DIFFERENTIAL ANALYSIS OF FLUID FLOW

A: Mathematical Formulation (4.1.1, 4.2, (6.1-6.4)

B: Inviscid Flow: Euler Equation/Some Basic Plane Potential Flows (6 5 6 7)Basic, Plane Potential Flows (6.5-6.7)

C: Viscous Flow: Navier-Stokes EquationC: Viscous Flow: Navier Stokes Equation (6.8-6.10)

Introduction

There are situations in which the details of the flowDifferential Analysis

There are situations in which the details of the flow are important, e.g., pressure and shear stress variation l h ialong the wing….

Therefore, we need to develop relationship that apply , p p pp yat a point or at least in a very small region (infinitesimal volume) with a given flow field(infinitesimal volume) with a given flow field.This approach is commonly referred to as differential

l ianalysis.The solutions of the equations are rather difficult.e so u o s o e equ o s e e d cu .Computational Fluid Dynamic (CFD) can be applied t l fl blto complex flow problems.

PART APART AMathematical FormulationMathematical Formulation

(S ti 4 1 1 4 2 6 1 6 4)(Sections 4.1.1, 4.2, 6.1-6.4)

Fluid Kinematics (4 1 1 4 2)Fluid Kinematics (4.1.1, 4.2)

Kinematics involves position, velocity and l i facceleration, not forces.

kinematics of the motion: e at cs o t e ot o :velocity and acceleration of the fluid, and the description and visualization of its motiondescription and visualization of its motion.The analysis of the specific force necessary to

d th ti th d i f thproduce the motion - the dynamics of the motion.

4.1 The Velocity FieldyA field representation – representations of fluid parameters as functions of spatial coordinatep p

the velocity field

( ) ( ) ( )ktzyxwjtzyxvitzyxuV ,,,,,,,,, ++=A

Adr Vdt

=r uur

( )tzyxVV ,,,=

( ) 21222 wvuVV ++==

A change in velocity results in an acceleration whichA change in velocity results in an acceleration which may be due to a change in speed and/or direction.

4.1.1 Eulerian and Lagrangian g gFlow DescriptionsEulerian method: the fluid motion is given by completely prescribing the necessary properties as functions of space and time.

From this method, we obtain information about the flow in termsinformation about the flow in terms of what happens at fixed points in space as the fluid flows past those p ppoints.

Lagrangian method: follo ingLagrangian method: following individual fluid particles as they move about and determining how the fluidabout and determining how the fluid properties associated with these particles change as a function of time V4.3 Cylinder-velocity vectorsparticles change as a function of time. V4.4 Follow the particles

V4.5 Follow the particles

4.1.4 Streamlines, Streaklines and Pathlines

A streamline is a line that is everywhere tangent to the velocity fieldvelocity field.A streakline consists of all particles in a flow that have previously passed through a common pointpreviously passed through a common point.A pathline is a line traced out by a given flowing particle.

V4.9 streamlinesV4.10 streaklinesV4.1 streaklines

4.1.4 Streamlines, Streaklines and PathlinesFor steady flows, streamlines, streaklines and pathlines all coincide. This is not true for unsteady flows.y

Unsteady streamlines are difficult to generate y gexperimentally, but easy to draw in numerical computation.On the contrary, streaklines are more of a lab tool than anOn the contrary, streaklines are more of a lab tool than an analytical tool.How can you determine the unsteady pathline of a movingHow can you determine the unsteady pathline of a moving particle?

4.2 The Acceleration Field4.2 The Acceleration Field

The acceleration of a particle is the time rate h f i l ichange of its velocity.

For unsteady flows the velocity at a givenFor unsteady flows the velocity at a given point in space may vary with time, giving rise to a portion of the fluid acceleration.In addition a fluid particle may experience anIn addition, a fluid particle may experience an acceleration because its velocity changes as it flows from one point to another in space.

4 2 1 The Material Derivative4.2.1 The Material Derivative

Consider a particle moving along its pathline

( ) ( ) ( ) ( )V V r t V x t y t z t t= = ⎡ ⎤⎣ ⎦uur uur ur uur( ) ( ) ( ) ( ), , , ,A A A A A A AV V r t V x t y t z t t= = ⎡ ⎤⎣ ⎦

The Material DerivativeThe Material Derivative

Thus the acceleration of particle A,

( ) ( ) ( ) ( ), , , ,A A A A A A AV V r t V x t y t z t t= = ⎡ ⎤⎣ ⎦uur uur ur uur

( ) A A A A A AA

dV V V dx V dya tdt t x dt y dt

∂ ∂ ∂= = + +

∂ ∂ ∂

uur uur uur uuruur

A A

dt t x dt y dt

V dz

∂ ∂ ∂

∂+

uur

z dt

V V V V

+∂

∂ ∂ ∂ ∂uur uur uur uur

= A A A AA A A

V V V Vu v wt x y z

∂ ∂ ∂ ∂+ + +

∂ ∂ ∂ ∂

Acceleration

This is valid for any particle

V V V Va u v wt x y z

∂ ∂ ∂ ∂= + + +

∂ ∂ ∂ ∂

ur ur ur urr

t x y zu u u u

∂ ∂ ∂ ∂∂ ∂ ∂ ∂

+ + +xa u v wt x y z

= + + +∂ ∂ ∂ ∂

yv v v va u v wt x y z

∂ ∂ ∂ ∂= + + +

∂ ∂ ∂ ∂t x y zw w w wa u v w

∂ ∂ ∂ ∂∂ ∂ ∂ ∂

= + + +za u v wt x y z

= + + +∂ ∂ ∂ ∂

M t i l d i tiMaterial derivativeAcceleration:

ur ur ur ur ur ur

Associated with time variation

, DV DV V V V Va u v wDt Dt t x y z

∂ ∂ ∂ ∂= = + + +

∂ ∂ ∂ ∂

r

T t l d i ti t i l d i ti b t ti l

yAssociated with space variation

Total derivative, material derivative or substantial derivative

( ) ( ) ( ) ( ) ( )Du v w

D∂ ∂ ∂ ∂

= + + +∂ ∂ ∂ ∂

( ) ( )( )

Dt t x y z

V

∂ ∂ ∂ ∂

∂+ ∇

r( ) ( ) = ( )Vt

+ ⋅∇∂

Material derivativeThe material derivative of any variable is the rate at which that variable changes with timerate at which that variable changes with time for a given particle (as seen by one moving l i h h fl id h L ialong with the fluid – the Lagrangian

descriptions)p )If velocity is known, the time rate change of temperature can be expressed astemperature can be expressed as,

DT T T T T∂ ∂ ∂ ∂+ + +u v w

Dt t x y zT

= + + +∂ ∂ ∂ ∂

∂ = ( )T V Tt

∂+ ⋅∇

∂

r

Example: the temperature of a passenger experienced on a trainExample: the temperature of a passenger experienced on a train starting from Taipei on 9am and arriving at Kaohsiung on 12.

Acceleration along a gstreamline

( )3

0 3 1 , R V V u uV u x i V i a u u it t

⎛ ⎞ ∂ ∂ ∂ ∂⎛ ⎞= = + = + = +⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠

ur urur r r r r

( ) 3x t x t x⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠

33 4

3(1 ) [ ( 3 )] o oRa V V R x i−= + −

r r3o ox

4 2 2 Unsteady Effects4.2.2 Unsteady EffectsFor steady flow ( )/ 0 there is no change in flowt∂ ∂ ≡For steady flow ( )/ 0, there is no change in flow parameters at a fixed point in space.

t∂ ∂ ≡

For unsteady flow ( )/ 0.t∂ ∂ ≠

spatial (convective) derivative DT T

↓∂ DT T∂ DT T V

Dt t∂

= + ⋅∇∂

v (for an unstirred cup of coffee 0)

ti (l l) d i ti

DT TTDt t

∂→ <

∂↑ time (local) derivative

DV V V V

↑

∂= + ⋅∇

v vv v

local acceleration

V VDt t

= + ⋅∇∂↑ V4.12 Unsteady flowy

4.2.3 Convective Effects

DT T V TDt t

∂= + ⋅∇

∂

v

0

Dt tDT Tu

∂∂

= +0 suDt s

T T

= +∂

− =0+ out insT Tu

sΔ

4.2.3 Convective Effects

convective acceleration↓

DV V V VDt t

∂= + ⋅∇

∂

v vv v

local accelerationDt t

Du u

∂↑

∂0Du uuDt x

∂= +

∂

4.2.4 Streamline CoordinatesIn many flow situations it is convenient to use a coordinate system defined in terms of the streamlines of the flow.system defined in terms of the streamlines of the flow.

ˆV=V sˆˆ ˆ

VDV DV DV= = +v

V ssa s

ˆ

Dt Dt DtV V ds V dn∂ ∂ ∂⎛ ⎞= + +⎜ ⎟s

ˆ ˆ ˆt s dt n dt

ds dn

= + +⎜ ⎟∂ ∂ ∂⎝ ⎠∂ ∂ ∂⎛ ⎞

s

s s s ds dnVt s dt n dt

∂ ∂ ∂⎛ ⎞+ + +⎜ ⎟∂ ∂ ∂⎝ ⎠

s s s

ˆ̂ : change in tangential unit vector : change in tangential distances

δδ

s

V4.13 Streamline coordinates

g g

4.2.4 Streamline CoordinatesSteady flow

ˆˆ ˆVV V V∂ ∂⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟s

2 2

V V Vs s

V V V V

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠∂ ∂

a s

ˆ ˆ or ,

ˆ ˆ ˆ ˆ ˆ1

s nV V V VV a V as R s R

δδ δ δ

∂ ∂= + = =

∂ ∂⎛ ⎞∂

s n

0

ˆ ˆ ˆ ˆ1ˆ , or , limˆ s

sR s R s s Rδ

δδ δ δδδ δ→

⎛ ⎞∂= = = = =⎜ ⎟⎜ ⎟∂⎝ ⎠

Qs s s s ns

s

6 1 Fluid Element Kinematics6.1 Fluid Element Kinematics

Types of motion and deformation for a fluid element.

6.1.1 Velocity and Acceleration yFields Revisited

Velocity field representation ( ) ˆ ˆ ˆV V V i j k

Acceleration of a particle( ), , , or x y z t u v w= = + +V V V i j k

uuuu ∂∂∂∂cce e at o o a pa t c e

∂ ∂ ∂ ∂V V V Vzuw

yuv

xuu

tuax ∂

∂+

∂∂

+∂∂

+∂∂

=

vvvv ∂∂∂∂u v wt x y z

∂ ∂ ∂ ∂= + + +

∂ ∂ ∂ ∂V V V Va

zvw

yvv

xvu

tvay ∂

∂+

∂∂

+∂∂

+∂∂

=

wwww ∂∂∂∂zww

ywv

xwu

twaz ∂

∂+

∂∂

+∂∂

+∂∂

=

( )DDt t

∂= = + ⋅∇

∂V Va V V

( ) ( ) ( ) ( )ˆ ˆ ˆ

Dt t∂∂ ∂ ∂

∇ = + +i j k( )x y z

∇∂ ∂ ∂

j

6.1.2 Linear Motion and Deformation

variations of the velocity in the direction of ∂ ∂velocity, , , cause a linear stretching

deformation.xu

∂∂

yv

∂∂

zw

∂∂

deformation.Consider the x-component deformation:

Linear Motion and DeformationThe change in the original volume, , due to / :V x y z u x

uδ δ δ δ= ∂ ∂

∂

Rate change of per unit volume due to / :V u xδ ∂ ∂

Change in ( )( )( )uV x y z tx

δ δ δ δ δ∂=

∂

( )0

Rate change of per unit volume due to / :

/1 ( ) limt

V u x

u x td V uV dt tδ

δ

δδδ δ→

∂ ∂

⎡ ⎤∂ ∂ ∂= =⎢ ⎥ ∂⎣ ⎦0tV dt t xδδ δ→ ∂⎣ ⎦

If velocity gradient / and / are also present, thenv y w z∂ ∂ ∂ ∂1 ( ) d V u v wV dt x y z

δδ

∂ ∂ ∂= + + = ∇ ⋅ ←

∂ ∂ ∂V volumetric dilatation rate

The volume of a fluid may change as the element moves from oneas the element moves from one location to another in the flow field. For incompressible fluid, the volumetric dilation rate is zero.

6.1.3 Angular Motion and Deformation

Consider an element under rotation and angular deformationConsider an element under rotation and angular deformation

V6.3 Shear deformation

Angular Motion and Deformation

the angular velocity of OA is

Angular Motion and Deformation

the angular velocity of OA is

0OA tlim

tδ

δαωδ→

=

For small angles

0t tδ δ→

∂

tan

v x t vx tx x

δ δδα δα δ

δ

∂∂∂≈ = =∂x xδ ∂

so that

0

( / )OA t

v x t vlimt xδ

δ δ δ δωδ δ→

⎡ ⎤= =⎢ ⎥⎣ ⎦0t t xδ δ δ→ ⎣ ⎦

(if is positive then will be counterclockwise)v∂(if is positive then will be counterclockwise) xv

∂∂

OAω

Angular Motion and Deformation

the angular velocity of the line OB is

g

the angular velocity of the line OB is

lim δβω =0OB t

limtδ

ωδ→

=

tyu δδ∂

tyu

y

tyy δδ

δδδβδβ

∂∂

=∂

=≈tan

so that

⎡ ⎤( )0

/limOB t

u y t ut yδ

δω

δ→

⎡ ⎤∂ ∂ ∂= =⎢ ⎥ ∂⎣ ⎦

( if is positive, will be clockwise)yu

∂∂

OBωy∂ OB

Angular rotation

The rotation, , of the element about the z axis is definedzωV4.6 Flow past a wing

The rotation, , of the element about the z axis is defined as the average of the angular velocities and , if counterclockwise is considered to be positive, then,

z

OAω OBωp , ,

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∂∂

=yu

xv

z 21ω ⎟

⎠⎜⎝ ∂∂ yx2

similarly

⎟⎟⎞

⎜⎜⎛ ∂∂ vw1

⎟⎞

⎜⎛ ∂∂ wu1ω⎟⎟

⎠⎜⎜⎝ ∂

−∂

=zyx 2

ω ⎟⎠

⎜⎝ ∂

−∂

=xzy 2

ω,

h 11thus

ˆ ˆ ˆi j k

VVkjiω ×∇==++=21

21ˆˆˆ curlxyx ωωω

1 1 1 1 1ˆ ˆ ˆ2 2 2 2 2

w v u w v ux y z y z z x x y

⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞∇× = = − + − + −⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠

i j k

V i j k2 2 2 2 2x y z y z z x x y

u v w∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠

Definition of vorticity

D fi ti it ζ

y

Define vorticity ζ

2= = ∇×ζ ω VζThe fluid element will rotate about z axis as an undeformedblock ( ie: )ωω −=block ( ie: ) OBOA ωω −=

only when vu∂∂

−=∂∂y

xy ∂∂

Otherwise the rotation will be associated with an angular

vu ∂∂

deformation.

If or , then the rotation (and the vorticity ) xv

yu

∂∂

=∂∂

0∇× =V

are zero, and flow fields are termed irrotational.

Different types of angular motionsyp g

Solid body rotationSolid body rotation ru Ω=φ 0== zr uuφ

Ω= 2zω 0== φωω r

( )θ

ω φ ∂∂

−∂∂

= rz

ur

rurr

11

Free vortexkrku =φ 0== zr uu

0 00=φω 0=rω

( ) 01 ∂ω ru 0f( ) 0=∂

= φω rurrz 0≠rfor

Angular Deformation∂

Angular Deformation

Apart form rotation associated with these derivatives and , these derivatives can cause the element to

yu

∂∂

xv

∂∂

undergo an angular deformation, which results in a changein shape of the element.

x∂

The change in the original right angle is termed the shearing strain ,δγ

δβδαδγ +=

where is considered to be positive if the original right angle is decreasing.

δγ

Angular Deformation

R t f h i t i t f l d f ti

Angular Deformation

Rate of shearing strain or rate of angular deformation

v ut tδ δ∂ ∂⎡ ⎤+⎢ ⎥0 0

lim limt t

t tx y

t tδ δ

δ δδγγδ δ→ →

+⎢ ⎥∂ ∂= = ⎢ ⎥⎢ ⎥⎢ ⎥

&

v u

⎢ ⎥⎣ ⎦∂ ∂

= +x y

+∂ ∂

The rate of angular deformation is related to a corresponding g p gshearing stress which causes the fluid element to change in shape.

vu ∂∂If , the rate of angular deformation is zero and this condition indicates that the element is simply rotating as an

xv

yu

∂∂

−=∂∂

undeformed block.

6.2 Conservation of Mass

0=Dt

DM sysConservation of mass:Dt

In control volume representation (continuity equation):

ˆd dA∂∫ ∫

In control volume representation (continuity equation):

ˆ 0cv cs

dV dAt

ρ ρ∂+ ⋅ =

∂ ∫ ∫ V n (6.19)

To obtain the differential form of the continuity equation, E 6 19 i li d t i fi it i l t l lEq. 6.19 is applied to an infinitesimal control volume.

6.2.1 Differential Form of Continuity E tiEquation

zyxt

Vdt

δδδρρ∂∂

≡∂∂

∫ tt ∂∂ ∫Net mass flow in the direction zyx

xuzyx

xuuzyx

xuu δδδρδδδρρδδδρρ

∂∂

=⎥⎦⎤

⎢⎣⎡

∂∂

−−⎥⎦⎤

⎢⎣⎡

∂∂

+22

Net mass flow in the y direction zyxyv δδδ

ρρ∂

∂Net mass flow in the z direction zyxzw δδδ

ρρ∂

Net rate of mass out of flow wvu δδδρρρ⎥⎤

⎢⎡ ∂∂∂Net rate of mass out of flow zyx

zyxδδδρρρ

⎥⎦

⎢⎣ ∂

+∂

+∂

Diff ti l F f C ti it E tiDifferential Form of Continuity Equation

Thus conservation of mass become

0=∂

+∂

+∂

+∂ wvu ρρρρ

(continuity equation )

In vector form

0=∂

+∂

+∂

+∂ zyxt (continuity equation )

In vector form

0=⋅∇+∂∂ Vρρ

For steady compressible flow∂

ρt

0=∂

∂+

∂∂

+∂

∂zw

yv

xu ρρρ 0=⋅∇ Vρ

For incompressible flow∂∂∂ zyx

∂∂∂ wvu0=⋅∇ V0=

∂∂

+∂∂

+∂∂

zw

yv

xu

6.2.2 Cylindrical Polar Coordinates6.2.2 Cylindrical Polar Coordinates

The differential form of continuity equation

( ) ( )11 ∂∂∂∂ vvvr ρρρρ ( ) ( )011

=∂

∂+

∂∂

+∂

∂+

∂∂

zvv

rrvr

rtzr ρ

θρρρ θ

6.2.3 The Stream Function

For 2-D incompressible plane flow then, ∂∂ vu

D fi t f ti h th t

0=∂∂

+∂∂

yv

xu

( )Define a stream function such that( ),x yψ

ψ∂ ψ∂uyψ

=∂

vxψ

= −∂

2 2ψ ψ ψ ψ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂then 0

x y x y x y y xψ ψ ψ ψ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂

+ − = − =⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

For velocity expressed in forms of the stream function, the conservation of mass will be satisfied.function, the conservation of mass will be satisfied.

The Stream FunctionThe Stream Function

Li l t t t liψLines along constant are stream lines.Definition of stream line

ψv

ddy

=udx

Thus change of , from to ( , )x y ( , )x dx y dy+ +ψ

0dAl li f t t f h

udyvdxdyy

dxx

d +−=∂∂

+∂∂

=ψψψ

0dψ =

0=+− udyvdx vddy

=

ψAlong a line of constant of we have

yudx

which is the defining equation for a streamline.

Thus we can use to plot streamline.The actual numerical value of a stream line is not important but

ψp

the change in the value of is related to the volume flow rate.ψ

The Stream FunctionNote：Flow never crosses streamline, since by definition the velocity is tangent to the streamlines.

Volume rate of flow (per unit width perpendicular to the x-yplane)plane)

dq udy vdx

dy dx dψ ψ ψ

= −∂ ∂

= + =

2

2 1

dy dx dy x

q dψ

ψ

ψ ψ ψ

+∂ ∂

= = −∫1

2 1qψ

ψ ψ ψ∫

2 1ψ ψ>If then q is positive and vice versa.In cylindrical coordinates the incompressible continuity equation becomes,

q p

( )011

=∂

+∂ θvrvrequation becomes, 0=

∂+

∂ θrrr

Then,1

rv ψθ

∂=

∂vθ

ψ∂= −

∂, r θ∂ r∂

Ex 6.3 Stream function

6.3 Conservation of Linear Momentum6.3 Conservation of Linear Momentum

Li t tiLinear momentum equationD dm= ∫F V

sysdm

Dt ∫F V

or dAVd nVVVF ˆ⋅+∂

= ∫∫∑ ρρ

C id diff ti l t ith d

or

δ Vδ

dAVdt CSCV

nVVVF volumecontrol

theofcontents ⋅+∂

= ∫∫∑ ρρ

Consider a differential system with andmδ Vδ

( )D δVthen ( )D mDt

δδ =

VF

Using the system approach then Dδ δ δVF Dm mDt

δ δ δ= =VF a

6.3.1 Descriptions of Force Acting on theDifferential Element

Two types of forces need to be considered surface forces：which action the surfaces of thesurface forces which action the surfaces of the

differential element.b d f ： hi h di t ib t d th h t thbody forces：which are distributed throughout the

element.

For simplicity the only body force considered is theFor simplicity, the only body force considered is the weight of the element,

b mδ δ=F g

or xbx mgF δδ = yby mgF δδ = zbz mgF δδ =

Surface force act on the element as a result of itsSurface force act on the element as a result of its interaction with its surroundings (the components depend on the area orientation) )

nFδ Aδ 1Fδ 2FδWhere is normal to the area and and are parallel to the area and orthogonal to each other.g

The normal stress is defined asσThe normal stress is defined as,nσ

Fnδσ lim=

AAn δσ

δ 0lim

→=

and the shearing stresses are define as

AF

A δδ

τδ

1

01 lim→

=AF

A δδ

τδ

2

02 lim→

=

Si f t

we use for normal stresses and for shear stresses.σ τ

Sign of stressesPositive sign for the stress as

i i di di ipositive coordinate directionon the surfaces for which the outward normal is in theoutward normal is in the positive coordinate direction.

Note：Positive normal stresses are tensile stresses, ie, they tend to stretch the material.

Thus yxxx zxFτσ τδ δ δ δ

∂⎛ ⎞∂ ∂+ +⎜ ⎟Thus yxx zx

sx

xy yy zy

F x y zx y z

F

δ δ δ δ

τ σ τδ δ δ δ

= + +⎜ ⎟∂ ∂ ∂⎝ ⎠∂ ∂ ∂⎛ ⎞

⎜ ⎟xy yy zy

sy

yx

F x y zx y z

F

δ δ δ δ

ττ σδ δ δ δ

= + +⎜ ⎟∂ ∂ ∂⎝ ⎠∂⎛ ⎞∂ ∂

⎜ ⎟yxxz zz

szF x y zx y z

τ σδ δ δ δ⎛ ⎞∂ ∂

= + +⎜ ⎟∂ ∂ ∂⎝ ⎠ˆ ˆ ˆF F Fδ δ δ δ= + +F i j ks sx sy sz

s b

F F Fδ δ δ δ

δ δ δ

= + +

= +

F i j k

F F F

6.3.2 Equation of Motion6.3.2 Equation of Motion

amFamFamF δδδδδδ ===

zyxm δδρδδ =

zzyyxx amFamFamF δδδδδδ === ,,

zyxm δδρδδ =

ThusThusy xx x z x

xu u u ug u v w

x y t x yτσ τρ ρ

∂ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂+ + + = + + +⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠

x y y y zyy

x y z t x y z

v v v vg u v wt

τ σ τρ ρ

⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠∂ ∂ ∂ ⎛ ⎞∂ ∂ ∂ ∂

+ + + = + + +⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠(6.50)y

y zx z z zz

x y z t x y z

w w w wg u v wt

ττ σρ ρ

⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠∂ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂

+ + + = + + +⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠

( )

z x y z t x y zρ ρ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠

PART BInviscid Flow:

Euler Equation/Some Basic PlaneEuler Equation/Some Basic, Plane Potential FlowsPotential Flows

(Sections 6 5-6 7)(Sections 6.5-6.7)

6.4 Inviscid Flow

For an inviscid flow in which the shearing stresses are all 6.4.1 Euler’s Equation of Motion

gzero, and the normal stresses are replaced by -p, thus the equation of motion becomes

⎛ ⎞x

p u u u ug u wx t x y z

ρ ρ υ⎛ ⎞∂ ∂ ∂ ∂ ∂

− = + + +⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠

ypg u wy t x y z

υ υ υ υρ ρ υ⎛ ⎞∂ ∂ ∂ ∂ ∂

− = + + +⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠

zp w w w wg u wz t x y z

ρ ρ υ

⎝ ⎠⎛ ⎞∂ ∂ ∂ ∂ ∂

− = + + +⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠or

z t x y z∂ ∂ ∂ ∂ ∂⎝ ⎠

∂⎡ ⎤V

The main difficulty in solving the equation is the nonlinear

( )pt

ρ ρ ∂⎡ ⎤− ∇ = + ⋅∇⎢ ⎥∂⎣ ⎦Vg V V

The main difficulty in solving the equation is the nonlinear terms which appear in the convective acceleration.

6.4.2 The Bernoulli Equation For steady flow

q

( )( )(up being positive)

pg z

ρ ρ− ∇ = ⋅∇

= − ∇

g V Vg

( ) ( ) ( )12

⋅∇ = ∇ ⋅ − × ∇×V V V V V V

thus the equation can be written as, 2

( ) ( )g z p ρρ ρ− ∇ − ∇ = ∇ ⋅ − × ∇×V V V V( ) ( )2

or

g z pρ ρ∇ ∇ ∇ × ∇×V V V V

T k h d d f h i h diff i l l h

( ) ( )212

p V g zρ

∇+ ∇ + ∇ = × ∇×V V

Take the dot product of each term with a differential length ds along a streamline

1∇ ( ) ( )212

p d V d g z d dρ

∇⎡ ⎤⋅ + ∇ ⋅ + ∇ ⋅ = × ∇× ⋅⎣ ⎦s s s V V s

Since ds and V are parallel, therefore

( ) 0d⎡ ⎤× ∇× ⋅ =⎣ ⎦V V s

Sinceˆ ˆ ˆd dx dy dz= + +s i j kd dx dy dz

p p pp d dx dy dz dp

= + +∂ ∂ ∂

∇ ⋅ = + + =

s i j k

sp d dx dy dz dpx y z

∇ = + + =∂ ∂ ∂

s

Thus the equation becomes q

( )21 02

dp d V g dzρ

+ + =( )2ρ

where the change in p, V and z is along the streamline g p, g

Equation after integration become2

constantdp V gz+ +∫ constant2

gzρ

+ + =∫which indicates that the sum of the three terms on the left side of the equation must remain a constant along a given streamline.

For inviscid incompressible flow the equation becomeFor inviscid, incompressible flow, the equation become, 2

constp V gz+ + =※ (1) i i id fl2

or

gρ ※ For (1) inviscid flow

(2) steady flow(3) incompressible flow2 2

1 1 2 21 22 2

p V p Vz zg gγ γ

+ + = + +(3) incompressible flow(4) flow along a streamline

g gγ γ

6.4.3 Irrotational FlowIf the flow is irrotational, the analysis of i i id fl bl i f h i lifi dinviscid flow problem is further simplified.The rotation ω of the fluid element is equal

1to , and for irrotational flow field,V×∇21

0∇× =VSince , therefore for an irrotational

flow field the vorticity is zero

∇× =V ζ

flow field, the vorticity is zero.The condition of irrotationality imposes specific relationships among these velocity gradientsamong these velocity gradients. For example, 1 0

2zuυω

⎛ ⎞∂ ∂= − =⎜ ⎟∂ ∂⎝ ⎠2

, ,

z x yu w u wυ υ

⎜ ⎟∂ ∂⎝ ⎠∂ ∂ ∂ ∂ ∂ ∂

= = =∂ ∂ ∂ ∂ ∂ ∂

A general flow field would not satisfy these three equations. x y y z z x∂ ∂ ∂ ∂ ∂ ∂

Can irrotational flow hold in a viscous fluid?

According to the 2-D vorticity transport equation (cf. Problem 6.81)Problem 6.81)

2zz

DDζ ν ζ= ∇

Vorticity of an fluid element grows along with its motion

zDtζ

Vorticity of an fluid element grows along with its motion as long as ν is positive. So, an initially irrotatioal flow will eventually turn into rotational flow in a viscouswill eventually turn into rotational flow in a viscous fluid.On the other hand, an initially irrotatioal flow remains irrotational in an inviscid fluid, if without external ,excitement.

6.4.4 The Bernoulli Equation for Irrotational Flow

In Section 6.4.2, we have obtained along a streamline that,

⎡ ⎤( ) 0d⎡ ⎤× ∇× ⋅ =⎣ ⎦V V s

0∇× =VIn an irrotational flow, , so the equation is zero reg

Consequently, for irrotational flow the Bernoulli equation is

0∇ V, , q gardless of the direction of ds.

q y, f qvalid throughout the flow field. Therefore, between any flow points in the flow field,points in the flow field,

2

constant2

dp V gzρ

+ + =∫

2 21 1 2 2

orp V p V1 1 2 2

1 22 2p pz z

g gγ γ+ + = + +

※ For (1) Inviscid flow (2) Stead flow※ For (1) Inviscid flow (2) Stead flow (3) Incompressible flow (4) Irrotational flow

6.4.5 The Velocity PotentialFor irrotational flow since

y

0 th φ∇ ∇V V0, thus

, ,u w

φφ φ φυ

∇× = = ∇∂ ∂ ∂

= = =

V V

, ,u wx y z

υ∂ ∂ ∂

so that for an irrotational flow the velocity is expressible as

The velocity potential is a consequence of the irrotationalitythe gradient of a scalar function φ . The velocity potential is a consequence of the irrotationalityof the flow field (only valid for inviscid flow), whereas the stream function is a consequence of conservation of mass (valid for inviscid or viscous flow)(valid for inviscid or viscous flow).

Velocity potential can be defined for a general threeVelocity potential can be defined for a general three-dimensional flow, whereas the stream function is restricted to two-dimensional flows.

Thus for irrotational flow

2

0 , further with 0 for incomp. flow,0

φ

φ

∇ × = = ∇ ∇ ⋅ =

⇒ ∇

V V V0φ⇒ ∇ =

In Cartesian coordinates,2 2 2

2 2 2 0x y zφ φ φ∂ ∂ ∂

+ + =∂ ∂ ∂

Thus, inviscid, incompressible, irrotational flow fields are governed by Laplace’s equation.Cylindrical coordinate

( ) ( ) ( ) ( )1ˆ ˆ ˆ∂ ∂ ∂

∇ = + +e e e( )1ˆ ˆ ˆ

r z

r z

r r zθ

θ

θφ φ φφ

∇ = + +∂ ∂ ∂

∂ ∂ ∂∇ = + +

e e e

e e e

( )where , ,

r zr r zr z

θφθ

φ φ θ∂ ∂ ∂

=

ˆ ˆ ˆSince r r z zθ θυ υ υ= + +V e e eThus for an irrotational flow with φ= ∇V

2 2

2 2 2

1 1 0rr r r r z

φ φ φθ

∂ ∂ ∂ ∂⎛ ⎞ + + =⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

Example 6.4p

22 sin 2rψ θ=

1 φ∂ ∂ ( )

( )

21

2

1 4 cos 2 2 cos 2

14 sin 2 2 cos 2

r r r fr r

r r f r

ψ φυ θ φ θ θθψ φυ θ φ θ

∂ ∂= = = = +

∂ ∂∂ ∂

+ ( )2

2

4 sin 2 2 cos 2

2 cos 2

r r f rr r

Thus r C

θυ θ φ θθ

φ θ

= − = − = = +∂ ∂

= +

The specific value of C is not important, therefore 22 cos 2rφ θ=

( ) ( )2 22 2

2 21 1 2 2

4 cos 2 4 sin 2 16V r r r

p V p V

θ θ= + − =

+ +1 1 2 2

2 2p p

g gγ γ+ = +

6.5 Some basic, plane potential flows

Since the Laplace equation is a linear differential equation,Since the Laplace equation is a linear differential equation, various solutions can be added to obtain other solutions.i.e. φ φ φ

The practical implication is that if we have basic solutions we

1 2φ φ φ= +

The practical implication is that if we have basic solutions, we can combine them to obtain more complicated and interesting solutionssolutions.In this section several basic velocity potentials, which describe some relatively simple flows will be determinedsome relatively simple flows, will be determined.

F i li i l di i l fl ill bFor simplicity, only two-dimensional flows will be considered.

1 , or , :potentialvelocity θφφφφ

θ ∂∂

=∂∂

=∂∂

=∂∂

=r

vr

vy

vx

u ry

, 1or , :function streamr

vr

vx

vy

u r ∂∂

−=∂∂

=∂∂

−=∂∂

=ψ

θψψψ

θ

Defining the velocities in terms of the stream function

rrxy ∂∂∂∂ θ

Defining the velocities in terms of the stream function, conservation of mass is identically satisfied.Now impose the condition of irrotationality,Now impose the condition of irrotationality,

Thus xv

yu

∂∂

=∂∂

Thus xy ∂∂

022 ∂∂

⎟⎞

⎜⎛ ∂∂⎟

⎞⎜⎛ ∂∂ ψψψψ 0or 22 =

∂∂

+∂∂

⎟⎠⎞

⎜⎝⎛

∂∂

−∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

yxxxyyψψψψ

Thus for a two-dimensional irrotational flow the velocityThus for a two dimensional irrotational flow, the velocity potential and the stream function both satisfy Laplace equation.equation.It is apparent from these results that the velocity potential and the stream function are somehow relatedand the stream function are somehow related.Along a line of constant ψ, dψ =0

d dx dy vdx udy x yψ ψψ ∂ ∂

= + = − +∂ ∂

, dy vudy vdxdx u

∴ = =

Along a line of constant φ, dφ =0∂∂ φφ

d

vdy udxdyy

dxx

d =+=∂∂

+∂∂

= 0 φφφ

vu

dxdyvdyudx −=−=∴ ,

Therefore the equations indicate that lines of constant φTherefore, the equations indicate that lines of constant φ(equipotential lines) are orthogonal to lines of constant ψ(stream line) at all points where they intersect.(stream line) at all points where they intersect.

Q: Why V2 > V1? How about p1 and p2?p1 p2

6.5.1 Uniform FlowThe simplest plane flow is one for which the streamlines are all straight and parallel and the magnitude of the velocity isall straight and parallel, and the magnitude of the velocity is constant – uniform flow.

0U 0

0

u U v

Uφ φ= =

∂ ∂= =, 0U

x yUx Cφ

= =∂ ∂

= +Ux Cφ = +

Thus, for a uniform flow in the positive x direction, , p ,

The corresponding stream function can be obtained in aUx=φ

The corresponding stream function can be obtained in a similar manner,

UU ∂∂ ψψ 0 Uyx

Uy

=→=∂∂

=∂∂ ψψψ 0,

Th l it t ti l d t f ti fThe velocity potential and stream function for a uniform flow at an angle α with the x axis,

)sincos( ααφ yxU +)sincos()sincos(

ααψααφ

xyUyxU

−=+=

6.5.2 Source and Sink- purely radial6.5.2 Source and Sink purely radial flow

Consider a fluid flowing radially outward from a line through the origin perpendicular to the x-y plane.g p p y pLet m be the volume rate of flow emanating from the line (per unit length).unit length).

Conservation of mass

( ) 2 or 2r rmr v m v

rπ

π= =

Also, since the flow is purely radial,velocity potential becomes, 0vθ =

1 , 02m

r r rφ φ

π θ∂ ∂

= =∂ ∂

ln2m rφπ

=

Source and Sink flows

If m is positive, the flow is radially outward, and the flow is considered to be a source flow.If m is negative, the flow is toward the origin, and theIf m is negative, the flow is toward the origin, and the flow is considered to be a sink flow.The flow rate m is the strength of the source or sinkThe flow rate, m, is the strength of the source or sink.The stream function for the source:

θψψψ 01 mm=→=

∂=

∂ θπ

ψπθ 2

0,2 rrr

=→=∂

=∂

Note: At r=0, the velocity becomes infinite, which is of course physically impossible and is a singular point.

6.5.3 Vortex-streamlines are6.5.3 Vortex streamlines are concentric circles (vr=0)

Consider a flow field in which the streamlines are concentric circles i e e interchange the elocit potential and streamcircles. i.e. we interchange the velocity potential and stream function for the source.Th lThus, let

and lnK ψ K rφ θ= = −

where K is a constant.

and lnK ψ K rφ θ= =

t )(f1 K∂∂ ψφ vortex)(free rrr

v =∂

−=∂

=ψ

θφ

θ

Free and Forced vortexRotation refers to the orientation of a fluid element and not th th f ll d b th l tthe path followed by the element.

Free vortex Forced vortex

If the fluid were rotating as a rigid body, such that , this type of vortex motion is rotational and can not be described b a elocit potential

v Krθ =

described by a velocity potential.

F b h b flFree vortex: bathtub flow. Forced vortex: liquid contained in a tank rotating about its axis.

V6.4 Vortex in a beaker

Combined vortexCombined vortex

Combined vortex: a forced vortex as a central core and a free orte o tside the corevortex outside the core.

0v r r rθ ω= ≤ 0

0

v r r rKv r r

θ

θ

ω ≤

= >

where K and r are constant and r corresponds to the radius of

0rθ

where K and r are constant and r0 corresponds to the radius of central core.

CirculationA mathematical concept commonly associated with vortex motion is that of circulation.

The integral is taken around curve C in the counterclockwise∫ ⋅=ΓC

dsV (6.89)The integral is taken around curve, C, in the counterclockwise direction.Note: Green’s theorem in the plane dictatesNote: Green’s theorem in the plane dictates

ˆ( ) CR

dxdy d∫∫ ∫∇× ⋅ = ⋅V k V sFor an irrotational flow

φφφ ddd =⋅∇=⋅→∇= ssVVtherefore,

φφφ

0==Γ ∫ dφFor an irrotational flow, the circulation will generally be zero.However, if there are singularities enclosed within the curve,

∫C φ

However, if there are singularities enclosed within the curve, the circulation may not be zero.

Circulation for free vortexCirculation for free vortexK

For example, the free vortex with vrθ =

2 Kπ Γ∫ ( )

2

02

2K rd K Kr

πθ π

πΓ

Γ = = =∫Note: However Γ along any path which does not include the singular point at the origin will be zero.

The velocity potential and stream function for the free vortex are commonly expressed in terms of circulation asare commonly expressed in terms of circulation as,

(6.90)θφ Γ= (6.90)

(6 91)rln

2

ψ

θπ

φ

Γ=

=

(6.91)rln2π

ψ −=

Example 6.6Determine an expression relating the surface shape to theDetermine an expression relating the surface shape to the strength of the vortex as specified by circulation Γ.

Γ2

φ θπΓ

=

For irrotational flow, the Bernoulli equation

2 21 1 2 2

1 2 1 2 02 2

p V p Vz z p pg gγ γ

+ + = + + = =

2 21 2

1 02 2s

g gV Vz z

g g

γ γ

= + =

1

2 21 V 0

g g

vθφ∂ Γ

= = ≈1

2

V 02

vr r

z

θ θ π∂Γ

= 2 28szr gπ

= −

6.5.4 Doublet6.5.4 DoubletConsider potential flow that is formed by combining a source and a sink in a special way.Consider a source-sink pair

tantan2 θθπψ ⎞⎛m

21

212121 tantan1

tantan)tan(2tan)(2 θθ

θθθθπψθθπ

ψ+

−=−=⎟

⎠⎞

⎜⎝⎛−→−−=

mm

arr

arr

+=

−=

θθθ

θθθ

cossin tanand

cossintan Since 21 arar +θθ coscos

⎟⎞

⎜⎛→⎟

⎞⎜⎛ −1 sin2tsin22tTh armar θθπψ

⎟⎠⎞

⎜⎝⎛

−−=→

−=⎟

⎠⎞

⎜⎝⎛− 22

122 tan

2tan Thus

ararm πψψ

DoubletDoubletFor small values of a

)(sinsin2

2 2222 armar

ararm

−−=

−−=

πθθ

πψ (6.94))(

0→aDoublet is formed by letting the source and sink approach one another ( ) while increasing the strength m ( ) so∞→m

22

0→aanother ( ) while increasing the strength m ( ) so that the product ma/π remains constant.

→m

rarra /1)/(,0 As 22 →−→

E 6 94 d tEq. 6.94 reduces to:

rK θψ sin

−= (6.95)

where K = ma/π is called the strength of the doublet. The corresponding velocity potential isp g y p

rK θφ cos

= (6.96)

Doublet-streamlines

2cos1 K

rrVr

θθψφ

=∂∂

=∂∂

= 2

sin1 KV

rrrθψφ

θ

θ −=∂

−=∂

=

∂∂

2rrrV

θθ =∂

=∂

=

Streamlines for a doublet

Summary of basic, plane potential flows

6.6 Superposition of Basic, Plane p p ,Potential Flows M th d f itiMethod of superposition

A li i i i id fl fi ld bAny streamline in an inviscid flow field can be considered as a solid boundary, since the conditions along a solid boundary and a streamline are the same-no flow through the boundary or the streamline.o ow oug e bou d y o e s e e.Therefore, some basic velocity potential or stream f ti b bi d t i ld t li th tfunction can be combined to yield a streamline that corresponds to a particular body shape of interest.This method is called the method of superposition.

6.6.1 Source in a Uniform Stream- Half-BodyConsider a superposition of a source and a uniform flow.The resulting stream function isThe resulting stream function is

ψψψ sourceflowuniform +=

θπ

θ2

sin mUr += (6.97)

and the corresponding velocity potential is

rmUr lncosθφ += rUr ln2

cosπ

θφ += V6.5 Half-body

For the source aloneFor the source alonemvr 2

=rr π2

Let the stagnation point occur at x=－b, where mU =g p ,

sob

Uπ2

Umbπ2

=

The value of the stream function at the stagnation point can be b i d b l i b d θ hi h i ld f

Uπ2

obtained by evaluating x at r=b and θ=π, which yields from Eq. 6.97

bUm πψ

Thus the equation of the streamline passing through the

bUπψ ==2stagnation

Thus the equation of the streamline passing through the stagnation point is,

( )b θ( )sin or sin

bbU Ur bU r π θπ θ θθ

−= + = (6.100)

The width of the half body asymptotically approaches 2πbThe width of the half-body asymptotically approaches 2πb.This follows from Eq. 6.100, which can be written as

)( θbso that as θ→0 or θ→2π, the half-width approaches ±bπ.

)( θπ −= by

With the stream function (or velocity potential) known, the ( y p ) ,velocity components at any point can be obtained.

θψ1 mU∂

θψπ

θθψ

i

2cos

U

rU

rvr

∂

+=∂

=

θψθ sinU

rv −=

∂∂

−=

Thus the square of the magnitude of the velocity V at any point is,

2 2 2 2 2cosUm mθ2 2 2 2 2cos ( )2r

Um mV v v Ur r

m

θθ

π π= + = + +

2

since 2

mbUπ

=

⎛ ⎞22 2

21 2 cosb bV Ur r

θ⎛ ⎞

= + +⎜ ⎟⎝ ⎠

(6.101)

With the velocity known, the pressure distribution can be determined from the Bernoulli equation,dete ed o t e e ou equat o ,

2 20

1 1p U p Vρ ρ+ = + (6 102)0 2 2p U p Vρ ρ+ +

Note: the velocity tangent to the surface of the body is not zero;

(6.102)

Note: the velocity tangent to the surface of the body is not zero; that is, the fluid slips by the boundary.

Example 6.7p2

2 2 1 2 cosb bV U θ⎛ ⎞

= + +⎜ ⎟

( )

21 2 cos

h f /

V Ur r

b b

θ

π θ πθ

= + +⎜ ⎟⎝ ⎠

−( )

2 2

on the surface = / 2 sin 2

4Th V 1

br

U

πθ πθ

= =

⎛ ⎞⎜ ⎟

2 22

4Thus V 1U

bπ

π

⎛ ⎞= +⎜ ⎟⎝ ⎠

2 2by π

=

2 21 1 2 2

1 22 2p V p Vy y+ + = + +

( ) ( )

1 2

2 21 2 2 1 2 1

2 2y y

g g

p p V V y y

γ γρ γ− = − + −( ) ( )1 2 2 1 2 12

p p V V y yγ+

6.6.2 Rankine OvalsConsider a source and a sink of equal strength combined with a uniform flow to form the flow around a closed bodywith a uniform flow to form the flow around a closed body.The stream function and velocity potential for this combination are,combination are,

)(2

sin 21 θθθψ −−=mUr )ln(ln

2cos 21 rrmUr −−= θφ)(

2 21πψ )(

2 21πφ

As in Section 6.5.4

⎟⎠⎞

⎜⎝⎛−= −

221 sin2tan

2sin armUr θ

πθψ

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎠⎝ −

−222

1

22

2tan2

or

2

aymUy

ar

ψ

π

⎟⎠

⎜⎝ −+ 2222 ayx

yπ

ψ

The stream line ψ=0 forms a closed body.Since the body is closed, all of the flow emanating from the source

flows into the sink.• These bodies have an oval shape and are termed Rankine ovals.p

• The stagnation points correspond to the points where the uniform velocity the source velocity and the sink velocity all combine tovelocity, the source velocity, and the sink velocity all combine to give a zero velocity.

Th l ti f th t ti i t d d th l f• The location of the stagnation points depend on the value of a, m , and U.

The body half length:The body half length:1 1

2 22 or 1ma l ml a⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟ or 1l a

U a Uaπ π= + = +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

source:2rmv

rπ=

h fTherefore

0m mU − + =( ) ( )

02 2

2 0

Ur a r a

m aU

π π+ =

− +

2 2

2 2

02

1

m aUr a

m mπ

− =−

2 22 2

1

11 0 or m mr aU r a Uπ π

− = − =−

⎛ ⎞ 22ml r a

Uπ⎛ ⎞= = +⎜ ⎟⎝ ⎠

The body half width, h, can be obtained by determining y y gthe value of y where the y axis intersects the ψ=0 streamline. Thus, from Eq. 6.105 with ψ=0, x=0, and y=h, It f ll th tIt follows that

⎟⎞

⎜⎛⎟

⎞⎜⎛ −− 11 2t02t ahmUhaymU ⎟

⎠⎞

⎜⎝⎛

−−=→⎟⎟

⎠⎜⎜⎝ −+

−= 221

2221 tan

20tan

2 ahUh

ayxyUy

ππψ

12 2

2 2tan ah Uhh a m

π− ⎛ ⎞ =⎜ ⎟⎝ ⎠

2 2

2 2tan

h a mah Uh

hπ

−⎝ ⎠

=2 2

2 2 2tan2

h a mh a Uhh π

−−

=

2 2

21 2 11 tan 1 tan 2

a mh h Uh h Ua hπ π⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = −⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥1 tan 1 tan 2

2 2a a m a m a= =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Both l/a and h/a are functions of the dimensionlessBoth l/a and h/a are functions of the dimensionless parameter Ua/m.As l/h becomes large, flow around a long slender body isAs l/h becomes large, flow around a long slender body is described, whereas for small value of parameter, flow around a more blunt shape is obtained.Downstream from the point of maximum body width the surface pressure increase with distance along the surface. I t l i fl d di t illIn actual viscous flow, an adverse pressure gradient will lead to separation of the flow from the surface and result in a large low pressure wake on the downstream side ofin a large low pressure wake on the downstream side of the body.However, separation is not predicted by potential theory., p p y p yRankine ovals will give a reasonable approximation of the velocity outside the thin, viscous boundary layer and y y ythe pressure distribution on the front part of the body.

V6.6 Circular cylinderV6.8 Circular cylinder with separationV6.9 Potential and viscous flow

6.6.3 Flow around a circular cylinderWhen the distance between the source-sink pair approaches zero the shape of the Rankine oval becomes more blunt andzero, the shape of the Rankine oval becomes more blunt and approach a circular shape.A combination of doublet and uniform flow will representA combination of doublet and uniform flow will represent flow around a circular cylinder.

KK θi ⎞⎛

K

rrKU

rKUr

θ

θθθψ

cos

sinsinsin :function stream 2 ⎟⎠⎞

⎜⎝⎛ −=−=

rKUr θθφ coscos :potentialvelocity +=

to determine K with ψ=0 for r=a,

22 0 UaK

aKU =→=−

Thus the stream function and velocity potential for flowThus the stream function and velocity potential for flow around a circular cylinder are

2 ⎞⎛θψ sin1 2

2

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

raUr

θφ cos1 2

2

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

raUr

The velocity components are⎠⎝ r

ψφ 1 2⎟⎞

⎜⎛∂∂ a

φ

θθψφ

1

sin11

2

2

⎞⎛∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

∂∂

=∂∂

=raU

rrvr

θψθφ

θ sin112

2

⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

∂∂

−=∂∂

=raU

rrv

On the cylinder surface (r=a):Potential flow around a circular cylindera circular cylinder

θθ sin2 and 0 Uvvr −==

h f h i l i hTherefore the maximum velocity occurs at the top and bottom of the cylinder θ = ±π/2 and has a magnitude of twice the upstream velocity U.The pressure distribution on the cylinder surface is p yobtained from the Bernoulli equation,

2 20

1 12 2 θρ ρ+ = +s sp U p v

( )2 20

2 21 1 4sin2

ρ θ= + −sp p U ( )2

where p and U are pressure andwhere p0 and U are pressure and velocity for point far from the cylindercylinder.

The figure reveals that gonly on the upstream part of the cylinder is there yapproximate agreement between the potential flow and the experimental results.

The resulting force (per unit length) developed on the g (p g ) pcylinder can be determined by integrating the pressure over the surface.

2π

∫0

cos 0θ θ= − =∫x sF p ad

2

sin 0π

θ θ= − =∫y sF p ad0∫

Both the drag and lift as predicted by potential theory for a fixed cylinder in a uniform stream are zero. since the pressure distribution is symmetrical around the cylinder.In reality, there is a significant drag developed on a cylinder when it is placed in a moving fluid. (d’Alembert paradox)

Ex 6.8 Potential flow--cylinder

By adding a free vortex to the stream function or velocity i l f h fl d li d hpotential for the flow around a cylinder, then

θψ lnsin12 Γ

−⎟⎟⎞

⎜⎜⎛

−= raUr (6 119)πθψ ln

2sin1

2

2

Γ⎟⎞

⎜⎛

⎟⎟⎠

⎜⎜⎝

=

a

rr

Ur (6.119)

where Γ is the circulation

θπ

θφ2

cos1 2Γ

+⎟⎟⎠

⎞⎜⎜⎝

⎛+=

raUr

(6.120)

Tangential velocity on the surface (r=a):

Γ∂a

Ur

var

s πθψ

θ 2sin2 Γ

+−=∂∂

−==

(6.121)

This type of flow could be approximately created by placing a rotating cylinder in a uniform stream. Because the presence of viscosity in any real fluid, the fluid in contact with the rotating cylinder would rotate with the same velocity as the cylinder, and the resulting flow field would resemble that developed by thethe resulting flow field would resemble that developed by the combination of a uniform flow past a cylinder and a free vortex.

Location of the stagnation pointg p

4sin

2sin20 stag π

θπ

θθΓ

=→Γ

+−==Uaa

Uv s

or 0 0 if42

stag

g

πθππ

=→=ΓUaa

cylinderthefromawaylocatedis14/if

surfaceon thelocation other someat is 14/1 if stag

θπ

θπ

→>Γ

→≤Γ≤−

Ua

Ua

cylinderthefromaway locatedis 14/ if stagθπ →>Γ Ua

Force per unit length developed on the cylinderForce per unit length developed on the cylinder2

20

1 1 2 sin2 2 2sp U p U

aρ ρ θ

πΓ⎛ ⎞+ = + − +⎜ ⎟

⎝ ⎠2

2 20 2 2 2

2 2 2

1 2 sin1 4sin2 4s

a

p p UaU a U

π

θρ θπ π

⎝ ⎠⎛ ⎞Γ Γ

= + − + −⎜ ⎟⎝ ⎠

2

2 4

cos 0x s

aU a U

F p adπ

π π

θ θ

⎝ ⎠

= − =∫0

2 22sin siny s

UF p ad d Uπ πρθ θ θ θ ρ

πΓ

= − = − = − Γ∫ ∫0 0π

For a cylinder with circulation, lift is developed equal to the product of the fluid density the upstream velocity and theproduct of the fluid density, the upstream velocity, and the circulation.

yF Uρ= − Γ

( )( ) +, counterclockwise the is downwardy

yU F+ Γ

The development of this lift on rotating bodies is called the Magnus effect.

6.7 Other Aspects of Potential Flow6.7 Other Aspects of Potential Flow Analysis

Exact solutions based in potential theory will usually provide at best approximate solutions to real fluid problems.Potential theory will usually provide a reasonable approximation in those circumstances when we are dealing with a low viscosityfl id i l i l hi h l i i i f h flfluid moving at a relatively high velocity, in regions of the flow field in which the flow is accelerating.O t id th b d l th l it di t ib ti d thOutside the boundary layer the velocity distribution and the pressure distribution are closely approximated by the potential flow solutionflow solution.In situation when the flow is decelerating (in the rearward portion of the bluff body expanding region of a conduit) andportion of the bluff body expanding region of a conduit), and adverse pressure gradient is reduced leading to flow separation, a phenomenon that are not accounted for by potential theory.p y p y

V6.10 Potential flow

PART CViscous Flow:

Navier-Stokes Equationq(Sections 6.8-6.10)(Sections 6.8 6.10)

6.8 Viscous FlowEquation of Motion

x xmaδ δ=F y ymaδ δ=F z zmaδ δ=Fδδδδ zyxm δδρδδ =

Thus

yxxx zxx

u u u ug u v wτσ τρ ρ

∂ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂+ + + = + + +⎜ ⎟

⎝ ⎠x

xy yy zy

gx y z t x y z

v v v vg u v w

ρ ρ

τ σ τρ ρ

⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠∂ ∂ ∂ ⎛ ⎞∂ ∂ ∂ ∂

+ + + + + +⎜ ⎟y yy y

yg u v wx y z t x y z

w w w w

ρ ρ

ττ σ

+ + + = + + +⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠∂ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂yzxz zz

zw w w wg u v w

x y z t x y zττ σρ ρ

∂ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂+ + + = + + +⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠

Wh h i li d fl id6.8.1 Stress-Deformation Relationships

When a shear stress is applied on a fluid:• Fluids continuously deform (stress τ ~ rate of strain) • Solids deform or bend (stress τ ~ strain)

strain rate ~ velocity gradient

dd

strain rate ~ velocity gradient

dydu

dtd =α

from Fox, McDonald and Pritchard, Introduction to Fluid Mechanics.

6.8.1 Stress-Deformation RelationshipsF i ibl N t i fl id it i k th tFor incompressible Newtonian fluids it is known that the stresses are linearly related to the rate of deformationdeformation.

V1.6 Non-Newtonian behavior

F i ibl N t i fl id th i t

2 yx x xzvv v vvμ μ μ

⎡ ∂⎛ ⎞∂ ∂ ∂∂⎛ ⎞+ +⎢ ⎜ ⎟ ⎜ ⎟⎤⎥

For incompressible, Newtonian fluids, the viscous stresses are:

,visc

2

2xx xy xz

y y yx z

x y x x z

v v vv v

μ μ μ

σ τ τ

+ +⎢ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠⎢⎡ ⎤ ⎢ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞∂ ∂⎢ ⎥ ⎢ ⎜ ⎟ ⎜ ⎟

⎥⎥⎥⎥

visc, ,visc

,visc

2y y yx zij yx yy yz

zx zy zz

y x y z y

vvv v v

τ σ τ μ μ μτ τ σ

⎢ ⎥ ⎢≡ ≡ + +⎜ ⎟ ⎜ ⎟⎢ ⎥ ∂ ∂ ∂ ∂ ∂⎢ ⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢⎣ ⎦ ∂⎛ ⎞∂∂ ∂ ∂⎛ ⎞⎢

τ ⎥⎥⎥⎥2yxz z z

vvv v vx z z y z

μ μ μ∂⎛ ⎞∂∂ ∂ ∂⎛ ⎞⎢ + +⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎣

⎥⎢ ⎥

⎦

6.8.1 Stress-Deformation RelationshipsBut in normal stresses, there is additional contribution of pressure p, where

( )13 xx yy zzp σ σ σ− = + +

p p,

Consequently,

u υτ τ μ⎛ ⎞∂ ∂

+⎜ ⎟

for normal stresses

2 upσ μ ∂+

for shearing stresses

xy yx y x

w

τ τ μ

υ

= = +⎜ ⎟∂ ∂⎝ ⎠⎛ ⎞∂ ∂

+⎜ ⎟

2

2

xx px

p

σ μ

υσ μ

= − +∂∂

= − +yz zy z y

w u

τ τ μ

τ τ μ

= = +⎜ ⎟∂ ∂⎝ ⎠∂ ∂⎛ ⎞+⎜ ⎟

2

2

yy pywp

σ μ

σ μ

= +∂∂

= − +zx xz x z

τ τ μ= = +⎜ ⎟∂ ∂⎝ ⎠2zz p

zσ μ= − +

∂

Can you figure out why the normal viscous stress σ can beCan you figure out why the normal viscous stress σxx,visc can be expressed as ?2 u

xμ ∂∂

6.8.1 Stress-Deformation Relationships

For viscous fluids in motion the normal stresses are not necessarily the same in different directions, thus, the need to define the pressure as the average of the three normal stresses. Stress-strain relationship in cylindrical coordinate

2 rrr p

rυσ μ ∂

= − +∂

1 rr r r

r r rθ

θ θυ υτ τ μ

θ⎛ ⎞∂∂ ⎛ ⎞= = +⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠⎝ ⎠

12 r

r

pr r

θθθ

υ υσ μθ

∂∂⎛ ⎞= − + +⎜ ⎟∂⎝ ⎠

1 zz z

r r r

θθ θ

θ

υ υτ τ μθ

∂ ∂⎝ ⎠⎝ ⎠∂ ∂⎛ ⎞= = +⎜ ⎟∂ ∂⎝ ⎠

2 zzz

r r

p

θυσ μ

∂⎝ ⎠∂

= − +∂

z z

r z

z rθ θ μθ

υ υτ τ μ

⎜ ⎟∂ ∂⎝ ⎠∂ ∂⎛ ⎞= = +⎜ ⎟zz p

zμ

∂ rz zr z rτ τ μ= = +⎜ ⎟∂ ∂⎝ ⎠

Note: Notation x: plane perpendicular to x coordinateτNote: Notation x: plane perpendicular to x coordinatey: direction

xyτ

6 8 2 The Navier-Stokes Equations6.8.2 The Navier-Stokes Equations

The Navier Stokes equations are considered to beThe Navier-Stokes equations are considered to be the governing differential equations of motion for incompressible Newtonian fluids

2 2 2⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

incompressible Newtonian fluids

2 2 2

2 2 2xu u u u p u u uu w gt x y z x x y z

ρ υ ρ μ⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

+ + + = − + + + +⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠2 2 2

2 2 2ypu w g

t x y z y x y zυ υ υ υ υ υ υρ υ ρ μ

⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂+ + + = − + + + +⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

2 2 2

2 2 2z

y y y

w w w w p w w wu w gt x y z z x y z

ρ υ ρ μ

⎝ ⎠ ⎝ ⎠⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

+ + + = − + + + +⎜⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎟t x y z z x y z∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

The Navier-Stokes Equations

In terms of cylindrical coordinate

The Navier Stokes Equations

y2

r r r rθ θυ υυ υ υ υρ υ υ⎛ ⎞∂ ∂ ∂ ∂

+ + − +⎜ ⎟

2 21 1 2

r z

r r r r

t r r r z

p g r θ

ρ υ υθ

υυ υ υ υρ μ

+ + +⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠⎛ ⎞∂∂ ∂ ∂∂ ∂ ⎛ ⎞= − + + − + − +⎜ ⎟⎜ ⎟ 2 2 2 2 2r

r

g rr r r r r r r z

θ θ θ θ θ θ

ρ μθ θ

υ υ υ υ υ υ υρ υ υ

= − + + − + − +⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠∂ ∂ ∂ ∂⎛ ⎞+ + + +⎜ ⎟

21 1 1

rr zt r r r z

p

θ θ θ θ θ θ

θ θ θ

ρ υ υθ

υ υ υ

+ + + +⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

∂ ∂∂ ∂ ⎛ ⎞⎜ ⎟

22 r θυυ⎛ ⎞∂∂⎜ ⎟2 2 2

p g rr r r r r r

θ θ θθρ μ

θ θ⎛ ⎞= − + + − +⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ 2 2

r

r zθ

θ

θ

υυ υ υ υ

− +⎜ ⎟∂ ∂⎝ ⎠∂ ∂ ∂ ∂⎛ ⎞

2 21 1

z z z zr zt r r z

p

θυυ υ υ υρ υ υθ

υ υ υ

∂ ∂ ∂ ∂⎛ ⎞+ + +⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠⎛ ⎞∂ ∂ ∂∂ ∂ ⎛ ⎞

2 2 2

1 1z z zz

p g rz r r r r z

υ υ υρ μθ

⎛ ⎞∂ ∂ ∂∂ ∂ ⎛ ⎞= − + + + +⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠

6 9 Some Simple Solutions for6.9 Some Simple Solutions forViscous, Incompressible Fluids

There are no general analytical schemes for solving nonlinear partial differential equations and eachnonlinear partial differential equations, and each problem must be considered individually.

2 2 2⎛ ⎞⎛ ⎞ 2 2 2

2 2 2xu u u u p u u uu w gt x y z x x y z

ρ υ ρ μ⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

+ + + = − + + + +⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠2 2 2

2 2 2ypu w g

t x y z y x y zυ υ υ υ υ υ υρ υ ρ μ

⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂+ + + = − + + + +⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

2 2 2

2 2 2z

y y y

w w w w p w w wu w gt

ρ υ ρ μ

⎝ ⎠ ⎝ ⎠⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

+ + + = − + + + +⎜⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎟2 2 2t x y z z x y z⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

Nonlinear termsNonlinear terms

6.9.1 Steady Laminar Flow BetweenFixed Parallel plates

u

0, 0wυ = =

maxu

,

Thus continuity indicates that u∂ 0ux

∂=

∂

for steady flow ( )u u y=2 2 2u u u u p u u uu w gρ υ ρ μ

⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂+ + + = + + + +⎜ ⎟⎜ ⎟for steady flow, ( )u u y=

0,x y zg g g and g o= = − =

2 2 2

2 2 2

2 2 2

x

y

u w gt x y z x x y z

pu w gt x y z y x y z

ρ υ ρ μ

υ υ υ υ υ υ υρ υ ρ μ

+ + + = − + + + +⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

+ + + = − + + + +⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠y

2 2 2

2 2 2z

y y y

w w w w p w w wu w gt x y z z x y z

ρ υ ρ μ

⎝ ⎠ ⎝ ⎠⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

+ + + = − + + + +⎜⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎟

Steady Laminar Flow Between Fixed P ll l l tParallel platesThus us

2

20 p ux y

μ∂ ∂= − +

∂ ∂

( )10 p g p gy f xy

ρ ρ∂= − − → = − +

∂

2

0

1

pz

d

∂= −

∂∂

2 2 2

2 2 2xu u u u p u u uu w gt x y z x x y z

ρ υ ρ μ⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

+ + + = − + + + +⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

2

1

1

d u pdy xdu p

μ∂

=∂

∂⎛ ⎞

2 2 2

2 2 2

2 2 2

ypu w g

t x y z y x y z

w w w w p w w w

υ υ υ υ υ υ υρ υ ρ μ⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

+ + + = − + + + +⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

1

2

1

1

du p y Cdy x

pμ

∂⎛ ⎞= +⎜ ⎟∂⎝ ⎠∂⎛ ⎞

2 2 2zw w w w p w w wu w gt x y z z x y z

ρ υ ρ μ⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

+ + + = − + + + +⎜⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎟

21 2

12

pu y C y Cxμ

∂⎛ ⎞= + +⎜ ⎟∂⎝ ⎠

p∂px

∂∂( is treated as a constant since it is not a function of y)

Steady Laminar Flow Between Fixed P ll l l tthe constants are determined from the boundary

Parallel platest e co sta ts a e dete ed o t e bou da yconditions.

BCs : 0u for y h= = ±V6.11 No-slip boundary conditions

1

BCs : 0Thus 0

u for y hC

= = ±=

⎛ ⎞ 22

1 2

pC hxμ

∂⎛ ⎞= − ⎜ ⎟∂⎝ ⎠

Thus the velocity distribution becomes,

( )2 212

pu y hxμ

∂⎛ ⎞= −⎜ ⎟∂⎝ ⎠2 xμ ∂⎝ ⎠

which indicates that the velocity profile between th t fi d l t i b lithe two fixed plates is parabolic.V6.13 Laminar flow

Steady Laminar Flow Between Fixed

The volume rate of flow

Parallel platesThe volume rate of flow

( )2 21h h pd h d∂⎛ ⎞⎜ ⎟∫ ∫ ( )2 2

3

2

1

h h

h

pq u dy y h dyx

p y

μ− −

⎛ ⎞= = −⎜ ⎟∂⎝ ⎠⎡ ⎤∂

∫ ∫

212 3 h

p yq h yxμ

−

⎡ ⎤∂⎢ ⎥= −

∂ ⎢ ⎥⎣ ⎦3 3

3 312 3 3

p y hh hxμ

⎡ ⎤∂= − + −⎢ ⎥∂ ⎣ ⎦

323

h pxμ

∂= −

∂μ

The pressure gradient is negative, since the pressure d i th di ti f th fl decreases in the direction of the flow.

Steady Laminar Flow Between Fixed

f h d b

Parallel platesIf Δp represents the pressure drop between two points a distance apart, thenl

p px

Δ ∂= −

∂l3 3 22 2 ,

3 3 2 3h p h p q h pq V

x hμ μ μ∂ Δ Δ

= − = = =∂ l lμ μ μ

The maximum velocity umax , occurs midway y=0 between athe two plates, thus

Vuxphu

23 or

2 max

2

max =∂∂

−=μ x 22 ∂μ

Steady Laminar Flow Between Fixed

The pressure field

Parallel platesThe pressure field

( )p gy f xρ= − +

( )1 0pf x x px

∂⎛ ⎞= +⎜ ⎟∂⎝ ⎠⎝ ⎠where is a reference pressure at x=y=0 0p

Thus the pressure variation throughout the fluid can be obtained from

⎛ ⎞

The above analysis is valid for0

pp gy x px

ρ ∂⎛ ⎞= − + +⎜ ⎟∂⎝ ⎠ 2R V hρThe above analysis is valid for remains below about 1400

Re ρμ

=

Problem 6.88: 10 tons on 8psi

6.9.2 Couette FlowTherefore

21 2

12

pu y C y Cxμ

∂⎛ ⎞= + +⎜ ⎟∂⎝ ⎠

boundary conditionsu=o at y=0, u=U at y=by , y

( )212

b pu U y byy xμ

∂⎛ ⎞= + −⎜ ⎟∂⎝ ⎠

or in dimensionless form 2

1u y b p y y∂⎛ ⎞⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟

The actual velocity profile will depend on the

12

y p y yU b U x b bμ

⎛ ⎞⎛ ⎞⎛ ⎞= − −⎜ ⎟⎜ ⎟⎜ ⎟∂⎝ ⎠⎝ ⎠⎝ ⎠

dimensionless parameter 2b pP ∂⎛ ⎞= − ⎜ ⎟

This type of flow is called Couette flow.

2P

U xμ ⎜ ⎟∂⎝ ⎠

Couette flowThe simplest type of Couette flow is one for which the pressure gradient is zero i.e. the fluid motion is causedpressure gradient is zero i.e. the fluid motion is caused by the fluid being dragged along by the moving boundary. 0p∂y

Thus

0px

yu U

=∂

Thus u Ub

=

which indicates that the velocity varies linearly y ybetween the two plates.

e g ： Journal bearinge.g. ： Journal bearingro-ri << ri

Th fl i l d dThe flow in an unloaded journal bearing might be approximated by this simpleapproximated by this simple Couette flow.

Example 6.9( )0 0u w xυ υ υ∂

= = = = ( )0 0

0

u w xy

p p

υ υ= = = =∂

∂ ∂= = 0

x zx h p atmospheric pressure

d d

∂ ∂= =

0 0dp dpdx dz

∴ = ∴ =

Th f 2

20 dgdx

υρ μ= − +Therefore

2

2

ddx

υ γμ

=

1d x Cdxυ γ

μ= +

on the film surface x=h we assume that the shearing stress is on the film surface x=h, we assume that the shearing stress is zero

0xy xyd at x hdxυτ μ τ⎛ ⎞= = =⎜ ⎟

⎝ ⎠

1hC γ

μ= −

2nd integration2nd integration2

22hx x Cγ γυ

μ μ= − +

0 2 0

2

20x V C V

h

μ μυ

γ γ= = ∴ =

20

2

2h h

hx x V

h

γ γυμ μ

γ γ

= − +

⎛ ⎞∫ ∫ 2

00 0

3

2h h hq dx x x V dx

h

γ γυμ μ

γ

⎛ ⎞= = − +⎜ ⎟

⎝ ⎠∫ ∫

0 3hq V h γμ

= −

The average film velocityThe average film velocity2

0q hV V γ

= = −0 3V V

h μ2

0 3hV γ

>Only if , will there be a net upward flow of liquid.0 3μy , p q

Q: Do you find anything weird in this problem?

6.9.3 Steady, Laminar flow in6.9.3 Steady, Laminar flow in Circular Tubes

Hagen–Poiseuille flow or Poiseuille flow steady, laminar flow through a straight circular tube y g gof constant cross sectionConsider the flow through a horizontal circular tube gof radius R

0Assume the flow is parallel 0rv vv

θ= =∂ 0zv

z∂

=∂

( )z zv v r∴ =

Steady, Laminar flow in Circular Tubes

Thus θsinggr −=θρ ∂

−−=pg 1cos0

rpg

∂∂

−−= θρ sin0

ggrθρ

∂rg

θθ cosgg −=r∂

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+∂∂

−=rvr

rrzp z10 μ

⎦⎣ ⎠⎝ ∂∂∂ rrrz

Integration of equations in the r and θ directions( )

( )1

1

sinp gr f z

gy f z

ρ θ

ρ

= − +

= − + ( )1gy fρ

which indicate that the pressure is hydrostatically distributed at any particular cross section and the zdistributed at any particular cross section and the zcomponent of the pressure gradient, , is not a function of r or θ

zp ∂∂ /function of r or θ.

Steady, Laminar flow in Circular Tubesthe equation of motion in the z direction

1 1zv pr ∂∂ ∂⎛ ⎞ =⎜ ⎟⎝ ⎠

21

12

z

rr r r z

v pr r C

μ⎜ ⎟∂ ∂ ∂⎝ ⎠∂ ∂⎛ ⎞= +⎜ ⎟∂ ∂⎝ ⎠

21 2

21 ln

4z

r zpv r C r Cz

μ

μ

⎜ ⎟∂ ∂⎝ ⎠∂⎛ ⎞= + +⎜ ⎟∂⎝ ⎠

Boundary conditions 4 zμ ∂⎝ ⎠

At r=0, vz is finite at the center of the tube, thus C1=0

At r=R v =0 then 21 pC R∂⎛ ⎞= − ⎜ ⎟At r R, vz 0, then 2 4C R

zμ= − ⎜ ⎟∂⎝ ⎠

Thus the velocity distribution becomes,Thus the velocity distribution becomes,

( )22

41 Rr

zpvz −⎟

⎠⎞

⎜⎝⎛

∂∂

=μ4 z ⎠⎝ ∂μ

That is, at any cross section, the velocity distribution is parabolic.

Steady, Laminar flow in Circular TubesVolume flow rate

drrvdQ z= π )2(

zpRrdrRr

zprdrvQ

RRz

z

∂∂

−=−⎟⎠⎞

⎜⎝⎛

∂∂

== ∫∫ μπ

μππ

8)(

4122

4

0

22

0 zz ∂⎠⎝ ∂∫∫ μμ 8400

Let , thenpp ∂

−=Δ

l

π8

4 pRQ Δ= Poiseuille’s law ,

z∂l lμ8Q Poiseuille s law

mean velocity

lμπ 8

2

2

PRRQV Δ

==

y

2 2R p R pv ∂ Δ⎛ ⎞= =⎜ ⎟ so Vv 2=

maximum velocity

max 4 4v

zμ μ⎜ ⎟∂⎝ ⎠ lso Vv 2max =

the velocity distribution in terms of vmax2⎞⎛

max

1 ⎟⎠⎞

⎜⎝⎛−=

Rr

vvz

6.9.4 Steady, Axial, Laminar Flow in an Annulusan Annulus

212 ln

41 CrCrpvz ++⎟

⎠⎞

⎜⎝⎛

∂∂

=

B.Cs：vz=0 at r=ro and r=ri

214 zz ⎟⎠

⎜⎝ ∂μ

thus( ) ( )2 2

2 2 00 0

1 ln /iz i

r rpv r r r r⎡ ⎤−∂⎛ ⎞= − +⎢ ⎥⎜ ⎟

⎝ ⎠ ⎢ ⎥( ) ( )0 004 ln /z i

iz r rμ ⎢ ⎥⎜ ⎟∂⎝ ⎠ ⎢ ⎥⎣ ⎦

volume rate of flow( )0

22 24 42

ro ir rpQ d π ⎡ ⎤−∂⎛ ⎞ ⎢ ⎥⎜ ⎟∫

( )( )

4 4

0

28 ln /

i

o iz o i

ir

pQ v rdr r rz r r

πμ

⎛ ⎞ ⎢ ⎥= = − − −⎜ ⎟ ⎢ ⎥∂⎝ ⎠⎣ ⎦

∫

( )22 24 4 o ir rp r rπ ⎡ ⎤−Δ ⎢ ⎥= − − −

⎢ ⎥( )08 ln /o ii

r rr rμ

⎢ ⎥=⎢ ⎥⎣ ⎦

l

The maximum velocity occur at the , rr = 0v r∂ ∂ =The maximum velocity occur at the , mrr 0zv r∂ ∂

122 2⎡ ⎤

( )22 2

02ln /o i

mi

r rrr r

⎡ ⎤−= ⎢ ⎥

⎢ ⎥⎣ ⎦( )02ln / ir r⎢ ⎥⎣ ⎦

The maximum velocity does not occur at the mid point of theThe maximum velocity does not occur at the mid point of the annulus space, but rather it occurs nearer the inner cylinder.

T d t i R ld b it i ti tTo determine Reynolds number, it is common practice to use an effective diameter “hydraulic diameter” for on circular tubes.

i ttt darea sctional-cross4×

=hD

ρ VDR hf f

perimeterwettedh

μρ

R he =Thus the flow will remain laminar if remains below 2100.

6.10 Other Aspects of Differential Analysis

0∇ V

∂⎛ ⎞V

0∇ ⋅ =V

2( ) pt

ρ ρ μ∂⎛ ⎞+ ⋅∇ = −∇ + + ∇⎜ ⎟∂⎝ ⎠

V V V g V

The solutions of the equations and not readily available.

6.10.1 Numerical MethodsFinite difference methodFi it l t ( fi it l ) th dFinite element ( or finite volume ) methodBoundary element method

V6.15 CFD example