and endurance can be obtained by numerical integration of Eqs.(7.8),
(7.8a),(7.9) and (7.9a).
iii) Appendix ‘A’ section 6 considers the range and endurance performance of a
piston engined airplane at an altitude of 8000 feet (2438 m) in constant velocity
flights at different speeds. The variations in propeller efficiency and fuel
consumption are taken into account. It is seen that the endurance is maximum
around flight speed of 135 kmph. The range is maximum for flight speeds
between 165 to 185 kmph.
iv) Section 6 of Appendix ‘B’ considers the range and endurance performance of
a jet transport at an altitude of 36000 feet (10973 m) in constant velocity flights at
different speeds. The endurance is near its maximum value in the speed range of
684 to 828 kmph. The maximum range occurs around 240 m/s (864 kmph). The
corresponding Mach number is 0.82, which is slightly higher than the Mach
number beyond which the CDOand K begin to increase due to compressibilityeffects.
7.6 Cruising speed and cruising altitude
The cruising speed (Vcr
) and the cruising altitude (hcr
) together constitute
the combination at which the maximum range is obtained. To arrive at the values
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
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of Vcr
and hcr
the range is calculated at various speeds at a number of altitudes
and the plots as shown in Fig.7.5 are obtained. The dotted line in Fig.7.5 is the
envelop of all the curves. The speed and altitude at which the maximum of this
envelop occurs is called the most economical cruising speed and altitude. In
some cases this speed is rather low and a higher cruising speed may be chosen
from other considerations like, shorter flight time and speed appeal. i.e. a faster
airplane may be more appealing to the passengers even if it consumes more fuel
per kilometer of travel.
Fig.7.5 Determination of cruising speed and cruising altitude
7.7 Cruise climb
To prepare the back ground for the analysis of the cruise climb, consider
Eq.(7.8a) which gives the range of a jet airplane. i.e.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 9
2
1
W
W D
L
-3.6 V dWR =
CTSFC W
C
(7.24)
The TSFC is generally assumed to be constant during the flight. Furthersimplifications are needed to carry out the integration in Eq.(7.24). In the
constant altitude - constant CL flight considered in subsection 7.4.2, as the name
suggests, the lift coefficient (CL) is assumed constant during the flight. In this
case, to satisfy the requirement of L = W = 2
L
1ρV SC
2, the flight velocity is
decreased as the weight of the airplane decreases due to consumption of fuel
(see example 7.2). In the constant altitude – constant velocity flight, considered
in section 7.5, the flight speed (V) is held constant during the flight. In this case,
CL decreases as the fuel is consumed.
Equation (7.24) suggests a third possibility, other than the above two cases, of
both V and CL being held constant during the flight.In this case, to satisfy
L = W = L21
ρV SC2
, it has been suggested that the airplane be allowed to climb
slowly such that the decrease of atmospheric density ρ with altitude
compensates for the decrease of airplane weight due to consumption of fuel.
With these simplifications Eq.(7.24) gives :
1 2
D L D L
2
1
W
W
-3.6V dW 3.6VR = = ln W /W
TSFC C /C W TSFC C /C (7.25)
The flight is called ’Cruise climb’ as the altitude continuously increases during the
flight.
Remarks:
(i) Exercise 7.3 would show that for a jet airplane with Wf / W1 = 0.2 and starting
the cruise climb at h = 11 km, the range would be 5141 km and the change of
flight altitude between the end and the start of cruise climb would be only
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 10
1.415 km. Thus, it is observed that the change in the altitude between the start
and end of cruise climb is very small as compared to the distance covered and
the level flight equations (L = W and T = D) are valid.
(ii) It can be shown (Ref.1.1, chapter 5) that the range in a cruise climb is higherthan that in level flight at the altitude where the cruise climb begins.
(iii) In actual practice continuous increase in altitude may not be permitted by Air
Traffic Regulations. As an alternative, a stepped climb approximation may be
used i.e. the flight path is divided into segments of constant altitude flights with
stepped increase in altitude after certain distance.
(iv)In a cruise climb the thrust required would be
T = D = (1/2) ρV2 SC
D Since, the flight velocity and C
L (and hence C
D) are held constant, the thrust
required will be proportional to ambient density (ρ). It may be pointed out that in
lower stratosphere the engine output (thrust available) is also proportional to the
ambient density. Thus, in a cruise climb in lower stratosphere the thrust setting
required is also constant and it becomes a very convenient flight – the pilot has
just to set the Mach number and then the autopilot will take care of the flight.
7.8 Effect of wind on range and endurance In the foregoing discussion, it was assumed that the airplane moves in a
mass of air which is stationary with respect to the ground. However, in many
situations the air mass has a velocity with respect to the ground and the airplane
encounters head wind or tail wind. (see subsection 7.2.2 for definition of head
wind and tail wind). The wind velocity is denoted by VW . When VW is non-zero,
the velocity of the airplane with respect to the ground (Vg) and that with respect
to air (V
a
) are different. To analyze the effect of wind on airplane performance, it
may be pointed out that the aerodynamic characteristics of the airplane (lift, and
drag) and the engine characteristics depend on the velocity with respect to air
(Va), whereas the distance covered in the flight depends on the velocity with
respect to the ground (Vg). In the presence of head wind the velocity of the
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 11
airplane with respect to the ground will be lower than its velocity with respect to
air and the range decreases. For example, in a hypothetical case of head wind
being equal to the stalling speed, the airplane, in principle, can remain airborne
without moving with respect to the ground. The fuel will be consumed as enginewould produce thrust to overcome the drag, but no distance will be covered as
the airplane is hovering! When there is tail wind the range increases.
An expression for range with effect of wind can be derived as follows.
Consider a jet airplane. Let Rg be the range in the presence of wind.
Equation(7.8a) can be used to calculate Rg, but the quantity ‘V’ in that equation
should be replaced with Vg i.e. :
2 2
1 1
W
W W3.6 V dW 3.6 (V - V ) dWg aR = - =gTSFC × T TSFC × T
W W , VW in m/s
2 2
1 1
W W3.6 V dW dWaR = - - 3.6 V = R - 3.6V Eg w a wTSFC × T TSFC × T
W W (7.26)
where Ra is the range in still air = -2
1
W 3.6 V dWaTSFC × T
W
and E is the duration of flight in hours. Thus, with head wind the range decreases
by 3.6 Vw E. In example 7.1 the range is 2667 km and the endurance is 3.33
hours. If a head wind of 15 m/s is encountered then the range would decrease
by 15 x 3.6 x 3.33 = 180 km.
Remarks:
i) Before a flight takes- off, the information about head wind, likely to occur on
the route is gathered from weather reports, and adequate amount of fuel is
provided to take care of the situation.
ii) The maximum endurance (Emax) is not affected by the presence of wind,
because Emax depends on airspeed only. The airspeed indicator in the cockpit, as
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 12
the name suggests, indicates airspeed and the pilot only needs to fly at airspeed
corresponding to Emax.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 7
Reference
Riley , K.F., Hobson, M.P. and Bence, S.J. “Mathematical methods for physics
and engineering” Cambridge University press Cambridge, U.K. (1998).
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-7
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 7
Exercises
7.1 A jet airplane is flying in level flight at a constant velocity (V). Show that when
the drag polar is parabolic the endurance (E) is given by :
2E E ζ-1max 1E = tanTSFC 2E 1 - KC E ζmax L1 1
where, = Wfuel
/ W1
W1= Weight of airplane at the beginning of the flight; W2 = W1 (1 -ζ )
E1 = W1/D1, D1 = drag at the beginning of the flight
CL1 = Lift coefficient at the beginning of the flight.
TSFC = Specific fuel consumption (assume as constant).
7.2 Define safe range and gross still air range. Obtain the gross still air range
in steady level flight for a turboprop airplane flying at a constant speed of
400 kmph at an altitude where = 0.65, given that:
CD = 0.021 + 0.06C
L
2 ; W1 = 176, 600 N, W
fuel = 35, 300 N,
S = 90 m
2
, p = 0.82, BSFC = 3.90 N/kW - hr.
(Answer: R = 2104 Km).
7.3 Consider a jet airplane with 20% of its weight as fuel fraction. It starts the
cruise climb at an altitude of 11km. What will be the altitude at the end of cruise
climb (hf )? Assuming V = 240 m/s, TSFC = 0.6 and CL/ C
D = 16, estimate the
range in cruise climb (Rcc
). What is the angle of climb ( γcc
) in cruise climb?
(Answers: hf = 12415 m, Rcc = 5141 m, γcc = 0.0157o).
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 8
Performance analysis IV – Accelerated level flight and climb
(Lecture 27)
Keywords: Accelerated level flight; accelerated climb; energy height.
Topics
8.1 Introduction
8.2 Accelerated level flight
8.2.1 Equations of motion in accelerated level flight
8.2.2 Time taken and distance covered in accelerated level flight
8.3 Accelerated climb
8.3.1 Equations of motion in accelerated climb
8.3.2 Effect of acceleration on rate of climb
8.3.3 Performance in accelerated climb from energy point of view
8.3.4 Energy height
Exercise
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
Chapter 8
Lecture 27
Performance analysis IV – Accelerated level fl ight and climb
Topics
8.1 Introduction
8.2 Accelerated level flight
8.2.1 Equations of motion in accelerated level flight
8.2.2 Time taken and distance covered in accelerated level flight
8.3 Accelerated climb
8.3.1 Equations of motion in accelerated climb
8.3.2 Effect of acceleration on rate of climb
8.3.3 Performance in accelerated climb from energy point of view
8.3.4 Energy height
8.1 Introduction
The last three chapters dealt with the performance airplane in steady
flights. The flights with acceleration are considered in this and the next two
chapters. The accelerated flights could be along a straight line e.g. accelerated
level flight and accelerated climb or along curved paths like loops and turn. In
this chapter the accelerated level flight and climb are discussed.
8.2 Accelerated level flight
When an airplane moves along a straight line at a constant altitude but its
velocity changes with time, then it is said to execute an accelerated level flight.
This type of flight occurs in the following situations.
(i) The take-off speed of an airplane is about 1.15 to 1.3 times the stalling speed.
However, the speed corresponding to the best rate of climb is generally much
higher than this speed (see Figs.6.3a and c). Hence the airplane may accelerate
from the take-off speed to the speed corresponding to the desired rate of climb.
Similarly, the speed , at the end of the climb to the cruising altitude, is lower than
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
the cruising speed (Figs.6.3a and c) and an airplane would accelerate at the
cruising altitude to attain the desired cruising speed.
(ii) The airplane may also accelerate in the transonic flight range to quickly pass-
over to the supersonic speeds (see Fig.5.11)(iii) The airplane may decelerate during a combat or when the pilot notices the
possibility of over-shooting a target.
8.2.1 Equation of motion in accelerated level flight
The forces acting on an airplane in an accelerated level flight are shown in
Fig.8.1. It may be recalled that the equations of motion are obtained by applying
Newton’s second law. For this purpose, the forces acting on the airplane are
resolved along and the perpendicular to the flight path. Sum of the components
of the forces in each of these directions, is equated to the product of the mass of
the airplane and the component of the acceleration in that direction.
The flight path in this case is a horizontal line. Hence, the equations of motion
are :
WT - D = m a = a
g (8.1)
L - W = 0 (8.2)
where ‘a’ is the acceleration.
Fig.8.1 Forces in accelerated level flight
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 4
8.2.2 Time taken and distance covered in an accelerated level flight
As regards the analysis of performance in an accelerated flight, it is of
interest to obtain the time taken and the distance covered for a given change invelocity. The accelerated or decelerated flights last only for a short duration and
the weight of the airplane can be assumed to remain constant during such flights.
However, in a level flight L = W = (1/2) ρ V2
S CL, should be satisfied. Hence, the
value of CL and consequently of C
D change continuously as the flight velocity
changes. From Eq.(8.1), the acceleration ‘a’ is given by:
a = g (T-D) / W
Substituting for D as (1/2)ρ V2
S CD , gives:
1
2
g 2a = (T - ρ V S C )DW
(8.3)
Note that :ds dV dV ds dV
V = and a = = = Vdt dt ds dt ds
dV V dV
Consequently, dt = and ds = 8.3aa a
Let the distance covered and the time taken for velocity to change from V1
to V2
be denoted by ‘s’ and ‘t’ respectively, Integrating expressions in Eq.(8.3a) gives:
2 2
1 1
V VVdV dV
s = and t =a a
V V (8.4)
Substituting for ‘a’ from Eq.(8.3) yields:
2 2
1 1
V VW V dV W dV
s = and t =2 21 1
g (T- ρ
V S C ) g (T- ρ
V S C )D DV V2 2
(8.5)
The expressions in Eq.(8.5) can be directly integrated if T and D are simple
functions of velocity. Otherwise a numerical integration as illustrated in the
following example can be carried out.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
Example 8.1
An airplane with a weight of 156,960 N and a wing area of 49 m2 has a
drag polar given by CD = 0.017+0.06C
L2. It accelerates under standard sea level
conditions from a velocity of 100 m/s to 220 m/s. Obtain the distance covered
and the time taken during the acceleration, assuming the thrust output to remain
roughly constant at 53,950 N.
Solution:
1
2
1 12 2
1
2
2L = W = ρ V SCL
22 KW2 2D = ρ V SC = ρ V S C +D D0 2ρ SV
22× 0.06 × 1569602Or D = × 1.225 ×49×0.017 × V +21.225 × 49 × V
74.9225 × 102Or D = 0.5102 V +2V
To carry out the numerical integration, the integrands in Eq.(8.5) are evaluated
for several values of V and the methods like trapezoidal rule or Simpson’s rule
are used. Books on numerical analysis be consulted for further details of these
methods. Simpson’s rule gives accurate results with a small number of points
and is used here. For this purpose the range between V1 and V
2 is divided into
six intervals, each of 20 m/s. The values are tabulated below:
V (m/s) 100 120 140 160 180 200 220
D (N) 10042 10771 12518 14999 18050 21660 25731
W
g(T-D) 0.3644 0.3705 0.3861 0.4107 0.4456 0.4954 0.5669
W V
g(T-D) 36.44 44.46 54.06 65.72 80.21 99.09 124.72
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
Using Simpson’s rule,
s = (20/3) {36.44 + 4 (44.46 + 65.72 + 99.09 ) + 2(54.06 + 80.21) + 124.72}
= 8445 m = 8.445 kmt = (20/3) {0.3644 + 4 (0.3705 + 0.4107+0.4954) + 2 (0.3861 + 0.4456 ) + 0.5669}
= 51.34 s.
Answers:
Distance covered = 8.445 km ; time taken = 51.39 s.
8.3 Accelerated Climb
In this case, the flight takes place along a straight line inclined to the
horizontal at an angle as shown in Fig.8.2. The flight velocity increases or
decreases along the flight path. Figure 8.2 also shows the forces acting on the
airplane.
Fig.8.2 Accelerated climb
8.3.1 Equations of motion in accelerated climb
The equations of motion are:
WT - D - Wsin = a
g (8.6)
L - W cos = 0 (8.7)
8.3.2 Effect of acceleration on rate of climb
From Eq.(8.6), the acceleration can be expressed as:
g (T - D - W sin )a =
W
(8.8)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
Note that : ;dV dV dh dh
a = = . but = V = R/CCdt dh dt dt
dVConsequently, a = Vc
dh (8.9)
Substituting for ‘a’ in Eq.(8.6) and noting that sin = Vc / V,
V W dV (T-D)VcT - D - W - V = 0 or V =c cV g dh V dV
W 1+g dh
(8.10)
From Eq.(6.4), (T-D) V / W is the rate if climb in steady flight. Denoting it by Vco
,
Eq.(8.10) reduces to:
VcoV =c V dV
1+g dh
(8.11)
Remark:
The term (dV/dh) in Eq.(8.11) represents the rate of change of velocity with
altitude. This quantity would be positive if the flight velocity increases with
altitude. Thus, in an accelerated climb, the rate of climb, for given values of
thrust, speed and altitude, will be lower than that in a steady climb. This has
relevance to the flight with shortest time to climb, i.e., to calculate the shortest
time required to achieve desired altitude.
From Fig.6.3c it is observed that the flight speed for maximum rate of climb
(VR/Cmax) increases with altitude. Thus, in a climb which attempts to fly the
airplane at speeds corresponding to the maximum rate of climb (V(R/C)max) at
different altitudes, would not be a steady climb but an accelerated climb.
Consequently, the values of (R/C)max given in Fig.6.3e may need to be corrected
for the effect of acceleration.
8.3.3 Performance in accelerated climb from energy point of view
The performance of an airplane in an accelerated flight can also be viewed from
the energy point of view. Multiplying Eq.(8.6) by V gives:
W dVT V - DV - W V sin = V
g dt
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 8
Or2dh W d V
TV = DV + W +dt g dt 2
(8.12)
In Eq.(8.12) the term ‘TV’ represents the available energy provided by the
propulsive system. The term ‘DV’ represents the energy dissipated in overcoming
the drag. The term ‘W (dh / dt)’ represents the rate of change of potential energy
and (W/g) {d(V2
/ 2) / dt} represents the rate of change of kinetic energy. Thus,
the total available energy can be utilized in three ways viz. overcoming drag,
change of potential energy and change of kinetic energy. If the flight takes place
at Vmax
or (Vmin
)e in level flight, then entire energy is used in overcoming the
drag and no energy is available for climb or acceleration. Only at speeds inbetween (V
min)e and V
max, can an airplane climb or accelerate and the excess
power (T-D)V has to be shared for increase of potential energy or kinetic energy
or both. If climb takes at V(R/C)max
then no acceleration is possible.
8.3.4 Energy height
Equation (8.12) can be rewritten as:
2(T-D)V d V= h +
W dt 2g
(8.13)
The term (h + V2
/2g) is denoted by he and is called ‘Specific energy or Energy
height’. It is called specific energy because it is equal to the sum of potential
energy and kinetic energy divided by the weight. It is called energy height
because this term has the dimensions of height. It may be noted that
(dhe / dt) = (T-D)V/ W (8.14)
The energy height concept is used in optimization of climb performance.
Reference 1.9 chapter 7 and Ref.1.12 chapter 2 may be referred to for details.
The quantity (dhe/ dt) is called specific excess power and denoted by ‘Ps’.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 9
Example 8.2
An airplane climbs at constant equivalent air speed in troposphere. Obtain an
expression for the correction to be applied to the value of rate of climb calculated
with the assumption of the steady climb(the denominator in Eq.8.11).
Solution:
In a climb with Ve as constant, the true air speed (V) is given by:
1
2
12V = V / σ ,e
dV dσ-3/2Consequently, = - V σedh dh
In troposphere the variation of with h is given as follows (Eq.2.7):g
-1T -λho λRσ =To
where, TO = Temperature at sea level,
λ = Temperature lapse rate and
R = gas constant.
1
2
-(g+λR)
2(g-λR)
e o
dV λ g-λR
Hence, = V σdh T λR
(8.15)
In I.S.A., λ = 0.0065 K/m. Using g = 9.81 m/s and R = 287.05 m/s2 K, the
correction factor in Eq.(8.11) is:
V dV -6 2 -1.2351+ = 1+ 4.894 × 10 V σeg dh
(8.16)
It is seen that the correction required depends on Ve and . Typical values of the
correction factor at sea level ( = 1) and at 11 km altitude ( = 0.2971) are given
in Table E8.1.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 10
Ve (m/s) 50 100 200
V dV1+
g dh
at s.l1.01224
1.0489 1.1958
V dV1+
g dh
at 11 km1.0548
1.2191 1.8766
Table E8.1 Correction factor in climb at constant equivalent air speed in
troposphere.
It is worth noting that at 11 km altitude the actual rate of climb, in constant Ve
flight at 200 m/s, is reduced to about half of its value in a steady climb.
Remark:
In a constant Mach number flight in troposphere, the flight velocity
decreases with altitude. Hence, the term (dV / dh) is negative and the rate of
climb in constant Mach number flight is more than that in a steady climb. See
exercise 8.1.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-8
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 8
Exercise
8.1 A jet trainer is climbing in troposphere at a constant Mach number of 0.6.
Obtain the rate of climb when it is climbing at an altitude of 5 km. The airplane
has the following data.
W = 54,000 N, S = 17 2m , CD = 0.017 + 0.055 2
LC , and thrust available at 5 km
altitude = 13,000 N.
[Hint: Show that in a constant Mach number flight :
dV λ R V dV 1 λ 2 = - M and 1+ = 1- RMdh 2 T -λh g dh g 2o
where, = ratio of specific heats,
λ = temperature lapse rate
R = Gas constant
Taking = 1.4, λ = 0.0065 K / m , R = 287.05 2m -2s K and g = 9.81 m / 2s
gives:
V 1c =2Vco 1-0.1331M
]
(Answers: Vco = 1798 m/min, Vc = 1888 m/min)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter- 9
Performance analysis – V- Manoeuvres
(Lectures 28 to 31)
Keywords : Flights along curved path in vertical plane – loop and pull out ;
load factor ; steady level co-ordinated-turn - minimum radius of turn, maximum
rate of turn; flight limitations ; operating envelop; V-n diagram.
Topics
9.1 Introduction
9.2 Flight along a circular path in a vertical plane (simplified loop)
9.2.1 Equation of motion in a simplified loop
9.2.2 Implications of lift required in a simplified loop
9.2.3 Load factor
9.2.4 Pull out
9.3 Turning flight
9.3.1 Steady, level, co-ordinated-turn
9.3.2 Equation of motion in steady, level, co-ordinated-turn
9.3.3 Factors limiting radius of turn and rate of turn
9.3.4 Determination of minimum radius of turn and maximum rate of turnat a chosen altitude
9.3.5 Parameters influencing turning performance of a jet airplane
9.3.6 Sustained turn rate and instantaneous turn rate
9.4 Miscellaneous topics – flight limitations, operating envelop and V-n
diagram
9.4.1 Flight limitations
9.4.2 Operating envelop
9.4.3 V-n diagram
Exercises
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
Chapter 9
Lecture 28
Performance analysis V – Manoeuvres – 1
Topics
9.1 Introduction
9.2 Flight along a circular path in a vertical plane (simplified loop)
9.2.1 Equation of motion in a simplified loop
9.2.2 Implications of lift required in a simplified loop
9.2.3 Load factor
9.2.4 Pull out
9.3 Turning flight
9.3.1 Steady, level, co-ordinated-turn
9.3.2 Equation of motion in steady, level, co-ordinated-turn
9.1 Introduction
Flight along a curved path is known as a manoeuvre. In this flight the
radial acceleration is always present even if the tangential acceleration is zero.
For example, from particle dynamics (Ref.1.2) we know that when a body moves
with constant speed along a circle it is subjected to a radial acceleration equal to
(V2 / r) or 2
ω r where, V is the speed, r is the radius of curvature of the path and
is the angular velocity ( = V / r). In a general case, when a particle moves
along a curve it has an acceleration along the tangent to the path whose
magnitude is equal to the rate of change of speed ( V ) and an acceleration along
the radius of curvature whose magnitude is (V2 / r). Reference 1.1, chapter 1 may
be referred to for details. In order that the body has these accelerations a net
force, having components along these directions, must act on the body. For
example, in the simpler case of a body moving with constant speed along a
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
circle, there must be a centripetal force of magnitude m 2ω r in the radially inward
direction; m is the mass of the body.
For the sake of simplicity, the motions of an airplane along curved paths
confined to either the vertical plane or the horizontal plane, are only consideredhere. The flight along a closed curve in a vertical plane is refered to as loop and
that in the horizontal plane as turn. Reference 2.1 and Ref.1.12, chapter 2, may
be referred to for various types of loops and turns. However, the simpler cases
considered here illustrate important features of these flights.
9.2 Flight along a circular path in vertical plane (simplified loop)
Consider the motion of an airplane along a circular path of radius r with
constant speed V. The forces acting on the airplane at various points of the flight
path are shown in Fig.9.1. Note also the orientation of the airplane at various
points and the directions in which D and L act; in a flat earth model W always
acts in the vertically downward direction.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 4
Fig.9.1 Flight along a loop with constant radius and speed
(Note: The quantity2W V
g r is the magnitude of the inertia force at various points)
9.2.1 Equations of motion in a simpl ified loop
The equations of motion, when the airplane is at specified locations, can be
written down as follows.
At point A :2WV
T - D = 0 ; L - W =g r
(9.1)
At Point B:2W V
T - D - W = 0 ; L =g r
(9.2)
At point C :2W V
T - D = 0 ; L + W =g r
(9.3)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
At point D :2WV
T - D + W = 0 ; L =gr
(9.4)
At a general point G the equations of motion are:
2WVT - D - W sin = 0 ; L + W cos =gr
(9.5)
Note that the Eqs.(9.1) to (9.4) for points A, B, C and D can be obtained from
Eqs.(9.5) by substituting as 180o, 90o, 0o and 270o respectively.
Remarks:
i) If the tangential velocity is not constant during the loop then the first equation of
Eqs.(9.5) would become:
T - D - W sin = (W / g) a, where a = dV / dt (9.6)ii) From Eqs.(9.1 to 9.5) it is observed that the lift required and the thrust required
during a loop with constant ‘r’ and ‘V’ change rapidly with time. It is difficult for the
pilot to maintain these values and the actual flight path is somewhat like the one
shown in Fig.9.2.
Fig.9.2 Shape of a normal loop
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
9.2.2 Implications of lift required during simplified loop
It is observed, that at the bottom of the loop i.e. point ‘A’ in Fig.9.1, the lift
required is equal to2WV
W +
gr
or2V
L = W 1+
gr
. The term (V2
/ gr) could be
much larger than 1 and the lift required in a manoeuvre could be several times
the weight of the airplane. As an illustration, let the flight velocity be 100 m/s and
the radius of curvature be 200 m, then the term (V2
/ gr) is equal to 5.1. Thus the
total lift required at point ‘A’ is 6.1 W. In order that an airplane carries out the
manoeuvres without getting disintegrated, its structure must be designed to
sustain the lift produced during manoeuvres. Secondly, when lift produced is
high, the drag would also be high and the engine must produce adequate output.
Further, lift coefficient cannot exceed CLmax, and as such no manoeuvre is
possible at V= Vstall.
9.2.3 Load factor
The ratio of the lift to the weight is called ‘Load factor’ and is denoted by ‘n’ i.e.
n = (L / W) (9.7)
A flight with a load factor of n is called ‘ng’ flight. For example, a turn (see
example 9.2) with load factor of 4 is referred to as a 4g turn. In level flight, nequals 1 and it is a 1g flight.
Higher the value of n, greater would be the strength required of the structure and
consequently higher structural weight of the airplane. Hence, a limit is prescribed
for the load factor to which an airplane can be subjected to. For example, the civil
airplanes are designed to withstand a load factor of 3 to 4 and the military
airplanes to a load factor of 6 or more. The limitation on the military airplane
comes from the human factors namely, a pilot subjected to more than 6g may
black out during the manoeuvre which is an undesirable situation.
To monitor the load factor, an instrument called ‘g-meter’ is installed in the
cockpit.
9.2.4 Pull out
The recovery of an airplane from a dive or a glide is called a pull out
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
(Fig.9.3). The dive is an accelerated descent while the pull out phase can be
regarded as a flight along an arc of a circle (See example 9.1).
Fig.9.3 Pull out from dive
Example 9.1
An airplane with a wing area of 20 m2 and a weight of 19,620 N dives with
engine switched off, along a straight line inclined at 60o to the horizontal. What is
the acceleration of the airplane when the flight speed is 250 kmph? If the airplane
has to pull out of this dive at a radius of 200 m, what will be the lift coefficient
required and the load factor? Drag polar is given by: CD = 0.035 + 0.076 2
LC and
the manouevre takes place around an altitude of 2 km.
Solution:
From Fig.9.3 the equations of motion in the dive can be written as follows.
WL - Wcos = 0; Wsin - D = a
g
= 60o, Hence, cos = 0.5 and sin = 0.866
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 8
Consequently, L = 19620 x 0.5 = 9810 N.
The drag of the airplane(D) can be obtained by knowing CD which depends on
CL .
2LC =L 2
ρSV
V = 250 kmph = 69.4 m/s, at 2 km = 1.0065 kg / m3
Hence,
CL =
2 × 9810
21.0065 × 20 × 69.4 = 0.2024
Consequently, CD = 0.035 + 0.076 0.20242
= 0.03811
The drag D = LC 0.03811D = 9810 × = 1847.3NC 0.2024L
Hence, (W/g) a = W sin - D = 19620 x 0.866 - 1847.3 = 15144.1 N
Or a = 15144.1 9.81/ 19620 = 7.57 m/s2.
To obtain the lift required during the pull out, let us treat the bottom part of the
flight path during the pull out as an arc of a circle.
From Eqs.(9.1) to (9.5), the lift required is maximum at the bottom of the loop and
is given by:
2 2WV 1 69.4L= W+ or L = 19620 × 1+ ×
gr 9.81 200
Or L = 19620 x 3.45 Then,
19620 × 3.45 × 2C = = 1.396L 21.0065 × 20 × 69.4
Remarks:
i) The maximum load factor in the above pull out is 3.45. The value of lift
coefficient required is 1.396. This value may be very close to CLmax and the
parabolic drag polar may not be valid.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 9
ii) Since CL cannot exceed C
Lmax, a large amount of lift cannot be produced at low
speeds. Thus maximum attainable load factor (nmaxattainable
) at a speed is:
nmaxattainable
= (1/2) ρ V2 S CLmax / W
At stalling speed the value of n is only one.
9.3 Turning flight
When an airplane moves along an arc of a circle about a vertical axis then
the flight is called a turning flight. When the altitude of the airplane remains
constant in such a flight, it is called a level turn. In order that a turning flight is
possible, a force must act in the direction of the radius of curvature. This can be
done by banking the airplane so that the lift vector has a component in the
horizontal direction. It may be added that the side force produced by deflecting
the rudder is not large. It also causes considerable amount of drag, which is
undesirable.
9.3.1 Steady, level, co-ord inated-turn
If there is no tangential acceleration i.e. the flight speed is constant, then
the flight is called a steady turn. If the altitude remains constant then the flight is
called a level turn. When the airplane executes a turn without sideslip, it is called
co-ordinated-turn. In this flight the X-axis of the airplane always coincides withthe velocity vector. The following two aspects may also be noted regarding the
steady, level, co-ordinated-turn.
(a) The centripetal force needed to execute the turn is provided by banking the
wing. The horizontal component of the lift vector provides the centripetal force
and the vertical component balances the weight of the airplane. Hence, the lift in
a turn is greater than the weight.
(b) An airplane executing a turn, does produce a sideslip.
Because of the aforesaid two factors, a pilot has to apply appropriate deflections
of elevator and rudder to execute a co-ordinated-turn.
A co-ordinated-turn is also called ‘Correctly banked turn’. In this chapter,
the discussion is confined to the steady level, co-ordinated-turn.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 10
9.3.2 Equations of motion in steady level co-ordinated-turn
The forces acting on an airplane in steady, level, co-ordinated-turn are
shown in Fig.9.4. The equations of motion in such a flight can be obtained by
resolving the forces in three mutually perpendicular directions.
Fig.9.4 Turning flight
As the turn is a steady flight: T – D = 0 . (9.8)
As the turn is a level flight: W – L cos = 0. (9.9)
As the turn is co-ordinated which implied that, there is no unbalanced sideforce.
2W VL sin =
g r (9.10)
where is the angle of bank and r is the radius of turn.
Remarks:
i) From the above equations it is noted that L = W / cos . Hence, in a turn L is
larger than W. Consequently, drag will also be larger than that in a level flight at
the same speed. The load factor n is equal to 1/ cos and is higher than 1.
ii) From Eqs.(9.9) and (9.10), the radius of turn r is given by:
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 11
2 2W V Vr = =
g Lsin g tan (9.11)
Noting that,1
cos =
n
gives 2tan = n -1 and
2
2
Vr =
g n -1 (9.11a)
The rate of turn, denoted by (ψ ), is given by:
2V V g tanψ = = V / =
r gtan V
(9.12)
Noting 2tan = n -1 gives :
2g n -1ψ =
V (9.12a)
(iii) In some books, the radius of turn is denoted by ‘R’. However, herein the letter
‘R’ is used to denote range, and to avoid confusion, the radius of turn is denoted
by ‘r’.
Example 9.2
An airplane has a jet engine which produces a thrust of 24,525 N at sea
level. The weight of the airplane is 58,860 N. The wing has an area of 28 m2,
zero-lift angle of – 2.2o and a slope of lift curve of 4.6 per radian. Find (a) the
radius of a correctly banked 4g level turn at the altitude where = 0.8 and the
wing incidence is 8o, (b) time required to turn through 180o and (c) thrust
required in the manoeuvre if the drag coefficient at this angle of attack be 0.055.
Solution:
The given data are: W = 58860 N, S = 28 m
2
, = 8
o
,0L = -2.2
o
,
dCLdα
= 4.6 per radian =4.6
180 x 2 per degree = 0.083 per degree,
allowable n = 4 and T = 24525 N at sea level.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 12
Consequently,dCLC = (α - α )L oLdα
= 0.0803 (8 + 2.2) = 0.82
In a 4g turn L = 4W = 1/2 ρV2 S CL
Hence, V =
1/22 ×4 × 588601/2(2L /ρ SC ) =L
1.225 × 0.8 × 28 × 0.82
= 144.6 m/s.
0 '1 1cos = = or = 75 31
n 4
Hence, tan = 3.873
Consequently, 22 144.6V
r = = = 550.3mgtan 9.81× 3.873
Rate of turn = V 144.6ψ = =r 550.3 = 0.2627 rad /s
Hence, time to turn through 180o is equal to = 11.95s0.2627
The thrust required = Tr = 1/2 ρ V2 S CD
= (1/2) x 1.225 x 0.8 x 144.62 x 28 x 0.055 = 15786 N
Answers : (a) Radius of correctly banked turn = 550.3 m, (b) time required to turn
through 1800
= 11.95 s and (c) thrust required during turn = 15,786 NRemark:
The thrust available is given as 24525 N at sea level. If the thrust available is
assumed to be roughly proportional to (σ0.7), the thrust available at the chosen
altitude would be 24525 x 0.80.7 = 20978 N. This thrust is more than the thrust
required during the turn and the flight is possible.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 9
Lecture 29
Performance analysis V – Manoeuvres – 2
Topics
9.3.3 Factors limiting radius of turn and rate of turn
9.3.4 Determination of minimum radius of turn and maximum rate of turn
at a chosen altitude
9.3.3 Factors limi ting radius of turn and rate of turn
Turning flight is a very important item of performance evaluation,
especially for the military airplanes. Minimum radius of turn and maximum rate of
turn are important indicators of the manoeuverability of an airplane. From
Eqs.(9.11) and (9.12) it is observed that, at a given altitude and flight velocity, a
small radius of turn and a high rate of turn are achieved when the bank angle ( )
has the highest possible value. Equations (9.11a) and (9.12a) indicate that at a
given altitude, the minimum radius of turn (r min) and the maximum rate of turn
( maxψ ) are obtained when ‘V’ is low and ‘n’ is high. The following considerations
limit the achievable values of r min and maxψ .
(I)Limitation due to CLmax
: From the above discussion we observe that the lift
coefficient in a turning flight is higher than the lift coefficient required at the same
speed in level flight. Let CLT
be the lift coefficient in the turning flight and CLL
be
the lift coefficient in the level flight at the same speed.
Then, CLT = n W / ( ½ V2 S ) = n CLL
However, CLT
cannot be more than CLmax
and this imposes limitations on the
attainable values of load factor (n) and the bank angle ( ). Let these two values
be denoted by max CLmaxn and max CLmax . They can be expressed as :
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
(nmax
)CLmax = /Lmax LLC C (9.13)
,-1Noting that = cos 1/n max CLmax =cos-1( /LL LmaxC C ) (9.14)
It may be noted that, at stalling speed (Vs), the value of CLL
equals CLmax
or n =
1. Hence, turn is not possible at stalling speed .
(II) Limitation due to allowable load factor from structural consideration : The
bank angle and the load factor in a turn are related by:
cos = 1/n .
However, n cannot exceed the value permitted by the structural design of the
airplane. Let this value be denoted by (nmax)str . Hence, max is limited to
cos-1 {1/(nmax)str }.
(III) The drag coefficient in a turning flight is higher than that in a level flight at the
same speed. However, in a steady turn the thrust required cannot exceed the
thrust available (Ta). This also imposes limitations on the attainable values of
and n. Let these two values be denoted as ( max
)Ta
and (nmax
)Ta
.
It may be noted that, at V = Vmax
and (Vmin
)e the entire engine output is used in
overcoming the drag in level flight. Hence, the steady level turn is not possible atthese two speeds.
The lowest of the above three values viz max CLmaxn , (nmax)str and (nmax
)Ta
is
the permissible value of nmax
. Let this value be denoted by (nmax)perm .
Substituting this value in Eqs.(9.11a) and (9.12a) gives r and ψ .
9.3.4 Determination of minimum radius of turn and maximum rate of turn at
a chosen altitude
In a general case, the drag polar and the thrust available are functions of
Mach number. In such a case, the minimum radius of turn (r min) and the
maximum rate of turn (ψmax
) at an altitude, can be obtained by using the
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
following steps. The limitations stated in the previous subsection, are taken into
account during the procedure.
(i) Choose an altitude. Obtain Vmax
and Vmin
at this altitude. Note that a steady
level, co-ordinated-turn is possible only within this speed range.(ii) Choose a flight speed (V) in between V
max and V
min and obtain CLL
as:
CLL
= 2W / ( SV2)
Obtain Mach number (M) corresponding to the chosen V and the speed of sound
at chosen altitude.
(iii) Obtain the CLmax at the chosen flight Mach number. It may be recalled from
subsection 3.7.4, that for airplanes flying at high speeds, the CLmax depends on
Mach number. Obtain the ratio CLmax / CLL
.
The ratio CLmax
/ CLL
gives the quantity max CLmaxn defined above. If this value
is smaller than the allowable load factor from structural consideration viz.
(nmax)str , then the turn may be limited by CLmax
. In this situation, choose
CLT1
= CLmax
. It may be mentioned that the procedure presented here, aims at
obtaining the value of lift coefficient in the turn (CLT) which satisfies all the three
limitations on the turn mentioned above. The quantity CLT1
is the value of CLT
as
limited by CLmax
. This will be modified in the subsequent steps.
If CLmax
/ CLL
is more than (nmax)str , then the turn may be limited by (nmax)str. In
this situation, choose CLT1 as (nmax)str x C
LL.
(iv) Obtain from the drag polar, the drag coefficient CDT1
, corresponding to CLT1
and the chosen Mach number. Calculate the drag DT1
from:
DT1
= 1/2 V2 S CDT1
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 4
If D T1
is greater than the available thrust (Ta), then the turn is limited by engine
output. In this situation, obtain the maximum permissible value of drag coefficient
in turning flight (CDT
) as limited by Ta . It is given as : C
DT = Ta / (1/2 V
2
S)
Corresponding to this value of CDT
, obtain the lift coefficient CLT
by referring to
the drag polar.
If DT1
is smaller than Ta , then the turn is not limited by the engine output. In this
situation, the turn is limited by CLmax
or (nmax)str. Consequently, CLT
is the
smaller of the two values obtained in step (iii).
(v) Once CLT
is known, is given by:
= cos-1 (CLL
/CLT
). Knowing and V, the radius of turn (r) and rate of turn
(ψ ), at the chosen speed, can be calculated using Eqs.(9.11) and (9.12).
(vi) The previous steps should be repeated at various values of flight speeds
ranging between Vmin
and Vmax
. Plotting these results, the values of r min and
ψ max and the corresponding speeds Vrmin and Vψ max
can be determined at the
chosen altitude.
(vii) Repeat steps (i) to (vi) at different altitudes.
The procedure is illustrated, at a chosen altitude, in example 9.3.
Example 9.3
A passenger airplane has a gross weight of 176,400 N and a wing area of
45 2m . Obtain the variations of r and ψ with velocity at an altitude of 8 km from
the following data.
CLmax
= 1.4, (nmax)str = 3.5, CD = 0.017 + 0.05 2
LC
Vmin
= 103 m/s, Vmax
= 274 m/s and the thrust output (Ta) varies as given in the
table below.
V (m/s) 105 115 125 145 165 185 205
Ta (N) 21100 21125 21150 21480 21580 21980 22270
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
Solution:
At 8 km altitude the value of ρ is 0.525 kg/m3. The minimum radius of turn and
ψmax
at various speeds are worked out in a tabular manner using the procedure
outlined above.
V (m/s) 105 115 125 145 165 185 205
CLL
1.354 1.129 0.955 0.710 0.548 0.436 0.355
CLmax / CLL
1.034 1.240 1.466 1.972 2.553 3.21 3.94
CLT1
1.4 1.4 1.4 1.4 1.4 1.4 1.243*
CDT1
0.115 0.115 0.115 0.115 0.115 0.115 .0942
DT1(N) 15000 17993 21258 28601 37042 46568 46852
Ta (N) 21100 21125 21150 21480 21580 21980 22270
CDT
** ** 0.1114 0.0864 0.067 0.0543 0.0448
CLT
1.4$ 1.4$ 1.396£ 1.178£ 1.08£ 0.863£ 0.745£
LT
LL
C=n
C 1.034 1.240 1.461 1.659 1.824 1.98 2.10
(degrees)14.75 36.25 46.9 52.93 56.76 59.63 61.6
r (m) 4273 1838 1491 1619 1819 2043 2321
ψ (rad/s) 0.0246 0.0626 0.0838 0.0896 0.0907 0.0906 0.0883
The symbols in the above table have the following meanings:
* Turn is limited by load factor (nmax)str hence CLT1
= (nmax)str CLL.
** Thrust available is more than thrust required. Hence, CLT
= CLT1
$ Turn is limited by CLmax
£ Turn is limited by Ta
Table E9.3 Variations of radius of turn (r) and rate of turn ψ with
flight velocity (V) for airplane in example 9.3
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
Fig.E9.3 Variations of radius of turn (r) and rate of turn ψ with flight
velocity (V) for the airplane in example 9.3
The plots of r vs V and ψ vs V are shown in Fig.E9.3. From these plots
r min = 1.490 m and maxψ = 0.0907 rad/s, Vrmin = 124 m/s andψmaxV
= 165 m/s
Answers:
Minimum radius of turn (r min) = 1490 m at Vrmin = 124 m/s
Maximum rate of turn ( maxψ ) = 0.090 rad/s atψmaxV
= 165 m/s
Remarks:
i) Turning performance of a jet airplane :
Section 7 of Appendix B presents the turning performance of a jet airplane.
Figures 9.5a and b show the variations of ψ and r with velocity at different
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
altitudes for that airplane. Figures 9.5c and d present the variations ofψmaxV
and Vrmin with altitude.
Fig.9.5a Turning performance of a jet transport – rate of turn (ψ )
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 8
Fig.9.5b Turning performance of jet transport – radius of turn (r)
Fig.9.5c Turning performance of jet transport - variation of Vψ max
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 9
Fig.9.5d Turning performance of jet transport - variation of Vrmin
Note:
Some curves in Figs.9.5a,c and d show discontinuity in slope at certain points.
This occurs when the criterion limiting the turning performance changes from
(nmax)str to (nmax
)Ta
.
ii) Turning performance of a piston engined airplane :
Section 7 of Appendix A presents the turning performance of a piston engined
airplane. Figures 9.6a and b show the variations of r and maxψ with velocity at
different altitudes for that airplanes. Figures 9.7c and d present the variations of
r min and maxψ with altitude. Figure 9.6e presents the variations ofψmaxV
and
Vrmin with altitude. Both these speeds increase with altitude. The two speeds
come close to each other as absolute ceiling is approached. Minimum radius of
turn (r min) increases with altitude and maxψ decreases with altitude. At absolute
ceiling, the rate of turn becomes zero and the radius becomes infinite.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 10
Fig.9.6a Turning performance of a piston engined airplane
- variation of rate of turn (ψ )
Fig.9.6b Turning performance of a piston engined airplane –
variation of radius of turn ( r)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 11
Fig.9.6c Turning performance of a piston engined airplane –variation of r
min with altitude
Fig.9.6d Turning performance of piston engined airplane –
variation of maxψ with altitude
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 12
Fig.9.6e Turning performance of piston engined airplane –
variations of Vrmin and
ψmaxV
with altitude
iii) In many situations the minimum radius of turn in level flight is limited by the
available engine output. This can be overcome and a smaller radius of turn can
be obtained by allowing the airplane to descend during the turn. In this manner a
loss of potential energy is used to increase the available energy during turn.
Reference 1.12, chapter 2 may be consulted for additional details. See also
subsection 9.3.6 for further information.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 9
Lecture 30
Performance analysis V – Manoeuvres – 3
Topics
9.3.5 Parameters influencing turning performance of a jet airplane
9.3.6 Sustained turn rate and instantaneous turn rate
9.3.5 Parameters influencing turning performance of a jet airplane
The steps, described in subsection 9.3.4 , to determine r min, maxψ
, Vrmin and
ψmaxV
constitute a general procedure which is applicable to all types of airplane.
However, the influence of the wing loading (W/S) and the thrust loading (Ta/W)
can be examined by the following simplified analysis. It is based on the following
two assumptions.
(a)Thrust available (Ta) is constant.
(b)The drag polar is parabolic with DOC and K as constants.The following relationships are observed in a steady, level, co-ordinated-turn.
T = D, L = nW ; n =1
cos ,
2
2
Vr =
g n -1,
2g n -1ψ =
V
Hence, 2 2 2a D LDO
1 1T = ρV SC = ρV S C + KC
2 2
Or
2
2a DO 2
1 2nWT = ρV S C +K
2 ρSV
(9.15)
Solving Eq.(9.15) for n2, gives :
2 2DO2 a
1 1ρV ρV C
T2 2n = -K W/S W W/S
(9.16)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
Let the free stream dynamic pressure be denoted by ‘q’ :
Or q = 21ρV
2 (9.17)
Consequently, Eq.(9.16) can be rewritten as :
2 a DOqCTq
n = -K(W/S) W W/S
(9.18)
From Eq.(9.11a)
2
2Vr =
g n -1
Or2
2qr =
gρ n -1 (9.19)
From Eq.(9.19) it is observed that ‘r’ is a function of q and n. However, when the
constraint of thrust available is taken into account, then n and q are related by
Eq.(9.18).
The value of q which would give minimum radius of turn (r min ), can be obtained in
two stages.
(a) Substitute the expression for ‘n’ as given by Eq.(9.18) in Eq.(9.19).
(b) Differentiate the resultant equation for ‘r’ obtained in step (a), with respect to
‘q’, and equate it to zero.However, the resulting expression is complicated. An
alternate way is as follows.
(i) Differentiate Eq.(9.19) with respect to q and equate it to zero.
-1/22 2
2 2 2
2gρ n -1 - 2gρqn n -1 dn/dqdr =
dq g ρ n - 1= 0
Or 2 dnn - 1 - qn = 0dq
(9.20)
(ii)The quantity (dn/dq) is obtained by differentiating Eq.(9.18) with respect to q
i.e.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
DO
2aT /W qCdn
n = -dq 2K(W/S) K W/S
(9.21)
(iii)Substituting for n2 and n (dn/dq) in Eq.(9.20) yields :
2 2aa DO DO
2 2
q C q Cq T /WTq- - 1 - + = 0
K W/S W 2K W/SK W/S K W/S
Simplifying :
aq T /W
= 12K W/S
(9.22)
Equation (9.22) yields the value of q which gives minimum radius of turn. This
value is denoted by ‘qrmin ‘ i.e. :
qrmin =
a
2K W/S
T /W (9.23)
Using Eq.(9.17), Vrmin is given as :
Vrmin =
a
4K W/S
ρ T /W (9.24)
Substituting qrmin in Eq.(9.18) gives :
222 DOrmin 2 2
a
a a
2K W/S T /W 4K W/S Cn = -
T /W K W/S T /W K W/S=
DO
2a
4 KC2 -
T /W
Or
2DO
rmin
a
4K Cn = 2 -
T /W (9.25)
Substituting from Eqs.(9.24) and (9.25) in Eq.(9.11a) gives :
2rmin
min 2rmin
Vr =
g n -1
DO2
a
a
4K(W/S) 1=ρ T /W 4KC
g 2- -1T /W
2
a aDO
4K W/S=
gρ T /W 1 - 4K C / T / W
(9.26)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 4
Proceeding in a similar manner, the values ofψmax
V
,ψmax
n
and maxψ , which
take into account the constraint of thrust available, can be derived. The final
expressions are given below.
DO
1/21/4
ψmax
2(W/S)V = K/C
ρ
(9.27)
DO
1/2
ψmaxaT /W
n = - 1KC
(9.28)
1/2
DOmax
1/2a CT /Wρ
ψ = g -W/S 2K K
(9.29)
Remarks:
(i)From Eqs.(9.26) and (9.29) it is observed that for a jet airplane to have a low
value of r min and a high value ofψmax
V
, the value of (Ta/W) should be high and
that of (W/S) should be low. However, as stated in section 7.4.3 the wing loading
(W/S) is a compromise between various considerations like range, take-off and
landing. Consequently, the general practice is to select (Ta/W) to give the desired
value of maxψ , taking into account the wing loading chosen from other
considerations.
(ii) The constraints of (nmax)str and CLmax have not been taken into account in
the above analysis. Also the variation of thrust available with flight speed has
been ignored.
Equation (9.25) shows that the load factor for minimum radius of turn (nrmin) is
less than 2 . However, the load factor for maximum rate of turn (ψmax
n
), as
given by Eq.(9.28), could be high, especially near the sea level where (Ta/W) is
at its highest. In this situation the constraint of (nmax)str needs to be taken into
consideration.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
(iii)The constraint of CLmax is likely to affect the value of r min. Example 9.4
illustrates such a situation.
(iv) A simplified analysis of the turning performance of an airplane with engine
propeller combination can be carried out by assuming that (a) THP in constant
with flight velocity and (b)DO
C and K are constants. However, the resulting
expression has the following form.
4rmin r min A V +B V +C = 0
This equation does not have an analytical solution and a graphical or numerical
procedure is needed. Reference 1.12 chapter 2 can be consulted for details.
It can be inferred from the analysis of Ref.1.12, that if it is desired to increase
maxψ or decrease r min of a given airplane, then the wing loading (W/S) should
be reduced and / or the ratio (BHP/W) should be increased.
Example 9.4
Consider the airplane in example 9.3 with the simplification that the thrust
remains constant with flight velocity and has the value of 21685 N. Obtain the
values of Vrmin,ψmax
V
, nrmin, ψmaxn
, r min and maxψ as given by the analysis in
subsection 9.3.5.
Solution :
The given data are :
W = 176, 400 N, S = 45 m2 ,DO
C = 0.017, K = 0.05, h = 8000 m or
3ρ = 0.525 kg/m , Ta = 21685 N .
The constraints are : CLmax = 1.4, (nmax)str = 3.5.
Consequently, W/S = 176400 /45 = 3920 N/m
2
&Ta/W = 21685/176400 = 0.1229.
Based on the analysis of subsection 9.3.5, which considers only the constraint of
thrust available, the following expressions are obtained.
rmina
4K(W/S)V =
ρ T /W
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
2
DOrmin
a
4K Cn = 2 -
T /W
minr
2DOa a
4K W/S=
gρ T /W 1-4KC / T /W
DO
1/21/4
ψmax
2(W/S)V = K/C
ρ
DO
1/2
ψmaxaT /W
n = - 1KC
1/2
DOmax
1/2
a CT /Wρψ = g -W/S 2K K
Accordingly ,
rmin
4 × 0.05 × 3920V = = 110.23 m/s
0.525 × 0.1229
rmin 2
4 ×0.05 × 0.017n = 2 - = 1.332
0.1229
2
min 2 2
2rminV 110.23r = = = 1407.6 m
g n -1 9.81 1.332 -1
1/2 1/4
ψmax
2×3920 0.05V = = 160.04 m/s
0.525 0.017
1/2
ψmax
0.1229n = - 1 = 1.793
0.05 × 0.017
2ψmax
max
ψmax
g n - 1ψ =V
29.81× 1.793 -1= = 0.0912 rad/s
160.04
The values of lift coefficients corresponding to Vrmin andψmax
V
are:
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
2rmin
Lrmin2rmin
n W 1.332 × 176400C = = = 1.637
1 0.5 × 0.525 × 110.23 × 45ρV
2S
2
ψmax
Lψmax 2ψmax
n W 1.793 × 176400
C = = = 1.0451 1ρV S × 0.525 × 160.04 × 45
2 2
It is observed that in case of maxψ the values of n and CL are 1.793 and 1.045.
These values are lower than the prescribed values of (nmax)str and CLmax.
Hence, this turn is possible and maxψ of 0.0912 rad/s at V = 160.04 m/s is
possible. However, the value of CLrmin is 1.637 which is higher than CLmax and
this turn is not possible. In this situation, a new value of flight velocity (V) is to be
obtained at which the values of load factor (n) given by the two constraints of
thrust available and CLmax, are equal.
The value of n from the constraint of thrust available can be denoted by ‘Ta
n ’. It
is given by Eq.(9.16):
2
2 DO
1/2
Ta
a
1ρV CT 12n = - ρV
K W/S W 2 W/S
(9.30)
The value of n from the constraint of CLmax can be denoted by ‘CLmax
n ’. It is
given by :
L = 2LmaxCLmax
1n W = ρV SC
2
Or 2 LmaxCLmax
C1n = ρV
2 W/S
(9.31)
Equating Eqs.(9.30) and (9.31) gives the value of ‘V’ which satisfies both the
constraints i.e.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 8
2
2DO Lmax
1/2
2a
1ρV CT C1 12 - ρV = ρV
K W/S W 2 W/S 2 W/S
Simplifying yields :
DO
22Lmax
2 2aT /W CC 1
= + ρVK W/S 2K W/S W/S
Substituting various values gives :
22
2 2
0.1229 0.017 1.4 1= + ×0.525×V
0.05×3920 20.05×3920 3920
Or V = 126.32 m/s
Consequently,
22Lmax
1ρV C
0.5×0.525×126.32 ×1.42n = = = 1.496W/S 3920
2
min 2
126.32r = = 1461.9 m
9.81 1.496 -1
The value of V = 126.32 m/s satisfies the constraints of Ta and CLmax. The
corresponding value of n = 1.496 is also less than (nmax)str of 3.5. Hence, all
constraints are satisfied.
Answers : Based on the simplified analysis at 8000 m altitude the following
values are obtained.
Vrmin = 126.32 m/s, nrmin = 1.496, r min = 1461.9 m,
ψmaxV = 160.04 m/s
,
ψmaxn = 1.793
, maxψ = 0.0912 rad/s
Remark :
The values by exact analysis are :
Vrmin = 124 m/s, nrmin = 1.451 , r min = 1490 m ,
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 9
ψmaxV = 165 m/s
,
ψmaxn = 1.824
maxψ = 0.0907 rad/s .
The agreement between the two results is seen to be reasonable. The reasons
are that (Ta/W) is rather low and the variation of Ta with V is not large.
9.3.6 Sustained turn rate and instantaneous turn rate
The maximum rate of turn in a steady level co-ordinated-turn is called
‘Maximum sustained turn rate(MSTR)’ (Ref. 1.12 chapter 2). An airplane can
maintain this turn rate continuously for some time. However, as explained in
subsections 9.3.3 and 9.3.4 this turn rate is generally limited by the thrust
available. A rate of turn higher than MSTR can be obtained if the airplane is
allowed to descend or slow down. In this manner, the loss of potential energy or
kinetic energy can be utilized to increase the available energy during turn and
increase the rate of turn. This rate of turn is called ‘Instantenous rate of turn’. The
maximum instantenous rate of turn will be limited by other two factors viz. CLmax
and (nmax)str . See also item (iv) in subsection 9.4.3.
General Remark:
In the foregoing sections various types of flight situations of practical
interest have been analyzed. To analyze any other flight situation one can begin
by writing down the equations of motion along and perpendicular to the flight
path. From these equations, the lift required, thrust required and accelerations in
tangential and radial directions can be worked out.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 9
Lecture 31
Performance analysis V – Manoeuvres – 4
Topics
9.4 Miscellaneous topics – flight limi tations, operating envelop and V-n
diagram
9.4.1 Flight limitations
9.4.2 Operating envelop
9.4.3 V-n diagram
9.4 Miscellaneous topics
A flight is called free flight when the airplane is away from the influence of
the ground i.e. it is at a height more than a few wing spans above the ground.
The performance in level flight, climb, turn etc. come under this category. In
contrast, the analysis of take-off and landing requires consideration of the
influence of proximity of ground. The discussion of performance in free flight is
concluded in this section by describing aspects like flight limitations, operating
envelop and V-n diagram. Chapter 10 describes the performance in take-off and
landing.
9.4.1 Flight Limi tations:
In chapters 5 to 8 and the previous subsections of this chapter, the
performance of an airplane in free flight has been discussed under categories of
level flight, climb, accelerated flights and manoeuvres. The important aspects of
these analyses are generally brought out in a diagram which is called here as
‘Variations of characteristics velocities’. Figure 9.7a shows the variations of Vmax
,
(Vmin
)e, Vs, V
(R/C) max V
max, Vrmin and ψmaxV
with altitude for a typical subsonic jet
airplane. Figure 9.7b presents similar plots for a piston engined airplane.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
It has been pointed out earlier that (a) the maximum lift coefficient limits the
minimum speed in level flight (Vs), the minimum radius of turn (r min
) and the
maximum rate of turn ( maxψ ) , (b) the power output limits the maximum speed
(Vmax), the minimum speed (Vmin)e , the maximum angle of climb (max), the
maximum rate of climb (R/C)max
, r min
and maxψ , (c) the maximum allowable load
factor, (nmax)str , limits r min
and maxψ . In addition to these, the performance of the
airplane may also be limited by considerations like buffeting, sonic boom,
maximum dynamic pressure (q) limit and aerodynamic heating . These limitations
are briefly described below.
Fig.9.7a Variations of characteristic velocities – Jet transport
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
Fig.9.7b Variations of characteristic velocities – Piston engined airplane
(i)Buffeting is an irregular oscillation of a part of an airplane, caused by the
passing of separated flow from another component. For example, the horizontal
tail experiences buffeting when the separated flow from the wing passes over it
(horizontal tail). This happens when the wing is at a high angle of attack or the
shock stall takes place on it in the transonic flow regime. To prevent buffeting,
the permissible value of CLmax
may be limited. This in turn would affect Vs, r min
and ψmax
.
(ii)The sonic boom problem is encountered when an airplane flies at supersonic
speed at low altitudes. The shock waves created by an airplane, when it is flying
at a supersonic speed, coalesce and form two waves across which there is a
finite pressure rise (overpressure). When these waves reach the ground each of
them is perceived as an explosive like sound called sonic boom or sonic bang.
The intensity of the boom depends on the size and shape of the airplane, its flight
altitude and the atmospheric conditions. It increases with the increase in the size
of the airplane and decreases with the increase of the altitude of the flight. An
overpressure in excess of about 100 N/m2 is quite annoying and may cause
0
1000
2000
3000
4000
5000
6000
0 20 40 60
Velocity (m/s)
A l t i t u d e ( m )
V for minimum radius of turn
V for maximum rate of turn
Stalling speed
Vmin from engine output
V max
V for maximum angle of climb
V for maximum rate of climb
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 4
vibrations of buildings and rattling of window panes. To keep the overpressure on
the ground within socially acceptable limits, the supersonic transport airplanes
are generally not permitted to cross Mach number of one below tropopause and
they cruise at altitudes of 16 to 20 km.
(iii)The airplanes are generally designed for a dynamic pressure (q = 21ρV
2) of
100,000 N/m2. This limit would not permit attainment of high supersonic Mach
number at low altitudes.
(iv)As the flight Mach number increases, the stagnation temperature (Ts) on the
surface increases. It is given by:
-1 2T = T (1+ε M )s amb2
(9.32)
where, Tamb is the ambient temperature and ε is the recovery factor which has a
value of around 0.9 for turbulent boundary layer on the surface. The maximum
stagnation temperature (Ts) may be limited from the consideration of material
used for the fabrication of the airplane. This would limit the maximum permissible
Mach number.
(v) Reference 3.9, chapter 17, mentions about other limits like engine relight limit,
pilot ejection altitude and duct pressure limit. The minimum speed from engine
relight limit is encountered in some cases at high altitudes where enough air may
not be available to restart the engine in the event of flame-out. The highest
altitude may be limited to about 15 kms which is the the highest altitude at which
ejection by the pilot is permitted.
9.4.2 Operating envelop
The maximum speed and the minimum speed of the airplane can be
calculated from the level flight analysis. However, the attainment of maximum
speed may be limited by the considerations mentioned in the previous
subsection. A diagram which indicates the range of flight speeds permissible for
an airplane at different altitudes is called ‘Operating envelope’. Typical operating
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
envelope for a military airplane is shown in Fig.9.8. Explanation of the curves in
this figure is as follows.
(i)The curve ABCDE is the level flight boundary based on the engine output. The
portion ABC is the Vmax or (Mmax) boundary. The portion CDE is the (Vmin)e or(M
min) boundary, limited by the engine output. It may be mentioned that for these
Fig.9.8 Operating envelope of a military airplane - Schematic
curves (ABC and CDE) the engine output is with the afterburner on. On this
boundary (ABCDE) the specific excess power (Ps) is zero.
(ii)The curve FG is the line representing stalling speed (Vs).
2WV = ; C without flaps Lmax
ρ S CLmax
Recalling that when Mach number exceeds 0.5, the maximum lift coefficient
(CLmax
) decreases due to shock stall or buffetting. The line FG includes this effect
when Mach number corresponding to Vs is more than 0.5.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
(iii)The line HJK represents the dynamic pressure (q) limit corresponding to q of
100,000 N/m2.
(iv)The line LMNOP represents the boundary corresponding to stagnation
temperature (Ts) of 400K. It may be pointed out that Tamb and hence the speed
of sound change with altitude in troposphere. They are constant in lower
stratosphere. Hence, the allowable flight Mach number, for stagnation
temperature to be below allowable value, changes with altitude.
The flight envelope taking into account the above limits is the curve FDCONMJH.
Remark:
Figure 9.8 also shows zones marked as : (I) advantageous for interceptor role,
(II) advantageous for aerial combat and (III) suitable for high speed low altitudeflight.
It may be added that for the interceptor role, it is advantageous if the airplane
flies at high altitude and high speed (zone I in Fig.9.8).
For aerial combat the manoeuverability, which is measured mainly by the rate of
turn, is important. It may be recalled from subsection 9.3.3 that the rate of turn is
low at (a) altitudes near the ceiling and (b) flight speeds close to Vmax and Vmin.
Further, the aerial combat cannot take place at very low altitudes. Hence, theaerial combat zone is the region marked as (II) in Fig.9.8.
For airplanes used as ground attack fighter, the ability to fly at high speed and at
low altitude is important. Zone (III) in Fig.9.8 is appropriate for these airplanes.
9.4.3 V-n diagram
The load factor (n) has already been defined as the ratio of lift and weight
i.e. n = L / W. In level flight n = 1. However, as pointed out in subsections 9.2.3
and 9.3.3 the value of ‘n’ during a manoeuver is greater than one. Hence, the
structure of the airplane must be designed to withstand the permissible load
factor. Further, when an airplane encounters a gust of velocity Vgu (see Fig.7.1b)
the angle of attack of the airplane would increase by Δα = Vgu /V. This increase
in angle of attack, would increase the lift by ΔL, given by :
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
ΔL = ½ ρV2 S C
Lα Δα = ½ ρV
2 S CLα
Vgu / V
= ½ ρVSCLα
Vgu (9.33)
Hence, Δn = ΔL / W = ½ ρVSCLα
Vgu / W (9.34)
From Eqs.(9.33) and (9.34), ΔL increases with Vgu. Further, for a given Vgu, the
values of ΔL & Δn increase with flight velocity. An airplane must be designed to
withstand the gust loads also.
In aeronautical engineering practice, the load factors due to manoeuver
and gust are indicated by a diagram called ‘Velocity-load factor diagram or V-n
diagram’. A typical V-n diagram is shown in Fig.9.9. This diagram can be
explained as follows.
(i) Curves OIA and OHG : The lift (L) produced by an airplane is given by
L = ½ ρV2S CL. It should be noted that (i) C
L ≤ C
Lmax and (ii) at stalling
speed(Vs), L = W and n = 1. However, if the airplane is flown with CL = C
Lmax at
speeds higher than Vs, then (a) L will be more than W and (b) L or n would be
proportional to V2. This variation is a parabola and is shown by curve OIA in
Fig.9.9. In an inverted flight the load factor will be negative and the V vs n curve
in such a flight is indicated by the curve OHG in Fig.9.9. It may be mentioned that
an airplane can fly only at V ≥ Vs and hence the portions OI and OH in Fig.9.9
are shown by chain lines.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 8
Fig.9.9 A typical V-n diagram
(ii) Positive and negative manoeuver load factors : An airplane is designed to
withstand a certain permissible load factor. Higher the permissible load factor,
heavier will be the weight of airplane structure. Hence, for actual airplanes the
manoeuver load factor is limited depending on its intended use. Federal Aviation Administration (FAA) in USA and similar agencies in other countries, prescribe
the values of permissible manoeuver limit load factors (npositive and nnegative) for
different categories of airplanes. Table 9.1 gives typical values. A limit load is
obtained by multiplying the limit load factor with the weight (W). The airplane
structure is designed such that it can withstand the limit load without yielding.
The ultimate load factor, in aeronautical practice, is 1.5 times the limit load factor.
The ultimate load is obtained by multiplying the ultimate load factor with theweight (W). The airplane structure is designed such that it can withstand the
ultimate load without failing, though there may be permanent damage to the
structure.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 9
Type of airplane npositive nnegative
General aviation-non aerobatic 2.5 to 3.8 -1
Transport 3 to 4 -1
Fighter 6 to 9 -3
Table 9.1 Typical limit load factors
In Fig.9.9, npositive
= 3 and nnegative
= -1.2 have been chosen; the actual values
depend on the weight of the airplane and its category. Reference 3.18 part
V,chapter 4 may be consulted for details.
(iii) Lines AC, GF and FD : The positive manoeuver load factor is prescribed to
be constant upto the design diving speed (Vd); line AC in Fig.9.9. According to
Ref.3.9, chapter 14, the design diving speed could be 40 to 50% higher than the
cruising speed (Vc) for subsonic airplanes. For supersonic airplanes, the Mach
number corresponding to Vd could be 0.2 faster than the maximum level flight
Mach number. The negative manoeuver load factor is prescribed to be constant
upto design cruising speed (line GF in Fig.9.9) and then increases linearly to zero
at V = Vd (line FD in Fig.9.9).
(iv) Manoeuvre load diagram : The diagram obtained by joining the points
OACDFGO is called ‘Manoeuvre load diagram’.
(v) Manoeuvre point and Corner speed : The point ‘A’ in Fig.9.9 is called
’Manoeuvre point’. The flight speed at this point is denoted by V* and is called
’Corner speed’. At point ‘A’ the lift coefficient equals CLmax and the load factor
equals npositive. This combination would result in the maximum instanteneous
turn rate at the speed V*. See subsection 9.3.6 for definition of instanteneous
turn rate.
(vi) Positive and negative gust load factors : From Eq.(9.33) it is observed that
the gust load factor varies linearly with velocity. The regulating agencies like FAA
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 10
prescribe that an airplane should be able to withstand load factors corresponding
to Vgu = 30 ft/s (9.1 m/s) upto cruising speed (Vc) and Vgu = 15 ft/s (4.6 m/s)
upto design diving speed (Vd). Lines JB and JF’ in Fig.9.9 show the gust lines
corresponding to Vgu = 30 ft/s (9.1 m/s) and Lines JC’ and JE in the same figure
show the gust lines corresponding to Vgu = 15 ft/s (4.6 m/s).
It may be pointed out that a gust in real situation, is not a sharp edged gust as
shown in Fig.7.1b and the velocity Vgu is attained in a gradual manner. This
causes reduction in the gust load factor. To take care of this reduction the gust
load factor is multiplied by a quantity called ‘Gust alleviation factor’. Reference
3.9 chapter 14 may be referred for details.
(vii) Gust load diagram : The diagram obtained by joining the points JBC’EF’J is
called ‘Gust load diagram’.
(vi) Final V-n diagram : For its safe operation, an airplane must be designed to
withstand load factors occuring at all points of the gust and manouever load
diagrams. Hence, the final V-n diagram is obtained by joining the parts of these
two diagrams representing the higher of the manoeuver and gust load factors.
The final V-n diagram in the case presented in Fig.9.9, is given by the solid curve
obtained by joining the points IAB’BB’’CEF’’FGHI.It may be pointed out that the gust load line JB’ is above the curve IA in the
region IK. However, along the curve IK the airplane is already operating at CLmax
and any increase in angle of attack due to gust cannot increase CL beyond C
Lmax.
Hence, the portion JK of the line JB is not included in the final V-n diagram.
It may also be pointed out that the angles of attack of the airplane are different at
various points of the V-n diagram. Consequently, the components of the resultant
aerodynamic force along and perpendicular to the chord of the wing (N and C inFig.3.7) would be different at different angles of attack. The structural analysis
needs to take this into account. For example, the angle of attack is positive and
high at point A and it is positive and low at point C. At points G and E the angles
of attack are negative. Books on Airplane structures may be consulted for details.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-9
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 9
Exercises
9.1 Define steady level co-ordinated-turn. An airplane having a weight of 11,000
N has a wing area of 15 m2 and drag polar of CD = 0.032 + 0.06C
L
2. Obtain the
radius of turn in a steady level coordinated turn at a speed of 160 kmph at sea
level from the following data.
CLmax
= 1.4, (THP)available
= 90 kW, maximum load factor = 3.5.
What is the time taken to turn through 180o?
[Answers: r = 124.6 m; t = 8.81 s]
9.2 Define load factor. What are its values in (a) level flight (b) free fall (c) in a
turn of radius 200 m at a speed of 100 m/s and (d) at the bottom of a loop of
radius 200 m at a speed of 100 m/s?
[Answers: (a) 1 (b) 0 (c) 5.19 (d) 6.097]
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 10
Performance analysis VI – Take-off and landing
(Lectures 32-34)Keywords: Phases of take-off flight — take-off run, transition and climb; take-
off distance; balanced field length; phase of landing flight; landing distance.
Topics
10.1 Introduction
10.2 Defini tions of take-off run and take-off distance
10.3 Phases of take-off f light
10.3.1 Take-off ground run10.3.2 Transition and climb phases
10.4 Estimation of take-off performance
10.4.1 Distance covered and time taken during ground run
10.4.2 Various speeds during take-off run
10.4.3 Distance covered and time taken during transition phase
10.4.4 Distance covered and time taken during climb phase
10.4.5 Parameters influencing take-off run
10.4.6 Effect of wind on take-off run
10.4.7 Guidelines for estimation of take-off distance
10.4.8 Balanced field length, its estimation and effect of number of
engines on it.
10.5 Landing performance
10.5.1 Definition of landing distance
10.5.2 Phases of landing flight
10.5.3 Estimation of landing distance
10.6 Flap settings during take-off and landing
References
Exercises
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
Chapter 10
Lecture 32
Performance analysis VI – Take-off and landing –1
Topics
10.1 Introduction
10.2 Defini tions of take-off run and take-off distance
10.3 Phases of take-off f light
10.3.1 Take-off ground run
10.3.2 Transition and climb phases
10.4 Estimation of take-off performance
10.4.1 Distance covered and time taken during ground run
10.4.2 Various speeds during take-off run
10.1 Introduction
An airplane, by definition, is a fixed wing aircraft. Its wings can produce lift
only when there is a relative velocity between the airplane and the air. In order to
be airborne, the lift produced by the airplane must be at least equal to the weight
of the airplane. This can happen when the velocity of the airplane is equal to or
greater than its stalling speed. To achieve this velocity called ‘Take-off
velocity(VTO
)’ the airplane accelerates along the runway. Thus, an airplane
covers a certain distance before it can take-off. Similarly, when an airplane
comes in to land, the lift produced must be nearly equal to the landing weight.
Hence, the airplane has a velocity, called ‘Touch down speed (VTD)’, when it
touches the ground. It then covers a certain distance before coming to halt.
The estimation of take-off distance and landing distance are the topics
covered in this chapter.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
10.2. Definitions of take-off run and take off d istance
The horizontal distance covered along the ground, from the start of take-
off till the airplane is airborne is called the take-off run. However, to decide the
length of the runway required for an airplane, it is important to ensure that the
airplane is above a certain height before it leaves the airport environment. This
height is called ‘Screen height’ and is equal to 15 m (sometimes 10 m), which is
above the height of common obstacles like trees and electricity poles. The take-
off distance is defined as the horizontal distance covered by an airplane from the
start of the run till it climbs to a height equal to the screen height. It is assumed
that the weight of the airplane during take-off is the gross weight for which it is
designed and that the take-off takes place in still air.
10.3 Phases of take off flight
The take-off flight is generally divided into three phases namely (i) ground
run (ii) transition (or flare) and (iii) climb (see Fig.10.1a).
Fig.10.1a Phases of take-off flight
10.3.1 Take-off ground run
During the ground run the airplane starts from rest and accelerates to the
take-off speed (VT0
or V1). The flaps and engine(s) are adjusted for their take-off
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 4
settings. In the case of an airplane with tricycle type of landing gear, all the three
wheels remain in contact with the ground till a speed of about 85% of the VT0
is
reached. This speed is called ‘Nose wheel lift off speed’. At this speed the pilot
pulls the stick back and increases the angle of attack of the airplane so as to
attain a lift coefficient corresponding to take-off (CLT0
). At this stage, the nose
wheel is off the ground (Fig.10.1b) and the speed of the airplane continues to
increase. As the speed exceeds the take off speed the airplane gets airborne and
the main landing gear wheels also leave the ground.
When the airplane has a tail wheel type of landing gear, the angle of attack is
high at the beginning of the take-off run (Fig.10.1c). However, the tail wheel is
lifted off the ground as soon as some elevator effectiveness is gained
(Fig.10.1d). This action reduces the angle of attack and consequently the
drag of the airplane during most of the ground run. As the take-off speed is
approached the tail wheel is lowered to get the incidence corresponding to CLT0
.
When VT0
is exceeded, the airplane gets airborne.
The point at which all the wheels have left the ground is called ‘Unstick point’
(Fig.10.1a).
Fig.10.1b Nose wheel lift-off
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
Fig.10.1c Tail wheel type airplane at start of take-off run
Fig.10.1d Tail wheel type of airplane during middle part of take-off run
10.3.2 Transition and cl imb phases
During the transition phase the airplane moves along a curved path
(Fig.10.1a) and the pilot tries to attain a steady climb. As soon as the airplane
attains an altitude equal to the screen height, the take-off flight is complete. For
airplanes with high thrust to weight ratio the screen height may be attained during
the transition phase itself.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
10.4 Estimation of take-off performance
From the point of view of performance analysis, the following two
quantities are of interest.
(i) The take-off distance (s) (ii) The time (t) taken for it.
Since the equations of motion are different in the three phases of take-off flight,
they (phases) are described separately in the subsequent subsections.
10.4.1 Distance covered and time taken during ground run
The forces acting on the airplane are shown in Fig.10.1a. It is observed
that the ground reaction (R) and the rolling friction, μR, are the two additional
forces along with the lift, the drag, the weight and thrust ; μ is the coefficient of
rolling friction between the runway and the landing gear wheels. The equations of
motion are :
WT- D - μ R = a
g (10.1)
L + R – W = 0 (10.2)
Hence, R = W - L and
T - D - μ (W-L)
a = W/ g (10.3)
Further,
dV dV ds dVa = = = V
dt ds dt ds
Hence, ground run (s1) is given by:
1 1V V
V dV W V dVs = =1
a g T- D - μ(W-L)o o (10.4)
The time taken during ground run (t1) is given by:
1 1V V
dV W dVt = =1
a g T - D - μ (W-L)o o (10.5)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
Equations (10.4) and (10.5) can be integrated numerically, when the variations
of T, D and L are prescribed and µ is known. The value of µ depends on the type
of surface. Typical values are given in Table 10.1.
Type of surface Coefficient of
rolling friction (µ)
Concrete, wood or asphalt 0.02
Hard turf 0.04
Average field-short grass 0.05
Average field-long grass 0.1
Soft ground 0.1-0.3
Table 10.1 Coefficient of rolling friction
The thrust during take-off run can be approximated as T = A1 – B
1V
2
. The angle
of attack and hence, the lift coefficient (CL ) and the drag coefficient (CD ) can be
assumed to remain constant during the take-off run. With these assumptions, theleft-hand side of Eq.(10.1) becomes :
12 2T - D -μ (W -L) = A - B V - μ W - ρV S (C - μ C )1 1 D L2
= A – BV2
where A = A1 - µ W and
1B = B + ρS (C - μC )1 D L
2
Substituting in Eqs.(10.4) and (10.5) gives :
1
VW V dV W 2s = = ln A/(A - BV )1 12g 2gB A - BVo
(10.6)
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 8
and1
V A + B VW dV W 1t = = ln1 2g 2g AB A - B V A-BV 1o
(10.7)
Remarks:
i) The denominator in the integrands of Eqs.10.4 and 10.5, i.e. [T- D - µ (W - L)],
is the accelerating force during the take-off run. A good approximation to s1 and
t1 is obtained by taking an average value of the accelerating force (Fa) to be its
value at V = 0.7 V1 i.e.
Fa = [T- D - µ (W - L)]V = 0.7 V1
Consequently,
1V 2W VW V dV 1s = =1
g F 2g Fa ao (10.8)
and1
VWVW dV 1t = =1
g F g Fa ao (10.9)
ii) Generally the flaps are kept in take-off setting (partial flaps) right from the
beginning of the take-off run. Hence, CD
during the take-off run should include
the drag due to flaps and landing gear.
Reference 3.6, section 3.4.1 may be consulted for increase in CDO due to the flap
deflection and the landing gear. See also section 2.9 of Appendix A. The
proximity of the ground reduces the induced drag. As a rough estimate, the
induced drag with ground effect can be taken to be equal to 60% of that in free
flight at the same CL.
(iii) The take-off speed (VTO or V1) is (1.1 to 1.2) Vs ; where Vs is the stalling
speed with W = WTO and CL = CLTO . As mentioned in subsection 3.7.4, CLTO is
0.8 times CLland.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 9
10.4.2 Various speeds during take-off run
In the subsection 10.3.1 the nose wheel lift-off speed and take-off speed have
been explained. Section 6.7 of Ref.1.10 mentions additional flight speeds
attained during the ground run. A brief description of the speeds, in the sequence
of their occurance, is as follows.
(a) Stalling speed (Vs) : It is the speed in a steady level flight at W = WTO and
CL = CLTO.
(b) Minimum control speed on ground (Vmcg): At this speed, the deflection of full
rudder should be able to counteract the yawing moment due to failure of one
engine of a multi-engined airplane when the airplane is on ground.
(c) Minimum control speed in air (Vmca) : At this speed, the deflection of full
rudder should be able to counteract the yawing moment, due to failure of one
engine of a multi-engined airplane if the airplane was in air.
(d) Decision speed (Vdecision) : This speed is also applicable to a multi-engined
airplane. In the event of the failure of one engine, the pilot has two options. ( I) If
the engine failure takes place during the initial stages of the ground run, the pilot
applies brakes and stops the airplane. (II) If the engine failure takes place after
the airplane has gained sufficient speed, the pilot continues to take-off with one
engine inoperative.
If the engine failure takes place at decision speed (Vdecision), then the distance
required to stop the airplane is the same as that required to take-off with one
engine inoperative. See subsection 10.4.8 for additional details.
(e) Take-off rotation speed (VR): At this speed the elevator is powerful enough to
rotate the airplane to attain the angle of attack corresponding to take-off.
(f) Lift-off speed (VLO) : This is the same as unstick speed mentioned in
subsection 10.3.1. This speed is between (1.1 to 1.2) VS.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 10
It is mentioned in Ref.1.10, chapter 6, that Vmcg, Vmca, Vdecision, VR lie
between VS and VLO.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 10
Lecture 33
Performance analysis VI – Take-off and landing –2
Topics
10.4.3 Distance covered and time taken during transition phase
10.4.4 Distance covered and time taken during climb phase
10.4.5 Parameters influencing take-off run
10.4.6 Effect of wind on take-off run
10.4.7 Guidelines for estimation of take-off distance
10.4.3 Distance covered and time taken during transition phase
From Fig.10.1a it is observed that during the transition phase the airplane
changes the direction of flight and its speed would increase from V1 to V
2. The
height attained during this phase and the horizontal distance traversed can be
obtained by treating the flight path as part of a circle. However, according to the
procedure given in Royal Aeronautical Data sheets (now called Engineering
Sciences Data Unit, ESDU for short), the increase in height during the transition
phase is small and the horizontal distance (s2) can be obtained by assuming that
the work done by the engine is used in overcoming the drag and in increasing the
kinetic energy of the airplane i.e.
)2 22 2 2 1
WT s = D s + (V - V
2g
Or 2 2
2 12
V - VWs =2g T-D
(10.10)
T and D in Eq.(10.10) are evaluated at the mean speed during this phase i.e., at
(V2 + V
1) / 2. The time taken (t
2) in transition is given by:
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
2 1
22
st =
0.5 (V + V ) (10.11)
V1 generally lies between (1.15 to 1.2) Vs and V
2 is (1.05 to 1.1) V
1.
10.4.4 Distance covered and time taken during climb phase
Since the vertical height covered during the transition has been ignored,
the horizontal distance covered in climb phase (s3) is the distance covered while
climbing to screen height i.e.
s3 = (Screen height) / tan (10.12)
where, is the angle of climb at velocity V2 :
sin γ =T-D
W
where, T and D are evaluated at V2.
The time taken in climb phase (t3) is:
t3 = (Screen height) / V
2 sin (10.13)
Hence, the take-off distance (s) and the time taken for it (t) are given by :
s = s1 + s
2 + s3 (10.14)
t = t1 + t2 + t3 (10.15)
Example 10.1
A jet airplane with a weight of 441, 450 N and wing area of 110 m2
has a
tricycle type landing gear. Its CLmax with flaps is 2.7. Obtain the take-off distance
to 15 m screen height and the time taken for it. Given that:
(i) V1 = 1.16 Vs
(ii) V2 = 1.086 V
1
(iii) CL during ground run is 1.15
(iv) Drag polar with landing gear and flaps deployed is CD = 0.044 +0.05C
L
2
(v) Thrust variation during take-off can be approximated as :
T = 128,500 – 0.0929 V2
; where V is in kmph and T is in Newton
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
(vi)Take-off takes place from a level, dry concrete runway (µ = 0.02) at sea level.
Solution:
0
Lmax LmaxT
C = 0.8 C = 0.8 ×2.7 = 2.16
1/2 1/22W 2 × 441450
V = = = 55.08 m/ssρS C 1.225 × 110 × 2.16Lmax
Hence,
V1 = 1.16 x 55.08 = 63.89 m/s
and V2 = 1.086 x 63.89 = 69.38 m/s.
For CL = 1.15,
CD = 0.044 + 0.05 x 1.152 = 0.1101
Hence, 2D L
1T - D - μ (W-L) = T- μW - ρV S {C - μC }
2
= 128500 - 0.0929 (3.6V)2 - 0.02 x 441450
– 0.5 x 1.225 x V2
x 110 (0.1101 - 0.02 x 1.15)
= 119671 – 7.0752 V2
Using Eqs.(10.6) and (10.7) the ground run (s1) and time taken for it (t1) are:
s1 =
441450
2 × 9.81× 7.0752 ln [119671/(119671-7.0752 x 63.89 x 63.89)] = 878.32 m
1 12 2
1 1 12 2 2
1
441450 (119671) + (7.0752) × 63.89t = ln
2 × 9.81(119671× 7.0752) 119671 - 7.0752 × 63.89
= 26.34 s.
The distance covered during transition (s2) is obtained as follows.
63.89 + 69.38
Average speed during transition = m/s = 66.635 m/s = 239.9 kmph2
Hence, thrust during this phase = 128500 – 0.0929 x 239.92 = 123,153 N
To get the drag during this phase it is assumed that CL equals C
LTOand it is given
by :
CLTO
= CLmax
(Vs / V1)2
= 2.16 / (1.16)2
= 1.605
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 4
Assuming the same drag polar as in the ground run gives:
CD = 0.044 + 0.05 x 1.6052 = 0.1728
Hence, D = 0.5 x1.225 x (66.635)2
x 110 x 0.1728 = 51695 NUsing Eqs.(10.10) and (10.11) gives :
2
2 2441450 69.38 - 63.89s = = 230.4 m
2 × 9.81 123159 - 51695
and t2 = 230.4 / 66.635 = 3.46 s.
During the climb phase, V = 69.38 m/s = 249.77 kmph.
Hence, T = 128500 - 0.0929 x 249.772 = 122704 N
To get the drag in the climb phase the lift coefficient should be known. For this
purpose L is taken roughly equal to W.
Hence,1
12
2
L
4414502C = W/ ρV S = = 1.362× 1.225 × 110 × 69.38
Consequently, CD = 0.044 + 0.05 x 1.362 = 0.1365
and D = 0.5 x 1.225 x (69.38)2 x 110 x 0.1365 = 44269 N
Hence, 122704 - 44269sin = = 0.1777441450
Consequently, tan = 0.1805.
Using Eqs.(10.12) and (10.13) gives:
s3 = 15/0.1805 = 83.1 m
and t3 = 15/(69.38 x0.1805) = 1.20 s.
Finally, s = s1 + s
2 + s
3 = 878.32 + 230.4 + 83.1 = 1192 m
and t = t1 + t
2 + t3 = 26.34 + 3.46 + 1.20 = 31.0 s.
Answers:
Take off distance = 1192 m ; Time taken for take-off = 31 s.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
10.4.5 Parameters influencing take-off run
The major portion of the take-off distance is the ground run. Hence if
ground run is reduced, the take-off distance is also reduced. From Eq.(10.8), it is
observed that the distance s1 is given by :
2
11
avg
VWs =
g T-D-μ W -L (10.16)
Let V1 = 1.1 VS. Recalling,S
Lmax
2WV =
ρSC, Eq.(10.16) can be rewritten as :
1
2
Lmax avg
1.21× 2Ws =
2gρSC T -D-μ W -L
=
Lmax avg
1.21 W/S
gρC T/W - D/W -μ 1-L/W (10.17)
The following observations can be made from Eq.(10.17).
i) The ground run increases when the wing loading (W / S) increases.
ii)The ground run also increases when decreases. Since decreases with
altitude, the take-off distance will be more when the altitude of the airport
increases.
iii)The ground run decreases as CLmax
increases. Hence, the high speed
airplanes which have high wing loading from consideration of cruise, employ
elaborate high lift devices to increase CLmax
.
iv)The take-off run decreases by increasing the accelerating force which mainly
depends on (T/W). It may be recalled from subsection 4.3.5 that the thrust of a jet
engine can be increased temporarily by using an afterburner. The thrust can also
be augmented by using an auxiliary rocket fired during the take-off run. In
shipboard airplanes a catapult is used to augment the accelerating force.
10.4.6 Effect of w ind on take-off run
While discussing the range performance it was shown, with the help of a
derivation in section 7.8, that the distance covered with respect to the ground
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
decreases when the flight takes place in the presence of head wind. Same effect
occurs during the take-off and the take-off distance reduces in the presence of
head wind. In a hypothetical case of head wind being equal to the stalling speed
(Vs), the airplane can get airborne without having to accelerate along the ground.
A quantitative estimate of the effect of wind velocity (Vw) on s
1, can be obtained
from Eq.(10.4), by replacing the limits of integration from (0 to V1) by (V
w to V
1)
i.e. :
1
w
1 with wind
VW V dV
s =g T- D - μ(W-L)
V
Thus, the head wind, though bad for range, is beneficial during take-off as it
reduces the take-off distance.
Airports have a device to indicate the direction of wind. The take-off flight takes
place in such a manner that the airplane experiences a head wind. This is
referred to as ‘Take-off into the wind’.
10.4.7 Guidelines for estimation of take-off distance
In subsections 10.4.1, 10.4.3 and 10.4.4, a procedure to estimate the
take-off distance has been presented. However, it is based on several
assumptions and consequently has significant amount of uncertainty. In actual
practice, there would be further uncertainty due to factors like condition of the
runway surface (wet or dry), and piloting technique. Hence, for the purpose of
preliminary design of airplane, the following guidelines can be used.
For airplanes with engine-propeller combination, the Federal Aviation
Regulations designated as FAR-23 (Ref.10.1) are used. Under these regulations,
the take-off distance to attain 50 feet (or 15 m) is obtained under certain
prescribed conditions. This distance is denoted here by ’sto23
’. Reference 10.2
has estimated sto23
for several airplanes. It is observed (Ref.10.2) that sto23
is
related to the following parameter.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
T0 T0
LT0
W W( ) ( )
S Pσ C
where,
(W/S)T0
= wing loading based on take-off weight.
(W/P) T0
= power loading based on take-off weight and sea level static power
output.
σ = density ratio = ρ/ρ0
CLT0
= Lift coefficient in take off configuration (about 80% of CLmax
in landing
configuration)
The above quantity is called take-off parameter for FAR-23 and denoted by
‘TOP23’ i.e.
T0 T0
23LT0
W W( ) ( )
S PTOP =σ C
(10.18)
Based on the data of Ref.10.2, the following relationship has been deduced in
Ref.3.18, pt.I, chapter 3.
sto23 (in ft) = 8.134 TOP23 + 0.0149 223TOP (10.19)
where, W/S is in lbs / ft2, W in lbs and P in hp.
When SI units are used this relationship takes the following form.
sto23 (in m) = 8.681x10-3 TOP
23+5.566x10-8 2
23TOP (10.20)
where W / S is in N / m2 , W in N and P in kW.
Example 10.2
Consider an airplane with the following features.
W/S = 2400 N / m2
, W/P = 24 N /kW , CLTO
= 1.6 and σ = 1.
Estimate the take-off distance for this airplane.
Solution :
The parameter ‘TOP23
’ in this case is :
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 8
TOP23
= 2400 x 24/(1 x 1.6) = 36000 N2/( m2
kW)
Using Eq.(10.20) gives,
sto23 = 8.681x10-3 x 36000 +5.566x10-8 x 360002 = 385.9 m or 1260 ft.
Answer : Take-off distance = 385.9 m or 1260 ft.
As regards the airplanes with jet engines, the take-off parameter (TOP) is defined
as :
T0
LT0 T0
W( )
STOP =T
σC ( )W
(10.21)
where, T = sea level static thrust.
Reference 3.9 chapter 5, gives a curve as guideline for sto in feet and TOP in
lbs / ft2
. However, when a second order equation is fitted to that curve, the
relationship can be expressed in SI units in the following form.
sto (in m) = 0.1127 TOP +1.531 x 10-6 TOP2
(10.22)
Example 10.3
Consider a jet airplane with the following features.
W/S = 5195 N/m2
, T / W = 0.3, CLT0
= 2.16 and σ = 1.
Estimate the take-off distance.
Solution :
In this case TOP is :
TOP = 25195= 8017 N/m
1×2.16×0.3
From Eq.(10.22) :
sto = 0.1127x8017 +1.531 x 10-6 x 80172 = 1002 m.
Answer : Take-off distance = 1002 m.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 10
Lecture 34
Performance analysis VI – Take-off and landing – 3
Topics
10.4.8 Balanced field length, its estimation and effect of number of
engines on it.
10.5 Landing performance
10.5.1 Definition of landing distance
10.5.2 Phases of landing flight
10.5.3 Estimation of landing distance
10.6 Flap settings during take-off and landing
10.4.8 Balanced field length and i ts estimation
Take-off is a critical phase of flight operation and various eventualities are
taken into account to arrive at the length of the runway required for the operation
of the airplane. In the case of multi-engined airplane, the possibility of the failureof one of the engines during take-off is an important consideration. If the engine
failure takes place during initial stages of ground run, then the pilot can apply the
brakes and bring the airplane to halt. If the engine failure takes place after the
airplane has gained sufficient speed, then the following two alternatives are
available.
(a) Apply brakes and stop the airplane, but this may need much longer runway
length than in the case of take-off without engine failure.
(b) Instead of applying brakes, continue to fly with one engine inoperative and
take-off; but the take-off distance would be longer than when there is no engine
failure.
These two alternatives indicate the possibility of a speed, called “Decision
speed”. If the engine failure occurs at the decision speed, then the distance
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 2
required to stop the airplane is the same as that required to take-off with one
engine inoperative. The take-off distance required when engine failure takes
place at the decision speed is called ‘Balanced field length (BFL)’. It is estimated
as follows.
FAR 25 (see Ref.10.1) is used as a set of regulations for obtaining the take-off
distance of jet airplanes. The regulations also prescribe a procedure to calculate
the balanced field length (BFL). Reference 10.2 has estimated BFL for many jet
airplanes and observed that BFL is a function of TOP defined in Eq.(10.21).
Based on this data, the BFL in feet, when W/S in lbs / ft2
is given as (Ref.3.18,
Pt.I, chapter 3):
BFL (in ft) = 37.5 TOP (in lbs / ft2) (10.23)
When SI units are used, Eq.(10.23) takes the following form.
TO
LT0 TO
W( )
SBFL (inm) = 0.2387T
σ C ( )W
(10.24)
where W / S is in N / m2
.
Remark :
(i) Effect of number of engines on BFL :
The data in Ref.10.2, on which Eq.(10.23), is based, shows some scatter
(Fig.3.7 of Ref.10.2). However, the data for airplanes with two, three and four
engines show some definite trend; the BFL is more as the number of engines
decrease. This is expected, as for a two engined airplane, when one engine is
inoperative, the thrust available would decrease to half of the full thrust, whereas
for an airplane with four engines, with one engine inoperative, the thrust available
would be three fourth of the full thrust. Consequently, BFL would be less for a
four engine airplane as compared to that for a two engined airplane. Perhaps,
based on this argument, Ref.3.9, chapter 5, suggests three different lines for BFL
vs TOP curve for airplane with two three and four engines. In SI units these lines
can be expressed as:
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 3
For two engined airplane: BFL (in m) = 0.2613 TOP (in N / m2
) (10.25)
For three engined airplane: BFL (in m) = 0.2387 TOP (in N / m2
) (10.26)
For four engined airplane: BFL (in m) = 0.2196 TOP (in N / m2
) (10.27)
Example 10.4
Consider the airplane of example 10.3 and obtain the balance field length.
Solution:
In this case :
W / S = 5195 N/m
2
, σ = 1.0 , CLTO = 2.16 and T/ W = 0.3.
Consequently, TOP is 8017 N/m2.
Using Eqs (10.25) to (10.27) the BFL would be (a) 2095 m for an airplane
configuration with two engines, (b) 1914 m for three engine configuration and
(c) 1761 m for four engine configuration. Comparing sto and BFL in examples
10.3 and 10.4, it is seen that is BFL is nearly twice of sto .
(ii) See Appendices A and B for calculation or take-off distance for a piston
engined airplane and a jet airplane respectively.
10.5 Landing performance
10.5.1 Defini tion of landing distance
While describing the take-off distance it was mentioned that the airplane should
clear the screen height before it leaves the airport environment. For the same
reason, the landing flight begins when the airplane is at the screen height. The
landing distance is defined as the horizontal distance that the airplane covers in
descending from the screen height and to come to halt. In actual practice, the
airplane does not halt on the runway. After reaching a sufficiently low speed the
pilot takes the airplane to the allotted parking place.
10.5.2 Phases of landing
Figure 10.2 shows the phases of landing flight for an airplane with tricycle type
landing gear.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 4
Fig.10.2 Phases of landing flight
During the final approach phase, the airplane performs a steady descent. The
flight velocity in this phase is called approach speed and denoted by V A
. During
the flare, the pilot makes the flight path almost horizontal. In the float phase the
pilot gently touches the main wheels to the ground. This is done gradually so that
the vertical velocity of the airplane is not more than about 4 m/s. The flight speed
at the point of touch down is denoted by VT. It is about 90% of V A. After the
touch down, the airplane rolls for a period of about 3 seconds during which the
nose wheel is gently lowered to touch the ground. Brakes are not applied in this
phase as their application would produce a large decelerating force which would
cause a large nose down moment and the nose wheel may hit the ground with a
bang. After the three wheels have touched the ground, the brakes are applied
as well as other devices like reverse thrust or reversed pitch of propeller are
deployed. The ground run is said to be over when the airplane comes to halt or
attains a low speed when it can turn off the runway and go to the parking place.
10.5.3 Estimation of landing distance
This can be done in a way similar to the estimation of the take-off distance
i.e., by writing down equations for each phase of the flight. However, the
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5
estimation cannot be done accurately as the flare and float phases depend very
much on the judgment of the pilot.
Royal Aeronautical Society Data sheets (presently called Engineering Science
Data Unit or ESDU) have given a simple method which amounts to assuming a
constant deceleration and calculating the distance to decelerate from V A and to
come to a halt i.e.
sland
= - (V A
)2
/ 2a (10.28)
where, a = -1.22 m/s2 (or 4ft/s
2) for simple braking system
= -1.52 m/s2 (or 5 ft/s
2) for average braking system.
= - 1.83 m/s2 (or 6 ft/s2) for modern braking system and
= - 2.13 to 3.0 m/s2 (or 7 to 10 ft /s
2) for airplane with modern braking
system and reverse thrust or reverse pitch propellers.
The approach speed (V A
) depends on factors like stalling speed under
approach conditions, minimum speed at which adequate control is possible and
the type of approach viz. visual landing or instrumented landing system or aircraft
carrier deck approach. As a first estimate V A
can be taken as 1.3 Vs.
Example 10.5
Obtain the landing distance for the airplane in example 10.1. Assume that
the airplane has modern braking system with reverse thrust and that V A
= 1.3 Vs.
Solution:
From example 10.1, W = 441, 450 N, S = 110 m2 ,
CLmax during landing = 2.7.
Hence,
1/2
s 2 ×441450V =1.225×110×2.7
= 49.24 m/s
Consequently, V A = 1.3 x 49.24 = 64.01 m/s.
Taking a = - 2.13 m/ 2s , the estimate of landing distance is :
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6
land
264.01s = - = 961.9 m
2 × (-2.13)
Answer : Landing distance = 961.9 m
Remarks:
i) Appendix A also estimates the landing distance using Eq.(10.28). Appendix B
uses a different formula.
ii) The landing distance is proportional to (V A
)2 and consequently it is proportional
to (Vs)2
. The following observations can be made by noting that (Vs)2
equals
2W/(SCLmax
).
(a) The landing distance increases with increase of (W/S) and the altitude of
landing field. (b) The landing distance decreases with increase of CLmax
.
iii) The use of reverse thrust and reverse pitch propeller to reduce the landing
distance has been mentioned earlier. The landing run can also be decreased by
using (a) arresting gear, (b) drag parachute and (c) spoilers.
The arresting gear is used for airplane landing on the deck of a ship.
The drag parachute, when opened, increases the drag significantly and reduces
the landing run.
The spoilers are located on the upper surface of the wing. When deflected up,
the spoiler disturbs the flow, resulting in reduction of lift and increase of drag.
Spoiler ailerons are shown in Fig.1.2c. When used as a device to produce a
rolling moment, the spoiler aileron is deflected only on the left or the right wing
half. The lift on that wing half is reduced and the airplane rolls. Whereas, during
landing, the spoiler ailerons on both the wing halves are deployed
simultaneously. This results in a large reduction in lift and increase in drag. Boththese effects help in reducing the landing run.
iv) Like take-off distance the landing distance is also reduced by head wind.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7
10.6 Flap settings during take-off and landing
It is mentioned in subsection 10.4.1, that theLmax
C during take-off is 80% of that
during landing. The flap setting during take-off is lower than the setting during
landing. The reasons for this difference are as follows.
Equation (10.17) shows that the take-off run depends on ambient density ρ ,
wing loading (W/S), maximum lift coefficient (CLmax) and the average accelerating
force. Out of these parameters, as pointed out earlier, the values of (W/S) and
(T/W) are chosen based on considerations of cruise, maximum speed etc. In this
situation, the choices available to reduce the take-off distance are (a) CLmax and
(b) average accelerating force during the take-off.
It may be pointed out at this juncture that a high value of CLTO would reduce V1
and hence the take-off run (Eq.10.17). However, the high value of CLTO would
also result in high value of CD and consequently high value of drag and a lower
accelerating force. This would tend to increase the take-off run (Eq.10.17). On
account of these two opposing effects, there is an optimum value of C LTO and
the corresponding flap setting, that would result in lowest take-off run.
On the other hand, during landing the approach speed and the touch down
speed would be lowest when the CLmax is highest. Further, the high value of CD
associated with high value of CLmax would also increase the decelerating force
during landing run and consequently reduce it. Thus a high value of CLmax is
beneficial for reducing the landing run & distance.
Keeping these two aspects in view, the flap setting during the take-off is
lower than that during the landing. As a guideline it is mentioned in Ref.3.15,
chapter 5, that the flap deflection for take-off f TOδ is about half of that during
landing f Landδ .The deflection of the leading edge slat during take-off, is about
two-thirds of that during landing.
It may be further added that during landing run, after all the landing gear
wheels have touched ground, the lift is not needed. Hence, in airplanes with
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 8
provision of spoilers, they (spoilers) are deployed during the landing run to
reduce the lift and increase the drag.
Acknowledgements
The major portion of the lecture material was prepared when the author was an
AICTE Emeritus fellow at IIT Madras. Support of AICTE and IIT Madras is
greatfully acknowledged. He is also grateful to Prof.J.Kurian, Prof.P.Sriram and
Prof.K.Bhaskar, the Heads of the department of Aerospace engineering, IIT
Madras and to Prof. K. Mangala Sunder, Co-ordinator NPTEL, and Prof.S.R.
Chakravarthy, Co-ordinator for Aerospace Engineering, NPTEL, IIT Madras for
providing facilities to carry-out the work.
The lecture material in powerpoint format was reviewed by Prof. K.
Sudhakar, Dept.of Aerospace Engg. , IIT Bombay, Prof.C.V.R. Murti, formerly of
IIT Kanpur and now at Institue of Aeronautical Engg. near Hyderabad,
Prof. B.S.M. Augustine, Sathyabama University, Prof.K.Elangovan ,Dept.of
Aeronautical Engg., M.I.T., Chennai, Prof. R.Rajasekhar, Park college of
Engg.&Technology, Coimbatore and Mr.K.Ibrahim , former chief deisgner, HAL.
The author is indebted to them for their comments which helped in considerably
improving the text. Prof.C.V.R. Murti made detailed comments and even went
through the revised draft. Special thanks are due to him.
The lecture material in the running matter format was reviewed by two reviewers
selected by NPTEL. The comments by the reviewers, helped in adding new
topics and giving explanatory notes. Author’s wife, Mrs. Suniti, also went through
the lecture material and her comments helped in refining the text. The author is
thankful to these persons.
The help of Mr. Amudan Arun Kumar and Mr.S.Ananth former B.Tech
students, Mr.Aditya Sourabh, Dual Degree student, Mr.M.Mahendran, M.S.
scholar, Mr.S.Gurusideswar, Ph.D. scholar and Sandip Chajjed, Project staff
Department of Aerospace Engineering and Ms. K. Sujatha of NPTEL Web studio,
IIT Madras is gratefully acknowledged.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 9
The author would like to record appreciation of Mr. G.Manikandasivam of
NPTEL Web studio, IIT Madras for painstakingly keying in several revisions of
the lecture material and also for preparing figures suitable for conversion to PDF
format.
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 10
References
10.1 Federal Aviation Regulations (FAR), Federal Aviation Administration,
Washington D.C. USA.
10.2 Loftin , Jr. L.K. “Subsonic aircraft evolution and the matching of size to
performance” NASA Reference publications ,1060, August 1980. This report can
be downloaded from the site “NASA Technical Report Server (NTRS)“ .
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Flight dynamics-I Prof. E.G. TulapurkaraChapter-10
Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1
Chapter 10
Exercises
10.1 Describe the various phase of take-off flight, Write down the equations of
motion during take-off run. Taking CD, C
L and T as constant during take-off run
show that the ground run (s1) is given by:
W Γs = ln1
S g ρ (C -μ C ) Γ -qD L 1
where W, S, g, and have the usual meanings, q1 = dynamic pressure at the
unstick point and
T-μWΓ =
S (C -μ C )D L.
10.2 A rocket motor firing for a short duration of say 10 s is proposed to be used
to reduce the take of run. Explain that a larger reduction in the take-off distance
would be achieved by using the rocket motor in the later part of the take-off run
than in the beginning of the take-off run.
10.3 Describe various methods to reduce the take-off distance and landing
distance.
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APPENDIX - A
PERFORMANCE ANALYSIS OF A PISTON ENGINEDAIRPLANE – PIPER CHEROKEE PA-28-180
(Lectures 35 - 37)
E.G. TULAPURKARA
S. ANANTHTEJAS M. KULKARNI
REPORT NO: AE TR 2007-1
FEBRUARY 2007(REVISED OCTOBER 2011)
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Performance Analysis of a piston enginedairplane – Piper Cherokee PA-28-180
E.G.Tulapurkara*, S Ananth$ and Tejas M Kulkarni$
ABSTRACT
The report is intended to serve as an example of performance calculation of a typical piston
engined airplane.
Problem statement: Obtain the following for the prescribed airplane:
Information about the airplane.
Drag Polar at cruising speed and during take-off condition.
Engine Characteristics. Variation of stalling speed with altitude for flaps up and flaps down
conditions.
Variations of the maximum speed (Vmax) and minimum speed (Vmin) with altitude.
Variations of maximum rate of climb (R/C)max and maximum angle of climb (γmax) with
speed and altitude. Variation of VR/Cmax and Vγmax with altitude. Values of absolute
ceiling and service ceiling.
Variations of range and endurance with flight speed in constant velocity
flights at cruising altitude. Speeds corresponding to R max and Emax .
Variation of minimum radius of turn (r min ) and maximum rate of turn
( maxψ ) at selected altitudes and variations of (Vrmin ) and ( ψmaxV
) with
altitude.
Take-off and landing distances.
* AICTE Emeritus Fellow, Department of Aerospace Engineering, IIT Madras
$ Third year B.Tech students, Department of Aerospace Engineering, IIT Madras
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Contents
1 Information about the airplane
1.1 Overall dimensions
1.2 Power plant
1.3 Weights
1.4 Wing geometry
1.5 Fuselage geometry
1.6 Horizontal tail geometry
1.7 Vertical tail geometry
1.8 Landing gear
1.9 Flight condition
1.10 Performance of PA-28-181 as given in Ref.3*
2 Estimation of drag polar
2.1 Estimation of DOWBC
2.2 Estimation of DOHC
2.3 Estimation of DOVC
2.4 Estimation of DLGC and DMISCC
2.5 Cooling drag and leakage drag
2.6 Estimation of parasite drag coefficient DOC
2.7 Estimation of induced drag coefficient DiC
2.8 Expression for drag polar during cruise
2.8.1 Slight modification of drag polar
2.9 Expression for drag polar during take-off condition
* Reference numbers in this Appendix relate to those given at the end of this appendix.
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3 Engine characteristics
3.1 Variation of engine BHP
3.2 Thrust horsepower available
4 Steady level flight
4.1 Variation of stalling speed with altitude
4.2 Variations of Vmax and Vmin with altitude
5 Steady climb performance
6 Range and endurance
6.1 Estimation of range in constant velocity flight
6.2 Calculation of BHP and fuel flow rate at different RPM’s and MAP’s at 8000 6.3 Sample calculations for obtaining optimum N and MAP for a chosen flight
velocity (V)
7 Turning performance
8 Take-off and landing distance estimates
8.1 Distance covered during take-off run
8.2 Distance covered during transition
8.3 Distance covered during climb phase
8.4 Landing distance estimate
9 Concluding remarks
Acknowledgements
References
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Appendix A
Lecture 35
Performance analysis of a piston engined airplane –1
Topics
1 Information about the airplane
1.1 Overall dimensions
1.2 Power plant
1.3 Weights
1.4 Wing geometry
1.5 Fuselage geometry
1.6 Horizontal tail geometry
1.7 Vertical tail geometry
1.8 Landing gear
1.9 Flight condition
1.10 Performance of PA-28-181 as given in Ref.3
2 Estimation of drag polar
2.1 Estimation of DOWBC
2.2 Estimation of DOHC
2.3 Estimation of DOVC
2.4 Estimation of DLGC and DMISCC
2.5 Cooling drag and leakage drag
2.6 Estimation of parasite drag coefficient DOC
2.7 Estimation of induced drag coefficient DiC
2.8 Expression for drag polar during cruise
2.8.1 Slight modification of drag polar
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2.9 Expression for drag polar during take-off condition
1. Information about the airplane
Airframe: Piper Cherokee PA-28-180
Type: Piston-engined propeller driven low speed recreational airplane.
Manufacturer and country of origin: The Piper Airplane Corporation, USA.
1.1 Overall dimensions*
Length : 7.148 m
Wing span : 9.144 m
Height above ground : 2.217 m
Wheel base : 1.897 m
Wheel track : 3.048 m
1.2 Power plant
Name : Lycoming O-360-A3A
Rating : 180BHP (135 kW) at 2700 RPM
Weight : 129 kgf (1265.5 N)
Number : 1
Propeller : 1.88 m diameter, fixed pitch.
1.3 Weights
Maximum take-off weight : 1088 kgf (10673.28 N)
Empty weight : 558 kgf (5473.98 N)
Fuel capacity : 50 US gallons (189 litres) usable 178.63 litres
Payload : 468.1 kgf (4592.06 N)
Maximum wing loading : 73.2 kgf/m2 (718.1 N/m2)
Maximum power loading (P/W) : 0.1241 kW/kgf (0.01265 kW/N)
1.4 Wing geometryPlanform shape : Trapezoidal near root, rectangular afterwards
and elliptical fillets at the tip.
Span (b) : 9.144 m
Reference area (S or SRef ) : 14.864 m2
* The dimensions / areas are based on Fig.1 and the additional details given in Ref.2.
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Flap area : 1.384 m2
Aileron area : 1.003 m2
Airfoil : NACA 652 – 415, t/c = 15 %, Clopt = 0.4
Root chord : 2.123 m
Tip chord : 1.600 m
Quarter chord Sweep : 1.480
Dihedral : 60
Twist : -20
Incidence : 4.620 at root, 2.620 at tip
High lift devices : Simple flaps having 3 different settings : 100 ,
250 and 400
Derived parameters of wing:
(i) Aspect ratio (A ) :
A = b2/ S = 9.144 2 / (14.864) = 5.625
(ii) Root chord of equivalent tropazoidal wing (creq) :
req t
bS= (c +c )
2
Or 14.864 = req
9.144(c +1.60)
2
creq= 1.651 m(iii) Root chord of exposed wing (cre):
From Fig.1, the maximum fuselage width is 1.168 m. Hence semi span of the exposed wing
(be / 2) is:
e b 1= (9.144-1.168)=3.988m
2 2
(iv) The root chord of exposed equivalent wing (cre) is obtained as follows.
An expression for the chord of the equivalent wing is
yc=1.651- (1.651-1.600)
b/2
Hence,
re
0.584c =1.651- (1.651-1.600)=1.644 m
9.144/2
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(v) Taper ratio of the exposed wing (λ e) is:
λ e= 1.6 / 1.644 = 0.9732
(vi) Mean aerodynamic chord of the exposed wing ( ec )
2 2
e ee
ree
(1+λ +λ )2 2 (1+0.9732+0.9732 )
c = c = ×1.6443 1+λ 3 1+0.9732
=1.622 m
(vii) Planform area of the exposed wing (Se) is:
Se = 3.988 (1.644+1.6) = 12.937 m2
(viii) Wetted area of exposed wing (Swet)e is :
(Swet)e = 2 Se {1+1.2 x (t/c) } = 2 x 12.937 { 1 + 1.2 x 0.15} = 30.53 m2
1.5 Fuselage geometry
Length (l b) : 6.547 m (measured from Fig.1)
Frontal area (S b) : 1.412 m2 (Ref.2 p.179)
Maximum width : 1.168 m
Derived parameters for fuselage:
(i) Equivalent diameter (de) of fuselage :
2
e e
πd =1.412 ord =1.341m
4
(ii) Height of maximum cross section (hmax)
hmax = 1.412 / 1.168 = 1.209 m.
(iii) Rough estimate of wetted area of fuselage (Ss)e is :
(Ss)e = 0.75 x ( perimeter of the maximum cross section ) x l b
= 0.75 (1.209 + 1.168) x 2 x 6.547 = 23.34 m2.
(iv) Fineness ratio of fuselage (Af ) :Af = l b/ de = 6.547 / 1.341 = 4.882
1.6 Horizontal tail geometry
Plan-form shape : Rectangular with elliptical fillets at tips.
Span : 3.048 m
Area : 2.267 m2
Root chord and tip chord : 0.762 m
Airfoil : NACA 0012.
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Derived parameters of horizontal tail:
(i)Aspect ratio = At = 3.0482 / 2.267 = 4.098
(ii)Exposed area of horizontal tail = area of h.tail – area inside fuselage ≈ 2.15 m2 Hence wetted area of h.tail (Swet)h is :
(Swet)h : = 2 x 2.15 [1+1.2 x 0.12] = 4.919 m2
1.7 Vertical tail geometry
Span : 1.219 m
Area : 1.059 m2
Root chord : 1.182 m
Tip chord : 0.517 m
Quarter chord sweep : 21.80
Airfoil : NACA 0010.
Derived parameters of vertical tail:
(i) Taper ratio : 0.4374
(ii) Aspect ratio : 1.403
(iii)Exposed area of vertical tail : same as area of v.tail = 1.059 m2
(iv) Wetted area of v.tail (Swet)v is :
(Swet)v = 2 x 1.059 { 1+ 1.2 x 0.1} = 2.372 m2
(v) Mean aerodynamic chord of vertical tail is :
Vtc = (2/3) x 1.182 x (1+0.4374 + 0.43742 ) /(1+0.4374) = 0.893 m.
1.8 Landing gear
Type : Non-retractable, nose wheel type with fairing.
Number of wheels : Nose 1, main 2, all same size.
Thickness : 0.135 m
Diameter : 0.4547 m
Wheel base : 1.897 m
Wheel track : 3.048 m
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1.9 Flight condition
Altitude : 2438 m (8000 )
Density : 0.9629 kg/m3
Speed of sound : 330.9 m/s
Kinematic viscosity ( υ ) : 40.17792 10 (m2/s)
Flight speed : 237 km/hr (65.83 m/s)
Mach number : 0.1992
Weight of the airplane : 1088 kgf (10673.28 N)
1.10 Performance of PA-28-181$ as given in Ref.3.
Maximum take-off weight : 1157 kgf (2550 lbf)
Power plant rating : 135 kW (180 BHP)
Wing loading : 73.3kgf/ m2
Maximum level speed : 246 kmph
Cruising speed : 237 kmph
Stalling speed : 86 kmph, with flaps down condition
Maximum rate of climb : 203 m/min at sea level
Service ceiling : 4035 m
Take-off run : 350 m
Take-off to 15m : 488 m
Landing run : 280 m
Landing distance from 15m : 427 m
Range with allowance for taxi, take-off, climb, descent and 45 min reserves at 6000 feet
(1830 m) : 924 km at 55 % power ; 875 km at 65 % power ; 820 km at 75 % power.
$Remark: The performance calculations are being done for PA-28-180 as a large
amount of data on the airplane, the engine and the propeller are available in Ref.2. However,
information on actual performance of this airplane is not given there. Ref.3 (which is easily
accessible) contains information about PA-28-181 which is only slightly different from
PA-28-180.
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Fig.1. Three-view drawing of Piper Cherokee PA-28-180
Dimensions in m
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2. Estimation of drag polar
Following Ref.1, the drag polar is assumed to be of the following form.
22L
D Do Do L
CC =C + = C +KC
πAe (1)
Do DoWB DoV DoH DoMiscC =C +C +C +C (2)
where suffixes WB, V, H and Misc denote wing-body combination, vertical tail, horizontal tail
and miscellaneous items respectively.
2.1 Estimation of CDOWB
From Ref.1, section 3.1.1, at low subsonic Mach number, DoWBC is given by the following
expression.
4 wet eDoWB fw LS
Ref
S eB BfB WB Db3
f Ref Ref
(S )t tC ={C [1+L( )+100( ) ]R +c c S
(S )l S60C [1+ +0.0025( )] }R +C
(l /d) d S S
(3)
Cfw = skin friction drag coefficient of wing (see below).
L = 1.2 when (t/c)max of the airfoil used on the wing is located at (x/c) 0.3, which is the case
here.
t/c = 0.15.
R LS = 1.07 from Fig. 3.3 of Ref.1; note M < 0.25 and = 0.
(Swet)e = 30.53 m2
Sref = 14.864 m2
CfB = skin friction drag of fuselage (see below)
l b/de= 4.882
(Ss)e = 23.34 m2
R WB = wing - body interference correction factor (see below)
CDb = base drag coefficient.Base drag contribution is neglected as the fuselage
gradually tapers down to zero width.
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Skin friction drag of wing (Cfw):
This quantity depends on the lower of the two Reynolds numbers viz.
(i) Reynolds number based on mean aerodynamic chord of exposed wing e(c ) and (ii) cut-offReynolds number (Recut-off ) based on the roughness of the surface.
Reynolds number based on ec is :
R e = (1.622 x 65.83) / (0.17792 x 10-4) = 6 x 106
The roughness parameter is (l/k) where l is the reference chord, here 1.662 m.
The value of k corresponding to mass production point, from Ref.1, is :
3.048 x 10-5 m. Hence, l/k = 1.622/(3.048 x 10-5) = 53215
Corresponding to this value of (l/k), Recut-off from Fig 3.2 of Ref.1 is 4 x 106.
Since Recut-off is lower, Cfw depends on it. Corresponding to Recutoff , the value of Cfw from
Fig .3.1 of Ref.1 is 0.00348
Skin friction coefficient of fuselage (CfB):
The Reynolds number based on length of the fuselage (R lB) is:
R lB = 6.547 x 65.83 / (0.17792 x 10-4
) = 24.22 x 106
In this case l/k = 6.547/(3.048 x 10-5
) = 2.14 x 105
Recut-off in this case,Fig 3.2 of Ref.1, is : 18 x 106
Hence CfB, from Fig 3.1 of Ref.1, is 0.00272
R WB : From Fig 3.5 of Ref.1, for M < 0.25 and R lB = 24.22 x 106, R WB = 1.06.
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Hence,
4
DWB
3
30.53C ={0.00348[1+1.2(0.15)+100(0.15) ]×1.07×
14.864
60 23.34+0.00272[1+ +0.0025×4.882] }1.06+04.882 14.864
30.53 23.34={0.00348[1+0.18+0.051]×1.07× +0.00272[1+0.5156+0.0122]× }1.06
14.864 14.864
={0.00941+0.006525}×1.06=0.009975+0.006917=0.01689
2.2 Estimation of CDoH
The drag coefficient of horizontal tail is given by (Ref.1) as:
wet4 hDoH fH LS
ref
(S )t tC = C [1+L( )+100( ) ]R
c c S (4)
The tail has NACA 0012 airfoil. Hence, t/c = 0.12
The wetted surface area of horizontal tail (Swet)h = 4.919 m2
Sref = 14.864 m2
The mean aerodynamic chord of exposed horizontal tail is taken equal to the root chord of the
horizontal tail i.e. etc = 0.762 m.
Reynolds number based on etc is :
0.762 x 65.83 / (0.17792 x 10 -4) = 2.52 x 106
The value of l/k in this case is:
0.762/(3.048 x 10-5) =25000
Hence, Recutoff =1.5 x 106
Consequently, CfH = 0.00414
For = 0 and M< 0.25, R LS =1.07
Finally, CDoH = 0.00414[1+1.2 x 0.12 + 100(0.12)4]1.07x4.919/14.864 = 0.00171
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2.3 Estimation of CDoV
The drag coefficient CDoV is given by:
wet4 vDoV fV LS
ref
(S )t tC = C [1+L( )+100( ) ]R
c c S (5)
The vertical tail has NACA 0010 airfoil; Hence, t/c=0.10
Wetted surface area of vertical tail = 2.372 m2
Sref = 14.564 m2
Reynolds number based on Vtc is:
0.893 x 65.83 /(0.17792 x 10
-4
)=3.30 x 10
6
The value of l/k is 0.893/(3.048 x 10-4) = 29298
Recutoff = 1.9 x 106
Consequently, CfV = 0.00394
Corresponding to M < 0.25 & = 21.8°, R LS=1.07
Finally, CDOV = 0.00394[1+1.2 x 0.1 + 100(0.1)4
] x1.07x2.372/14.864
= 0.00076
2.4 Estimation of CDOLG and CDOMisc
The landing gear drag coefficient can be obtained from Ref.1. However, the values for Piper
Cherokee given in Ref.2 are used as guidelines. Table 4.3 of Ref.2 indicates that parasite area of
landing gears components would be (a) wheel strut 0.19 ft2, (b) wheels 0.44 ft2 (c) wheel pants
0.40 ft2
(see remark on p.180 of Ref.2). Thus, parasite drag area of landing gear would be:
0.19 + 0.44 + 0.4 = 1.03 ft2= 0.0957 m
2
Again from Table 4.3 of Ref.2 The sum of the parasite drag areas of miscellaneous items like
beacon, antennas etc is 0.52 ft2 or 0.0483 m
2. Thus,
CDOLG + CDOMisc = (0.0957 + 0.0483)/14.864 = 0.00645+0.00325 = 0.00970
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Remarks:
i) Reference 7, chapter 5 mentions that the drag of landing gear (CDLG)without fairing is
about 35% of the sum of the drags of major components viz. wing-body, horizontal
tail and vertical tail. For landing gear with fairings, CDLG would be about 25% of the
aforesaid sum. In the present case :
CDWB + CDHT + CDVT is (0.01689 + 0.00171 + 0.00076 = 0.01936).Thus CDLG of
0.00645, estimated above appears reasonable.
ii)
The value of CDmisc of 0.00325 is about 17% of the aforesaid sum and appears
reasonable.
2.5 Cooling drag and leakage drag
These drags are important for piston engined airplanes. Appendix A of Ref.7 gives some
guidelines. However, Ref.2, p.179 mentions that the sum of the two drags could be
approximately taken into account by multiplying the sum of all the other drags by a factor of 1.2.
2.6 Estimation of parasite drag coefficient (CDO)
In light of the above discussion CDo can be expressed as:
CDO = 1.2 (CDOWB + CDOHT + CDOVT + CDOLG + CDOMisc) (7)
In the present case,
CDO = 1.2(0.01689+0.00171+0.00076+0.00645+0.00325)
= 1.2 x 0.02905 = 0.0349 (8)
Remark:
For comments on the above value of CDO see remark at the end section 2.8
2.7 Estimation of induced drag coefficient (CDi)
The quantity K in Eq.(1) is given by:
1K =
πAe
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where, A = Aspect ratio of the wing, e = Oswald efficiency factor
Following Ref.1 section 2.3.1 the Oswald efficiency factor is expressed as:
wing fuselage other
1 1 1 1= + +
e e e e
(9)
Figure 2.4 of Ref.1 presents ewing for unswept wings of rectangular and tapered planforms. In the
present case the taper ratio ( ) is almost unity. The value of ewing for rectangular wing of
A = 5.625 is 0.845.
Further, for a fuselage of rectangular cross section and wing of aspect ratio 5.625, Fig.2.5 of
Ref.1 gives:
b b
fuse ref
S1( ) =1.6 ; S = frontal area of fuselage
e S
Orfuse
1 1.412=1.6× = 0.152
e 14.864
Again from Ref 1,other
1=0.05
e
Consequently,
1 1= +0.152+.05=1.3854
e 0.845
e = 0.722
Hence, 1
K = = 0.0784π×5.625×0.722
CDi = 0.0784 2LC (10)
2.8 Expression for drag polar during cruise
Combining Eqs.(8) and (10) gives the drag polar in cruise condition as:
CD = 0.0349 + 0.0784 2
LC (11)
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The value of (L/D)max is given by
max
DO
1(L/D) =
2 C K
Substituting the values of CDO and K from Eq.(11) gives:
max
1(L/D) = =9.56
2 0.0349×0.0784
2.8.1 Slight modification of the expression for drag polar
The value of (L/D)max is an indication of the aerodynamic efficiency of the airplane.
From Ref.7 chapter 3 it is observed that the value of (L/D)max for Piper Cherokee is slightly more
than 10. Thus, the estimated value of 9.56 is lower than that of the actual airplane and suggests
need for slight modification. References 8 and 9 give the values of CDO and e for similar
airplanes, with non-retracted landing gear, made by manufacturers of Piper, Cessna and Beech
aircraft. These values are presented in Table 1.
Airplane A CDO L/D e
Piper Cherokee 6.02 0.0358 10 0.758
Piper J-3 cub 5.81 0.0373 9.6 0.75
Cessna Skyhawk 7.32 0.0317 11.6 0.747
Beechcraft D17S 6.84 0.0348 10.8 0.76
Table 1 Values of A, CDO, (L/D) and e for similar airplane
From Table 1 it is seen that the estimated value of CDO in the present case appears reasonable.
However, the value of e should perhaps be higher, say 0.76 . With CDO of 0.0349 and e = 0.76
the drag polar becomes :
2
D L
1C = 0.0349+ C
π×5.625×0.76
Or 2
D LC = 0.0349+0.0755C (12)
The expression in Eq.(12) would give (L/D)max of 9.81.
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Remarks:
i) It may be added that Piper Cherokee is an airplane famous in its class but is of
older design. The current trend is to have (a) smoother surfaces which would
reduce CDO to about 0.032 and (b) wing of larger aspect ratio of 8 and above,
which would give K of around 0.053. These would give (L/D)max of in excess of
12.
ii) For subsequent calculations, the following expression for drag polar is used.
2
D LC =0.0349+0.0755C
2.9 Expression for drag polar during take-off condition
To obtain the drag polar under take-off condition, the flight velocity is taken as 1.2 Vs, where Vs
is the stalling speed with flaps in take-off condition (δf =100). In the present case, CLmax with 100
flap deflection, from Ref.2 is 1.42. Hence,
s
2×10673.28V = = 28.73m/s
1.42×1.225×14.864
Consequently, VTo = 1.2 28.73 = 34.47m/s
Reynolds number based on mean aerodynamic chord of the exposed wing in take-off condition
is:
6
-6
1.622×34.47 = 3.83×10
14.6×10
We notice that this Reynolds number is very close to the cutoff Reynolds number for the wing
(4 106) obtained in Section 2.1. Thus, the value of C f and other calculations will remain almost
the same. Hence, DoC for the airplane in take-off condition, without the flap, can be taken as
0.0349.
Similarly K, without the flap, can be taken as 0.0755.
The correction to the drag polar for flap deflection, is carried-out using the following steps.
The flap type is plain flap.
From Fig.1, the ratio of flap chord to wing chord is 0.16 and flap deflection is 100 .
The ratio of the area of the flapped portion of the wing to the wing plan-form area is 0.4827.
The ratio of the span of the flapped portion of the wing (including the fuselage width) to the total
span is 0.604.
The ratio of the fuselage width to the wing span is 0.127; the wing aspect ratio is 5.625.
Following Ref.1, section 3.4.1
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Dflap = Dp Di DintΔC ΔC + ΔC + ΔC ,
where, DpΔC = increase in profile drag coefficient due to flaps,
DiΔC = increase in induced drag coefficient due to flaps and
DintΔC = increase in interference drag due to flaps.
The increment in CLmax due to 100 flap deflection, Δ CLmax, as noted earlier, is 0.09.
Using these data and interpolating the curves given in Ref.1, section 3.4.1, gives dpC , the
increment in the drag coefficient of airfoil due to flap deflection, as 0.008. Hence,
DpΔC = dpΔC x (area of flapped portion of the wing/ wing area)
= 0.008 x 0.4827 = 0.0038
According to Ref.1, the increase in induced drag coefficient ( DiΔC ) due to flap deflection is
2 2Lmaxf ΔK ×ΔC . Using Ref.1, section 3.4.1 f ΔK is estimated as 0.163.
Consequently, 2 2DiΔC = 0.163 ×0.09 = 0.00022
The interference drag due to deflection, of plain flaps is negligible.
Thus, the parasite drag coefficient in take-off condition is
DoC = 0.0349+0.0038+0.00022= 0.03892 0.0389
Hence, the drag polar in take-off condition is given by:
2D LC =0.0389+0.0755C (13)
Remarks:
i) In the approach just presented, to estimate the drag polar in take-off condition, the change in
the induced drag coefficient is included in the parasite drag coefficient. When the flap
deflections are large, the change in the induced drag can be accounted for by reducing the value
of the Ostwald efficiency factor (e) by 0.05 for take-off condition and 0.1 for landing condition
(Ref.4 section 3.4.1). Equations 12 and 13 are the drag polars for cruise condition and take-off
condition respectively. The polars are presented in Fig.2.
ii) It may be pointed out that the parabolic drag polar is not valid beyond CLmax. It is only
approximate near CL = 0 and CL = CLmax.
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0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
0 0.05 0.1 0.15 0.2
Drag coefficient
L i f t c o e f f i c i e n t
Cruise condition
Take-off condition
Fig.2 Drag polars at cruise and take-off conditions
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Appendix A
Lecture 36
Performance analysis of a piston engined airplane – 2
Topics
3 Engine characteristics
3.1 Variation of engine BHP
3.2 Thrust horsepower available
4 Steady level flight
4.1 Variation of stalling speed with altitude
4.2 Variations of Vmax and Vmin with altitude
5 Steady climb performance
6 Range and endurance
6.1 Estimation of range in constant velocity flight
6.2 Calculation of BHP and fuel flow rate at different RPM’s and MAP’s at 8000
6.3 Sample calculations for obtaining optimum N and MAP for a chosen flight
velocity (V)
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3 Engine characteristics
Model : Lycoming O-360-A3A.
Type : Air-cooled, carbureted, four-cylinder, horizontally opposed
piston engine.
Sea level power : 180 BHP (135 kW)
Propeller : 74 inches (1.88 m) diameter
The variations of power output and fuel consumption with altitude and rpm are shown in Fig.3.
For the present calculations, the values will be converted into SI units.
Fig.3 Characteristics of Lycoming O-360-A
(with permission from Lycoming aircraft engines )
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3.1 Variation of engine BHP
The variation of available engine output (BHPa) with altitude is assumed to be of the form:
aBHP = sealevelBHP (1.13σ -0.13)
where σ is the density ratio = sLρ/ρ .
The power outputs of the engine at select altitudes are given in Table 2 and plotted in Fig.4.
Note: At a given altitude, the variation of engine BHP with flight speed is very slight and is
generally neglected.
h(m) σ BHPa (kW)
0 1 135.00
1000 0.9075 120.89
2000 0.8217 107.80
3000 0.7423 95.69
4000 0.6689 84.49
5000 0.6012 74.16
5500 0.5691 69.27
6000 0.538 64.52
6500 0.5093 60.14
7000 0.4812 55.86
Table 2 Variation of BHP with altitude
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Fig.4 Variation of BHP with altitude at maximum power condition
3.2 Thrust horsepower available
The available thrust horsepower is obtained as product of a pBHP × η , where η is the propeller
efficiency. The propeller efficiency (η ) depends on the flight speed, rpm of the engine and the
diameter of the propeller. It can be worked out at different speeds and altitudes using the
propeller charts. However, chapter 6 of Ref.2 gives an estimated curve of efficiency as a function
of the advance ratioV
(J = )nD
for the fixed pitch propeller used in the present airplane. This
variation is shown as data points in Fig.5.
It may be added that this variation of η with J is used in chapter 6 of Ref.2, to estimate the drag
of Piper Cherokee airplane from measurements in flight. In another application, in Ref.10,
chapter 17, the same variation is used to compare the performance of fixed pitch and variable
pitch propellers. Based on these two applications, it is assumed here that the variation of η with
J shown in Fig.5, can be used at all altitudes and speed relevant to this airplane.
For the purpose of calculating the airplane performance, an equation can be fitted to the η vs J
curve in Fig.5. A fourth degree polynomial for η in terms of J is as follows.
4 3 2
pη (J) = -2.071895J +3.841567J -3.6786J +2.5586J-0.0051668 (14)
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It is seen that the fit is very close to the data points. The dotted portions are extrapolations.
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0 0.2 0.4 0.6 0.8 1 1.2
Advance ratio (J)
P r o p e l l e r e f f i c i e n c y
Fig.5 Variation of propeller efficiency with advance ratio
For the calculation of maximum speed, maximum rate of climb and maximum rate of turn, it is
convenient to have maximum power available a p(THP =η ×BHP) as a function of velocity. The
maximum power occurs at 2700 rpm (45 rps). Noting the propeller diameter as 1.88m, the ηP vs J
curve can be converted to η vs V curve (Fig.6).
The expression for ηP in terms of velocity is as follows.
-8 4 -6 3 -4 2 -2
pη = -4.0447×10 V + 6.3445 ×10 V - 5.1398×10 V +3.0244×10 V - 0.0051668 (15)
Curve fit (Eq.14)
Data from Ref.2
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Fig.6 Variation of propeller efficiency with velocity at 2700rpm
Making use of the power available at different altitudes as given in Table 2 and the values of the
propeller efficiency at different speeds given by Eq.(15), the maximum available thrust
horsepower a p(THP = η ×BHP) can be obtained at different speeds and altitudes. The variations
are plotted in Fig.7.
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0.00
20.00
40.00
60.00
80.00
100.00
120.00
0 10 20 30 40 50 60 70 80
Velocity (m/s)
T H P a ( k W )
Sea level
1000 m
2000 m
3000 m
4000 m
5000 m
5500 m
Fig.7 Variations of THPa with altitude
4 Steady level flight
4.1 Variation of stalling speed with altitude
Fig.8 Forces on an airplane in steady level flight
In steady level flight, the equations of motion are:
T - D = 0 (16)
L - W = 0 (17)
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Further,
2
L
1L= ρV SC =W
2 (18)
2
D
1T = D = ρV SC
2 (19)
L
2WV=
ρSC
Since CL cannot exceed CLmax , there is a flight speed below which the level flight is not
possible. The flight speed at which CL equals CLmax is called the stalling speed and is denoted
by Vs.
Hence,
sLmax
2W
V = ρSC
Since density decreases with altitude, the stalling speed increases with height.
In the present case, W = 1088 9.81 = 10673.28 N and S = 14.864 m2.
As regardsLmax
C , Reference 2 gives the values ofLmax
C as 1.33, 1.42, 1.70 and 1.86 for flap
deflections of o0 , o10 , o25 and 40o respectively.
Using these data, the variations of stalling speeds with altitude are presented in Table 3 and
plotted in Fig.9.
H(m) σ Vs (δf = 0o)
(m/s)
Vs (δf =10o)
(m/s)
Vs (δf =25o)
(m/s)
Vs (δf = 40o)
(m/s)
0 1.000 29.69 28.73 26.26 25.10
1000 0.908 31.16 30.16 27.57 26.35
2000 0.822 32.75 31.70 28.97 27.69
3000 0.742 34.46 33.35 30.48 29.14
4000 0.669 36.30 35.13 32.11 30.70
4500 0.634 37.28 36.08 32.97 31.52
5000 0.601 38.29 37.06 33.87 32.38
5500 0.569 39.36 38.09 34.81 33.28
6000 0.538 40.46 39.16 35.79 34.22
Table 3 Stalling speeds for various flap settings
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Fig.9 Variations of stalling speed with altitude for different flap settings
4.2 Variations of Vmax and Vmin with altitude
With a parabolic drag polar and the engine output given by an analytical expression, the
following procedure gives Vmax and Vmin. Available power is denoted by Pa and power required
to overcome drag is denoted by Pr. At maximum speed in steady level flight, required power
equals available power.
a pP =BHP×η (20)
3
r D
D×V 1P = = ρV SC
1000 2000
The drag polar expresses CD in terms of CL. Writing CL as2
2W
ρSV
and substituting in the above
equations we get:
23
p DO
1 KWBHP×η = ρV SC +
2000 500ρSV. (21)
The propeller efficiency has already been expressed as a fourth order polynomial function of
velocity and at a chosen altitude, BHP is constant with velocity. Their product (η BHP) gives
0
1000
2000
3000
4000
5000
6000
0 10 20 30 40 50
Stalling speed (m/s)
No flap
Flap deflection 10 degrees
Flap deflection 25 degrees
Flap deflection 40 degrees
A l t i t u d e ( m )
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an analytical expression for power available. Substituting this expression on the left hand side of
Eq.(21) and solving gives maxV and min e(V ) at at the chosen altitude. Repeating the procedure at
different altitudes, we get maxV and min e(V ) at various heights. Sample calculations and the plot
for sea level conditions are presented in Table 4 and Fig.10.
V
(m/s)
ηp Pa
(kW)
Pr(kW)
0 0.000 0.000 -
5 0.134 18.086 188.983
10 0.252 33.995 94.789
15 0.352 47.549 64.053
20 0.438 59.185 49.778
25 0.513 69.259 42.753
30 0.578 78.045 40.069
35 0.635 85.735 40.615
40 0.685 92.438 43.953
45 0.727 98.184 49.947
50 0.762 102.918 58.611
55 0.789 106.503 70.040
60 0.805 108.724 84.376
65 0.809 109.280 101.792
70 0.798 107.790 122.480
Table 4 Steady level flight calculations at sea level
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0
20
40
60
80
100
120
140
160
180
200
0 10 20 30 40 50 60 70 80
Velocity (m/s)
P o w e r ( k W )
Pa
Pr
Fig.10 Sample plot for Pa and Pr at sea level
It may be noted that
The minimum speed so obtained corresponds to that limited by power min e(V ) .
If this minimum speed is less than the stalling speed, a level flight is not possible at this
speed. The minimum velocity is thus higher of the stalling speed and min e(V ) .
The results for VS , min e(V ) , Vmin and max(V ) at various altitudes are tabulated in Table 5 and
plotted in Fig.11. It may be noted that at h = 5200 m,max
V andmin e
(V ) are same. This altitude is
the maximum height attainable by the airplane and will be referred later as absolute ceiling.
A
B
Point A: (Vmin)e
Point B: Vmax
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Table 5 VS , (Vmin)e, Vmin and maxV at various altitudes
Fig.11 Variations of maximum and minimum flight velocities with altitude
Remark:
The calculated value of maxV of 240.6 kmph at sea level is fairly close to the value of
246 kmph of the actual airplane quoted in section 1.10.
h
(m)
Vs(no flap)
(m/s)
(Vmin)e
(m/s)
Vmin
(m/s)
Vmax
(m/s)
Vmax
(kmph)
0 29.7 18 29.7 66.84 240.624
1000 31.2 20.4 31.2 65.75 236.7
2000 32.75 23.3 32.75 64.3 231.48
3000 34.46 27 34.46 62.3 224.28
4000 36.3 32 36.3 59.15 212.94
5000 38.29 41 41 52.7 189.72
5200 38.73 46.5 46.5 46.5 167.4
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5 Steady climb performance
Fig.12 Forces on an airplane in steady climb
Calculation of rate of climb:
In this flight, the C.G of the airplane moves along a straight line inclined to the horizontal at an
angle γ . The velocity of flight is assumed to be constant during the climb.
Since the flight is steady, acceleration is zero and the equations of motion can be written as:
T-D-Wsin γ = 0 (22)
L-Wcos γ = 0 (23)
Noting that CL = 2L/ρV2S =
2
2Wcos γ
ρSV, gives:
2
D Do 2
2 W cos γC = C +K ( )
ρSV
Also cV =Vsin γ , or cVsin γ =
V
2
c
2
Vcos γ = 1-
V
Substituting in Eq.(22) gives :
222 c c
a DO2
V V1 KWT = ρV S C + 1- + W
12 V VρV S
2
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Or
2
c cV VA + B( ) + C = 0
V V
(24)
where,
22
a Do2
KW 1A = , B = - W and C= T - ρV SC -A
1 2ρV S
2
;
Ta = available thrust = 1000 x Pa/V.
The available thrust horsepower is given by the following expression:
Pa = sealevel pBHP (1.13σ -0.13)η
Equation 24 gives 2 values of cV
V
. The value which is less than 1.0 is chosen as appropriate.
Consequently,
-1 cVγ=sin
V (25)
cV = V sin γ (26)
The climb performance is calculated using following steps.
(i) Choose an altitude.
(ii) Choose a velocity between Vmin and Vmax and obtain A, B and C in Eq.(24).
(iii) Solve for cV
V, obtain γ and cV .
(iv) Repeat calculations, at chosen altitude, at various velocities in the range of V min to
Vmax.
(v) Repeat steps (i) to (iv) at various altitudes.
Sample calculations at sea level are presented in Table 6.
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V
(m/s)η p THPa
(kW)
T
(N)
A C Vc/V γ
(deg.)
Vc
(m/s)
Vc
(m/min)
30 0.578 78.04 2601.49 1049.68 1265.85 0.120 6.894 3.60 216.03
35 0.635 85.73 2449.56 771.19 1289.14 0.122 7.000 4.26 255.89
40 0.685 92.43 2310.96 590.44 1212.13 0.114 6.563 4.57 274.29
45 0.727 98.18 2181.86 466.52 1071.92 0.101 5.790 4.53 272.36
50 0.762 102.91 2058.35 377.88 886.123 0.083 4.777 4.16 249.80
55 0.789 106.50 1936.42 312.30 662.971 0.062 3.568 3.42 205.35
60 0.805 108.72 1812.06 262.42 405.797 0.038 2.181 2.28 137.00
65 0.809 109.28 1681.23 223.60 115.193 0.011 0.619 0.70 42.10
Note: B = - W = -10673.28 N
Table 6 Steady climb calculations at sea level.
Repeating similar calculations at various altitudes gives the variations of and Vc with velocity
at different altitudes. The results are plotted in Figs.13 and 14. From these figures the variations
of maxγ , Vcmax or (R/C)max, γmaxV and R/CmaxV at various altitudes are obtained. The results are
presented in Table 7 and in Figs.15, 16 and 17.
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Fig.13 Variations of angle of climb with flight velocity at different altitudes
Fig.14 Variations of rate of climb with flight velocity at different altitudes
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h (m)max
γ (deg) Vcmax (m/min)γmax
V (m/s)R/Cmax
V (m/s)
0 7 276 34.1 41.7
1000 5.4 219.7 35 42.6
2000 3.83 165.8 38 43.6
3000 2.5 111.7 40.9 45
4000 1.28 60.5 44 45.9
5000 0.2 10 46 46.5
5200 0 0 46.5 46.5
Table 7 Climb performance
Fig.15 Variation of maximum angle of climb with altitude
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Fig.16 Variation of maximum rate of climb with altitude
Fig.17 Variations of Vγmax and V(R/C)max with altitude
Remark:
It is observed that the maximum rate of climb and maximum angle of climb decrease with
altitude, but the velocity at which the rate of climb and angle of climb are maximum increase
slightly with height.
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Service ceiling and absolute ceiling
The altitude at which the maximum rate of climb becomes 100 ft/min (30.5 m/min) is called the
service ceiling and the altitude at which the maximum rate of climb becomes zero is called the
absolute ceiling of the airplane. These can be obtained from Fig.16. It is observed that the
absolute ceiling is 5200 m and the service ceiling is 4610 m. It may be pointed out that the
absolute ceiling obtained from maxR/C consideration and that from maxV consideration are same
(as they should be). Further, the service ceiling of 4610 m is close to the value of 4035 m for the
actual airplane quoted in section 1.10.
6 Range and endurance
6.1 Estimation of range in a constant velocity flight
It is convenient for the pilot to cruise at constant velocity. Hence, the range performance in
constant velocity flights is considered here. In such a flight at a given altitude, the range (R) of a
piston-engined airplane is given by the following expression (Eq.7.23 of the main text of the
course).
p p-1 -1 -11 1 2max
max L1 1 1 2 1 2 1 2
7200 η 3600ηE ζ W WR = E tan = [ tan - tan ] (27)
BSFC 2E (1-KC E ζ) BSFC k k k /k k /k
where, 2
1 Do 2 1 22
1 2K k = ρV SC ,k = , W and W
2 ρSV are the weights of the airplane at the start and
end of the cruise, L1 1 2max 1 L1 2
D1 1DO
C 2W W1E = , E = , C = , ζ =1-
C ρSV W2 C K ,
D1C = DC corresponding to L1C .
From this expression the range and endurance in constant velocity flights, can be obtained at
different flight speeds, at the cruising altitude. From this information, the flight speeds which
would give the maximum range and endurance can be arrived at. It may be pointed that the
above expression gives the gross still air range as defined in section 7.2.3 of the main text of the
course.
The following values are taken as the common data for the subsequent calculations.
Cruising altitude = 8000 (2478 m)
W1 = Weight at the start of range flight = maximum gross weight = 10673.28 N
Usable fuel = 178.63 litre = 1331.78 N of petrol
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W2 = Weight at the end of the flight = 10673.28 – 1331.78 = 9341.5 N
Wing area (S) = 14.864 m2, CDO = 0.0349, K = 0.0755
ρ= density at 8000 = 0.9629 kg/m3
The power required during a constant velocity flight varies as the fuel is consumed. However,
for the purpose of present calculations the power required is taken as the average of power
required at the start and end of cruise. It is denoted as THP avg . It is noted that the power required
(THPavg) can be delivered by the engine operating at different settings of RPM (N) and manifold
air pressure (MAP). But, for each of these settings the propeller efficiency (η ) and fuel flow
rate would be different. The optimum setting, which would give the maximum range, can be
arrived at by using the following steps.
(a) Select a value of N and calculate J (= V/nd); n = N/60 .
(b) Obtain η corresponding to this value of J from Eq.(14).
(c) Then, BHP required (BHPr ) = THPavg / η
(d) The left hand side of Fig 4.2 of the main text, shows the BHP vs MAP and fuel flow rate vs
MAP curves with rpm as parameter. Similar curves are generated for h = 8000 .
(e) From the curves in step (d) the sets of N and MAP values which would give desired BHPr can
be obtained.
(f) Obtain fuel flow rate for each set of MAP and N. Calculate BSFC. Subsequently Eq.(27)
gives the range for chosen set of N and MAP.
(g) Repeat calculations at different value of N.
(h) The combination of N and MAP which gives longest range is the optimum setting.
The aforesaid steps are carried-out in the next three subsections.
6.2 Calculation of BHP and fuel flow rate at different RPM’s and MAP’s at 8000
Example 4.2 of the main text illustrates the procedure to obtain BHP and fuel flow rate at
N = 2200 and MAP of 20 of Hg. Similar calculations are repeated at N = 2700, 2600, 2400,
2200 and 2000 and at MAP = 15, 16, 17, 18, 19, 20, 21 and 21.6
of Hg. It may be pointed outthat the atmosphere pressure at 8000 is 21.6 of Hg. (see also right hand side of the engine
characteristics shown in Fig.3 of this Appendix). The values so obtained are plotted and
smoothed. Figure 18 shows the calculated values by symbols and the smoothed variations by
curves.
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Fig.18 Variations of BHP and fuel flow rate with MAP
6.3 Sample calculations for obtaining optimum N and MAP for a chosen flight velocity (V)
For the purpose of illustration V is chosen as 50 m/s or 180 kmph
I) Calculation of THPavg :
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CL1 = Lift coefficient at start of range 1
2
10673.280.5966
0.5 0.9629 50 14.8642
W
1ρV S
2
CL2 = Lift coefficient at the end of cruise 1
2
9341.50.5220
0.5 0.9629 50 14.8642
W
1ρV S
2
CD1 = CD corresponding to CL1 = 0.0349 + 0.0755 20.5966 = 0.06177
CD2 = CD corresponding to CL2 = 0.0349 + 0.0755 20.5220 = 0.05547
CDavg = (0.06177 + 0.05547)/2 = 0.05862
THPavg =2
Davg
1ρV SC /1000
2= 0.5 x 0.9629 x 50
3 x 14.864 x 0.05862/1000 = 52.43 kW
= 70.31 HP
II) Steps to obtain highest η /BSFC or the range
(a) Choose ‘N’ from 2700 to 2000
(b) Calculate, J = V/nd ; n = N/60 ; d = 1.88 m
(c) Corresponding to the value of J in step (b), obtain η from Eq.(14)
(d) Obtain BHPr = THPavg/η ; THPavg in HP
(e) From upper part of Fig.18, obtain MAP which would give BHP r at chosen N. For these values
of N and MAP obtain the fuel flow rate (FFR) in gallons/hr , from the lower part of Fig.18.
(f) Convert FFR in gallons per hour to that in N/hr and BHPr in HP to kW.
Obtain BSFC =FFR in N/hr
BHPinkW
(g) Obtain η /BSFC and also range from Eq.(27).
The above calculations at different values of N are presented in the table below.
N
(RPM)
J η p BHP
(HP)
MAP FFR
(gal/hr)
FFR
(N/hr)
BHP
(kW)
BSFC
(N/kW-
hr)
η p/BSFC Range
(km)
2700 0.591 0.762 92.22 15.90 8.32 234.58 68.77 3.410 0.223 1023.82600 0.613 0.773 90.88 16.10 7.92 223.44 67.77 3.297 0.234 1074.8
2400 0.664 0.794 88.54 16.47 7.38 208.07 66.02 3.151 0.251 1154.2
2200 0.725 0.807 87.03 17.04 6.95 196.06 64.90 3.020 0.267 1224.9
2000 0.797 0.806 87.22 18.25 6.97 196.56 65.04 3.021 0.266 1221.8
Table 8 Sample calculation at V = 180 kmph
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It is observed from the above table that at the chosen value of V =180 kmph, the range is
maximum for the combination of N = 2200 and MAP of 17.04 of Hg. The value of R is
1224.9 km.
III Obtaining range and endurance at different flight speeds
Repeating the calculations indicated in item (II), at different values of flight speeds in the
range of speeds Vstall from Vmax at 8000 , yield the results presented in Table 8a. Since the flight
speed is constant, the endurance (E) is given by the following expression.
E (in hours) =
Range in km
V inkmph
V
(m/s)
V
(kmph)
THPavg
(kW) RPMopt MAP
FFR
(gal/hr)
FFR
(N/hr)
ηp BHP
(kW)
BSFC
(N/kW
‐hr)
Range
(km)
Endur‐
nce(hr)
34 122.4 41.01 2000 16.61 6.254 176.26 0.734 55.859 3.155 929.6 7.59
36 129.6 41.13 2000 16.39 6.16 173.62 0.753 54.597 3.18 999.2 7.66
38 136.8 41.64 2000 16.29 6.12 172.52 0.77 54.069 3.19 1061.1 7.76
40 144 42.51 2000 16.32 6.13 172.79 0.784 54.198 3.19 1114.3 7.74
43 154.8 44.53 2000 16.58 6.24 175.81 0.8 55.643 3.16 1176.5 7.6
46 165.6 47.37 2000 17.1 6.46 182.07 0.808 58.573 3.11 1214.5 7.33
50 180 52.43 2200 17.04 6.95 196.06 0.807 64.904 3.02 1225 6.81
52 187.2 55.53 2200 17.7 7.25 204.4 0.81 68.576 2.98 1222.1 6.52
54 194.4 58.98 2200 18.49 7.63 215.12 0.808 72.97 2.95 1205.5 6.2
56 201.6 62.81 2200 19.43 8.13 228.99 0.803 78.23 2.93 1174.1 5.82
58 208.8 67.03 2200 20.56 8.77 247.02 0.793 84.5 2.92 1127.1 5.4
60 216 71.63 2400 20.42 9.37 264.14 0.806 88.87 2.97 1090.2 5.05
Table 8a Range and endurance in constant velocity flights at 8000 (2438 m)
The results are plotted in Figs.19 and 20.
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Fig.19 Range in constant velocity flights at h = 8000 (2438 m)
Fig.20 Endurance in constant velocity flights at h = 8000 (2438 m)
Remarks:
i) It is seen that the maximum endurance of 7.7 hours occurs in the speed range of
125 to 145 kmph.
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ii) The range calculated in the present computation is the Gross Still Air Range (GSAR).
The maximum range is found to be around 1220 km which occurs in the speed range of
165 to 185 kmph.
iii) The range quoted in Section 1.10 for Cherokee PA – 28 - 181 accounts for taxi, take-off,
climb, descent and reserves for 45 min. This range can be regarded as safe range. This
value is generally two-thirds of the GSAR. Noting that two-thirds of GSAR (1220 km) is
813 km, it is seen that the calculated value is within the range of performance given in
Section 1.10.
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Appendix A
Lecture 37
Performance analysis of a piston engined airplane – 3
Topics
7 Turning performance
8 Take-off and landing distance estimates
8.1 Distance covered during take-off run
8.2 Distance covered during transition
8.3 Distance covered during climb phase
8.4 Landing distance estimate
9 Concluding remarks
Acknowledgements
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7 Turning performance
Fig.21 Forces on an airplane in turning flight
In this section, the performance of the airplane in a steady level co-ordinated-turn is studied.
The forces acting on the airplane are shown in Fig.21.
The equations of motion in this flight are:
T – D = 0, as it is a steady flight (28)
W – L cos = 0, as it is a level flight (29)
2W VLsin =
g r , as it is a co-ordinated-turn (30)
These equations give:
Radius of turn = 2 2W V Vr = =g Lsin g tan
(31)
Rate of turn =2V V g tan
ψ = = V =r g tan V
(32)
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Load Factor = n =L 1
=W cos
(33)
In the following calculations, LmaxC =1.33 and nmax = 3.5 are assumed ; where maxn is the
maximum load factor for which the airplane is designed. The following procedure is used to
obtain minr and maxψ .
1. The flight speed and altitude are chosen. The lift coefficient in level flight ( LLC ) is
obtained as :
LL 2
2(W/S)C =
ρV
2. Obtain Lmax
LL
C
C. If Lmax
max
LL
C nC
, then the turn is limited by LmaxC and LT1 LmaxC = C .
However, if Lmax LL maxC /C > n , then the turn is limited by maxn , and LT1C = maxn LLC .
3. From the drag polar, DT1C is obtained corresponding to LT1C . Then,
2
T1 DT1
1D = ρV SC
2
If T1D > aT , where aT is the available thrust at chosen speed and altitude, then the turn is
limited by the engine output. The maximum permissible value of DC in this case is found
from:
aDT 2
2TC =
ρV S
From the drag polar, the value of LTC is calculated as:
DT DOLT C - CC =
K
However, if T1D < aT , then the turn is not limited by the engine output and the value of CLT1
calculated in step (2) is taken as CLT.
4. Once LTC is known, the load factor n, which satisfies the three constraints namely of
LmaxC , maxn and aT , is given by:
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LT
LL
Cn =
C
5. Knowing n, the values of the radius of turn (r) and the rate of turn ( ψ ) can be calculated
from Eqs.(31), (32) and (33).
6. The above steps are repeated for various speeds at the same altitude and subsequently the
procedure is repeated at various altitudes.
Sample calculations of turning performance at sea level are represented in Table 9. Figures
22 and 23 present turning performance at various altitudes.
V(m/s)
CLLLmax
LL
C
C CLT1 CDT1 DT1
(N) THP1
(kW) η p THPa
(kW)CLT n
(deg)r(m)
ψ
(rad/s)
30 1.30 1.021 1.33 0.168 1380 41.4 0.578 78.0 1.33 1.02 11.6 445 0.067
35 0.96 1.390 1.33 0.168 1879 65.8 0.635 85.7 1.33 1.39 44.0 129 0.270
38 0.81 1.638 1.33 0.168 2215 84.2 0.666 89.9 1.33 1.64 52.4 113 0.335
40 0.73 1.815 1.33 0.168 2454 98.2 0.685 92.4 1.28 1.75 55.1 114 0.351
45 0.58 2.297 1.33 0.168 3106 139.8 0.727 98.2 1.05 1.82 56.6 136 0.330
50 0.47 2.836 1.33 0.168 3834 191.7 0.762 102.9 0.86 1.83 56.9 166 0.300
55 0.39 3.432 1.33 0.168 4639 255.2 0.789 106.5 0.69 1.77 55.5 212 0.260
60 0.33 4.084 1.14 0.133 4359 261.5 0.805 108.7 0.52 1.60 51.2 295 0.203
65 0.28 4.793 0.97 0.106 4082 265.3 0.809 109.3 0.34 1.23 35.7 600 0.108
Table 9 Turning performance calculations at sea level
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Fig.22 Variations of radius of turn with velocity at various altitudes
Fig.23 Variations of rate of turn with velocity at various altitudes
From Figs.22 and 23 the values of r min , maxψ , Vrmin , ψmaxV
can be obtained at various
altitudes. The variations are presented in Table 10 and Figs.24, 25 and 26.
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h (m) r min(m)maxψ (rad/s) Vrmin(m/s)
ψmaxV
(m/s)
0 110 0.351 38 40.0
1000 135 0.301 39 41.2
2000 163 0.248 39.5 41.5
3000 198 0.194 40.5 41.7
4000 324 0.128 41 44.0
5000 918 0.048 45.7 46.0
Table 10 Turning performance
Fig.24 Variation of minimum radius of turn with altitude
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Fig.25 Variation of maximum rate of turn with altitude
Fig.26 Variations of rminV andψmaxV
with altitude
Remark:
The minimum radius of turn at sea level is about 110 m at flight speed of about 38 m/s. The
maximum rate of turn at sea level is about 0.35 rad/sec at flight speed of about 40 m/s.
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8 Take-off and landing distance estimates
Take-off flight can be divided into three phases: take-off run or ground run, transition and
climb (Fig.27).
Fig.27 Phases of take-off flight
8.1 Distance covered during take-off run (s1)
The equations of motion during the take off run are:
W dVT- D-μR =
g dt and R = W - L (34)
where R is the ground reaction. The acceleration can be written as:dV g
= ×[T-D-μ(W-L)]dt W
WritingdV
dtas
dV ds×
ds dt, gives :
W VdVds =
g T -D-μ(W-L)
Further, at sea level, BHP = constant = 135kW at 2700 rpm. Thrust is given by :
T = BHP x pη / V .
The distance covered during the take-off run (s1) can be expressed as:
TOV
1
0
W Vs = dV
g F (35)
where F is the accelerating force given by:
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pBHP ηF = - D-μ (W-L)
V
Since η is a function of velocity, an accurate way to estimate 1s is to evaluate the integrand in
Eq.(35) at several values of V and carry-out a numerical integration. Simpson ’s rule is used for
this purpose. Various quantities needed for the purpose are estimated below.
TOV =1.2 sV , where sV is stalling speed, given by :
s
Lmax
2WV =
ρSC
During the take-off, flap deflection ( f δ ) is10 , hence LmaxC =1.42. It is assumed that the
coefficient of friction is 0.02.
The take-off weight is W =10673.28 N
S = wing planform area =14.864 2m
Density ρsl =1.225 kg/m3
Thus, sV = 28.73m/s and TOV = 34.48 m/s
To estimate CL and CD during take -off run it is noted that the airplane has a nose wheel type of
landing gear and hence the airplane axis can be considered as horizontal and the wing produces
lift corresponding to the wing setting angle (see Section 10.3.1 of the main text of the course
material).From Sec. 1.4, the average wing incidence is the average of incidence at root (4.62
o) and that at
tip (2.62o) i.e. 3.62
o. The slope of the lift curve of the wing ( LαWC ) is approximately given by:
LαW
A 5.625C = 2π = 2π = 4.63/ rad = 0.0808/ deg
A+2 7.625
The angle of zero lift ( 0Lα ) for the airfoil NACA 652 – 415, from Ref. 5, is -2.6o.
Hence, lift coefficient during take-off run due to wing lift is :
0.0808(3.62+ 2.6)= 0.502
Since, the flaps are deflected during the take-off run the lift coefficient will be increased by (1.42
– 1.33 = 0.09). Hence, CL during take-off run ( Ltr C ) is :
Ltr C =0.502+0.09= 0.592
The drag coefficient during take-off run (CDtr ) , using the drag polar corresponding to take-off,
is:
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2
Dtr C =0.0389+0.0755×0.592 =0.0654
For applying Simpson’s rule in this case, the various quantities are evaluated at seven points in
the speed range of 0 to 34.48. The calculations are shown in Table 11.
V
(m/s) η
T
(N)
D
(N)
L
(N)
F
(N)
WV/gF
(s)
0 0 * 0 0 - 0.000
5.75 0.153 3589.94 19.68 132.9 3364.45 1.859
11.49 0.283 3329.77 78.60 711.5 3051.9 4.096
17.24 0.392 3072.83 176.9 1601.3 2714.5 6.910
22.99 0.484 2843.74 314.7 2848.7 2372.5 10.54
28.73 0.562 2642.65 491.4 4448.2 2026.7 15.42
34.48 0.629 2464.68 707.8 6407.0 1671.6 22.42
* The value of thrust (T) at V = 0 is not zero. It can be evaluated using propeller charts.
However, it is not needed in the present calculation, as the integrand is zero when V is zero.
Table 11 Evaluation of integrand in Eq.(35).
Using the values of integrand in Table 11 and employing Simpson’s rule the ground run (s1) is
given by :
1 5.747s = 0+4(1.859+6.91+15.42)+2(4.096+10.54)+22.42 =284.43
m
8.2 Distance covered during transition (s2)
The entire power of the engine is assumed to be used to overcome the drag and to accelerate to a
velocity V2 given by V2 = 1.1 VT0. The height attained during the transition phase is ignored.
Hence, 2 2
2 2 2 TO
WTs = Ds + (V -V )
2g
2 2
2 TO
2
(V -V )W
s = 2g T -D
where T and D are evaluated at a speed which is mean (Vavg) of 2V and TOV
V2 = 1.1 x 34.48 = 37.93 m/s
TO 2avg
V + V 34.48+37.93V = = = 36.71m/s
2 2
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p
avg
η ×BHP×1000T =
V
From Eq.(15), η at a speed of 36.71 m/s is 0.65285
Hence,
0.65285×135×1000
T = = 2400.5N36.71
L 2
2×10673.28Further, C = = 0.87
1.225×14.864×36.71
2
DC =0.0389+ 0.0755×0.87 =0.096
21D = ×1.225×36.71 ×14.864×0.096=1178.4 N
2
Hence, 2 2 2 2
2 TO2
V -VW 10673.28(37.93 -34.48 )s = = =111.3m
2g T-D 2×9.81(2400.5-1178.4)
8.3 Distance covered during climb phase (s3)
The airplane is assumed to climb to screen height (15m) at an angle of climb γ,
where the climb angle γ is given by:
T-Dγ = ( )
W
For the climb phase, T and D are evaluated at V2 which is equal to 37.93 m/s
From Eq.(15), η at a speed of 37.93 m/s is 0.665. Hence,
0.665×135×1000T = = 2366.86 N
37.93
2×10673.28C = = 0.82
L 21.225×14.864×37.93
2
DC =0.0389+ 0.0755×0.82 =0.0897
21
D = ×1.225×36.71 ×14.864×0.0897=1174.5N2
oT - D 2366.86 - 1174.5sinγ = = = 0.1117 or γ = 6.41
W 10673.28
3
15 15Hence, s = = = 133.4m
tan γ 0.1124
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Total takeoff distance is given by:
1 2 3s = s +s +s = 284.4+111.3+133.4 = 529.1 m
Remarks:
i) The above estimation of take off distance is based on several assumptions. Reference 8 has
compiled data on take-off distances of many propeller driven airplanes. This take-off distance is
based on FAR 23 specifications and can be denoted by sto23 . Based on this data the following
formula is obtained for sto23 in terms of a parameter called take-off parameter and denoted by
TOP23 . In SI units the relationship is given as (See Guidelines for take-off distance in
Section 10.4.7 of the main text of the course) .
STo23 = 8.681 x 10-3
x TOP23 + 5.566 x 10-8
x 2
23TOP
where, 2
23
LTO
W W( )×( )
S PTOP = ; (W/S) is in N/m , Win N and Pin kW.σC
σ is density ratio at the altitude of take-off.
In the present case:
W/S = 10673.28 / 14.864 = 718.1 N/m2, W/P = 10673.28 / 135 = 79.06 N/ kW
σ = 1.0 and CLTO = 1.42.
Consequently,
23
718.1×79.06TOP = =39981
1.42
-3 -8 2
to23s = 8.681 x 10 x 39981+ 5.566 x 10 x39981 =347+89=436 m
ii) The estimated take-off distance of 530 m is somewhat higher than the actual take-off distance
of 488 m (section 1.10). This may be because the height attained during the transition phase has
been ignored.
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8.4 Estimation of landing distance
The landing distance can be calculated in a manner similar to that for take-off distance.
However, due to uncertainty associated with piloting techniques during landing, the following
formula is used.
2
aland
Vs = -
2a
where, aV = 1.3 s×V in landing configuration
The weight of the airplane during the landing is taken same as that during the take-off. However,
LmaxC with landing flap setting is 1.86. The stalling speed in this configuration is 25.1 m/s.
Hence, approach speed is 32.6 m/s.Taking a = -1.22m/ s2 for a simple braking system yields:
sland = 436 m,
which is close to the value of 426 m given in Section 1.10.
9 Concluding remarks
1. The performance of a piston-engined airplane has been estimated for stalling speed, maximum
speed, minimum speed, steady climb, range, endurance, turning, take-off and landing.
2. A reasonable agreement has been observed between the calculated performance and the actual
performance of the airplane (PA – 28 – 181).
3. Figure 28 presents the variations, with altitude, of the characteristic velocities corresponding
to:
Stalling speed Vs
Maximum speed Vmax
Minimum speed as dictated by power (Vmin)e
Maximum rate of climb VR/C max
Maximum angle of climb Vγmax
Maximum rate of turn ψmaxV
Minimum radius of turn Vr min
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0
1000
2000
3000
4000
5000
6000
0 20 40 60
Velocity (m/s)
A l t i t u d e ( m )
V for minimum radius of turn
V for maximum rate of turnStalling speed
Vmin from engine output
V max
V for maximum angle of climb
V for maximum rate of climb
Fig.28 Variations of characteristic velocities with altitude
Acknowledgements
The first author (EGT) thanks AICTE for the fellowship which enabled him to carry out the work
at IIT Madras. He is grateful for the support given by Prof.J.Kurian, Prof.P.Sriram,
Prof.K.Bhaskar, the Heads of the department of Aerospace engineering, IIT Madras. The helprendered, while carrying out the revision, by Mr.Aditya Sourabh, Dual Degree student,
Mr.S.Gurusideswar, Ph.D. scholar and Sandip Chajjed, Project staff Department of Aerospace
Engineering and Ms. K. Sujatha and Mr. G. Manikandasivam of NPTEL Web studio is gratefully
acknowledged.
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Appendix A
References
1.Roskam, J. “Methods for estimating drag polars of subsonic airplanes” Roskam aviation
engineering corporation, Kansas,U.S.A,(1983).
2. McCormick B.W. “Aerodynamics, aeronautics and flight mechanics”, John Wiley,
New York, (1979 First edition, 1995 Second edition).
3. Jackson, P. (editor-in-chief) “Jane’s all the world’s airplane (1999-2000)” Jane’s
information group ltd, Surrey, U.K.
4. Roskam, J. “Airplane design Vol. I”, Roskam aviation engineering corporation, Kansas,
U.S.A, (1989).
5. Perkins C.D. and Hage R.E. “Airplane performance stability and control”, John Wiley,
(1960).
6. Torenbeek. E. “Synthesis of subsonic airplane design” Delft University Press (1981).
7. Raymer, D.P. “Aircraft design: a conceptual approach” AIAA` educational series fourth
edition, (2006).
8. Loftin, Jr. L.K. “Subsonic aircraft evolution and the matching of size to performance”
NASA Reference publications, 1060, August 1980. This report can be downloaded from
the site “NASA Technical Report Server (NTRS)”.
9. Samoylovitch, O. and Strelets, D. “Determination of the Oswald efficiency factor at
airplane design preliminary stage”, Aircraft Design, Vol. 3, pp. 167-174, (2000).
10. Nicholai, L.M. and Carichner, G.E. “Fundamentals of aircraft and airship design Vol. I –
Aircraft design” AIAA educational series (2010).
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APPENDIX- B
PERFORMANCE ESTIMATION OF A TYPICAL
SUBSONIC JET TRANSPORT AIRPLANE
(Lectures 38 – 40)
E.G.TULAPURKARA
V.GANESH
REPORT NO: AE TR 2007-2
FEBRUARY 2007
(REVISED OCTOBER 2011)
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Performance estimation of a typical subsonic
jet transport airplane
E.G.Tulapurkara*V.Ganesh $
February 2007
(Revised November 2009)
Abstract
This report contains details of the performance estimation of a medium range jet
airplane similar to B737. The following aspects are considered.
Drag polar estimation
Engine characteristics
Level flight performance - stalling speed, maximum and minimum speeds
Steady climb performance – maximum rate of climb, maximum angle of climb,
service ceiling and absolute ceiling
Range and endurance
Steady level co-ordinated turn - minimum radius of turn, maximum rate of turn
Take-off and landing distances
The report is intended to serve as an example of performance calculation of a
typical jet airplane.
* AICTE Emeritus Fellow, Department of Aerospace Engineering, IIT Madras
$ Dual Degree Student, Department of Aerospace Engineering, IIT Madras
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Contents
1 Airplane details
1.1 Overall dimensions
1.2 Engine details
1.3 Weights1.4 Wing geometry
1.5 Fuselage geometry
1.6 Nacelle geometry
1.7 Horizontal tail geometry
1.8 Vertical tail geometry
1.9 Other details
1.10 Flight condition
Three-view drawing of the airplane
2 Estimation of drag polar
2.1 Estimation of (CDo)WB
2.2 Estimation of (CDo)V and (CDo)H
2.3 Estmation of misc drag – nacelle
2.4 CDo of the airplane
2.5 Induced drag
2.6 Final drag polar
3 Engine characteristics
4 Level flight performance
4.1 Stalling speed
4.2 Variations of Vmin and Vmax with altitude
5 Steady climb
6 Range and endurance
7 Turning performance
8 Take-off distance
9 Landing distance
10 Concluding remarks
Acknowledgements
References
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Appendix B
Lecture 38
Performance analysis of a subsonic jet transport –1
Topics
1 Airplane details
1.1 Overall dimensions
1.2 Engine details
1.3 Weights
1.4 Wing geometry
1.5 Fuselage geometry
1.6 Nacelle geometry1.7 Horizontal tail geometry
1.8 Vertical tail geometry
1.9 Other details
1.10 Flight condition
Three-view drawing of the airplane
2 Estimation of drag polar
2.1 Estimation of (CDo)WB
2.2 Estimation of (CDo)V and (CDo)H
2.3 Estmation of misc drag – nacelle
2.4 CDo of the airplane
2.5 Induced drag
2.6 Final drag polar
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1 Airplane Details
1.1 Overall Dimensions
Length : 34.32 m
Wing span : 32.22 m
Height above ground : 11.17 m
Wheel base : 13.2 m
Wheel track : 5.8 m
1.2 Engine details
Similar to CFM 56 - 2B
Seal level static thrust : 97.9 kN per engine
By pass ratio : 6.5 (For which the engine characteristics are given
in Ref.3*)
SFC : 0.6 hr -1
at M = 0.8 and h = 10973 m (36000 ft)
1.3 Weights
Gross weight : 59175 kgf (580506.8 N)
Empty weight : 29706 kgf (291415.9 N)
Fuel weight : 12131 kgf (119005.1 N)
Payload : 17338 kgf (170085.8 N)
Maximum landing weight : 50296 kgf (493403.8 N)
1.4 Wing Geometry
Planform shape : Cranked wing
Span : 32.22 m
Area (Sref ) : 111.63 m2
Airfoil : NASA - SC(2) series, t/c = 14%,
Clopt = 0.5
Root chord : 5.59 m (Equivalent trapezoidal wing)
Tip chord : 1.34 m (Equivalent trapezoidal wing)
Root chord of cranked wing : 7.44 m
Portion of wing with straight
trailing edge : 11.28 m
* Reference numbers in this Appendix relate to those given on page 40.
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Mean aerodynamic chord : 3.9 m
Quarter chord sweep : 27.69o
Dihedral : 5 o
Twist : 3 o
Incidence : 1.4 o
Taper ratio : 0.24 (Equivalent trapezoidal wing)
Aspect ratio : 9.3
1.5 Fuselage geometry
Length : 33 m
Maximum diameter : 3.59 m
1.6 Nacelle geometry
No. of nacelles : 2
Nacelle diameter : 1.62 m
Cross sectional area : 2.06 m2
Length of nacelle : 3.3 m (based on B737 Nacelle)
1.7 Horizontal tail geometry
Span : 11.98 m
Area : 28.71 m2
Mean aerodynamic chord : 2.67 m
Quarter chord sweep : 32 o
Root chord : 3.80 m
Tip chord : 0.99 m
Taper ratio : 0.26
Aspect ratio : 5
1.8 Vertical tail geometry
Span : 6.58 m
Area : 25.43 m2
Root chord : 5.90 m
Tip chord : 1.83 m
Mean aerodynamic chord : 4.22 m
Quarter chord sweep : 37 o
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Taper ratio : 0.31
Aspect ratio : 1.70
1.9 Other details
CLmax without flap : 1.4
CLmax with landing flaps : 2.7
CLmax with T.O flaps : 2.16
Maximum load factor (nmax ) : 3.5
1.10 Flight condition
Altitude : 10973 m (36000 ft)
Mach number : 0.8
Kinematic viscosity : 3.90536 x10-5
m2/s
Density : 0.3639 kg/m3
Speed of sound : 295.07 m/s
Flight speed : 236.056 m/s
Weight of the airplane : 59175 kgf (580506.8 N)
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Fig.1 Three-view drawing of the airplane
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2 Estimation of drag polar
The drag polar is assumed to be of the form:
2
LD Do
CC =C +
πAe
The quantity CDO is assumed to be given by:
Do Do WB Do V Do H Do MiscC = (C ) + (C ) + (C ) + (C ) (1)
where suffices WB, V, H, Misc denote wing-body combination, vertical tail,
horizontal tail, and miscellaneous contributions respectively.
2.1 Estimation of (CDo)WB
Initially, the drag polar is obtained at a Mach number of 0.6 as suggested in
Ref.5, section 3.1.2. (CDo)WB is given as :
BDo WB Do w Do B
ref
S(C ) = (C ) + (C )
S
The suffix B denotes fuselage and SB is the maximum frontal area of fuselage.
(CDO)W is given as :
wetDo W fW wing
ref
St(C ) =C 1+L( ) ( )
c S
where, Cfw is the turbulent flat plate skin friction coefficient. The Reynolds number usedto determine it (Cfw) is lower of the two Reynolds numbers viz. Reynolds number based
on the mean aerodynamic chord of the exposed wing (R e) and R ecuttoff based on surface
roughness. Further, (Swet)e is the wetted area of the exposed wing.
In the present case, cr = 5.59m, ct= 1.34m, b/2 = 16.11m and dfus = 3.59m. Hence,
Root chord of exposed wing = cre =5.59 1.34 3.59
5.5916.11 2
= 5.116 m
e
1.34λ = =0.262
5.116
Hence, mean aerodynamic chord of exposed wing ( ec ) is :
2
e
2 1+0.262+0.262c = [5.116( )]
3 1+0.262 = 3.596 m
Span of exposed wing = (b/2)e = 16.11 – 1.795 = 14.315m
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Further, M = 0.6, a = 295.07m/s. Hence, V = 177.12m/s.
Also ν = 3.90536 X 10-5 m2/s.
Hence,
R e 5
177.12 3.596
3.90536 10
16.31 x 106
The height of roughness corresponding to the standard camouflage paint, average
application, is k = 1.015 x 10-5
m (Ref.5, table 3.1). Hence, l/k in this case is:
5
5
l 3.5963.543 10
k 1.015 10
The R ecutoff corresponding to the above l/k is 30 x 106. Consequently, fwC corresponding
to R e = 16.31 x 106 is obtained from Fig.3.1 of Ref.5, as :
fwC = 0.00265.
(t/c)avg = 14% and (t/c)max occurs at x/c > 0.3 Hence, L = 1.2 and
Sexposedplanform =5.116 1.341
14.314( ) 22
= 92.41m2
wetWS = 2 x 92.41(1+1.2 x 0.14) = 215.8m2
Hence,
(CDf )w = 0.00265 (1+1.2 x 0.14)215.8
111.63 = 0.00598
(CDo)B is given as:
(CDo)B = (CDf )B + (CDp)B + CDb
(CDo)B = wet basefB fus Db3
B ref
l S S60C [1+ +0.0025( )]( ) +C
(l /d) d S S
In the present case , lf = 33.0m , dmax = 3.59m ,
R eb = 5
177.12 33
3.905 10
= 149.6 x 106
5l 33k 1.015 10
= 32.51x105
The R ecutoff corresponding to the above l/k is 2.6 x 108. The Cfw corresponds to
R eb = 149.6 x 106 measured from the graph in Ref.5, Fig.3.1 is:
Cfw = 0.0019
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(Swet)fus = 0.75 x x 3.59 x 33 = 279m2
SB =4
x 3.592 = 10.12m2
Hence,
(CDf )B = 0.0019 x279
10.12 = 0.0524
Dp B 3
60 279(C ) =0.0019[ +0.0025×(33/3.59)]
(33/3.59) 10.12=0.00524
Since, base area is almost zero, CDb is assumed to be zero. Hence,
(CDo)B = 0.0524 + 0.00524 + 0 = 0.0576
D canopy(ΔC ) is taken as 0.002. Hence, (CDo)B = 0.0596
Finally,
(CDo)WB = 0.00598 + 0.059610.12
111.63 = 0.01138
2.2 Estimation of (CDo)V and (CDo)H
The estimation of (CDo)H and (CDo)V can be done in a manner similar to that for the wing.
However, the details regarding the exposed tail area etc. would be needed. In the absence
of the detailed data on the shape of fuselage at rear, a simplified approach given in Ref.5,
section 2.2 is adopted, wherein CDf = 0.0025 for both horizontal and vertical tails.
SW = 2(Sh + Sv)
Hence,
(CDo)HV = 0.0025(28.71 + 25.43)2
111.63 = 0.0024 (2)
2.3 Estmation of misc drag - nacelle
For calculating drag due to the nacelles the short cut method is used i.e.:
(CDo)nacelle = 0.006 x wet
ref
S
S
where, Swet is the wetted area of nacelle. Here, Swet = 16.79m2. Since, there are two
nacelles, the total drag will be twice of this. Finally,
(CDo)nacelle = 0.006 x16.79
111.63 x 2 = 0.0018
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2.4 CDo of the airplane
Taking 2% for miscellaneous roughness and protuberances(Ref.5, section 3.4.6 ), the CDo
of the airplane is:
CDo = 1.02 [0.01138 + 0.0024 + 0.0018] = 0.0159 (3)
2.5 Induced drag
The induced drag component has the Oswald's efficiency factor e which is estimated by
adding the effect of all the airplane components on induced drag (Ref.5, section 2.3).The
rough estimate of e can be obtained as follows :
Figure 2.4 of Ref.5 is useful only for estimating ewing of unswept wings of low speed
airplanes. For the present case of swept wing, the following expression given in Ref.2 ,
chapter 7 is used.
ewing = (ewing)Λ=0 cos(Λ - 5)
where Λ is the quarter chord sweep. Ref.1, chapter 1 is used to estimate (ewing)Λ=0. In
the present case, with A = 9.3 and = 0.24, the value of (ewing)Λ=0 is 0.97.
Hence, ewing = 0.97 x cos (27.69 - 5) = 0.8948.
From Ref.5, section 2.3, fus
f
1/e
(S /S) = 0.8 for a round fuselage. Hence,
fus
1 10.122= 0.8 ×
e 111.63
= 0.0725
Further, from Ref.5, section 2.3,other
1= 0.05
e
Finally,
e =-1
1= 0.8064
0.8948 + 0.0725 + 0.05
Hence,
1 1K = =πAe π × 9.3 × 0.8064
= 0.04244
Remark:
Based on Ref.7, a detailed estimates of ewing and efuselage are given in Ref.5, section 3.3.
For an untwisted wing the value of ewing is given as:
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Lαwwing
Lαw
1.1(C /A)e =
CR( )+(1-R)π
A
where,
LαW2
12 2
2
2 2
2πAC =tan
A β2+ 1+ +4
κ β
LαWC = slope of lift curve of wing per radian
A = aspect ratio of wing
R = a factor which depends on (a) Reynolds number based on leading edge radius, (b)
leading edge sweep (ΛLE), (c) Mach number (M), (d) wing aspect ratio (A) and (e) taperratio (λ ).
2β = 1-M
Λ1/2 = sweep of semi-chord line
= ratio of the slope of lift curve of the airfoil used on wing divided by 2π . It is
generally taken as unity.
In the present case,
M= 0.6, h= 10973 m (36000 ft), V= 177.12 m,-5 2
=3.90536×10 m /s , S = 111.63 m2
b = 32.22 m, cre = 5.59 m, ct = 1.34 m, Λ1/4 = 27.69 deg,
Hence, A = 9.3, =0.24, β =0.8 , 1/ 2tan = 0.4589, LEcos 0.8609 ,
Average chord = 3.615 m
The airfoil is NASA – SC(2) with 14 % thickness. From Ref.8 the leading edge radius is
3 % of the chord.
From these data:
LαWC = 5.404
R Ler = Reynolds number based on leading edge radius = 4.974 x 104
R Ler x cot ΛLE x 2 2
LE1-M cos = 7.198 x 105
LE
Aλ
cos= 2.592
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Corresponding to these data, R = 0.943 is obtained from Ref.5, Fig.3.14. Consequently,
wing
1.1 5.404/ 9.3e
5.4040.943 1 0.943
9.3
= 0.8793
This value of ewing is close to the value of 0.8948 obtained by the simpler approach.
However, detailed approach is recommended for wings with sweep of above 35o.
Reference 7, section 4.5.3 contains guidelines for estimating drag of wing-body-tail
combination with allowance for trim drag.
2.6 Final drag polar
CD = 0.0159 + 0.04244 2
LC (4)
The drag polar is presented in Fig.2.
Fig.2 Drag polar at sub-critical Mach numbers
Remarks:
i) The polar given by Eq.(4) is valid at subcritical Mach numbers. The increase in CDO
and K at higher Mach numbers is discussed in section 4.2.
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ii) The maximum lift to drag ratio ((L/D)max) is given by:
max
Do
1(L/D) =
2 C K
Using CDO and K from Eq.(4), (L/D)max is 19.25, which is typical of modern jet
transport airplanes.
iii) It may be noted that the parabolic polar is an approximation and is not valid beyond
CLmax. It is also not accurate close to CL = 0 and CL = CLmax.
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Appendix B
Lecture 39
Performance analysis of a subsonic jet transport –2
Topics
3 Engine characteristics
4 Level flight performance
4.1 Stalling speed
4.2 Variations of Vmin and Vmax with altitude
5 Steady climb
3 Engine characteristics
To calculate the performance, the variations of thrust and SFC with speed and altitudes
are needed. Chapter 9 of Ref.3 contains these variations for turbofan engines with various
bypass ratios. The thrust variations versus Mach number with altitude as parameter are
given, in non-dimensional form, for take-off, cruise and climb ratings. The values were
read from those curves, interpolated and later smoothed. The values multiplied by
97.9 kN, the sea level static thrust rating for the chosen engine, are shown in Figs.3 and 4.
Figure 3 also contains (a) the variation of thrust with Mach number at sea level with take-
off rating and (b) variations of climb thrust with Mach number at various altitudes. The
values at h = 38000 ft and 39000 ft are obtained by interpolating the values at 36000 ft
and 40000 ft and are used for computation of performance.
The SFC variation is also given in Ref.3, but is taken as 0.6 hr -1
under cruise
conditions based on the trend shown in Fig.3.3 of Ref.4.
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Fig.3 Output for single engine – take-off thrust at sea level and climb
thrust at various altitudes.
Fig.4 Output of single engine – cruise thrust at various altitudes
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4 Level flight performance
Forces on an airplane in steady level flight
In steady level flight, the equations of motion, in standard notation, are:
T - D = 0 (5)
L - W = 0 (6)
2
L
1L = W = ρV SC
2 (7)
2
D
1D = ρV SC =T
2 (8)
4.1 Stalling speed
In level flight,
L
2WV =
ρSC (9)
Since, CL cannot exceed CLmax, there is a flight speed below which level
flight is not possible. The flight speed at CL = CLmax is called the stalling
speed and is denoted by Vs
s
max
2WV =
ρSC (10)
Since, ρ decreases with altitude, Vs increases with height. It may be noted that
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W/S = 5195 N/m2, CLmax = 2.7 with landing flaps and CLmax = 1.4 without flaps. The
values of stalling speed at different altitudes and flap settings are tabulated in Table 1 and
shown in Fig.5.
h
(m)
ρ
(kg/m3)
Vs
(CLmax = 1.4)
(m/s)
Vs
(CLmax = 2.7)
(m/s)
0
2000
4000
6000
8000
10000
11000
12000
1.225
1.006
0.819
0.659
0.525
0.412
0.363
0.310
77.83
85.86
95.18
106.06
118.87
134.09
142.80
154.52
56.04
61.83
68.54
76.37
85.59
96.56
102.83
111.27
Table 1 Variation of stalling speed with altitude
Fig.5 Stalling speed vs altitude
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4.2 Variations of Vmin and Vmax with altitude
To determine the Vmin and Vmax at each altitude, the following procedure is adopted. The
engine thrust as a function of velocity at each altitude is obtained from the smoothed data.
The drag at each altitude is obtained as a function of velocity using the drag polar and the
level flight formulae given below.
L 2
2 (W/S)C =
ρV
(11)
2
D Do LC C K C (12)
Thrust required = Drag =2
D
1ρV SC
2 (13)
Thrust available = Ta = f(M) (14)
where, CDo = 0.0159 and K = 0.04244.
However, the cruise Mach number (Mcruise) for this airplane is 0.8. Hence, CDo and K are
expected to become functions of Mach number above Mcruise. To get some guidelines
about variations of CDo and K, the drag polars of B-727 given in Volume VI, Chapter 5 of
Ref.6 are considered. These drag polars are shown in the Fig.6 as discrete points.
Fig.6 Drag polars at different Mach numbers for B727-100; Symbols are data from Ref.6
and various lines are the parabolic fits.
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These polars were approximated by the parabolic polar expression namely
2
D Do LC = C + KC . The values of CDo and K at various Mach numbers, obtained by least
square method, are given in the Table 2. The parabolic fits are also shown in Fig.6.
M CDo K0.7
0.76
0.82
0.84
0.86
0.88
0.01631
0.01634
0.01668
0.01695
0.01733
0.01792
0.04969
0.05257
0.06101
0.06807
0.08183
0.10300
Table 2 Variations of CDo and K with Mach number (Parabolic fit)
The variations of CDo and K with Mach number are plotted in Figs.7 and 8. It is seen that
there is no significant increase in CDo and K upto M = 0.76. This is expected to be the
cruise Mach number for the airplane (B727-100). Following analytical expressions have
been found to closely represent the changes in CDo and K from M = 0.76 to M = 0.86.
CDo = 0.01634 – 0.001 x (M – 0.76) + 0.11x (M – 0.76)2
(15)
K = 0.05257 + (M – 0.76)2 + 20.0 x (M – 0.76)
3 (16)
Fig.7 Variation of CDo with Mach number
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Fig.8 Variation of K with Mach number
In the case of the present airplane, the cruise Mach number is 0.8. The variations of CDo
and K above Mcruise and upto M = 0.9, based on the B727-100 data are taken as follows.
CDo = 0.0159 – 0.001 x (M – 0.80) + 0.11 x (M – 0.80)2 (17)
K = 0.04244 + (M – 0.80)2 + 20.0 x (M – 0.80)
3 (18)
The thrust available and thrust required curves are plotted at each altitude as a function of
velocity. The points of intersection give the (Vmin)e and Vmax at each altitude from thrustavailable consideration (Figs.9 – 14).
However, to arrive at the minimum speed (Vmin), the stalling speed (Vs) also needs to be
taken in to account. Since, the drag polar is not valid below V s, in the Figs.9 to 14, the
thrust required curves are plotted only for V ≥ Vs. Stalling speed is taken for CLmax
without flaps.
The calculations are carried out for h = 0, 10000, 15000, 25000, 30000 and 36000 ft, i.e
S.L, 3048, 4572, 7620, 9144 and 10972.8 m using Ta as both climb thrust (Tclimb) and as
cruise thrust (Tcr ). Results in Figs.9 – 14 are presented only for climb thrust case. The
variations of Vs, (Vmin)e and Vmax are tabulated in Table 3 and presented in Fig.15.
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Fig.9 Available and required thrust at S.L
Fig.10 Available and required thrust at h = 3048 m
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Fig.11 Available and required thrust at h = 4572 m
Fig.12 Available and required thrust at h = 7620 m
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Fig.13 Available and required thrust at h = 9144 m
Fig.14 Available and required thrust at h = 10973 m
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h(in ft)
h(in m)
Vs (m/s)
(Vmin)e (m/s)
T = Tcr
(Vmin)e (m/s)
T=Tclimb
Vmax(m/s)T = Tcr
Vmax(m/s)T=Tclimb
Vmax(kmph)T=Tclimb
S.L
10000
15000
25000
30000
36000
38000
38995
39220
0
3048
4572
7620
9144
10973
11582
11884
11954
77.833
90.579
98.131
116.292
127.278
142.594
149.557
153.159
153.950
< Vs
< Vs
< Vs
< Vs
< Vs
176.054
217.386
235.48
----
< Vs
< Vs
< Vs
< Vs
< Vs
169.071
200.896
229.865
236.40
258.711
272.060
275.613
272.929
267.854
253.671
243.676
235.48
------
269.370
280.595
283.300
279.291
271.755
258.154
248.630
238.649
236.40
969.7
1010.1
1019.9
1005.4
978.3
929.4
895.1
859.1
851.04
Table 3 Variations of Vs, (Vmin)e , Vmin and Vmax
Fig.15 Variations of Vmin and Vmax with altitude
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5 Steady climb
Forces on an airplane in a steady climb
In this flight, the C.G of the airplane moves along a straight line inclined to the horizontalat an angle γ . The velocity of flight is assumed to be constant during the climb. Since,
the flight is steady, the acceleration is zero and the equations of motion can be written as:
T - D - W sin = 0 (19)
L - W cos = 0 (20)
To calculate the variation of rate of climb with flight velocity at different altitudes, the
following procedure is adopted.
Choose an altitude.
Choose a flight speed.
Noting that CL = 2W cos γ / SV2, gives :
2
D Do 2
2WcosγC = C + K
ρSV
Also,Vc = V sin
Hence,
2
c
2
Vcosγ = 1-
V
Substituting various quantities in Eq.(19) yields :
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12
222 c c
a DO2
V V1 KWT = ρV S C + 1- + W
12 V VρV S
2
Or 2c cV VA( ) + B( ) +C = 0V V
(21)
2
2
KWA =
1ρV S
2
; B = -W;2
2
a Do a2
1 2KWC = T - ρV SC - ,T Thrust available
2 ρV S (22)
Since, altitude and flight velocity have been chosen, the thrust available is read from the
climb thrust curves in Fig.3. Further, the variation of CDo and K with Mach number is
taken as in Eqs.17 and 18. Equation 21 gives 2 values of Vc/V . The value which is less
than 1.0 is chosen, as sin γ cannot be greater than unity. Hence ,γ = sin
-1(Vc/V) (23)
and Vc = V sin γ (24)
This procedure is repeated for various speeds between Vmin and Vmax. The entire
procedure is then repeated for various altitudes. The variations of (R/C) and γ with
velocity and with altitude as parameter are shown in Figs.16 and 18. The variations of
(R/C)max and γmax with altitude are shown in Figs.17 and 19. The variations of V(R/C)max
andmax
V
with altitude are shown in Figs.20 and 21. A summary of results is presented in
Table 4.
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h
(ft)
h
(m)
(R/C)max
(m/min)
V(R/C)max
(m/s)
γmax
(degrees)
maxV
(m/s)
0
1000015000
25000
30000
36000
38000
38995
39220
0.0
3048.04572.0
7620.0
9144.0
10972.8
11582.4
11885.7
11954.0
1086.63
867.34738.16
487.41
313.43
115.57
41.58
1.88
0
149.7
167.5174.0
198.2
212.2
236.1
236.9
236.5
236.40
8.7
6.04.7
2.6
1.5
0.5
0.2
0.0076
0
88.5
111.6125.7
164.1
188.0
230.2
234.0
236.0
236.40
Table 4 Climb performance
Fig.16 Rate of climb vs velocity for various altitudes
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Fig.17 Maximum rate of climb vs altitude
Fig.18 Angle of climb vs velocity for various altitudes
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Fig.19 Maximum angle of climb vs altitude
Fig.20 Velocity at maximum rate of climb vs altitude
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Fig.21 Velocity at maximum angle of climb vs altitude
Remarks:
i) The discontinuity in slope in Figs.20 and 21 at high velocities are due to the change in
drag polar as the Mach number exceeds 0.8.
ii) From Fig.17, the absolute ceiling (at which (R/C)max is zero) is 11.95 km. The service
ceiling at which (R/C)max equals 100 ft /min (30.5 m/min) is 11.71 km.
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Appendix B
Lecture 40
Performance analysis of a subsonic jet transport – 3
Topics
6 Range and endurance
7 Turning performance
8 Take-off distance
9 Landing distance
10 Concluding remarks
Acknowledgements
6 Range and endurance
In this section, the range of the airplane in a constant altitude and constant
velocity cruise is studied. The range is given by the following formula.
1
-1max 1
max L 1
7.2 E V E ζR = tan
TSFC 2E (1-KC E ζ)
(25)
where,
O
max
D
1E =
2 K C; K and CDo are at Mach number corresponding to V.
f 2
1 1
W Wζ = = 1-
W W
L11
D1
CE =
C , 1
L12
WC =
1ρV S
2
,
CD1 = Drag coefficient at CL1 and Mach number corresponding to V.
W1 is the weight of the airplane at the start of the cruise and W2 is the weight of the
airplane at the end of the cruise.The cruising altitude is taken as h = 10973 m (36000 ft). TSFC is taken to be constant as
0.6hr -1
. The variation of drag polar above M = 0.8 is given by Eqs.17 and 18.
W1 = Wo = 59175 x 9.81 = 580506.8 N , Wf = 0.205 x W1
Allowing 6% fuel as trapped fuel, W2 becomes
W2 = W1 – 0.94 x Wf or ζ = 0.94 x 0.205 = 0.1927
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The values of endurance (in hours) are obtained by dividing the expression for range by
3.6V where V is in m/s. The values of range (R) and endurance (E) in flights at different
velocities are presented in Table 5 and are plotted in Figs.22 and 23.
M V
(m/s)
CDo K Emax CL1 CD1 E1 R
(km)
E
(hr)
0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.81
0.82
0.83
0.84
0.85
0.86
0.870.88
147.53
162.29
177.04
191.79
206.54
221.30
236.05
239.00
241.95
244.90
247.85
250.80
253.75
256.71259.66
0.0159
0.0159
0.0159
0.0159
0.0159
0.0159
0.0159
0.0159
0.01592
0.01597
0.01604
0.01613
0.01624
0.016370.01652
0.04244
0.04244
0.04244
0.04244
0.04244
0.04244
0.04244
0.04256
0.04300
0.04388
0.04532
0.04744
0.05036
0.054200.05908
19.25
19.25
19.25
19.25
19.25
19.25
19.25
19.22
19.11
18.89
18.54
18.08
17.48
16.7916.00
1.312
1.085
0.911
0.777
0.670
0.583
0.513
0.500
0.488
0.476
0.465
0.454
0.444
0.4330.424
0.089
0.066
0.051
0.041
0.035
0.030
0.027
0.02654
0.02616
0.02591
0.02584
0.02591
0.02617
0.026530.02714
14.75
16.48
17.82
18.72
19.17
19.23
18.95
18.84
18.65
18.37
18.00
17.52
16.97
16.3215.62
2979.0
3608.0
4189.6
4691.7
5095.6
5396.5
5599.8
5619.7
5621.6
5597.7
5544.1
5460.4
5349.3
5210.15051.1
5.61
6.18
6.57
6.80
6.85
6.77
6.59
6.53
6.45
6.35
6.21
6.05
5.86
5.645.40
Table 5 Range and endurance in constant velocity flights at h = 10973 m (36000 ft)
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Fig.22 Range in constant velocity flights at h = 10973 m
Fig.23 Endurance in constant velocity flights at h = 10973 m
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Remarks:
i) It is observed that the maximum range of 5620 km is obtained around a velocity of
240 m/s (864 kmph). Corresponding Mach number is 0.82 which is slightly higher than
the Mach number beyond which CDo and K increase. This can be explained based on two
factors namely (a) the range increases as the flight speed increases and (b) after Mcruise is
exceeded, CDo and K increase thus, reducing (L/D)max.
ii) The range calculated above is the gross still air range. The safe range would be about
two-thirds of this. In the present case, the safe range would be around 3750 km.
iii) The maximum endurance of 6.85 hours occurs in a flight at V = 206 m/s. (742 kmph).
It is observed that the endurance is roughly constant over a speed range of 190 m/s to 230
m/s (684 to 828 kmph).
7 Turning performance
Forces acting on an airplane in turning flight
In this section, the performance of the airplane in a steady level, co-ordinated-turn is
studied. The equations of motion in this case are:
T – D = 0
W - L cos = 0
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L sin =W
g
2V
r
where is the angle of bank.
These equations give:
r =
2V
gtan
V g tanψ = =
r V
Load factor = n =L 1
=W cos
where, is the rate of turn and r is the radius of turn.
The following procedure is used to obtain r min andψ max .
1) A flight speed and altitude are chosen and the level flight lift coefficient
CLL is obtained as :
CLL =2
2(W/S)
ρV
2) If CLmax/CLL < nmax, where nmax is the maximum load factor for which the airplane is
designed, then the turn is limited by CLmax and CLT1 = CLmax. However, if
CLmax/CLL > nmax, then the turn is limited by nmax, and CLT1 = nmaxCLL.
3) From the drag polar, CDT1 is obtained corresponding to CLT1 . Then,
DT1 =2
DT1
1ρV SC
2 .
If DT1 > Ta, where Ta is the available thrust at that speed and altitude, then the turn is
limited by the engine output. In this case, the maximum permissible value of CD in
turning flight is found from
aDT
2
TC =
1ρV S
2
From drag polar, the value of CLT is calculated as
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DT DoLT
C -CC =
K
However, if DT1 < Ta, then the turn is not limited by the engine output and the value of
CLT calculated in step (2) is retained.
4. Once CLT is known, the load factor during the turn is determined as
LT
LL
Cn =
C
Once n is known, the values of , r and can be calculated using the equations given
above.
The above steps are repeated for various speeds and altitudes. A typical turning flight
performance estimation is presented in Table 6. In these calculations, CLmax = 1.4 and
nmax = 3.5 are assumed. The variation of turning performance with altitude is shown in
Table 7. Figures 24, 25, 26 and 27 respectively present (a) radius of turn vs velocity with
altitude as parameter, (b) Vrmin vs altitude, (c) rate of turn vs velocity with altitude as
parameter and (d) ψmaxV
vs altitude.
V(m/s)
CLL Lmax
LL
C
C
CLT1 CDT1 T1
(N)Ta
(N)CDT CLT n r
(m)
ψ
(rad/s)
78.8 1.365 1.026 1.4 0.0991 42106 126250 0.0991 1.4 1.026 12.9 2768 0.0285
98.8 0.868 1.612 1.4 0.0991 66182 118125 0.0991 1.4 1.612 51.7 787 0.1255
118.8 0.602 2.331 1.4 0.0991 95678 113750 0.0991 1.4 2.331 64.6 684 0.1738
138.8 0.440 3.181 1.4 0.0991 130595 106611 0.0809 1.238 2.813 69.2 747 0.1858
158.8 0.336 4.164 1.177 0.0747 128778 101539 0.0589 1.006 2.993 70.5 912 0.1742
178.8 0.265 5.279 0.928 0.0525 114709 97041 0.0444 0.819 3.089 71.1 1115 0.1603
198.8 0.215 6.527 0.751 0.0398 107635 92606 0.0343 0.661 3.080 71.1 1384 0.1437
218.8 0.177 7.905 0.620 0.0322 105461 89483 0.0273 0.519 2.930 70.0 1772 0.1235
238.8 0.149 9.415 0.521 0.0274 106860 86229 0.0221 0.383 2.573 67.1 2452 0.0974
241.8 0.145 9.655 0.508 0.0268 107282 85779 0.0215 0.362 2.494 66.4 2609 0.0927
Table 6 A typical turning flight performance at sea level
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6
Fig.24 Radius of turn vs velocity at various altitudes
Fig.25 Velocity at r min vs altitude
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Fig.26 Rate of turn (ψ ) vs speed at various altitudes
Fig.27 Velocity at maxψ vs altitude
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h
(m)
r min
(m)
Vrmin
(m/s)
ψ max
(rad/s)
Vψ max
(m/s)
0.0
3048.0
4572.0
7620.0
9144.0
10972.8
666
945
1155
1971
3247
8582
126.8
132.6
135.1
138.3
151.3
211.0
0.1910
0.1410
0.1170
0.0731
0.0513
0.0256
127.8
133.6
136.1
165.3
187.3
231.0
Table 7 Turning flight performance
Remarks:
i) The maximum value of is 0.191 and occurs at a speed of 127.8m/s at sea level.
ii) The minimum radius of turn is 666 m and occurs at a speed of 126.8m/s at sea level.
iii) The various graphs show a discontinuity in slope when the criterion which limits the
turn, changes from nmax to thrust available.
8 Take-off distance
In this section, the take-off performance of the airplane is evaluated. The take-off
distance consists of take-off run, transition and climb to screen height. Rough estimates
of the distance covered in these phases can be obtained by writing down the appropriate
equations of motion. However, the estimates are approximate and Ref.4 chapter 5
recommends the following formulae for take-off distance and balance field length based
on the take-off parameter.
This parameter is defined as:
Take-off parameter =LTo
W/S
σC (T/W) (26)
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where W/S is wing loading in lb/ft2, CLTO is 0.8 x CLand and is the density ratio at
take-off altitude. In the present case:
W
S = 5195N/m
2= 108.2lb/ft
2; CLTO = 0.8 x 2.7 = 2.16; = 1.0(sea level)
andT 2×97900
=W 59175×9.81
= 0.3373
Hence, take-off parameter =108.2
1.0 2.16 0.3373 = 148.86 (27)
From Ref.4, chapter 5, the take-off distance, over 50', is 2823' or 861m. The balance
field length for the present case of two engined airplane is 6000' or 1829m.
Remark:
It may be noted that the balance field length in this case, is more than twice the take-off
distance.
9 Landing distance
In this section the landing distance of the airplane is calculated. From Ref.4, chapter 5,
the landing distance for commercial airliners is given by the formula:
land
Lmax
W 1s = 80 ( ) +1000 ft
S σC (28)
where W/S is in lbs/ft2. In the present case:
(W/S)land = 0.85 x (W/S)takeoff = 0.85 x 108.5 = 92.225 lb/ft2
CLmax = 2.7 , = 1.0
Hence,
sland = 80 x 92.2251
10001.0 2.7
= 3732 ft =1138 m (29)
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10 Concluding remarks
1. Performance of a typical commercial airliner has been estimated for stalling speed,
maximum speed, minimum speed, steady climb, range, endurance, turning, take-off
and landing.
2. The performance approximately corresponds to that of B737-200.
3. Figure 28 presents the variations with altitude of the characteristic velocities
corresponding to :
stalling speed, Vs
maximum speed, Vmax
minimum speed as dictated by thrust, (Vmin)e
maximum rate of climb, V(R/C)max
maximum angle of climb, V max
maximum rate of turn, V max
minimum radius of turn, Vrmin
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