Basic Governing Differential Equations

Basic Governing Differential Equations

CEE 331

April 18, 2023

CEE 331

April 18, 2023

OverviewOverview

Continuity Equation Navier-Stokes Equation

(a bit of vector notation...) Examples (all laminar flow)

Flow between stationary parallel horizontal plates

Flow between inclined parallel plates Pipe flow (Hagen Poiseuille)

Continuity Equation Navier-Stokes Equation

(a bit of vector notation...) Examples (all laminar flow)

Flow between stationary parallel horizontal plates

Flow between inclined parallel plates Pipe flow (Hagen Poiseuille)

Conservation of Mass in Differential Equation Form

Conservation of Mass in Differential Equation Form

v x z

xx

yy

z z

FHG

IKJy

y

FHG

IKJy

y

t

y x z t

y x z

Mass flux into differential volume

Mass flux out of differential volume

Rate of change of mass in differential volume

vvy

y

FHG

IKJv

vy

y

FHG

IKJ x z x z

Continuity EquationContinuity Equation

vy

vy t

vy

vy t

Mass flux out of differential volume

vvy

y vy

yvy y

y x z

FHG

IKJ

2 vvy

y vy

yvy y

y x z

FHG

IKJ

2 Higher order termHigher order term

outout inin Rate of mass decreaseRate of mass decrease

vy t

0

vy t

0 1-d continuity equation1-d continuity equation

vvy

y vy

y x z v x zt

y x z

FHG

IKJ

vvy

y vy

y x z v x zt

y x z

FHG

IKJ

u, v, w are velocities in x, y, and z directions

u, v, w are velocities in x, y, and z directions

Continuity EquationContinuity Equation

0tr

r¶

+Ñ× =¶

V

t

u

x

v

y

w

z

af af a f0 3-d continuity equation3-d continuity equation

If density is constant...

ux

vy

wz

0

Vector notationVector notation

0Ñ× =Vor in vector notation

True everywhere! (contrast with CV equations!)True everywhere! (contrast with CV equations!)

Continuity IllustratedContinuity Illustrated

ux

vy

wz

0

x

y

What must be happening?

ShearShear

GravityGravityPressurePressure

Navier-Stokes EquationsNavier-Stokes Equations

momentummomentum

Derived by Claude-Louis-Marie Navier in 1827

General Equation of Fluid Motion Based on conservation of ___________ with forces…

____________ ___________________ ___________________

U.S. National Academy of Sciences has made the full solution of the Navier-Stokes Equations a top priority

Derived by Claude-Louis-Marie Navier in 1827

General Equation of Fluid Motion Based on conservation of ___________ with forces…

____________ ___________________ ___________________

U.S. National Academy of Sciences has made the full solution of the Navier-Stokes Equations a top priority

Navier-Stokes Equation

Inertial forces [N/m3], a is Lagrangian acceleration

Pressure gradient (not due to change in elevation)

Shear stress gradient

h is vertical (positive up)

Navier-Stokes EquationsNavier-Stokes Equations

V r =åa FV r =åa F

( ) 2h pr g m=-Ñ + + Ña V( ) 2h pr g m=-Ñ + + Ña V

a a

ph ph

2mÑ =V2mÑ =Vdudx

t m=dudx

t m=

What is in a static fluid? _____( )h pg- Ñ +( )h pg- Ñ + Zero!

Is acceleration zero when dV/dt = 0?

x2 1VVV =D - =2 1VVV =D - =

tV

a V V¶

= + ×ѶtV

a V V¶

= + ×Ѷ

Over what time did this change of velocity occur (for a particle of fluid)? x

tVD

D =x

tVD

D =

space tV

a =DD

=space tV

a =DD

=V

Vx

DDV

Vx

DD

1V1V 2V2V

VDVD

111111 VVM QAV 111111 VVM QAV

222222 VVM QAV 222222 VVM QAV

( )1 2 QM M Vr+ = D( )1 2 QM M Vr+ = D

Why no term?tV¶¶tV¶¶

Lagrangian accelerationLagrangian acceleration

Notation: Total DerivativeEulerian Perspective

Notation: Total DerivativeEulerian Perspective

u v wt x y z

¶ ¶ ¶ ¶= + + +¶ ¶ ¶ ¶V V V V

a u v wt x y z

¶ ¶ ¶ ¶= + + +¶ ¶ ¶ ¶V V V V

a

( , , , )D

t x y z u v wDt t x y za a a a a¶ ¶ ¶ ¶

= + + +¶ ¶ ¶ ¶

( , , , )D

t x y z u v wDt t x y za a a a a¶ ¶ ¶ ¶

= + + +¶ ¶ ¶ ¶

( , , , )D

t x y z u v wDt t x y z

¶ ¶ ¶ ¶= + + +¶ ¶ ¶ ¶

V V V V V( , , , )

Dt x y z u v w

Dt t x y z¶ ¶ ¶ ¶

= + + +¶ ¶ ¶ ¶

V V V V V

tV

a V V¶

= + ×ѶtV

a V V¶

= + ×Ѷ

kjizyx

()()()

() kjizyx

()()()

()

( ) () () ()() u v w

x y z¶ ¶ ¶

×Ñ = + +¶ ¶ ¶

V( ) () () ()() u v w

x y z¶ ¶ ¶

×Ñ = + +¶ ¶ ¶

V

( , , , )D dt dx dy dz

t x y zDt t dt x dt y dt z dta a a a a¶ ¶ ¶ ¶

= + + +¶ ¶ ¶ ¶

( , , , )D dt dx dy dz

t x y zDt t dt x dt y dt z dta a a a a¶ ¶ ¶ ¶

= + + +¶ ¶ ¶ ¶

Total derivative (chain rule)

Material or substantial derivative

N-S

Application of Navier-Stokes Equations

Application of Navier-Stokes Equations

The equations are nonlinear partial differential equations

No full analytical solution exists The equations can be solved for several

simple flow conditions Numerical solutions to Navier-Stokes

equations are increasingly being used to describe complex flows.

The equations are nonlinear partial differential equations

No full analytical solution exists The equations can be solved for several

simple flow conditions Numerical solutions to Navier-Stokes

equations are increasingly being used to describe complex flows.

2h pr g m=- Ñ - Ñ + Ña V2h pr g m=- Ñ - Ñ + Ña V

ph 0 ph 0

Navier-Stokes Equations: A Simple Case

Navier-Stokes Equations: A Simple Case

No acceleration and no velocity gradients No acceleration and no velocity gradients

constantp yg=- +constantp yg=- +

hp hp

zh

zp

yh

yp

xh

xp

zh

zp

yh

yp

xh

xp

0 0p p px y z

g¶ ¶ ¶

= =- =¶ ¶ ¶

0 0p p px y z

g¶ ¶ ¶

= =- =¶ ¶ ¶

xyz could have any orientation

Let y be vertical upward

1

For constant

Infinite Horizontal Plates: Laminar Flow

Infinite Horizontal Plates: Laminar Flow

2h pr g m=- Ñ - Ñ + Ña V

20 h pg m=- Ñ - Ñ + Ñ V

Derive the equation for the laminar, steady, uniform flow between infinite horizontal parallel plates.

2

20p ux y

mæ ö¶ ¶

=- + ç ÷è ø¶ ¶

0py

g¶

=- -¶

0 0

02

2

2

2

2

2

FHG

IKJ h

ypy

vx

vy

vz

02

2

2

2

2

2

FHG

IKJ h

zpz

wx

wy

wz

02

2

2

2

2

2

FHG

IKJ h

xpx

ux

uy

uz

11

y

x

Hydrostatic in yHydrostatic in y

0v =

0w =

x

y

z

Infinite Horizontal Plates: Laminar Flow

Infinite Horizontal Plates: Laminar Flow

2

20p ux y

mæ ö¶ ¶

=- + ç ÷è ø¶ ¶

2

20p ux y

mæ ö¶ ¶

=- + ç ÷è ø¶ ¶

2

2

dp d udx dy

mæ ö

= ç ÷è ø

2

2

dp d udx dy

mæ ö

= ç ÷è ø

2

2

dp d udy dy

dx dym

óóôôô ôõ õ

æ ö= ç ÷è ø

2

2

dp d udy dy

dx dym

óóôôô ôõ õ

æ ö= ç ÷è ø

dp duy A

dx dymæ ö

+ = ç ÷è ødp du

y Adx dy

mæ ö

+ = ç ÷è ø

dp duy A dy dy

dx dym

óóôô

ô ôõ õ

æ öé ù+ = ç ÷ê ú è øë û

dp duy A dy dy

dx dym

óóôô

ô ôõ õ

æ öé ù+ = ç ÷ê ú è øë û

2

2y dp

Ay B udx

m+ + =2

2y dp

Ay B udx

m+ + =

Pressure gradient in x balanced by shear gradient in yPressure gradient in x balanced by shear gradient in y

dudy

t m=dudy

t m=

ddyt

=ddyt

=

No a so forces must balance!

Now we must find A and B… Boundary Conditions

negativenegative

Infinite Horizontal Plates: Boundary Conditions

Infinite Horizontal Plates: Boundary Conditions

No slip condition

u = 0 at y = 0 and y = au = 0 at y = 0 and y = a a

B 0

a dpdx

Aa2

20 A

a dpdx

2

( )2

y y a dpu

dxm-

=2a dp

ydx

t æ ö= -è ø2

a dpy

dxt æ ö= -

è ø

y

du dpy A

dy dxmæ ö

= +ç ÷è ødu dp

y Ady dx

mæ ö

= +ç ÷è ø

dpdx

dpdx

let be___________

uu

2

2y dp

Ay B udx

m+ + =

What can we learn about ?

Laminar Flow Between Parallel Plates

Laminar Flow Between Parallel Plates

2h pa Vr g m=- Ñ - Ñ + Ñ

20 h pg m=- Ñ - Ñ + Ñ V

2

20h p ux x y

g mæ ö¶ ¶ ¶

=- - + ç ÷è ø¶ ¶ ¶

02

2

2

2

2

2

FHG

IKJ h

xpx

ux

uy

uz

U

a

u

y

x

No fluid particles No fluid particles are acceleratingare accelerating

Write the x-componentWrite the x-component

Flow between Parallel Plates Flow between Parallel Plates

2

20h p ux x y

g mæ ö¶ ¶ ¶

=- - + ç ÷è ø¶ ¶ ¶

( ) 2

20d h p d u

dx dy

gm

+ æ ö=- + ç ÷è ø

General equation describing laminar General equation describing laminar flow between parallel platesflow between parallel plates

( )2

2

d h pd udy dx

gm

+æ ö=ç ÷è ø

u is only a function of y

h and p are only functions of x

Flow Between Parallel Plates: Integration

Flow Between Parallel Plates: Integration

( )2

2

d p hd udy dx

gm

+=

( )2

2

d p hd udy dx

gm

+=

( )du dy p h A

dy dxm g= + + =( )du d

y p h Ady dx

m g= + + =

( )2

2

d u ddy p h dy

dy dxm g= +

ó óôô õõ( )

2

2

d u ddy p h dy

dy dxm g= +

ó óôô õõ

( )du ddy y p h A dy

dy dym g

æ ö= + +ç ÷è øóó

ô ôõ õ( )du d

dy y p h A dydy dy

m gæ ö

= + +ç ÷è øóó

ô ôõ õ

( )2

2y d

u p h Ay Bdx

m g= + + +( )2

2y d

u p h Ay Bdx

m g= + + +

U

a

u

y

x

tt

u = U at y = a

Boundary ConditionsBoundary Conditions

( )2

2y d

u p h Ay Bdx

m g= + + +( )2

2y d

u p h Ay Bdx

m g= + + +

Boundary conditions

B 000 B 000

Boundary conditions

( )2

2a d

U p h Aadx

m g= + +( )2

2a d

U p h Aadx

m g= + + ( )2

U a dA p h

a dxm

g= - +( )2

U a dA p h

a dxm

g= - +

( )2

2Uy y ay d

u p ha dx

gm-

= + +( )2

2Uy y ay d

u p ha dx

gm-

= + +

u = 0 at y = 0

Discharge per unit width!

DischargeDischarge

( )2

00

2

aa y y ay d

Q udy U p h dya dx

gm

æ ö-= = + +ç ÷è ø

óôõ

ò ( )2

00

2

aa y y ay d

Q udy U p h dya dx

gm

æ ö-= = + +ç ÷è ø

óôõ

ò

( )3

2 12Ua a d

Q p hdx

gm

= - +( )3

2 12Ua a d

Q p hdx

gm

= - +

( )2

2y y ay d

u U p ha dx

gm-

= + +( )2

2y y ay d

u U p ha dx

gm-

= + +

Example: Oil SkimmerExample: Oil Skimmer

An oil skimmer uses a 5 m wide x 6 m long moving belt above a fixed platform (=30º) to skim oil off of rivers (T=10 ºC). The belt travels at 3 m/s. The distance between the belt and the fixed platform is 2 mm. The belt discharges into an open container on the ship. The fluid is actually a mixture of oil and water. To simplify the analysis, assume crude oil dominates. Find the discharge and the power required to move the belt.

An oil skimmer uses a 5 m wide x 6 m long moving belt above a fixed platform (=30º) to skim oil off of rivers (T=10 ºC). The belt travels at 3 m/s. The distance between the belt and the fixed platform is 2 mm. The belt discharges into an open container on the ship. The fluid is actually a mixture of oil and water. To simplify the analysis, assume crude oil dominates. Find the discharge and the power required to move the belt.

30º

hl

= 1x10-2 Ns/m2

= 8430 N/m3

Example: Oil SkimmerExample: Oil Skimmer

dpdx

=dpdx

= sin(30) 0.5dhdx

= =sin(30) 0.5dhdx

= =

m 0.002a m 0.002a m/s 3U m/s 3U

)N/m 84305.0s/mN 1x1012

m) (0.0022

m) m/s)(0.002 (3 322-

3

Q )N/m 84305.0

s/mN 1x1012m) (0.002

2m) m/s)(0.002 (3 3

22-

3

Q

In direction of beltQ = 0.0027 m2/s (per unit width)

Q = 0.0027 m2/s (5 m) = 0.0136 m3/s

( )3

2 12Ua a d

Q p hdx

gm

= - + 00

dominatesdominates

How do we get the power requirement? ___________________________

What is the force acting on the belt? ___________________________

Remember the equation for shear? _____________ Evaluate at y = a.

How do we get the power requirement? ___________________________

What is the force acting on the belt? ___________________________

Remember the equation for shear? _____________ Evaluate at y = a.

Example: Oil Skimmer Power Requirements

Example: Oil Skimmer Power Requirements

( )du dy p h A

dy dxm g= + +( )du d

y p h Ady dx

m g= + + ( )2

U a dA p h

a dxm

g= - +( )2

U a dA p h

a dxm

g= - +

( )2a d U

y p hdx a

mt gæ ö= - + +

è ø( )

2a d U

y p hdx a

mt gæ ö= - + +

è ø

Power = Force x Velocity [N·m/s]

Shear force (·L · W)

=(du/dy)

Example: Oil Skimmer Power Requirements

Example: Oil Skimmer Power Requirements

( )2a d U

y p hdx a

mt gæ ö= - + +

è ø( )

2a d U

y p hdx a

mt gæ ö= - + +

è ø

a

Ua sin2 a

Ua sin2

( ) sind

p hdx

g g q+ =( ) sind

p hdx

g g q+ =

2

22

3 mN

9.21m 0.002

msN

1x10sm 3

0.5m

N 84302

m 0.002

2

22

3 mN

9.21m 0.002

msN

1x10sm 3

0.5m

N 84302

m 0.002

Power LWUt=Power LWUt=

sm 3

m 5m 6mN

19.2Power 2

sm 3

m 5m 6mN

19.2Power 2

(shear by belt on fluid)

= 3.46 kW

FVFV

Lifting the oil (potential energy)

Heating the oil (thermal energy)

Example: Oil Skimmer Where did the Power Go?

Example: Oil Skimmer Where did the Power Go?

Where did the energy input from the belt go?

Where did the energy input from the belt go?

h = 3 m

hQP hQP

m 3s

m0.0136

mN

84303

3

P m 3

sm

0.0136mN

84303

3

P

W443P W443P

Example : Oil Skimmer Was it Really Laminar Flow?

Example : Oil Skimmer Was it Really Laminar Flow?

We assumed that the flow was laminar (based on the small flow dimension of 2 mm)

We need to check our assumption!!!!

We assumed that the flow was laminar (based on the small flow dimension of 2 mm)

We need to check our assumption!!!!

VD

R

VDR

2

2-

2

3

1x10

m 0.002_______9.81m/s

8430N/m

m

sNR

2

2-

2

3

1x10

m 0.002_______9.81m/s

8430N/m

m

sNR 234R 234R Laminar

A

QV

A

QV = 1.36 m/s

1.36 m/s

0.0136 m3/s0.002 m* 5 m

Example: No flowExample: No flow

Find the velocity of a vertical belt that is 5 mm from a stationary surface that will result in no flow of glycerin at 20°C (= 0.62 Ns/m2 and =12300 N/m3)

Find the velocity of a vertical belt that is 5 mm from a stationary surface that will result in no flow of glycerin at 20°C (= 0.62 Ns/m2 and =12300 N/m3)

Laminar Flow through Circular Tubes

Laminar Flow through Circular Tubes

Different geometry, same equation development (see Munson, et al. p 367)

Apply equation of motion to cylindrical sleeve (use cylindrical coordinates)

Different geometry, same equation development (see Munson, et al. p 367)

Apply equation of motion to cylindrical sleeve (use cylindrical coordinates)

Max velocity when r = 0

Laminar Flow through Circular Tubes: Equations

Laminar Flow through Circular Tubes: Equations

( )2 2

4z

p hr Rv

l

gm

D +-=

( )2 2

4z

p hr Rv

l

gm

D +-=

( )2

max 4

p hRv

l

gmD +

=-( )2

max 4

p hRv

l

gmD +

=-

( )2

8

p hRV

l

gmD +

=-( )2

8

p hRV

l

gmD +

=-

( )4

8

p hRQ

l

gpmD +

=-( )4

8

p hRQ

l

gpmD +

=-

Velocity distribution is paraboloid of revolution therefore _____________ _____________

Q = VA =

average velocity (V) is 1/2 vmax

VR2

R is radius of the tubeR is radius of the tube

Laminar Flow through Circular Tubes: Diagram

Laminar Flow through Circular Tubes: Diagram

Velocity

Shear

( )2

p hdv rdr l

gmD +

=( )

2

p hdv rdr l

gmD +

=

( )2

p hdv rdr l

gt m

D += =

( )2

p hdv rdr l

gt m

D += =

l

hr l

2

l

hr l

2

l

dhl

40

l

dhl

40

True for Laminar or Turbulent flow

Shear at the wallShear at the wall

Laminar flow

( )2 2

4z

p hr Rv

l

gm

D +-=

( )2 2

4z

p hr Rv

l

gm

D +-=

cv pipe flow

The Hagen-Poiseuille EquationThe Hagen-Poiseuille Equation

lhzp

zp 2

2

21

1

1

lhzp

zp 2

2

21

1

1

2

2

21

1

1 zp

zp

hl

2

2

21

1

1 zp

zp

hl

hp

hl

hp

hl

4

128lhD

Ql

gpm

=4

128lhD

Ql

gpm

=2

32lhD

Vl

gm

=2

32lhD

Vl

gm

=

Constant cross section

Laminar pipe flow equations

CV equations!CV equations!

h or zh or z

pz

Vg

Hp

zV

gH hp t l

1

11 1

12

2

22 2

22

2 2

( )lh p hg g=- D +( )lh p hg g=- D +

( )4

8

p hRQ

l

gpmD +

=-( )4

8

p hRQ

l

gpmD +

=- From Navier-Stokes

Example: Laminar Flow (Team work)

Example: Laminar Flow (Team work)

Calculate the discharge of 20ºCwater through a long vertical section of 0.5 mm ID hypodermic tube. The inlet and outlet pressures are both atmospheric. You may neglect minor losses. What is the total shear force?

What assumption did you make? (Check your assumption!)

Calculate the discharge of 20ºCwater through a long vertical section of 0.5 mm ID hypodermic tube. The inlet and outlet pressures are both atmospheric. You may neglect minor losses. What is the total shear force?

What assumption did you make? (Check your assumption!)

SummarySummary

Navier-Stokes Equations and the Continuity Equation describe complex flow including turbulence

The Navier-Stokes Equations can be solved analytically for several simple flows

Numerical solutions are required to describe turbulent flows

GlycerinGlycerin

QUa a d

dlp h

2 12

3

a f

02 12

3

Ua a

ddl

p h a f

Ua

2

6

Um N m

Ns mm s

0 005 12300

6 0 620 083

2 3

2

. /

. /. /

a fc hc h

RVl kg m m s m

Ns m

1254 0 083 0 005

0 620 8

3

2

/ . / .

. /.

c ha fa f

Example: Hypodermic Tubing Flow

Example: Hypodermic Tubing Flow

QD h

Ll

4

128

QN m m

x Ns m

9806 0 0005

128 1 10

3 4

3 2

/ .

/

c hafa fc h

Q x m s 158 10 8 3. /

VQd

4

2V m s 0 0764. /

RVd

R

m s m kg m

x Ns m

0 0764 0 0005 1000

1 10

3

3 2

. / . /

/

a fa fc hc h

R 38

Q L s 158. /ldhl

40

ldhl

40

ldhrl

F lshear 4

2 l

dhrlF l

shear 42

2shear lF r hp g=- 2shear lF r hp g=-

= weight!

Euler’s Equation Along a Streamline

Euler’s Equation Along a Streamline

va 2 phg va 2 phg

xp

xh

gxu

utu

xp

xh

gxu

utu

phg a phg a

Inviscid flow (frictionless)

x along a streamline

v = u = velocity in x direction

Velocity normal to streamline is zero

u v w g h pt x y z

r r¶ ¶ ¶ ¶é ù

+ + + =- Ñ - Ñê ú¶ ¶ ¶ ¶ë û

V V V Vu v w g h p

t x y zr r¶ ¶ ¶ ¶é ù

+ + + =- Ñ - Ñê ú¶ ¶ ¶ ¶ë û

V V V V

Euler’s EquationEuler’s Equation

xp

xh

gxu

u

1

xp

xh

gxu

u

1

01

dxdp

dxdh

gdxdu

u

01

dxdp

dxdh

gdxdu

u

0 udugdhdp

0 udugdhdp

(Multiplying by dx converts from a force balance equation to an energy equation)

xp

xh

gxu

utu

xp

xh

gxu

utu

We’ve assumed: frictionless and along a streamline

SteadySteady

Euler’s equation along a streamline

x is the only independent variable

Bernoulli EquationBernoulli Equation

0 udugdhdp

0 udugdhdp

Euler’s equation

01 udugdhdp

01 udugdhdp

constant2

2

vgh

p

constant2

2

vgh

p

constant2

2

g

vh

p

constant2

2

g

vh

p

The Bernoulli Equation is a statement of the conservation of ____________________

constant2

2

vgh

p

constant2

2

vgh

p

Integrate for constant density

Bernoulli Equation

Mechanical Energy p.e. k.e.

Hydrostatic Normal to Streamlines?

Hydrostatic Normal to Streamlines?

phgz

wy

vx

ut

vvvvphg

zw

yv

xu

t

vvvv

xp

xh

gxu

utu

xp

xh

gxu

utu

y

p

y

hg

0y

p

y

hg

0 dhdp dhdp

y

p

y

hg

z

vw

y

vv

x

vu

t

v

y

p

y

hg

z

vw

y

vv

x

vu

t

v

y, v perpendicular to streamline (v = 0)

x, u along streamline

Laminar Flow between Parallel Plates

Laminar Flow between Parallel Plates

U

l

a

uy

lyy

lyy

yllp

p

yl

lp

p

ypyp

ll

yl yl

sinyl sinyl

y

h

l

Equation of Motion: Force Balance

Equation of Motion: Force Balance

lyy

lyy

yllp

p

yl

lp

p

ypyp

ll

yl yl

sinyl sinyl

ypyp

yllp

p

yl

lp

p

ll

lyy

lyy

maF maF

sinyl sinyl

00

pressure

shear

gravity

acceleration

+

-

-

+

+

= l

Equation of MotionEquation of Motion

0sin

ylp

0sin

ylp

lh

sinlh

sin

lh

lp

y

lh

lp

y

hp

ly

hp

ly

hp

dld

dyd hp

dld

dyd

dydu dydu 2

2

dyud

dyd 2

2

dyud

dyd

But

h

l

Laminar flow assumption!Laminar flow assumption!

0sin

ylly

ylyl

lp

pyp 0sin

ylly

ylyl

lp

pyp

U

a

u

Limiting casesLimiting cases

( )2

2y y ay d

u U p ha dx

gm-

= + +( )2

2y y ay d

u U p ha dx

gm-

= + +

Both plates stationary

Hydrostatic pressure ( ) 0d

p hdx

g+ =( ) 0d

p hdx

g+ =

aUy

u a

Uyu

0U 0U

( )2

2y ay d

u p hdx

gm-

= +( )2

2y ay d

u p hdx

gm-

= +

Linear velocity distribution

Parabolic velocity distribution

Motion of plate

Pressure gradient

y

x