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
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