Chapter 5

1

PERFORMANCE ANALYSIS

Prof.E.G.Tulapurkara

Chapter 5

2

5.1 INTRODUCTION:

During its normal operation an airplane takes –off, climbs to the cruising altitude, cruises at almost constant altitude, descends and lands. It may also perform flights along curved paths like turns, loops etc. (flight along a curved path is called a manouvre).

Analysis of these flights is the subject matter of performance analysis which is dealt with in this chapter.

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Performance analysis covers the following.

I) Unaccelerated flights:

(a) Steady level flight. This analysis would give the maximum level speed and minimum level speed at different altitudes.

(b) Steady climb. This analysis would provide information on the maximum rate of climb, maximum angle of climb and maximum attainable altitude (ceiling).

(c) Descent and glide. This analysis would give the minimum rate of sink and time to descend from an altitude.

(d) Range and endurance. This analysis would provide information about the maximum distance

4

the airplane can cover or and the maximum time for which an airplane can remain in air with a given amount of fuel .

II) Accelerated flights :

(a) Accelerated level flight. This analysis provides information about time required and distance covered during acceleration over a specified velocity range.

(b)Accelerated climb. This analysis gives information about the change in the rate of climb when the flight velocity changes during the climb as compared to that in a steady climb.

(c) Manoeuvres like loop and turn. This analysis would give information about the maximum rate of

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turn and minimum radius of turn. These items indicate the maneuverability of an airplane.

(d) Take-off .This analysis would give information about the take-off distance required.

(e) Landing. This analysis would provide information about the distance required for landing.

Remarks:

i) The performance analysis is important to asses the capabilities of an airplane. Further from the point of view of airplane designer, this analysis would give the thrust or power required, maximum lift coefficient required etc. to achieve a desired performance. Performance analysis also points out as to what new developments in airplane aerodynamics or engine

6

performance are needed to attain better specifications.

ii) As mentioned in chapter 1 our approach is to apply is the Newton’s laws and arrive at equations of motion. The analysis of the equations would give the performance.

iii) References 1.1,1.3 to 1.8 may be referred to supplement the analysis described in the subsequent sections.

iv) Let us recapitulate the following:

(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).

7

(b) The external forces acting on a rigid airplane are: (I) Aerodynamic forces (lift and drag)

(II) Gravitational force

(III) Propulsive force (thrust)

(c) Forces produced due to control deflection, needed to balance the moments, may be 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).

8Fig.5.1 Flight path

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5.2 STEADY 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 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.

Equations of motion in steady level flight:

The equations of motion are obtained by resolving, along and perpendicular to the flight direction, the forces acting on the airplane.

T-D = m ax

L-W = m az

10Fig. 5.2 Steady level flight

11

Where m is the mass of airplane and ax, and az, are the components of acceleration along and perpendicular flight path.

Since the flight is steady i.e. no acceleration along tangent to the flight path, ax = 0.

Further the flight is straight and level i.e. no change in altitude, az=0.

Hence the equations of motion reduce to:

T – D = 0, L – W = 0 (5.1)

Since W=L = (1/2)ρV2 SCL

We get V= (2W/ρSCL)1/2 (5.2)

and (1/2)ρV2S = W/CL

We get,

12

Further the thrust required (Tr) is given by:

Tr = D =(1/2) ρV2SCD

Substituting for (1/2) ρV2S as W/CL, gives

Tr = W (CD/CL) (5.3)

The power required (Pr), in kilowatts, is given by

Pr = TrV/1000 ,

Substituting for V and Tr from Eqs. (5.2) and (5.3)

where Tr is in Newton and V in m/s.

13

21000

Dr

L L

CW WPC S Cρ

= × ×

3

3 / 2

1 21 0 0 0

Dr

L

CWPS Cρ=

Remark :

i) Equations (5.1) to (5.4) are the basic equations for steady level flight and would be used in subsequent analysis of this case.

ii) To fly in a steady level flight at chosen values of h and V, the pilot should adjust (a) the angle of attack of the airplane to get the desired lift coefficient so that L equals W and (b) engine setting so that thrust

gives :

(5.4)Or

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equals drag at desired angle of attack. He 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 force produced due to elevator deflection is neglected.

5.2.1 Stalling speed:

From Eq.(5.2) we note that for an airplane with a given weight (W) , wing area (S) and flying at an altitude (h), the flight velocity (V) is proportional to 1/CL

1/2 . Thus the value of CL required wouldincrease as the flight speed decreases.

Since CL cannot exceed CLmax, there is a flight speed below which level flight is not possible.

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The flight speed at which CL equals CLmax is called the stalling speed and is denoted by Vs. Then

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. Variations of Vs with h for a typical piston engined airplane and a typical jet airplane are presented in Figs. 5.3 (a) and (b). See Appendices A & B for details of calculations.

Remark:

The maximum lift coefficient (CLmax) depends on 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.

16Fig.5.3(a) Variation of stalling speed with altitude for a low speed airplane

17Fig.5.3(b) Variation of stalling speed with altitude

for a jet transport

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5.2.2 Equivalent air speed

2 21 12 2 o eV Vρ ρ=

1 / 2

0

2e

L

WV VSC

σρ

= =

Equivalent air speed (Ve) is defined as :

Noting , σ=ρ/ ρ0,

(5.6)

Remarks:

i) From Eq.(5.6) it is evident that for a given wing loading (W/S), the equivalent air speed in steady level flight is proportional to 1/CL

1/2 and is

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independent of altitude. Thus the stalling speed, for a given airplane configuration, when expressed as equivalent speed 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.

iii) The significance of equivalent airspeed in reducing the work involve in performance estimation will be clear from subsequent sections.

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5.2.3 Thrust Required and Power Required in Level Flight:

From Eqs. (5.3) and (5.4) we get:

CD depends on CL and Mach number. The relationship between CD and CL, the drag polar, is known from the estimation of the aerodynamic characteristics of the airplane.

3

3 / 2

1 21 0 0 0

Dr r

L

D

L

CWT W and PS C

CC ρ

= =

21

Thus, when the drag polar, the weight of the airplane and the wing area are prescribed, 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 chosen values of V and h, and given values of the weight of airplane (W) and the wing area (S) calculate CL as :

where ρ corresponds to density at the chosen ‘h’.

2

2L

WCS Vρ

=

22

v) For the values of CL and M, calculated in steps (iii) and (iv), obtain CD from the drag polar. It (drag polar) may be given in the form of Eqs. (3.10) or (3.12). 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).

Remark:

When the Mach number is less than about 0.7, the drag polar is generally independent of Mach number. In this case, plotting CD/CL versus CL and

iv) Calculate Mach number from M = V/a, where ‘a’is the speed of sound at the chosen altitude.

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CD/CL3/2 versus CL,we get the curves shown in

Figs.5.4.(a) & (b).

From these curves we find 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/CL3/2

is minimum. The CL at which this occurs is denoted by CLmp .

Thus in steady level flight.

Trmin=W(CD/CL)min (5.7)

min

3

3/ 2min

2 Dr

L

CWPS Cρ

⎛ ⎞= ⎜ ⎟

⎝ ⎠(5.8)

24Fig. 5.4a Variation of CD/CL with CL

25Fig.5.4b CD/CL3/2 vrs CL

26

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:

Note:

i) CLmd and CLmp are not same and the corresponding

speeds are different. Equation (5.9) shows that

these speeds increase with altitude.

(ii) Since for Mach numbers lower than about 0.7,the

drag polar is assumed to be independent of Mach

number, the values of CLmd, CLmp , (CD/CL) min

2 2,md mp

md mpL L

W WV VS C S Cρ ρ

= = (5.9)

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and (CD/CL3/2)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) As seen in Fig.5.4a, 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 trust, in general, flight is possible at two speeds (VA and VB). These speeds are:

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VA= (2W/ρSCLA)1/2 , VB= (2W/ρSCLB)1/2 (5.8a)

Similarly from Fig.5.4.b we observe that with the same power, in general, level flight is possible at two values of lift coefficient viz. CLA and CLB and correspondingly two flight speeds viz. VA and VB .

iii) Typical variations of thrust required with flight speed and altitude are shown in Fig. 5.5. Following interesting observations can be made. From Eq.(5.7) the minimum drag depends only on W and (CD/CL)min and is independent of altitude when the drag polar is independent of Mach number. This is the case for subsonic airplanes especially at CL=CLmd. However the speed corresponding to minimum drag (Vmd)

29Fig.5.5 Thrust required and thrust available for

subsonic jet airplane

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increases with altitude (Eq.5.9). Hence the thrust required curves at various altitudes have the same minimum at all altitudes and the curves have a horizontal line, corresponding to T=Tmin , as common tangent (See Fig.5.5). This feature should be kept in mind when thrust required curves for subsonic airplanes are plotted.

iv) Typical variations of power required with flight speed and altitude are shown in Fig. 5.6a. Interesting observations can be made in this case also. From Eq.(5.8) the minimum power required (Prmin) depends on W3/2 , (CD/CL

3/2)min and ρ-1/2 . From Eq.(5.9) we observe 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

31Fig 5.6a Power required and power available curves

32Fig 5.6b Power required and power available at an

altitude near ceiling

33

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 Pr/V or Tr . However Tr has a minimum which is 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 brought 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.

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5.2.4 Consideration of Parabolic Drag Polar

The earlier discussion was with reference to a general drag polar given in the form of a table or a curve. Now we consider the parabolic polar given by

CD=CD0+KCL2 (5.10)

Since we now have an equation for the drag polar,

it is possible to obtain mathematical expressions for the power required and thrust required. Presently we will assume that CD0 and K are constant with Mach number.Substituting for CD in expression for thrust required gives :

Tr= D= (1/2)ρV2SCD= (1/2) ρV2S(CD0+KCL2) (5.11)

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Substituting for CL as W/{(1/2)ρV2S} in Eq. (5.11)yields:

In Eq. (5.12) the first term(½)ρV2SCD0 is called parasite drag. The second term 2KW2/(ρV2S) is called induced drag. The variations of parasite drag and induced drag and the total drag are shown in Fig.5.7.

The parasite drag, being proportional to V2, increases rapidly with speed.

2

212 21

2

2Or D

WT V S C K

V Sρ

ρ= +

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

0

2 2 212

2 /( )r DT V SC KW V Sρ ρ= + (5.12)Or

36Fig 5.7 Variation of Drag with flight speed.

37

The induced drag being proportional to 1/V2 is high at low speeds but decreases rapidly as speed increases. The total drag, the 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 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.

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Substituting Vmd in Eq.(5.12) gives minimum thrust required i.e.

Trmin= W(CD0K)1/2+W(CD0K)1/2= 2W(CD0K)1/2 (5.14)

Expression for power required with parabolic polar:

Substituting for CD from Eq.5.10 gives :

12

1 / 42

O

m dD

W KVS Cρ

⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

31 11000 1000 2

rr D

T VP V SCρ= =

2

0 3

1 2 ( 2)2 02

rmd D

md

dT KWV SCdV S V

ρρ

−= + =

(5.13)or

39

231 1

2000 500Or DKWP V S C

VSρ

ρ= +

The first term in Eq. (5.15) is called the parasite power and the second term is called the induced power. The variations of induced power, parasite power and the total power required are shown in Fig. 5.8.

3 20

3 20 21

2

1 1 [ ]1000 2

1 1 [ ( ) ]1000 2

r D L

r D

P V S C KC

WP V S C KV S

ρ

ρρ

= +

= +or

or (5.15)

40Fig 5.8 Variation of power required with flight speed

41

It is seen that the minimum power occurs at a speed, Vmp, at which the induced power is three times the parasite power. This verification, which can be done by differentiating Eq. (5.15) with respect to V and equating it to zero, is left as an exercise to the student.

12

1 / 42

3O

m pD

W KVS Cρ

⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

12

min

1/ 4331 2 256

1000 27 Or DWP C KSρ

⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

(5.16)

(5.17)

42

Remarks:

i) The expressions given in Eqs. (5.13) and (5.14)

can be obtained in the following alternative way:

Tr=W(CD/CL)

Hence, Trmin= W (CD/CL)min (5.18)

But for a parabolic polar

(5.19) ODD

LL L

CC K CC C

= +

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The value of CL at which (CD/CL) is minimum i.e.

(CLmd) is given by

2

( / ) 0 0ODD L

L L m d

Cd C C o r Kd C C

= − + =

This gives CLmd as:

CLmd=(CD0/K)1/2 (5.20)

The corresponding drag coefficient, CDmd is

(5.21)

Equation (5.21) shows that when Tr equals Tmin, both parasite drag coefficient and induced drag coefficient are equal to CD0. Hence under this condition the parasite drag and induced drag are each equal to (1/2)ρV2SCD0.

2O

md O O

DD D D

KCC C C

K= + =

44

Further,

Hence Tmd and Vmd are

( )12

12

min

22 ( )

/md O

O

mdO

D DDD

L L D

C CC C KC C C K

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

( ) ( )121

21 / 4

min

22 / ,

O OD DmdW

T W C K and V K CSρ

= =⎛ ⎞⎜ ⎟⎝ ⎠

(5.22)

which are same as Eqs.(5.14) & (5.13).

ii) Similarly, expressions given in Eqs. (5.16) and

(5.17) can be obtained in the following alternate

manner. 123

3 / 2

1 21 0 0 0

Dr

L

CWPS Cρ

⎛ ⎞= ⎜ ⎟

⎝ ⎠

45

Hence, Prmin occurs when CD/CL3/2 is minimum.

For a parabolic polar

Therefore,

Equating the R.H.S. to zero, gives the , the value

of CL at which the power required is minimum i.e.

CLmp as:

12

3/ 2 3/ 2ODD

LL L

CC K CC C

= +

( )12

3/ 2

5/ 2

/ 3 12 2

OD L D

L L L

d C C C Kd C C C

= − +

46

CLmp=(3CD0/K)1/2 (5.23)

Then the drag coefficient , corresponding to CLmp

is given by:

Equation (5.24) shows that when Pr equals Prmin the

parasite drag coefficient is equal to CD0 and the

induced drag coefficient is equal to 3CD0.

Consequently the parasite power is (1/2)ρV3SCD0

and induced power is 3 times of that.

34O

mp O O

DD D D

K CC C C

K= + =

Hence,1/ 4

33/ 2 3/ 4

min

4 256(3 / ) 27

O

O

O

DDD

L D

CC C KC C K

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

(5.24)

47

The above expression for Vmp is same as in

Eq.(5.16).

iii) The manner in which the performance analysis

helped in evolution of airplane configuration, can

be appreciated from the following discussion.

(a) The low speed airplanes are powered by engines

delivering BHP/ESHP. In this case the major

portion of the power required is induced power,

1122

1/ 42 2

3mp O

mpL D

W W KVSC S Cρ ρ

⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟= = ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠

1 / 4

1 0 .7 63 m d m dV V= ≈

48

power, which depends on the factor K in drag polar (Eq.5.10). This factor is given as 1 / (π Ae) where A is the aspect ratio of the wing and ‘e’ is the Oswald’s efficiency factor (see Eq. 3.7). 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 a consideration. However medium speed commercial airplanes have aspect ratio between 10 to 12. Gliders have aspect ratio has high as 16 to 20.

(b) For high subsonic airplanes most of the drag is parasite drag which depends on CD0 (Eq.5.12).

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Hence high speed airplanes have features like smooth surfaces, thin wings, streamlined fuselage, smooth fairings at wing-fuselage joint and retractable landing gear. They also have high wing loading (W/S) to reduce the wing area. Manufacturing techniques have been improved to achieve smooth surface finish. At supersonic speeds thin wings ( thickness ratio around 3 to 5%) are used to reduce CD0. Table 3.2 may be referred to for typical values of CD0 , A and e of different airplanes.

The reciprocal of (CD/CL) is (CL/ CD). 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

50

12 to 20 for a subsonic airplanes and between 5 to 8 for supersonic airplanes.

iv) 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 (see Eqs. 5.3 and 5.4). Hence airplane design bureaux have a group which keeps a close watch on any increase in the weight of the airplane.

51

5.2.5 Steady Level Flight Performance with a given Engine

In steady level flight the thrust must be equal to drag (Eq.5.1). The thrust is provided by the engine or the engine-propeller combination and from chapter 4 we know that the thrust or power output varies with engine RPM, flight speed and altitude.

Typical variations, with altitude and speed, of the maximum thrust available (Ta) and the maximum thrust horse power available (THP)a are shown in Figs. 5.5 and 5.6.

For airplanes with piston engine or turboprop engine the output is the power available at the engine shaft. Hence to estimate the performance such airplanes we work in terms of BHP or THP.

52

For airplanes with turbofan or turbojet engines the output is in terms of engine thrust and we work with that quantity for estimation of their performance.

The thrust required and power required are also shown in the Figs.(5.5) & (5.6a). It is seen that near the sea level 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. 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

53

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.5.6 b).

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 -points D’ in Fig.5.5 and point C’ in Fig.5.6 b. It may be added that Fig.5.6b has been drawn separately from Fig.5.6a to show the points C and C’ clearly.

54

Thus, the minimum speed can be limited by available output. We denote it by (Vmin)e. However in level flight the lift must be equal to weight and hence level flight is not be possible below stalling speed. Thus two factors viz. the 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 speeds limited by the engine output (Vmin)e and the stalling speed VS.

From the thrust available (Ta) and thrust required (Tr) charts, the maximum speeds at various altitudes can be obtained as the intersection of the Ta and Tr curves for a given altitudes (Fig. 5.5).

55

Typical variations of Vmax, (Vmin)e and Vstall 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.

It is seen that for a jet airplane the Vmax may slightly increase with altitude initially 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.

56Fig. 5.9 Vmax and Vmin for jet airplane

57Fig. 5.10 Vmax , (Vmin)e and Vs for airplane with

engine propeller combination

58

The minimum speed at low altitudes is limited by stalling. However near the ceiling the minimum level flight speed is limited by thrust available.

In the case of 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. The ceiling in this case is lower than in the case of jet airplane because the piston engine looses power 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.

59

5.2.6 Steady level flight with 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 out put consideration, can be calculated analytically. i.e. by solving an equation. It may be noted that from Figs. 5.5 & 5.6 the assumption of Ta or Pa as constant with V appears reasonable near the speeds where Vmax

occurs.

I) Airplane with jet engine:

Choose an altitude ‘h’. Let Ta be the thrust available in the range of speeds where Vmax is likely to occur. Then,

60

2 0O

aL L D

TKC C CW

− + =

0a DDL

L L

T CC K CW C C

= = +

Or (5.25)

Equation (5.25) is a quadratic in CL . Its solution gives two values of CL at which level flight with the given thrust is possible. Let these values of CL be CL1 and CL2. Then, the corresponding flight speeds, V1 and V2, are given as:

1 12 2

1 2

1 22 2

L L

W WV a n d VS C S Cρ ρ

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(5.26)

Tr=Ta=W(CD/CL)

Hence,

61

The same results can be obtained by using Eq. (5.12), i.e.

or AV4 – BV2 + C = 0 (5.27)

Where

For given value of thrust (Ta), Eq.(5.27) also gives two solutions for level flight speeds V1 and V2. Let V1

be the higher among V1 and V2.Then V1 is the maximum speed and V2 is the minimum speed, based on engine output (Vmin)e. The higher of (Vmin)e and the stalling speed (Vs) will be the minimum speed at

221

2 2

2Oa r D

WT T V S C KV S

ρρ

⎛ ⎞= = + ⎜ ⎟

⎝ ⎠

212

2,OD a

KWA SC B T and CS

ρρ

= = =

62

the chosen altitude (see example 5.1). Remarks:

i) Calculate the Mach number corresponding to V1. If it is more than the critical Mach number then CD0 and K would need correction and revised calculation would be needed.

ii) Obtain from the engine charts the thrust available at V1 and let us denote it as Ta1 .If the thrust available (Ta), assumed at the start of the calculation is significantly different from Ta1, then calculations would have to be revised with new value of Ta. However it is expected that the calculations would converge in a few iterations.

63

II. Airplane with engine propeller combination

Assume an altitude ‘h’. Let Pa be the THP available

in kW.

In this case, from Eq. (5.15)

or A1V4 – B1V + C1 = 0 (5.28)

Where

1 11 ,

2000 OD aA S C B Pρ= =

231 1

2000 500Or a DWP P V SC KVS

ρρ

= = +

2

1

1,

500

KWC

Sρ=

64

Equation (5.28) is not a quadratic. An iterative method of solving Eq. (5.28) is given in example 5.2. Equation (5.28) has two solutions V1

and V2. The higher of the two gives Vmax and the lower value gives (Vmin)e. The higher of (Vmin)e and the stalling speed (Vs) will be the minimum speed at chosen altitude (see example 5.2).

Remark:

Obtain power available at V1 calculated above and denote it by Pa1 . If Pa assumed at the beginning of the calculation is significantly different from Pa1, then the calculations would need to be revised with new value of Pa. However it is expected that the calculations would converge in a few iterations.

65

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 25m2

calculate (a) stalling speed at sea level and at 10 km altitude, (b) (CD/CL)min, (CD/CL

3/2)min ,Tmin, Pmin, Vmd

and Vmp under sea level conditions and (c) maximum and minimum speeds in steady level flight.

Assume CLmax = 1.5, CD = 0.016 + 0.064 CL2.

Solution:

Here W = 100,000 N, T = 20,000 N,

CD = 0.016 + 0.064 CL2 .

S = 25m2, CLmax =1.5

66

a)

at s.l. ρ = 1.225 kg/m3,

at 10 km ρ = 0.413 kg / m3

= 66 m/s at sea level

=113.6m/s at 10 km altitude.

b)

2 1000001.225 25 1.5sV ×

∴ =× ×

max

2 ,sL

WVS Cρ

=

2 1000000.413 25 1.5sV ×

=× ×

0 .016 / .064 0.5DLmd

CCK

= = =

02 0 .032D m d DC C= =

67

∴ (CD / CL)min = 0.032/0.5 = 0.064 and

Trmin = W (CD/ CL)min = 100000 x 0.064= 6400 N

3/ 2min( / )D LC C∴

3 / 0.866mp OL DC C K= =

04 0.064

mpD DC C= =

= 0.064/0.8663/2 = 0.0794

2 2 100000 114.5 /1.225 25 .5

md

mdL

WV m sS Cρ

×= = =

× ×

68

c) In this case T/W = 20000 / 100000 = 0.2= CD/CL

( )3 3

3/2min min

1 2 1 2 100000/ .0794 641.51000 1000 1.225 25D L

WP C C kWSρ

×= = × =

×

2 100000 86.30 /1.225 25 .866mpV m s×

= =× ×

Note: Vmp=Vmd/31/4

2.0160.2 0.064 .064 0.2 0.016 0L L LL

C or C CC

∴ = + − + =

69

Solving CL = 3.04 and 0.0822

The corresponding speeds are

max2 100000 281.8 /

1.225 25 .0822V m s×

= =× ×

min2 100000( ) 45.4 /

1.225 25 3.04eV m s×= =

× ×and

Since VS is larger than (Vmin)e the minimum speed is decided by VS and equals 66.0 m/s. Remark: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

70

Mcruise =0.8 and obtain ∆CD0 and ∆K from Eqs.(3.13a) & (3.14a).Consequently

∆CD0 = -0.001 (M-0.8) + 0.11 (M-0.8)2 and ∆K = (M-0.8)2 + 20(M-0.8)3

For M=0.828, ∆CD0 = 0.000055 and ∆K = 0.00122

Hence the polar at M=0.828 likely to be

CD= (0.016 + 0.000055) + (0.064+0.00122)CL2

= 0.016055 + 0.06522 CL2

Using this polar and revising the calculations we get Vmax= 281.3 m/s

Remark:

This revised value of Vmax is very close to the value of 281.8 m/s obtained earlier and further revision is not needed.

71

Results: Vmax= 281.3 m/s =1012.7 kmph

Vmin= 66 m/s = 237.6 kmph

Example 5.2

A piston-engined airplane has the following characteristics. W = 11,000 N, S = 11.9 m2 ,

CD = 0.022 + 0.055 CL2, CLmax =1.4. Obtain the

maximum and minimum speeds in level flight at an altitude of 3 km assuming that the engine BHP is 103 kW and propeller efficiency is 83 percent.

72

Solution:

W = 11,000 N , S = 11.9 m2, CD = 0.022 +

0.055 CL2

CLmax = 1.4, ρ at 3km altitude = 0.909 kg/m3 ,

Pa = η x BHP = 0.83 x 103 = 85.5 kW2

31 22000 1000Oa D

K WP V S CSV

ρρ

= +Hence

3185.5 0.909 11.9 .0222000

V= × × × ×22 .0 5 5 1 1 0 0 0

1 0 0 0 0 .9 0 9 1 1 .9 V×

+ ×× ×

4 3 1230.51.19 10x VV

−= + (5.28a)

73

Equation (5.28a) is not a quadratic. However it can be solved for Vmax and (Vmin)e by trial and error. Solution for Vmax:When solving for Vmax, by an iterative procedure, we assume that the first approximation (Vmax1) is obtained by retaining only the term containing highest power of V in Eq. (5.28a) i.e.

1st approximation: 85.5 = 1.19 x 10-4 V3max1

This gives Vmax1= 89.6 m/s

To obtain the 2nd approximation, we substitute Vmax1

in the second term on RHS of Eq. (5.28a). Note that this term was ignored in the first approximation i.e.

74

or Vmax2 = 84.1 m/s

To obtain the 3rd approximation, we substitute Vmax2 in the second term on RHS of Eq. (5.28a), i.e.

-4 3max 2

1230.585.5 = 1.19 x 1089.6

V +

4 3m a x 3

1 2 3 0 .58 5 .5 1 .1 9 1 08 4 .1

X V−= +

or Vmax3 = 83.8 m/sTo obtain the 4th approximation, we substitute

Vmax3 in the second term on RHS of Eq. (5.28a), i.e.

4 3max 4

1230.585.5 1.19 1083.8

X V−= +

75

or Vmax4 = 83.75 m/sSince the 3rd and 4th approximations are close to

each other, Vmax = 83.75 m/s.

Solution for (Vmin)e :When solving for (Vmin)e, by an iterative

procedure, we assume that the first approximation (Vmin)e1, is obtain by retaining only the term containing the lowest power of V in Eq. (5.28a) i.e

m in 1

1230 .585 .5( ) eV

=

76

Or (Vmin)e1= 14.4 m/s.

To obtain the 2nd approximation, we substitute (Vmin)e1 in the first term on RHS of Eq. (5.28a). Note that this term was ignored in the first approximation i.e.

or (Vmin)e2 = 14.45 m/s

Since the second approximation is very close so

the first one, (Vmin)e = 14.45 m/s

4 3

min 2

1230.585.5 1.19 10 14.4( )e

xV

−= × +

77

The stalling speed at 3 km altitude is :

Since Vs is greater than (Vmin)e, the minimum

speed is 38.2 m/s.

m ax

2 2 11000 38.2 /0.909 11.9 1.4s

L

WV m sS Cρ

×= = =

× ×

Results: At 3 km altitude: Vmax= 83.75 m/s=301 kmph

Vmin = 38.2 m/s = 137.4 kmph

78

Remark:

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).

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,C in Fig. 5.11). It is interesting to note

that if the airplane can go past with Mach number

represented by point B in Fig.5.11, then it can fly

79

up to Mach number represented by point C with the

same engine. To overcome the rapid drag rise in

transonic region (Fig.5.11), afterburning operation

of the engine is resorted to. This gives additional

thrust for a short duration. The thrust with

afterburner is shown schematically by a dotted line

in Fig. 5.11. Now 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 run on the normal

engine operation. It may be pointed

80Fig. 5.11 Level flight performance at high speeds

81

out that with afterburner the specific fuel

consumption is very high and its (afterburner)

use is restricted to a short duration of operation.

82

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 = 48 m2;

CD = 0.021 + 0.065CL2

BHP = 400 kW; Propeller efficiency = 82%;

CLmax = 1.45

[Answers: Vmax = 280.2 kmph;

Vmin = 104.6 kmph]

83

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

CD0 = 0.017, A = 6.5, e = 0.80, CLmax = 1.4.

[Answer: Vmax = 877 kmph,

Vmin = 283.6 kmph]

84

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 is same at the two altitudes.

[Answers: M=0.27, Ve= 68 m/s, (Tr)s.l= (Tr)5km as CL is same]

85

5.4 Show that the thrust required in steady level flight at a speed V for an airplane with parabolic drag polar is given by:

where Vmd =speed for minimum drag, W= weight of airplane and A= (CD0K)1/2.

22( )

( / )rmd md

V AWT D AWV V V

= = +

86

5.3 STEADY CLIMB

During a steady climb the center of gravity of the

airplane moves along a straight line inclined to

the horizontal at an angle γ (Fig.5.12) at constant

velocity. The forces acting on the airplane are

shown in Fig. 5.12.

87Fig. 5.12 Steady climb

88

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 γ = 0 (5.29)

L – W. cosγ = 0 (5.30)

∴ Sin γ = (5.31)T D

W−

From the velocity diagram in Fig. 5.12 we see that the vertical component of the flight velocity (Vc) is given by:

Vc = = V sin γ = V T D

W−

89

The vertical component of the velocity (Vc) is called rate of climb and denoted by R/C. It is also the rate of change of height and denoted by (dh/dt). Hence

Vc = R/C =dh/dt = (5.32)Rate of climb is generally quoted in m/min.Remarks :

i) Multiplying Eq.(5.29) by flight velocity V, we get:

T.V= D.V + W.V sin γ = D.V + W.Vc (5.33)

In Eq. (5.33) the term TV represents the power available, DV represents the energy dissipated in overcoming the drag and WVc is the rate of increase of potential energy. Thus, when the airplane climbs, its potential energy increases

V sin =γ T D VW−

90

and this is possible only if the available engine output is more than the energy required to overcome the drag. Consequently at Vmax and (Vmin)e , where the power available is equal to power required to overcome the drag, the rate of climb will be zero. The climb is possible only in between these two speeds and there will be a speed at which the rate of climb will be a maximum. This flight speed is denoted by V(R/C)max and the maximum rate of climb is denoted by (R/C)max. The speed at which the angle of climb γ is maximum, is denoted by Vγmax.

91

ii) In steady level flight, the lift is equal to weight but in climb the lift is less than weight as cos γ is less than one, when γ is not zero. Note that when an airplane climbs vertically, its attitude is as shown in Fig.5.13. It is seen that in this flight the resolution of forces along and perpendicular flight direction gives:

L = 0 , T = D + W which are consistent with Eqs.(5.29) and (5.30) when γ =900 is substituted in them. Note that in this flight the thrust is more than the weight.

92

Fig 5.13 Airplane in vertical climb

93

5.3.1 Thrust and Power required for a prescribed rate of climb at a given forward speedHere we assume 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 the following steps.

i) Calculate the angle of climb from:

= sin-1 (Vc / V)ii) The lift required is W cos (Eq. 5.30).

The lift coefficient CL = 212

co sWV S

γρ

γγ

γ

212

L

V Sρ=

94

iii) Obtain the flight Mach number; M =V/a

iv) Corresponding to these values of CL and M, obtain CD from the drag polar. The drag ( D) is given by:

D =(1/2ρV2SCD)

v) The Thrust required in climb (Trc) is then given by:

Trc = W sinγ + D

and the power required in climb (Prc) is :

Prc = (kW).

1000rcT V

95

5.3.2 Climb Performance with a given engineIn this case the engine output is prescribed at a given speed and altitude, and we should find the rate of climb (Vc) and the angle of climb(γ) .

The solution to this problem is not straight forward as sinγ depends on (T-D) and the drag (D) depends on lift (L), which in turn depends on W cos γ . Hence we obtain the solution in an iterative manner. This is explained later in this section. However, if the drag polar is parabolic with constant coefficients then an exact solution can be obtained using Eqs. (5.29) to (5.32). The procedure is as follows:

96

Sin γ = Vc/V

L = W cosγ = W (1-sin2γ)2

= W

∴ CL =

Now, D = (1/2)ρV2SCD

= (1/2)ρV2S( CD0+KCL2)

1221 ( / )cV V⎡ ⎤−⎣ ⎦

122

212

1 ( / )cW V VV Sρ

⎡ ⎤−⎣ ⎦21

2

L

V Sρ=

212 V Sρ=

22

212

1O

cD

VK WCV S Vρ

⎡ ⎤⎛ ⎞+ −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

97

From Eq.(5.29),T = D + W sin γ =

2

212

,KWAV Sρ

=

.cW VDV

+

Substituting for D, we get,22

212 21

2

1O

c cD

V W VK WT V S CV S V V

ρρ

⎡ ⎤⎛ ⎞= + − +⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

2

0c cV VA B CV V

⎛ ⎞ ⎛ ⎞− + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Where,

or (5.33)

B=W

22

0 2

1 2 .2 D

KWC T V SCV S

ρρ

= − −

98

Equation (5.33) is a quadratic in . It has two

solutions. The solution(s) which is (are) less than or

equal to one, is (are) the valid solution(s)

because Vc/V equals sinγ and sinγ cannot be more

than one .

Once (Vc/V) is known, the angle of climb and the

rate of climb can immediately calculated (see

example 5.3).

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 or THP.

⎛ ⎞⎜ ⎟⎝ ⎠

CVV

99

Such a procedure is necessary because, to calculate the drag coefficient we need the lift coefficient and the lift is equal to W cosγ and cosγ is not known in the beginning. Hence, to start the calculation we assume that L ≈ W and calculate CL1= W/(1/2)ρV2S

From CL1 and the flight Mach number we obtain CD1

from the drag polar and calculate the first approximation of drag (D1) as:

D1= (1/2)ρV2SCD1

Then, the first approximation to the rate of climb (Vc1) is given by:

11

ac

T DV VW−

=

100

The first approximation to the angle of climb (γ1) is given by:

γ1 = sin-1

In the next iteration we put L = W cos γ1 and carry out the calculations and get a second approximation of the angle of climb (γ2).

The calculations are repeated till the values of γafter consecutive iterations are almost the same.

It is found that the convergence is fast and correct values of γ and Vc are obtained within two or three iterations. This is due to the fact that when γ is small ( i.e. less than 10O), cos γ is

1cVV

⎛ ⎞⎜ ⎟⎝ ⎠

101

almost equal to one, and when γ is large the lift dependent drag is only a small fraction of the thrust required. (Example 5.4 illustrates the procedure).

5.3.3 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.

Figure 5.14a to f present typical climb performance of a jet transport. Figure 5.15a to d present the climb performance of a piston engined airplane. Details of the calculations are presented in Appendices B and A.

102Fig. 5.14a Climb performance of a jet transport-

rate of climb

103

Fig.5.14b Climb performance of a jet transport-angle of climb

104

Fig 5.14c Climb performance of a jet transport-V(R/C) max

105Fig.5.14d Climb performance of a jet transport-Vγmax

106

Fig.5.14e Climb performance of a jet transport -variation of (R/C)max with altitude

1075.14f Climb performance of a jet transport- variation

of γmax with altitude

108Fig. 5.15a Climb performance of a piston engined

airplane- rate of climb

109Fig. 5.15b Climb performance of a piston engined

airplane- angle of climb

110Fig. 5.15 c Climb performance of a piston engined airplane- V(R/C)max, and Vγmax

111Fig. 5.15d Climb performance of a piston engined

airplane - Variation of (R/C)max with altitude

112

Remarks:i) At V = Vmax the available thrust or THP is equal to the thrust required or power required in level flight. Hence no climb is possible at this speed. Similar is the case at (Vmin)e limited by engine output (Fig.5.14a and 5.15a). For the same reasons at Vmax and (Vmin)e the angle ofclimb (γ ) is also zero (Fig.5.14b and 5.15b). It may be recalled that at low altitudes the minimum speed is decided by stalling and hence the calculations of rate of climb and angle of climb are restricted 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 is maximum is called γmax . Figures 5.14c and d

γ

113

and Fig.5.15c show the variation of these speeds with altitudes for a jet transport and a piston engined airplane respectively. V(R/c)max and Vγmax are different from each other. For jet airplane V(R/C)max is higher than Vγmax at low altitudes . The two velocities approach each other as the altitude increases. For piston engined airplane V(R/C)max is lower than Vγmax at low altitudes . The two velocities approach each other as the altitude increases. These trends can be explained as follows: From Eqs.(5.31) and (5.32), we see that γ is proportional to the excess thrust i.e. (Ta-D) and the rate of climb is proportional to the excess power i.e. (TaV – DV). It may be recalled that for a piston engined airplane the power available remains roughly constant with velocity and hence the thrust available will

114

decrease with velocity. For a jet engined airplane the thrust available is roughly constant with velocity and the power available will increase linearly with velocity (see exercise 5.7).

iii) (R/C)max and γmax decrease with altitude as the excess power and excess thrust decreases with altitude.

iv) Climb hodograph

When we plot the vertical velocity (Vc) vs the horizontal velocity (Vh) for climb performance at a chosen altitude, the resulting curve is called a hodograph. In this plot 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

115

horizontal axis (Vh-axis) is the angle of climb (γ ). Figure 5.15 e shows various quantities. A line from origin which is tangent to the curve gives γmax and the velocity corresponding to it. Actually a climb hodograph gives complete information about climb performance especially γmax , Vγmax , (R/C)γmax , (R/C)max, V(R/C)max, γ(R/C)max and Vmax (Fig.5.15e).5.3.4 Absolute Ceiling and Service Ceiling:Figures 5.14e and 5.15d present the variations of maximum rate of climb, (R/C)max, with altitude. (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 hmax. At this altitude level flight is possible only at one speed.

116Fig. 5.15e Climb hodograph(Adapted from Ref.1.2c, chapter 13 )

117

(See sec. 5.2.2). Near the ceiling altitude the rate of climb is very small and the time to climb becomes very large. It is not possible to reach the absolute ceiling. Hence for practical purposes an altitude at which the maximum rate of climb is 50 m /min is used as ‘service ceiling’.

From Fig.5.14e the absolute ceiling and service ceiling for a jet transport are 11.88 and 11.55 km respectively. From Fig.5.15d the values of these ceilings for a piston engined airplane are 6.34 and 5.32 km respectively.

118

5.3.5 Time to climb:

From the knowledge of the variation of rate of climb with altitude, the time to climb (t) from altitude h1 to h2 can be calculated as follows:

Hence t = (5.34)

The rate of climb (Vc) in Eq. (5.34) depends on the speed and altitude at which the climb takes place. The appropriate values can be taken from plots similar to those given in Figs. 5.14e or 5.15d.

cc

dh dhV or dtdt V

= =

2

1

h

ch

dhV∫

119

However 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 and as such is not a steady climb. The value of Vc needs to be appropriately corrected for acceleration (see section 5.7 on accelerated climb).

120

Example 5.3

An airplane weighing 180,000N has a wing area of 45 m2 and drag polar given by

CD = 0.017 + 0.05 CL2

(a) 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.

(b) Obtain the rate of climb at 400 kmph at sea level if the trust available is 45,000 N.

121

Solution:

W = 180,000 N, S = 45 m2, CD = 0.017 + 0.05 CL2

ρ at 3 km altitude = 0.909 kg/m3

a) Vc = 2,000 m/min = 33.33 m/s,

V = 540 kmph = 150 m/s.Sin γ = Vc / V = 33.33/150 = 0.2222or γ = 12o-50’ , Cos γ = 0.975L = W cos γ = 180000 x0.975

or CL = 2

2W C osV S

γρ

122

Or CL= = 0.381

∴ CD = 0.017 + 0.05 X0.3812 = 0.02426

D = CD

=(1/2) X 0.909 X 150 X 150 X 45 X0.02426

= 11163 N

∴ Trc = W Sin γ + D = 180000 X 0.2222 + 11163

= 51160 N

180000 0.975 20.909 150 150 45

× ×× × ×

212 V Sρ

123

Prc= TrcV/1000 = 51160 X 150/1000

= 7674 kW

b) V = 400 kmph = 111.1 m/s, T = 45000N

T = W sin γ + D, but sin γ = Vc / V , then

cos γ =

122

1 cVV

⎛ ⎞⎛ ⎞⎜ ⎟− ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

D = 2 2 21 12 2 ( )

OD D LV S C V S C K Cρ ρ= +

CL =

2

2

2 2 21 1 12 2 2

1 cVWL W Cos VV S V S V S

γρ ρ ρ

−= =

124

∴ T = W sinγ + 22

2102 2 21

2

1 cD

VK WV S CV S V

ρρ

⎛ ⎞+ −⎜ ⎟

⎝ ⎠Substituting for various quantities gives:

45000 = W X

+

Simplifying, + 7.24 = 0

Or = 37.62, 0.192.

212 1.225 111.1 45 0.017CV

V+ × × × ×

22

2 212

0.05 180000 11.225 111.1 45

CVV

⎛ ⎞×× −⎜ ⎟× × × ⎝ ⎠

2

37.82c cV VV V

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

cVV

125

Since sin γ cannot be larger than unity the first value is not admissible.

∴ Vc / V = sin γ = 0.192 or γ = 11O 4’

Vc = 0.192 × 111.1 = 21.33 m / sec

= 1280 m/min.

Example 5.4

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.

1260.1161.20.0751.00.0630.90.0550.80.0470.70.0400.60.0340.50.0300.40.0260.30.0240.20.02250.10.0220.0CDCL

127

Solution:Here we have :W = 60,330 N, S = 64m2,

V=180 kmph = 50 m/s, THP = 500 kW

We obtain the values of γ and Vc by the iterative procedure explained in section 5.3.2.

1st approximation : L ≈ W =

= 0.615

CD1: By interpolating between the values given in the above table, the value of CD1 is 0.041, corresponding to CL1 of 0.615.

∴ D1 = (1/2) × 1.225 × 50 × 50 × 64 X0.041

= 4030 N

1

212 LV S Cρ

1

60330 21.225 50 50 64LC ×

∴ =× × ×

128

From Eq. (5.32) : Vc1 =

= 4.95 m/s

sin γ1 = (4.95/50) = 0.099 or γ1 = 5o41’

∴ cos γ1 = 0.995

2nd approximation :

L = W cos γ1 = 60330 × 0.995 = 60036 N

= 0.612

1 110001000

T D D VV THPW W− ⎛ ⎞= −⎜ ⎟

⎝ ⎠

1

1000 4030 5050060330 1000CV ×⎛ ⎞∴ = −⎜ ⎟

⎝ ⎠

2

60036 21.225 50 50 64LC ×

=× × ×

129

From above table CD2 is 0.0408 corresponding to CL2

of 0.612.

∴ D2 = (1/2) × 1.225 × 50 × 50 × 64× 0.0408

= 4010 N

= 4.96 m/s

The two approximations , Vc1 and Vc2 are fairly close to each other. Hence the iterations are stopped.

Vc = 4.96 m/s = 298 m/min.

2

1000 4010 5050060330 1000CV ×⎛ ⎞= −⎜ ⎟

⎝ ⎠

130

Remark :

In the present example γ is small (5O41’) hence 2nd iteration itself gives the correct value. For an interceptor airplane which has very high rate of climb(about 15000 m/min) larger number of iterations may be needed.

5.4 Steady Descent and Glide:

Figure 5.16 shows an airplane in a descent. In such a flight thrust is less than drag. The equations of motion can be written down as follows:

T + W sin γ - D = 0 (5.35)

L – W cos γ = 0 (5.36)

131Fig 5.16 Descent or glide

132

Hence, sin γ = (5.37)

Rate of descent (Vd) = (5.38)

The rate of descent is also called rate of sink and denoted by (R/S).

5.4.1 Glide

In a glide the thrust is zero. This may happen due to engine failure. However gliders are airplanes which do not have an engine.

D TW−

D T VW−

133

With thrust equal to zero, the following equations of motion for glide, can be obtained from Eqs. (5.35) and (5.36).

W sinγ - D =0 (5.39)

L-W cos γ =0 (5.40)

Hence,

sin γ = D/W (5.41)

and,

Vd = V sin γ = (5.42)

The angle of glide, γ, is generally small. Hence, L ≈ W and one can write,

sin γ = tan γ = γ = (5.43)

D VW

D

L

CD DW L C

= =

134

and (5.44)

Remarks:

i) Multiplying Eq. (5.39) by V we get:

W V sin γ -D V = 0

or W Vd – D V = 0

Noting that Vd is the rate of descent and equals dh/dt,

12 1 / 2

2 2

L L

L WVS C S Cρ ρ

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1 / 2

3 / 2

2 Dd

L

CD V D V WVW L S Cρ

⎛ ⎞= = = ⎜ ⎟

⎝ ⎠

0d hW D Vd t

− = (5.45)

135

From Fig.5.16 we note that ‘V’ is along the glide path and hence in the downward direction. Consequently in Eq (5.45) dh/dt is negative as the altitude is decreasing. As a result the potential energy of the glider is decreasing with time. This loss in potential energy is utilized to provide for the energy required to overcome the drag (the second term in Eq. 5.45). Hence in order that a glider may stay aloft, it must be brought to a certain height and speed before it can carry out the glide. This can be done by launching the glider by a winch or by towing the glider by another powered airplane.

136

(ii) From Eq. (5.43) the minimum angle of glide

occurs when CD/CL is minimum or at CL = CLmd. From Eq. (5.44) the minimum rate of sink occurs when CD/CL

3/2 is minimum or at CL=CLmp.

This can be understood by the following alternative explanation. When a glider sinks it is expending energy to overcome 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.

137

iii) Gliders with very low rate of sink (around 0.5 m/s) are called ‘Sail planes’.

From Eq.(5.44) we note that this low rate of sink is achieved by (a) low wing loading (b) low CD0 with smooth surface finish and (c ) large aspect ratio (16 to 20) to reduce K. Note from Eq. (5.24a ) that (CD/CL

3/2)min depends on CD01/4 and K3/4.

iv) If a glider is left at a height ‘h’ above the ground, then the horizontal distance (s) covered in descending to the ground is given by:

(5.46)L

D

Ch hs htan Cγ γ

= ≈ =

138

The time to descend is (h/Vd) when Vd is assumed to be constant. It is evident from the above discussion that the flight speeds at which γmin or (R/S)min occur are different .

v) Like the climb hodograph (Fig.5.15 e) a glide hodograph can be plotted with Vh on the x-axis and Vd on the y-axis. Such a hodograph would give information about γmin , Vγmin , (R/S)γmin , (R/S)min, V(R/S)min, γ(R/S)min . See Ref.1.2c , chapter 13 and Ref.1.4, chapters 9 and 11 for further details.

139

Example 5.5:

A glider weighing 4905 N has an area of 25 m2, CD0= 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 from a height of 300m. Neglect the changes in density during glide.

140

Solution:

(a)

= 0.012 + 0.023 CL2

22

010.012

3.14 16 0.87L

D D LCC C C

Aeπ= + = +

× ×

12

min (0.012 / 0.023) 0.721mdL LC Cγ = = =

m in2 0 .0 2 4

OD DC Cγ

= =

min( / ) 3 1.25mp mdL R S L LC C C= = =

min( / ) 4 0.048OD R S DC C= =

141

Therefore,

=0.0332 radians or 1.9O

At sea level ρ = 1.225 kg/m3 and hence,

min min( / ) 0.024 / 0.721D LC Cγ = =

( )12

3 / 2min min

2( ) /d D LWV C CSρ

⎛ ⎞= ⎜ ⎟⎝ ⎠

( )12

3 / 2min

2 4905 0.048 0.615 /1.225 25 1.25dV m s⎛ ⎞× ⎛ ⎞= =⎜ ⎟ ⎜ ⎟× ⎝ ⎠⎝ ⎠

142

(b) Greatest distance, in descending from 300 m to sea level is:

300 /0.0332 = 9040 m.

Longest time taken in descending from 300 m to sea level (Eglide)max is:

300/0.615 = 487 s = 8 min 7s.

12

m in

2 4905 21 .05 / .1 .225 25 0 .721

V m sγ⎛ ⎞×

= =⎜ ⎟× ×⎝ ⎠1 12 2

m in( / )2 2 4905 16 /

1 .225 25 1 .25m p

R SL

WV m sS Cρ

⎛ ⎞ ⎛ ⎞×⎜ ⎟= = =⎜ ⎟⎜ ⎟ × ×⎝ ⎠⎝ ⎠

Now,

143

Note :

The rate of sink in the flight when the greatest distance is covered is higher than the minimum rate of sink . Hence the time of flight will be shorter in the former case though the distance covered is more.

Remark

If the glide takes place from a sufficiently high altitude (as may happen for an airplane having an engine failure in cruise) then, the rate of sink (R/S) cannot be taken as constant during the descent. Equation (5.44) could be used to calculate the rates of sink at various altitudes.

144

The time elapsed during glide (Eglide) is given by:

2

1

h

glidedh

dhEV

= ∫

145

Exercises

5.5 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 + 0.048CL

2 .

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]

146

5.6 A sailplane having a wing loading of 220 N/m2

has the following drag polar

CL CD CL CD

0 0.014 0.8 0.039

0.2 0.015 1.0 0.064

0.4 0.019 1.2 0.103

0.6 0.026

Obtain the minimum rate of sink, angle of flattest

glide and corresponding speeds at sea level.

(Hint: Obtain CD/CL and CD/CL3/2 , plot them, obtain

(CD/CL)min, (CD/CL3/2)min and proceed.)

[Answer: (R/S)min = 1.17 m/s, γmin = 2.45o,

V(R/S)min = 79.3 kmph, Vγmin= 90.3 kmph]

147

5.7. Consider a subsonic jet transport. Assume that (a) thrust available (Ta) is roughly constant, (b) L≈W in climb or drag in climb (D) is roughly equal to drag in level flight and ( c ) drag polar is parabolic. With these assumption and from exercise 5.4 which gives:

show that (V/Vmd) for (R/C)max i.e. (V/Vmd) (R/C)max is given by:

202( ) , ,

( / ) Dmd md

V AWD AW A C KV V V

= + =

m a x

2 2

( / )

( ) 1 2( )

6

a a

R Cm d

T T AV W WV A

± +=

148

Further taking CD0= 0.016 and K= 0.05625 or A=0.03 obtain the following table.

Notice that as altitude increases Ta/W decreases as a consequence (V/Vmd)(R/C)max tends to 1.

At absolute ceiling (V/Vmax)(R/C)max =1 but (R/C)max is zero !.

1.01.161.361.54(V/Vmd)(R/C)max

0.060.10.150.2Ta/W

149

5.5 Range and Endurance:

Range (R) is the horizontal distance covered, with respect to a given point on the ground, for a given amount of fuel. It is measured in km. Endurance (E) 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 we use the following definitions which include some details of the flight plan.

Safe range: It is the maximum distance between two destinations between which an airplane can carry out a safe, reliably regular service with a given amount of fuel.

150

This flight involves take-off, acceleration to the speed corresponding to desired rate of climb, climb to the cruising altitude, cruise according to chosen flight plan, descent and landing.

Allowance is also given for the extra fuel requirement due to factors like (i) head winds normally encountered en-route (ii) possible navigational errors (iii) need to remain in air before permission to land is granted at the destination (iv) diversion to alternate airport in case of landing being refused at the scheduled destination.

151

Remark :

i) Generally the performance of an airplane is carried out assuming that the flight takes place in still air. However the airmass may move in different directions. Three cases are especially important.

(a) Head wind and tail wind: In this case the direction of motion of air (Vw) is parallel to the flight direction. If Vw is opposite to that of flight direction it is called head wind. When Vw is in the same direction as the flight direction, it is called tail wind (Fig.5.17a). In the presence of wind, the velocities of the airplane with respect to air (Va) and that with respect to ground (Vg) will be different. For head

152Fig. 5.17a Head wind and tail wind

153Fig. 5.17b Gust

154Fig.5.17c Side wind

155

wind case, Vg = Va-Vw, and for tail wind case, Vg=Va+Vw.

(b) gust: When the velocity of the air mass is perpendicular to flight path and along vertical direction it is called gust. Here we denote velocity of gust by Vgu (Fig.5.17b). This air movement would change the angle of attack of the airplane.

(c ) Cross wind: When the velocity of the air mass is perpendicular to flight path and parallel to the side ward direction it is called cross wind. Here we denote it by ‘v’ (Fig.5.17c).

Gross still air range (G.S.A.R.)

The calculation of safe range depends on the route

156

on which the flight takes place and other practical aspects. It is not suited 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 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 the ‘gross still air range’. In the subsequent discussion the range will mean gross still air range.Remark: As a rough guideline G.S.A.R. is roughly equal to one and half times the safe range.

157

5.5.1 Rough estimates of range and endurance

If the weight of the fuel available (Wf in N ) and the average rate of fuel consumption is known, then rough estimates of range (R) and endurance(E) are:

R = Wf x (km /N of fuel)average (5.46)

E = Wf x (hrs /N of fuel)average (5.47)

To illustrate the procedure let us consider a simple example.

Example 5.6 :

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

158

800 kmph the TSFC of the engine is 0.8 and L/D ratio is 12. Obtain rough estimates of the range and endurance.

Solution:W1=Weight at the start of the flight

=180,000NWf=Weight of the fuel =0.2x180,000 = 36,000N

W2=Weight of the airplane at the end ofthe flight= 180,000-36,000 = 144,000N.

159

∴ Average weight of the airplane during the flight

Wa =

Hence average thrust (Tavg) required in flight is:

Tavg = Wa / (L/D) = 162000/12 = 13500 N

Average fuel consumed per hour

= 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. The fuel consumed in 1 hr is 10,800 N.

Hence (km / N of fuel) average = 800/10800.

180000 144000 1620002

N+=

160

Consequently, R = 36000 x

=2667 km

and the endurance E = 36000 x

5.5.2 Accurate estimates of range and endurance

For accurate estimates of range and endurance the continuous variation of the weight of the airplane and consequent changes in the thrust required (or power required), TSFC (or BSFC) and flight velocity and lift coefficient must be taken into account. This is done as follows.

80010800

1 3.33 .10800

hrs=

161

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 (5.48)Note: W1 = weight of the airplane at the start of

flight.Let dR and dE be the distance covered in km and the time interval in hours during which a smallquantity of fuel dWf is consumed. Then

dR = dWf x (km/N of fuel) (5.49)

162

//f

km hrdR dWN of fuel hr

⎛ ⎞= ⎜ ⎟

⎝ ⎠

And dE = dWf (hrs /N of fuel) (5.51)

or dE = dWf {1/(N of fuel/hr)} (5.52)Now, km/hr = 3.6 x V , where V is the flight speed in m/s. The fuel/hr in Newtons is equal to: SFC x BHP for engine propeller combination (E.P.C.) and TSFC x T for jet airplane (J.A.). Note that In case of Engine propeller combination the engine could be a piston engine or a turboprop engine and in the case of jet airplane the engine could be a turbofan or a turbojet engine. Hence ,

or (5.50)

163

3.6f

VdR dWBSFC BHP

=×

fdWdE

BSFC BHP=

×

For J.A. (5.53a)3 .6

fVd R d W

T S F C T= ×

×

fdWdE

TSFC T=

×

For E.P.C. (5.53)

For E.P.C. (5.54)

For J.A. (5.54a)

Further in level flight

21&2

DL

L

CT D W L W V S CC

ρ= = = = (5.55)11222 2

L O L

W WVS C S Cρ σρ

⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

164

using ρ0 = 1.225 kg/m3

(5.56)

BHP = (5.57)

where ηp is the propeller efficiency.

During the analysis of range, the rate of change weight of the airplane is only due to consumption of fuel. Hence

dWf=-dW

Substituting for V, BHP, T and dWf in Eqs. (5.53) and (5.54) we get,

( )123/ 2 3/ 21 1 / ( ) /

1000 782.6 p D Lp

T V W S C Cη ση

⎡ ⎤= ⎣ ⎦

1 / 21 .278( )L

WVSCσ

=

165

Let W2 be the weight at the end of flight. Then the range and endurance are given by:

1 12 2

4.6( ) ( / )D L

dWdRTSFC S W C Cσ

−=

( )

12

3 / 2 3 / 2

782.6 ( )/

p

D L

S dWdE

BSFC W C Cη σ

= −×

( )/D L

dWdETSFC W C C

−=

×

For E.P.C. (5.59)

2 2

1 1

3600( / )

W Wp

D LW W

dWR dR

BSFC W C Cη

= = −×∫ ∫

For J.A. (5.58a)

For J.A (5.59a)

For E.P.C. (5.60)

3600( / )p

D L

dWdR

BSFC W C Cη−

=× For E.P.C. (5.58)

and

or

166

2

1( / )

W

D LW

dWRTSFC W C C

−=

×∫

12 2 2

1 1

3 / 2 3 / 2

782.8 ( )( / )

W Wp

D LW W

S dWE dE

BSFC W C Cη σ

= = −×∫ ∫

2

1( / )

W

D LW

d WET S F C W C C

−=

×∫ For J.A.(5.61a)or

or

Remarks:

i) Equations (5.60) and (5.61) or (5.53) and (5.54) when integrated give the range and endurance.

For J.A (5.60a)

For E.P.C. (5.61)

and

167

However, while doing this, one should keep in mind that the weight of the aircraft decreases continuously as the fuel is consumed. Further we are considering 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 way velocity changes with time.

Three types of flight plans can be cited as examples.

(a) Level flight at constant velocity . In this flight the lift coefficient decreases gradually as the weight decreases (Eq.5.55). The thrust required also decreases continuously.

168

(b) Level flight with constant lift coefficient (or constant angle of attack) . In accordance with Eq.(5.55)the flight velocity and thrust required decrease continuously.

(c) Level Flight with constant thrust. In this case the flight velocity and CL have to be adjusted continuously so that the thrust balances the drag and lift balances the weight.

ii) The airplanes are commercial vehicles to transport men and materials. Hence maximization of range and endurance are important requirements. However the right hand sides of Eqs.(5.60),(5.60a),(5.61) and (5.61a) involve an integral. The optimization of an integral is different from the optimization of an expression. The latter

169

is done by taking the derivative of the expression and equating it to zero. Where as in the case of an integral it is to be noted that the value of integral depends on how the integrand varies with independent variable. This variation is called a path. For example, as mentioned in remark (i) above, the range will depend on the flight plan viz. constant angle of attack flight, constant velocity flight or constant thrust flight. 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 see

170

Ref.5.1. 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.(5.60),(5.60a),(5.61) and (5.61a) become easy to integrate. The expressions for range and endurance with these assumptions are called ‘Breguet formulae’. These are derived in the next section. It may be pointed out that Breguet was a French pioneer in aeronautical engineering.

5.5.3 Breguet formulae

The assumptions made to derive these formulae are that during the flight:

171

(i) SFC or TSFC is constant

(ii) ηp 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. Then, from Eqs. (5.60) and (5.60a).

3600( / )

p

D L

dWRBSFC C C W

η= − ∫

or

110

2

8289.3log

( / )p

D L

WRBSFC C C W

η ⎛ ⎞= ⎜ ⎟

⎝ ⎠

For E.P.C

2

1

1/ 2 1/ 2 1/ 2

4.6( ) ( / )

W

D L W

dWRTSFC S C C Wσ

−= ∫ For J.A.

Hence

For E.P.C.(5.62)or

172

Similarly, from Eqs. (5.61) and (5.61a)

( )

12 2

1

3 / 23 / 2

782.6 ( )/

Wp

WD L

S dWEWBSFC C C

η σ= −

× ∫

( )1 12 21/ 2

2 13/ 21

1565.2[ ]

/p

D L

SE W WWBSFC C C

η σ ⎡ ⎤= −⎣ ⎦×

1

2

2 .303 log( / )D L

WETSFC C C W

⎛ ⎞= ⎜ ⎟

⎝ ⎠For J.A. (5.63a)

For J.A. (5.62a)

For E.P.C.

2

1

1( / )

W

D L W

d WET S F C C C W

= − ∫or For J.A.

For E.P.C. (5.63)

or

1/ 21/ 21 2

1/ 21

9.2 ( ) 1( / )D L

W WRTSFC C C S Wσ

⎛ ⎞⎛ ⎞⎜ ⎟= −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

or

Hence

173

Remarks :From the above expressions (Eqs. 5.62, 5.62a, 5.63 and 5.63a) we can draw the following conclusions:

(i) For range and endurance to be high, ηp should be high and the TSFC and BSFC should be low.

(ii) For an airplane with engine-propeller combination the endurance is maximum when the lift coefficient is such that CD/CL

3/2 is minimum, i.e., CL = CLmp.

For range to be maximum, in this case, CD/CL

should be minimum or CL = CLmd. This follows from the fact that range is proportional to V/THP or L/D hence, range is maximum when CL

corresponds to minimum drag.

174

(iii) For a jet engined airplane the endurance is maximum when CD/CL is minimum or CL= CLmd. The range is maximum when CD/CL

1/2 is minimum. The CL corresponding to (CD/CL

1/2)min is denoted by CLmrj. Further for range to be high, the wing loading (W/S) should be high and σshould be low or altitude should be high. This is the reason why 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. It may be added that wing loading of

175

an airplane is compromise between requirements of cruise, climb and landing. The landing distance increases in direct proportion as wing loading increases (see section 5.12)

(iv) If the drag polar is parabolic, an expression for CLmrj can be worked out as follows.

2OD D LC C K C= +

01 12 2

3/ 2DDL

L L

CC K CC C

∴ = +

( )12

0 3/ 2 1/ 2/ 3 0

2 2D L D

Lmrj LmrjL

d C C CC K C

dC−= + =

or CLmrj = (CDo / 3K)1/2

176

(v) The points on the drag polar at which CL is equal to CLmax, CLmp, CLmd and CLmrj are shown in Fig. 5.18.

(vi) In section 5.2.4 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/ 2

1 1/D LTSFC C C 1/ 2

1 1L

D L

CTSFC C Cor

177Fig 5.18 Important points on a drag polar

178

Noting that 1/CL1/2 is proportional to flight speed (V),

Since high speed airplanes fly in stratosphere, where speed of sound is constant,

The quantity can be referred to as figureof merit (FM) for the following reasons (a) A low values 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

1 L

D

CR VTSFC C

α

1 L

D

CR MT S F C C

α

1 L

D

C MT S F C C

179

early 1960’s.The contemporary subsonic jets like Boeing707 would fly around M =0.8, have (L/D)max

around 16 and TSFC around 0.9. Then FM would be 0.8x16/0.9 or 14.2. If Concorde were to compete with subsonic jets, it should have a similar value of FM. The contemporary fighter airplanes flying at Mach number of two had TSFC of 1.5 and (L/D)max

of 5. This would give FM of (2/1.5)2x5 = 6.66 which was far too low as compared to 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 and TSFC of 1.2. This would give FM of 2.2(1/1.2)x7.5 = 13.75. Which was comparable to FM of subsonic airplanes. However to achieve a TSFC of 1.2 at M=2.2, a large amount of

180

research was carried out and the Olympus engine used on Concorde was developed jointly by Rolls-Royce and SNECMA. Similarly to achieve (L/D)max

at 7.5 at M=2.2 needed a large amount of computational and experimental effort. A picture of Concorde, a technological marvel, is shows in Fig.5.19a.

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 aerodynamics heating would cause surface temperatures around 300oC at which the strength and modulus of elasticity of aluminum will be significantly reduced.

181

Fig 5.19a Concorde

(Adapted from: Google.com)

182

The B787 (Fig.5.19b) 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. The values of (L/D)max

and TSFC indicate steady improvements in aerodynamics and engines over the last five decades .

183

Fig 5.19b Boeing 787 Dream liner (Adapted from: Google.com)

184

5.5.4 Range in constant velocity- constant altitude flight (Rh,v)

The assumption of constant CL during cruise gives the longest range. However it is more convenient for the pilot to fly the airplane at 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), let us assume that SFC is constant and consider an airplane with jet engine. From Eq.(5.53), when flight velocity (V) is constant, we can write: 2

1

1

3.6 w

h vrw

V dWRTSFC T

= ∫ (5.63b)

185

Tr = Thrust required

Assuming a parabolic polar,

Tr =

Note: The dynamics pressure (q), is constant in a constant velocity and constant altitude flight.

Substituting for Tr in Eq.(5.63b) gives:

2 22 21

22

1 2 ;2 O OD D

K W KWV S C q S C q VV S q S

ρ ρρ

+ = + =

2

1

1

2

3 .61

O

w

h vD w

V dWRTSFC q S C aW

−=

+∫

2 2OD

Kq S C

1

1 11 2[tan tan ]

O

h vD

VR a W a Wq S C TSFC a

− −∴ = − (5.63c)

Where a =

186

Let , where Wf = weight of fuel

Then W2 = W1 (1-ζ) ;

Further let E1 = = initial lift drag ratio,

1

fWW

ζ =

1

1

WD

CL1 = CL at start of flight =

12

ODK C

121

2

WV Sρ

where W2= weight of the airplane at the end of the flight.

Emax =

187

1

1

1max 1

max 1

7.2 tan2 (1 )h v

L

E V ERTSFC E KC E

ζζ

−⎡ ⎤

= ⎢ ⎥−⎢ ⎥⎣ ⎦

For an engine – propeller combination, the range at constant speed and constant altitude (Rh,v) is given as:

Assuming BSFC and ηp to be constant and the drag polar as parabolic i.e.

2

1

1

3600Wp

h vW

dWR

BSFC Tη

= ∫

Substituting in Eq.(5.63c) and simplifying we get:

(5.63d)

(5.63e)

188

and substituting in Eq.(5.63e) gives:

1

1max

7200tanp

h vR EBSFC

η −=1

max 1 12 (1 )L

EE KC E

ζζ

⎡ ⎤⎢ ⎥−⎣ ⎦

221

2 2

2OD

KWT V S CSV

ρρ

= +

(5.63f)

Remarks:

i) Comparing the ranges in 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 is a constant CL flight.

189

ii) In actual practice SFC 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.(5.53) and (5.54).

iii) Appendix ‘A’ section 6 considers range and endurance performance of a piston engined airplane at an in altitude of 6000 feet (1830 m) in constant velocity flights at different values of velocities . The variations in propeller efficiency and fuel consumption are taken into account. It is seen that the endurance is nearly constant in the speed range of 33 to 45 m/s. the range is maximum around flight speed of 64m/s. It may be added that the

190

Vmp and Vmd for this airplane at 6000 feet altitude are 46.67 m/s and 61.42 m/s respectively.

iv) Section 6 of Appendix ‘B’ considers range and endurance performance of a jet transport at an altitude of 36000 feet (10973 m) in constant velocity flights at different values of velocities. The endurance is near the maximum value in the speed range of 196 to 230 m/s. The maximum range occurs at M=0.81, which is slightly higher than the Mach number beyond which the CD0 and K begin to increase due to compressibility effects.

191

5.5.5 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 of Vcr and hcr the range is calculated at various speeds at a number of altitudes and the plots as shown in Fig. 5.20 are obtained.

The dotted line in the Fig.5.20 is the envelop of all the curves. The speed and altitude at which the maximum of this envelope 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, speed appeal etc.

192Fig. 5.20 Determination of cruising speed and

cruising altitude

193

5.5.6 Cruise climb

In this flight both the lift coefficient and the velocity are kept constant. Assuming TSFC to be constant the integration of Eq. (5.53) gives:

2

1

1 23.6 3.6 log ( / )( / ) ( / )

W

eD L D LW

V dW VR W WTSFC C C W TSFC C C

−= =∫ (5.64)

In this flight CL and V are both maintained constant, but the weight continuously decreases owing to the consumption of fuel. Hence to satisfy the condition of L=W=(1/2)ρV2SCL, the density (ρ) should decrease during flight or the flight altitude must increase as the flight progresses. The airplane climbs gradually and the flight is called cruise climb. However the rate of climb is very small (γ less than 1/2 degree)

194

and level flight equations can be used without any appreciable error (see exercise 5.10).

It can be shown that (see Ref.1.1) the range in cruise climb is higher than that in level flight at the altitude where the cruise climb begins. In actual practice continuous increase in altitude may not be permitted by Air Traffic Regulations. 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.

195

Remark:

In a cruise climb the thrust required would be

T=D=(1/2)ρV2SCD.

Since the flight velocity and CL (and hence CD) are held constant, the thrust required will be proportional to ambient density (ρ). Further in stratosphere the engine output ( thrust available) is also proportional to the ambient density. Thus, in a cruise climb the thrust setting required is also constant and it becomes a very convenient flight in stratosphere – the pilot has just to set the Mach number and then the autopilot will take care of the flight.

196

5.5.7 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 air mass has a velocity with respect to the ground and airplane encounters head wind or tail wind. (see section 5.5 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 (Va) are different.

To analyze the effect of wind on airplane performance, it may be pointed out that the

197

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 airplane with respect to the ground will be lower than the 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 engine would be producing thrust to overcome the drag, but no distance will be covered as the airplane is hovering! When there is tail wind the range increases.

198

2 2

1 1

3.6 3.6 3.6W W

ag W a w

W W

V dW dWR V R V ETSFC T TSFC T

= − − = −× ×∫ ∫

An expression for range with effect of wind can be derived as follows:

Consider a jet engined airplane. Let Rg be the range in the presence of wind, then from Eq.(5.53a):

2 2

1 1

3.6 3.6 ( )W Wg a W

gW W

V dW V V dWRTSFC T TSFC T

−= − =

× ×∫ ∫ ; VW in m/s

where Ra is the range in still air =2

1

3.6Wa

W

V dWTSFC T×∫

and E is the duration of flight in hours. Thus with head wind the range decreases by 3.6 Vw E. In

199

example 5.6 the range is 2667 km and the endurance is 3.33 hours. If a head wind of 15 m/s is encounter 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 whether reports, and adequate amount of fuel is provided to take care of the situation.

ii) The maximum endurance is not affected by the presence of wind, because Emax depends on airspeed only.

200

Example 5.7

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 + 0.059CL2. 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 ηp = 85%.

Note: Along with the fuel the lubricating oil is also consumed and the latter is taken into account here.

201

Solution:

W1 = 88290 N,

W2 = 88290 – 15450 = 72840 N

Since BSFC, ηp and CL are constant, the maximum range and endurance occur when CL

values equal CLmd and CLmp respectively.

CLmd = (0.022/0.059)1/2 = 0.6106,

CDmd = 2 CD0 = 0.044

CLmp= (3 x 0.022/0.059)1/2

= 1.058, CDmp = 4CD0 = 0.088

∴ (CD /CL)min = 0.044/0.6106 = 0.0721

202

and

= 0.0808

From Eq. (5.62)

= 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

3/ 2 3/ 2min( / ) 0.088 /(1.058)D LC C =

110 10

min 2

8289.3 8289.3 0.85 88290log log( / ) 2.67 0.0721 72840

p

D L

WRBSFC C C W

η ⎛ ⎞ × ⎛ ⎞= = ×⎜ ⎟ ⎜ ⎟× × ⎝ ⎠⎝ ⎠

203

following results illustrate the changes.

Velocity at the beginning of the flight :

=

= 72.41 m/s. = 260.7 kmph.

Velocity at the end of flight :

= (2 x 72840/1.225 x 45 x 0.6106)1/2

= 65.8 m/s = 236.8 kmph.

( )1 12 22 / (2 88290 /1.225 45 0.6106)LmdW S Cρ = × × ×

204

Power required in the beginning of the flight:

=

Power required at the end of flight :

=

88290 0.044 72.41 460.71000 1000 0.6106T V kW× ×

= × =×

72840 0.044 65.8 345.51000 0.6106

kW×× =

×

Maximum endurance:From Eq. (5.64) the maximum endurance is obtained as:

( )

1 12 2

1max 3/ 2

1 2min

1565.2 1/

p

D L

WSEBSFC W WC C

η σ ⎡ ⎤⎡ ⎤ ⎛ ⎞⎢ ⎥= −⎜ ⎟⎢ ⎥⎢ ⎥⎣ ⎦ ⎝ ⎠⎣ ⎦

205

1 12 21565.2 0.85 1.00 45 88290 1

2.67 0.0808 88290 72840

⎡ ⎤× ×⎡ ⎤ ⎛ ⎞= − −⎢ ⎥⎜ ⎟⎢ ⎥× ⎣ ⎦ ⎝ ⎠⎢ ⎥⎣ ⎦

= 14.06 hrs.

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 Emax flight are 197.8 kmph and 179.7 kmph respectively. The power outputs required at the beginning and the end of the flight are 402.8 kW and 302.0 kW respectively.

206

Example 5.8

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

CD = 0.017 +0.0663 CL2

Obtain the maximum range in constant CL flight at an altitude of 10 km assuming the TSFC to be 0.95 hr-1.

207

Solution

In a flight with constant CL the maximum range occurs when

= 0.292

CDmrj = 0.017 + 0.0663 × (0.292)2

= 0.02265

12( / 3 )

OL Lmrj DC C C K= =120.017

3 0.0663LmrjC⎛ ⎞

= ⎜ ⎟×⎝ ⎠

σ at 10 km = 0.3369

( )12

12/ 0.02265 / 0.292 0.04192D LC C = =

208

from Eq. (5.62a)

=

= 5317 km

Remark

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.

( )11 22

12

1 2max

1

9.2 1/D L

W WRS WTSFC C C σ

⎡ ⎤⎛ ⎞⎡ ⎤ ⎢ ⎥= − ⎜ ⎟⎢ ⎥ ⎢ ⎥⎣ ⎦× ⎝ ⎠⎣ ⎦1 12 29.2 922140 922140 2943001

0.95 0.04192 0.3369 158 922140

⎡ ⎤⎛ ⎞ −⎛ ⎞× −⎢ ⎥⎜ ⎟ ⎜ ⎟× × ⎝ ⎠⎝ ⎠ ⎢ ⎥⎣ ⎦

209

The speed of sound at 10 km is 299.5 m/s. Thus the Mach number at this speed would be 311.1/299.6 = 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.

Let us assume that the critical Mach number is 0.85 and work out the range in a flight at constant CL which begins at this Mach number.

210

V = 0.85 x 299.6 = 254.5 m/s. ∴ CL = (2 x 922140/0.413 x 158 x 254.52)

= 0.436

∴ CD = 0.017 + 0.0663 x (0.436)2

= 0.0296

and

= 0.0448

The range at this CL is :

1 12 2/ 0.0296 / 0.436D LC C =

1 12 29.2 922140 922140 2943001

0.95 0.0448 0.3369 158 922140

⎡ ⎤⎛ ⎞ −⎛ ⎞= × −⎢ ⎥⎜ ⎟ ⎜ ⎟× × ⎝ ⎠⎝ ⎠ ⎢ ⎥⎣ ⎦= 4975 km.

211

Exercises

5.8 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

Where ζ = Wfuel /W1, ui = V/Vmd

W1= Weight of airplane at the beginning of flight

Vmd= Speed for minimum drag.

21max

4

2( / ) tan1

i

i

L D uETSFC u

ζζ

− ⎛ ⎞= ⎜ ⎟− +⎝ ⎠

212

5.9 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 σ = 0.65, given that:

CD = 0.021 + 0.06CL2 ; W1 = 176, 600 N,

Wfuel = 35, 300 N,

S = 90 m2,

ηp = 0.82,

BSFC = 3.90 N/kW/hr.

[Answer: R = 2104 Km].

213

5.10 Consider a jet airplane having 20% of its weight as fuel fraction. It starts 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/CD = 16, estimate the range in cruise climb (Rcc). What is the angle of climb ( γcc ) in cruise climb?

[Answer: hf = 12416 m, Rcc = 5041 m,

γcc = 0.015o].

214

5.6 Accelerated level FlightIn the last four sections of this chapter we dealt with airplane performance in steady flights. Now we consider the accelerated flights. Such flights could be along a straight line e.g. accelerated level flight and accelerated climb or along curved paths like loops and turn. We first consider the accelerated level flight.

When an airplane moves along a straight line at a constant altitude but its velocity changes, 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

215

speed corresponding to the best rate of climb is generally much higher than this speed (see Fig.5.14e). 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 the cruising speed (Fig.5.14e) 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

216

or when the pilot notices the possibility of over-shooting a target.

The forces acting on an airplane in accelerated level flight are shown in Fig. 5.21.

The equations of motion are obtained by resolving the forces along and perpendicular to the flight path which in this case is a horizontal line.

L-W = 0 (5.66)

Where ‘a’ is the acceleration.

As regards the analysis of performance in an

WT D ma ag

− = = (5.65)

217Fig.5.21 Forces in accelerated level flight

218

accelerated flight we are interested in finding out the time taken and the distance covered for a given change in velocity. The accelerated or decelerated flights last only for a short duration and the weight of the airplane can be assumed to remain constant during this flight. However, since it is a level flight i.e. L=W=(1/2)ρV2SCL, the value of CL and consequently of CD change continuously as the flight velocity changes.

From Eq. (5.65), the acceleration ‘a’ is given by:

a=g (T-D)/W

Substituting for D as (1/2)ρV2SCD , we get,

219

Now,

ds dV dV ds dVV and a Vdt dt ds dt ds

= = = =

dV V dVdt and dsa a

∴ = =

The distance covered (s) and time taken (t) for velocity to change from V1 to V2 are given by:

Substituting for ‘a’ from Eq.(5.67) yields:

(5.68)

2 2

1 1

V V

V V

V dV dVs and ta a

= =∫ ∫

212( )D

ga T V S CW

ρ= − (5.67)

220

The expressions in Eq.(5.69) can be directly integrated if T and D are simple function of velocity. Otherwise a numerical integration as illustrated in the following example can be carried out.

(5.69)2 2

1 1

2 21 12 2( ) ( )

V V

D DV V

W V dV WdVs and tg T V S C g T V S Cρ ρ

= =− −∫ ∫

221

Example 5.9

An airplane having a weight of 156,960 N and a

wing area of 49 m2 has a drag polar given by

CD=0.017+0.06CL2. 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.

222

Solution:

or

or

212 LL W V S Cρ= =

22 21 1

2 2 2

2OD D

KWD V S C V S CSV

ρ ρρ

= = +

221

2 2

2 0.06 1569601.225 49 0.0171.225 49

D VV

× ×= × × × × +

× ×

72

2

4.9225 100.5102D VV

×= +

223

To carry out the numerical integration we evaluate the integrands in Eq. (5.69) for several values of V and can use methods like trapezoidal rule or Simpson’s rule. Books on numerical analysis be consulted for details of the 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 V2 is divided into six intervals, each of 20 m/s. The values are tabulated below:

224

( )W

g T D−

( )W V

g T D−124.7299.0980.2165.7254.0644.4636.44

0.56690.4954

0.4465

0.4107

0.3861

0.3705

0.3644

2573121660180501499912518107714

10042D (N)

220200180160140120100V (m/s)

225

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 km

t= (20/3){0.3644 + 4 (0.3705 + 0.4107+0.4954)

+ 2 (0.3861 + 0.4456 ) + 0.5669 }

= 51.34 s.

226

5.7 Accelerated Climb

In this case the flight takes place along a straight line inclined to the horizontal at an angle γ and the velocity increases or decreases along the flight path. Fig. 5.22 shows the forces acting on the airplane.

The equations of motion are:

(5.70)

L-W cos γ = 0 (5.71)

sin WT D W ag

γ− − =

( sin )g T D WaW

γ− −∴ = (5.72)

227Fig.5.22 Accelerated climb

228

Substituting for ‘a’ in Eq. (5.70) and noting that

sin γ = Vc / V,

From Eq.(5.32), (T-D)V/W is the rate if climb in

steady flight. Denoting it by Vco, Eq.(5.74) reduces

to:

1

coc

VVV d Vg d h

=⎛ ⎞

+⎜ ⎟⎝ ⎠

( )01

cc c

V W dV T D VT D W V or V aV g dh V dVW

g dh

−− − − = =

⎛ ⎞+⎜ ⎟

⎝ ⎠

(5.74)

(5.75)

. , /CdV dV dh dha but V R Cdt dh dt dt

= = = =

cd Va Vd h

∴ = (5.73)

Further,

229

Remarks

i) The term (dV/dh) in Eq.(5.75) 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 at a given 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. The values given for the rate of climb in Fig. 5.14e have to be corrected for the effect of change in velocity with attitude; in Fig.5.14c one notices that the speed for maximum rate of climb increases with altitude.

230

ii) The performance of an airplane in an accelerated

flight can also be viewed from the energy point of

view. Multiplying Eq. (5.70) by V gives:

(5.76)

In Eq.(5.76) 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

s in .W d VT V D V W V Vg d t

γ− − =

2

2d h W d VT V D V Wd t g d t

⎛ ⎞= + + ⎜ ⎟

⎝ ⎠or

231

‘W(dh/dt)’ represents the rate of change of

potential energy and (W/g){d/dt(V2/2)}

represents the rate of change of kinetic energy.

Thus the total available energy is utilized in these

three ways – 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 in between (Vmin)e

and Vmax, can an airplane climb or accelerate and

the excess power (T-D)V has to be shared for

232

increase of potential energy or kinetic energy or both.If climb takes at V(R/C)max then no acceleration is possible.

iii) Equation (5.76) can be rewritten as:

The term (h+V2/2g) is denoted by he and is called specific energy or the 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

2( )2

T D V d VhW dt g

⎛ ⎞−= +⎜ ⎟

⎝ ⎠

233

(dhe/dt) = (T-D)V/W (5.78)

The energy height concept is used in optimization of climb performance. See Ref. 1.7 & 1.9 for details. The quantity (dhe/dt) is called specific excess power and denoted by ‘ps’.

234

Example 5.10

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 in steady climb.

Solution:

In a climb with Ve as constant, the true air speed

(V) is given by :

12 3 / 21

2/ ,e edV dV V Vdh dh

σσ σ −= ∴ = −

235

In troposphere the variation of σ with h is given

as follows ( Eq.2.7):

where, TO = Temp. at sea level,

= Temperature lapse rate and

R = gas constant.

1gO R

O

T hT

λλσ−⎛ ⎞−

= ⎜ ⎟⎝ ⎠

( )2 ( )1

2

g Rg R

eO

dV g RVdh T R

λλλ λ σ

λ

− +−−⎛ ⎞∴ = ⎜ ⎟

⎝ ⎠(5.79)

λ

236

In I.S.A., = 0.0065oC/m.

Using g = 9.81 m/s2 and

R = 287.05 m/s2 K, we get the correction factor

in Eq. (5.75) as:

(5.80)

It is seen that the correction required depends on

Ve and σ, typical values at sea level (σ=1) and at

11 km attitude (σ = 0.2971) are given in table 5

6 2 1 .2 3 51 1 4 .8 9 4 1 0 eV d V Vg d h

σ− −+ = + ×

λ

237

1 V dVg dh

⎛ ⎞+⎜ ⎟

⎝ ⎠

1.87661.21911.0548at 11 km

1.19581.04891.01224at s.1.

20010050Ve (m/s)

Table5.1 Correction factor in climb at constant

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.

1 V dVg dh

⎛ ⎞+⎜ ⎟

⎝ ⎠

equivalent air speed in troposphere.

238

Remark

In a constant Mach number flight in troposphere the flight velocity will decrease with altitude. Hence the term (dV/dh) will be negative and the rate of climb in constant Mach number flight will be more than that in steady climb. See exercise 5.8.

Exercise 5.8

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 m2 , CD= 0.017+0.055CL

2 , and thrust available at 5km altitude= 13,000 N.

239

Hint: [Show that in a constant Mach number flight

Where γ = ratio of specific heats, = temperature lapse rate

R = Gas constantTaking γ = 1.4, = 0.0065oC/m

R= 287.05 m2 s-2 K and g= 9.81 m/s2 gives:

]

02d V RM a n dd h T h

λ γλ

= −−

211 12

V d V R Mg d h g

λ γ+ = −

λ

20

11 0.1331

c

c

VV M

=−

λ

240

[Answer: VCo = 29.89 m/s, Vc= 31.39 m/s= 1883.7 m/min].

241

5.8 Symmetric flight Along curved Path in Vertical Plane

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 ω2 r where V is the speed, r is the radius of curvature of the path and ω is the angular velocity (ω = 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 acceleration along the

242

radius of curvature which is equal to (V2/r) see Ref. 1.1.

In order that the body has these accelerations a net force must act along these directions.

Let us 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. (5.23a). Note also the attitude of the airplane at various points and the directions in which D and L act; in a flat earth model W always acts in vertically downward direction.

The equations of motion at various points can be

written down as follows:

243Fig 5.23a Flight along a loop with constant

radius and speed

D

A

B

CG

244

2

0; WVT D W Lgr

− + = =

At point A : (5.81)

At Point B: (5.82)

At point C : (5.83)

At point D :

At a general point G the equations of motion are

2

0 ; WVT D L Wg r

− = − =

2

0 ; W VT D W Lg r

− − = =

2

0 ; W VT D L Wg r

− = + =

(5.84)

2

sin 0 ; cos WVT D W L Wgr

γ γ− − = + = (5.85)

245

Note that the Eqs. (5.81) to (5.84) for points A,B,C and D can be obtained from Eqs. (5.85) by substituting γ as 180o, 90o, 0o and 270o

respectively.

Remarks:

i) If the tangential velocity is not constant then the first equation of (5.85) would become:T-D-W sin γ =(W/g) a , where a = dV/dt (5.86)ii) Flight along a closed path in a vertical plane is called ‘Loop’. From Eqs. (5.81 to 5.85) it is seen that the lift required and the thrust required change rapidly with time. It is difficult for the pilot to maintain all these values and the actual flight is somewhat like the one shown in Fig. 5.23 (b).

246Fig 5.23b Shape of a normal loop

247

iii) It is seen that at the bottom of the loop i.e.

point ‘A’ in Fig. 5.23 (a), the lift required is equal

to or

The term (V2/gr) could be much larger than 1 and

the lift required in a manoeuvre would 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.

2W VWgr

+2

1 VL Wg r

⎛ ⎞= +⎜ ⎟

⎝ ⎠

248

Thus the total lift required is 6.1 W. In order that an

airplane carries out the manoeuvres without getting

disintegrated its structures must be designed to

sustain the lift produced during manoeuvres.

iv) Load factor

The ratio of the lift to the weight is called load factor

and is denoted by ‘n’ i.e.

n = (L/W)

A flight with a load factor of n is called ‘ng’ flight. For

example a loop with load factor of 4 is a 4g loop

(5.88)

249

In level flight n = 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 civil airplanes are designed to withstand a load factor of 3 to 4 and military airplanes to a load factor of 6 or more.

The limitation on military aircraft comes from the human factors namely a pilot subjected to more than 6g may black out during the manoeuvre.

To monitor the load factor, an instrument called g-meter is installed in the cockpit.

250

(v) Pullout

The recovery of an airplane from a dive or a glide is called a pullout (Fig. 5.24). The dive is an accelerated descent while the pullout phase can be regarded as a flight along an arc of a circle (See example 5.11).

251Fig 5.24 Pull out from dive

252

Example 5.11

An airplane having a wing area of 20 m2 and a weight of 19620 N dives with power off along a straight line inclined at 60o to the horizontal. What is the acceleration of the airplane when the flight speed is 300 kmph? If the airplane has to pullout of this dive at a radius of 180 m, what will be the lift coefficient required and the load factor? Drag polar is given by:

CD = 0.02 + 0.06CL2 and the manouevre takes

place around an altitude of 2 km .

253

Solution:

From Fig. 5.24 the equations of motion in the dive can be written as follows:

os 0 ; sin WL W c W D ag

γ γ− = − =

γ = 60o, ∴ cos γ = 0.5 and sin γ = 0.866

Consequently L = 19620 x 0.5 = 9810 N

Now or

V = 300 kmph

= 83.3 m/s, ρ at 2 km = 1.0065 kg/m3

212 LL V SCρ=

2

2L

LCS Vρ

=

254

Hence

CL = = 0.1403

∴ CD = 0.02 + 0.06 × 0.14032

= 0.0212

The drag D = L

Hence, a =W sin γ - D = 15682 N

or a = 15682 × 9.81/ 19820

= 7.76 m/s2

2

2 98101.0065 20 83.33

×× ×

0.02129810 14820.1403

D

L

C NC

= × =

Wg

255

To obtain the lift required during the pullout let us treat the bottom part of the flight path during the pullout as an arc of a circle.

Then from Eqs. (5.81) to (5.85), the lift required is maximum at the bottom of the loop and is given by:

or L = 19820 x 4.92 Then,

2 21 83.319820 1 9.81 180

WVL W or Lgr

⎛ ⎞= + = × + ×⎜ ⎟

⎝ ⎠

2

19820 4.92 2 1.391.0065 20 83.3LC × ×

= =× ×

256

Remarks:

i) The maximum load factor in the pullout is 4.92.

The value of lift coefficient required is 1.39.

This value may be very close to CLmax and the

parabolic drag polar may not be valid.

ii) Since CL cannot be exceeds CLmax , a large

amount of lift cannot be produced at low

speeds. Thus maximum attainable load factor

nmaxattainable at a speed is:

nmaxattainable = (1/2)ρV2SCLmax/W

At stalling speed the value of n is only one.

257

5.9 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.

258

If there is no tangential acceleration i.e. the flight speed is constant then the flight is called steady turn. If the altitude remains constant then the flight is called level turn. When the airplane executes a turn without sideslip, it is called coordinated turn. In this flight the X-axis of the airplane always coincides with the velocity vector. 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. Further a turning airplane does produce a sideslip. Because of these two factors a pilot has to apply appropriate deflections of elevator and rudder to execute a coordinated turn .

259

A coordinated turn is also called a correctly banked turn. Herein we will confine our attention to the steady coordinated level turn.

The forces acting on an airplane in steady, level, coordinated turn are shown in Fig.5.25. The equations of motion in such a flight can be obtained by resolving the forces in three mutually perpendicular directions:

260Fig 5.25 Turning flight

261

T – D = 0 as the turn is steady flight (5.89)

W – L cos φ = 0 as the turn is level flight

as the turn is correctly banked

Where φ is the angle of bank and r is the radius of turn.

2

sin W VLg r

φ =

Remarks:

i) From the above equations we find that

L = 1 / cos φ. 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 φ and is higher than 1.

ii) From Eqs. (5.90) and (5.91), the radius of turn r is given by:

(5.91)

(5.90)

262

(5.92)

and the rate of turn (ψ ) is given by:

(5.93)

2 2

sin tanW V Vrg L gφ φ

= =

2 ta n/ta n

V V gVr g V

φψφ

= = =

.

.

263

Example 5.12

An airplane has a jet engine which produces a thrust of 24525 N at sea level. The weight of the airplane is 58860 N. The wing has an area of 28 m2, zero-lift angle of – 2.2O and a slope of life curve of 4.6 per radian. Find (a) the radius of a correctly banked 4g level turn at the altitude where σ = 0.8 and the wing incidence is 8o,

(b) time required to turn through 180o and(c) thrust required in the manoevre if the drag coefficient at this angle of attack be 0.055.

264

Solution:

Given, W = 58860 N,

S = 28 m2, α = 8o,αOL = -2.2o,

= 4.6 per radian

= x 2π per degree

= 0.083 per degree,

allowable n = 4 and

T= 24525 N at sea level

LdCdα

4.6180

265

= 0.0803 (8 + 2.2) = 0.82

In a 4g turn L = 4W =

Hence V =

= 144.6 m/s.

( )LL O L

dCCd

α αα

= −

212 LV S Cρ

12

12

2 4 58860(2 / )1.225 0.8 28 0.82LL SCρ

⎛ ⎞× ×= ⎜ ⎟× × ×⎝ ⎠

Hence tan φ = 3.873

0 '1 1c o s 7 5 3 14n

φ φ= = ∴ =

266

Rate of turn = ψ = = 0.2627 rad /s

( )22 144.6550.3

tan 9.81 3.873Vr m

g φ∴ = = =

×

144.6550.3

Vr=

.

Hence time to turn through 1800 is equal to

= 11.95 s.

The thrust required = Tr=

= (1/2)x1.225x144.62x28x0.055

= 15786 N

Remark: The thrust available is given as 24525 N at sea

level. If we assume that the thrust is roughly

proportional to density ratio (σ), then the thrust

0.2627π

212 DV S Cρ

267

available at the chosen altitude would be

24525x0.8 = 19620 N. This thrust is more than

the thrust required during the turn and the flight

is possible.

5.9.1 Minimum radius of turn and maximum

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 indices of

the maneuverability.

268

From Eqs.(5.92) and (5.93) we find that the a small radius of turn and a high rate of turn are achieved when φ is high. The value of φ is limited by the factors discussed below.

(I) From the above discussion we notice that the lift coefficient in a turning flight is higher than the lift co-efficient 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 = nW / ½ ρ V2 S = nCLL

However, CLT cannot be more than CLmax and this imposes a limitation on the attainable value of n because

269

Further at stalling speed (Vs), CLL equals CLmax. Hence turn is not possible at Vs.

(II) 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. This also imposes a limitation on attainable value of φ.

(III) cos φ = 1/n . But n cannot exceed the value permitted by the structural design of the airplane, hence φmax is limited to cos-1 .

At V = Vmax or (Vmin)e the entire engine output is used in overcoming the drag in level flight. Hence

m ax1cos

L

LL

Cn

Cφ= =

max

1n

⎛ ⎞⎜ ⎟⎝ ⎠

270

sustained level turn is not possible at these two speeds.

Determination of minimum radius of turn and maximum rate of turn:

In a general case where the drag polar and thrust are functions of Mach number, the minimum radius of turn (rmin) and the maximum rate of turn (ψmax) at an altitude can be obtained from the following procedure.

i) Choose an altitude. Obtain Vmax and (Vmin) at this altitude.

(ii) Choose a flight speed (V) in between Vmax and Vmin and obtain CLL as :

CLL = 2W / ρ SV2

..

271

Obtain Mach number (M) corresponding chosen V.

(iii) Obtain the CLmax at the chosen flight Mach number and the ratio CLmax/CLL. It may be recalled that for airplanes flying at high speeds the CLmax depends on Mach number. If CLmax/CLL is smaller than the allowable load factor (nmax), then the turn may be limited by CLmax. Hence choose CLI1 = CLmax. If CLmax / CLL is more than nmax then the turn may be limited by nmax and choose CLT1 as nCLL.

(iv) Obtain from the drag polar, the drag coefficient CDTI corresponding to CLT1 and the chosen Mach number. Calculate the drag DT1 from:

DT1 = ½ ρ V2 S CDT1

272

If DT1 is greater than the available thrust (Ta) , then the turn is limited by engine output. In this case, obtain the maximum permissible value of CD in turning flight (CDT) from:

CDT = Ta / ( ½ ρ V2 S)

Corresponding to 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 and CLT is the smaller of the two values obtained in step (iii).

(v) Once CLT is known, φ is given by:

φ = cos-1 (CLL/CLT).

Knowing φ and V, the radius of turn (r ) and rate of turn (ψ) at the chosen speed can be calculated

.

273

from Eqs. (5.92) and (5.93).

(vi) The previous steps should be repeated at various values of flight speeds ranging between Vmin and Vmax. Plotting these results the values of rmin and ψmax 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 in example 5.13.

.

.

274

Example 5.13

A passenger airplane having a gross weight of 176,400 N has a wing area of 45 m2. Obtain the variations of r and with velocity at an altitude of 8 km from the following data. CLmax = 1.4, nmax = 3.5,

CD = 0.017 + 0.05CL2

Vmin = 103 m/s, Vmax = 274 m/s and the thrust output (Ta) varies as given below.

ψ.

275

22270205

21980185

21580165

21480145

21150125

21100105

Ta (N) V (m/s)

276

Solution:

ρ at 8 km 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.

.

277

LT

LL

C nC

=

0.08830.09060.09070.08960.08380.0246(rad/s)232120431819161914914273r (m)61.659.6356.7652.9346.914.75φ (degrees)

2.101.981.8241.6591.4611.0340.745£0.863£1.08£1.178£1.396£1.4$CLT

0.04480.05430.0670.08640.1114**CDT

222702198021580214802115021100Ta(N)468524656837042286012125815000DT1(N).09420.1150.1150.1150.1150.115CDT1

1.243*1.41.41.41.41.4CLT1

3.943.212.5531.9721.4661.034CLmax/CLL

0.3550.4360.5480.7100.9551.354CLL

205185165145125105V (m/s)

ψ.

278

The symbols in the table have the following meanings:

* Turn is limited by load factor (nmax) hence CLT1= nmaxCLl.

** Thrust available is more than thrust required. Hence CLT=CLT1

$ Turn is limited by CLmax

£ Turn is limited by Ta

Remarks:

i) Section 7 of Appendix B presents the turning performance of a jet engined airplane Fig. 5.26 a and b show the variations of and r with velocity at different altitudes. Figures 5.26 c and d present the variations of V max and Vrmin

.ψ

ψ.

279

.

Fig 5.26a Turning performance of jet transport- rate of

turn ( )ψ.

.

280Fig 5.26b Turning performance of jet transport- radius of

turn (r )

281Fig 5.26c Turning performance of jet transport- variation

of V max.ψ

282Fig 5.26d Turning performance of jet transport- variation of Vrmin

283

Section 7 of Appendix A presents the turning performance of a piston engined airplane Fig. 5.27 a and b show the variations of and r with velocity at different altitudes. Figures 5.27 c and d present the variations of V max and Vrmin.

It is seen that at low altitudes Vrmin and Vψmax

are close to each other at higher altitudes Vψmax is higher than Vrmin. However, both these speeds increase with altitude. The two speeds come close to each other as absolute ceiling is approached. Minimum radius of turn (rmin) increases with altitude and ψmax decreases with altitude. At absolute ceiling the rate of turn becomes zero and radius becomes infinite.

.ψ.

.

.

.

.

.ψ

.

284Fig 5.27a Turning performance of a piston engined

airplane - variation of rate of turn ( )ψ.

285

Fig 5.27b Turning performance of a piston engined airplane - variation of radius of turn ( r)

286Fig 5.27c Turning performance of a piston engined

airplane - variation of V max.ψ

287Fig 5.27d Turning performance of piston engined

airplane- variation of Vrmin

288

(ii) In many situations the minimum radius of turn in level flight is limited by the available engine output. A smaller radius of turn can be obtained by allowing the airplane to descend during the turn. This way a loss of potential energy is used to increase the available energy during turn (see Ref. 1.9).

General remark:

In the foregoing sections various types of flight situations have been analyzed which are of practical interest. To analyze any other situation one has only to write 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.

289

5.10. Miscellaneous topics

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, then the flight is called a free flight. The performance in level flight, climb, turn etc.. come under this category. We conclude this discussion of performance in free flight by describing aspects like flight limitations, operating envelope and V-n diagram. In sections 5.11 & 5.12 we describe the performance in take-off and landing which are influenced by the proximity of the ground.

290

5.10.1 Flight Limitations:

In the previous subsections, the performance of an airplane in free flight has been discussed under conditions of level flight, climb, accelerated flight, turn etc. Fig. 5.28a shows the variations of Vmax, (Vmin)e, Vs, V(R/C) max Vγ max, Vr min and Vψ max with altitude for a typical jet airplane. Figure 5.28b presents a similar plot for piston engined airplane.

It was pointed out earlier that stalling limits the minimum speed; power output limits Vmax, (Vmin)e, γmax, (R/C)max, rmin and ψmax.

.

.

291Fig 5.28a Variations of characteristic velocities –

jet transport

( )e

292Fig 5.28b Variations of characteristic velocities – Piston engined airplane

293

In addition to these, the performance may also be limited by buffeting and sonic boom. Buffeting is an irregular oscillation of a part of an airplane caused by the separated flow from some other component, passing over it, e.g., the horizontal tail experiences buffeting when the wing is at high angle of attack or in the transonic flow regime and the separated flow from wing passes over it (horizontal tail).

To prevent buffeting, the permissible value of CLmax

may be limited. This in turn would affects Vs, rmin

and ψmax.

Sonic boom problem is encountered when an airplane flies at supersonic speed at low altitudes.

.

.

294

.

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 they are perceived as two explosive like sounds called sonic boom or sonic bang. The intensity of the boom depends on the size and shape of the airplane, its flight altitude and 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 vibrations of

buildings and rattling of window panes. To keep

295

the overpressure on the ground within socially acceptable limits the supersonic transport airplanes are generally not permitted to cross Mach one below tropopause and they cruise at altitudes of 16 to 20 km.

5.10.2 Operating envelope :

The maximum speed and minimum speed can be calculated from the level flight analysis. However the attainment of maximum speed may be limited by other considerations. The operating envelope for an airplane is the range of flight speeds permissible at different altitudes. Typical operating envelope for a fighter airplane is shown in Fig. 5.28c. Explanation of the terms in

296Fig 5.28c Operating envelope -level flight(Adapted from Ref.3.7, Chapter 17)

297

this figure is as follows.

The Ps = 0 curve in Fig.5.28c refers to zero specific excess energy i.e. (TV-DV)/W . ‘T’ is the thrust available, ‘D’ is the drag, ‘V’ is the flight velocity and W is the weight of the airplane. The level flight operating envelope is determined by the stall limit and the engine thrust operating limit. Ps = 0 lines correspond to maximum or minimum speed based on engine output.

Military thrust and maximum thrust in Fig.5.28c refer to the maximum thrust without and with afterburner. The minimum speed from engine relight limit is encountered in some cases at high altitudes where enough air may not be available to

298

restart the engine in the event of flame-out.

At absolute ceiling the rate of climb or Ps is zero and at service ceiling it is 100 ft/min for propeller airplane and 500 ft/min for jets . The ceiling may further be limited by the highest altitude at which ejection by pilot could be permitted (about 50,000 ft).

The other limits shown in Fig.5.28c arise from structural considerations. The maximum q-limit relates to permissible maximum free stream dynamic pressure on the structure (1/2ρV∞

2 ).This limit may be around 2000lb/ft2 (96000 N/m2) . This limit would not permit attainment of high supersonic Mach number at low altitudes.

299

The duct pressure limit is imposed by the restriction on pressure inside the inlet duct of the engine.

The temperature limit is due to the aerodynamic heating which increases the surface temperature and limits are imposed depending on the material used for construction of the airframe.

300

5.10.3 V-n diagram

The load factor of (n) is defined as the ratio of lift and weight i.e. n= L/w . In level flight n=1. However as pointed out in sections 5.8 and 5.9 the value of ‘n’during a maneuver 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.5.17b) the angle of attack of the airplane would increase by ∆α = Vgu /V. This increase in angle of attack would increase lift (∆L) by:

∆L = ½ ρV2S CLα ∆α = ½ ρV2S CLα Vgu/V

= ½ ρVSCLα Vgu (5.93a)

Thus ∆L increases with Vgu and for a given Vgu

301

they ( ∆L & ∆n ) are higher at higher flight velocity. An airplane must be designed to withstand these gust loads also.

In aeronautical engineering practice the load factors due to maneuver 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.5.29. This diagram can be explained as follows.

(i) The lift (L) produced by an airplane is given by L = ½ ρV2S CL . Noting that CL ≤ CLmax, we observe that at staling speeds (Vs), L=W and n=1. However if the airplane is flown with CL = CLmax at speeds higher than Vs, then L will be more than W and L or n would be proportional to V2. This variation is a parabola and is shown by curve OIA in Fig.5.29. In an inverted

302Fig.5.29 A typical V-n diagram (Adapted from Ref.5.2 , chapter 2)

303

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.5.29. It may be mentioned that an airplane can fly only at V ≥ Vs and hence the portions IA and HG are shown by solid lines.

(ii) Higher the permissible load factor, heavier will be the airplane structure. Hence for actual airplanes the maneuver load factor is limited depending on its intended use. Federal Aviation Administration (FAA) and other agencies prescribe values of nmax for differentcategories of airplanes (see for example chapter 7 of Ref.1.7). Table 5.1a gives typical limit load factors. The limit load is the load that can be supported by the structure without yielding. The ultimate load factor, in aeronautical practice, is 1.5 times the limit load factor.

304

-3 to -66.5 to 9Fighter

-1 to -23 to 4Transport

-36General aviation-Aerobatic

-1 to -1.52.5 to 3.8General aviation- NormalnnegativenpositiveType of airplane

Table 5.1a Typical limit load factors (Adapted from Ref.3.7 , chapter 14)

305

In Fig.5.29, npositive = 3 and nnegative = -1.2 have been chosen; the actual values depend on the weight of the airplane and its category.

(iii) The positive maneuver load factor is prescribed to be constant upto the design diving speed (Vd); line AC in Fig.5.29. According to Ref.3.7, chapter 14, the design diving speed could be 40 to 50% higher than the cruising speed 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 maneuver load factor is prescribed to be constant upto design cruising speed (line GF in Fig.5.29) and then increases linearly to zero at V=Vd

(line FD in Fig.5.29).

306

(iv) As per Eq.(5.93a) the gust load factor varies linearly with velocity. It is prescribed that an airplane should be able to withstand Vgu=30 ft/s (9.1 m/s) upto Vc and Vgu = 15 ft/s (4.6 m/s) upto Vd (see lines JB’, JC’ , JF’ and JE in Fig.5.29). It may be added that a gust in reality is not a sharp edged gust as shown in Fig.5.17b and the gust load factor is reduced by “Gust alleviation factor” (see Ref.3.7 chapter 14 for details).

(v) For safe operation the airplane must be designed to withstand load factors at all points of the gust and manouever load curves. It may be pointed out that angles of attack are different at various points of the V-n diagram and the structure analysis needs to take this into account. For example the angle of

307

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.

(vi) The final V-n diagram , satisfying gust and manouever loads, is given by the solid line 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 CLmax.

308

5.9 Define steady level coordinate turn. An airplane having a weight of 11,000 N has a wing area of 15 m2 and drag polar of CD = 0.02 + 0.06CL

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 = 4.

What is the time taken to turn through 180O?

[Answers: r= 113.3 m ; ψ = 0.3912 rad /s,

t = 8.006 s].

.

Exercises

309

5.10 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) in a loop of radius 200 m at a speed of 100 m/s? Are these turn and loop permissible.

[ Answers: (a) 1 (b) 0 (c) 5.19 (d) 6.09]

310

5.11 Take off performance

An airplane 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 must be at least equal to its weight which can happen when the velocity of the airplane is equal to or more than its stalling speed. To do this the airplane accelerates along the runway and reaches take-off speed (VTO).

The distance covered along the ground is called the take-off run. However to decide the length of the runway required for an airplane, one should ensure that the airplane is above a certain height

311

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. Thus the take off distance is defined as the horizontal distance covered by an airplane from the start of the run and climb to a height equal to the screen height.

5.11.1 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. 5.30a). Ground run:During the ground run the airplane starts from rest and accelerates to the take-off speed (VTO or V1).

312Fig. 5.30a Phases of take-off flight

313

The flaps and engine (s) are adjusted for the take-off setting. 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 VTO is reached. After this 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 (CLTO). At this stage the nose wheel is off the ground (Fig.5.30b) and airplane speed continues to increase. As the airplane exceeds the take off speed it gets airborne and the main landing gear wheels also leave the ground.

When the airplane has a tail wheel type of

314Fig. 5.30b Nose wheel lift-off

315

landing gear, the angle of attack is high at the beginning of the take-off run (Fig.5.30c). However, the tail wheel is lifted off the ground as soon as some elevator effectiveness is gained (Fig.5.30d). 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 CLTO. When VTO is exceeded, the airplane gets airborne.

The point at which all the wheels have left the ground is called the unstick point (Fig.5.30a).

316

Fig. 5.30c Tail wheel type airplane at start of take-off run (Adapted from Ref.5.1)

317

Fig. 5.30d Tail wheel type of airplane during middle part

of take-off run (Adapted from Ref.5.1)

318

5.11.2 Estimation of Take-off performance

In this flight we are interested in obtaining the take-off distance (s) and the time (t) taken for it. Since the equations of motion are different in the three phases of take-off flight, we consider them separately.

Transition and climb:During the transition phase the airplane moves along a curved path (Fig.5.30a) 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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Ground run:

The forces acting on the airplane are shown in Fig.5.30a. In addition to lift, drag, weight and thrust, we have the ground reaction. The force due to ground friction is µR, where µ is the coefficient of rolling friction between the runway and the landing gear wheels, and R is the

reaction at the ground. The equations of motion are

WT D R ag

µ− − = (5.94)

L + R – W = 0 (5.95)

320

Hence, R = W-L and

Further,

( )/

T D W LaW gµ− − −

=

d V d V d s d Va Vd t d s d t d s

= = =

Hence, ground run(s1) is given by:

1 1

1 ( )

V V

o o

V dV W V dVsa g T D W Lµ

= =− − −∫ ∫ (5.97)

(5.96)

321

and the time taken during ground run (t1) is given by:

(5.98)

1

1

V

o

dVta

= ∫

1

( )

V

o

W dVg T D W Lµ

=− − −∫

Equations (5.97) and (5.98) can be integrated numerically or graphically when variations of T, D and L are prescribed and µ is known. The value of µdepends on the type of surface. Typical values are given in Table 5.2.

322

0.1-0.3Soft ground

0.1Average field-long grass

0.05Average field-short grass

0.04Hard turf

0.02Concrete, wood or asphalt

Coefficient of rolling friction (µ)

Type of surface

Table 5.2 Coefficient of rolling friction

323

The thrust during take-off run can be approximated as T= A1–B1V2. The angle of attack can be assumed to remain constant during the take-off run. With these assumptions we get:

2 21 1

1( ) ( )2 D LT D W L A B V W V S C Cµ µ ρ µ− − − = − − − −

11 ( )2 D LB B S C Cρ µ= + −

Substituting in Eqs.(5.97)and (5.98) we get

(5.99){ }1

21 12 /( )

2

V

o

W VdV Ws ln A A BVg A BV gB

= = −−∫

= A – BV2 where A = A1 - µW and

324

and2

11 2

12

V

o

A B VW dV Wt ln ag A BV g AB A B V

⎧ ⎫+⎪ ⎪= = ⎨ ⎬− −⎪ ⎪⎩ ⎭∫ (5.100)

Remarks:i) A good approximation to s1 and t1 is obtained by

taking an average value of accelerating force (Fa)

to be the value of [T-D-µ(W-L)] at V = 0.7 V1.

Then

(5.101)

and

(5.102)

1 21

1 2

V

a ao

W VW VdVsg F g F

= =∫

1

11

V

a ao

WVW dVtg F g F

= =∫

325

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 drag due to flaps and landing gear. The proximity of the ground reduces 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.

Transition:From Fig. 5.30a we see that during the transition phase the airplane changes the direction of flight and its speed increases from V1 to V2. The height attained during this flight and the horizontal distance traversed can be obtained by treating the

326

( )2 22 2 2 12

WTs Ds V Vg

= + −

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 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.

or( )2 2

2 12 2

V VWsg T D

−=

−(5.103)

327

T and D in Eq.(5.103) are evaluated at the mean speed during this phase i.e., at (V2 + V1)/2. The time taken in transition (t2) is given by:

(5.104)

V1 generally lies between (1.15 to 1.2) Vs and V2

is (1.05 to 1.1) V1.

22

2 10.5 ( )st

V V=

+

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 in climbing to screen height i.e.

328

s3 = (Screen height) / tan γ (5.105)

where γ is the angle of climb at velocity V2

where T and D are evaluated at V2.

sin T DW

γ −=

The time taken in climb phase (t3) is:

t3 = (Screen height) / V2 sin γ (5.106)

Hence the take-off distance (s) and the time taken for it (t) are given by

s = s1 + s2 + s3 (5.107)t = t1 + t2 + t3 (5.108)

or

329

Example 5.14

A jet airplane having 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 V1

(iii) CL during ground run is 1.15

(iv) drag polar with landing gear and flaps is

CD = 0.044 +0.05CL2

330

(v) Thrust variation during take-off can be

approximated as :

T = 128,500 – 0.0929 V2

where V is the kmph

(vi) Take-off takes place from a level, dry concrete

runway (µ=0.02) at sea level.

331

Solution:

Hence

V1 = 1.16 x 49.14

= 57.12 m/s and

V2 = 1.086 x 57.12 = 62.04 m/s.

For CL = 1.15,

CD = 0.044 + 0.05 x 1.152 = 0.1101

1 12 2

max

2 2 441450 49.24 /1.225 110 2.7s

L

WV m sS Cρ

⎛ ⎞ ⎛ ⎞×= = =⎜ ⎟ ⎜ ⎟× ×⎝ ⎠⎝ ⎠

332

Hence

= 128500-0.0929 (3.6V)2 - 0.02 x 441450 x 0.5 X

1.225 x V2 x 110 (0.1101 - 0.02 x 1.15)

= 119671 – 7.0752 V2

21( ) { }2 D LT D W L T W V S C Cµ µ ρ µ− − − = − − −

Using Eqs. (5.99) and (5.100) we get the ground

run (s1) and time taken for it (t1) as:

s1= [119671/(119671-7.0752 x 57.12 x 57.12)]

4414502 9.81 7.0752× ×

ln

= 681.5 m

333

= 23.05 s.

To calculate s2 we proceed as follows:

Average speed during transition =

= 59.58 m/s.

= 214.49 kmph

( ) ( )

1 12 2

1 1 12 2 2

1441450 (119671) (7.0752) 57.12

2 9.81 (119671 7.0752) 119671 7.0752 57.12t ln

⎡ ⎤+ ×⎢ ⎥=× × ⎢ ⎥− ×⎣ ⎦

57.12 62.042+

Average thrust = 128,500 – 0.0929 x (214.49)2 = 124,226 N

To get the drag during this phase let us assume

that CL = CLTO. Now

334

CLTO = CLmax (Vs/V1)2 = 2.7/(1.16)2 = 2.007

Assuming the same drag polar as in the ground

run we get,

CD = 0.044 + 0.05 x 2.0072 = 0.2454

or D = 0.5 x1.225 x (59.58)2 x 110 x 0.2454

= 58742 N

Using Eqs. (5.103) and (5.104) we get

2 2

2441450 62.04 57.12 201.42 9.81 124226 58742

s m⎛ ⎞−

= =⎜ ⎟× −⎝ ⎠

335

and

t2 = 201.4 / 59.58 = 3.38 s.

During the climb phase V = 62.04m/s

= 223.34 kmph.

T = 128500 - 0.0929 x 223.342

= 123866 N

To get the drag we should know the lift produced.Taking L ≈ W we get

( )212 21

2

441450/ 1.71.225 110 62.04LC W V Sρ= = =

× × ×

336

Hence CD = 0.044 + 0.05 x 1.72 = 0.1885

D = 0.5 x 1.225 x (62.04)2 x 110 x 0.1885

= 48942 N

Hence123866 48942sin 0.1697

441450γ −= =

Hence tanγ = 0.1722.Using Eqs. (5.105) and (5.106) we get

s3 = 15/0.1722 = 87.1 m. and t3 = 15/(62.04 x0.1697) = 1.42 s.

Finally, s = s1 + s2 + s3= 681.5 + 201.4 + 87.1= 970 m

337

Remarks :

i) 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. (5.101),

it is seen that the distance s1 is proportional to V12 or

in turn to VS2 which is given by:

2

max

2s

L

WVS Cρ

=

and t = t1 + t2 + t3= 23.05 + 3.38 + 1.42 = 27.85 s.

338

Hence the ground run increases when wing loading (W/S) increases, and when ρ decreases or take-off altitude increases. The ground run decreases as CLmax increases. Hence the high speed airplanes which have high W/S, from cruising flight considerations, have elaborate high lift devices to increase CLmax.

The take-off run can be decreased by increasing thrust. This is done by using after burner in jet engines or augmenting thrust by an auxiliary rocket fired during take-off run. In shipboard airplanes a catapult is used to augment the accelerating force.

339

ii) Effect of wind: While discussing the range performance it was shown in section 5.5.7 that the distance covered with respect to ground decreases when the flight takes place in the presence of head wind. Same effect occurs during take-off in the presence of head wind and the take-off distance reduces. In a hypothetical case of head wind being equal to Vs, the airplane can simply take-off without having to accelerate on the ground. A quantitative estimate of the effect of wind of velocity Vw, on s1

can be obtained by replacing the limits of integration in Eq.(5.97) from 0 to V1 by Vw to V1. Thus head wind, though bad for range, is beneficial during take-off run. Airports have a device, earlier it used to be wind sock, to indicate the direction of

340

wind. Advantage in taken of the wind to reduce take-off run.

(iii) Balanced field length:

Take-off is a critical phase of flight operation and various eventualities are taken into account. In the case of multi-engined airplane, the possibility of the failure of one of the engines during take-off, needs to be considered. If the engine failure takes place during initial stages of ground run, then the pilot can apply brakes and bring the airplane to halt. If the engine failure takes place after airplane has gained sufficient speed, then there are two possibility viz. (a) apply brakes and stop the airplane, but this may need much longer runway

341

length than in the case of take-off without engine failure. (b) instead of applying brakes, continue to fly with one engine of 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 engine failure occurs at decision speed then the distance required to stop the airplane is same as that required to take-off with one engine failed. The take of distance required when engine failure takes place at decision speed is called balanced field length. See next remark for an estimate of balance field length .

342

(iv) In section 5.11.2 a procedure to estimate take-off distance has been presented. However it is based on several assumptions and hence has significant amount of uncertainty. In actual practice there would be further uncertainty due to factors like condition of runaway surface (wet or dry), and piloting technique. Hence for the purpose of preliminary design , use of charts based on statistical data is regarded as adequate. A typical chart is presented in Fig.5.31. The take-off run, take-off distance over 50’ (15m), and balance field length are plotted against take of parameter which depends on W/S ,σ, CLTo and T/W (or BHP/W). The ground run and take-off distance depend

343Fig 5.31 Estimation of take-off distance using

chart (Adapted from Ref. 3.7, Chapter 5)

Take-off distance in 1000 ft

Take-off parameter-LTO LTO

W/S W/SorC T/ W C BHP/ Wσ σ

344

on the type of engine used. The balance field length (BFL) depends also on number of engines. BFL is larger for two engined airplane than for four engined airplane as failure of one engine in the case of four engines would reduce the thrust by 25%, where as for a two engined airplane the reduction would be 50%.

Section 8 Appendix A presents calculation of take-off distance for piston engined airplane. Where as

Appendix B presents results based on the take-off chart. It is interesting to note that the balance field length is almost double the normal take-off distance.

345

5.12 Landing Performance

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 reasons 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 screen height and to come to halt. In actual practice after reaching a sufficiently low speed the pilot will taxy the airplane to the allotted parking place.

Fig. 5.32 shows the phases of landing flight for an airplane with tricycle type landing gear.

346Fig 5.32 Phases of landing flight

347

During the final approach phase, the airplane performance a steady descent. The flight velocity in this phase is called approach speed and denoted by VA.

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% or VA.

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

348

5.12.1 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 estimation cannot be done accurately as the flare and float

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.

349

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 VA and to come to a halt i.e.

sland = - (VA)2 / 2a (5.109)

where a = -1.22 m/s2 (or 4ft/s2) for simple

braking system

= -1.52 m/s2 (or 5 ft/s2) for average

braking system.

= - 1.83 m/s2 (or 6 ft/s2) for modern

braking system and

350

Remark:

The approach speed (VA) 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 VA can be taken as 1.3 Vs.

= -2.13 to 3.0 m/s2 (or 7 to 10 ft /s2)for airplane with modern braking system and reverse thrust or reverse pitch propellers

351

Example 5.15:

Obtain the landing distance for the airplane in example 5.14. Assume that the airplane has modern braking system with reverse thrust and that VA = 1.3 Vs.

Solution:

From the solution to problem 5.14,

Vs=49.24 m/s.

Hence VA = 1.3 x 49.240= 64.01 m/s.

Taking a = -2.13 m/s2, the first estimate of landing distance is :

264.01 961.92 ( 2.13)lands m= − =× −

352

Remarks

i) Appendix A also estimates the landing distance using Eq.(5.109). Appendix B uses different formula.

ii) The landing distance is proportional to (VA)2 and hence it is proportional to (Vs)2 which equals2W/(ρSCLmax). Hence this distance increases with increase of W/S and landing altitude and decreases with increase of CLmax.

iii) In addition to reverse thrust mentioned earlier, the landing run can be decreased by using arresting gear, drag parachute and spoilers.

iv) Like take-off distance the landing distance is also reduced by head wind.

353

References:

5.1 Jackson, P “Jane’s all the world’s aircraft-1999-2000”, Janes’s information group, Surrey, UK 1999.

5.2 Barton M.V. “Fundamentals of aircraft structures” Prentice-Hall, New York, 1948.

354

Exercises

5.11 Describe the various phase of take-off flight, Write down the equations of motion during take-off run. Taking CD, CL and T as constant during take-off run show that the ground run, s, is given by:

Where W, S, g, ρ and µ have the usual meanings, q1 = dynamic pressure at the unstick point and

1

ln( )D L

WsS g C C qρ µ

⎛ ⎞Γ= ⎜ ⎟− Γ −⎝ ⎠

( )D L

T WS C C

µµ

−Γ =

−

355

5.12 A rocket motor firing for short duration of say 10s is used to reduce the take of run. Explain that more benefit i.e. larger reduction in take-off distance would be achieved by using the rocket motor in the latter part of the take-off run.