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Outline and AnnouncementsOutline:-
Introduction – Course instructor and website.Assessment.Lectures, Class Notes, and DownloadsAssignments, Tutorials, and Exams.Role of the Textbook.Teaching Style.Course objectives.Interesting Flows.The “First Question”.
Important announcements:-1. Tutorials start the week of Sept 15.
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Assessment
Two-term tests (October and November) = 30%.
Assignments = 15% .
Final examination = 55% .
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LecturesClass discussions might include points that are not necessarily included in the textbook.
All exams will include questions on theory and concepts covered in lectures and class discussions.
Assignments might include questions on theory and concepts covered in lectures and class discussions.
Attending lectures is very important!!
Laptops and Cell Phones are not allowed during lectures, tutorials, and exams.
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Class Notes and Downloads
Class notes are print out of my PowerPoint presentations.
Notes taken by students during lectures are very important.
Lecture notes and other material will be posted on the course website in a password protected section.
All material is copyright protected and should not be shared with and/or distributed to others.
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Assignments
Assignments – will be assigned regularly, roughly every week.
Assignments might include problems from the textbook and from lectures and class discussions.
Due dates are posted on the web.
A late penalty of 10% per day will be applied.
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Tutorials and Exams
Tutorials are provided to:-Address any unclear points.Help you solve assignments.
All exams will include questions on theory and concepts covered in lectures and class discussions.
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Role of the Textbook
The textbook will be used to assign problems.
It supplements class discussions.
It is not a substitute for lectures!!
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Teaching StyleI love and enjoy teaching. So, my teaching style is dynamic and interactive.
Everybody is encouraged to interact (ask questions, inquire, answer questions, etc.)
Bonus cards will awarded to individuals based on their level of interaction and understanding of the material.
Each card is worth 0.25%. There is no limit on how many cards you can get!!
This bonus is added to your total. So, your total could be more than 100%!!
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Important Announcements
Please check the ““Important AnnouncementsImportant Announcements”section on the web on a regular basis.regular basis.
Tutorials will start the week of Sept 15Sept 15.
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Course Objectives
Introduce special “vocabularyvocabulary” and “basic concepts” used in fluid mechanics.
Develop a good understanding of these concepts.
Use them to analyze and understand fluid flows in “real (practical)real (practical)” problems.
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Why Fluid Mechanics is A Core Course?
Please try to think of or name:
An industry,A piece of machinery, orAny engineering system,
where fluid mechanics does not play an important role!
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Industries where Concepts of Fluid Mechanics Play a Vital Role
••Automobile EngineeringAutomobile Engineering••Aerospace EngineeringAerospace Engineering••Oil & Gas EngineeringOil & Gas Engineering••Power GenerationPower Generation••Thermal ManagementThermal Management••Environmental ControlEnvironmental Control••BiotechnologyBiotechnology••Energy ConversionEnergy Conversion••Process Engineering Process Engineering ▪▪
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Some Interesting Flows
Airfoil zero angle Airfoil 25° angle Wing Vortex
Personal PlumeAir flow over a CarReference: Multimedia Fluid Mechanics CD-ROM
Cambridge University Press ISBN-10: 0521604761
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OK, So what would be the first question we should address in this class?
What is Fluid Mechanics ?
Forms of Matter:
• Solid.• Liquid.• Gas. Motion and its “Cause”
Force
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Scope of Fluid Mechanics
Fluid Mechanics is the study of the behavior of fluids at restrest and in motionin motion.
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What is the Definition of a ““FluidFluid”” ??
areaforce tangential
Definition of A Fluid:
Fluid is a substance that deforms continuouslyunder the application (or the effect) of a shear stress.
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When a shear stress is applied:Solids deform or bend, then stop!Fluids continuously deform ⇒ i.e., they Flow
Definition of a Fluid – cont’d
Fig. 1.1
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Q. What makes a fluid deform?
A. The “no-slip condition”.
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Two Famous Flows in Fluid Mechanics
1. Couette flow.
2. Poiseuille flow.
Poiseuille Flow
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Two Famous Flows in Fluid Mechanics
1. Couette Flow2. Poiseuille Flow
Poiseuille Flow
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Basic Equations
In the analysis of any fluid mechanics problem, we need to use a set of equations called ““Governing EquationsGoverning Equations””.
These equations can be classified into:1. Basic or General Laws.2. Particular or Special Laws.
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Basic or General Laws
Conservation of mass.Newton’s second law of motion.The principle of angular momentum.
The first law of thermodynamics.The second law of thermodynamics.
From Mechanics
From Thermodynamics
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Basic or General Laws – Cont’d
Note that these laws do not depend on the type of fluid involved in the problem. That’s why they are called “General Laws”.
Not all these laws are required every time.
In many problems, it is necessary to bring into the analysis additional relations ⇒ “Particular Laws”
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Particular or Special LawsThese laws describe the behavior of physical properties of specific types of fluids.
Example: Ideal Gas Law ⇒
This law can be used only for ideal gases.It relates absolute pressure and temperature to gas density.P in Pa = N/m2, ρ in kg/m3, and T in K.R = gas constant, for air = 286.9 J/kg.K ■
TRP ρ=
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Methods of Description
1. The system approach (the Lagrangianapproach), which follows one specific particle or system (e.g., following the motion of a falling object).
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Methods of Description - Continue
2. The control volume approach (the Eurlerianapproach), which focuses on a certain region not on a certain particle (e.g., flow in a pipeline).
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Methods of Analysis
System(or “Closed System”)
Control Volume(or “Open System”)
in Thermodynamics
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Methods of Analysis in Fluid Mechanics
1. Infinitesimal or finite control volume.2. Large control volume.
Differential control volume Large Control volume
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Methods of Analysis in Fluid Mechanics
1. Infinitesimal or finite system.2. Control volume.
In each case the equations will look different.
In (1), the resulting equations are differential equations, which provide details of the flow.
In (2), the resulting equations are integral, which give a global or overall behavior of the flow. ■
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Why Use Infinitesimal Control Volume (Differential) Approach?
Differential approach allows us to determine flow details, e.g., velocity distribution:-
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Why Use Control Volume (Integral) Approach?
Integral approach allows the determination of global or overall values, such as, average velocities, forces, etc..
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Units and Dimensions
A unit is a specific quantitative measure of a physical quantity. Example, the foot and the meter, which are used to measure the physical quantity ‘length’.
A physical quantity, such as length, is called a dimension.
Dimensions can classified as basic or primarydimensions and secondary dimensions:-
1. Length and time are basic dimensions.2. Velocity is a secondary dimension because it can be
represented using primary dimensions (velocity = length / time).
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Systems of Basic Dimensions
1. [M], [L], [t], and [T].
2. [F], [L], [t], and [T].
3. [F],[M], [L], [t], and [T].
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Systems of Units
1. MLtT - SI (kg, m, s, K).
2. FLtT - British Gravitational (lbf, ft, s, oR).
3. FMLtT - English Engineering (lbf, lbm, ft, s, oR).
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Preferred Systems of Units
SI (kg, m, s, K)
British Gravitational (lbf, ft, s, oR)
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Dimensional Consistency
All equations and formulas must have consistent dimensions.
I.e., all terms in any equation must have the same dimension.
What is the dimension of each term of Bernoulli’s equation?
ρρ2
2
221
1
21
22pzgVpzgV
++=++rr
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Dimensional Consistency – Cont’dIn your engineering studies and practice, you might deal with two types of equations having inconsistent dimensions:-1. Semi-empirical equations, e.g., the Manning equation,
used to calculate the velocity of flow in an open channel (such as a canal):-
Where V is the velocity in m/s, R is the hydraulic radius of the channel in m, and S is the channel slope (ratio). For unfinished concrete, n = 0.014.
What is the unit of V? Does n have a unit?
If we use n =0.014, and R in ft, can we get the correct value of V in ft/s?
nSRV n
2/10
3/2
=
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Dimensional Consistency – Cont’d2. In the second type the dimensions of an equation are
consistent but the use of units is not.
For example, the commonly used Energy Efficiency ratio (EER) of an air conditioner is:-
The equation is dimensionally consistent, with EER being dimensionless (ratio). However, it is used in an inconsistent way.
A good A/C has EER = 10, which means 10 Btu/hr for each 1 W of electrical power.
One must say 10 (Btu/hr)/W because it is not dimensionless.
timeenergytimeenergyEER
//
input electricalrate cooling
==
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Exercise
A sky diver with a mass of 75 kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be FD = kV2, where k = 0.228 N.s2/m2.
Determine the maximum speed of free fall for the sky diver and the speed reached after 100 m of fall.
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Vocabulary List1. A Fluid.2. Mechanics.3. Shear force.4. Normal force (pressure).5. Flow = continuous deformation.6. Scope of Fluid Mechanics.7. Basic Equations.8. Methods of Description.9. Methods of Analysis.10. No-slip condition.11. Lagrangian approach.12. Eulerian approach.13. Infinitesimal or finite system = infinitesimal control volume
(C.V.).14. Infinitesimal C.V. = Differential approach.15. Control volume or integral approach.16. Basic dimensions.
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A flow is described by the velocity field,
Where a = 1 1/s and b = 2 m/s2. t in seconds.
(a) Plot the pathline of the particle that passed point (1,2) at t = 2.
(b) Streakeline at t = 3 of the particles that passed point (1,2).
jtbiyaV ˆˆ+=Problem
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Note the difference between the meaning of t and to
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Note the difference between the meaning of t and to
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Problem – time permitFluids of viscosities μ1 = 0.1 N.s/m2 and μ2 = 0.15 N. s/m2 are contained between two plates (each plate is 1 m2 in area). The thicknesses are h1 = 0.5 mm and h2 = 0.3 mm. respectively. Find the force F to make the upper plate move at a speed of 1 m/s. What is the fluid velocity at the interface between the two fluids?
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Chapter 2 - Fundamental ConceptsOutline:-
Fluid as a Continuum.Scalar quantities.Vector quantities - Velocity Field.Steady Vs Unsteady Flow Fields.Uniform Flow Fields.One-, Two-, and Three-Dimensional Flows.Visual Representation of flow fields:
Timeline.Streamlines. Pathlines. Streaklines.
Types of Forces acting on a fluid element.Stresses (name and sign convention).Stress Field (stress at a point).Shear Stress and rate of deformation of a fluid element.Viscosity.
1. Newtonian and 2. Non-Newtonian Fluids.Surface Tension.Classification of Fluid Motions.Compressibility effect.
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Nature of Fluids (Matter)
Fluids (gases or liquids) are forms of matter and they consist of molecules with atoms and space in between.
So, generally speaking, mass of a fluid is not continuously distributed in space:
(mass – space – mass – space – mass – etc…)
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Fluid as a Continuum Continuum Hypothesis: We can assume that fluids are continuous
medium.
1. What does it mean?It means that a fluid regardless of its molecular nature, can be treated as a continuous medium.
2. What is the result or benefitbenefit of this assumption?a) Each fluid property is assumed to have a definite value at every point
in space, thus
b) Fluid properties are considered to be continuous function of position and time.
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3. When it is not possible to make such an assumption?We can not use this assumption if the mean free path of the molecules is of the same order of magnitude as the smallest significant characteristic dimension of the problem.
Molecules always vibrate in space.
Mean free path is the distance the molecules travel as it vibrate.
Continuum Hypothesis – Cont’d
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Examples, for air:-
1. At STP (15º C and 101.3 kPa) ⇒ Lm = 6 × 10-8 m = 0.06 μm = 60 nm.
2. At P = 1.33 × 10-7 kPa (Rarefied Gas) ⇒ Lm = 0.9 m = 90 cm.
Applications = air flow between parallel plates with a = 10 cm at atmospheric pressure i.e., P = 101.3 kPa.
Mean Free Path, Lm
Can we assume that air is a continuous mediumin this case ?
?⇐
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Quantities of interest
In fluid mechanics we are interested in quantities that are:
1. Scalar quantities, such as: density, temperature, pressure, etc.
2. Vector quantities, such as: velocity, acceleration, stress, ect.
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Scalar Quantities – e.g., Density ρ
Since ρ is a scalar quantity, we need only the specification of its magnitude.
The complete or field representation of ρ is given by:
Equation (2.1) represents a scalar field.
(2.1)
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Density ρ - Cont’dDensity is sometimes expressed in terms of specificgravity, SG for liquids:-
where, ρwmax = max density of water = 1000 kg/m3 at 4 ºC.
For gases, where ρa= air density.
Or, in terms of specific weight, γ:
maxw
SGρρ
=
ggm ργ =∀
==volumeweight
a
SGρρ
=
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Similar to ρ, we can represent the velocity as:-
Eqn. 2.2 represents the velocity field, which is a vector field.
A Vector has magnitude and direction.
Vector Quantities – e.g., Velocity Field,
(2.2)
V
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Velocity can be written in terms of its 3 scalar components (u, v, w) in the x, y, and z directions:
where = unit vectors in x, y, and z directions, respectively.
In general each component u, v, and w is a function of x,y, z, and t, e.g., u = u(x,y,z,t)
Velocity Field, - Cont’dV
kji ˆ and,ˆ,ˆ
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Types of Flows - Steady Flow
If properties at every point in a flow field do not change with time (t), the flow is called steady.
If η is any fluid property in a steady flow field, mathematically:
Remember : if f = f(x), then rate of change of f w.r.t. x
If f=f(x,y), then rate of change of f w.r.t. x = ■
),,(or,0 zyxt
ηηη==
∂∂
dxdf
=
xf∂∂
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Types of Flows - One-, Two-, and Three-Dimensional Flows
Depending on the number of space coordinates required to specify the velocity field, a flow can be described as 1D, 2D or 3D.
⇐ 3D, unsteady
⇐ 2D, unsteady
⇐ 1D, steday
),,,( tzyxVV =
),,( tyxVV =
)(xVV =
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Examples – Steady Flow
1D 2D
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Uniform Flow
Sometimes we can neglect the no-slip condition at the wall and assume that the velocity is uniform across the whole cross section, as shown below:
u = u(x,y) u=u(x)
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Visual Representation of Flow Fields
Sometime we want to have a visual representation of flow fields. Such a representation is provided by:
Timelines.
Streamlines.
Path lines.
Streak lines.
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TimelinesA timeline is a line connecting the positions of a set of fluid particles at a given instant.
This is a timeline
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StreamlinesA streamline is the line drawn in the flow field so that at any instant in time it is tangent to the direction of the flow, i.e., tangent to the velocity vector.
,tan.,.dxdy
uvei == α 0=−⇒ dxvdyu
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Streamlines – Cont’d
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Pathlines
A path line is the line traced out by a given particle as it flows from one point to another.
Path lines are useful in studying , for example, the trajectory of a contaminant leaving a smokestack.
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Pathlines – Cont’d
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Streaklines
A streakline is the locus of all particles that at an earlier instant in time, passed through a prescribed point in space.
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Streaklines – Cont’d
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What type of Lines are these?
The redlines, andThe white lines?
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Important points to note
A streamline and a timeline are instantaneous lines, e.g., snap shots.
While streaklines and pathlines are generated by the passage of time, e.g., video recording.
In steady flow, the velocity at each point in the flow field remains constant with time and consequently, in a steady flow, pathlinespathlines, streaklinesstreaklines, and streamlinesstreamlines are identical.
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Example Problem
A two-dimensional unsteady velocity field is given by u = x (1 + 2 t), v = y. Find:-
1. The time-varying streamlines which pass through some reference point (xo,yo). Sketch some for the case of xo=1, yo = 1.
2. Find the equation of the pathline which passes through the point (xo, yo) at t = 0. Sketch this pathlinefor the case of xo=1, yo = 1.
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Streamlines at different times
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Pathline for t < 0 to t > 0
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Types of Forces Acting on a Fluid Element
1. Surface Forces.2. Body Forces.
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Types of Forces Acting on a Fluid ElementA. Surfaces Forces:- e.g., Pressure and Friction
These are forces generated on the surface of the fluid element by contact with other fluid particles or a solid surface.
B. Body Forces:- e.g., Gravity and Electromagnetic Field
These forces are not concentrated at the surface of the fluid element; they are rather experienced throughout the particle.
Example: gravitational body force acting on a fluid element of volume, =
gdgmd ∀=×= ρforce nalgravitatio
∀d
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Surface Force on a Fluid Element
δAyδAx
δAz
z
x
y
Δx
Δy
Δz
F
Fluid Element in Cartesian Coordinates
xAδ
Surface Forces cause StressesStresses on the surfaces of the fluid element
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Stress Caused by Surface Forces
σ (sigma) is used to denote normal stresses.τ (Tao) is used to denote tangential stresses.
Fluid element
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Stress Caused by Surface Forces –Cont’d
Normal stress at point C =
Tangential stress at point C = Or,
x
x
xAxx A
Fδδσ
δlim
0→=
x
y
xAxy A
Fδδ
τδlim
0→=
x
z
xAxz A
Fδδτ
δlim
0→=
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Name Convection - Stress needs two directions to be define (force and area).
First letter on the left refers to the plan on which the stress acts.
Second letter refers to the direction in which the stress acts.
Note that the subscript inrefers to the direction
normal to the surface of interest.
xAδ
xAδ
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xAδ
Sign Convention
Velocity or force is positive if it is in the positive direction of the axis.
The plan is considered positive if the normal to it is in the positive direction of the axis.
Sign of stress is determined by the sign of its plan and force.
What is the sign of What is the sign of σσxx xx ??
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Stress Field
Note: all stresses shown here are positive.Note: all stresses shown here are positive.
Stress at a point = ⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
zzzyzx
yzyyyx
xzxyxx
σττ
τστ
ττσ
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Shear Stress in a Fluid Element Exposed to a Shear Force
Fluid continuously deforms (i.e., flows) under the effect of a shear force.
Shear Stress =y
x
y
x
yAyx dAdF
AFlim ==
→ δδτ
δ 0
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Rate of Deformation of a Fluid Element Exposed to a Shear Force
Rate of deformation = (1)dtd
tlimt
αδδα
δ=
→ 0
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Rate of Deformation of a Fluid Element Exposed to a Shear Force
Because δα is very small we can say that: δl = δα. δy (2)
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Rate of Deformation of a Fluid Element Exposed to a Shear Force
Due to the no slip condition, point M’ will be moving at δu, thus : δl = δu. δt (3)
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Rate of Deformation of a Fluid Element Exposed to a Shear Force
From (2) and (3), Rate of deformation =
dydu
dtd
yu
t==
αδδ
δδα ,or ,
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Relation Between Shear Stress and Rate of Deformation
Because every fluid element exposed to shear stress will deform, there must be a relationship between the shear stressshear stress and the rate of rate of deformationdeformation.
This relationship depends on the type of fluid:-1. Newtonian, or2. Non-Newtonian.
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Fluid ViscosityNewtonian Fluids:-
Most of the common fluids (water, air, oil, etc.)“Linear” fluids
For Newtonian fluids, the rate of deformation is in direct proportiondirect proportion with the shear stress, i.e.,
The constant of proportionality is the fluid viscosity, μ, i.e.,
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Non-Newtonian FluidsSpecial fluids (e.g., most biological fluids, toothpaste, some paints, etc.)“Non-linear” fluids
In this case, In this case, ηη is called the apparent viscosityis called the apparent viscosity
In this case the relation between Shear stress and rate of deformationTakes the form:
Which can also be written as :
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Non-Newtonian Fluids – Cont’d
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Dynamic and Kinematic Viscosity
This equation is called Newton’s law of viscosity for a one-dimensional flow (note note only y appears in the equationonly y appears in the equation).
μ (mu) is called the Dynamic Viscosity of the fluid and it is a physical property of the fluid.
ν (nu) =μ/ρ is called the Kinematic viscosityand it is another physical property.
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Dynamic and Kinematic Viscosity
Units of μ : Pa.s = N.S/m2 = kg/(m.s),
1 Poise = 1 gm/(cm.s).
Units of ν: m2/s,
1 Stoke = 1 cm2/s.
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ProblemA block of mass M slides on a thin film of oil. The film thickness is h and the area of the block is A. When released, mass m exerts tension on the cord, causing the block to accelerate. Neglect friction in the pulley and air resistance. Develop an expression for the viscous force that acts on the block when it moves at speed V. Obtain an expression for the block speed as a function of time. If mass M = 5 kg, m = 1 kg, A = 25 cm2 , and h = 0.5 mm. If it takes 1 s for the speed of the block to reach 1 m/s, find the oil viscosity μ.
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Surface Tension, σSurface tension is a force that appears along any common surface (interface) between two fluids.
Units of σ is force per unit length (length of the interface), e.g., N/m or lbf/ft.
interface
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Surface Tension – Cont’d
As shown below, if the two fluids are in contact with a solid surface, a contact angle, θ, develops.
Both of values of σ and θ depend on the type of fluids in contact.
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Capillary Rise (CR) and Capillary Depression (CD).
CR occurs when θ < 90°.
CD occurs when θ > 90°.
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Calculation of Δh
Note the sign of Δh when θ > 90º and θ < 90º.
0=∑ zF
0cos =∀Δ−=∑ gDFz ρθπσ (1)
hDΔ=∀Δ
4
2πQ (2)
From (2) in (1):-Dg
hρ
θσ cos4=Δ∴
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Description and Classification of Fluid Motions
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Viscous and Inviscid FlowsUnder some special circumstances, the effect of fluid viscosity can be ignored (neglected).
Example:- in region of flow away from solid surfaces.
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Laminar and Turbulent FlowsA laminar flow is a flow in which the fluid particles move in smooth layers, laminas.laminas.
A Turbulent Flow is a flow in which the fluid particles rapidly mix as they move due to random velocity fluctuations.
The flow in a pipe is considered laminar if Re < 2300, where,
Where, ρ,μ, V, & D are the density, viscosity, velocity, and diameter, respectively.
number ReynoldsRe ==μ
ρ DV
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Compressible and Incompressible Flows
Incompressible flows are those in which variations in density are negligible.
When variations in density are not negligible, the flow is called compressible.
Variations in density are due to changes in pressure and/or temperature.
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Compressible and Incompressible Flows
Mostly, liquids can be regarded as incompressible fluids.
Pressure and density changes in liquids are reflected by the bulk compressibility modulus, or modulus of elasticitymodulus of elasticity:
For water at 15°C, Ev = 2010 kPa = 2.92 x 105 psi.
ρρ /ddpE v =
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Can we treat a gas flow as an Incompressible flow?
For Mach number M < 0.3, the maximum density variation is less than 5%.
Thus, gas flows with M < 0.3 can be treated as incompressible.
Mach number = V/c, where V is the flow velosity and c is the speed of sound.
The speed of sound in an ideal gas is given by:
Where k = Cp/Cv, R = gas constant, and T is the absolute temperature.
TRkc=
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Problem
At what minimum speed (in mph) would an automobile have to travel for compressibility effects to be important? Assume the local air temperature is 60°F.
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Problem – time permitFluids of viscosities μ1 = 0.1 N.s/m2 and μ2 = 0.15 N. s/m2 are contained between two plates (each plate is 1 m2 in area). The thicknesses are h1 = 0.5 mm and h2 = 0.3 mm. respectively. Find the force F to make the upper plate move at a speed of 1 m/s. What is the fluid velocity at the interface between the two fluids?
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Vocabulary List
1. Continuum hypothesis.2. Mean free path.3. Scalar quantity.4. Vector quantity.5. Uniform flow.6. Multi-dimension flow.7. Flow Visualization.
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Vocabulary List
1. Timeline.2. Streamline.3. Pathline.4. Streakline.
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Vocabulary List1. Body force.2. Surface force.3. Stress field.4. Shear stress.5. Normal stress.6. Rate of deformation.7. Viscid and Inviscid flows. 8. Laminar and Turbulent flows.9. Compressible and Incompressible flow.
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Outline:- Fluid Statics – Chapter 3
The Basic Equation of Fluid StaticsTypes of Pressures.Pressure Variation in a Static Fluid.Example Problem.Hydrostatic Force on Submerged Surfaces.Example Problems.
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Statics means that the fluid is not moving, i.e., its velocity =0; and its acceleration = 0.
Fluid velocity = 0 means that it does not flowit does not flow.
If a static fluid does not flow, how much shear stress the fluid is exposed to?
no flow = no deformation = no shear
What does Static Fluid mean?
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What does Static Fluid mean?
In this case, fluid can be exposed to only normal forces and behaves as
““a rigid bodya rigid body”” –– no deformationno deformation
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The Basic Equation of Fluid StaticsConsider the following fluid element:-
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The Basic Equation of Fluid StaticsConsider the following fluid element:-
Newton’s 2nd law:-
Divide both sides by gives:0. ==∑ admFd rr
∀d∑ ==
∀0.a
dFd rr
ρ
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The Basic Equation of Fluid Statics – Cont’d
Forces affecting on the fluid element = surface + body forces, i.e.,
Body Force =
or,
∑ += Bs FdFdFdrrr
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Surface Force in y- direction, dFys
dzdydxyp
dzdxdyyppdy
yppFd yS
..
.22
∂∂
−=
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+−⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
−=r
Surface force in y-direction =
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Surface Force, dFs
Combining the other two directions, we get:
Where, ∇p = gradient of p.
dzdydxp
dzdydxkzpj
ypi
xpFd S
...
..ˆˆˆ
−∇=
⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+∂∂
−=r
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Basic Equation of Fluid Statics
∑ ==∀
0.adFd rr
ρ
∑ += Bs FdFdFdrrr
dzdydxpFd S ...−∇=r
(2)
(1)
From (2) in (1):
Basic Equation of Fluid Statics
Recall – Newton’s 2nd Law:-
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Basic Equation of Fluid Statics
Note that equation above is not really one equation, it is rather three equationsthree equations in the three directions x, y, and z.
0,0,0 =+∂∂
−=+∂∂
−=+∂∂
− zyx gzpg
ypg
xp ρρρ
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Knowing that
Assuming that z is the vertical direction, we can say that gx = gy = 0, gz = - g, the three equations:
Basic Equation of Fluid Statics
kgjgigg zyxˆˆˆ ++=
r
0,0,0 =+∂∂
−=+∂∂
−=+∂∂
− zyx gzpg
ypg
xp ρρρ
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Can be written as :-
i.e.,
Basic Equation of Fluid Statics
gzp
yp
xp ρ−=
∂∂
=∂∂
=∂∂ ,0,0
γρ −=−= gdzdp
i.e., pressure is not function of x i.e., pressure is not
function of y
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Important Restrictions:-
In order to use the following equation, three three conditions must be satisfiedconditions must be satisfied.
1. Fluid must be static, i.e., velocity = acceleration = 0.
2. Gravity is the only body force.3. The Z axis is vertical and upward.
γρ −=−= gdzdp
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Types of Pressures
Fig. 3.2Pabs = Patm + Pgage
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Pressure Variation in a Static Fluid
gdzdp ρ−=
∫ ∫−=p
op
z
ozdzgdp ρ
For Incompressible Fluid: Manometers
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Pressure Variation in a Static Fluid
∫ ∫−=p
op
z
ozdzgdp ρ
( ) ( )hgpp
zzgzzgpp
o
ooo
ρρρ
+=−=−−=− or,
For Incompressible Fluid: Manometers
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Simple Rules to Analyze Multiple-Liquid Manometer Problems
1. Any two points at the same elevation in a continuous in a continuous volume of the same liquidvolume of the same liquid are at the same pressure.
2. Pressure increases as one goes downincreases as one goes down a liquid column.
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Example- ProblemDetermine the gage pressure in psig at point ““aa””, if liquid A has SG = 0.75 and liquid B has SG = 1.20. The liquid surrounding point ““aa”” is water and the tank on the left is open to the atmosphere.
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Hydrostatic Force on Submerged Surfaces.
This equation allows us to determine how pressure varies in a static fluid.
We would like to determine the force due to that pressure on a surface submerged in a liquid.
hgpp o ρ+=
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Hydrostatic Force on Submerged Surfaces – Cont’d
In order to fully determine the force on a surface submerged in a liquid, we must determine the following:-
1. The magnitude of the force;
2. The direction of the force; and
3. The line of action of the force.
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1- Direction of the Force on a Plane Submerged SurfaceSince fluid is not moving (static), there is no shear, i.e., only normal forces might exist.
Since this force is caused by pressure of fluid, it will always be normal to the surfacenormal to the surface.
This determines the direction of the force.
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2- Magnitude of the Force on a Plane SubmergedSurface
dAPdF .=
hgPP o ρ+=
( )∫ ∫ +==∴A A
oR dAhgPdFF ρ
(1)
(2)
From (2) in (1)
(3)
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2- Magnitude of the Force on a Plane Submerged Surface
( )∫ ∫ +==∴A A
oR dAhgPdFF ρ
∫+=A
oR dAygAPF .sinθρ
θsinyh =
(3)
but,
(4)
From (4) in (3)
Where, Po is the pressure at the fluid surface, A is the surface area.
Note how Note how θθ is measuredis measured
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3- Line of Action of the Force on a Plane Submerged Surface
∫ ∫==∴A A
R ydAPydFyF .... '
( ) dAygyPyFA
oR .sin.. 2' ∫ +=∴ θρ
hgPP o ρ+=
θsinyh =
but,
and
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3- Line of Action of the Force on a Plane Submerged Surface
dAygF
dAyPF
yA AR
oR
.sin.1...1 2' ∫ ∫+=∴ θρ
Solving for y’
( ) dAygyPyFA
oR .sin.. 2' ∫ +=∴ θρ
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Hydrostatic Force on Submerged Surfaces – Cont’d
1. The magnitude of the force;
2. The direction of the force = normal to the surface.normal to the surface.3. The line of action of the force.
∫+=A
oR dAygAPF .sinθρ
dAygF
dAyPF
yA AR
oR
.sin.1...1 2' ∫ ∫+=∴ θρ
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Example ProblemThe pressure in the air gap is 8000 Pa gage. The tank is cylindrical. Calculate the net hydrostatic force
(a) On the bottom of the tank; (b) On the cylindrical sidewall CC; (c) On the annular plane panel BB.
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Example ProblemGate AB is a homogeneous mass of 180 kg, 1.2 m wide into the paper, resting on smooth bottom B. All fluids are at 20°C. For what water depth h will the force at point B be zero? Assume specific gravity of Glycerin = 1.26.
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Vocabulary List
1. Static fluid 2. Manometer.3. Hydrostatic pressure.4. Gauge pressure.5. Vacuum.6. Hydrostatic force on a submerged surface.
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Outline Chapter 4- Basic Equations in Integral Form for a Control Volume.
Definition of a control volume.Dot product of Two Vectors.Volume and Mass Rate of Flow through a C.V.
Volume Flux.Mass Flux.Sign Convention.
Conservation of Mass.Special cases.
Example Problem.Extensive and Intensive Fluid Properties.Reynolds Transport Theorem.Momentum Equation.Sign Convention of Terms in the Momentum Equation.Types of forces.Example problem.
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Basic EquationsBasic Equations in Integral Formfor a Control Volume
Chapter 4
Conservation of Mass.Conservation of Mass.Conservation of Momentum Conservation of Momentum
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Why do we need Basic Equations in integral Form ?
In some applications we do not need do not need detailsdetails of the flow field, we rather need need some global valuessome global values, such as:-
1. Average velocity at a certain section.2. Force due to fluid flow.3. Mass or Volume flow rates.
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Examples:-
Given velocity at the exit section, fine mass flow rate and average velocity at the inlet section.
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Examples:-Find velocity V3 and force on the scale due to fluid flow.
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Examples:-
Find force on the 90° elbow.
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Examples:-
Find flow rate and force on the gate in the open and closed positions.
gate
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Q1 How do we obtain these equations?
A1 By applying conservation lawsconservation laws on a controlvolume.
Q2 Which conservation laws?A2 1- Conservation of mass.
2- Conservation of momentum.
Q3 What is a control volume?
Basic Equations in Integral Form
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Definition of a Control Volume1. A control volume is an arbitrary volume in space
through which fluid flows. The geometric boundary of the control volume is called the control surfacecontrol surface.
2. The control surface may be real or imaginary.
Fig 3.1
Real surface
Imaginary surface
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Background - Dot Product of Two Vectors
Dot product of V and dA = V dA cos α = projection of V on dA.
Fig 4.3
α
α
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Sign Convention
dAVdAVAdV +== 0cos.rr
Fig 4.3
dAVdAVAdV −== 180cos.rr
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Volume Flow Rate through a C.V.
Volume flow rate = volume flux through dA= dQ = V cosα . dA = V dA cos α
Total volume flow rate through control surface (CS)=
AdVrr
.=
∫∫ ==CSCS
AdVQdQrr
&& .
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Mass Flow Rate through a C.V.
Mass flow rate = Mass flux through dA= ρ dQ = dm
Total mass flow rate through control surface (CS)=
AdVrr
.ρ=
∫∫ ==CSCS
AdVmdmrr
&& .ρ
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Sign Convention
dAVAdV +==rr
.flux volume
Fig 4.3
dAVAdV −==rr
.flux volume
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Conservation of Mass of any System
M = mass of the system = constant.
Or, the time rate of change of the mass of the system = 0
0 i.e.,system
=⎟⎠⎞
dtdM
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Conservation of Mass of a Control Volume
Time rate of decreasedecrease of mass within the control volume = net mass outflowoutflow rate from the control volume.
Time rate of decreasedecrease of mass within the control volume =
net mass outflowoutflow rate from the control volume =
∫ ∀∂∂
−CV
dt
ρ
∫CS
AdVrr
.ρ
0. =+∀∂∂
=⎟⎠⎞
∫∫CSCVsystem
AdVdtdt
dM rrρρ
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Special Cases 1. Unsteady Incompressible Flow, i.e., ρ = c
0. integrals, outside take =+∀∂∂
∫∫CSCV
AdVdt
rrρρρ
0. ,by divide =+∀∂∂
∫∫CSCV
AdVdt
rrρ
C.V. theof volumethe but, =∀=∀∫CV
d
0.thus, =+∂∂∀
∫CS
AdVt
rr
0. =+∀∂∂
=⎟⎠⎞
∫∫CSCVsystem
AdVdtdt
dM rrρρ
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Special Cases 1. Unsteady Incompressible Flow, i.e., ρ = c
0.thus, =+∂∂∀
∫CS
AdVt
rr
0. =∫CS
AdVrr
thus,0 C.V., (fixed) deformable-non afor =∂∂∀
t
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Special Cases
2. Unsteady Incompressible flow through a fixed C.V.
0. =+∀∂∂
∫∫CSCV
AdVdt
rrρρ
0. =+∀∂∂
=⎟⎠⎞
∫∫CSCVsystem
AdVdtdt
dM rrρρ
0. ,by devide =+∀∂∂
∫∫CSCV
AdVdt
rrρ
0or c, c, =∂∂∀
=∀=t
ρ
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Special Cases
2. Unsteady Incompressible flow through a fixed C.V.
0. =+∀∂∂
=⎟⎠⎞
∫∫CSCVsystem
AdVdtdt
dM rrρρ
0.thus, =+∂∂∀
∫CS
AdVt
rr
0. =∫CS
AdVrr thus,0but =
∂∂∀
t
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Special Cases
3. Steady Compressible flow
(1) 0. =∫CS
AdVrr
ρ
0. =+∀∂∂
=⎟⎠⎞
∫∫CSCVsystem
AdVdtdt
dM rrρρ
∑∫ == mAdVAdVmdCS
&rrrr
& . thus,,.Recall, ρρ
c leCompressib ,0 Steady ≠⇒=∂∂∀
⇒ ρt
outinCS
mmAdVm &&rr
& === ∫∑ i.e., ,0. (1),in ρ
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Example Problem
Water flows steadily through a pipe of length L and radius R = 3 in. Calculate the uniform inlet velocity, U, if the velocity distribution across the outlet is given by:
ft/s. 10 and , 1 max2
2
max =⎥⎦
⎤⎢⎣
⎡−= u
Rruu
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Solution
1. Water = incompressible fluid → ρ = constant.
2. Flowing steadily = steady flow, i.e.,
Conservation of mass equation:-
Under conditions 1 and 2, equation (1) can be
written as:
.0=∂∂t
(1) 0. =+∀∂∂
∫∫CSCV
AdVdt
rrρρ
0. =∫CS
AdVrr
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Extensive Vs Intensive Fluid Properties
An extensiveextensive property is one that depends on the size of the C.V. Example:- volume or mass.
An intensiveintensive property is one that does not depend on the size of the C.V.
Example:- Specific volume, specific enthalpy.
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Reynolds Transport TheoremLet N = any extensive property.
the intensive property corresponding to N.
For any C.V., Reynolds Transport Theorem is:
==mNη
(I) .∫∫ +∀∂∂
=⎟⎠⎞
CSCVsystemAdVd
tdtdN rr
ρηρη
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Reynolds Transport Theorem
If N = mass, M ⇒ in (I),
Conservation of mass equation:
0. =+∀∂∂
=⎟⎠⎞
∫∫CSCVsystem
AdVdtdt
dM rrρρ
1==mNη
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Interpretation of each term in R.T.T.
time rate of change of any extensive property of the system.
time rate of change of N within the C.V.
mass of an element contained in the C.V.
amount of N in that element.
=∀∂∂∫
CVd
tρη
=⎟⎠⎞
systemdtdN
=∀dρη
=∀dρ
(I) .∫∫ +∀∂∂
=⎟⎠⎞
CSCVsystemAdVd
tdtdN rr
ρηρη
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the net rate of flux of N out through the C.S.
the rate of mass flux exiting dA = mass flowrate.
=AdVrr
.ρ
=∫CS
AdVrr
.ρη
Interpretation of each term in R.T.T.
(I) .∫∫ +∀∂∂
=⎟⎠⎞
CSCVsystemAdVd
tdtdN rr
ρηρη
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Momentum Equation for an Inertial (not accelerating) C.V.
VMPrr
momentumlinear ==
VMPrr
NLet == VMN r
==∴η
∫∫ +∀∂∂
=⎟⎠⎞
CSCVsystemAdVd
tdtdN rr
.ρηρη
∫∫ +∀∂∂
=⎟⎟⎠
⎞
CSCVsystem
AdVVdVtdt
Pd rrrrr
.ρρ
In R.T.T.
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Momentum Equation “Newton’s Second Law” for an Inertial C.V.
∑ ==dtPdFr
r momentum of change of rate time
∫∫∑ +∀∂∂
=∴CSCV
AdVVdVt
Frrrrr
.ρρ
∫∫ +∀∂∂
=⎟⎟⎠
⎞
CSCVsystem
AdVVdVtdt
Pd rrrrr
. R.T.T., from ρρ
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Momentum Equation – meaning of each term
∫∫∑ +∀∂∂
=∴CSCV
AdVVdVt
Frrrrr
.ρρ
(A) = sum of all forces acting on the fixed C.V.
(B) = time rate of change of momentum inside the C.V.
(C) = net rate of flux of momentum out through the C.S.
(A) = (B) + (C)
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Types of Forces ∑∑ ∑ += BS FFFrrr
∫∑ −=⇒=CS
S dAPF sF pressure e.g., forces, surfacerr
∫∑ ∀=⇒=CV
B dgF ρBF gravity e.g., forces,body rr
∫∫∑ +∀∂∂
=CSCV
AdVVdVt
Frrrrr
.in sub ρρ
(I) .∫∫ +∀∂∂
=+CSCV
BS AdVVdVt
FFrrrrrr
ρρ
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Scalar Equations of the Momentum Equation
(I) .∫∫ +∀∂∂
=+CSCV
BS AdVVdVt
FFrrrrrr
ρρ
Equation (I) is a vector equation. Therefore, it may be written as three scalar component equations.
In Cartesian coordinates (x, y, z), equation (I) can be written as:
. :equationdiretion -z
. :equationdiretion -y
. :equationdiretion -x
∫∫
∫∫
∫∫
+∀∂∂
=+
+∀∂∂
=+
+∀∂∂
=+
CSCVBzsz
CSCVBysy
CSCVBxsx
AdVwdwt
FF
AdVvdvt
FF
AdVudut
FF
rr
rr
rr
ρρ
ρρ
ρρ
NoteNote NoteNote
NoteNote
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Sign Convention of Terms in the Momentum Equation.
(I) .∫∫ +∀∂∂
=+CSCV
BS AdVVdVt
FFrrrrrr
ρρ
F is positive if it is in the positive direction of the coordinate.
V is positive if it is in the positive direction of the coordinate.
Sign of depends on the relative directions ofand .
AdVrr
. Vr
Adr
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ProblemA jet of water issuing from a stationary nozzle at 15 m/s (Aj = 0.05 m2) strikes a turning vane mounted on a cart as shown. The vane turns the jet through angle θ = 50º. Determine the value of mass, M, required to hold the cart stationary. If the vane angle θ is adjustable, plot the mass, M, needed to hold the cart stationary versus θfor 0 ≤ θ ≤ 180°.
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Solution Procedure for This Type of Problems
1. Identify your control volume, C.V.
2. Identify its C.S., and number important sections to consider.
3. Identify your coordinates and draw them.
4. Identify which equation (s) will be used to solve the problem.
5. Identify which assumptions you can make to simplify your equations.
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Solution Procedure for This Type of Problems
6. Common assumptions are:-a. Steady.b. Incompressible, i.e., density, ρ = constant .c. Uniform flow.d. Body force = 0.e. Effect of atmospheric pressure is negligible.
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Solution
1. Identify your control volume, C.V.2. Identify its C.S., and number important sections to
consider.
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Solution
1. Identify your coordinates and draw them.
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Solution – Cont’d
4. Identify which equation (s) will be used to solve the problem.
. :equationdiretion -z
. :equationdiretion -y
. :equationdiretion -x
-:Equation Momentum
∫∫
∫∫
∫∫
+∀∂∂
=+
+∀∂∂
=+
+∀∂∂
=+
CSCVBzsz
CSCVBysy
CSCVBxsx
AdVwdwt
FF
AdVvdvt
FF
AdVudut
FF
rr
rr
rr
ρρ
ρρ
ρρ
Rx
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Solution – Cont’d
5. Identify which assumptions you can make to simplify your equations.
6. Common assumptions are:-a. Steady.b. Incompressible, i.e., density, ρ = constant .c. Uniform flow.d. Body force = 0.e. Effect of atmospheric pressure is negligible.
Rx
.∫∫ +∀∂∂
=+CSCV
Bxsx AdVudut
FFrr
ρρ
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Rx
Solution – Cont’d
( )θρ cos1M2
−=g
AV
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Vocabulary List
1. Basic equations (conservation laws).2. Control volume.3. Dot product of two vectors.4. Volume flux.5. Extensive Property.6. Intensive Property.7. Reynolds Transport Theorem
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Exercise
Water flows steadily through a pipe of length L and radius R = 3 in. Calculate the uniform inlet velocity, U, if the velocity distribution across the outlet is given by:
ft/s. 10 and , 1 max2
2
max =⎥⎦
⎤⎢⎣
⎡−= u
Rruu
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Solution
1. Water = incompressible fluid → ρ = constant.
2. Flowing steadily = steady flow, i.e.,
Conservation of mass equation:-
Under conditions 1 and 2, equation (1) can be
written as:
.0=∂∂t
(1) 0. =+∀∂∂
∫∫CSCV
AdVdt
rrρρ
0. =∫CS
AdVrr
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Solution – cont’d
(a) 0.. 0.21
=+⇒= ∫∫∫ AdVAdVAdVCS
rrrrrr
02)1(2. (a),in 0
2
2
max0
=−++− ∫∫RR
drrRrurdrU ππ
drrπdARr- u V 2 ,1 (2)at constant,UV (1)at 2
2
max =⎥⎦
⎤⎢⎣
⎡===
rr
Note the sign of the first and the second integrals.
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Solution – cont’d
(b) 042
2.0
2
42
max2 =⎥
⎦
⎤⎢⎣
⎡−+−
R
RrruRU ππ
ft/s.052
:for U (b) Solving max .uU ==
02)1(2. (a),in 0
2
2
max0
=−+− ∫∫RR
drrRrurdrU ππ
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Outline Chapter 5- Differential Analysis of Fluid Motion.
Why differential.Region of interest.Conservation of Mass for an infinitesimal C.V. (fluid particle) -Special cases.Lagrangian and Eulerian descriptions of the motion of a fluid particle.Material, substantial, or particle derivative.Example problem.Particle acceleration.Types of motion of a fluid particle. Rotation and vorticity vectors.Fluid deformation:1. Angular deformation.2. Linear deformation.Differential momentum equation.
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Differential Analysis of Fluid Motion
Why differential?
To obtain detailed knowledge, we must apply the basic equations of fluid motion in differential form.
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Region of interestSince we are interested in formulating differential equations, our analysis will be in terms of infinitesimal systems or infinitesimal control volumes (i.e., differential C.V.)
Fig 5.1
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Conservation of Mass for an infinitesimal C.V. (fluid particle)
(I) =Time rate of decreasedecrease of mass inside C.V. = (II) = net rate of mass flux (mass flow rate) outout through the C.S.
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Conservation of Mass for an infinitesimal C.V. (fluid particle)
(I) = Time rate of decreasedecrease of mass inside C.V. =
∀∂∂
−= dtρ dzdydx
t∂∂
−=ρ
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(II) = net rate of mass flux (mass flow rate) outout through the C.S. in the x-direction =
( ) dzdyudzdydxxuu ρρρ −⎟
⎠⎞
⎜⎝⎛
∂∂
+=)(
dzdydxxu
∂∂
+=)(ρ
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Conservation of Mass for an infinitesimal C.V. (fluid particle)
(II) = net rate of mass flux (mass flow rate) outout through the C.S. in the three directions =
dzdydxzw
yv
xuII ⎥
⎦
⎤⎢⎣
⎡∂
∂+
∂∂
+∂
∂+=
)()()()( ρρρ
0)()()(=
∂∂
+∂
∂+
∂∂
+∂∂
∴zw
yv
xu
tρρρρ
dzdydxt
dt
I∂∂
−=∀∂∂
−=ρρ)(
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Conservation of Mass - Special Cases
1. Incompressible Flow:- ρ = constant
2. Steady Flow:-
0)()()(=
∂∂
+∂
∂+
∂∂
+∂∂
zw
yv
xu
tρρρρ
0=∂∂t
0 =∂∂
⇒tρ
0. .,.
0 0)()()(
=∇
=∂∂
+∂∂
+∂∂
⇒=∂
∂+
∂∂
+∂
∂
Vei
zw
yv
xu
zw
yv
xu
r
ρρρ
0 =∂∂
⇒tρ
0)(. .,. 0)()()(=∇⇒=
∂∂
+∂
∂+
∂∂ Vei
zw
yv
xu r
ρρρρ
⇐ Compressible , UnsteadyCompressible , Unsteady
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Example Problem
Which of the following sets of equations represent possible two-dimensional incompressible flow cases?
a)
b)
)2(2
23
222
yyxxvyxyxu
−+=
−+=
22
2
22
xyxyvyxxyu
+−=
+−=
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Lagrangian and Eulerian Descriptions of the Motion of a Fluid Particle
1. Lagrangian Description:-In the Lagrangian description, any fluid property, “F”, is function of the position vector and time (t), i.e., X
r
),( tXFFr
=
dtXd Vr
r=∴ velocity,particle
2
2
on,accelerati particle anddt
XddtVd a
rrr
==
Xr
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Lagrangian and Eulerian Descriptions of the Motion of a Fluid Particle
2. Eulerian Description:-In the Elurian description, any fluid property, “F”, is function of the space coordinates x, y, z, and time (t), i.e., ),,,( tzyxFF =
). velocityparticlecertain a(not
direction - xin the velocity local=∂∂
∴tx
location) z y,certain x, a(at F of change local =∂∂
tF
EulerianEulerian description is used in most fluid mechanics problems.description is used in most fluid mechanics problems.
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121 2 3
ProblemThe temperature, T, in a long tunnel is known to vary approximately as:
where To , α, L, and τ are constants, and x is measured from the entrance of the tunnel.
A particle moves into the tunnel with a constant speed, U. Obtain an expression for the rate of change of temperature experienced by the particle.
)/2sin( τπα teTT Lx
o−
−=
? toequal change of rate required theIstT∂∂
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131 2 3
Substantial Derivative
If we want to find the time rate of change of any fluid property,”F”, following a certain particlecertain particle and still use Eulerian descriction, we have to use what is called a “substantialsubstantial” or “particleparticle” or “materialmaterial””derivativederivative.
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141 2 3
Substantial Derivative
In the Eulerian description:-
If we are following a certain particle:-
),,,( tzyxFF =
dzzFdy
yFdx
xFdt
tFdF
∂∂
+∂∂
+∂∂
+∂∂
=
(1) dt
dt,by devidedtdz
zF
dtdy
yF
dtdx
xF
tFdF
∂∂
+∂∂
+∂∂
+∂∂
=
wdtdzv
dtdyu
dtdx
=== , ,
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151 2 3
Substantial Derivative
(1) dt
dtdz
zF
dtdy
yF
dtdx
xF
tFdF
∂∂
+∂∂
+∂∂
+∂∂
=
wdtdzv
dtdyu
dtdx
=== , ,
FVtF
zFw
yFv
xFu
tFDF
∇+∂∂
=
∂∂
+∂∂
+∂∂
+∂∂
=
.DtDF or,
Dt (1)in
r
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161 2 3
Substantial Derivative – meaning of each term
FVtF
zFw
yFv
xFu
tFDF
∇+∂∂
=
∂∂
+∂∂
+∂∂
+∂∂
=
.DtDF or,
Dt
r
particle.certain aby
dexperienceor seen F of change of rate time total=DtDF
F. of change of rate timelocal=∂∂
tF
particle. theofmotion the todue change of rate change of rate convective.
==∇FV
r
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171 2 3
ProblemThe temperature, T, in a long tunnel is known to vary approximately as:
where To , α, L, and τ are constants, and x is measured from the entrance of the tunnel.
A particle moves into the tunnel with a constant speed, U. Obtain an expression for the rate of change of temperature experienced by the particle.
)/2sin( τπα teTT Lx
o−
−=
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181 2 3
Particle Acceleration
In the Eulerian description:- ),,,( tzyxVVrr
=
paDt
VD onaccelerati particle ==r
onaccelerati local=∂∂
tVr
VVtVa
zVw
yVv
xVu
tVVDa
p
p
rrr
rrrrr
∇+∂∂
=
∂∂
+∂∂
+∂∂
+∂∂
==∴
. or,
(I) Dt
Note:Note:-- Equation (I) is a vector equation, so three equations can be written in the three coordinates.
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191 2 3
Particle Acceleration
VVtVa
zVw
yVv
xVu
tVVDa
p
p
rrr
rrrrr
∇+∂∂
=
∂∂
+∂∂
+∂∂
+∂∂
==∴
. or,
(I) Dt
zuw
yuv
xuu
tu
DtDuapx ∂
∂+
∂∂
+∂∂
+∂∂
==∴ :direction-in x
zvw
yvv
xvu
tv
DtDvapy ∂
∂+
∂∂
+∂∂
+∂∂
==∴ :direction-yin
zww
ywv
xwu
tw
DtDwapz ∂
∂+
∂∂
+∂∂
+∂∂
==∴ :direction-zin
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Types of Motion of A Fluid Particle (Kinematics)
1. Translational.2. Rotation.3. Linear Deformation.4. Angular Deformation.
Fig 5.5
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Deformation and Rotation of a Fluid Element
Rate of Deformation =
Rate of rotation = average rotational speed =
( )tΔΔ+Δ βα
( )tΔΔ−Δ βα
21
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221 2 3
Rotation of a Fluid Element
(1) ,yx Δ
Δ=Δ
ΔΔ
=Δξβηα
tyyututy
yuu ΔΔ
∂∂
=Δ−Δ⎟⎟⎠
⎞⎜⎜⎝
⎛Δ
∂∂
+=Δ ...ξ
txxvtvtx
xvv ΔΔ
∂∂
=Δ−Δ⎟⎠⎞
⎜⎝⎛ Δ
∂∂
+=Δ ...η
(2) , (1)in yu
xv
∂∂
=Δ∂∂
=Δ∴ βα
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231 2 3
Rotation Vector
tΔΔ−Δ
==βαω
21 velocity rotational Average z
⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
−∂∂
=∴yu
xv
21 (2) from zω
ωz = component of the rotation vector about the z-axis.
⎟⎠⎞
⎜⎝⎛
∂∂
−∂∂
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
−∂∂
=xw
zu
zv
yw
21 ,
21
yx ωω
wvuzyx
kji
kωjωiω wyx ∂∂
∂∂
∂∂
==×∇=++==
ˆˆˆ
21V curl
21 V
21ˆˆectorrotation v
rr)rω
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241 2 3
Vorticity Vector
wvuzyx
kji
ω∂∂
∂∂
∂∂
=×∇===
ˆˆˆ
V2vectorvorticity rrr
ζ
wvuzyx
kji
kωjωiω wyx
∂∂
∂∂
∂∂
==
=×∇=++==
ˆˆˆ
21V curl
21
V21ˆˆectorrotation v
r
r)rω
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251 2 3
Fluid Deformation: (1) Angular Deformation
tΔΔ+Δ
=βα planey -in xn deformatioangular of Rate
, but,yu
xv
∂∂
=Δ∂∂
=Δ βα
⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
=∴yu
xvplaney -in xn deformatioangular of Rate
⎟⎠⎞
⎜⎝⎛
∂∂
+∂∂
=∴zu
xwplane z-in xn deformatioangular of Rate
⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
=∴zv
ywplane z-yin n deformatioangular of Rate
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Types of Motion of A Fluid Particle (Kinematics)
1. Translational.2. Rotation.3. Linear Deformation.4. Angular Deformation.
Fig 5.5
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Fluid Deformation: (2) Linear Deformation
tx
tutxΔ
∂∂
=Δ
Δ−Δ⎥⎦⎤
⎢⎣⎡ Δ
∂∂
+=
xu x
uudirection-in xn deformatioLinear
xudirection-in xn deformatiolinear of Rate∂∂
=
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281 2 3
Fluid Deformation: (2) Linear Deformation
xudirection-in xn deformatiolinear of Rate∂∂
=
yv∂∂
=direction-yin n deformatiolinear of rate Similarly,
zw∂∂
=direction-zin n deformatiolinear of rate and
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291 2 3
Rate of Volume Deformation or Dilation
V
zw
yv
r.
xu
dilation or n deformatio volumeof Rate
∇=
∂∂
+∂∂
+∂∂
=
=
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301 2 3
0. .,.
0 0)()()(
=∇
=∂∂
+∂∂
+∂∂
⇒=∂
∂+
∂∂
+∂
∂
Vei
zw
yv
xu
zw
yv
xu
r
ρρρ
Recall - Conservation of Mass - Special Cases
Incompressible Flow:- ρ = constant
0)()()(=
∂∂
+∂
∂+
∂∂
+∂∂
zw
yv
xu
tρρρρ
0 =∂∂
⇒tρ
⇐⇐Compressible Compressible ⇐⇐ , Unsteady, Unsteady
i.e., for an incompressible fluid, rate of volume deformation = 0
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Differential Momentum Equation.
For an incompressible fluids with constant density and viscosity, momentum equations are:-
1) x–direction,
2) y-direction,
3) z-direction,
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ME3O04 – Chapter 6http://mech.mcmaster.ca/~hamedm/me3o04/
11 2
Outline Chapter 6- Incompressible Inviscid Flow
Momentum equation - special cases:Incompressible flow with constant viscosity.Inviscid (frictionless flow), i.e., μ = 0 - Euler’s equation.
Euler’s equations along a streamline.Bernoulli’s Equation. Hydrostatic, Static, Dynamic, and Stagnation Pressures.Applications of Bernoulli’s equation.Energy Grade Line (EGL) and Hydraulic Grade Line (HGL).Example problem.
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21 2
Differential Momentum Equation.
For an incompressible fluids with constant density and viscosity, momentum equations are:-
1) x–direction,
2) y-direction,
3) z-direction,
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Incompressible Inviscid Flow
Incompressible means ρ = constant.
Inviscid means that viscosity μ = 0.
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Differential Momentum Equation For an incompressible fluid (incompressible fluid (ρρ = c)= c) with constant viscosity, momentum equation is:-
1) x–direction,
2) y-direction,
3) z-direction,
VpgDtVD rr
2∇+∇−= μρρOr,
Inviscid ⇒ μ = 0
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51 2
Euler’s EquationFor an incompressible (incompressible (ρρ = c= c)) , inviscid flow inviscid flow ((μμ = 0)= 0), momentum equation is:-
1) x–direction,
2) y-direction,
3) z-direction,
Or,pg
DtVD
∇−= ρρr
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61 2
Euler’s Equation in Streamline Coordinates
Euler’s equations shown in the previous slide are written using x-y-z coordinates.
In steady flow a fluid particle will move along a streamline because, for a steady flow, pathlinesand streamlines coincide.
Thus, in describing the motion of a fluid particle in a steady flow, the distance along a streamline is a logical coordinate to use in writing the equations of motion.
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71 2
Fluid Particle Moving along a Streamline
Fig 6.1
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81 2
Euler’s Equation in Streamline Coordinates1. Apply Newton’s 2nd Law in the s-direction
∑ =+= sBS amFFF
saddgdxdndssppdxdnds
spp ∀=∀−⎥⎦
⎤⎢⎣⎡
∂∂
+−⎥⎦⎤
⎢⎣⎡
∂∂
− ρβρ sin.2
.2
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91 2
Euler’s Equation in S-direction
saddgdxdndssppdxdnds
spp ∀=∀−⎥⎦
⎤⎢⎣⎡
∂∂
+−⎥⎦⎤
⎢⎣⎡
∂∂
− ρβρ sin.2
.2
sVV
tV
DtVDa
szg
sp
s ∂∂
+∂∂
===∂∂
−∂∂
−r
rrr
ρ1 (1)in
dsdz
=βsin but,
(1)
If we neglect the body force and consider only steady flow:-
1 sVV
sp
∂∂
=∂∂
−r
r
ρ
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101 2
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111 2
Euler’s Equation in Streamline Coordinates2. Apply Newton’s 2nd Law in the n-direction
∑ =+= nBS amFFF
naddgdxdsdnnppdxdsdn
npp ∀=∀−⎥⎦
⎤⎢⎣⎡
∂∂
+−⎥⎦⎤
⎢⎣⎡
∂∂
− ρβρ cos.2
.2
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121 2
Euler’s Equation in n-direction
RVa
nzg
np
n
21 (2)in r
−==∂∂
−∂∂
−ρ
dndz
=βcos but,
(2)
an = centripetal acceleration, R = radius of curvature.If we neglect the body force and consider only steady flow:-
1 2
RV
np
r
=∂∂
ρ
naddgdxdsdnnppdxdsdn
npp ∀=∀−⎥⎦
⎤⎢⎣⎡
∂∂
+−⎥⎦⎤
⎢⎣⎡
∂∂
− ρβρ cos.2
.2
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131 2
Euler’s Equation in n-direction
This equation indicates that pressure increases in the direction outwards from the center of curvature of the streamline.
In case of flow in a straight line:-
i.e., there is no variation in pressure in the n-direction.
1 2
RV
np
r
=∂∂
ρ
0 =∂∂
∴∞=npR
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141 2
Measurement of Static Pressure
1 2
RV
np
r
=∂∂
ρ
0 =∂∂
∴∞=npR
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151 2
Measurement of Static Pressure
Can we put the pressure gage at the elbow?Can we put the pressure gage at the elbow?
1 2
RV
np
r
=∂∂
ρ
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161 2
Bernoulli’s Equation
Bernoulli’s equation results from the integration of Euler’s equation along a streamline for a steady flow.
Euler’s equation in s-direction (along a streamline):-
For steady flow:-
sVV
tV
szg
sp
∂∂
+∂∂
=∂∂
−∂∂
−r
rr
ρ1
(3) 1sVV
szg
sp
∂∂
=∂∂
−∂∂
−r
r
ρ
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171 2
Bernoulli’s Equation – Cont’dSince we are moving a long a streamline, thus,
p=p(s,t)ds
spdt
tpds
spdp
∂∂
=∂∂
+∂∂
=∴
dsszdt
tzds
szdz
∂∂
=∂∂
+∂∂
=and,
dssVdt
tVds
sVVd
∂∂
=∂∂
+∂∂
=rrr
rand,
(3) 1 (3) RecallsVV
szg
sp
∂∂
=∂∂
−∂∂
−r
r
ρ
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181 2
Integration of (4) along S:-
For an incompressible fluid ρ = c, in (5):
( )4 0 or,
1 (3)in
=++
=−−
ρ
ρdpdzgVdV
VdVdzgdp
rr
rr
( )5 constant 2
2=++ ∫ ρ
dpzgVr
constant2
2=++
ρpzgV
r⇐ Bernoulli’s equation
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191 2
Restrictions of The Application of Bernoulli’s Equation
Bernoulli’s equation is a very powerful tool, but is has to be
used very carefullyvery carefully. Because it has very strict applicability
limitations.
1. Incompressible flow, i.e., ρ = c, Mach number, Ma < 0.3.
2. Inviscid flow, i.e., μ = 0.
3. Steady flow, i.e.,
4. Along a streamline.
constant2
2=++
ρpzgV
r
0=∂∂t
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201 2
Hydrostatic, Static, Dynamic, and Stagnation Pressures1. Hydrostatic pressure is pressure resulting from weight
of a fluid column. P = ρ g h.
2. Ps=Static pressure is pressure due to the thermodynamic state of the fluid.
3. Pd=Dynamic pressure is pressure due to the velocity of the fluid.
4. Po=Stagnation pressure is pressure exerted by a moving fluid when brought from motion to rest.
Po=Ps + Pd
2
2VPd
rρ
=
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ME3O04 – Chapter 6http://mech.mcmaster.ca/~hamedm/me3o04/
211 2
Measurement of Static Pressure, Ps
Fig 6.2
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221 2
Measurement of Stagnation Pressure
Po = Ps + Pd
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231 2
Simultaneous Measurement of Static and Stagnation Pressures
gzVP
gzVPo
oo ++=++22
sBernoulli’ From22rr
ρρ
2 z,z and 0
2
o
VPPV oo
rr
ρ+=∴==
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241 2
Applications of Bernoulli’s Equation
Bernoulli’s equation can be written for any two points along a same stream line as:
ρρ2
2
221
1
21
22pzgVpzgV
++=++rr
constant2
2=++
ρpzgV
r
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251 2
Measurement of Fluid Velocity at a Point
Apply Bernoulli’s equation between points A and B:
(1) 22
22
BBB
AAA zgVPzgVP
++=++rr
ρρ
o
B
Pzz
===
BB
A
P thus,0V and :Note
2
(1)in 2
ρρoAA PVP
=+r
( )ρ
AoA
PPV −=∴
2
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261 2
Applications – Nozzle Flow
ρρ2
2
221
1
21
22pzgVpzgV
++=++rr
One can relate info at section 2 to those at section 1 using:
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271 2
Flow through a Siphon One can relate info at sections 1, A, and 2 using:
ρρρ2
2
22
21
1
21
222pzgVpzgVpzgV A
AA ++=++=++
rrr
.21 :Note atmPPP ==
0 then,
areaarea Since
1
pipereservoir
≈
>>
VIf one uses Patm =0, this means that pressures are gage.If one uses Patm = 101.3 kPa, this means that pressures are absolute.
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281 2
Hzg
Vg
p=++
2 :gby devide
2r
ρ
Meaning of Each Term in Bernoulli’s Equation
constant2
2=++ zgVp
r
ρ
=∀
=gm
pg
pρ flow energy per unit weight of the flowing fluid=
head due to local static pressure.
2V
2V
22
==mg
m
g
rr
kinetic energy per unit weight = head due to localdynamic pressure.
==gmzgmz Potential energy per unit weight = head
due to elevation.H = total mechanical energy per unit weight
= total head of the flow
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291 2
Example ProblemA tank with a reentrant orifice called a Bordamouthpiece is shown. The fluid is inviscid and incompressible. The reentrant orifice essentially eliminates flow along the tank walls, so the pressure there is nearly hydrostatic. Calculate the contraction coefficient, Cc = Aj/Ao.
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301 2
Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)
EGL is a line representing the total head.
HGL is a line representing the sum of elevation heads and static pressure head.
The difference EGL-HGL = dynamic head.
Hg
Vg
pEGL
HGL
=++
4484476 r
3212
z 2
ρ
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Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)
At point (1) p1 = patm.=0 (gage) and V1 = 0, thus H1 = z1.
At point 4, p4=patm. = 0, thus the height of HGL = z4.
Hg
Vg
pEGL
HGL
=++
4484476 r
3212
z 2
ρ
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Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)
Flow through a constant cross section will have a horizontal HGL (bec. V = c).
Hg
Vg
pEGL
HGL
=++
4484476 r
3212
z 2
ρ
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Example 1A tank with a reentrant orifice called a Bordamouthpiece is shown. The fluid is inviscid and incompressible. The reentrant orifice essentially eliminates flow along the tank walls, so the pressure there is nearly hydrostatic. Calculate the contraction coefficient, Cc = Aj/Ao.
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+
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21 ==∴
o
jc A
AC
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Heavy loads can be moved with relative ease on air cushions by using a load pallet as shown. Air is supplied from the plenum through porous surface AB. It enters the gap vertically at uniform speed, q. Once in the gap, all air flows in the positive x direction (there is no flow in across the plane at x = 0) Assume air flow in the gap isincompressible and uniform at each cross section, with speed u(x) as shown in the enlarged view. Although the gap is narrow (h<<L), neglect frictional effects as a first approximation. Use a suitably chosen control volume to show that u(x) = qx/h in the gap. Calculate the acceleration of a fluid particle in the gap. Evaluate the pressure gradient dp/dx, and sketch the pressure distribution within the gap. Be sure to indicate the pressure at x=L.
Example 2
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Water is admitted at the center pipe of the platform shown below at a rate of 1 m3/s and discharged into the air around the periphery. The upper circular plate in the figure is horizontal and is fixed in position to the ceiling. The lower annular plate is free to movevertically and is not supported by the pipe. The annular plate weighs 30 N, and the weight of water on it should be considered. If the distance, d, between the two plates is to be maintained at 3.5 cm, what is the total weight W that this platform can support?
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Water is discharged from a narrow slot in a 150 mm diameter pipe. The resulting horizontal two-dimensional jet is 1 m long and 15 mm thick, but of non-uniform velocity. The pressure at the inlet section is 30 kPag. Calculate (a) the volume flow rate at the inlet section and (b) the forces required at the coupling to hold the spray pipe in place. Neglect the mass of the pipe and the mass of water it contains.
Coupling
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1 2
Outline Chapter 7- Dimensional Analysis and Similitude
Meaning of Similitude.Dimensionless numbers.Methods of dimensional analysis:-
1. Nondimensionalizing the basic differential equations.2. Using Buckingham PI Theorem.
Buckingham PI Theorem. Procedure to determine the PI groups (illustrative example - Drag force on a sphere).Significant Dimensionless Groups in Fluid Mechanics.Significant Dimensionless Groups in Fluid Mechanics.Flow Similarity and Model Studies.Scaling with Multiple Dependent Parameters.Scaling with Multiple Dependent Parameters.Example Problem
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Similitude
????
ghV 2
One of the important dimensionlessdimensionless
numbers in this problem =
mms
sm
ghV 12
2
22=
= dimensionless
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Similitude
????V2
2
22
1
21
ghV
ghV
=
21
1
222 V
hhV =
VVV 22 i.e., 12 ==2
12
2 2 VV =
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Roasting Time of a TurkeyWhat are the parameters affecting roasting time of a turkey (t) ?
1. Mass, m2. Density, ρ3. Thermal conductivity, k4. Specific heat, cp
Using dimensional analysis we can show that:-Or,
kmc
t p3 2
constant ρ
=
3 2mckt
p ρ=constant = dimensionless number
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Roasting Time of a Turkey
2 no. bird3 2
1 no. bird3 2 ⎟
⎟
⎠
⎞
⎜⎜
⎝
⎛=⎟
⎟
⎠
⎞
⎜⎜
⎝
⎛
mckt
mckt
pp ρρ
So, if we know the cooking time of bird no.1, we can calculatethe cooking time of bird no. 2 from this equation.
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Drag Force on a Sphere Drag force on a sphere, Fdepends on:-1. Diameter, D2. Fluid density, ρ3. Fluid viscosity, μ4. Fluid velocity, U
Dimensional analysis shows that this problem is governed by two dimensionless numbers :
f that And 22 ⎟⎟⎠
⎞⎜⎜⎝
⎛=
μρVDρ
DVF
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Does the size of sphere (i.e., D) matter in using this figure?Does the type of fluid (i.e., ρ and μ) matter in using this figure?
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Similitude
So, if we know which dimensionless parameters are important in a problem, all other similar problems can be dealt with easily, or we should say, similarlysimilarly.
The “QUESTION” now is how can we determine important “Dimensionless NumbersDimensionless Numbers” in any problem of interest?
Methods to do that:1. Nondimensionalizing basic differential equations.2. Using Buckingham PI Theorem.
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Nondimensionalizing the Basic Differential EquationsExample: Steady, incompressible, two-dimensional
flow, of a Newtonian fluid, with constant viscosity.
Assumptions:
1. Steady
2. Incompressible
3. Two-dimensional
4. Newtonian fluid with constant μ.
0=∂∂
⇒t
c=⇒ ρ
0==∂∂
⇒ wz
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For an incompressible fluid with constant viscosity, momentum equations are:-
1) x–direction,
2) y-direction,
3) z-direction,
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Momentum Equations after Applying all Other Assumptions
(3) direction y
(2) direction x
2
2
2
2
2
2
2
2
⎥⎦
⎤⎢⎣
⎡
∂∂
+∂∂
+−∂∂
−=⎥⎦
⎤⎢⎣
⎡∂∂
+∂∂
−
⎥⎦
⎤⎢⎣
⎡
∂∂
+∂∂
+∂∂
−=⎥⎦
⎤⎢⎣
⎡∂∂
+∂∂
−
yv
xvg
yp
yvv
xvu
yu
xu
xp
yuv
xuu
μρρ
μρ
1) x–direction,
2) y-direction,
3) z-direction,
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Non- Dimensionalizing Equations 1 and 2
Non-dimensionalizing these equations means :
1. To divide all lengths by a reference length, L;
2. To divide all velocities by a reference velocity, V∞;
Note: dimensionless quantities are denoted with asterisks:
Lyy
Lxx == ** ,
∞∞
==Vvv
Vuu ** ,
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Non- Dimensionalizing Equations 1 and 2
3. To divide pressure by twice the dynamic head = ρV∞
2.
Note: dimensionless quantities are denoted with asterisks:
2*
∞
=Vpp
ρ
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(3)
:directiony Momentum,
(2)
:direction xMomentum,
2*
*2
2*
*2
2*
*
*
**
*
**
2*
*2
2*
*2
*
*
*
**
*
**
⎥⎦
⎤⎢⎣
⎡
∂∂
+∂∂
+−∂∂
−=∂∂
+∂∂
−
⎥⎦
⎤⎢⎣
⎡
∂∂
+∂∂
+∂∂
−=∂∂
+∂∂
−
∞∞
∞
yv
xv
LVVgL
yp
yvv
xvu
yu
xu
LVxp
yuv
xuu
ρμ
ρμ
Equations in Dimensionless Form
number ReynoldsRe == ∞
μρ LV
number Froud2== ∞
LgVFr
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RecallRecall - Similitude
So, if we know which dimensionless parameters are important in a problem, all other similar problems can be dealt with easily, or we should say, similarly.
The “QUESTION” now is how can we determine these important “Dimensionless NumbersDimensionless Numbers” in any problem of interest?
Methods to do that:1. Nondimensionalizing the basic differential equations.2. Using Buckingham PI Theorem.
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Methods to find the Dimensionless Numbers Relevant to a Certain Problem of Interest
The method of non-dimensionalizing differential equations depends on knowing which equations to use, which is not always the case.
If we do not know the equations, we use the second method known as the ““Buckingham PI TheoremBuckingham PI Theorem”” .
Before we discuss the Buckingham PI Theorem, we need first to discuss two important concepts:-
1. Fundamental, or Independent, or Primary Dimensions.2. Dimensional homogeneity.
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Fundamental, or Independent, or Primary Dimensions
Dimensions of all physical quantities can be expressed in terms of a group of ““independent independent or primary primary dimensionsdimensions””.
These primary dimensions are: 1. Mass, M;2. Length, L;3. Time, t;4. Temperature, T.Examples:-
1. Velocity −=== tLtLV
,. 22
−=== tLtL
tVonAccelerati
22 ... −=== tLM
tLMmaForce
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Dimensional HomogeneityAny valid equation that relates physical quantities must be dimensionally homogeneousdimensionally homogeneous, i.e., each term in the equation must have the same dimensions.
Example: Newton’s Law of Viscosity
(1) yu∂∂
= μτ
LtM
LtLM
.1..
areaforce (1) of L.H.S. 222 ===τ
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Dimensional Homogeneity
(1) yu∂∂
= μτ
LtM
LtL
Lt
tLM
mN.sμ
.1... (1) of R.H.S. 2222 ==∴=Q
0y)u,,,(f i.e.,
,01
yu :as written becan (1)Equation
=
=−
∂∂
μτ
μ
τ
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Buckingham PI TheoremThis theorem states that: Given a relation among nn
parameters, q1, q2, …, qn, of the form:
The nn parameters may be grouped into (nn--mm) independent dimensionless groups or ratios (πparameters), expressible in functional form by:
where m m = rr = the minimum number of independent dimensions required to specify the dimensions of the n n parameters.
0),.....,,( 21 =nqqqg
0),.....,,( 21 =−mnG πππ
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Illustrative example: Drag force on a sphere.Step 1
List all dimensional parameters involved in the problem and determine n. Where n is the number of dimensional parameters.Drag force on a sphere, FD depends on:-1. Diameter, D2. Fluid density, ρ3. Fluid viscosity, μ4. Fluid velocity, U
Procedure to Determine the π Groups
0),,,,( =ρμUDFg D ∴ n = 5
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Procedure to Determine the π Groups –Cont’d
Step 2 Select a set of primary dimensions = M, L, t, TM, L, t, TStep 3 Construct the ““Dimensional MatrixDimensional Matrix”” by listing all
parameters in terms of the primary dimensions.
000000110231111
11001
−−−−−
TtLM
FD D U μ ρ
““Dimensional MatrixDimensional Matrix””33
11
1
212
,
,
,
−
−−
−
−−
==
==
==
=
==
LMLM
tLMtL
M
tLtLU
LD
tLMtL
MF
ρ
μ
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Procedure to Determine the π Groups –Cont’d
Step 4 Determine the rank of the dimensional matrix, r.where the rank, r, is the order of the largest non-zero determinant in the matrix.
000000110231111
11001
−−−−−
TtLM
FD D U μ ρ
∴ r = 3
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Procedure to Determine the π Groups –Cont’dStep 5 Select a set of r dimensional parameters that
includes all the primary dimensions used in step 3 (i.e., M, L, and t).
Since r = 3, then select [ U, D, ρ ]
These parameters are called the repeatingrepeating parameters.Note: It is common to select a velocitya velocity, a dimensiona dimension, and a a
fluid propertyfluid property.
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Procedure to Determine the π Groups –Cont’dStep 6 Set dimensional equations using the repeating
parameters and one of the other parameters, one-at-a-time.
Solving for the exponents:of t: -a -2 = 0 ⇒ a = -2of M: c + 1 = 0 ⇒ c = -1of L: a + b -3c + 1 = 0 ⇒ b = -2
...1 == Dcba FDU ρπ
221 DUFD
ρπ =∴ = drag coefficient
000 .. tLM=23 ... t
LMLML
tL c
ba
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
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Procedure to Determine the π Groups –Cont’dStep 6 Set dimensional equations using the repeating
parameters and one of the other parameters, one-at-a-time.
Solving for the exponents:of t: -a -1 = 0 ⇒ a = -1of M: c + 1 = 0 ⇒ c = -1of L: a + b -3c -1 = 0 ⇒ b = -1
...2 == μρπ cba DU
Re1
2 ==∴DUρ
μπ
000 .. tLM=tLM
LML
tL c
ba
.... 3 ⎟⎠
⎞⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
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Procedure to Determine the π Groups –Cont’d
Step 7 Check that each π group is dimensionless.
The use of Buckingham theorem in the problem of drag force on a sphere shows there are two two important dimensionless numbers in this problem, which are:
But how these two groups relate, needs to be determined experimentally.
22 DUFD
ρdrag coefficient =
μρ DUand, Reynolds number =
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Relationship between Drag Coefficient and Reynolds number
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Buckingham PI Theorem
Important points to note:-
Buckingham theorem allows us to determine:-1. The number of π groups (dimensionless numbers)
involved in the problem.2. The form of each one of these dimensionless
numbers (π’s).
But, it does not allow us to determine the form of the function, , which has to be determined experimentallyexperimentally.
0),.....,,( 21 =−mnG πππ
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Forces encountered in Fluid MechanicsForces encountered in flowing fluids include forces
due to:1. Inertia2. Viscosity3. Pressure4. Gravity5. Surface tension6. Compressibility.The ratio of any two forces will be dimensionless, and defines a significant dimensionless number in Fluid Mechanics.
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Dimensions of These Forces
1. Inertia force =
2. Viscous force =
3. Pressure force =
( ) , VLbut t ,.. 3 =⎟
⎠⎞
⎜⎝⎛=
tVLam ρ
LVLLVA
yuA μμμτ ==∂∂
= 2...
( ) 222
3 .m.a Thus, VLL
VL ρρ =⎟⎟⎠
⎞⎜⎜⎝
⎛=
2.. LpAp Δ=Δ
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Dimensions of These Forces –Cont’d4. Gravity force =
5. Surface tension force = , where, σ = surface tension = force per unit length.
6. Compressibility force = , where Ev is the compressibility modulus, or modulus of elasticity.
2.. LEAE vv =
L.σ
stressareaforce
==⎟⎠⎞⎜
⎝⎛
=
ρρddpEv
( ) gLgm .. 3ρ=
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Significant Dimensionless Groups in Fluid Mechanics
1.
2.
3.
4.
number Reynolds Reforce viscousforce inertia 22
====μ
ρμρ LV
LVVL
numberEuler Eu
21force inertia
force pressure2
22
2
==Δ
=Δ
=V
pVLLp
ρρ
( )222
3
22
number Froudforcegravity force inertia
==== FrgLV
gLVL
ρρ
numberWeber Weforce tension surface
force inertia 222
====σ
ρσ
ρ VLLVL
1.
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Significant Dimensionless Groups in Fluid Mechanics – Cont’d
5.
and C = speed of sound =
Note:- in case of incompressible flow, ρ = constant, i.e., dρ = 0. Thus, C = ∞ ⇒ Ma = 0.
222
2
22
forceility compressibforce inertia Ma
EV
EV
LEVL
vvv
====ρ
ρρ
numberMach Ma where, ====CV
ddpV
EV
v
ρρ
ρddp
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Significant Dimensionless Groups in Fluid Mechanics – Cont’d
6.
where, p = pressure in liquid stream.pv = vapor pressure of liquid.
Note:- As Ca ↓, the more likely cavitation to occur.
number Cavitation
21force inertia
force pressure2
22
2
==−
=Δ
= CaV
ppVLLp v
ρρ
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Flow Similarity and Model Studies
A model test must yield data that can be scaled to obtain information of interest on the full-scale prototype.
To be able to do that, there are conditions that have to be met to ensure similarity of model and prototype flow.
Model and prototype must have:-1. Geometrical similarity.2. Kinematic similarity.3. Dynamic similarity.
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Geometrical Similarity
Geometrical similarity requires that:-
1. The model and prototype be of the same shape.
2. All linear dimensions of the model related to corresponding dimensions of the prototype by a constant scale factorconstant scale factor.
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Kinematic SimilarityTwo flows are kinematically similar when the velocities at corresponding points are in the same direction and differ by a constant scale factor.
Thus two kinematically similar flows have streamlines related by a constant scale factor.
Since the boundaries form the boundary streamlines of the flow, flows that are kinematically similar must be geometrically similar.
i.e., geometrical similarity is a prerequisite for kinematic similarity.
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Dynamic Similarity
Dynamic similarity requires that Identical types of forces to be :-1. Parallel, and2. Related in magnitude by a constant scale.
For condition 1, dynamic similarity requires kinematic similarity, hence geometric similarity too.
For condition 2, each independent dimensionless group (force ratio) must have the same value in the model and the prototype.
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Example – Drag Force on a Sphere
),,,( ρμUDfF =
⎟⎟⎠
⎞⎜⎜⎝
⎛==
μρ
ρDVff
DUF
2222 (Re)Drag coefficient =
Dynamic similarity is achieved if we use a model sphere(could be smaller or bigger) and keep:-
prototypemodelprototypemodel i.e., ,ReRe ⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
μDVρ
μDVρ
prototype22
model22 and, ⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛
DUF
DUF
ρρ
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1 2
Example – Drag Force on a Sphere –Cont’d
Equations (1) and (2) state that:-
1. We do not need to use the same fluid to test the model.
2. The resulted drag force in the model will not be equal to the drag force in the prototype. However, the drag coefficients are the same.
(1) prototypemodel⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛μ
DVρμ
DVρ
(2) prototype
22model
22 ⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛
DUF
DUF
ρρ
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1 2
Scaling with Multiple Dependent Parameters
In the example of drag force on a sphere, we were interested in one dependent parameterdependent parameter, which is the drag force.
In some practical applications there might be more than one dependent parameter of interest.
In such cases, dimensionless groups must be formed separately for each one of those dependent parameters.
Example – performance of a typical centrifugal pump.
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1 2
Centrifugal Pump or Fan
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1 2
Example for Scaling with Multiple Dependent Parameters
Performance of a typical centrifugal pump.
In this case, dependent parameters of interest are:-1. Pressure rise or head developed by the pump, h.2. Power input required to derive the pump, P.
We are interested to know how h and P depend on:-1. Volume flow rate, Q.2. Angular speed, ω.3. Impeller diameter, D.4. Fluid properties ρ and μ.
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1 2
Performance of a Typical Centrifugal Pump
If we apply Buckingham Theorem, we can find out that:-
),,,,( ),,,,(
2
1
DQf PandDQfh
μρωμρω
==∴
⎟⎟⎠
⎞⎜⎜⎝
⎛=
μωρ
ωω
2
3122 , DDQf
Dh
⎟⎟⎠
⎞⎜⎜⎝
⎛=
μωρ
ωωρ
2
3253 , and, DDQf
DP
22 where,Dh
ω= head coefficient
=53 ,D
Pωρ
power coefficient
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1 2
Performance of a Typical Centrifugal Pump
If we apply Buckingham Theorem, we can find out that:-
),,,,( ),,,,(
2
1
DQf PandDQfh
μρωμρω
==∴
⎟⎟⎠
⎞⎜⎜⎝
⎛=
μωρ
ωω
2
3122 , DDQf
Dh
⎟⎟⎠
⎞⎜⎜⎝
⎛=
μωρ
ωωρ
2
3253 , and, DDQf
DP
=3DQ
ωflow coefficient,
( )μ
ρμωρ
μωρ DVDDD
≡=2
and is a form of Re number
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1 2
Performance of a Typical Centrifugal Pump – Cont’d
From practice it has been found that viscous effects are relatively unimportant w.r.t. inertial effects.
So, we can exclude Re number, and thus:
⎟⎟⎠
⎞⎜⎜⎝
⎛= 3
'122 D
QfDh
ωω ⎟⎟⎠
⎞⎜⎜⎝
⎛= 3
'253 and,
DQf
DP
ωωρ
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1 2
Performance Curves of a Typical Centrifugal Pump
⎟⎟⎠
⎞⎜⎜⎝
⎛= 3
'122 D
QfDh
ωω
⎟⎟⎠
⎞⎜⎜⎝
⎛= 3
'253 D
QfD
Pωωρ
HorsepowerShaft Efficiency P
=Fig. 7.5
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1 2
Similarity in Pump Performance
Complete similarity in pump performance test would require:-
222
22
11
1
DQ
DQ
ωω=
52
322
25
13
11
1
DP
DP
ωρωρ=
22
22
22
12
1
1
Dh
Dh
ωω=
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1 2
Example ProblemThe drag of an airfoil at zero angle of attack is a function of density, viscosity, and velocity, in addition to a length parameter.
A 1/10-scale model of an airfoil was tested in a wind tunnel at a Reynolds number of 5.5 X 106, based on chord length. Test conditions in the wind tunnel air stream were 15°C and 10 atmospheres absolute pressure. The prototype airfoil has a chord length of 2 m, and it is to be flown in air at standard conditions.
Determine the speed at which the wind tunnel model was tested, and the corresponding prototype speed.
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1 2
α = angle of attack.
α = 0
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Outline Chapter 8- Internal Incompressible Viscous FlowClassification of Continuum Fluid Mechanics.Internal Incompressible Viscous Flow.Flow Regimes – Laminar and Turbulent.Boundary Layer and meaning of fully developed flow. Fully developed, Laminar Flow between Infinite Parallel Plates.Flow in Pipes and Ducts:-1. Velocity Profiles in fully developed pipe flow.2. Turbulent velocity profiles in fully developed pipe flow -
“Power Law” .3. Calculation of head loss:-
a. Major losses.b. Minor losses.
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1 2
Classification of Continuum Fluid Mechanics
To be coveredin 4th year
Chapters 8 and 9
444 3444 21 Chapter 6
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Internal Incompressible Viscous Flow
Internal means that the flow is completely bounded by solid surfaces.
Examples:- flow in nozzles, ducts, diffusers, pipes, etc.
Flow Regimes:- internal flows can be classified, based on the flow regime, into:-
1. Laminar flow: fluid flows in layers or laminas.
2. Turbulent flow: flow is characterized by high-frequency velocity fluctuations.
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a) Laminar Flow (Re < 2300). b) Turbulent Flow (Re > 2300).
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Osborne Reynolds Experiments
a) Laminar Flow (Re < 2300). b) Turbulent Flow (Re > 2300).
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Boundary LayerViscosity is responsible for what we call the no-slip condition.
Therefore, as fluid approaches a solid surface and due to the no-slip condition, a region of significant deformation, i.e., significant velocity gradient is formed.
This region is called the boundary layer.
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Entrance Regionthickness of boundary layer
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Fully Developed FlowWhen the boundary layer reaches its maximum thickness the velocity distribution in the direction of the velocity distribution in the direction of flow does not change anymoreflow does not change anymore, at which case, the flow is said to have been fully developedfully developed.
thickness of boundary layer
u(r)u i.e., ,0.,. ==∂∂
xuei x)u(r,u i.e., ,0.,. =≠
∂∂
xuei
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Entrance length = Development length, Lfor Flow in a Pipe.
1. For Laminar Flow:
e.g., at Re = 2300, L = 138 D. where D = pipe diameter
2. For Turbulent Flow: L = 80 D
Re06.0=DL
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Fully-Developed Laminar Flows
Common cases are:-
1. Fully developed laminar flow between infinite parallel plates –two cases:-
a. Both plates stationary.b. Upper plate moving with constant speed , U.
2. Fully developed laminar flow in a pipe.
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Fully Developed Laminar Flow Between Infinite Parallel Plates –upper plate moving with constant speed U.
Assumptions:-
1. Steady flow, i.e.,
2. Incompressible flow, i.e., ρ = constant.
3. Constant viscosity, i.e., μ = constant.
4. If L>> a, then flow is fully developed, i.e.,
where, L = length of plates and a = height of gap between plates.
5. Infinite plates means , i.e., flow is two-dimensional.
6. Neglect body forces in x and y directions.
0=∂∂t
0 u(y)uxu
=≡=∂∂
0=∂∂z
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Fully Developed Laminar Flow Between Infinite Parallel Plates – upper plate moving with constant speed U.
Boundary conditions:-
1. at y = 0, u = v = 0.
2. at y = a, u = U, v = 0.
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Analysis
For this type of flow, we would like to determine the following :-
1. Velocity distribution, 2. Pressure distribution.3. Shear stress distribution.4. Volume flow rate.5. Average velocity.6. Point of maximum velocity.
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1 2
Analysis – Differential Equations
1. Conservation of mass:-
Applying assumptions number 1, 5 , and 2:-
0)()()(=
∂∂
+∂
∂+
∂∂
+∂∂
zw
yv
xu
tρρρρ
0=∂∂
+∂∂
yv
xu 0 =
∂∂
⇒yv
constant
)(
=
=
⇒or
xf
v
)( any x at 0 Since xfvv ≠⇒=.everywhere 0constant ==∴ v
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Analysis – Differential Equations – cont’dFor an incompressible fluid (incompressible fluid (ρρ = c)= c) with constant viscosity, momentum equation is:-
Momentum equation in y - direction:-
⎥⎦
⎤⎢⎣
⎡
∂∂
+∂∂
+∂∂
++∂∂
−=⎥⎦
⎤⎢⎣
⎡∂∂
+∂∂
+∂∂
+∂∂
2
2
2
2
2
2
zv
yv
xvg
yp
zvw
yvv
xvu
tv
y μρρ
VpgDt
VD rr
2∇+∇−= μρρ
constant
)(0
=
=⇒=
∂∂
∴ orxp
pyp
Can not be constant. If it is constant, there will be no flow.
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Analysis – Differential Equations – cont’dMomentum equation in x-direction
⎥⎦
⎤⎢⎣
⎡
∂∂
+∂∂
+∂∂
++∂∂
−=⎥⎦
⎤⎢⎣
⎡∂∂
+∂∂
+∂∂
+∂∂
2
2
2
2
2
2
zu
yu
xug
xp
zuw
yuv
xuu
tu
x μρρ
(1) 2
2
yu
xp
∂∂
=∂∂
∴ μ
)( and )( Recall yuuxpp ==⇒
constant (1)equation of R.H.S. and or
f(y)
constant (1)equation of L.H.S. or
f(x)∴
)()( yfxf ≠
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1 2
(2) say constant, 12
2c
yu
xp
==∂∂
=∂∂
∴ μ
(3) . 211 cxcpcxp
+=⇒=∂∂
∴
2
1
at 0at
p L p xp px
====
Lp
Lpppc Δ
−=−
−==∴ 21112 c and
1 (3)in pxLpp +Δ
−=∴ ⇐ Pressure distribution
Pressure Distribution
Note that p is function of x.
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Velocity Distribution
Lpc
xp Δ
−==∂∂
1 but
(1) -Recall 2
2
yu
xp
∂∂
=∂∂ μ
L p- (1)in 2
2
μΔ
=∂∂yu
3CyL p- +
Δ=
∂∂
∴μy
u
U a u u
====
yat 00yat
( ) ( )22
21
2yya
xpu
aUyay
Lpy
aUu −
∂∂
−=−Δ
+=μμ
(4) CyL 2
p- u and 432 Cy ++
Δ=
μ
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Shear Stress Distribution
yu
yx ∂∂
= μτQ
( )yaLp
aU 2
2−
Δ+=∴ μτ
( )ya−Δ= 2y
L 2p-y
aU u and
μ
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Volume Flow Rate, Q
WdyyuAdVQa
A∫∫ ==0
.)(.rr
( ) dyWyayL
pyaUQ
a.
20
2∫ ⎥⎦
⎤⎢⎣
⎡−
Δ−=
μ
WL
apUaQ ⎥⎦
⎤⎢⎣
⎡ Δ+=∴
μ122
3
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Average Velocity
aWW
LapUa
AQuav
1122
3×⎥
⎦
⎤⎢⎣
⎡ Δ+==
μ
LapUuav μ122
2Δ+=∴
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1 2
Point of Maximum VelocityPoint of maximum velocity is defined by:-
0 when i.e., ,0 yx ==∂∂ τ
yu
( )pa
LUayyaLp
aU
Δ+=⇒=−
Δ+=
μμτ2
022
Note: - Point of maximum velocity is not at y =a/2.
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Velocity Distribution
Velocity distribution will change as pressure gradient changes.
( )yayxp
−∂∂
+= 2
21y
aU u
μ
Pressure decreases Pressure decreases to the right. to the right.
Pressure increases Pressure increases to the right, i.e., decreasesto the right, i.e., decreasesto the left. to the left.
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Velocity distribution will change if the upper plate is not moving
Pressure decreases Pressure decreases to the right. to the right.
( )yayxp
−∂∂
+= 2
21y
aU u
μ
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1 2
For an incompressible, InviscidInviscid (i.e., μ = 0) flow, Bernoulli’s equation can be applied between points 1 and 2 as:-
Flow in Pipes and Ducts
(1) constant 22 2
222
1
211 =++=++ zgVpzgVp
rr
ρρ
inviscid flow
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Due to friction, i.e., due to shear stress at the wall, we do not expect the right hand side of Bernoulli’s equation to remain constant for an incompressible viscid flow.
Flow in Pipes and Ducts
(2) 22 2
222
1
211
lThzgVpzgVp+++=++
rr
ρρ
viscid flow
where hlT = total energy loss per unit mass.
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Velocity Profiles for Fully Developed Pipe Flow
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Velocity Profile in Fully Developed Pipe Flow – Laminar Flow
1. Laminar Flow:-
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−
∂∂
−==22
14
)(Rr
xpRruu
μ
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Velocity Profile in Fully Developed Pipe Flow – Turbulent Flow
2. Turbulent Flow:- velocity vector in case of fully developed turbulent flow in a pipe can be represented by:
time-mean velocity.u’ and v’ are fluctuating velocity components in x- and y-directions, respectively
jviuuV ˆˆ)( ′+′+=r
=u where,
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Velocity Profile in Fully Developed Pipe Flow – Turbulent FlowThe velocity profile for turbulent flow through a smooth pipe may be approximated by the empirical power-law
where, U = maximum velocity = velocity at the centerline.
for ReU > 2x104, n = -1.7+1.8 log ReU
ReU = Reynolds number calculated using U =
(1) 1/1 n
Rr
Uu
⎟⎠⎞
⎜⎝⎛ −=
μρ DU
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Energy Equation for Viscid Flow in Pipes
Equation (I) is not the known Bernoulli’s equation. This equation has two major differences:-
1. hlT = total energy loss per unit mass, which takes into account energy loss due to friction between points 1 and 2.
2. Kinetic energy coefficients α1 and α2, which allow us to use average velocities V1 and V2.
α = 2 in case of laminar flow, and = 1 in case of turbulent flow.
(I) 22 2
22
22
1
21
11
lThzgVpzgVp+++=++
rr
αρ
αρ
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Solution of Pipe Flow Problems
The energy conditions relating conditions at any two points 1 and 2 for a single-path pipe system (i.e., no branching) is:
where hlT = total energy loss per unit mass.
Δhpump= head caused by a pump=
α = kinetic energy coefficient = 2.0 for laminar flow, Re < 2300.= 1.0 for turbulent flow Re ≥ 2300.
(II) 2
h2 2
22
22
pump1
21
11
lThzgVpzgVp+++=Δ+++
rr
αρ
αρ
mWp pumppump
&
&==
Δ
rate flow Masspower Pump
ρ
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1 2
Calculation of Head Loss, hlT
hlT = total energy loss per unit mass due to friction, is calculated from:
where∑hl = sum of major losses due to frictional effects in fully
developed flow in constantconstant--area sectionsarea sections.
∑hlm=sum of minor losses due to changes in flow direction flow direction and cross section areaand cross section area, e.g., entrances, elbows, contractions, etc.
∑ ∑+= lmllT hhh
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1 2
Calculation of Major Losses, ∑hl
For each straight part of the pipe system
Where, D = pipe diameter, L = pipe length (length of straight part), f = friction coefficient, which can be calculated from:-1. Moody chart, or2. The following equations:
2
2av
lV
DLfh =
Re64 2300 Re i.e., flow,laminar for =< f
⎥⎦
⎤⎢⎣
⎡+−=≥ 5.0Re
51.27.3
/log0.21 2300 Re i.e., flow,Turbulent for f
Def
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Moody Chart – Fig 8.12
e = surface roughness
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Surface Roughness, e
Depends on pipe material – Table 8.1
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1 2
Calculation of Major Losses, ∑hlm
These are additional losses encountered, primarily as a result of flow separation (due to change in flow direction and/or change in cross section area) in pipe fittings.Depending on the type of fitting, minor losses are computed in one of two ways:-1. Using k = the loss coefficient, which is determined
experimentally.
2. Or, using Le = the equivalent length of straight pipe, which is also determined experimentally.
2
2av
lmVkh =
2
2ave
lmV
DLfh =
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1 2
Inlets and Exits – Table 8.2
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Sudden Enlargements and Contractions –Fig 8.14
Note the difference in the velocity to be used in each case
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1 2
Gradual Contractions – Nozzles
2
22av
lmVkh =
Value of k given in Table 8.3
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1 2
Gradual Enlargements - Diffusers
Where, Cp is the pressure recovery coefficient – from Fig 8.15. Cpi is the ideal pressure recovery coefficient.
AR = area ratio, to calculated as shown below
( )2
21av
ilmVCpCph −=
211
ARCpi −=
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1 2
Fig 8.15
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1 2
Bends – Fig 8.16
2
2ave
lmV
DLfh =
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1 2
Valves and Other Fittings – Table 8.4
2
2ave
lmV
DLfh =
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Pumps, Fans, and Blowers
ρpump
pumpp
hΔ
=Δ
QW
p pumppump &
&=Δ
mW
QW
h pumppumppump &
&
&
&==Δ∴
ρ
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46ME3O04 – Chapter 8http://mech.mcmaster.ca/~hamedm/me3o04/
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Non-Circular Ducts
Instead of using D, we use the hydraulic diameter, Dh, defined by:-
Where, A = cross-section area and P = wetted perimeter.
For a rectangle with two sides a and b, A = a x b, P = 2(a+b), thus:
For a square a = b, thus:
PADh
4=
)(2
)(24
baba
babaDh +
=+
=
aDh =
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47ME3O04 – Chapter 8http://mech.mcmaster.ca/~hamedm/me3o04/
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Example ProblemWater is pumped at the rate of 2 ft3/S from a reservoir 20 ft above a pump to a free discharge 90 ft above the pump. The pressure on the intake side of the pump is 5 psig and the pressure on the discharge side is 50 psig. All pipes are commercial steel of 6 in. diameter. Determine (a) the head supplied by the pump and (b) the total head loss between the pump and point of free discharge.
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ME3O04 - Chapter 9http://mech.mcmaster.ca/~hamedm/me3o04/
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Outline Chapter 9- External Incompressible Viscous Flow
The Concept of Boundary LayerEffect of Boundary Layer on a Blunt BodyEffect of Boundary Layer on a Streamlined Body
Boundary Layer Thicknesses:1. Disturbance Thickness, δ992. Displacement Thickness, δ*Example Problem.Fluid Flow about Immersed BodiesDrag and Lift.Types of Drag.CD and CL.
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External Flows
External flows are flows over bodies immersed in an unbounded fluid.
Examples of external flows are the flow fields around such objects as airfoils, automobiles, and airplanes.
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Definition of Boundary Layer
The boundary layer is the region adjacent to a solid surface in which viscous stresses are present.
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Boundary Layer Thicknesses
1. Disturbance Thickness, δ99
2. Displacement Thickness, δ*
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Disturbance Thickness, δ99
δ99 is the boundary layer thickness at which u equals to 99% of the free stream velocity, U.
In other words, it is the distance from the surface at which the velocity is within 1 % of the free stream
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Displacement Thicknesses , δ*
What do we do to make both mass flow rates equal?
h h
∫=h
AdU0
flow massr
ρ ∫=h
Adu0
flow massr
ρ>
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Displacement Thicknesses , δ*
Raise the plate a distance δ* so that mass flow rates are equal.The displacement thickness, δ*, is the distance the plate would be moved so that the loss of mass flux (due to reduction in uniform flow area) is equivalent to the loss the boundary layer causes.
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Displacement Thicknesses , δ*
∫ ∫∞ ∞
==0 0
W.dy UU B.L noflux with mass ρρ Adr
∫ ∫∞ ∞
==0 0
W.dy B.Lflux with mass uAdu ρρr
( ) ( ) (1) W.dy flux massin difference0 0∫ ∫∞ ∞
−=−= uUAduU ρρr
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Displacement Thicknesses , δ*
(2)W flux massin difference *δρ U=
∫∫ ⎟⎠⎞
⎜⎝⎛ −=⎟
⎠⎞
⎜⎝⎛ −=∴=
∞ δ
δ00
* dy 1dy 1 (2)(1) fromUu
Uu
Note: W is depth normal to paper.
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The Use of Displacement Thicknesses , δ*, in Practical Applications
One can use Bernoulli’s equation to design or determine the pressure drop in a duct by reducing duct dimensions by 2δ* as shown above.
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Example Problem
Laboratory wind tunnels have test sections 1 ft square and 2 ft long. With nominal air speed U1 = 80 ft/s at the test section inlet, turbulent boundary layers form on the top, bottom, and side walls of the tunnel. The boundary-layer thickness is δ1 = 0.8 in. at the inlet and δ2 = 1.2 in. at the outlet from the test section. The boundary-layer velocity profiles are of power-law form, with:
a) Evaluate the freestream velocity, U2, at the exit from the wind-tunnel test section.b) Determine the change in static pressure along the test section.
7/1
⎟⎠⎞
⎜⎝⎛=δy
Uu
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Fluid Flow about Immersed Bodies
Source: Fluid Mechanics by Douglas et.al., Prentice Hall, 2001.
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Forces from the Surrounding Fluid on a Two-Dimensional Object
(a) Pressure force (b) Viscous (shear or friction) force
Source: Fundamentals of Fluid Mechanics by B. R. Munson et. al., Wiley, 1994.
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Resultant Force
Source: Fluid Mechanics by Douglas et.al., Prentice Hall, 2001.
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Drag and Lift
Source of fig: Fluid Mechanics by Douglas et.al., Prentice Hall, 2001.
Net force F is resolved into:-1. The drag force, FD, defined as the component of the force parallel to
the direction of motion, and
2. The lift force, FL, defined as the component of the force perpendicular to the direction of motion
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Types of Drag1. Pressure or Form Drag:
This type of drag is due to pressure difference in front of and at the back of the object. The formation of a low pressure wake behind the object depends on the shape or “formform” of the object. This is why this type of drag is called “form” drag.
In case of a streamlined object, the total drag will be due to friction, as in (a). In (b), due to this large wake (region of low pressure), form drag will be significant.
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Types of Drag
2. Friction or Skin Friction Drag:
This type of drag results from viscous (shear) stresses at the surface of the object.
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Drag Coefficient
with
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ME3O04 - Chapter 9http://mech.mcmaster.ca/~hamedm/me3o04/
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Flow over a Flat Plate Parallel to the Flow: only friction drag, no pressure drag
Boundary Layer can be 100% laminar, partly laminar and partly turbulent, or essentially 100% turbulent; hence several different drag coefficients are available
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Flow over a Flat Plate Parallel to the Flow: Only Friction Drag
Turbulent BL:
Laminar BL:
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ME3O04 - Chapter 9http://mech.mcmaster.ca/~hamedm/me3o04/
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Drag Coefficient for a Smooth Flat Plate –Fig 9.8
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Flow over a Flat Plate Perpendicular to the Flow: Pressure Drag, No Friction Drag.
Drag coefficients are usually obtained empirically
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Flow over a Flat Plate Perpendicular to the Flow: Pressure Drag (Continued)
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Flow over a Sphere: Friction and Pressure Drag
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Drag Coefficient - Typical Values
Source: Principles of Fluid Mechanics by A. Alexandrou, Prentice Hall, 2001.
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Effect of flow Regime on Form or Pressure Drag - flow over a cylinder
Source: Fluid Mechanics by F.M. White, Wiley, 2003.
< 1.0
> 1.0
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StreamliningUsed to Reduce Wake and hence reduce pressure drag
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Important Note Regarding the Area, A
1. In case of pure friction drag, the area, A, is the total surface area in contact with the fluid (i.e., the wetted the wetted areaarea).
2. In case of pure pressure drag, the area A is the frontal frontal area or projected areaarea or projected area of the object.
3. For combined cases, drag coefficient for flow over an immersed object usually is based on the frontal area or frontal area or projected area of the objectprojected area of the object, except for airfoils and wings.
4. for airfoils and wings, use the planformplanform area.area.
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Airfoils and Wings Planform Area
Planform area is the maximum projected area of the wing.Source: Fluid Mechanics by F.M. White, Wiley, 2003.
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How de calculate Lift? Lift Coefficient, CL
Note: Ap is the planform area = maximum projected area.
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CL is Function of Re and Angle of Attack, αExamples: NACA 23015; NACA 662-215
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Minimum Flight Speed, Vmin
At steady-state flight conditions, lift force, FL, must be equal to aircraft weight, W, thus:
V = Vmin = Minimum flight speed when CL=CLmax
The question is how to maximize CL?
AVCWF LL2
21 ρ==
ACWV
Lmaxmin
2ρ
=
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Increase of CL using Winglets (Flaps)
Figure 9.23