1

Chapter 6

Flow Analysis Using Differential Methods(Differential Analysis of Fluid Flow)

2

• In the previous chapter--• Focused on the use of finite control volume for the

solution of a variety of fluid mechanics problems.• The approach is very practical and useful since it

doesn’t generally require a detailed knowledge of the pressure and velocity variations within the control volume.

• Typically, only conditions on the surface of the control volume entered the problem.

• There are many situations that arise in which the details of the flow are important and the finite control volume approach will not yield the desired information

3

• For example --• We may need to know how the velocity varies over the cross

section of a pipe, or how the pressure and shear stress vary along the surface of an airplane wing.

we need to develop relationship that apply at a point,

or at least in a very small region ( infinitesimal volume)

within a given flow field.

involve infinitesimal control volume (instead of finite

control volume)

differential analysis (the governing equations are

differential equation)

4

• In this chapter—

• (1) We will provide an introduction to the differential equation that describe (in detail) the motion of fluids.

• (2) These equation are rather complicated, partial differential equations, that cannot be solved exactly except in a few cases.

• (3) Although differential analysis has the potential for supplying very detailed information about flow fields, the information is not easily extracted.

• (4) Nevertheless, this approach provides a fundamental basis for the study of fluid mechanics.

• (5) We do not want to be too discouraging at this point, since there are some exact solutions for laminar flow that can be obtained, and these have proved to very useful.

5

• (6) By making some simplifying assumptions, many other analytical solutions can be obtained.

for example , μ small 0 neglected inviscid flow.• (7) For certain types of flows, the flow field can be conceptually

divided into two regions— (a) A very thin region near the boundaries of the system in

which viscous effects are important. (b) A region away from the boundaries in which the flow is

essentially inviscid.• (8) By making certain assumptions about the behavior of the fluid

in the thin layer near the boundaries, and using the assumption of inviscid flow outside this layer, a large

class of problems can be solved using differential analysis .

• the boundary problem is discussed in chapter 9.• Computational fluid dynamics (CFD) to solve differential eq.

6

)(.)((.)(.)(.)(.)(.)(.)

.

.

problem. particular afor t,z, y, on x, dependly specifical components

velocity thesehow determine tois analysis aldifferenti of goals theof One

vtz

wy

vx

utt

D

a

a

z

uw

y

uv

x

uu

t

ua

dt

vd

z

vw

y

vv

x

vu

t

va

z

y

x

7

elementtheofndeformatioangularx

v

y

u

elementtheofndeformatiolinearz

w

y

v

x

uNote

v

ratedilationvolumetricvz

w

y

v

x

u

dt

d

,

,,:

fluid, ibleincompressan for 0

)(1

8

flow alIrrotation0

zero is axis-z thearound Rotation

only when ) (i.e.block undeformed an as axis-zabout Rotation

)(2

1 as such (6.12) Eq.From

2 Define

vor

y

u

x

v

x

v

y

uww

y

u

x

vw

Vcurlvwvorticity

OBOA

z

Vcurlv

wvuzyx

kji

ky

u

x

vj

x

w

z

ui

z

v

y

w

kwjwiw

vectorrotationtheW

zyx

2

1)(

2

1

2

1

})()(){(2

1

9

§ 6.2.1 Differential Form of Continuity Equation

inout

cv

vAvA

dAnv

zyxt

dt

dt

zyxd

)()(

element theof surfaces the through flow mass of rate The)(

)(

0

10

s

yx

xx

xxx

x

uuu

y

x

uuu

)(,)( as such , terms

order high neglecting --expansion seriesTaylor

2

)(|

2

)(|

direction-x theinflow The

2

2

2

zyxz

w

zyxy

v

zyxx

uzy

x

x

uuzy

x

x

uu

)(direction -zin rateNet

)24.6()(

direction -yin rateNet

similarly

)23.6()(

]2

[]2

[

direction-in x outflow mass of rateNet

11

. form aldifferenti in equation continuity The

)27.6(0

outflow mass of rateNet :

0][

0)( Since

z

w

y

v

x

u

t

zyxz

wzyx

y

vzyx

x

uNote

zyxz

wzyx

y

vzyx

x

uzyx

t

dAnvdt cscv

12

)31.6(0

)30.6(0

0

flow ibleincompressFor --

)29.6(0

0)(

fluid lecompressib of flowsteady For --

)28.6(0

form In vector

mechanics fluid of equations lfundamenta theof One --

z

w

y

v

x

uor

v

tconst

z

w

y

v

x

uor

v

vt

13

equation. continuity hesatisfy t torequired , w: Determine

?

flow ibleincompressan For

2 6. Example

222

w

zyzxyv

zyxu

),(2

3n Integratio

3)(2

0)()(

0

continuity ofequation thefrom

:Solution

2

222

yxcz

xzw

zxzxxz

w

z

wzyzxy

yzyx

x

z

w

y

v

x

u

14

§ 6.2.2 Cylindrical Polar Coordinates

15

01)(1

) flowunsteady or steady ( flow ibleincompressFor

0)()(1

)(1

flow lecompressib steady,For

scoordinate lcylindricain

equation y continuall theof form aldifferenti theis This

)33.6(0)()(1)(1

z

vv

rr

rv

r

vz

vr

vrrr

z

vv

rr

vr

rt

zr

zr

zr

16

§ 6.2.3 The Stream Function

)36.6(0

)2(0)(0

flow D-2 & plane, ible,incompress steady, of equation continuity For the

0

equation Continuity

y

v

x

u

flowDz

wcte

twhere

z

w

y

v

x

u

t

0)()(

eq. continuity thesatisfiesit that so ; where

function, stream the),(function a Define

xyyxy

v

x

u

xv

yu

yx

satisfied be willmass of onconservati

unknow one unknows two functionstream using

v

u

17

6.3. Example

)42.6(1

01)(1

0)()(1)(1

flow. D-2 place, , ibleIncompress

for equation y continuall the, scoordinate lcylindricaIn

rv

rv

v

rr

rv

rz

vv

rr

vr

rt

r

rzr

18

Example 6.3 Stream FunctionExample 6.3 Stream Function

• The velocity component in a steady, incompressible, two The velocity component in a steady, incompressible, two dimensional flow field aredimensional flow field are

Determine the corresponding stream function and show on a Determine the corresponding stream function and show on a sketch several streamlines. Indicate the direction of glow sketch several streamlines. Indicate the direction of glow along the streamlines.along the streamlines.

4xv2yu

19

Example 6.3 Example 6.3 SolutionSolution

(y)fx2(x)fy 22

12

Cyx2 22

From the definition of the stream functionFrom the definition of the stream function

x4x

vy2y

u

For simplicity, we set C=0For simplicity, we set C=0

22 yx2 Ψ=0Ψ=0

ΨΨ≠≠0012/

xy 22

20

§ 6.3 Conservation of Linear Momentum

amF

amFordt

vdmFEq

VCsmallFor

FdAnvvdvtdt

vmD

dmvdvP

Pdt

Ddv

dt

D

dt

vmD

sys

cvcv

csvc

cv

sys

syssys

sys

sys

systema for law 2nd Newtons The

)44.6(

..

)44.6()()(

momentumlinear for the t theorem transpor Reynolds theFrom

where

)(

momentumlinear For the

..

21

Figure 6.9 (p. 287)Components of force acting on an arbitrary differential area.

22

Figure 6.10 (p. 287)Double subscript notation for stresses.

23

Figure 6.11 (p. 288)Surface forces in the x direction acting on a fluid element.

24

§ 6.3.2 Equation of Motion

Velocities stresses ----- Unknowns

.rest at or motion in fluid)or (solid continuumany toapplicable also areThey

fluid. afor motion ofequation aldifferenti General

)50.6()(

)50.6()(

)50.6()(

using

cz

ww

y

wv

x

wu

t

w

zyxg

bz

vw

y

vv

x

vu

t

v

zyxg

az

uw

y

uv

x

uu

t

u

zyxg

dzyxm

zzyzxzz

zyyyxyy

zxyxxxx

zszbzzz

ysybyxx

xsxbxxx

maFFmaF

maFFmaF

maFFmaF

25

§ 6.4 Inviscid Flow

stress normal ecompressiv

0&

0 flow invisicidFor

. ssfrictionleor , nonviscous , inviscid be tosaid are

negligible be toassumed are stresses shearing thein which field Flow

zzyyxxP

0&0&

& , water andair assuch , fluidcommon Some

waterair

waterair small

26

§ 6.4.1 Euler’s Equations of Motion

)51.6()(

)51.6()(

)51.6()(

00 with (6.50c) & (6.50b) (6.50a) Eq From

cz

ww

y

wv

x

wu

t

w

z

Pg

bz

vw

y

vv

x

vu

t

v

y

Pg

az

uw

y

uv

x

uu

t

u

x

Pg

P

and

z

y

x

zzyyxx

(6.52) Eq solve to ) (usingSimplify

. solve toDifficulty

)52.6(])([

motion of equations Eulers as toreferredCommonly are equations These

vvt

vPg

27

§ 6.4.2 The Bernoulli Equation

equation Eulers equation Bernoulli section In this

law 2nd Newtons equation Bernoulli 2 3.section In

)(2

1

)()(2

becomes (6.53) Eq

)()(2

1)(

identity vector and ng Usi

)53.6()(

) statesteady ( 0 where

)52.6(])([

equation Eulers Form

2 vvzgvp

vvvvpzg

vvvvvv

zgg

vvpg

t

v

vvt

vpg

28.2

02

1

)]([

./)(0)]([

)()(

)(

Similarly

state.)steady if (

,,,,

)]([)()(2

1

streamline alonglength aldifferenti aLet

)(2

1

22

22

2

2

constgzvdp

gdzdvdp

sdvv

papeyofoutinvvbecausesdvv

dzkdzjdyidxkz

zdsz

dvsdv

dpdzz

pdy

y

pdx

x

p

dzdydxz

p

y

p

x

pdsp

sdvvsdzgsdvsdp

kdzjdyidx

ds

dsvvzgvp

29

streamline a along Flow

flow ibleIncompress

flowSteady

flow Inviscid

2

fluid ibleincompress Inviscid,For

2

constgzvp

30

§ 6.4.3 Irrotational Flow

flow alirrolationan of exampleAn

0

0

.)(

does. flow uniform a However,equations. threehesesatisfy t

not could field flow generalA

Vorticity)( 0Vorticity00)(2

1

flow alIrrotation

w

v

constUu

x

w

z

uz

v

y

wy

u

x

v

VorVorVw

31

§ 6.4.5 The Velocity Potential

mass. ofon conservati of econsequenc a --

function stream The

. field flow theofally irrotation theof econsequenc a --

potential velocityThe :Note

flow D-2 torestricted is

flow D-3 general afor defined becan

potential velocity function scale a is),,( where

,,

0

00)(2

1 flow alirrotationFor --

kwjviuv

zyx

zw

yv

xu

x

w

z

uz

v

y

wy

u

x

v

wvuzyx

kji

v

vorvw

32

flow. potential a calledcommonly is flow of typeThis

. field flow alIrrotation

, ibleincompress , Inviscid

)66.6(00

thatfollowsit ,)( flow alirrotation and

)0( fluid ibleincompressan For

2

2

2

2

2

22

equationLaplace

zyxor

v

v

�

pressures calculate Toequation Bernoulliwith

determined becan

conditionsboundary

withEq.(6.66) from

known is If

vor

w

v

u

33

)71.6(01

)(1

)70.6(;1

;

)69.6( Since

),,(

)68.6(1

)67.6((.)(.)1(.)

(.)

,, s,Coordinate

2

2

2

2

22

zrrr

rr

zv

rv

rv

evevevv

zrwhere

ez

er

er

ez

er

er

zrlCylindricaIn

zr

zzrrr

zr

zr

34

2133

1

22

,/10,30 if

(2)point at pressure (b)

potential velocity(a)

:e Determin

right on re Figu

&/2sin2 : Given

function stream 6.4 Example

zzmkgkpaP

mrsmr

)1()(2cos2)(2cos4

;1

;

2sin42sin2

2cos42)2(cos21

2sin211

)(

:

12

1

2

22

CrCdrrdrvdrv

zv

rv

rv

conditonslcylindricainpotentialVelocity

flowalIrrotation

rrr

v

rrr

rr

v

rv

rv

massofonconservati

conditionslcylindricainfunctionstreama

Solution

rr

zr

rr

35

)(2cos2(2) & (1) Eq

)2(.)(0)(

)(2)2sin(22sin4

)](2cos2[1

2sin4

1 Since

)1()(2cos2

)1(

2

11

122

12

12

AnsCr

constCCorC

Crr

Crr

r

rv

Cr

EqFrom

36

)(36/)416(/102

11030

(3) Eq From

/245.01616

5.0 (2),point At

/41616

1 (1),point At

16)2sin4()2cos4(

sin

)3(2

1

2

1

)(2

1

2

1

as written becan equation Bernoulli the

fluid, ibleincompress ,nonviscous a of flow alirrotationan For (b)

223332

2222

2

122

1

2222

222

22

2112

21222

21

21

1

Anskpasmmkgpap

smvrv

mr

smvrv

mr

rrrv

vvvevevvce

vvpp

zzgzvp

gzvp

rrr

37

§ 6.5 Some Basic, Plane Potential Flows

00

0)()(0

flow alIrrotation , ibleincompressfor (6.66) Eq From

0)()(

(6.74) Eq Usingflow) nal(Irrotatio

(6.72) Eq From

2

2

2

2

2

2

2

2

2

2

2

2

2

2

yxzyx

yyxxy

v

x

u

yxxxyy

x

v

y

u

flowplaneyx

38

§ 6.8 Viscous Flow

es. velociti& stresses hebetween t iprelationsh a establish tonecessary isIt

equations. than unknowns more are There

. fluid afor motion ofequation aldifferenti General

)50.6()(

)50.6()(

)50.6()(

. assuch

Eq.6.50, motion, of equations general derived previously thereturn tomust we

motion, fluid of analysis aldifferenti theinto effects viscouseincorporat To

cz

ww

y

wv

x

wu

t

w

zyxg

bz

vw

y

vv

x

vu

t

v

zyxg

az

uw

y

uv

x

uu

t

u

zyxg

zzyzxzz

zyyyxyy

zxyxxxx

39

§ 6.8.1 Stress - Deformation Relationships

)125.6(2

)125.6(2

)125.6(2

2

2

2

n.deformatio of rate the torelatedlinearly are

stresses t theknown tha isit , fluids Newtonian , ibleincompressFor

cz

wP

by

vP

ax

uP

z

wP

y

vP

x

uP

zz

yy

xx

zz

yy

xx

)125.6()(

)125.6()(

)125.6()(

fx

w

z

u

ey

w

z

v

dx

v

y

u

xzzx

zyyz

yzxy

40

)126.6(][

)126.6(]1

[

)126.6(]1

)([

)126.6(2

)126.6()1

(2

)126.6(2

fluids ibleincompress Newtonian,for stresses The

scoordinatepolar lcylindricaIn

fr

V

z

V

eV

rz

V

dV

rr

V

rr

cz

VP

br

VV

rP

ar

VP

zrrzzr

zzz

rrr

zzz

r

rrr

41

§ 6.8.2 The Navier–Stokes Equations

)127.6()()(

direction-z

)127.6()()(

direction -y

)127.6()()(

direction -x

(6.31), Eq ,continuity of Eq. and

(6.125f) ~ (6.125a) with (6.50c) ~ Eq.(6.50a) From

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

cz

ww

y

wv

x

wu

t

w

z

w

y

w

x

w

z

Pg

bz

vw

y

vv

x

vu

t

v

z

v

y

v

x

v

y

Pg

az

uw

y

uv

x

uu

t

u

z

u

y

u

x

u

x

Pg

z

y

x