유체역학및열전달 chapter 4. basic equations of fluid flow...

부산대학교현규유체역학및열전달 1

유체역학및열전달

Chapter 4. Basic Equations of Fluid Flow (3)

-solving velocity profile

부산대학교 화공생명공학부현 규 (Kyu Hyun)

부산대학교현규

Balance Equation – Mass and Momentum

0=×Ñ+¶¶ )( Vrrt

l Continuity equation: Microscopic Mass balance

0=×Ñ+ )( VrrDtD

l Equation of motion: Microscopic Momentum balance

gτVVV rrr +×Ñ--Ñ=×Ñ+¶¶ pt

)()(

gτV rr +×Ñ--Ñ= pDtD

gVVVV rmr +Ñ+-Ñ=÷øö

çèæ Ñ×+¶¶ 2pt

0=×Ñ+ )( VrrDtD

( )( )TVVτ Ñ+Ñ-=-= mgm &

l Navier-Stokes Equation

Continuity equation

Newtonian fluid equation

0=×Ñ VIncompressible fluid

유체역학및열전달 2


Newton’s law of viscosity


( )T)( VVτ Ñ+Ñ-= m

÷÷÷÷÷÷÷

ø

ö

ççççççç

è

æ

¶¶

¶¶

¶¶

¶¶

¶¶

¶¶

¶¶

¶¶

¶¶

=¶¶

¶¶

¶¶

=Ñ

yw

zv

zu

yw

yv

yu

xw

xv

xu

wvuzyx

),,)(,,(V

÷÷÷÷÷÷÷

ø

ö

ççççççç

è

æ

¶¶

¶¶

¶¶

¶¶

¶¶

¶¶

¶¶

¶¶

¶¶

=Ñ

yw

yw

xw

zv

yv

xv

zu

yu

xu

T)( V

( )

÷÷÷÷÷÷÷

ø

ö

ççççççç

è

æ

¶¶

¶¶

+¶¶

¶¶

+¶¶

¶¶

+¶¶

¶¶

¶¶

+¶¶

¶¶

+¶¶

¶¶

+¶¶

¶¶

=Ñ+Ñ

yw

yw

zv

xw

zu

zv

yw

yv

xv

yu

zu

xw

yu

xv

xu

T

2

2

2

VV

÷÷÷÷÷÷÷

ø

ö

ççççççç

è

æ

¶¶

¶¶

+¶¶

¶¶

+¶¶

¶¶

+¶¶

¶¶

¶¶

+¶¶

¶¶

+¶¶

¶¶

+¶¶

¶¶

-=

yw

yw

zv

xw

zu

zv

yw

yv

xv

yu

zu

xw

yu

xv

xu

2

2

2

mτ

• Stress tensor should be symmetric

( )TVV Ñ+Ñ=g&


Equation of Motion

• Basic Equation of motion using Momentum Balance

• Mass Balance Equation (Continuity Eq’n) 적용 (Incompressible or compressible)

• Navier Stokes Equation (Newtonian Fluid) = μ (점도)가 일정

gτVVV rrr +Ñ-×Ñ-×-Ñ=¶

¶ pt

][)()(

gτV rr +×Ñ--Ñ= ][pDtD

gVV rmr +Ñ+-Ñ= 2pDtD


( )T)( VVτ Ñ+Ñ-= m


Step for finding velocity profile

Step 1. Draw a schematics of the system

Step 2. Make a list of assumptions (ex. z-direction only)

Step 3. Mathematical formulation in a proper coordinate

Step 4. Get Boundary conditions and initial condition

Step 5. Get Velocity profile (O.D.E. or sometimes P.D.E)


xgzu

yu

xu

xp

zuw

yuv

xuu

tu rmr +÷÷

ø

öççè

æ¶¶

+¶¶

+¶¶

+¶¶

-=÷÷ø

öççè

æ¶¶

+¶¶

+¶¶

+¶¶

2

2

2

2

2

2


Boundary Conditions (BCs)

• BCs are commonly encountered in momentum transfer

1. Fixed Boundary: constant value of velocity at the surface (No-slip)

2. Free Boundary: no stress at the surface

3. Max velocity : velocity gradient should be zero

4. Two fluid : same stress at the interface between two liquids

HzatVu == 00 == zatu

( ) ( )IIFluidxyIFluidxy

tt =

0=÷÷ø

öççè

æ¶¶

+¶¶

-==xv

yu

yxxy mtt


0=¶¶yu


Couette flow (Plane Couette flow) –문제 1


0u

),,(),,( 00uwvu ==V

B

x

y

z

)(yu

lx 방향으로만흐름이존재, 그러므로velocity u, v=w=0 그리고 u=u(y)

l No Force of gravity in x direction

lNo Pressure difference also in x-direction

Drag flow

• When a Newtonian fluid is confined between two broad parallel plates, separated by B, as shown in below figure. The upper plate is moving at a constant velocity.


Couette flow (Plane Couette flow) –문제 1• When a Newtonian fluid is confined between two broad parallel plates,

separated by B, as shown in below figure. The upper plate is moving at a constant velocity.

• x comp.

xgzu

yu

xu

xp

zuw

yuv

xuu

tu rmr +÷÷

ø

öççè

æ¶¶

+¶¶

+¶¶

+¶¶

-=÷÷ø

öççè

æ¶¶

+¶¶

+¶¶

+¶¶

2

2

2

2

2

2

lx방향으로만 흐름이 존재, 그러므로 velocity u, v=w=0 그리고 u=u(y)

l Force of gravity may be neglectedl Pressure difference also may be neglected

lx방향으로만 흐름이 존재, 그러므로 velocity u, v=w=0 그리고 u=u(y)

l Force of gravity may be neglectedl Pressure difference also may be neglected

2

2

0yu

¶¶

=\ 21 cycu +=

0

00uuBy

uy====

,,

Byuu 0=

0uB

AF

dyduss ==\

/tm


gVVVV rmr +Ñ+-Ñ=÷øö

çèæ Ñ×+¶¶ 2pt


Example 4.3 –문제 2• A Newtonian fluid confined between two parallel vertical plates• The plate on the left is stationary• That on the right is moving vertically at a constant velocity v0.• Assuming that flow is laminar, find the equation for the steady-

state velocity.

• Continuity equation

• Navier-Stokes equation

-y방향으로만흐름이존재, 그러므로 u=w=0 그리고 v=f(x)

0=¶¶

+¶¶

+¶¶

zw

yv

xu 0=

¶¶yv

ygzv

yv

xv

yp

zvw

yvv

xvu

tv rmr +÷÷

ø

öççè

æ¶¶

+¶¶

+¶¶

+¶¶

-=÷÷ø

öççè

æ¶¶

+¶¶

+¶¶

+¶¶

2

2

2

2

2

2

Steady state Continuitygr-

022

=-¶¶

-¶¶

\ gyp

xv rm



Example 4.3

.)( constgyp

xv

=+¶¶

=¶¶ rm 22

상수v는 x만의 함수

p는 y만의 함수

21

2

1

2CxCg

ypxv

Cgypx

dxdv

++÷÷ø

öççè

æ+

¶¶

=

+÷÷ø

öççè

æ+

¶¶

=

rm

rm

Bxatvvxatv====

0

00

-위의 식을 적분하면

-한번더 적분하면

BxvxBxg

ypv 0

2

21

+-÷÷ø

öççè

æ+

¶¶

-=\ )(rm

-경계조건을 적용하면



Layer flow with free surface (Gravity driven) (1)-교과서 page 83 (문제 3)

Assumptions and conditions• No end effect

• Velocity

• Pressure ,중력보다 작아서 무시

• Newton’s law of viscosity

d>>LW &

0=== vuxww ),(

)(zpp =

)(xw



z

x

Layer flow with free surface (2) –문제 3

)(xw



Layer flow with free surface (3)

br cosgzzzyzxz gzyxz

pzww

ywv

xwu

tw rtttr +÷÷

ø

öççè

æ¶¶

+¶¶

+¶¶

-¶¶

-=÷÷ø

öççè

æ¶¶

+¶¶

+¶¶

+¶¶

brt cosgxxz =

¶¶

\

1Cxgxz +=\ )cos( brt

-B.C. 1 1 00 == xatxzt xgxz )cos( brt =

÷øö

çèæ

¶¶

+¶¶

-=÷÷ø

öççè

æ¶¶

+¶¶

-=¶¶

-=xw

zu

yw

zv

zw

xzyzzz mtmtmt ,,2

xgxw

mbr cos

-=¶¶

\

- 마찰이 없어서 전단응력이 없다.


-Governing Equation (지배방정식)


Layer flow with free surface (4)

-B.C. 2 1 d== xatw 0

xgxw

mbr cos

-=¶¶

22

2Cxgw +-=

mbr cos

úúû

ù

êêë

é÷øö

çèæ-=\

22

12 dm

bdr xgw cos

-Total mass flow rate1

mbrdr

d

3

23

0

cosbgwbdxm == ò&31

2

3/

cos ÷÷ø

öççè

æ=

brmdgbm&



Flow through a Parallel Plate (Pressure driven)- 교과서 page 129 (Problems 5.1 & 5.3)


• Velocity

• Pressure , x방향 중력은 없음


BLW >>&

0=== wvyuu ),(

)(xpp =

Lower Plate

Upper Plate

x

y

B-

B+W

L



Flow through a Parallel Plate (2)

xgzu

yu

xu

xp

zuw

yuv

xuu

tu rmr +÷÷

ø

öççè

æ¶¶

+¶¶

+¶¶

+¶¶

-=÷÷ø

öççè

æ¶¶

+¶¶

+¶¶

+¶¶

2

2

2

2

2

2

xp

yu

¶¶

=¶¶

\ 22

m

02

2

Cconstxp

yu

==¶¶

=¶¶

\ .m

LppLxppxat

====

,, 00

0

00

==

=¶¶

=

uByxuyat

,

,

0Cxp=

¶¶

10 CxCp += 00 px

Lppp L +-=

02

2

Cyu=

¶¶m 20 Cy

Cyu

+=¶¶

m 320

2Cy

Lppu L +-=

m

úúû

ù

êêë

é÷øö

çèæ-

-=

220 12 B

yLBppu L

m)(


-Governing Equation

-B.C.


Flow through a Parallel Plate (3)

úúû

ù

êêë

é÷øö

çèæ-=

úúû

ù

êêë

é÷øö

çèæ-

D=

úúû

ù

êêë

é÷øö

çèæ-

-=

222220 11

21

2 Byu

By

LpB

By

LBppu L max)(

mm

-Average velocity

ò= udSSV1

max

max

max

u

yyBBu

WdyBy

BWu

uWdyBW

V

B

B

B

B

B

B

32

31

2

12

21

322

2

=

úûù

êëé -=

úúû

ù

êêë

é÷øö

çèæ-=

=

+

--

-

ò

ò

ò

Lower Plate

Upper Plate

W

WdydS =

dy

BWS 2=

32

=maxuV -주황색면으로흘러

들어가는유체에대해서

B-

B+



Flow through a circular tube (1)-page 102~103


• Velocity

• Pressure , 중력은 무시


RL >>

0=== quuzruu r),,(

)(zpp =

zr




• Continuity equation011 =

¶¶

+¶¶

+¶¶

zuu

rru

rr r)()( qq

),,(),,( uuuur 00== qV

• Equation of motion

rrrrrrr

rr g

rp

zuu

ru

rrru

rrru

zuuuu

ruu

tu r

qqm

qqr qqq +

¶¶

-úû

ùêë

é¶¶

+¶¶

-¶¶

+÷øö

çèæ

¶¶

¶¶

=÷÷ø

öççè

æ-

¶¶

+¶¶

+¶¶

+¶¶

2

2

22

2

2

2 211 )(

qqqqqqqqqq r

qqqm

qr gp

rzuu

ru

rrru

rrruu

zuuu

ru

ruu

tu rr

r +¶¶

-úû

ùêë

é¶¶

+¶¶

+¶¶

+÷øö

çèæ

¶¶

¶¶

=÷øö

çèæ +

¶¶

+¶¶

+¶¶

+¶¶ 1211

2

2

22

2

2

)(

zr gzp

zuu

rrur

rrzuuu

ru

ruu

tu r

qm

qr q +

¶¶

-úû

ùêë

é¶¶

+¶¶

+÷øö

çèæ¶¶

¶¶

=÷øö

çèæ

¶¶

+¶¶

+¶¶

+¶¶

2

2

2

2

2

11

)(ruu =\

zp

rur

rr ¶¶

=÷øö

çèæ¶¶

¶¶

\1m




q¶¶

=¶¶

=p

rp0

01 Cconst

zp

rur

rr==

¶¶

=÷øö

çèæ¶¶

¶¶

\ .m

-p는 z만의함수라는것을알수있다.

• Equation of motion에서 r과 θ 요소 방정식에서 다음과 같은결과가 나온다.

LppLzppzat

====

,, 00

10 CzCp += 00 pz

Lppp L +-=

0

00

==

=¶¶

=

uRrrurat

,

,0

1 Crur

rr=÷øö

çèæ¶¶

¶¶m




00 =¶¶

=rurat ,

rCrur

r m0=÷

øö

çèæ¶¶

¶¶

01 C

rur

rr=÷øö

çèæ¶¶

¶¶m 2

20

2CrC

rur +=¶¶

m

rCrC

ru 20

2+=

¶¶

m

-위의경계조건을만족시키려면 C2가 0이되어야한다

320

4CrCu +=

m0== uRrat ,

[ ]

úúû

ù

êêë

é÷øö

çèæ-

-=

úúû

ù

êêë

é-÷

øö

çèæ-=

úúû

ù

êêë

é-÷

øö

çèæ=-=-=

220

220

2202202020

14

14

14444

Rr

LRpp

Rr

LRpp

RrRCRrCRCrCu

LL

mm

mmmm

)()(




úúû

ù

êêë

é÷øö

çèæ-=

úúû

ù

êêë

é÷øö

çèæ-

D=

úúû

ù

êêë

é÷øö

çèæ-

-=

222220 11

41

4 Rru

Rr

LpR

Rr

LRppu L max)(

mm

l Average velocity

ò= udSSV1

( )

242

2

21

21

4

4

0

324

0

2

2

2

maxmax

max

max

uRRu

drrrRRu

rdrRr

Ru

rdruR

V

R

R

=×=

-=

úúû

ù

êêë

é÷øö

çèæ-=

=

ò

ò

ò

pp

pp

21

=maxuV

2RS p=

rdrdS p2=


부산대학교현규유체역학및열전달 23

유체역학및열전달

Chapter 4. Basic Equations of Fluid Flow

using Shell Balance

부산대학교 화공생명공학부현 규 (Kyu Hyun)


Flow through a Parallel Plate (Pressure driven)- 교과서 page 129 (Problems 5.1 & 5.3)


• Velocity

• Pressure , 중력은 무시


BLW >>&

0=== wvyuu ),(

)(xpp =

Lower Plate

Upper Plate

x

y

B-

B+W

L



Flow through a Parallel Plate (1)-Shell Balance를이용한방법

Lower Plate

Upper PlateW

L

x

y

xDyD

-두개의평판안에있는유체의 volume element를생각하자. 이러한계에서 z방향으로는특별한변화는없고단지길이W만가진다고생각하자. -빨간색으로표시된작은 volume element에대해서 momentum balance를생각하자.



Flow through a Parallel Plate (2)-shell Balance를이용한방법

gΦgτVV

gτVVV

rrr

rrr

+Ñ-×-Ñ=+Ñ-+×-Ñ=

+Ñ-×Ñ-×-Ñ=¶

¶

pp

pt

)(

][)()( τVVΦ += rConvective momentum과Molecular stress항을묶어서Total stress라고가정하자.

-Rate of x-momentum in across surface at x xxx

yW FD )(-Rate of x-momentum out across surface at x+Δx xxxxyW D+FD )(

-Rate of x-momentum in across surface at y yyx

xW FD )(

-Rate of x-momentum out across surface at y+Δy yyyx

xWD+

FD )(

-Pressure in across surface at xx

pyW )( D-Pressure out across surface at x+Δx xxpyW D+D )(



Flow through a Parallel Plate (3)-shell Balance를이용한방법

[ ] [ ] [ ] 0=D-D+FD-FD+FD-FDD+D+D+ xxxyyyxyyxxxxxxxx

pyWpyWxWxWyWyW )()()()()()(

- 이번 문제의 경우 이미 steady state를 가정했으므로 축적량(accumulation)은 없다. 옆에 보이는 빨간 volume element에걸리는 힘에 대해서 생각해보면 다음과 같다. 속도가 x 방향 쪽으로만존재하므로, x 방향의 힘만 고려하자.

0=D

-+

D

F-F+

D

F-FD+D+D+

xpp

yxxxxyyyxyyxxxxxxxx

0=¶¶

+¶

F¶+

¶F¶

xp

yxyxxx

τVVΦ += r

xuuuu xxxxxx ¶¶

-=+=F mrtr 22

÷÷ø

öççè

æ¶¶

+¶¶

-=+=Fxv

yuvuuu yxxyyx mrtr

02

=¶¶

+÷÷ø

öççè

æ¶¶

-¶¶

+¶¶

xp

yu

yxu mr

xp

yu

¶¶

=¶¶

\ 22

m유체역학및열전달 27

yxDD 로 나누기


Flow through a circular tube (1)-Shell balance를이용한방법


• Velocity

• Pressure


RL >>

0=== quuzruu r),,(

)(zpp =



Flow through a circular tube (2)-shell Balance를이용한방법

gΦgτVV

gτVVV

rrr

rrr

+Ñ-×-Ñ=+Ñ-+×-Ñ=

+Ñ-×Ñ-×-Ñ=¶

¶

pp

pt

)(

][)()( τVVΦ += rConvective momentum과Molecular stress항을묶어서Total stress라고가정하자.

-Rate of z-momentum in across annular surface at z zzz

rr FD )( p2-Rate of z-momentum out across annular surface at z+Δz

-Rate of z-momentum in across cylindrical surface at r-Rate of z-momentum in across cylindrical surface at r+Δr

-Pressure across surface at zz

prr )( Dp2-Pressure across surface at z+Δz

rz

zD

rD

zzzzrr

D+FD )( p2

rrzzr FD )( p2

rrrzzrr

D+FDD+ ))(( p2

zzprrπ

Δ+)Δ2(



Flow through a circular tube (3) -shell Balance를이용한방법



rz


zzzzzzzrrrr

D+FD-FD )()( pp 22

rrrzrrzzrrzr

D+FDD+-FD ))(()( pp 22

zzzprrprr

D+D-D )()( pp 22


Area= zrDp2

Area= rrDp2

벽면에서 걸리는 shear stress



[ ][ ] [ ] 02222

22

=D-D+FDD+-FD+

FD-FD

D+D+

D+

zzzrrrzrrz

zzzzzzz

prrprrzrrzr

rrrr

)()())(()(

)()(

pppp

pp

( ) ( )0=

D

-+

D

F-F+

D

F-FD+D+D+

zpp

rrrr

zr zzzrzrzrrzzzzzzzz

( )zpr

zrr

rzz

rz ¶¶

-¶F¶

-=F¶¶

τVVΦ += r

zuuuu zzzzzz ¶¶

-=+=F mrtr 22

÷øö

çèæ

¶¶

+¶¶

-=+=Fru

zuuuuu rrrzzrrz mrtr

zpr

rur

r ¶¶

=÷øö

çèæ¶¶

¶¶

\m

zrDDp2 로 나누어 주면

zpr

zur

rur

r ¶¶

-¶

¶-=÷

øö

çèæ

¶¶

-¶¶ )( 2rm

)(ruu =Q


유체역학및열전달 chapter 4. basic equations of fluid flow...

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