Gas Dynamics and Boundary Layer Flow

ENGR 466 Fall 2003

Gas Dynamics and Boundary Layer Flow

Class NotesInstructor: Dr. Jonathan [email protected]

650.723.2171

c© 2003 by Jonathan D. Posner, Ph.D.

Contents

1 Introduction 41.1 Classification of Fluid Motions . . . . . . . . . . . . . . . . . . . . . . . 51.2 Basic Laws Governing Fluid Motion . . . . . . . . . . . . . . . . . . . . 71.3 Ideal Gas Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.4 Integral vs. Differential Analysis . . . . . . . . . . . . . . . . . . . . . . 111.5 A Brief Note on Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Fluid Conservation Equations 122.1 Conservation of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.1.1 Differential Approach . . . . . . . . . . . . . . . . . . . . . . . . 142.2 Conservation of Momentum . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.2.1 Differential Analysis – Navier-Stokes Equations . . . . . . . . . . 25

3 Ideal Flows 343.1 Stream Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343.2 Velocity Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.3 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

4 Boundary Layers 394.1 Laminar Boundary Layers . . . . . . . . . . . . . . . . . . . . . . . . . . 414.2 Differential and Integral Analysis . . . . . . . . . . . . . . . . . . . . . . 424.3 Turbulent Boundary Layers . . . . . . . . . . . . . . . . . . . . . . . . . 43

5 Flow about Immersed Bodies 445.1 Lift and Drag . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

6 Introduction to Compressible Flows 496.1 Review of Relevant Thermodynamics . . . . . . . . . . . . . . . . . . . . 516.2 Classification of Compressible Flows . . . . . . . . . . . . . . . . . . . . 536.3 Stagnation Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

7 Steady 1-D Compressible Flows 577.1 Isentropic Flow in Ducts . . . . . . . . . . . . . . . . . . . . . . . . . . . 607.2 Real Nozzles and Diffusers . . . . . . . . . . . . . . . . . . . . . . . . . . 637.3 Normal Shock Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 677.4 Wind Tunnels and Shock Tubes . . . . . . . . . . . . . . . . . . . . . . . 70

8 Fluid Machinery 728.1 Centrifugal Pumps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 758.2 Performance Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

ENGR 466 Class Notes CONTENTS

Definitions

Symbols× cross product∇ gradient operator (1/m)· dot products scalarm mass kg

~a acceleration vector (m/s2)~U = ui + vj + wk velocity vector (m/s)U velocity magnitude |~U |(m/s)p pressure (kPa)ρ density (kg/m3)µ dynamic viscosity (kg/m · s)ν = µ/ρ kinematic viscosity/momentum diffusion coeff. (m2/s)M molecular mass (g/mol)T temperature (K)H enthalpy (J)h specific enthalpy (J/kg)V volume (m3)ψ stream function (1)φ potential functionN number of mols (mol)E total energy (J)e specific total energy J/kg

Q heat transfer (J/s W )S entropy (J/S ·K)W work (J/s W )

ConstantsNA Avogadro’s number 6.022E23 (#/mol)RU universal gas constant k ·NA = 8.314 (kj/kgmolK)k Boltzmann constant 1.38E-23 (J/K)1

1units are mostly in SI units. Equations in notes are not always in units listed here.

Fall 2003 3 Instructor: [email protected]

ENGR 466 Class Notes 1 INTRODUCTION

1 Introduction

A fluid is a substance that deforms continuously under the application of stress (force/area),no matter how small. Figure 1 shows the deformation of a solid which deforms a fixedamount for a given stress and a fluid which deforms continuously over all time for thesame stress. Examples of fluids are all gases and a large number of liquids: water, alco-hol, oil, and mercury. Knowledge and understanding of fluid mechanics allow analysis of

t0

(a) solid (a) fluid

t∞ t0t1 t2

Figure 1: Deformation of a solid and a liquid by a linear stress

of systems with fluids. All modes of transportation have fluid aspects: aerodynamics ofcars, boats, trains, submarines, the lift and drag on an airplane wing, combustion insidean engine, to name a few. Our weather is dominated by fluid mechanics in the oceancurrents and the atmosphere. The collapse of the Tacoma Narrows Bridge was due toresonant forcing of the suspension by pressure forces from a von Karman vortex street.Designers of power plants, architects of skyscrapers, and golfers all use fluid mechanics.The oil and water industries rely on the concepts of pressure, or “head loss”, to spec-ify pipes and the fluid machinery. Biomedical engineers use fluid mechanics to designbreathing aids, heart-lung machines, and artificial heart pumps and valves. Lubricationtheory is derived from fluid mechanics and is used to design virtually all bearings foundin rotating machinery. A modern application of fluid mechanics is its use in describingbioanalytical microfluidic systems. This list of fluid systems is just a snapshot of thelarge number of fluids applications. Clearly, this course, and these notes, will not at-tempt to cover all of these fluid systems, but it is important to keep consider the widespread application of fluid mechanics.

What you should have studied before:

• Inviscid, incompressible flow (stream function, Bernoulli equation)

• Viscous flows (laminar and turbulent flow internal flows (pipes))

• Basic thermodynamics (laws, internal energy, entropy)

• Basic calculus (derivative, integral)

What this course is about:

• External viscous flows (boundary layers)



• Compressible flows (shockwaves, supersonic flows)

• Fluid machinery (pumps)

A note from the author: Important terms and concepts will be denoted by italicsor bold face fonts. If a term is highlighted in this way, please be familiar with it.

1.1 Classification of Fluid Motions

In this course we will be dealing with fluids as a continuum. The continuum assumptionstates that each fluid property has a definite value at every point in space. Thus eachfluid property has a value of velocity, pressure, temperature, density, viscosity (not acomplete list) at each point in space x, y, z and time t.

δV

δmδV

δV’

ρ

Figure 2: Property of fluid as differential volume increases. Concept of continuum.

If we choose density for example (kg/m3) we can define a fluid continuum density asthe the limit of the volume where the density stabilizes

ρ ≡ limδV→δV ′

δm

δV(1.1)

Density is still a scalar and is generally a function of space and time

ρ = ρ(x, y, z, t)

For gases a fluid continuum is one in which the mean free path of molecules (10−7m

the distance they travel before colliding with another molecule)is much smaller thanthe characteristic dimension of the system under consideration. The Knudsen numberis a nondimensional number used to describe if a fluid problem can be treated as acontinuum

Kn ≡ λ

L



t

u

(b)t

u

(a)

Figure 3: Laminar and turbulent flow structure. Velocity of flow as measure by a singlepoint in time. (a) laminar flow (b) turbulent flow and average velocity (dotted line)

where L is the characteristic length and λ is the molecular mean free path. The Knudsennumber must be much less than 1 to be considered a continuum.In this course we will only be dealing with Newtonian fluids. A Newtonian fluid is onewhich stress (N/m2) is linearly related to strain (1/s) through viscosity (Ns/m2)

τyx ∝ du

dy

where the subscripts y and x are the shear plane and velocity coordinate, respectively.Some fluids pose a larger resistance to stress than others (such as honey to water). Wesay they have higher viscosity.

The first division of fluid motions is inviscid versus viscous flows. Inviscid flows neglectviscosity, µ = 0, and thus all terms that include viscosity can be dropped. Viscosity isphysically the diffusion of momentum in a fluid. Although fluids without viscosity donot exist, there are many problems that can be solved accurately assuming negligibleviscosity. Inviscid flows do not require that the no-slip boundary condition to be met.The no-slip condition is a common boundary condition for fluid problems. It describesthe fluid velocity near a solid boundary. The no-slip condition prescribes that thevelocity of the fluid immediately adjacent to the wall has the same velocity of the wall.This is the basis of viscous flow and will be used extensively in boundary layer theory.The concept of viscosity allows wakes and viscous drag on flow on immersed objects otbe described.Another classification of flows in incompressible versus compressible. Compressibilityrefers to the change in density, ρ in space and time. If the material derivative of densityis zero, the flow is incompressible

Dρ

Dt=

∂ρ

∂t+

∂ρ

∂x+

∂ρ

∂y+

∂ρ

∂z= 0 (1.2)

In general compressible flows are only seen in gas flows because liquids are incompress-ible (exceptions include cavitation and reactive flows with phase change). Compressiblegas flows are flows with considerable variations in pressure, temperature, or flows at



Continuum Fluid Mechanics

Viscous Inviscid µ=0

Incompressible Compressibleρ(T,p)M>0.3

Incompressible Compressibleρ(T,p)M>0.3

Turbulentf(Re)

Laminar

Figure 4: Classification of fluid flows in continuum regime

velocities that approach the speed of sound, c. The Mach number is a nondimen-sional number that describes the velocity of the fluid relative to the speed of sound inthe fluid,

M ≡ U

c(1.3)

flows that correspond to a Mach number less than M < 0.3 are incompressible.

Viscous flows have yet another classification, laminar versus turbulent. These flowsare described as laminar or turbulent based upon there flow structure. In the laminarregime, flow structure is characterized by smooth motion in laminae or layers. Flowstructure in turbulent regime is characterized by random, three-dimensional motionsof fluid particles in addition to the mean motion. Turbulent flows are described by avortices of all different length scales. Turbulence is one of the last untamed frontiers ofapplied physics and is the sole focus of a whole group of researchers.

1.2 Basic Laws Governing Fluid Motion

Fluid mechanics can be described by a few basic laws. These fundamental laws governfluid motion and are the basis of ones understanding of the subject:

1. Newton’s second law, conservation of momentum

2. Conservation of mass



Figure 5: Image of turbulent flow.

3. The 1st law of thermodynamics, Conservation of Energy

4. The 2nd law of thermodynamics

Although one may not need all of these laws for each fluid problem, these relations arethe basis of other formal relations the describe the physical behavior of fluids. Theselaws are the same ones used in thermodynamics and solid mechanics, but they will beformulated in such a way to describe fluid motion. When describing the fundamentallaws governing fluid systems, we must choose a way of describing a system. The mostcommon ways of describing a fluid systems are Lagrangian and Eulerian.

Conservation of mass, or the continuity equation as it is often called, states that themass of a closed system is fixed, Lagrangian description is given by

m = ρVsystem = const (1.4)

which can also be stated asdm

dt)

system= 0 (1.5)

The first law of thermodynamics is conservation of energy. This means that energycan neither be created nor destroyed; it can only change forms. The conservation ofenergy cannot be proved mathematically, but no process in nature is known to haveviolated the first law. From the first law we can draw two useful relations:

1. each closed system, fluid parcel, or control volume can be described by a totalenergy, E (J).

2. the change in energy of an open system can be described by Ein − Eout = ∆E



3. the change in internal energy is the heat added to the system plus the work doneon the system, ∆E = ∆Q + ∆W or in differential form dE = dQ + dW .

The first law can be used to describe the flow of heat from a hot object to a cooler one,i.e. a cup of hot coffee into a cool room, or the transfer of energy from a falling fluidto paddle wheel to produce useful work. Although the first law describes the energybalance of a system, it is not useful to determine if the the process will actually takeplace. A cup of hot coffee will lose its heat to the room, but a cool room will never heatup a cup of coffee. The second law of thermodynamics describes wether a process willoccur spontaneously and if it is reversible. A process will occur unless it satisfies boththe first and second law.

The second law of thermodynamics tracks the entropy of a system. Entropy is ameasure of useful work. Processes always occur in the direction of increasing entropy.Entropy is often considered as a measure of disorder with higher disorder correspondingwith greater entropy. A simple example of entropy is ones home. It is much easierto create disorder in your home then to clean it up. If the second law was not valid,your house would clean itself up. The Clausius formulation of the second law (Germanphysicist R.J.E Clausius 1822-1888) states:

∮δQ

T≤ 0 (1.6)

This holds true for any real process. Another way of stating the second law of thermo-dynamics is with the use of entropy S (J/s ·K),

∮δQ

T≤ dS (1.7)

which states that the entropy of an irreversible system always increases. All physicaland real systems must obey both the first and second law of thermodynamics.

Example

A hot cup of coffee at temperature TH is sitting in a cool room with temperature TC .TH > TC . Heat can transfer from the hot cup to the cool surroundings.Assumption: (1) There are no losses and all the heat from the cup goes to the surround-ings.



THTC

dQ

dS

first law

dE = 0

dE = −dQcup + dQroom = 0

dQcup = dQroom = dQ

the change entropy in the cup will be

dSH =−dQ

TH

the change in entropy in the room will be

dSC =dQ

TC

the total change in entropy is

dS = dSH + dSC = dQ

(TH − TC

TCTH

)

since TH > TC

dS > 0

From this example we can see that the entropy of the system increases. If we write theequations for the reverse process we can satisfy the first law, but not the second. Eachprocess in nature must satisfy both the first and second law.

1.3 Ideal Gas Law

Another equation that it useful in studying gas flows is the ideal gas equation forperfect gases:

p =ρRUT

M(1.8)

or pV = NRUT



is an equation of state which describes the volume of a mole of gas as a function oftemperature and pressure. Where the universal gas constant RU (8.314kj/kgmolK) isjust the multiplication of two familiar constants

RU = k ·NA

the Boltzmann constant (1.38 ·10−23J/K) and Avogadro’s number (6.022 ·1023#/mol).As you can see from the ideal gas equation, the volume of a single mole of gas onlydepends on it’s pressure and temperature and does not depend on its molecular weightor composition. A single mole of gas at standard temperature and pressure (T =273 K, P = 1 atm) has a volume of 22.4 liters.

Fluid Descriptions

Lagrangian method of description deals with fixed, identifiable element of mass. Anexample of a lagrangian description is the application Newton’s second law on a singleparticle with mass m. This example is shown in in Fox and McDonald (1.2)Newton’s second law of motion says that the sum of all the forces acting on a systemis equal to its mass multiplied by its acceleration, a Lagrangian description is given by

∑~F = m~a = m

d~U

dt= m

d2~r

dt2(1.9)

Later, we will see that Newton’s second law is the basis for an equation that describesfluid motion the momentum equation for fluids, Navier-Stokes equations.

One can also consider a fluid to be composed of a very large number of particleswhose motion must be described; keeping track of each fluid particle is a huge task.If one considers the number of gas molecules at standard temperature and pressure(T = 273 K, P = 1 atm) that are in a volume the size of your fist, it is a whop-ping 9 · 1021 molecules. Using a control volume to describe the bulk fluid motion, theEulerian method can be used to describe the property of a flow at a given point in spaceas a function of time. In the Eulerian formulation the properties of the flow field aredescribed as functions of space and time.

1.4 Integral vs. Differential Analysis

In applying the basic laws to a system of interest we ca employ a differential or integralapproach. In differential approach we apply equation to an “infinitesimal” unit of fluid.The result is a differential equation which can give continuous description of the behaviorof the system. In the integral formulation we are interested in the global behavior ofthe system. The integral approach is often used in thermodynamics.


ENGR 466 Class Notes 2 FLUID CONSERVATION EQUATIONS

The differential approach is most often used in the subsonic fluid systems while anintegral approach will be used for compressible flows.

1.5 A Brief Note on Dimensions

For this course the fundamental dimensions will be mass, length , time, and temperature(MLtT). Force, acceleration, power, energy, etc are secondary dimensions and shouldnot be used in dimensional analysis. More detail can be found in section 1-6.

Example:

1N ≡ 1kg m

s2

2 Fluid Conservation Equations

2.1 Conservation of Mass

dA

U

control surface (cs)

Figure 6: Depiction of control volume and control surface for mass continuity derivation.

The basic concept and equation of the conservation of mass (1.4) was introduced in thelast section. From the basic law expressed in equation 1.5 we can derive equations thatare more effective for fluid problems. In integral form, the mass of system is representedby

m =∫

VρdV (2.10)

Using an integral approach, we can consider the flow of mass (kg/s) across an area

m =dm

dt=

∫

Aρ~U · d ~A (2.11)



Conservation of mass for a fluid system is then

[mass out] - [mass in] + [mass storage] = 0

mass out - mass in is expressed in equation 2.11. The storage of mass in the system is∫

VρdV

and the rate of increase of mass is∂

∂t

∫

VρdV

and thus the integral equation for conservation of mass becomes∫

Aρ~U · d ~A +

∂

∂t

∫

VρdV = 0 (2.12)

Example

Air flows through a duct with variable area.Assume there is no density variation between (1) and (2).Find V2 as a function of V1, A1, A2.

A1

A2

1

2

U

Using conservation of mass∫

Aρ~U · d ~A +

∂

∂t

∫

VρdV = 0

and simplifying for steady state∫

Aρ~U · d ~A +

©©©©©©∂

∂t

∫

VρdV = 0



writing equation explicitly for mass flux in and out∫

A1ρ~U · d ~A−

∫

A2ρ~U · d ~A = 0

for uniform velocity (not always a good assumption)

ρ1U1A1 − ρ2U2A2 = 0

and with algebraic arrangement

U2 = U1ρ1A1

ρ2A2

2.1.1 Differential Approach

To apply this law to a fluid we can consider a differential volume of length dx, dy, dz

in cartesian coordinates. The density at the center of the volume is ρ and is gener-ally a function of space ρ(x, y, z). The fluid velocity at the center of the volume is~U = ui + vj + wk.

x

y

z

dx

dy

dz

u

v

w

control volume

Figure 7: Differential control volume in cartesian coordinates. The surfaces of thevolume make up the control surface.

The conservation of mass for a differential fluid volume can be written as,

[net rate of mass flux control surface] + [rate of change of mass inside control volume] = 0(2.13)



To evaluate the properties at each of the six faces on the control surface, we use a Taylorseries expansion about the center. For example, at the right face,

ρ)x+dx/2

= ρ +(

∂ρ

∂x

)dx

2+

(∂2ρ

∂x2

)12!

(dx

2

)2

+ . . .

neglecting higher order terms

ρ)x+dx/2

= ρ +(

∂ρ

∂x

)dx

2

The net rate of mass flux out through the control surface can be determined by summingthe mass flux out of each control surface face. For the right face the mass flux is

[u +

∂u

∂x

dx

2

] [ρ +

∂ρ

∂x

dx

2

]dy dz

Summing the mass fluxes through each of the six control surfaces (shown explicitly intable (5.1) in Fox and McDonald) yields,

[∂ρu

∂x+

∂ρv

∂y+

∂ρw

∂z

]dx dy dz (2.14)

The mass in the control volume at anytime is the product of the mass per unit volume,or density ρ, and the volume dx dy dz. The rate of change of mass in the volume is

∂ρ

∂tdx dy dz (2.15)

Substituting equations 2.14 and 2.15 into equation 2.13 we get the conservation of massfor a fluid in cartesian coordinates.

∂ρu

∂x+

∂ρv

∂y+

∂ρw

∂z+

∂ρ

∂t= 0 (2.16)

Another, more compact way of writing this equation uses the vector operator ∇, called“nabla”, which is defined as

∇ ≡ ∂

∂xi +

∂

∂yj +

∂

∂zk

then,

∇ · ρ~U =≡ ∂ρu

∂x+

∂ρv

∂y+

∂ρw

∂z

and the conservation equation 2.16 becomes

∇ · ρ~U +∂ρ

∂t= 0 (2.17)



For incompressible flow density is neither a function of time or space so the continuityequation becomes

∇ · ~U = 0 (2.18)

for steady flows ∂/∂t = 0 so the continuity equation becomes

∇ · ρ~U = 0 (2.19)

Examples of using the continuity equation are given in Fox and McDonald (Ex 5.1 &5.2). Similar relations can be derived for cylindrical and and spherical coordinates.



Example

ρU

x

Gas of initial density ρ0 in piston cylinder. The piston is initially at position x0 andat t = 0 spontaneously outward at velocity U . The fluid velocity in the cylinder is onedimensional and linear, zero at the fixed end x = 0 and u = U at the the piston surface.Assume: Density is uniform in cylinder at all time.Find: (1) u(x,t) (2) ρ(t) (3) rate of change of density at t = 0



2.2 Conservation of Momentum

The movement of fluids are due to the applied forces or forces per unit area calledstress. Our goal is to develop mathematical equations to express Newton’s second lawof motion for a fluid using an Eulerian control volume approach. Newton’s second lawfor a solid is ∑

~F = m~a = md~u

dt

for a fluid we need to be concerned about the change of mass as a function of time alsoso we write ∑

~F =dm~u

dt

the forces that act on the control volume are either body forces ~Fb or surface forces ~Fs.Body forces are gravity (buoyancy), electrostatic, or magnetic forces. Surface forces arepressure and viscous. The sum of external forces is

∑~F = ~Fs + ~Fs

The momentum in the control volume is just the velocity multiplied by the mass (givenin equation 2.10) ∫

V

~UρdV

the force is just its change in time

∂

∂t

∫

V

~UρdV (2.20)

the rate at which momentum enters the CV through the control surface (CS) is just thevelocity multiplied by the mass flux (given in equation 2.14)

∫

A

~Uρ~U · d ~A

setting the sum of the body and surface forces equal to the rate of change of the mo-mentum in the CV plus the flux of momentum through the CS we get an expression forthe conservation of momentum for a fluid in a defined controlled volume

∑~F = ~Fs + ~Fs =

∫

A

~Uρ~U · d ~A +∂

∂t

∫

V

~UρdV (2.21)



Note: For the expression momentum flux expression∫

A

~Uρ~U · dA

there are three dimensions, x, y, z

∫

Auρ~U · dA (2.22)

∫

Avρ~U · dA (2.23)

∫

Awρ~U · dA (2.24)

(2.25)

where u, v, w are the scalar velocity components of ~U in the x, y, z directions, respec-tively. Evaluating the sign for real problems depends on the vector product ~U · d ~A for

x

ydA U

α

this case the vector of velocity ~U is in the opposite direction as the area vector ~A so itis negative. U is the magnitude of ~U . For x-direction of momentum

uρ~U · d ~A = −u|ρUdA cos(α)| (2.26)∫

Auρ~U · dA =

∫

A−u|ρUdA cos(α)| (2.27)

to summarize U is the magnitude of ~U written as U = |~U |. U is also related to theindividual components of velocity ~U = ui + vi + wk by U =

√u2 + v2 + w2

External Forces

The sum of the forces∑ ~F are made up body forces and surface forces. Body forces

act equally everywhere in the CV (this is by design in of the CV) and are given by

~Fb =∫

V

~BρdV



where ~B is the body force per unit mass (m/s2), i.e. ~g is gravity (9.81 m/s2).Surface forces can be shear stress or pressure forces. At first we will only considerpressure forces

~Fs =∫

A−pdA

keep in mind the note above that one must consider the direction of the applied forcerelative to the CS vector and calculate the dot product. The momentum is a vectorequation and needs to be solved as three scalar equations.



Example

x

y

As

Rx

Find: horizontal support force, Rx

Given: from nozzle ~U = ai

Choice of CV changes how one formulates the problem:

Rx Rx

I II III

Assumptions:

1. steady flow

2. incompressible

3. uniform flow where fluid crosses CV

COMOM

∑~F = ~Fs + ~Fs =

∫

A

~Uρ~U · d ~A +©©©©©©©∂

∂t

∫

V

~UρdV

In this example we explore how no matter how we choose the CV the solution is thesame but approach changes. Remember that in applying the momentum equation theforce ~F represents all forces acting on the CV. We are concerned with the forces in thex-direction, gravity ~g acts in the y-direction so Fbx = 0CV I



Fsx =∫

Auρ~U · d ~A

Fsx = pA)left

− pA)right

+ Rx)bottom

Fsx = Rx

Rx =∫

Auρ~U · d ~A

fluid crossing top and bottom surfaces are moving in y-direction only, so for thosesurfaces (top and bottom) ui = 0

Rx =∫

Aleftuρ~U · d ~A +

∫

Atop0ρ~U · d ~A +

∫

Abottom0ρ~U · d ~A

Rx =∫

Aleftuρ~U · d ~A

apply vector dot product

Rx = −u|ρUAleft|Rx = −ρa2Aleft

CV IISame as CV I but support acts on CV with force Rx

CV IIIISurface forces are modified since the right side of the CV is not subject to atmosphericpressure. We are now concerned with the force of the object the force impinges on Sx.The surface forces become

Fsx = pA + Sx

pA + Sx =∫

Aleftuρ~U · d ~A

Sx = −u|ρUAleft| − pA

although this solution appears different we must consider the free body diagram of thesupport itself. Here we consider the forces that the flat area experiences itself and applyNewton’s second law. We solve for the force of the support that keeps the area in placeRx

∑~F = −Sx − pA + Rx = 0

Rx = Sx + pA

Rx = −u|ρUAleft| − pA + pA

Rx = −u|ρUAleft|which gives us the same solution for the force on the support as CV I.



x

y

Rx

-pASx

Example II

x

y

1

32

In this example we consider a tank that is being filled from the top at 1. The fluidescapes at 2 and 3 and is under the influence of gravity. If the tank is being weighedhow much does the scale read?Find: How much does the scale read?Given:an empty tank weighs W

steady height of fluid L

area of global tank AT~U1 = −aj

A2 = A3

Assumptions:

1. steady flow

2. incompressible

3. uniform velocity at control volume boundaries

here we are concerned with the y-momentum because we want to determine the forceon the scale in the vertical direction



COMOM

∑Fy =

©©©©©©∂

∂t

∫

VuρdV +

∫

Auρ~U · d ~A

∑Fy = Fsy + Fby

Fby = −ρgLAT −W

Fsy = pA− pA + Ry

∑Fy = = −ρLAT g − w + Ry =

∫

A1vρ~U · d ~A

−ρLAT g − w + Ry = −v1|ρU1A1|

but the remember that v1 = −a and U1 = v1 in this case, so the solution becomes forthe measured weight becomes

Ry = w + ρLAT g + ρa2A1



2.2.1 Conservation of Momentum Differential Analysis, Navier-Stokes Eqs.

In this section we will derive (not in its full glory) the Navier-Stokes equations, which isthe differential form of Newton’s second law for a fluid continuum. Before formulatingthe effects of forces on the fluid motion (dynamics), we should consider the motionof a fluid element, kinematics. Figure 2.2.1 shows an infinitesimal element of fluidwhich can undergo a variety of motions. A fluid element can simply move throughtranslation or rotation and it can deform through linear and angular deformations. Ingeneral, a fluid motion is a combination of translation, rotation, and linear and angulardeformation.

x

y

x

yx

y

x

y(a)

(c)

(b)

(d)

Figure 8: Pictorial representation of the components of fluid motion (a) translation (b)rotation (c) angular deformation (d) linear deformation

Acceleration of a Fluid Particle in a Velocity Field

As a first step in the ultimate goal of formulating the Navier-Stokes equations we willconsider a mathematical formulation of the acceleration of a fluid particle in a velocityfield, ~U = ~U(x, y, z, t). The power of this description is that it will describe the entireflow field in one equation. Since we are concerned with Newton’s second law, we arelooking for an expression that links the velocity field ~U(x,y,z,t) with the second law∑ ~F = md~U/dt = m/veca. Simply stated, the problem is: Given the velocity field~U = ~U(x, y, z, t), find the acceleration of a fluid particle ~ap. Using figure 2.2.1



x

y

z

r + dr

t+dt

particle path

t

r

Figure 9: Motion of a fluid particle in a velocity field.

as a graphical guide we write

at time t ~X = x, y, z ~Up)t= ~U(x, y, z, t)

at time t + dt ~X = x + dx, y + dy, z + dz ~Up)t+dt

= ~U(x + dx, y + dy, z + dz, t + dt)

d~Up =∂~U

∂xdx +

∂~U

∂ydy +

∂ ~U

∂zdz +

∂~U

∂tdt

~a =d~Up

dt=

∂~U

∂x

dx

dt+

∂~U

∂y

dy

dt+

∂ ~U

∂zdzdt +

∂~U

∂t

dt

dt

where

u =dx

dtv =

dy

dtw =

dz

dt

D~U

Dt= ~a = u

∂~U

∂x+ v

∂ ~U

∂y+ w

∂~U

∂z︸︷︷︸convective

+∂~U

∂t︸︷︷︸local

(2.28)

We use the capital D/Dt to denote the substantial derivative. The first term of theof equation 2.28 is the convective term. Physically the convective term deals with theacceleration of a fluid particle as it convects from one region of the flow to the next.



The local velocity, represented by the second term represents the change in velocity asa function of time. The substantial derivative, also called a material derivative can bewritten in a more compact manner using vector calculus notation:

(~U · ∇)~U +∂~U

∂t= u

∂~U

∂x+ v

∂ ~U

∂y+ w

∂~U

∂z+

∂~U

∂t

When using this equation it one must consider each component. The three componentsare written explicitly as:

x u∂u

∂x+ v

∂u

∂y+ w

∂u

∂z+

∂u

∂t(2.29)

y u∂v

∂x+ v

∂v

∂y+ w

∂v

∂z+

∂v

∂t(2.30)

z u∂w

∂x+ v

∂w

∂y+ w

∂w

∂z+

∂w

∂t(2.31)



Example

x

y

x=0 x=L

U

Given: Steady, 1-D, incompressible flow through converging channel

~U = U1

(1 +

x

L

)i

Find: x component of acceleration of a fluid particle

Solution:Start with full expression of particle acceleration in x-component

Du

Dt= u

∂u

∂x+ v

∂u

∂y+ w

∂u

∂z+

∂u

∂t

from given we know v = w = 0 and ∂/∂t = 0. The velocity field simplifies to

u = U1

(1 +

x

L

)

and the main equation can be simplified

u∂u

∂x+

¶¶¶

v∂u

∂y+

¡¡¡

w∂u

∂z+

¶¶¶

∂u

∂t

Du

Dt= u

∂u

∂x= U1

(1 +

x

L

)(U1

L

)

U21

L

(1 +

x

L

)



x

y

a

b

o

∆ζ

∆η∆x

∆y

Figure 10: Rotation of a fluid element. ∆η/∆x = tan(∆α) and ∆ζ/∆y = tan(∆β)

Fluid Rotation

Rotation of a fluid is defined as the average angular velocity of two mutually perpen-dicular line elements of a fluid element. Rotation is a vector quantity, ~ω, and has threecomponents

~ω = ωxi + ωy j + ωzk

use the right-hand-rule to get the positive sign of rotation. Consider rotation in thex−y plane ωz rotation of oa is due to variations of v in x-direction. Using Taylor SeriesExpansion

v(o) = v0 and v = v0 +∂v

∂x∆x + HOT · · ·

∆η = [v − v(o)]∆t =∂v

∂x∆x∆t

so using an approximation for small angles the angular velocity of oa is

ωoa = lim∆t→0

∆α

∆t= lim

∆t→0

∆η/∆x

∆t

ωoa =∂v

∂x

following the same procedure for ob

u(o) = u0 and u = u0 +∂u

∂y∆y + HOT · · ·

−∆ζ = [u− u(o)]∆t =∂u

∂y∆y∆t

ωob = lim∆t→0

∆β

∆t= lim

∆t→0

∆ζ/∆y

∆t

ωob = −∂u

∂y



from the definition of angular velocity

ωz =12

(∂v

∂x− ∂u

∂y

)(2.32)

and likewise

ωx =12

(∂w

∂y− ∂v

∂z

), ωy =

12

(∂u

∂z− ∂w

∂x

)(2.33)

and the angular velocity vector in all its glory is

~ω = ωxi + ωy j + ωzk =12

(∂w

∂y− ∂v

∂z

)i +

12

(∂u

∂z− ∂w

∂x

)j +

12

(∂v

∂x− ∂u

∂y

)k (2.34)

which can be written using vector notation as

~ω =12∇× ~U (2.35)



Example

Given: Flow Field with tangential motion only Ur, Uθ = f(r)Find: Rotation of fluid for rigid-body rotationSolution: Using equation for rotation in cylindrical coordinates (equation 5.16)

ωz =12

(1r

∂rUθ

∂r− 1

r

∂Ur

∂θ

)

the equation for solid-body rotation is simply Ur = ωr. through differentiation andsimplification (show this yourself) we get

ωz =12r

(2ωr) = ω

what if Uθ = C/r? (this is called a free vortex) show that ωz = 0.



Angular Deformation

Angular deformation of a fluid element involves changes in the angle between two mu-tually perpendicular line segments We can see that the rate of angular deformation of

a

b

o

∆ζ

∆η∆x

∆y

x

y

Figure 11: Angular deformation of a fluid element. ∆η/∆x = tan(∆α) and ∆ζ/∆y =tan(∆β)

a fluid element in the xy-plane is the rate of decrease of the angle between oa and ob.

lim∆t→0

∆α + ∆β

∆t

lim∆t→0

∆α

∆t= lim

∆t→0

∆η/∆x

∆t=

∂v

∂xand for ob

lim∆t→0

∆β

∆t= lim

∆t→0

∆ζ/∆y

∆t=

∂u

∂y

and the thus the angular deformation in the xy-plane is

∂u

∂y+

∂v

∂x

The shear stress is related to the rate of angular deformation through fluid viscosity.The presence of viscous forces means the flow is rotational. From the above formulationswe can observe something very fundamental to the study of fluid mechanics, namely,shear stress in a fluid system is due to velocity gradients (not velocity) and diffusesmomentum through viscosity

Linear Deformation

Linear deformation is described by a change in shape of a fluid element where theangles at the vertices remains unchanged, see figure 8. The element may change length,height, or width depending on the linear velocity gradients ∂u/∂x, ∂v/∂y, ∂w/∂z. Lineardeformation is the change in the shape without change in volume and has the same



expression as the conservation of mass for an incompressible fluid, equation 2.18.

∇ · ~U =≡ ∂u

∂x+

∂v

∂y+

∂w

∂z= 0

Navier-Stokes Equations

Here we will bring together all of the forces on a fluid element to derive an equationdescribing it’s motion using Newton’s second law. Having found the acceleration of afluid element of a fluid element moving in a velocity field we can write Newton’s secondlaw as

d~F = dmD~U

Dt= dm

[(~U · ∇)~U +

∂~U

∂t

]= dm

(u

∂~U

∂x+ v

∂ ~U

∂y+ w

∂~U

∂z+

∂~U

∂t

)

now we have to consider the forces that act on the fluid element. Figure 12 shows arrowsfor each fluid stress. The stresses are broken into normal and tangential components.for the x-direction the sum of the forces are

x

y

z

u

v

w

1

2

34

5

6

Figure 12: Stresses on a fluid element.

dFsx =(

σxx +∂σxx

∂x

dx

2

)dydz

︸︷︷︸1

−(

σxx − ∂σxx

∂x

dx

2

)dydz

︸︷︷︸3

+(

τyx +∂τyx

∂y

dy

2

)dxdz

︸︷︷︸2

−(

τyx − ∂τyx

∂y

dy

2

)dxdz

︸︷︷︸6

+(

τzx +∂τzx

∂z

dz

2

)dxdy

︸︷︷︸5

−(

τzx − ∂τzx

∂z

dz

2

)dxdy

︸︷︷︸4


ENGR 466 Class Notes 3 IDEAL FLOWS

3 Ideal Flows

A large number of external flows can be treated as viscous flows near the solid bound-ary and inviscid outside the boundary layer. Flows in which both viscosity andcompressibility are negligible are called ideal flows. The next sections will describe twofunctions that are used to describe ideal flows. The mathematics of ideal flows allowsthe velocity of pressure distributions. In this text only the two-dimensional functionswill be presented.

3.1 Stream Function

The stream function is a mathematical construct that satisfies both continuity (massconservation, ∇ · ~U) and irrotationallity (∇× ~U = 0). Graphically, the stream functionrepresents the streamlines of an inviscid, incompressible flow. If we consider a line thatis every where tangent to the local and instantaneous velocity vector,as shown in fig-ure 3.1, we can write

dy

dx=

v

uwhich can also be written as

dy

v=

dx

u(3.36)

Equation 3.36 equates the slope of the streamline with the local velocity vector.

x

y

Figure 13: Velocity streamline with local tangent vectors.



From conservation of mass we can write,

∇ · ~U = 0

for ~U(x, y),∂u

∂x+

∂v

∂y= 0

∂

∂x(u) +

∂

∂y(v) = 0

(3.37)

by introducing a function ψ(x, y) such that it immediately satisfies equation 3.37, wecan define it as,

u ≡ ∂ψ

∂yv ≡ −∂ψ

∂x(3.38)

substituting into 3.37,

∂

∂x(∂ψ

∂y)− ∂

∂y(∂ψ

∂x) = 0

we see that by our choice of the stream function we automatically satisfy mass conser-vation. The stream function is solenoidal.In our construction of the stream function we have prescribed that it should also beirrotational, namely ∇ × ~U = 0. For a two dimensional flow in cartesian coordinatesthe vorticity is,

∇× ~U(x, y) = 0 =∂v

∂x− ∂u

∂y(3.39)

substituting 3.38 (3.40)∂2ψ

∂x2+

∂2ψ

∂y2= 0

which is the Laplacian of ψ

∇2ψ = 0 (3.41)

For the arguments above we know that the stream function is both conservative andirrotational. To better understand what the stream function means physcially we canconsider the function for lines of constant ψ(x, y) = c. The function for lines of constantψ(x, y) can be determined by the definition of the derivative,

dψ =∂ψ

∂xdx +

∂ψ

∂ydy = 0 (3.42)

substituting equation 3.38 (3.43)

0 = −vdx + udy ordy

v=

dx

u(3.44)

which identical to equation 3.36 (3.45)



x

y ψ1

ψ2

ψ2

a

bc

d

a(x1,y1)b(x2,y2)c(x3,y3)d(x4,y4)

Figure 14: Plot of constant stream function, flow across ab,bc,cd are all equal.

Thus the stream function is everywhere tangent to the local velocity vector! The lines ofconstant stream function are the stream lines for an incompressible and irrotational flowfield. From the definition of the stream function we see that there is no flow across astreamline since the velocity is always tangent to the local vector. Figure 3.1 shows a fewstream lines, each representing constant ψ. From the definition of the stream functionthe flow across ab,bc,cd are all equal. The volume flow rate Q between ψ1andψ2 can beevaluated by integrating. From ab the flow rate is,

Q =∫ y2

y1

udy =∫ y2

y1

∂ψ

∂ydy (3.46)

and since x is constant along ab, from 3.42

dψ =∂ψ

∂ydy

so,

Q =∫ y2

y1

∂ψ

∂ydy =

∫ ψ2

ψ1

dψ = ψ2 − ψ1 (3.47)

The flow across bc is

Q =∫ x2

x1

vdx =∫ x2

x1

−∂ψ

∂xdx (3.48)

and since y is constant along bc, from 3.42

dψ =∂ψ

∂xdx

so,

Q =∫ x2

x1

−∂ψ

∂xdx =

∫ ψ2

ψ1

dψ = ψ2 − ψ1 (3.49)



This proves that the flow rate between any two streamlines can be written as the dif-ference between the constant values of ψ. From the definition of u in terms of ψ, it isclear that u is in the positive x direction when ψ increases with y.

3.2 Velocity Potential

The velocity potential is another mathematical construct that is used to visualize fluidflows. The potential, φ, is everywhere orthogonal to the stream function,

φ ⊥ ψ (3.50)

The slopes of constant ψ lines are the negative reciprocal of constant φ (compare toequation 3.38),

dy

dx= −u

v(3.51)

u =∂φ

∂xv =

∂φ

∂y

substituting the potential equation into the irrotational condition (3.39) we obtain,

∇× ~U = 0∂v

∂x− ∂u

∂y=

∂

∂x(∂φ

∂x)− ∂

∂y(∂φ

∂y) = 0 (3.52)

and in continuity

∇ · ~U = 0

∇ · ~U =∂u

∂x+

∂v

∂y=

∂2φ

∂x2+

∂2φ

∂y2= 0

By constructing the stream function and the potential function the way we did, theyboth satisfy continuity and the irrotational condition. The functions are orthogonal toeach other and can be used to visualize incompressible, inviscid flows. These functionscan also be used to calculate the pressure distribution in conjunction with the Bernoulliequation.



3.3 Example

Problem: Given the velocity field ~U = Axi − Ayj, with A = 10s−1, determine thestream function that will yield this velocity field.

Solution: We can start to attack this problem by looking at the individual componentsof velocity u and v,

u = Ax v = −Ay

and the definition of the velocities in respect to the stream function 3.38. If we want toget ψ(x, y) then we can try to integrate ∂ψ/∂y

ψ =∫

∂ψ

∂ydy + f(x) =

∫udy + f(x)

ψ =∫

Axdy + f(x) = Axy + f(x) (3.53)

then we differentiate in respect to x to use the information from v

∂ψ

∂x= Ay +

df

dx

from definition∂ψ

∂x= −v

∂ψ

∂x= −v = Ay +

df

dxthus df/dx = 0, f(x) = C that we set to 0

the stream function then becomes,

ψ(x, y) = Axy (3.54)

Using MATLAB the plot for constant ψ can be plotted as shown in figure 3.3. Thisplot represents the streamlines for inviscid, incompressible flow in a corner.


ENGR 466 Class Notes 4 BOUNDARY LAYERS

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

9

10

x

y

1

2

3

4

5

6

7

98

ψ1

ψ2

ψ3

Figure 15: Streamlines of constant ψ plotted for function ψ(x, y) = Axy, A = 10

4 Boundary Layers


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Gas Dynamics and Boundary Layer Flow