[2003] guide of fluid mechanics [j.e.shepherd, cit][ae.aph.101]

8/21/2019 [2003] Guide of Fluid Mechanics [J.E.shepherd, CIT][Ae.aph.101]

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Ae APh 101Skeleton Guide

Joseph E. Shepherd

1891

CA

LIFO

RN

IA

INSTIT

U TEOFT

EC

HN

OL

OGY

Graduate Aeronautical Laboratories

California Institute of Technology

Pasadena, CA 91125

February 8, 2003


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Contents

1 Fundamentals 11.1 Control Volume Statements . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Reynolds Transport Theorem . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Integral Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.3.1 Simple Control Volumes . . . . . . . . . . . . . . . . . . . . . . . 31.3.2 Steady Momentum Balance . . . . . . . . . . . . . . . . . . . . . 3

1.4 Vector Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4.1 Vector Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4.2 Curvilinear Coordinates . . . . . . . . . . . . . . . . . . . . . . . 41.4.3 Gauss Divergence Theorem . . . . . . . . . . . . . . . . . . . . . 51.4.4 Stokes Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4.5 Div, Grad and Curl . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4.6 Specific Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.5 Differential Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.5.1 Conservation form . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.6 Convective Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.7 Divergence of Viscous Stress . . . . . . . . . . . . . . . . . . . . . . . . . 81.8 Euler Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.9 Bernoulli Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.10 Vorticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.11 Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2 Thermodynamics 12

2.1 Thermodynamic potentials and fundamental relations . . . . . . . . . . . 122.2 Maxwell relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3 Various defined quantities . . . . . . . . . . . . . . . . . . . . . . . . . . 122.4 v(P, s) relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.5 Equation of State Construction . . . . . . . . . . . . . . . . . . . . . . . 14

3 Compressible Flow 163.1 Steady Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.1.1 Streamlines and Total Properties . . . . . . . . . . . . . . . . . . 163.2 Quasi-One Dimensional Flow . . . . . . . . . . . . . . . . . . . . . . . . 17

3.2.1 Isentropic Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.3 Heat and Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.3.1 Fanno Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.3.2 Rayleigh Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.4 Shock Jump Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.4.1 Lab frame (moving shock) versions . . . . . . . . . . . . . . . . . 22

3.5 Perfect Gas Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.6 Reflected Shock Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

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3.7 Detonation Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.8 Perfect-Gas, 2-Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.8.1 2-Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.8.2 High-Explosives . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.9 Weak shock waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.10 Acoustics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.11 Multipole Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.12 Baffled (surface) source . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.13 1-D Unsteady Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.14 2-D Steady Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.14.1 Oblique Shock Waves . . . . . . . . . . . . . . . . . . . . . . . . . 343.14.2 Weak Oblique Waves . . . . . . . . . . . . . . . . . . . . . . . . . 343.14.3 Prandtl-Meyer Expansion . . . . . . . . . . . . . . . . . . . . . . 353.14.4 Inviscid Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.14.5 Potential Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363.14.6 Natural Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 363.14.7 Method of Characteristics . . . . . . . . . . . . . . . . . . . . . . 37

4 Incompressible, Inviscid Flow 384.1 Velocity Field Decomposition . . . . . . . . . . . . . . . . . . . . . . . . 384.2 Solutions of Laplaces Equation . . . . . . . . . . . . . . . . . . . . . . . 384.3 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394.4 Streamfunction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

4.4.1 2-D Cartesian Flows . . . . . . . . . . . . . . . . . . . . . . . . . 404.4.2 Cylindrical Polar Coordinates . . . . . . . . . . . . . . . . . . . . 41

4.4.3 Spherical Polar Coordinates . . . . . . . . . . . . . . . . . . . . . 434.5 Simple Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444.6 Vorticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464.7 Key Ideas about Vorticity . . . . . . . . . . . . . . . . . . . . . . . . . . 484.8 Unsteady Potential Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . 484.9 Complex Variable Methods . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4.9.1 Mapping Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . 514.10 Airfoil Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514.11 Thin-Wing Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

4.11.1 Thickness Solution . . . . . . . . . . . . . . . . . . . . . . . . . . 54

4.11.2 Camber Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564.12 Axisymmetric Slender Bodies . . . . . . . . . . . . . . . . . . . . . . . . 584.13 Wing Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

5 Viscous Flow 615.1 Scaling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 625.2 Two-Dimensional Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . 625.3 Parallel Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

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5.3.1 Steady Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 635.3.2 Poiseuille Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . 665.3.3 Rayleigh Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

5.4 Boundary Layers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

5.4.1 Blasius Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 705.4.2 Falkner-Skan Flow . . . . . . . . . . . . . . . . . . . . . . . . . . 70

5.5 Karman Integral Relations . . . . . . . . . . . . . . . . . . . . . . . . . . 715.6 Thwaites Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 725.7 Laminar Separation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 725.8 Compressible Boundary Layers . . . . . . . . . . . . . . . . . . . . . . . 72

5.8.1 Transformations and Approximations . . . . . . . . . . . . . . . . 735.8.2 Energy Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 745.8.3 Moving Shock Waves . . . . . . . . . . . . . . . . . . . . . . . . . 765.8.4 Weak Shock Wave Structure . . . . . . . . . . . . . . . . . . . . . 77

5.9 Creeping Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

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1

1 Fundamentals

1.1 Control Volume Statements

is a material volume, V is an arbitrary control volume, indicates the surface of thevolume.mass conservation:

d

dt

dV = 0 (1)

Momentum conservation:

d

dt

u dV =F (2)

Forces: F=

G dV +

T dA (3)

Surface traction forces

T= Pn + n= T n (4)Stress tensor T

T= PI+ or Tik = P ik+ik (5)whereI is the unit tensor, which in cartesian coordinates is

I= ik (6)

Viscous stress tensor, shear viscosity , bulk viscosityv

ik = 2

Dik 13

ikDjj

+vikDjj implicit sum on j (7)

Deformation tensor

Dik =1

2

uixk

+ ukxi

or

1

2

u + uT

(8)

Energy conservation:

ddt

e+|u|22

dV = Q+ W (9)

Work:

W =

G u dV +

T u dA (10)

Heat:

Q=

q n dA (11)


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2 1 FUNDAMENTALS

heat fluxq, thermal conductivity k and thermal radiation qr

q= kT+ qr (12)Entropy inequality (2nd Law of Thermodynamics):

d

dt

s dV

q nT

dA (13)

1.2 Reynolds Transport Theorem

The multi-dimensional analog of Leibnizs theorem:

d

dt

V(t)

(x, t) dV =

V(t)

t dV +

V

uV ndA (14)The transport theorem proper. Material volume , arbitrary volume V.

d

dt

dV = d

dt

V

dV +

V(u uV) ndA (15)

1.3 Integral Equations

The equations of motions can be rewritten with Reynolds Transport Theorem to applyto an (almost) arbitrary moving control volume. Beware of noninertial reference framesand the apparent forces or accelerations that such systems will introduce.

Moving control volume:

d

dt V dV + V (u uV) ndA= 0 (16)d

dt

V

udV +

Vu (u uV) ndA=

V

G dV +

VT dA (17)

d

dt

V

e+

|u|22

dV +

V

e+

|u|22

(u uV) ndA=

VG u dV +

V

T u dA

Vq n dA (18)

d

dt

V sdV +

V s (u uV) ndA+ Vq

T n dA 0 (19)Stationary control volume:

d

dt

V

dV +

Vu ndA= 0 (20)

d

dt

V

udV +

Vuu ndA=

V

G dV +

VT dA (21)


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1.3 Integral Equations 3

d

dt

V

e+

|u|22

dV +

V

e+

|u|22

u ndA=

V

G u dV + V

T u dA V

q n dA (22)

d

dt

V

sdV +

Vsu ndA+

V

q

T n dA 0 (23)

1.3.1 Simple Control Volumes

Consider a stationary control volume V with i = 1, 2, . . ., I connections or openingsthrough which there is fluid flowing in and j = 1, 2, . . ., J connections through whichthe fluid is following out. At the inflow and outflow stations, further suppose that we

can define average or effective uniform properties hi, i, ui of the fluid. Then the massconservation equation is

dM

dt =

d

dt

V

dV =I

i=1

Ai mi J

j=1

Aj mj (24)

whereAi is the cross-sectional area of the ith connection and mi =iui is the mass flowrate per unit area through this connection. The energy equation for this same situationis

dE

dt = d

dt

V e+|u|

2

2 +gz dV =

Ii=1

Ai mi hi+|ui|2

2 +gzi

J

j=1

Aj mj

hj+

|uj|22

+gzj

+ Q+ W(25)

where Q is the thermal energy (heat) transferred into the control volume and W is themechanical work done on the fluid inside the control volume.

1.3.2 Steady Momentum Balance

For a stationary control volume, the steady momentum equation can be written asV

uu ndA+

VPn dA=

V

G dV +

V n dA+ Fext (26)

where Fext are the external forces required to keep objects in contact with the flow inforce equilibrium. Thesereaction forces are only needed if the control volume includesstationary objects or surfaces. For a control volume completely within the fluid,Fext =0.


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4 1 FUNDAMENTALS

1.4 Vector Calculus

1.4.1 Vector Identities

IfA and B are two differentiable vector fieldsA(x),B(x) and is a differentiable scalarfield(x), then the following identities hold:

(A B) = (B )A (A )B ( A)B + ( B)A (27)(A B) = (B )A + (A )B + B ( A) + A ( B) (28) () = 0 (29)

( A) = 0 (30) ( A) = ( A) 2A (31)

(A) = A + A (32)

1.4.2 Curvilinear Coordinates

Scale factors Consider an orthogonal curvilinear coordinate system (x1, x2, x3) definedby a triad of unit vectors (e1, e2, e3), which satisfy the orthogonality condition:

ei ek = ik (33)and form a right-handed coordinate system

e3 = e1 e2 (34)The scale factorshi are defined by

dr= h1dx1e1+h2dx2e2+h3dx3e3 (35)

or

hi

r

xi

(36)

The unit of arc length in this coordinate system is ds2 = dr dr:

ds2 =h21dx21+h

22dx

22+h

23dx

23 (37)

The unit of differential volume is

dV =h1h2h3dx1dx2dx3 (38)


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1.4 Vector Calculus 5

1.4.3 Gauss Divergence Theorem

For a vector or tensor field F, the following relationship holds:

V F dV V F n dA (39)This leads to the simple interpretation of the divergence as the following limit

F limV0

1

V

V

F n dA (40)

A useful variation on the divergence theorem isV

( F) dV

Vn F dA (41)

This leads to the simple interpretation of the curl as

F limV0

1

V

V

n F dA (42)

1.4.4 Stokes Theorem

For a vector or tensor field F, the following relationship holds on an open, two-sidedsurfaceSbounded by a closed, non-intersecting curve S:

S( F) n dA

S

F dr (43)

1.4.5 Div, Grad and Curl

The gradient operator or gradfor a scalar field is

= 1

h1

x1e1+

1

h2

x2e2+

1

h3

x3e3 (44)

A simple interpretation of the gradient operator is in terms of the differential of a functionin a directiona

da = limda0

(x + da) (x) = da (45)The divergence operator or div is

F= 1h1h2h3

x1

(h2h3F1) +

x2(h3h1F2) +

x3(h1h2F3)

(46)

The curl operator or curl is

F= 1h1h2h3

h1 e1 h2 e2 h3e3

x1

x2

x3

h1F1 h2F2 h3F3

(47)


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6 1 FUNDAMENTALS

The components of the curl are:

F =

e1

h2h3

x2(h3F3)

x3(h2F2)

+ e2h3h1

x3(h1F1)

x1(h3F3)

+ e3h1h2

x1(h2F2)

x2(h1F1)

(48)

The Laplacian operator2 for a scalar field is

2= 1h1h2h3

x1(

h2h3h1

x1) +

x2(

h3h1h2

x2) +

x3(

h1h2h3

x3)

(49)

1.4.6 Specific Coordinates

(x1, x2, x3) x y z h1 h2 h3

Cartesian

(x ,y ,z) x y z 1 1 1

Cylindrical

(r,,z) r sin r cos z 1 r 1

Spherical

(r,,) r sin cos r sin sin r cos 1 r r sin

Parabolic Cylindrical

(u ,v ,z) 12

(u2 v2) uv z u2 +v2 h1 1

Paraboloidal

(u ,v ,) uvcos uvsin 12

(u2 v2) u2 +v2 h1 uv

Elliptic Cylindrical

(u ,v ,z) a cosh u cos v a sinh u sin v z a

sinh2 u+ sin2 v h1 1

Prolate Spheroidal

( , , ) a sinh sin cos a sinh sin sin a cosh cos a

sinh2 + sin2 h1 a sinh sin

1.5 Differential Relations

1.5.1 Conservation form

The equations are first written in conservation form

tdensity + flux = source (50)


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1.6 Convective Form 7

for a fixed (Eulerian) control volume in an inertial reference frame by using the divergencetheorem.

t + (u) = 0 (51)

t(u) + (uu T) = G (52)

t

e+

|u|22

+

u

e+

|u|22

T u + q

= G u (53)

t(s) +

us+

q

T

0 (54)

1.6 Convective Form

This form uses the convectiveor materialderivative

D

Dt=

t+ u (55)

D

Dt = u (56)

Du

Dt = P+ +G (57)

D

Dt

e+

|u|22

= (T u) q +G u (58)

Ds

Dt qT (59)Alternate forms of the energy equation:

D

Dt

e+

|u|22

= (Pu) + ( u) q +G u (60)

Formulation using enthalpy h = e + P/

D

Dt

h+

|u|22

=

P

t + ( u) q +G u (61)

Mechanical energy equation

D

Dt

|u|22

= (u ) P+ u +G u (62)Thermal energy equation

De

Dt = PDv

Dt+v:u v q (63)


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8 1 FUNDAMENTALS

Dissipation

=:u= ikuixk

sum oni andk (64)

Entropy

Ds

Dt =

q

T

+

T +k

T

T

2(65)

1.7 Divergence of Viscous Stress

For a fluid with constant and v, the divergence of the viscous stress in Cartesiancoordinates can be reduced to:

=

2u + v+

1

3( u) (66)

1.8 Euler Equations

Inviscid, no heat transfer, no body forces.

D

Dt = u (67)

Du

Dt = P (68)

DDt

h+|u|22

= Pt

(69)

Ds

Dt 0 (70)

1.9 Bernoulli Equation

Consider the unsteady energy equation in the form

D

Dt

h+

|u|22

=

P

t + ( u) q +G u (71)

and further suppose that the external force field G is conservative and can be derivedfrom a potential as

G= (72)then if (x) only, we have

D

Dt

h+

|u|22

+

=

P

t + ( u) q (73)


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1.10 Vorticity 9

The Bernoulli constant is

H=h+|u|2

2 + (74)

In the absence of unsteadiness, viscous forces and heat transfer we have

u

h+|u|2

2 +

= 0 (75)

OrH = constant on streamlines

For the ordinary case of isentropic flow of an incompressible fluid dh= dP/in a uniformgravitational field = g(z z), we have the standard result

P+|u|2

2 +gz= constant (76)

1.10 Vorticity

Vorticity is defined as u (77)

and the vector identities can be used to obtain

(u )u= ( |u|2

2 ) u ( u) (78)

The momentum equation can be reformulated to read:

H=

h+|u|2

2 +

= u

t + u +Ts+

(79)


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10 1 FUNDAMENTALS

1.11 Dimensional Analysis

Fundamental Dimensions

L length meter (m)

M mass kilogram (kg)T time second (s) temperature Kelvin (K)I current Ampere (A)

Some derived dimensional units

force Newton (N) MLT2

pressure Pascal (Pa) ML1T2

bar = 105 Paenergy Joule (J) ML2T2

frequency Hertz (Hz) T1

power Watt (W) ML2T

3

viscosity () Poise (P) ML1T1

Pi Theorem Given n dimensional variables X1, X2, . . ., Xn, and f independent fun-damental dimensions (at most 5) involved in the problem:

1. The number of dimensionally independent variablesr is

r f

2. The number p = n - r of dimensionless variables i

i= XiX11 X

22 Xrr

that can be formed isp n f

Conventional Dimensionless Numbers

Reynolds Re U L/Mach M a U/cPrandtl P r cP/k = /Strouhal St L/UT

Knudsen Kn /LPeclet P e U L/Schmidt Sc /DLewis Le D/

Reference conditions: U, velocity;, vicosity;D, mass diffusivity; k, thermal conductiv-ity;L, length scale;T, time scale; c, sound speed; , mean free path;cP, specific heat atconstant pressure.


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1.11 Dimensional Analysis 11

Parameters for Air and Water Values given for nominal standard conditions 20 Cand 1 bar.

Air Water

shear viscosity (kg/ms) 1.8105

1.00103

kinematic viscosity (m2/s) 1.5105 1.0106thermal conductivity k (W/mK) 2.54102 0.589thermal diffusivity (m2/s) 2.1105 1.4107specific heat cp (J/kgK) 1004. 4182.sound speed c (m/s) 343.3 1484density (kg/m3) 1.2 998.gas constant R (m2/s2K) 287 462.thermal expansion (K1) 3.3104 2.1104isentropic compressibility s (Pa

1) 7.01106 4.51010

Prandtl number P r .72 7.1Fundamental derivative 1.205 4.4ratio of specific heats 1.4 1.007Gruneisen coefficient G 0.40 0.11


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12 2 THERMODYNAMICS

2 Thermodynamics

2.1 Thermodynamic potentials and fundamental relations

energy e(s, v)

de = Tds Pdv (80)enthalpy h(s, P) = e+P v

dh = Tds+v dP (81)

Helmholtz f(T, v) = e T sdf = s dT Pdv (82)

Gibbs g(T, P) = e T s+P vdg = s dT+v dP (83)

2.2 Maxwell relations

T

v

s

= Ps

v

(84)

T

P

s

= v

s

P

(85)

s

v

T

= P

T

v

(86)

s

P

T=

v

T

P(87)

Calculus identities:

F(x , y , . . .) dF = F

x

y,z,...

dx+ F

y

x,z,...

dy+. . . (88)

x

y

f

= fy

x

fx

y

(89)

xf

y= 1f

x

y

(90)

2.3 Various defined quantities

specific heat at constant volume cv eT

v

(91)


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2.3 Various defined quantities 13

specific heat at constant pressure cp hT

P

(92)

ratio of specific heats cpcv

(93)

sound speed c P

s

(94)

coefficient of thermal expansion 1v

v

T

P

(95)

isothermal compressibility KT 1v

v

P

T

(96)

isentropic compressibility Ks 1v

v

P

s

= 1

c2 (97)

Specific heat relationships

KT =Ks or P

v

s

= P

v

T

(98)

cp cv = T

P

v

T

v

T

2P

(99)

Fundamental derivative

c4

2v32v

P2

s

(100)

= v3

2c2

2P

v2

s

(101)

= 1 +c

c

P

s

(102)

= 1

2

v2

c2

2h

v2

s

+ 1

(103)

Sound speed (squared)

c2 P

s(104)

= v2 Pv

s

(105)

= v

Ks(106)

= v

Kt(107)


18/85

14 2 THERMODYNAMICS

Gruneisen Coefficient

G vcvKT

(108)

= v

P

e

v

(109)

= v

cpKs(110)

= vT

T

v

s

(111)

2.4 v(P, s) relation

dvv

= KsdP+ (KsdP)2 + Tdscp

+. . . (112)

= dPc2

+

dP

c2

2+G

Tds

c2 +. . . (113)

2.5 Equation of State Construction

Given cv(v, T) and P(v, T), integrate

de = cvdT+ T P

Tv Pdv (114)

ds = cv

T dT+

P

T

v

dv (115)

along two paths: I: variable T, fixed and II: variable , fixedT.Energy:

e= e+ T

Tcv(T, ) dT

I

+

P T P

T

d

2 II

(116)

Ideal gas limit 0,

lim0

cv(T, ) =cigv(T) (117)

The ideal gas limit of I is the ideal gas internal energy

eig(T) = T

Tcigv(T) dT (118)

Ideal gas limit of II is the residual function


19/85

2.5 Equation of State Construction 15

er(, T) = 0

P T P

T

d

2 (119)

and the complete expression for internal energy is

e(, T) =e+eig(T) +er(, T) (120)

Entropy:

s= s+ T

T

cv(T, )

T dT

I

+

P

T

d

2 II

(121)

The ideal gas limit 0 has to be carried out slightly differently since the ideal gasentropy, unlike the internal energy, is a function of density and is singular at = 0.Define

sig = T

T

cigv(T)

T dT R

d

(122)

where the second integral on the RHS is R ln /. Then compute the residual functionby substracting the singular part before carrying out the integration

sr(, T) = 0

R 1

P

T

d

(123)

and the complete expression for entropy is

s(, T) =s+sig(, T) +sr(, T) (124)


20/85

16 3 COMPRESSIBLE FLOW

3 Compressible Flow

3.1 Steady Flow

A steady flow must be considered as compressible when theMach

number M = u/c issufficiently large. In an isentropic flow, the change in density produced by a speedu canbe estimated as

s= c2P 1

2M2 (125)

from the energy equation discussed below and the fundamental relation of thermody-namics.

If the flow is unsteady, then the change in the density along the pathlines for inviscidflows without body forces is

1

D

Dt = u= u

u2

2c2 1

c21

2

u2

t 1

P

t (126)

This first term is the steady flow condition M2. The second set of terms in the squarebraces are the unsteady contributions. These will be significant when the time scaleT iscomparable to the acoustic transit timeL/c, i.e., T Lco.

3.1.1 Streamlines and Total Properties

Stream linesX(t; x) are defined by

dX

dt =u X= x when t= 0 (127)

which in Cartesian coordinates yields

dx1u1

=dx2

u2=

dx3u3

(128)

Total enthalpy is constant along streamlines in adiabatic, steady, inviscid flow

ht = h+|u|2

2 = constant (129)

Velocity along a streamline is given by the energy equation:

u= |u| =

2(ht h) (130)Total properties are defined in terms of total enthalpy and an idealized isentropic decel-eration process along a streamline. Total pressure is defined by

Pt P(s, ht) (131)Other total properties Tt,t, etc. can be computed from the equation of state.


21/85

3.2 Quasi-One Dimensional Flow 17

3.2 Quasi-One Dimensional Flow

Adiabatic, frictionless flow:

d(uA) = 0 (132)

udu = dP (133)h+

u2

2 = constant or dh= udu (134)

ds 0 (135)

3.2.1 Isentropic Flow

If ds= 0, then

dP =c2d+c2 ( 1)(d)2

+. . . (136)

For isentropic flow, the quasi-one-dimensional equations can be written in terms of theMach number as:

1

d

dx =

M2

1 M21

A

dA

dx (137)

1

c2dP

dx =

M2

1 M21

A

dA

dx (138)

1

u

du

dx = 1

1 M21

A

dA

dx (139)1

M

dM

dx = 1 + ( 1)M

2

1 M21

A

dA

dx (140)

1

c2dh

dx =

M2

1 M21

A

dA

dx (141)

At a throat, the gradient in Mach number is:

dM

dx

2=

2A

d2A

dx2 (142)

Constant-Gas If the value of is assumed to be constant, the quasi-one dimensionalequations can be integrated to yield:

t

=

1 + ( 1)M2 12(1) (143)

ctc

=

t

1=

1 + ( 1)M21/2

(144)


22/85


hth

=

1 + ( 1)M2

(145)

u = ct

M2

1 + (

1)M2

1/2(146)

A

A =

1

M

1 + ( 1)M2

2(1)

(147)

P Pttc2t

= 1

2 1

1 + ( 1)M2

212(1) 1

(148)

(149)

Ideal Gas For an ideal gasP = RT and e = e(T) only. In that case, we have

h(T) =e+RT =h+ T

T cv(T) dT, s= s+ T

T

cP(T)

T dT R ln(P/P) (150)and you can show that is given by:

ig =+ 1

2 +

12

T

d

dT (151)

Perfect or Constant-Gas Perfect gas results for isentropic flow can be derived fromthe equation of state

P =RT h= cpT cp= R 1 (152)

the value of for a perfect gas,

pg =+ 1

2 (153)

the energy integral,

Tt = T

1 + 1

2 M2

(154)

and the expression for entropy

s so= cpln TTo

R ln P/Po (155)

or

s so= cvln TTo

R ln /o


23/85

3.3 Heat and Friction 19

TtT

= 1 + 1

2 M2 (156)

PtP

=

TtT

1

(157)

t

=Tt

T

11(158)

Mach NumberArea Relationship

A

A =

1

M

2

+ 1

1 +

12

M2 +1

2(1)

(159)

Choked flow mass flux

M=

2

+ 1 +12(1)

cttA

(160)

or

M=

2

+ 1

+12(1) Pt

RTtA

Velocity-Mach number relationship

u= ctM

1 + 12 M2

(161)

Alternative reference speeds

ct= c

+ 1

2 umax= c

+ 1

1 (162)

3.3 Heat and Friction

Constant-area, steady flow with friction Fand heat additionQ

u = m= constant (163)

udu+ dP = Fdx (164)dh+udu = Qdx (165)

ds = 1

T

Q+

F

dx (166)

F is the frictional stress per unit length of the duct. In terms of the Fanning frictionfactorf


24/85


F = 2

Df u2 (167)

where D is the hydraulic diameter of the duct D = 4area/perimeter. Note that theconventional DArcy or Moody friction factor = 4 f.

Q is the energy addition as heat per unit mass and unit length of the duct. If theheat flux into the fluid is q, then we have

Q= q

u

4

D (168)

3.3.1 Fanno Flow

Constant-area, adiabatic, steady flow with friction only:


udu+ dP = Fdx (170)h+

u2

2 = ht= constant (171)

(172)

Change in entropy with volume along Fanno line, h + 1/2 m2v2=ht

T ds

dv

Fanno

= c2 u2v(1 + G)

(173)

3.3.2 Rayleigh Flow

Constant-area, steady flow with heat transfer only:


P+u2 = I (175)

dh+udu = Qdx (176)

(177)

Change in entropy with volume along Rayleighline,P+ m2v = I

T ds

dv

Rayleigh

=c2 u2

vG (178)


25/85

3.4 Shock Jump Conditions 21

3.4 Shock Jump Conditions

The basic jump conditions,

1w1 = 2w2 (179)

P1+1w21 = P2+2w

22 (180)

h1+w21

2 = h2+

w222

(181)

s2 s1 (182)

or defining [f]f2 -f1

[w] = 0 (183)P+w2 = 0 (184)h+

w2

2

= 0 (185)

[s] 0 (186)

The Rayleigh line:

P2 P1v2 v1 = (1w1)

2 = (2w2)2 (187)or

[P][v]

= (w)2 (188)

Rankine-Hugoniot relation:

h2 h1= (P2 P1)(v2+v1)/2 or e2 e1 = (P2+P1)(v1 v2)/2 (189)

Velocity-P v relation

[w]2 = [P][v] or w2 w1 =

(P2 P1)(v2 v1) (190)

Alternate relations useful for numerical solution

P2 = P1+1w21

1 1

2

(191)

h2 = h1+1

2w21

1

12

2 (192)


26/85


3.4.1 Lab frame (moving shock) versions

Shock velocity

w1 = Us (193)Particle (fluid) velocity in laboratory frame

w2 = Us up (194)Jump conditions

2(Us up) = 1Us (195)P2 = P1+1Usup (196)

h2 = h1+up(Us up/2) (197)

Kinetic energy:

u2p2

=1

2(P2 P1)(v1 v2)

3.5 Perfect Gas Results

[P]

P1=

2

+ 1

M21 1

(198)

[w]

c1 = 2

+ 1 M1 1

M1 (199)[v]

v1= 2

+ 1

1 1

M21

(200)

[s]

R = lnPt2

Pt1(201)

Pt2Pt1

= 1

2

+ 1M21

1+ 1

11

+ 1

2 M21

1 + 1

2 M21

1

(202)

Shock adiabat or Hugoniot:

P2P1

=

+ 1

1v2v1

+ 1

1v2v1

1(203)

Some alternatives


27/85

3.6 Reflected Shock Waves 23

P2P1

= 1 + 2

+ 1

M21 1

(204)

=

2

+ 1 M2

1

1

+ 1 (205)21

= + 1

1 + 2/M21(206)

M22 =M21 +

2

12

1M21 1

(207)

Prandtls relation

w1w2 = c2 (208)

wherec is the sound speed at a sonic point obtained in a fictitious isentropic process inthe upstream flow.

c =

2

1+ 1

ht, ht= h+w2

2 (209)

3.6 Reflected Shock Waves

Reflected shock velocityURin terms of the velocity u2and density2behind the incidentshock or detonation wave, and the density3 behind the reflected shock.

UR =

u2

32

1 (210)PressureP3 behind reflected shock:

P3= P2+ 3u

22

32

1(211)

Enthalpy h3 behind reflected shock:

h3= h2+u22

2

32

+ 1

32 1

(212)

Perfect gas result for incident shock waves:

P3P2

=(3 1) P2

P1 ( 1)

( 1) P2P1

+ (+ 1)(213)


28/85


3.7 Detonation Waves

Jump conditions:

1w1 = 2w2 (214)

P1+1w21 = P2+2w

22 (215)

h1+w21

2 = h2+

w222

(216)

s2 s1 (217)

3.8 Perfect-Gas, 2- Model

Perfect gas with energy release q, different values ofandR in reactants and products.

h1 = cp1T (218)h2 = cp2T q (219)P1 = 1R1T1 (220)

P2 = 2R2T2 (221)

cp1 = 1R1

1 1 (222)

cp2 = 2R2

2 1 (223)(224)

Substitute into the jump conditions to yield:

P2P1

=1 +1M

21

1 +2M22(225)

v2v1

=2M

22

1M21

1 +1M21

1 +2M22(226)

T2T1

=1R12R2

1

1 1+1

2M21 +

q

c211

2 1+

1

2M2

2

(227)

Chapman-Jouguet Conditions Isentrope, Hugoniot and Rayleigh lines are all tan-gent at the CJ point

PCJ P1vCJ V1 =

P

v

Hugoniot

= P

v

s

(228)

which implies that the product velocity is sonic relative to the wave


29/85

3.8 Perfect-Gas, 2- Model 25

w2,CJ=c2 (229)

Entropy variation along adiabat

ds= 12T

(v1 v)2 d m2 (230)

Jouguets Rule

w2 c2v2

=

1 G2v

(v1 v) P

v

Hug

Pv

(231)

whereG is the Gruniesen coefficient.The flow downstream of a detonation is subsonic relative to the wave for points above

the CJ state and supersonic for states below.

3.8.1 2-Solution

Mach Number for upper CJ (detonation) point

MCJ=

H +(1+2)(2 1)21(1 1) +

H +(2 1)(2+ 1)21(1 1) (232)

where the parameterH is the nondimensional energy release

H =(2 1)(2+ 1)q21R1T1

(233)

CJ pressure PCJP1

=1M

2CJ+ 1

2+ 1 (234)

CJ densityCJ1

=1(2+ 1)M

2CJ

2(1 +1M2CJ) (235)

CJ temperatureTCJT1

=PCJ

P1

R11R2CJ

(236)

Strong detonation approximation MCJ 1

UCJ

2(22 1)q (237)CJ 2+ 1

21 (238)

PCJ 12+ 1

1U2CJ (239)

(240)


30/85


3.8.2 High-Explosives

For high-explosives, the same jump conditions apply but the ideal gas equation of stateis no longer appropriate for the products. A simple way to deal with this problem isthrough the nondimensional slopesof the principal isentrope, i.e., the isentrope passingthrough the CJ point:

s vP

P

v

s

(241)

Note that for a perfect gas, s is identical to = cp/cv, the ratio of specific heats. Ingeneral, if the principal isentrope can be expressed as a power law:

P vk = constant (242)

thens = k. For high explosive products,s 3. From the definition of the CJ point,we have that the slope of the Rayleigh line and isentrope are equal at the CJ point:

P

v

s

=PCJ P1vCJ V1 =

PCJvCJ

s,CJ (243)

From the mass conservation equation,

vCJ= v1s,CJ

s,CJ+ 1 (244)

and from momentum conservation, with PCJ P1, we have

PCJ =

1U2CJ

s,CJ+ 1 (245)


31/85

3.9 Weak shock waves 27

3.9 Weak shock waves

Nondimensional pressure jump

=[P]

c2 (246)

A useful version of the jump conditions (exact):

= M1 [w]c1

= M21[v]

v1

[w]

c1=M1

[v]

v1(247)

Thermodynamic expansions:

[v]

v1= + 2 +O()3 (248)

= [v]

v1 + [v]

v12 +O ([v])3 (249)

Linearized jump conditions:

[w]c1

= 2

2 +O()3 (250)

M1 = 1 2

[w]

c1+O

[w]

c1

2(251)

M1 = 1 +

2 +O()2 (252)

M2 = 1 2

+O()2 (253)

[c]

c1= ( 1) +O()2 (254)

M1 1 1 M2 (255)

Prandtls relation

c w1+12

[w] or w2 12

[w] (256)

Change in entropy for weak waves:

T[s]

c21=

1

63 +. . . or = 1

6

[v]

v

3+. . . (257)


32/85


3.10 Acoustics

Simple wavesP =c2 (258)

P = cu (259)+ for right-moving waves, - for left-moving wavesAcoustic Potential

u = (260)

P = o t

(261)

= oc2o

t (262)

Potential Equation

2 1c2o

2

t2 = 0 (263)

dAlemberts solution for planar (1D) waves

= f(x cot) +g(x +cot) (264)

Acoustic Impedance The specific acoustic impedanceof a medium is defined as

z= P

|u| (265)For a planar wavefront in a homogeneous medium z=c, depending on the directionof propagation.

Transmission coefficients A plane wave in medium 1 is normally incident on aninterface with medium 2. Incident (i) and transmitted wave (t)

ut/ui = 2z1z2+z1

(266)

Pt /Pi = 2z

2

z2+z1(267)

Harmonic waves (planar)

= A exp i(wt kx) +B exp i(wt+kx) c= k

k=2

=

2

T = 2f

(268)


33/85

3.11 Multipole Expansion 29

Spherical waves

=f(t r/c)

r +

g(t+r/c)

r (269)

Spherical source strength Q, [Q] = L3T1

Q(t) = limr0

4r2ur (270)

potential function

(r, t) = Q(t r/c)4r

(271)

Energy flux

= Pu (272)

Acoustic intensity for harmonic waves

I== 1

T

T0

dt=P

2rms

c (273)

Decibel scale of acoustic intensity

dB= 10 log10(I/Iref) Iref= 1012 W/m2 (274)

or

dB= 20 log10(P

rms/P

ref) P

ref= 2 1010

atm (275)Cylindrical waves,qsource strength per unit length [q] = L2T1

(r, t) = 12

tr/c

q() d(t )2 r2/c2

(276)

or

(r, t) = 12

0q(t r/c cosh ) d (277)

3.11 Multipole ExpansionPotential from a distribution of volume sources, strength qper unit source volume

(x, t) = 14

Vs

q(xs, t R/c)R

dVs R= |x xs| (278)

Harmonic sourceq= f(x) exp(it)


34/85


Potential function

(x, t) = 14

Vs

f(xs)exp i(kR t)

R dVs (279)

Compact source approximation:

1. source distribution is in bounded region around the origin xs< a,and smalla r= |x|

2. source distribution is compactka


35/85

3.12 Baffled (surface) source 31

Harmonic source

un= f(x) exp(iwt)

Fraunhofer conditions|xs| a a

ar 1

Approximate solution:

= exp i(kr wt)2r

As

f(xs)exp i xsdA


36/85


3.13 1-D Unsteady Flow

The primitive variable version of the equations is:

t

+ (u) = 0 (289)u

t + (uu) = P (290)

t

e+

u2

2

+

u(h+

u2

2)

= 0 (291)

s

t+ (us) 0 (292)

(293)

Alternative version

1

D

Dt = u (294)

Du

Dt = P (295)

D

Dt

h+

u2

2

=

P

t (296)

Ds

Dt 0 (297)

The characteristic version of the equations for isentropic flow (s = constant) is:

d

dt(u F) = 0 on C : dx

dt =u c (298)

This is equivalent to:

t(u F) + (u c)

x(u F) = 0 (299)

Riemann invariants:

F = c

d =

dPc

= dc

1 (300)

Bending of characteristics:

d

dP (u+c) =

c (301)

For an ideal gas:


37/85

3.13 1-D Unsteady Flow 33

F = 2c

1 (302)

Pressure-velocity relationship for expansion waves moving to the right into state (1), final

state (2) with velocity u2 0.

[P]

P1=

(+ 1)

4

u2c1

2 1 +1 +

4

+ 1

c1u2

2 u2> 0 (304)

Shock Tube Performance

P4P1

=

1 c1

c4

4 1+ 1

Ms 1

Ms

2441

1 +

211+ 1

M2s 1

(305)

Limiting shock Mach number for P4/P1

Ms c4c1

1+ 1

4 1 (306)


38/85


3.14 2-D Steady Flow

3.14.1 Oblique Shock Waves

Geometry:

w1 = u1sin (307)

w2 = u2sin( ) (308)v = u1cos =u2cos( ) (309)

21

=w1w2

= tan

tan( ) (310)

Shock Polar

[w]c1

= M1tan

cos (1 + tan tan ) (311)

[P]

1c21=

M21tan

cot + tan (312)

Real fluid results

w2 = f(w1) normal shock jump conditions (313)

= sin1 (w1/u1) (314)

= tan1 w2

u21 w21

(315)

Perfect gas result

tan =2cot

M21 sin

2 1

(+ 1)M21 2

M21sin2 1

(316)Mach angle

= sin1 1

M (317)

3.14.2 Weak Oblique Waves

Results are all for C+ family of waves, take - for C family.


39/85

3.14 2-D Steady Flow 35

= 12

1

M211

[w]

c1+O

[w]

c1

2(318)

=

M21 1M21

[w]

c1+O

[w]

c1

2(319)

[P]

1c21=

M21M21 1

+O()2 (320)

T1[s]

c21=

16

M61(M21 1)3/2

3 +O()4 (321)

Perfect Gas Results

[P]P1

= M21

M21 1+O()2 (322)

3.14.3 Prandtl-Meyer Expansion

d =

M21 1d u1u1

(323)

Function , d = -d

d

M2

11 + ( 1)M2 d MM (324)

Perfect gas result

(M) =

+ 1

1tan1

1+ 1

(M2 1)

tan1

M2 1 (325)

Maximum turning angle

max =

2

+ 1

1 1 (326)

3.14.4 Inviscid Flow

Crocco-Vaszonyi Relation

u

t + ( u) u= TS(h+ u

2

2) (327)


40/85


3.14.5 Potential Flow

Steady, homoentropic, homoenthalpic, inviscid:

(u) = 0 (328) u = 0 (329)h+

u2

2 = constant (330)

or withu = = (x, y)

(2x c2)xx+ (2y c2)yy+ 2xyxy = 0 (331)Linearized potential flow:

u = U+x (332)v = y (333)

0 =

M2 1

xx yy (334)Wave equation solution

=

M2 1 =f(x y) +g(x +y) (335)Boundary conditions for slender 2-D (Cartesian) bodies y(x)

f() =

U

dy

dx y 0 (336)Prandtl-Glauert Scaling for subsonic flows

(x, y) =inc(x,

1 M2y) 2inc = 0 (337)Prandtl-Glauert Rule

Cp= Cincp

1 M2(338)

3.14.6 Natural Coordinates

x = cos

s sin

n (339)

y = sin

s+ cos

n (340)

u = Ucos (341)

v = Usin (342)


41/85

3.14 2-D Steady Flow 37

The transformed equations of motion are:

U

s

+U

n

= 0 (343)

UU

s +

P

s = 0 (344)

U2

s+

P

n = 0 (345)

z =U

s U

n = 0 (346)

Curvature of stream lines, R = radius of curvature

s=

1

R (347)

Vorticity production

z = 1U o

Pon

+(T To)

U

S

n (348)

Elimination of pressure dP =c2d

(M2 1)Us

Un

= 0 (349)

U

n U

s = 0 (350)

3.14.7 Method of Characteristics

s( ) + 1

M2 1

n( ) = 0 (351)

s(+) 1

M2 1

n(+) = 0 (352)

(353)

Characteristic directions

C dn

ds = 1

M2 1 = tan (354)

Invariants

J = = constant on C (355)


42/85

38 4 INCOMPRESSIBLE, INVISCID FLOW

4 Incompressible, Inviscid Flow

4.1 Velocity Field Decomposition

Split the velocity field into two parts: irrotational ue, and rotational (vortical)uv.

u= ue+ uv (356)

Irrotational Flow Define the irrotational portion of the flow by the following twoconditions:

ue = 0 (357)

ue = e(x, t) volume source distribution (358)

This is satisfied by deriving ue from a velocity potential

ue = (359)

2 = e(x, t) (360)

Rotational Flow Define the rotational part of the flow by:

uv = 0 (361) uv = (x, t) vorticity source distribution (362)

This is satisfied by deriving uv from a vector potential B

uv = B (363) B = 0 choice of gauge (364)2B = (x, t) (365)

4.2 Solutions of Laplaces Equation

The equation 2= e is known asLaplacesequation and can be solved by the techniqueofGreensfunctions:

(x, t) =

G(x|)e(, t)dV (366)


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4.3 Boundary Conditions 39

For a infinite domain, Greens function is the solution to

2G = (x ) (367)G =

1

4

1

|x |=

1

4r (368)

r = |r| r= x (369)

This leads to the following solutions for the potentials

(x, t) = 14

e(, t)

r dV (370)

B(x, t) = 1

4

(, t)

r dV (371)

The velocity fields are

ue(x, t) = 1

4

re(, t)

r3 dV (372)

uv(x, t) = 14

r (, t)r3

dV (373)

If the domain is finite or there are surfaces (stationary or moving bodies, free surfaces,

boundaries), then an additional component of velocity, u

, must be added to insurethat the boundary conditions (described subsequently) are satisfied. This additionalcomponent will be a source-free, u = 0, irrotational u = 0 field. The generalsolution for the velocity field will then be

u= ue+ uv+ u (374)

4.3 Boundary Conditions

Solid Boundaries In general, at an impermeable boundary, there is no relativemotion between the fluid and boundary in the local direction nnormal to the boundary

surface.

u n= u n on the surface (375)In particular, if the surface is stationary, the normal component of velocity must vanishon the surface

u n= 0 on a stationary surface (376)


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For an ideal or inviscidfluid, there is no restriction on the velocity tangential to theboundary, slipboundary conditions.

u

t arbitrary on the surface (377)

For a realor viscousfluid, the tangential component is zero, since the relative velocitybetween fluid and surface must vanish, the no-slipcondition.

u= 0 on the surface (378)

Fluid Boundaries At an internal or free surface of an ideal fluid, the normal compo-nents of the velocity have to be equal on each side of the surface

u1 n= u2 n= u n (379)

and the interface has to be in mechanical equilibrium (in the absence of surface forcessuch as interfacial tension)

P1 = P2 (380)

4.4 Streamfunction

The vector potential in flows that are two dimensional or have certain symmetries canbe simplified to one component that can be represented as a scalar function known asthe streamfunction. The exact form of the streamfunction depends on the nature of

the symmetry and related system of coordinates.

4.4.1 2-D Cartesian Flows

Compressible In a steady two-dimensional compressible flow:

u= 0 u= (u, v) x= (x, y) ux

+ v

y = 0 (381)

The streamfunction is:

u=1

y v=

1

x (382)

Incompressible The density is a constant

u= 0 u= (u, v) x= (x, y) ux

+ v

y = 0 (383)

The streamfunction defined by


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4.4 Streamfunction 41

u=

y v=

x (384)

will identically satisfy the continuity equation as long as

2

xy

2

yx= 0 (385)

which is always true as long as the function (x, y) has continuous 2nd derivatives.Stream lines (or surfaces in 3-D flows) are defined by = constant. The normal to

the stream surface is

n =

|| (386)

Integration of the differential of the stream function along a path L connecting pointsx1and x2 in the plane can be interpreted as volume flux across the path

d = u nLdl= v dx+u dy (387)L

d = 2 1=L

u nLdl= volume flux across L (388)

where1 =(x1) and2 =(x2). For compressible flows, the difference in the stream-function can be interpreted as the mass flux rather than the volume flux.

For this flow, the streamfunction is exactly the nonzero component of the vector

potential

B= (Bx, By, Bz) = (0, 0, ) u= B=x y

y x

(389)

and the equation that the streamfunction has to satisfy will be

2= 2

x2 +

2

y2 = z (390)

where the z-component of vorticity is

z = vx

uy

(391)

A special case of this is irrotational flow with z = 0.

4.4.2 Cylindrical Polar Coordinates

In cylindrical polar coordinates (r,,z) with u = (ur, u, uz)


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x = r cos (392)

y = r sin (393)

z = z (394)

u = urcos usin (395)v = ursin +ucos (396)

w = uz (397)

The continuity equation is

u= 0 =1r

rurr

+1

r

u

+ uz

z (398)

Translational Symmetry in z The results given above for 2-D incompressible flowhave translational symmetry in z such that /z= 0. These can be rewritten in termsof the streamfunction (r, ) where

B= (0, 0, ) (399)

The velocity components are

ur = 1

r

(400)

u = r

(401)

The only nonzero component of vorticity is

z =1

r

rur

1r

ur

(402)

and the stream function satisfies

1

r

r

r

r

+

1

r

1

r

= z (403)

Rotational Symmetry in If the flow has rotational symmetry in, such that /= 0, then the stream function can be defined as

B=

0,

r, 0

(404)

and the velocity components are:


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4.4 Streamfunction 43

ur = 1r

z (405)

uz =

1

r

r (406)

The only nonzero vorticity component is

=ur

z uz

r (407)

The stream function satisfies

z

1

r

z

+

r

1

r

r

= (408)

4.4.3 Spherical Polar Coordinates

This coordinate system (r,,) results in the continuity equation

1

r2

r

r2ur

+

1

r sin

u

+ 1

r sin

(usin ) = 0 (409)

Note that the r coordinate in this system is defined differently than in the cylindricalpolar system discussed previously. If we denote byr the radial distance from the z-axisin the cylindrical polar coordinates, then r =r sin . With symmetry in the direction/, the following Stokesstream function can be defined

B= 0, 0, r sin (410)Note that this stream function is identical to that used in the previous discussion of thecase of rotational symmetry in for the cylindrical polar coordinate system if we accountfor the reordering of the vector components and the differences in the definitions of theradial coordinates.

The velocity components are:

ur = 1

r2 sin

(411)

u = 1r sin

r

(412)

The only non-zero vorticity component is:

=1

r

rur

1r

ur

(413)

The stream function satisfies


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1

r

r

1

sin

r

+

1

r

1

r2 sin

= (414)

4.5 Simple FlowsThe simplest flows are source-free and irrotational, which can be derived by a potentialthat satisfies the Laplaceequation, a special case ofue

2= 0 u= 0 (415)In the case of flows, that contain sources and sinks or other singularities, this equation

holds everywhere except at those singular points.

Uniform Flow The simplest solution is a uniform flowU:

= U x u= U= constant (416)In 2-D cartesian coordinates with U = Ux, the streamfunction is

= U y (417)

In spherical polar coordinates, Stokes streamfunction is

=U r2

2 sin2 U= Uz (418)

Source Distributions Single source of strengthQ(t) located at point1. The meaningofQ is the volume of fluid per unit time introduced or removed at point 1.

limr10

4r21u r1 = Q(t) r1= x 1 e= Q(t)(x 1) (419)which leads to the solution:

= Q(t)4r1

u=r1Q(t)

4r31=

r1Q(t)

4r21(420)

For multiple sources, add the individual solutions

u= 14

ki=1

riQir3i

(421)

In spherical polar cordinates, Stokes stream function for a single source of strengthQat the origin is

= Q4

cos (422)


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4.5 Simple Flows 45

For a2-D flow, the source strength qis the volume flux per unit length or area per unittime since the source can be thought of as a linesource.

u= urr ur = q

2r

= q

2

ln r = q

2

(423)

Dipole Consider a source-sink pair of equal strengthQlocated a distanceapart. Thelimiting process

0 Q Q (424)defines a dipole of strength . If the direction from the sink to the source is d, then thedipole moment vector can be defined as

d= d (425)

The dipole potential for spherical (3-D) sources is

= d r4r3

(426)

and the resulting velocity field is

u= 1

4

3d r

r5 r d

r3

(427)

If the dipole is aligned with the z-axis, Stokes stream function is

= sin2

4r

(428)

and the velocity components are

ur = cos

2r3 (429)

u = sin

4r3 (430)

The dipole potential for 2-D source-sink pairs is

= 2

cos

r (431)

and the stream function is

=

2

sin

r (432)

The velocity components are


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ur =

2

cos

r2 (433)

u =

2

sin

r2 (434)

Combinations More complex flows can be built up by superposition of the flows dis-cussed above. In particular, flows over bodies can be found as follows:

half-body: source + uniform flowsphere: dipole (3-D) + uniform flowcylinder: dipole (2-D) + uniform flowclosed-body: sources & sinks + uniform flow

4.6 Vorticity

Vorticity fields are divergence free In general, we have ( A) 0 so that thevorticity = u, satisfies

0 (435)

Transport The vorticity transport equation can be obtained from the curl of the mo-mentum equation:

DDt

= ( )u ( u) + T s+

(436)The cross products of the thermodynamic derivatives can be written as

T s= P v= P 2

(437)

which is known as the baroclinic torque.For incompressible, homogeneous flow, the viscous term can be written 2 and

the incompressible vorticity transport equation for a homogeneous fluid is

DDt = ( )u +2 (438)

Circulation The circulation is defined as

=

u dl=

n dA (439)

where is is a simple surface bounded by a closed curve .


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4.6 Vorticity 47

Vortex Lines and Tubes A vortex line is a curve drawn tangent to the vorticityvectors at each point in the flow.

dx

x=

dy

y=

dz

z(440)

The collection of vortex lines passing through a simple curve C form a vortex tube. Onthe surface of the vortex tube, we have n =0.

A vortex tube of vanishing area is a vortex filament, which is characterized by acirculation . The contribution du to the velocity field due to an element dlof a vortexfilament is given by the Biot Savart Law

du= 4

r dlr3

(441)

Line vortex A potential vortex has a singular vorticity field and purely azimuthal

velocity field. For a single vortex located at the origin of a two-dimensional flow

=z(r) u =

2r (442)

For a line vortex of strength i located at (xi, yi), the velocity field at point (x, y) canbe obtained by transforming the above result to get velocity components (u, v)

u = i2

y yi(x xi)2 + (y yi)2 (443)

v =

i

2

x

xi

(x xi)2 + (y yi)2 (444)(445)

Or setting = z

ui=i ri

2r2i(446)

whereri = i -xi.The streamfunctionfor the line vortex is found by integration to be

i= i2ln ri (447)

For a system of n vortices, the velocity field can be obtained by superposition of theindividual contributions to the velocity from each vortex. In the absence of boundariesor other surfaces:

u=n

i=1

i ri2r2i

(448)


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4.7 Key Ideas about Vorticity

1. Vorticity can be visualized as local rotation within the fluid. The local angularfrequency of rotation about the direction n is

fn= limr0

u2r

= 1

2

| n|2

2. Vorticity cannot begin or end within the fluid.

= 0

3. The circulation is constant along a vortex tube or filament at a given instant intime

tube

n dA= constant

However, the circulation can change with time due to viscous forces, baroclinictorque or nonconservative external forces. A vortex tube does not have a fixedidentity in a time-dependent flow.

4. Thompsons or Kelvins theoremVortex filaments move with the fluid and the cir-culation is constant for an inviscid, homogeneous fluid subject only to conservativebody forces.

D

Dt = 0 (449)

Bjerknes theoremIf the fluid is inviscid but inhomgeneous, (x, t), then the circu-lation will change due to the baroclinic torque P :

D

Dt =

dP

=

P 2

ndA (450)

Viscous fluids have an additional contribution due to the diffusion of vorticity intoor out of the tube.

4.8 Unsteady Potential Flow

Bernoullis equation for unsteady potential flow

P P= t

( ) + U2

2 ||

2

2 (451)


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4.9 Complex Variable Methods 49

Induced Mass If the external force Fext is applied to a body of mass M, then theacceleration of the bodydU/dt is determined by

Fext = (m+M

)

dU

dt

(452)

whereMis the induced mass tensor. For a sphere (3-D) or a cylinder (2-D), the inducedmass is simply M=miI.

mi,sphere = 2

3a3 (453)

mi,cylinder = a2 (454)

(455)

Bubble Oscillations The motion of a bubble of gas within an incompressible fluid canbe described by unsteady potential flow in the limit of small-amplitude, low-frequencyoscillations. The potential is given by the 3-D source solution. For a bubble of radiusR(t), the potential is

= R2(t)

r

dR

dt (456)

Integration of the momentum equation in spherical coordinates yields the Rayleighequa-tion

Rd2R

dt2 +3

2dR

dt2 = P(R) P (457)

4.9 Complex Variable Methods

Two dimensional potential flow problems can be solved in the complex plane

z=x+iy= r exp(i) =r cos +ir sin

The complex potentialis defined as

F(z) =+i (458)

and the complex velocity w is defined as

w= u iv= dFdz

(459)

NB sign ofv-term! The complex potential is an analytic function and the derivativessatisfy the Cauchy-Riemannconditions


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x =

y (460)

y = x (461)

which implies that both2 = 0 and2 = 0, i.e., the real and imaginary parts of ananalytic function represent irrotational, potential flows.

Examples

1. Uniform flow u = (U, V)

F = (U

iV)z

2. Line source of strength qlocated at zo

F = q

2ln(z z)

3. Line vortex of strength located at z

F = i2

ln(z z)

4. Source doublet (dipole) at z oriented along +x axis

F = 2(z z)

5. Vortex doublet at z oriented along +x axis

F = i

2(z z)

6. Stagnation point

F =C z2

7. Exterior corner flow

F =C zn 1/2 n 1


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4.10 Airfoil Theory 51

8. Interior corner flow, angle

F =C zn 1 n=

9. Circular cylinder at origin, radiusa, uniform flow U at x =

F =U(z+a2

z)

4.9.1 Mapping Methods

A flow in theplane can be mapped into thezplane using an analytic functionz= f().An analytic function is a conformal map, preserving angles between geometric featuressuch as streamlines and isopotentials as long as df /dzdoes not vanish. The velocity inthe -plane is w and is related to the z-plane velocity by

w=dF

d =

wd

dz

or w=dF

dz =

wdz

d

(462)

In order to have well behaved values ofw, require w =0 at point wheredz/d vanishes.

Blasius Theorem The force on a cylindrical (2-D) body in a potential flow is givenby

D

iL=

i

2

body w2 dz (463)

For rigid bodies

D= 0 L= U (464)where the lift is perpendicular to the direction of fluid motion at . The moment offorce about the origin is

M= 12

bodyzw2 dz

(465)

4.10 Airfoil TheoryRotating Cylinder The streamfunction for a uniform flowUover a cylinder of radiusawith a bound vortex of strength is

= Ur sin

1

a

r

2

2ln(

r

a) (466)

The stagnation points on the surface of the cylinder can be found at


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sin s=

4Ua (467)

The lift L is given by

L= U (468)The pressure coefficient on the surface of the cylinder is

CP =P P

12U

2

= 1 4sin2 + 42aU

sin

2aU

2(469)

Generalized Cylinder Flow If the flow at infinity is at angle w.r.t. the x-axis, thecomplex potential for flow over a cylinder of radius a, center and bound circulation is:

F(z) =U

exp(i)(z ) + a

2 exp(i)

z

i

2ln(

z a

) (470)

Joukowski Transformation The transformation

z= +2T

(471)

is the Joukowski transformation, which will map a cylinder of radius T in the -planeto a line segmenty =0, 2T x 2T. Use this together with the generalized cylinderflow in the plane to produce the flow for a Joukowski arifoil at an angle of attack. Theinverse transformation is

=z

2

z

2

2 2T (472)

Kutta Condition The flow at the trailing edge of an airfoil must leave smoothlywithout any singularities. There are two special cases:

For a finite-angle trailing edge in potential flow, the trailing edge must be a stag-nation point.

For a cusp (zero angle) trailing edge in potential flow, the velocity can be finite butmust be equal on the two sides of the separating streamline.

Application to Joukowski airfoil: Locating the stagnation point at T=+a exp i,the circulation is determined to be:

= 4aUsin(+) (473)


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4.11 Thin-Wing Theory 53

and the lift coefficient is

CL = L12

U2c= 8

a

c

sin(+) (474)

4.11 Thin-Wing Theory

The flow consists of the superposition of the free stream flow and an irrotational velocityfield derived from disturbance potentials t and c associated with the thickness andcamber functions.

u = Ucos +ut+uc (475)

v = Usin +vt+vc (476)

ut = t (477)uc = c (478)

(479)

where is the angle of attack and2i = 0.

Geometry A thin, two-dimensional, wing-like body can be represented by two surfacesdisplaced slightly about a wing chord aligned with the x-axis, 0 x c. The upper (+)and lower () surfaces of the wing are given by

y = Y+(x) for upper surface 0 x c (480)y = Y(x) for lower surface 0 x c (481)

and can be represented by a thickness functionf(x) and a camber functiong(x).

f(x) = Y+(x) Y(x) (482)g(x) =

1

2[Y+(x) +Y(x)] (483)

The profiles of the upper and lower surface can be expressed in terms off and g as

Y+(x) = g(x) +1

2

f(x) upper surface (484)

Y(x) = g(x) 12

f(x) lower surface (485)

The maximum thicknesst = O(f), the maximum camber h = O(g), and the angle ofattack are all considered to be small in this analysis

tc h

c 1 and ui, vi


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Boundary Conditions The exact slip boundary condition for inviscid flow on thebody is:

dY

dx

= v

u (x,Y(x))(487)

The linearized version of this is:

dYdx

=+ limy0

vt+vcU

(x,y)

(488)

with cos 1, and sin . This can be written as

vt(x, 0+) +vc(x, 0+) = U

g +

1

2f

U (489)

vt(x, 0

) +vc(x, 0

) = U g

1

2

fU (490)

wheref = df /dx and g = dg/dx.The boundary conditions are then divided between the thickness and camber distur-

bance flows as follows:

vt = 12

Uf for y 0 (491)

vc = U(g ) for y 0 (492)

In addition, the disturbance velocities have to vanish far from the body.

4.11.1 Thickness Solution

The potential t for the pure thickness case, which can be interpreted as a symmetricbody at zero angle of attack, can be calculated by the superposition of sources of strengthqdx using the general solution for potential flow

t(x, y) = 1

2

c0

ln(y2 + (x )2)q() d (493)The velocity components are:

ut = 1

2 c

0

(x )q() dy2 + (x )2

(494)

vt = 1

2

c0

yq() d

y2 + (x )2 (495)

Apply the linearized boundary condition to obtain

12

Udf

dx = lim

y0

1

2

c0

yq() d

y2 + (x )2 (496)


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Delta Function Representation The limit of the integrand is one of the representa-tions of the Dirac delta function

limy0

1

y

y2 + (x )2 =

(x

) (497)

where

(x ) =

0 x = x=

+

f()(x ) d=f(x) (498)

Source Distribution This leads to the source distribution

q(x) =Udf

dx (499)

and the solution for the velocity field is

ut = t

x =

U2

c0

(x )f() dy2 + (x )2 (500)

vt = t

y =

U2

c0

yf() d

y2 + (x )2 (501)

The velocity components satisfy the following relationships across the surface of the wing

[u] = u(x, 0+) u(x, 0) = 0 (502)[v] = v(x, 0+) v(x, 0) =q(x) (503)

Pressure Coefficient The pressure coefficient is defined to be

CP =P P

12U

2

= 1 u2 +v2

U2(504)

The linearized version of this is:

CP 2ut+ucU (505)For the pure thickness case, then we have the following result:

CP 1

c0

f() d

(x ) (506)

The integral is to be evaluated in the sense of the Principal valueinterpretation.


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Principal Value Integrals If an integral has an integrand g that is singular at =x, the principal value or finite part is defined as

P a

0

g() d= lim0

x

0

g() d+ a

x+

g() d (507)Important principal value integrals are

P

c0

d

(x )= ln

x

x c

(508)

and

P

c0

1/2d

(x ) = 1

xln

c+

x

c x

(509)

A generalization to other powers can be obtained by the recursion relation

P c0

nd

(x ) =xP c0

n1d

(x )cn

n (510)

A special case can be found for the transformed variables cos = 1 - 2/c

P

0

cos nd

cos cos o =sin nosin o

(511)

4.11.2 Camber Case

The camber case alone accounts for the lift (non-zero) and the camber. The potentialc for the pure camber case can be represented as a superposition of potential vorticesof strength(x) dx along the chord of the wing:

c = 1

2

c0

()tan1

y

x

d (512)

The velocity components are:

uc = c

x = 1

2

c0

y() d

y2 + (x )2 (513)

vc = c

y =

1

2 c

0

(x )() dy2 + (x )2

(514)

Theucomponent of velocity on the surface of the wing is

limy0

uc(x, y) =u(x, 0) = (x)2

(515)

Apply the linearized boundary condition to obtain the following integral equation for thevorticity distribution


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U

dg

dx

=

1

2P

c0

() d

(x )d (516)

The total circulation is given by

= c0

() d (517)

The velocity components satisfy the following relationships across the surface of the wing

[u] = u(x, 0+) u(x, 0) = (x) (518)[v] = v(x, 0+) v(x, 0) = 0 (519)

Kutta Condition The Kutta condition at the trailing edge of a sharp-edged airfoilreduces to

(x= c) = 0 (520)

Vorticity Distribution The integral equation for the vorticity distribution can besolved explicity. A solution that satisfies the Kutta boundary condition is:

(x) = 2U

c xx

1/2 +1

P

c0

g()

x

c

1/2d

(521)

The pressure coefficient for the pure camber case is

CP = (x)U

for y 0 (522)

The integrals can be computed exactly for several special cases, which can be expressedmost conveniently using the transformation

z=2x

c 1 = 2

c 1 (523)

P

11

1

z

1 +

1 d= (524)

P

11

1 2z d= z (525)

P

11

11 2(z )d= 0 (526)

P

11

1 2(z )d= (527)


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P

11

21 2(z )d= z (528)

Higher powers of the numerator can be evaluated from the recursion relation:

P

11

n1 2(z )d= zP

11

n11 2(z )d

2[1 (1)n]1(3) (n 2)

2(4) (n 1)(529)

4.12 Axisymmetric Slender Bodies

Disturbance potential solution using source distribution onx-axis:

(x, r) = 14

c

0

f() d

(x )2 +r2(530)

Velocity components:

u = U+

x=

1

4

c0

(x )f() d[(x )2 +r2]3/2 (531)

v =

r =

1

4

c0

rf() d

[(x )2 +r2]3/2 (532)

(533)

Exact boundary condition on bodyR(x)

v

u

(x,R(x))

=dR

dx (534)

Linearized boundary condition, first approximation:

v(x, r= R) =UdR

dx (535)

Extrapolation to x axis:

limr0

(2rv) = 2RdR

dxU (536)

Source strength

f(x) =U2RdR

dx =UA

(x) A(x) =R2(x) (537)

Pressure coefficient

CP 2uU

dR

dx

2(538)


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4.13 Wing Theory 59

4.13 Wing Theory

Wing span isb/2 < y


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Di = 1

L2

12

U2b2

(549)

Induced drag coefficient

CD,i = C2LAR

AR= b2/S bc

(550)


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61

5 Viscous Flow

Equations of motion in cartesian tensor form (without body forces) are:

Conservation of mass:

t +

ukxk

= 0 (551)

Momentum equation:

uit

+ukuixk

= Pxi

+ ikxk

(i= 1, 2, 3) (552)

Viscous stress tensor

ik = ui

xk+ uk

xi

+ik ujxj

sum on j (553)

Lames constant

= v 23

(554)

Energy equation, total enthalpy form:

htt

+ukhtxk

=P

t +

kiuixk

qixi

sum on i and k (555)

Thermal energy form

h

t +uk

h

xk=

P

t +uk

P

xk+ik

uixk

qixi


or alternatively

e

t+uk

e

xk= Puk

xk+ik

uixk

qixi


Dissipation function

=ikuixk

(558)

Fouriers law

qi= k Txi

(559)


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62 5 VISCOUS FLOW

5.1 Scaling

Reference conditions are

velocity U

length Ltime T

density viscosity

thermal conductivity k

Inertial flow Limit of vanishing viscosity, 0. Nondimensional statement:

Reynolds number Re=UL

1 P U2 (560)

Nondimensional momentum equation

Du

Dt = P+ 1

Re (561)

Limiting case, Re , inviscid flow

Du

Dt = P (562)

Viscous flow Limit of vanishing density, 0. Nondimensional statement:

Reynolds number Re=

UL

1 PU

L (563)

Nondimensional momentum equation

ReDu

Dt = P+ (564)

Limiting case, Re0, creeping flow.

P = (565)

5.2 Two-Dimensional Flow

For a viscous flow in two-space dimensions (x, y) the components of the viscous stresstensor in cartersian coordinates are

=

xx xyyx yy

=

2ux +

ux

+ vy

uy

+ vx

uy

+ vx

2v

y+

ux

+ vy

(566)

Dissipation function


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5.3 Parallel Flow 63

=

2

u

x

2+ 2

v

y

2+

u

y+

v

x

2+

u

x+

v

y

2(567)

5.3 Parallel Flow

The simplest case of viscous flow is parallel flow,

(u, v) = (u(y, t), 0) u= 0 = (y) only (568)Momentum equation

u

t = P

x +

y

u

y

(569)

0 = Py

(570)

We conclude from the y-momentum equation that P = P(x) only.Energy equation

e

t+u

e

x=

u

y

2+

x

k

T

x

+

y

k

T

y

(571)

5.3.1 Steady Flows

In these flows t = 0 and inertia plays no role. Shear stress is either constant or variesonly due to imposed axial pressure gradients.

Couette Flow A special case are flows in which pressure gradients are absent

P

x = 0 (572)

and the properties strictly depend only on the y coordinate, these flows have x

= 0.The shear stress is constant in these flows

xy = uy =w (573)

The motion is produced by friction at the moving boundaries

u(y= H) =U u(y= 0) = 0 (574)

and given the viscosity (y) the velocity profile and shear stress w can be determinedby integration


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64 5 VISCOUS FLOW

u(y) =w

yo

dy

(y) w=U

H0

dy

(y)

1(575)

The dissipation is balanced by thermal conduction in the y direction.

u

y

2=

y

k

T

y

(576)

Using the constant shear stress condition, we have the following energy integral

u q= qw = constant q= k Ty

qw =q(y= 0) (577)

This relationship can be further investigated by defining the Prandtlnumber

P r=

cP

k =

=

k

cP (578)

For gases,P r0.7, approximately independent of temperature. The Eucken relation isa useful approximation that only depends on the ratio of specific heats

P r 47.08 1.80 (579)

For many gases, both viscosity and conductivity can be approximated by power laws Tn, k Tm where the exponents n and m range between 0.65 to 1.4 depending onthe substance.

Constant Prandtl Number Assuming P r = constant and using d h = cpd T, theenergy equation can be integrated to obtain the Crocco-Busemannrelation

h hw+P r u2

2 = qw

wP r u (580)

For constant cP, this is

T =Tw P ru2

2cPqw

w

P r

cPu (581)

Recovery Temperature If the lower wall (y = 0) is insulated qw = 0, then the tem-perature at y = 0 is defined to be the recovery temperature. In terms of the conditionsat the upper plate (y = H), this defines a recovery enthalpy

hr = h(Tr) h(TH) +P r 12

U2 (582)

If the heat capacity cP= constant and we use the conventional boundary layer notation,for which TH =Te, the temperature at the outer edge of the boundary layer


69/85


Tr =Te+P r1

2

U2

cP(583)

Contrast with the adiabatic stagnation temperature

Tt = Te+1

2

U2

cP(584)

The recovery factoris defined as

r=Tr TeTt Te (585)

In Couette flow, r = P r. The wall temperature is lower than the adiabatic stagnationtemperatureTt when P r 1, then viscous dissipation generatesheat faster than it can be conducted away from the wall and Tr > Tt.

Reynolds Analogy If the wall is not adiabatic, then the heat flux at the lower wallmay significantly change the temperature profile. In particular the lower wall temperature(forcp = constant) is

Tw =Tr+ qwcPw

P rU (586)

In order to heat the fluid qw > 0, the lower wall must be hotter than the recoverytemperature.

The heat transfer from the wall can be expressed as a heat transfer coefficient orStantonnumber

St = qw

UcP(Tw Tr) (587)

whereqw is the heat flux from the wall into the fluid, which is positive when heat is beingadded to the fluid. The Stanton number is proportional to the skin frictioncoefficient

Cf= w12

U2 (588)

For Couette flow,

St = Cf2P r

(589)

This relationship between skin friction and heat transfer is the Reynolds analogy.


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66 5 VISCOUS FLOW

Constant properties Ifand k are constant, then the velocity profile is linear:

w =U

H u=

w

y (590)

The skin friction coefficient is

Cf= 2

Re Re=

U H

(591)

5.3.2 Poiseuille Flow

If an axial pressure gradient is present, Px


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The skin friction coefficient is traditionally based on the mean speed u and using thepipe diameter d = 2Ras the scale length.

u= Q

R2

=

P

x

R2

8

(599)

and is equal to

Cf= w1/2u2

= 16

RedRed=

ud

(600)

In terms of the Darcy friction factor,

=8wu2

= 64

Red(601)

Turbulent flow in smooth pipes is correlated by Prandtls formula

1

= 2.0log Red

0.8 (602)or the simpler curvefit

= 1.02 (log Red)2.5 (603)

5.3.3 Rayleigh Problem

Also known as Stokes first problem. Another variant of parallel flow is unsteady flowwith no gradients in the x direction. The Rayleigh problem is to determine the motionabove an infinite ( < x < ) plate impulsively accelerated parallel to itself.

Thex-momentum equation (for constant ) is

u

t =

2u

y2 (604)

The boundary conditions are

u(y, t= 0) = 0 u(y= 0, t >0) =U (605)

The problem is self similarand in terms of the similarity variable , the solution is

u= U

[2003] guide of fluid mechanics [j.e.shepherd, cit][ae.aph.101]

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