case studies for fluid flow

8/7/2019 Case Studies for Fluid Flow

1/30

1

N-S equations are 2nd order , nonlinear , non-homogeneous

partial differential equations . They are very difficult to solve .

Only for very limited number of cases , they can be solved in

closed form .

CASE STUDIES: Application of Equations of Motion

(N-S Equations)


2/30

2

Flow of a Falling Film

The first example is the flow of a falling film . Consider the

Flow of a liquid at steady state along an inclined plate.

We are looking for (1) Volume flow rate

(2) Velocity profile perpendicular to the inclined plate


3/30

3

Take a control

volume from

the system

Assumptions:(1) Constant temperature Constant density , viscosity .

(2) Laminar Flow

(3) Neglect entrance & exit effects .


4/30

4

Start from the Navier-Stokes equation

For vz

:

vx=0 vz is not function of z vz is not function of y

0(steady state) vy=0 Assuming vz is not function of z

uniform pressure

Driving force-

Gravity only

GDE (Governing Differential Equation)

zzzzz

zz

yz

xz g

z

v

y

v

x

v

zz

vv

y

vv

x

vv

t

v

x

x

x

x

x

x

x

x!

x

x

x

x

x

x

x

x)(

2

2

2

2

2

2

RJ

02

2

!x

x@

z

zg

x

vR

0cos !x

x Fg

x

v

gz g

Fcos!g

gz


5/30

5

2nd order , needs 2 Boundary Conditions.

B.C.

,2

,

!!

!xx!

xx!!

z

zz

xz

vx

x

v

x

vx

H

QX


6/30

6

FVQ

F

VQ

cos1

cos

2

2

2

2

gx

v

g

x

v

z

z

!x

x

!xx

From GDE

,vz is a function of x only

Integrate

dx

dv

x

v zz !x

x@

1cos

1cxg

dx

dvz ! FVQ

From B.C.(1) c1=0

xgdx

dv! FV

Qcos

1


7/30

7

Integrate again

From B.C. (2)

Velocity Profile

2

2co

2

cxgvz

! FV

Q

])([H

HFVQ

HFV

Q

FV

Q

HFVQ

HFVQ

xgv

gxgv

gc

cgv

z

z

z

!

!@

!@

!!


8/30

8

When x=0 , vz=vz,max

For average velocity vz

For volume flow rate

0

3

00

co3

]3

co

|]3

co

]co

HFVQ

HHHFV

Q

HHFV

Q

HFHV

QHH

H

HH

!

!

!

!!

gv

g

xxg

dxx

gdxvv

z

zz

2

max,

cos2

1HFV

Q!@ gv

z

3cos3

1)( HFV

QH

wgwvQ z


9/30

9

Flow Between Parallel Plates

Consider the flow of fluid between parallel plates .

Assumption :

(1) Constant density , viscosity

(2) Laminar flow

(3) Neglect entrance effect


10/30

10

Start from the Navier-Stokes equation

For vz

:

vx is not function of x vz=0 vx is not function of z

steady state vy=0 vz is not function of x 0

Driving force-Pressure gradient

GDE

z

xxxxz

xy

xx

x gz

v

y

v

x

v

xz

vv

y

vv

x

vv

t

v

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x)(

2

2

2

2

2

2

RJ

xy

vx

x

x

!x

x J

R 2

2


11/30

11

B.C. 1

!!

!x

x!!

x

x

yx

vy

y

vy

H

X

L

PP

y

vL

PP

xLx

L

)(1

1

0

0

!

x

x

!

x

x

Q

V

J


12/30

12

2

20

0

1

10

)(

2

1

)(1

0)1.(.

)(1

cyL

PPv

y

L

PP

y

v

cCFromB

cy

L

PP

Y

V

Lx

Lx

LX

!

!

x

x

!

!

x

x

Q

Q

Q


13/30

13

From B.C. (2)

L

PPyv

L

PPyL

PPv

L

PPc

cLPP

Lx

LL

x

L

L

)()(

2

1

)(21)(

21

)(

2

1

)(21

22

22

2

2

22

!

!

!

!

HQ

HQQ

HQ

HQ


14/30

14

At y=0 , vx=vx,max

For average velocity


L

PP

v Lx 0max, 1 ! HQ

L

PPdyvv Lxx

0

2

03

1 !!

Q

H

H

H

L

PPwL 0

3

3

2

H


15/30

15

Flow through a Circular Tube

Consider the fully developed flow of a fluid in a along tube of

Length L and radius R .

Some assumptions

(1) Constant density and viscosity

(2) Laminar Flow

(3) Neglect entrance effect


16/30

16

Start from the Navier-Stokes eq in cylindrical coordinate

r

z L

Area=2rL Area=2(r+r)L

r

zzzzz

zzz

rz g

z

vv

rr

vr

rrzz

vv

v

r

v

r

vv

t

v

x

x

x

x

x

x

x

x

x

x!

x

x

x

x

x

x

x

x]

1)(

1[

2

2

2

2

2 UR

J

U

U

vr=0 vz not function of z

steady state v=0 vz not function of vz not function of z

gzr

vr

rr

gr

vr

rrz

z

z

xx

!x

x

xx

x

x

xx

xx

!

JR

RJ

)(1

)(1

0GDE , Driving force

gravity+pressure gradient


17/30

17

B.C. (1) 0,0 !x

x!

r

vr

z

0, !!z

vR


18/30

18

1

1

1

cgr

r

Lr

v

r

Integrate

gL

rr

vr

r

gz

Pr

r

vr

r

Lz

Lz

z

!

!

!

VL

VL

L

V

L

From B.C. (1) c1=0

2(

1 20

gr

r

L

PP

r

v

r

Lz

VL

!x

x

)2

(1

0 gr

L

PP

r

vLz

VL

!x

x


19/30

19

Integrate again

From B.C.(2)

2

2

)

4

(1

cgrr

L

v Lz

!

V

L

)4

(1

)4

(1

0

0

0

grRL

c

cgrR

L

L

L

VL

VL

!

!

])(1)[4

)((

)4

(1

)4

(1

22

0

2

0

2

0

R

rRgr

L

PPv

grR

L

PPgr

r

L

PPv

Lz

LLz

!

!

LV

VL

VL

Velocity Profile


20/30

20

At r=0 , vz=vz,max

For average velocity


Hagen-Poiseuille Law

related melt penetration into porous refractory

4((

2

0

max,LV

R

gL

PP

vL

z

!@

UT

T

8)(1

2

0

2

0 02

R

gL

PPd d

Rv L

R

zz

!!

)()(

4

LTU Rg !

LU

U

1

4

w

w R


21/30

21

Creeping flow around a solid sphere

Consider the flow of an incompressible fluid about a solid sphere.The fluid approaches the sphere upward along the z-axis with a

uniform velocity v

Assumptions

(1) Const ,

(2) V is 0

(3) Very slow flow ( , Reynolds number < 1)

Application : Inclusions Removal

Rg

Dv


22/30

22

Start from the Navier-Stokes equations in sphere coordinates(r, , )

For r-component

rrr

rrr

r

r

gv

rv

r

v

rv

rv

r

P

r

vvv

r

vv

r

v

r

vv

t

v

VJU

UU

L

JUUV

J

UU

JUJU

xx

xx

xx!

x

x

x

x

x

x

x

x

)sin

2cot222(

)sin

(

2222

2

22

Accelerationterms=0 due to very slow flow

0 v=0


23/30

23

: known as nabla or del is the vector differential operator .

For Cartesian coordinate

kz

jy

ix

x

x

x

x

x!

2 known as Laplacian operator

2

2

2

2

2

22

zyx xx

xx

xx!

For spherical coordinate

!

xx

xx

xx!

sinJU H

JUH

UH

rrrr


24/30

24

For the component

U

JUU

JUUJUUUU

VJU

U

UUL

U

U

JUUV

gv

rr

vv

rV

P

r

r

v

r

vvv

r

vv

r

v

r

vv

t

v

r

r

r

x

x

x

x

x

x

x

x

x

x

x

x

x

x

)sin

cos

sin(

1

)cot

sin(

Accelerationterms=0

V=0

Forcontinuity equation

0)(sin

1)sin(

sin

1)(

1!

x

x

x

x

x

x

x

xJ

VJ

VVV

vr

vr

vrrrt

r

const v=0


25/30

25

G.D.E.

)si

2(

)cot222

(

)si(si

)(

222

2

222

2

2

2

!xx

xx

!x

x

xx

!xx

xx

UU

U

UU

U

VUUU

VUU

UUU

gvvv

gvv

vv

vv

3 eqns , 3 unknowns (P , vr , v)


26/30

26

Solve the above equations , we obtain

U

U

UL

V

UL

X

U

U

sin]

os]22

os2

sin2

2

r

R

r

Rvv

r

R

r

Rvv

r

R

R

vgzPP

r

R

R

v

r

r

!

!

!

!

g

g

g

g


27/30

27

The normal force acting on the solid surface is due to the pressure

at the solid surface , where r=R , z=Rcos

JUUUL

UV

UL

UV

T TddR

R

vgRPF

R

vgRPP

n

Rr

si]s2

3s[

cos2

3cos|

22

0 0 0

0

g

g

!

!@

!

g! RvgRFn TLVT 2

Buoyant force when Buoyant force due to

the fluid is stegnant fluid movement

Fn : Buoyant force on form drag


28/30

28

As the surface , there is also shear stress acting tangentially , r

We are only interested in the z-component of the shear stress ,

rsin


29/30

29

at the surface

g

g

g

!

!

!

!

Rv

ddRR

vF

R

v

t

Rrr

TL

JL

L

XT T

U

4

ii2

3

i2

3

|

2

0 0

2

Ft=frictio drag


30/30

30

Total force acting upward

Stokes Law

Fs Fk

g! RvgRF TLVT 6

3

4 3

stegnant with fluid movement

R F

(inclusions rise fast)

case studies for fluid flow

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