Advanced Fluid Dynamics MAK 503E

Dr. Hasan Gunesguneshasa@itu.edu.tr

http://atlas.cc.itu.edu.tr/~guneshasa

DEFINITION OF A FLUIDA fluid is a substance that deforms continuously under the application of a shear(tangential) stress no matter how small the shear stress may be.

F/A≡ τ

t2 > t1 > t0

t0t1

t2

No-slip condition

Mechanics of fluids1. Inviscid flow: viscosity assumed to be zero –

simplify analyses but meaningful result needed2. Viscous flow: viscosity important

INTRODUCTION

Solid deforms when a shear stress is applied, but it does not deformcontinuously.

True fluids: all gases, common liquids, water, oil, gasoline, alcohol, ...Non-true fluids: high-polimer solutions, emulsions, toothpaste, egg white etc. Rheology: general study of flow and deformation of materials.

Actual molecular structure

Calculate/measuremacroscopic fluid properties, e.g. viscosity

Replace the actual molecularstructure by a hypotheticalcontinuous medium

Composed molecules in constant motionmolecules & gaps among them

Continuum Fluid Mechanics:

Continuous medium meansa. There are no gaps (in the fluid) or empty spacesb. Each fluid property is assumed to have a definite value at each point in space

( , , , ) ; ( , , , ) ; ( , , , ) ; p( , , , )x y z t T x y z t V x y z t x y z tρ ρ=

c. All the mathematical functions entering the theory are continuous functions exceptpossibly at a finite number of interior surfaces seperating regions of continuity.

ρ2 , µ2

ρ1 , µ1

ρ1 , µ1

d. Derivatives of the functions are also continuous too if they enter the theory

Knudsen number, Kd = λ / L << 1λ : mean free path of moleculesL : characteristics length

Ex.

Volume, Vof mass, m

Volume, δVof mass, δm

xx0

y0

z0

y

z

C

What is the density of fluid at point C ?.

mmean density = =ρ∀

lim mδ δ

δρδ′∀→ ∀

≡∀

In general mean density, ρ is not equal to density at point C.

∀

Still have largeenough moleculesfor consistent result

mδδ∀

δ ′∀δ∀

Volume so small, moleculescross into and out of CV

( , , , ) x y z tρ ρ=scalar fielde.g. 1m3 air ≈ 2.5x1025 molecules

Similarly, velocity at point C defined as the instataneous velocity of the fluid particlewhich, at a given instant, is passing through point C.Fluid particle: a small mass of fluid of fixed identity of volumeδ ′∀

: velocity field ; ( , , , ) : vector fieldV V x y z t

= , , , : components of velocity vector-scalar( , , , )( , , , )( , , , )

V ui v j wk u v wu u x y z tv v x y z tw w x y z t

+ +===

Steady/unsteady flow:Properties at every point in a flow field do not change with time.Time-independent flow, stationary flow.

0 , 0 , or ( , , )x y zt tη ρ ρ ρ∂ ∂

= = =∂ ∂

Incompressible flow: ρ = const.Compressible flow: ρ = variable

0.3 gases also incomp. V 100 m/s air at standart conditions

a = speed of sound

V Ma

= ≤ → → ≅

Re ref refV L ρµ

=

Small Re → laminar: no macroscobic mixing, fluid flow in layers on laminae (smooth flow pattern)Large Re → turbulent: time-dependent, 3-D , chaotic motion(random) macroscopic mixing

Basic Equations:1. Conservation of mass (continuity eq.)2. Conservation of momentum (Newton’s second law)3. Conservation of energy (First law of thermodynamics)4. 2nd law of thermodynamics5. Equation of state ρ= ρ(p,T)

Unknowns ρ,u,v,w,p,T,s total number of equations ?

VISCOSITYDensity : measure of the “heaviness” of a fluid.Viscosity (µ) : measure of the “fluidity” of a fluid.

Force, δFxVelocity, δu

No slipB C

A A’ D D’

δl

δyNo slip

y

x

e.g. water & oil approx. have same density but behave differently when flowing

Fluid element at time, t : ABCDFluid element at time, t + δt : A’BCD’

δα

•ability of a fluid to flow freely

directionplane that stressacts actson

y xτ0

lim x xyx Ay

y y

F dFA dAδ

δτδ→

= =

yAδ : area of fluid element in contact with the plate

0Deformation rate = lim

For small , tan

t

dt dt

l u tl l y u ty

u d dut y dt dy

δ

δα αδ

δ δ δδδα δα δα δ δαδ δ δδ

δα δ αδ δ

→=

=

≅ = ⇒ = =

= ⇒ =

Subject to shear stress, yxτfluid experiences a time rate of deformation (shear rate) as du/dy

yxdudy

τ ∼

yxdudy

τ µ=

dudy

0 0 ( 0)yxdudy

τ µ= ⇒ = ≠

Newtonian fluid, linear relationship

µ: [Pa.sec]=[kg/m-sec] (absolute viscosity)

: velocity gradient

No relative motion :

Example: Couette FlowU0

h y

x

0yu U ih

⎛ ⎞= ⎜ ⎟⎝ ⎠

Shear stressdirectionsurface

0 0y xUdu i

dy hτ µ τ µ= = = >

Stress sign conventionStress comp. Plane Direction+ + ++ - -- - +- + -

Which direction does the stress act?Depends on the surface we are looking at.

U0

y

x

(-)

(+)

(+)

(-)

(+)(+)

(-)(-)

Positive outward normal on y-surface

Fluid

Crude oil

Water (60 °F)Water (100 °F)

Air (60 °F)

du/dy0Time rate of deformation

du/dy0

τ τBingham plastic, toothpaste, mayonnaise

Shear thining, polimer solutions, latex paint, blood

Shear tickening, water-sandmixture

Bingham plastic: can withstand a finite shear stress without motion not fluidBut once the yield stress is exceeded it flows like a fluid not a solid

Non- true fluids

dudy

yxτ

: finite

finite ; u must vary continusly across the flow, no abrupt change between adjoining elements of fluids.Consider solid boundary : No-slip B.C.

With respect to size of fluid molecules, solid surfaces, no matter how well polished, haveirregularities, i.e cavities filled by fluid.Fluid immediately in contact with the boundary, has the same speed with it. Viscous fluid No-slip B.C.

Perfect Fluid: 0τ =No internal resistance to a change in shape (µ=0) Fails to predict drag of body

Seperation occurs due to APG 0dpdx

⎛ ⎞>⎜ ⎟⎝ ⎠

increasing pressure gradient in flow direction

U∞

Rr

Stagnation points

Vθ Vr

Flow past a circular cylinder: Inviscid theory2-D, ρ=constant (incompressible flow), irrotational flow Superposition of doublet & uniform flow

2 2

2 2radial azimuthal velocity velocity

1 ,

1 cos , 1 sin

r

r

V Vr r r

R RV U V Ur r

θ

θ

φ φ θ

θ θ∞ ∞

∂ ∂= − = −

∂ ∂⎛ ⎞ ⎛ ⎞

= − = − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

0( , ) ( ,0), ( , ) stagnation points

r rV V e V er R R

θ θ

θ π= + =

=

At the cylinder surface, r=R , Vr = 0 , Vθ = -2U∞sinθ≠0Violates the “no-slip” condition between solid & fluidPressure distribution at the cylinder surface, apply Bernoully eq. (neglect elevation dif.)

,max22

V U Vθ θπθ ∞= ⇒ = − =

Limititations:1. Steady2. ρ=constant3. frictionless flow: inviscid4. flow along a streamline5. gravity force

2

2P U gzρ∞ ∞+ +

2

2VP gz

ρ= + +

( ) ( )2 2 2 21 1 1 4sin2 2

P P U V Uρ ρ θ∞ ∞ ∞− = − = −

2

21 4sin1

2

pP Pc

Uθ

ρ

∞

∞

−= = − pressure distribution on cylinder, valid for inviscid flow

Cp: pressure coef. [-]

π/2 π0 θ

10

-1-2

One side of cylinderFD : drag (force): force component parallel to the freestream flow direction :

D xF dF= ∫: cos

xx

DdAA

F P dA θ−∫

( )cos sin

cos (projection in x-direction)sin

r

x

y

d A Rd i Rd i j

dA R ddA R d

θ θ θ θ

θ θθ θ

= = +

=

=

R

yx

θ

rd A Rd iθ=

2

2

12

32

p p U

p p U

ρ

ρ

∞ ∞

∞ ∞

− =

− = −

3 dynamic pressure units lower than p∞(atm. press.)

ri

2

0

( ) ( )2A

dA b R d b Rπ

θ π= =∫ ∫

( )2

2 2

02 2

2 2

00

1 1 4sin cos2

1 sin 3 sin 2 02 3

DF p u R d

p u R u R

π

π π

ρ θ θ θ

θρ θ ρ

∞ ∞

∞ ∞ ∞

⎡ ⎤= − + −⎢ ⎥⎣ ⎦

⎛ ⎞= − + + =⎜ ⎟⎝ ⎠

∫

0!DF =(Noviscosity) d’Alembert paradox: inviscid flow past immersed bodies, drag=0 , symmetric pressure distribution

0L yA

F pdA= − =∫ Lift is zero! !

In reality, large drag force! No symmetryWrt y-axis

R

y

θwake

γ

separationPressure drops

Karman Vortexstreet

Wake structure: depends on Re (gets complicated as Re ↑ ) cp = 0 → p = p∞cp = 1 → p = p∞ + ½* ρU2∞

γ

γ ≈ 60° (theory is valid up to)

Subcritical Re (laminar) /typical experimental trends)

Supercritical Re (turb.)180°90°

+1

0 -1

-2

-3

DRAG COEFFICIENT; CD :2

projectedtotat frontaldrag areaforce

12D DF C V Aρ=

Power ≡ Drag * Vel. Power, that we pay to move aircraft.Racing cars – unload tires, reduce drag & lift

W

FL

Inlet ducts to produce down force

Drag: due to1. Pressure forces 2. Friction forces (shear stress) flow over a flat

plate parallel to the flow

D wF dAτ= ∫From dimensional analysis,

1

2 22 2

23

( , , , )

(Re)

(Re) A:cross-sectional area

D

D

F f D V

F VDf fV D

A f

µ ρ

ρρ µ

=

⎛ ⎞= =⎜ ⎟

⎝ ⎠=

CD = CD (geometry,ReL) valid for ρ=const. over any body

Uρ,µ

Charact. length

Laminar

Theory stokes sol.

Friction drag %5 of total

Smooth surface

Rough surface

100 103 104 105 106 Re

~0.3

~1.2

Transition to turb. On cylinder causes “drag crises” at Rec≈3*105

Rec = f (roughness)

Drag coef. Of a circular cylinderRough surface: early turbulance due to roughness, drag crisses occurs earlier, dimpling on goal ballsFreestream turbulance: drag crisses occurs earlier, similar behaviour with sphere “dimples” to “trip” the B.L. to cause TBL.

Upsream of the cyl. midsection

Turbulent wake low pressure

S.P

L.BL

Rec ≈2*105 (smooth surface)Re< Rec

Net pressure force on cylinder is reduced CD

Smaller wake

S.P

L.BL

Re>Rec

Boundary-Layer Control: spinw:angular velocity (rd/sec)moving surface to reduce skin friction effects on BL

1 2 4

0.2

0.4

0.6

CD

CL

wake

p

pV

FL (lift force)

Spiln ratio,

Weak function of ReLow spin ratio , wD/2V ≤ 0.5 neg. lift!Flow pattern, lift and drag coef. for a smooth spinning sphere in uniform flow.The wake is not symmetric wrt incoming vel.

AIRFOIL:

s, span

thickness, t

Chord, L

angle of attack

α

U∞Thickness ratio, t/cAspect ratio , AR = s/L

LIFTTrailing edge

Thin wake

Thin B.L with no separation

Leading edgeStream lines

α=0α<5°

2-D infinite span

U∞ , p∞

reattachement

wake

sep. point

sep. pointRe↑

As long as flow reattaches its symmetric & no lift

Pstag

P∞

t↑

FPG

0dPdx

<

APG

0 possible seperationdPdx

> ⇒

21lift=2

: lift coefficient

L

L

C U A

C

ρ ∞

CL

α

slower

fasterU∞ , p∞

α>010-15°

Broad wake

B.L with seperation (stall)

CL

D

L

CC

Optimum cruise

Seperation , loss of lift increase drag

FORCE(LIFT)

Upper surface

Pressure distribution α≠0

Pstag

Lower surface

PROPERTIES OF A FLUID & FLOW FIELD:1. Kinematic properties: linear velocity, angular velocity, vorticity, acceleration, and

strain rate Flow field properties2. Transport properties: µ, k3. Thermodynamic properties: p, ρ, T, h, s, cp, Pr, β4. Other properties: surface tension, vapor pressure, etc

DESCRIPTION OF FLUID MOTION: A. Lagrangian description: useful in solid mech.B. Eulerian description: proper choice in fluid mechanics

A) specifies how an individual particle moves through space.formulation is always time-dependentB) specifies the velocity distribution in space and time. i.e. specifies how a particle at a point would be.

( , , , )V V x y z t= defines the motion at time t at all points of space occupied by the fluid

Advantages of Euler description of fluid motion:• no need to follow the path of particles• in some cases, unsteady flow can be considered as steady by appropriately

selecting coordinates

Ex:

Kinematics, Substantial derivative1) Acceleration of a Fluid Particle in a Velocity Field: pa

Problem: how to relate the local motion of a particle to the velocity field

Study velocity field as a function of position & time, not trying to follow any specificparticle paths. But conservation laws are formulated for particles (systems) of fixedidentity; i.e., they are Lagrangian in nature.

z

y

x

Particle pathParticle at time, t

Particle at time, t+dt

( , , , ) , ( , , )V x y z t ui v j wk r r x y z= + + =

( , , , )

at time t+dt ( , , , )

p t

p t dt

V V x y z t

V V x dx y dy z dz t dt+

=

= + + + +

For a particle moving in velocity field at time t.

pV V V VdV dx dy dz dtx y z t

∂ ∂ ∂ ∂= + + +

∂ ∂ ∂ ∂

total accel. localconvective accel.of particle accel.

pp

u v w

p

dV V dx V dy V dz Vadt x dt y dt z dt t

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

∂ ∂ ∂ ∂= = + + +

∂ ∂ ∂ ∂

∂ ∂ ∂ ∂= = + + +

∂ ∂ ∂ ∂

The change in particle velocity can be shown by differential calculus to be,

Or for total acceleration of particle,

(..)DDt

,x pDu u u u ua u v wDt x y z t

∂ ∂ ∂ ∂= = + + +

∂ ∂ ∂ ∂3 comp. Eq. etc.

Substantial derivativeMaterial derivativeParticle derivative

Generalization: let A represent any property of fluid, ρ (either scalar or vector)

( , , , )A A x y z t=

local time dependentchange in property due to change in the propertymotion through flow field

DA A A A Au v wDt x y z t

∂ ∂ ∂ ∂= + + +

∂ ∂ ∂ ∂

Substantial derivative: generalizes the rate of change of a local property of a materialparticle to the movement of the particle through the flow field.In vector form;

( )

( )

.

gradient operator

.

DA A V ADt t

i j kx y z

V u v wx y z

∂= + ∇∂

∂ ∂ ∂∇ = + +

∂ ∂ ∂∂ ∂ ∂

∇ = + +∂ ∂ ∂

ωD u v wDt x y z t

ω ω ω ω ω∂ ∂ ∂ ∂= + + +

∂ ∂ ∂ ∂

D u v wDt x y z t

ρ ρ ρ ρ ρ∂ ∂ ∂ ∂= + + +

∂ ∂ ∂ ∂

Ex1:Vorticity change, (vector)

Density change, ρ (scalar)

Ex2:

U0

h y

x

( )

0

0 0 0 0 0unsteady

motion

1 ( , )

yV U ihax bt x t

D y yU a b U a bDt h h

ρ ρ ρ

ρ ρ ρ ρ

⎛ ⎞= ⎜ ⎟⎝ ⎠

= + − =

⎡ ⎤⎛ ⎞ ⎢ ⎥= − = −⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎢ ⎥⎣ ⎦

Motion & Deformation of a Fluid:In order to develop differential equations of motion for a fluid we need to understand the

general type of motionExamine for 2-D Four different types of motion or deformation

1. Translation2. Rotation3. Distortion (i.e. deformation)

• Angular deformation - shear strain• Linear deformation (extensional strain) – dilatation

Consider each from of motion individually for time interval ∆t1)Translation: defined by displacements u ∆t & v ∆t,

that is, rate of translation is (u,v)

x

y

∆x

∆y u

v

u∆t

v∆t(displacement)

2) Rotation:

x

y

∆x

∆y∆α

∆β

Derivation of rotation: vorticity

Motion about centroiddue to motion normalto x-dir. with pozitif rotation

( )va x tx

∂∆ = ∆ ∆

∂

Rotation: average angular rotation of 2 perpendicular line elements (x&y)

12Z

d ddt dtα β⎛ ⎞Ω = −⎜ ⎟

⎝ ⎠

rotation about axis parallel to z-direction. i.e. angular velocity rate of rotationRate of translation : velocity , u, vRate of rotation : angular velocity,

( )

0 0

0

: rotation/ of line x= lim = lim

/= lim

Likewise, of line y:

t t

t

a xt t

v x t xd vxdt t x

d udt y

α

α

β

∆ → ∆ →

∆ →

Ω∆ ∆ ∆

Ω∆ ∆

∂∆ ∆ ∆ ∂∂ =

∆ ∂∂

Ω =∂

12Z

v ux y

⎛ ⎞∂ ∂Ω = −⎜ ⎟∂ ∂⎝ ⎠

ZΩ

ΩSimilarly, rotation about x & y axes,

1 1 , 2 2x y

w v u wy z z x

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

12x y zi j k VΩ = Ω + Ω + Ω = ∇×

For conventions sake, we define vorticity of a fluid particle as,

2 , , V curlVω ω ω= Ω = ∇ × =

3) Angular deformation: (shear strain)average decrease of the angle between two lines which are initially perpendicular

( )1 : shear strain increment2

d dα β+

1 12 2xy

d d v udt dt x yα β ⎛ ⎞∂ ∂⎛ ⎞∈ = + = +⎜ ⎟⎜ ⎟ ∂ ∂⎝ ⎠ ⎝ ⎠

Shear strain rate

1 1 , 2 2yz zx

w v u wy z z x

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

xx xy xz

ij yx yy yz

zx zy zz

∈ ∈ ∈∈ = ∈ ∈ ∈

∈ ∈ ∈

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

Strain rate tensor

x

y

∆x/2

∆y/2

2v x tx

∂ ∆⎛ ⎞ ∆⎜ ⎟∂ ⎝ ⎠

2u y ty

∂ ∆⎛ ⎞ ∆⎜ ⎟∂ ⎝ ⎠

4. Linear deformation: (extensional strain)Which motion(s) will result in stresses? Angular & linear deformation, relative change in dimensions of line element ∆x, ∆y : STRAINResult: Angular def. ~ shear stress

Linear def. ~ normal stress

x

y

∆x

2u x tx

∂ ∆⎛ ⎞ ∆⎜ ⎟∂ ⎝ ⎠

2v y ty

∂ ∆⎛ ⎞ ∆⎜ ⎟∂ ⎝ ⎠

22

xx

xx

xx

u xx t xxt

xux

t

ε

ε

ε

∂ ∆⎛ ⎞∆ + ∆ − ∆⎜ ⎟∂ ⎝ ⎠∆ =∆

∂=

∂∆ : extensional strain in x-direction

(extentional strain) dilatation or increase in volume is due to velocity derivatives

u v wx y z

∂ ∂ ∂∂ ∂ ∂

, ,

xx yy zzu v wx y z

ε ε ε∂ ∂ ∂= = =

∂ ∂ ∂ , ,

. xx yy zzV ε ε ε∇ = + +

( ) ( )2

.2

DV V V VV V V VDt t t

⎛ ⎞∂ ∂= + ∇ = + ∇ − × ∇×⎜ ⎟∂ ∂ ⎝ ⎠

0Vω = ∇× =φ

If

For irrotational flow, let be a continuous scalar function such that

in the flow field irrotational flow

V i j kx y zφ φ φφ ∂ ∂ ∂

= ∇ = + +∂ ∂ ∂

( , , , )x y z tφ φ=

( ) 0ω φ= ∇× ∇ = identically zero.

φV φ= ∇

• For irrotational flow there exist a scalar velocity potential function such that

Vorticity generators : boundaries (velocity gradients)• Real flows are always rotational

But Boundary Layer Theory vorticity effects are confined in a thin layer adjacent tothe boundaryOutside the boundary layer flow can be treated as irrotational.

0ω = V φ= ∇ ( )2 2 2

2 2 2. . 0Vx y zφ φ φφ ∂ ∂ ∂

∇ = ∇ ∇ = + + =∂ ∂ ∂

If & ρ=const.

2 0φ∇ = Laplace eq. (linear PDE)

.V divV∇ = change of (velocity) field at a point; indicates linear expansion of a field(i.e. changes parallel direction of interest)

. xx yy zzu v wVx y z

ε ε ε∂ ∂ ∂∇ = + + = + +

∂ ∂ ∂

Ex: V axi=

x

y.V i a l a

x∂

∇ = × =∂

a:expansion at a point

xxux

ε ∂=

∂extensional strain (linear deformation)

• If ρ=const. → conservation of mass → . 0V∇ = (no expansion)

dx

dy

A

BC

A D

CB

dy

dxv

ut t+dt dxdt

xu

∂∂

dydtyv

∂∂

dyyvv

∂∂

+

dxxuu

∂∂

+

Rectengular element under the influence of normal stresses

Rate of increase in unit area : .

u v u vdxdtdy dydtdx dxdt dydtu vx y x y V

dxdydt x y

∂ ∂ ∂ ∂+ +

∂ ∂∂ ∂ ∂ ∂ = + = ∇∂ ∂

Fluid behavior is a combination of these fluid motions, so we need to be able to expresscumulative behavior mathematically.

V ui v j= +

dr dxi dy j= +

Cauchy-Store Decomposition& fluid element moving from point P through aConsider 2-D velocity field,

distance

y

xP

( ).

=

p p

p pp

V V dV V dr V

V V u vV dx dy i dx dy j dx dyx y y x

u vx

uy

v

= + = + ∇

⎡ ⎤ ⎡ ⎤∂ ∂ ∂ ∂+ + = + + + + +⎢ ⎥ ⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦

∂ ∂

⎦∂ ∂⎣

Consider,

add&subst.split in half

1 1 1 12 2 2 2

u u u v vdy dy dy dy dyy y y x x

∂ ∂ ∂ ∂ ∂= + + −

∂ ∂ ∂ ∂ ∂

pV dV+pV

d r

Rearrange,

1 12 2

u u v u vdy dy dyy y x y x

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

Similarly,

1 12 2

v v u v udx dx dxx x y x y

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

angurotationlinear def.trans.

trans.linear def. rotation angular def.

1 1 1 1=2 2 2 2p p

u u v u v v v u v uV i u dx dy dy j v dy dxx y x y x y x y x y

⎡ ⎤⎢ ⎥⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

+ + − + + + + + − + +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

lar def.

dx

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

or in terms of tensors,

vorticityrate of tensorstraintensor

1 102 2

=1 1 02 2

= . .

p

p ij

u v u v ux x y x y

V V dr dru v v u vy x y y x

V dr E dr

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

+ + Ω

Normally, write tensors (2-D)

accounts for distorsion accounts for rotation

, xx xy xx xyij ij

yx yy yx yyE

∈ ∈ Ω Ω⎡ ⎤ ⎡ ⎤= Ω =⎢ ⎥ ⎢ ⎥∈ ∈ Ω Ω⎣ ⎦ ⎣ ⎦

Ex: Given a shear flow, yV U ih

= , determine components of deformations & rotation

U

h y

x

0xxux

ε ∂= =

∂

1 1 02 2 2

0

12 2

xy

yy

yx

v u U Ux y h h

vy

u v Uy x h

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

∈ = =∂

⎛ ⎞∂ ∂∈ = + =⎜ ⎟∂ ∂⎝ ⎠

0 always

1 1 02 2 20

12 2

xx

xy

yy

yx

v u U Ux y h h

u v Uy x h

Ω =

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

⎛ ⎞∂ ∂Ω = − =⎜ ⎟∂ ∂⎝ ⎠

x

y

x

y

rotation

rate of strain

( )yxΩ +

( )xyΩ −

yx∈

xy∈

Relation Between Stresses & Rate of Strainsij ijσ ∈∼

Strain rates : symmetric second-order tensor

xx xy xz

ij yx yy yz

zx zy zz

∈ ∈ ∈∈ = ∈ ∈ ∈

∈ ∈ ∈

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

, , xx yy zzu v wx y z

∂ ∂ ∂∈ = ∈ = ∈ =

∂ ∂ ∂

ij ji∈ =∈

12xy yx

v ux y

⎛ ⎞∂ ∂∈ =∈ = +⎜ ⎟∂ ∂⎝ ⎠

12yz zy

w vy z

⎛ ⎞∂ ∂∈ =∈ = +⎜ ⎟∂ ∂⎝ ⎠

12zx xz

u wz x

∂ ∂⎛ ⎞∈ =∈ = +⎜ ⎟∂ ∂⎝ ⎠

Remember:Transport properties of fluid; µ, k , α• Viscosity: a property of fluid: ability of a fluid to flow freely, relates applied stress to the resulting strain rate U

h y

xno-slip condition

( ) yu y Uh

= linear profile

( )yx yxfτ = ∈ general relations will be considered later!

For simple fluids such as water, oil, or gases relationship is linearOr Newtonian 2yx yx

U duh dy

τ µ µ µ= = ∈ =

µ: (coef. of) viscosity [N s /m2 = Pa.s] , µ= µ(T,p) different for liquids & gases

( )yx yxfτ = ∈ is not linear, the fluid is non-newtonian

time rate of deformation, Є inviscid flow (ideal fluid)

du/dy0

τshear stress

ideal bingham plastic

Pseudoplastic – shear thinning µ↓ as τ↑ Polimer solutions , toothpaste

dilatant – shear tickening, suspensions, starch, sand

Newtonian , µ=constant

yield stress

Rigid body

2 nyx yxKτ ≅ ∈ Power-law approx.

n=1 Newtonian K=µn<1 pseudoplasticn>1 dilatant (kaboran)K, n : material parameters

Power-Law:

01

Tn nT

yx yx yxkeτ −⎛ ⎞= ∈ ∈⎜ ⎟

⎝ ⎠(rheological models of blood)

DIFFERENTIAL EQUATIONS OF MOTION

Review of control volume (CV) versus differential eqs. approacha) CV eqs. (mass, momentum, energy)

1. Global characteristics2. Must know or assume inflow & outflow profile3. Easy & somewhat approximate

b) Differential eqs. (mass, momentum, etc.)1. Detailed profile charact.2. Only boundary & initial conditions3. Easy to very difficult

exact behaviour (for laminar flow)

Objectivies:1. Derive mass D.E. (continuity)2. Derive momentum D.E. • General• Navier-stokes (stress ~ strain)3. Solutions

Conservation of Mass: CONTINUITY EQUATION

Diff. Element

x

z

y d dxdydz∀ =

dm dρ= ∀

∀

∀

CV formulation

( ). 0cv cs

d V n dAt

ρ ρ∂∀ + =

∂ ∫ ∫

time rate of change NET mass of the fluid mass inside flow rate 0the CV through the CS

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥+ =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

( ) ( ) ( ) ( )u v wdiv V

x y z

t

ρ ρ ρρ

ρ

∂ ∂ ∂= + +

∂ ∂ ∂∂

∂∀∂ ( )

change of field at a point

div V dρ⎡ ⎤+ ∀⎣ ⎦

( )a scalar equation

0

0 , 0

div V

d

tρ ρ∂

+ =∂

= ∀ ≠

( ). 0Vtρ ρ∂

+ ∇ =∂

valid for any coordinate system

scalar

.u v wdivV Vx y z

V ui v j wk

∂ ∂ ∂= + + = ∇

∂ ∂ ∂

= + +

m constρ= ∀ =∀

II. method: particle derivative. Conservation laws are Lagrangian in nature, i.e. apply fixed systems (particles). If m is the mass and is volume of fixed particle,

Conservation of mass

( ) ( ) ( ). .. (.)

DV

Dt t∂

= + ∇∂

( )

1

0 0

0 0

DDmDt Dt

D D DDt Dt Dt

DDt

ρ

ρ ρρ ρ ∀∀

∀= ⇒ =

∀+ ∀ = ⇒ + =

Can relate DDt

∀to the fluid velocity by noticing total dilatation or normal strain-rate is

equal to the rate of volume increase of the particle.

( )( ) ( ) ( )

1.

0 or

0

. 0

xx yy zz

D d

u v w DV divVx y z Dt

u v wt

ivV Vt

x y z

Dtρ ρ

ε ε ε

ρ ρρ

ρ

ρ

ρ

∂ ∂ ∂ ∀+ + = + + = ∇ = =

∂ ∂ ∂ ∀

∂ ∂ ∂∂+ + + =

∂ ∂

∂+ = + ∇

∂

∂

∂

=

Notes pg.58 : cylindrical & spherical coordinates

Simplifications:ρ=const. : flow is said to be incompressible

. 0divV V= ∇ = rectangular coor. 0u v wx y z

∂ ∂ ∂+ + =

∂ ∂ ∂particles of constant volume,

but shape of volume can change.

STREAM FUNCTION ψ2-D , steady flow : continuity

( ) ( ) 0u v

x yρ ρ∂ ∂

+ =∂ ∂

steady compressible or unsteady incompressible

Define stream ψ function such that:

, u vy xψ ψρ ρ∂ ∂

= = −∂ ∂

2 2

0x y y xψ ψ∂ ∂

− =∂ ∂ ∂ ∂

continuity identically satisfied.

ψ : first & second order der. exist & continuous

Advantage • Continuity eq. discarded• # of unknowns (dependent variables) reduces by one.

Disadvantage • Remaining velocity derivatives are increased by one order???

( , ) d = .x y

x y dx dy vdx udy V d s dmψ ψψ ψ ψ ρ ρ ρ∂ ∂= + = − + = =

∂ ∂( )0dm =

Physical significance of ψ

• Lines of constant ψ (dψ=0) are lines across which mass flow

They are stream lines of flow

is zero

d s dxi dy j= +Along AB x=const.

d s dyi=

0

.B B B B

A A A A

y y y

B Ay y y

V ui v j

m V d s udy dy dy

ψ

ψ

ψρ ρ ψ ψ ψ

= + =

∂= = = = = −

∂∫ ∫ ∫ ∫

ρ=const.

- vdx

udy

dq

A

C

ψ+dψ

ψ

x

y

2

1

2 1 0

dq udy vdx dy dx dy x

dq d

q dψ

ψ

ψ ψ ψ

ψ

ψ ψ ψ

∂ ∂= − = + =

∂ ∂=

= = − >∫

q: volume flow rate between streamlines ψ1 & ψ2

q

ψ1

ψ2

If ψ1 > ψ2 : q= ψ1 - ψ2 (flow is to the left)

m=ρq• Difference between the constant values of ρ defining two stream lines is the mass flow rate (per unit depth) between the two streamlines.

Note: 2-D , ρ = const., in cylindrical coor. (r θ plane) continuity eq.

( )0

1( , , ) ,

show that continuity eq. is satisfied!

r

r

rV Vr

r t V Vr r

θ

θ

θψ ψψ θθ

∂ ∂+ =

∂ ∂∂ ∂

→ = = −∂ ∂

Uses of Continuity Eq.1. to simplify D.E. of momentum2. To relate changes in velocity in one direction to changes in another

Ex: 2-D Flow, ρ=const. 0u v v ux y y x

∂ ∂ ∂ ∂+ = ⇒ = −

∂ ∂ ∂ ∂

x

y

0 0 v is away from centeru vx y

∂ ∂< ⇒ > ⇒

∂ ∂

Ex:

A(x)

u(x)A0

U0

x

yConverging channel

0 0

0 0

( )

( )

U A UA xU AUA x

=

=

0 00 0 2

( )

1 /( ) ( ) ( )

( )( )

U x

U AU dA dA dxU Ax A x dx A x A x

U U x dAx A x dx

⎛ ⎞ ⎛ ⎞∂= − = −⎜ ⎟ ⎜ ⎟∂ ⎝ ⎠ ⎝ ⎠

∂ ⎛ ⎞= −⎜ ⎟∂ ⎝ ⎠

( ) 0 0 0dA u vA xdx x y

∂ ∂⇒ < ⇒ > ⇒ <

∂ ∂flow toward center

( ) 0 0 0dA u vA xdx x y

∂ ∂⇒ > ⇒ < ⇒ >

∂ ∂flow away from center

DERIVATION of MOMENTUM D.E

Newton’s Second Law applied to fluid element

applied resulting accelerationforce of particle of mass, dm

d F dm a=

velocity field terms+pressure & body forces

relates stresses acceleration

fluid deformation substantial derivative

↔⇓

↔

CLASSIFICATION OF FORCES ON A FLUID1.Body Forces: All external forces developed without physical contact. e.g. gravity, magnetic force

x

y

z

BdF gd gdxdydzρ ρ= ∀ =dx

dy

dz

d∀g→

body forces are distributed throughout

2.Surface Forces: All forces exerted on a boundary by surroundings through directcontact.

a) Normal forces – e.g. pressure forcesb) Tangential forces – e.g. shear forces

stress = force / unit area

1.Normal stresses

2.Shear stresses

0lim nnn A

F pressureA

σ ∆ →

∆= ⇒

∆

0lim sss A

F shearA

τ ∆ →

∆= ⇒

∆

Surface stresses havea) direction & b) surface that acts on

two shear forces on any surface

yx shear stressτ =

z y

x

nF

syF

sxF

F

Example: forces on a plane

y xτdirection that stress acts (x in this case)plane acts on (y in this case)

positive outward normal on y-surface

y

z

x

yyτ

yxτyzτ

negative outward

xzτzxτ

zyτ

xyτ

xzτ

Stress tensor

xx xy xz

ij yx yy yz

zx zy zz

τ τ τ

τ τ τ τ

τ τ τ

⎡ ⎤⎢ ⎥

= ⎢ ⎥⎢ ⎥⎣ ⎦

stress acting on z-plane

symmetric tensor

ij jiτ τ=

Derivation of Momentum Differential Equation

Begin by applying differential analysis to differential fluid element, (dx, dy, dz)

x

y

z

zxτ

xxσ

yxyx dy

yτ

τ∂

+∂

xxxx dx

xσσ ∂

+∂

zxzx dz

zττ ∂

+∂yxτ

b sDVd F d F dmDt

+ =

Consider x-direction forces & changes across element using truncated Taylor series

x-dir.x xb s

DudF dF dxdydzDt

ρ+ = xa

yx

xxx xx xx

yx zxyx yx zx

zx

xx zxx

DuDt

B dxdydz dx dydz dydzx

dy

Bx y z

u u u u

dxdz dxdz dz dxdyy z

dxdy

dxdydz

dxdyd u wt x y z

z v

σρ σ σ

τ ττ τ τ

τσ

τ

τρ

ρ

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

−

∂

=

⎡ ⎤∂ ∂+ + +⎢ ⎥∂ ∂ ∂⎣ ⎦

⎡ ⎤∂ ∂ ∂ ∂+ + +⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦

Similarly y-direction and z-direction

xy yy zyy

yzxz zzz

DvBx y z Dt

DwBx y z Dt

τ σ τρ ρ

ττ σρ ρ

∂ ∂ ∂+ + + =

∂ ∂ ∂∂∂ ∂

+ + + =∂ ∂ ∂

In vector form

. ijDVBDt

ρ τ ρ+ ∇ = orij i

ij

DVBx Dtτ

ρ ρ∂

+ =∂

Note: Above eqs. are general eqs. apply for any fluid.

General simplification Newtonien fluid

→ linear relationship between stress & rate of strain ij ijτ ε∼

From solid mechanics, the relation of stress tensor to strain-rate tensor yieldsa linear relationship of form,

1 2 3 4 5 6yx xx xy xz yy yz zzc c c c c cτ ε ε ε ε ε ε= + + + + +where each stress component depends on all of the six rate of strain components,

• We make assumption of an ISOTROPIC medium(i.e. material property independent of direction)

This reduces the number constants to two, since many of the coeefficientsare identically zero or related to each other.

Theory of elasticity Hookean Solid E - mod. of elasticityh - Poisson ‘s ratio

Newtonien Fluid µ: coef. of viscosityλ: 2nd coef. of viscosity

or bulk viscosity(associated only with

volume expansion)

General deformation law for Newtonian fluid

exp

2 ( )xx xx xx yy zz

linear stress volume ansion

pσ µε λ ε ε ε= − + + + +

( )2 ( ) 2 .yy yy xx yy zzpressure stresses due to

linear compressibilityrate ofstrain

vp p Vy

σ µε λ ε ε ε µ λ∂= − + + + + = − + + ∇

∂

2 ( )zz zz xx yy zzpσ µε λ ε ε ε= − + + + +

2xy yx xyv ux y

τ τ µε µ⎛ ⎞∂ ∂

= = = +⎜ ⎟∂ ∂⎝ ⎠

2xz zx xzu wz x

τ τ µε µ ∂ ∂⎛ ⎞= = = +⎜ ⎟∂ ∂⎝ ⎠

2yz zy yzw vy z

τ τ µε µ⎛ ⎞∂ ∂

= = = +⎜ ⎟∂ ∂⎝ ⎠

( ).jiij ij ij ij ij

j i

uup V px x

τ δ µ δ λ δ τ⎛ ⎞∂∂ ′= − + + + ∇ = − +⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠

thermodynamicpressure

viscousstresses

Note the inclusion of pressure → because if velocity vanishesnormal stress = - pressure (hydrostatic)

Fluid at rest ,0 ,

,ij

i jp i j

τ≠⎧

= ⎨− =⎩0B pρ − ∇ =

Stoke ‘s hypothesis23

λ µ= − (1845)

For air & most gas mixtures

For liquids . 0V∇ =

. .

3 2 3xx yy zz

V V

u v w u v wpx y z x y z

σ σ σ µ λ

∇ ∇

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

0

3 (2 3 ) .p Vµ λ=

= − + + ∇

Define: mean (mechanical) pressure, p

( )13 xx yy zzp σ σ σ= − + +

Mean pressure in a deforming viscous fluid is not equal to the thermodynamicpressure but distinction is rarely important

2 .3

normal viscous stresses

p p Vλ µ⎛ ⎞= − + ∇⎜ ⎟⎝ ⎠

usually small in typical flow problemscontroversial subject

Stokes‘ hypothesis 2 03

λ µ+ = (1845)

For liquids ; ρ = const. . 0V∇ =

3xx yy zzp

σ σ σ+ += −

Now, back to D.E. → substitute for stresses from above relationship,

Consider x – dir.

22 .3x x

u v u u wB p V ax x y x y z z x

ρ µ µ µ µ ρ⎡ ⎤⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ⎡ ∂ ∂ ⎤⎡ ⎤ ⎛ ⎞+ − + − ∇ + + + + =⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎢ ⎥∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎣ ⎦ ⎝ ⎠⎣ ⎦⎝ ⎠⎣ ⎦

Let µ = const.

22 .3x x

p u v u u wB V ax x x y x y z z x

ρ µ µ µ ρ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞− + − ∇ + + + + =⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠

( )2 2 2 2 2

2 2 2

4 51 2 3

22 .3x x

p u v u u wB V ax x x y x y z z x

ρ µ µ µ µ µ µ ρ∂ ∂ ∂ ∂ ∂ ∂ ∂− + − ∇ + + + + =

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

( )

2

2 2 2

2 2 2

1 4 1 53 2

.

2 .3x x

half of half of

Vu

p u u u u v wB V ax x y z x x y z x

ρ µ µ µ ρ

∇∇

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

− + + + + + + − ∇ =⎢ ⎥ ⎢ ⎥∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

( )2 1 .3x x

pB u V ax x

ρ µ ρ∂ ∂− + ∇ + ∇ =

∂ ∂Likewise

( )2 1 .3y y

pB v V ay y

ρ µ ρ∂ ∂− + ∇ + ∇ =

∂ ∂

( )2 1 .3z z

pB w V az z

ρ µ ρ∂ ∂− + ∇ + ∇ =

∂ ∂

NAVIER STOKES EQUATION

2 1 ( . ) ( . )3body deformationpressure convective

force stressesforce localdue to accelerationaccelerationcompressibility

VB p V V V Vt

ρ µ µ ρ ρ∂− ∇ + ∇ + ∇ ∇ = + ∇

∂

for ρ = const. . 0V∇ =

2 DVB p VDt

ρ µ ρ− ∇ + ∇ = General equationcoordinate independent

Cartesian coord. x – dir.2 2 2

2 2 2xp u u u u u u uB u v wx x y z t x y z

ρ µ ρ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

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

( )xg if B g gravitional acceleration=1

1 Force acting on the fluid element as a result of viscous stress distributionon the surface of element

Note: Cylindrical coord. pg. 60

• viscosity is constant →• ρ = const. . 0V∇ =

isothermal flow. For non-isothermal flows, esp. for liquids,viscosity is often highly temp. dependent. CAUTION

CONSERVATION OF ENERGY: THE ENERGY EQ.1st law of Thermodinamics for a system

tdE dQ dW= +3( / )J m

work done on systemheat added

increase of energy of the system

21 .2t

gz

E e V g rρ+

⎛ ⎞⎜ ⎟= + −⎜ ⎟⎝ ⎠

3( / )J m

x

z

y

r g

Et : total energy of the system (per unit volume)e : internal enregy per unit mass

: displacement of particler

moving system such as flowing fluid particle

need Material derivative : time rate of change, following the particle

t DQ DWDt

DED Dtt

= + (J/m3.s) Energy eq. for a flowing fluid

t De DVV gVDt D

DEDt t

ρ ⎛ ⎞+ −⎜ ⎟⎠

=⎝

Q & W in terms of fluid properties

xq

xw

dx

xx

qq dxx

∂+

∂

xx

ww dxx

∂+

∂

• assume heat transfer Q to theelement is given by Fourier ’slaw

xTq kx

∂= −

∂heat flows from positive to theneg. temp. (decreasing temperaturegradient)

(HEAT FLOW) q k T= − ∇

x

y

z

Heat flow (rate) into the element in x - dir. :

Heat flow (rate) out of the element in x - dir. :

The net heat transfer to the element in x - dir. :

Hence, the net heat transfer to the element =

xq dydz

xx

qq dx dydzx

∂⎛ ⎞− +⎜ ⎟∂⎝ ⎠

W/m2

xq dxdydzx

∂−

∂

. .x x xq q q d q dx y z

⎛ ⎞∂ ∂ ∂− + + ∀ = −∇ ∀⎜ ⎟∂ ∂ ∂⎝ ⎠

( ). .DQ q k TDt

= −∇ = ∇ ∇ [W/m3] neglect internal heat generation

Rate of work done to the element per unit area on the left facenegatif because work is done on the system

( )x xx xy xzW u v wσ τ τ= − + +

surface direction

derivation: force on the left face: ( )xx xy xzi j k dydzσ τ τ− + +

Rate of work done on the element by this force

( ) ( )( )

.xx xy xz

xx xy xz

i j k ui v j wk dydz

u v w dydz

σ τ τ

σ τ τ

= − + + + +

= − + +

Similarly, rate of work done by the right face stresses is

xx

WW dxx

∂⎛ ⎞= − +⎜ ⎟∂⎝ ⎠

• Net rate of work done on the element

( )

( )

( )

xx xy xz

yx yy yz

zx zy zz

u v wx

u v wy

u v wz

σ τ τ

τ σ τ

τ τ σ

∂+ +

∂∂

+ + +∂∂

+ + +∂

.DW divW WDt

= − = −∇ =

( ). . ijDW VDt

τ= ∇

indicial notation

( ). . ijVDWDt

τ= ∇

Using the indicial notation,

expression can be decomposed into

( ) ( ). . .. iij

ji ij j

uVx

V τ τ τ∇=∇∂

+∂

Remember Newton ’s 2nd law

.. i jj iDV DVg gDt Dt

ρ ρ τ ρτ⎛ ⎞

= + ⎯⎯→∇ ⎟⎝

∇ = −⎜⎠

hence,

( )&

.

. . .

kinetic potential energyterms in energy e

j

q

iDV g VV VDt

τ ρ ⎛ ⎞= −⎜ ⎟⎝ ⎠

∇

. iij

j

uDW DVV g VDt Dt x

ρ τ ∂⎛ ⎞= − +⎜ ⎟ ∂⎝ ⎠

Exercise: Show this!

Energy eq. becomes;

( ). iij

j

De k T uxDt

ρ τ ∂= ∇

∂∇ +

First law of thermodynamicsfor fluid motion

split the stress tensor into pressure and viscous terms using

ij

jiij ij ij

j i

uup divVx x

τ

τ δ µ δ λ

′

⎛ ⎞∂∂= − + + +⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠

deformation law givenby Stokes (1845)

' ij i

ii j

jj

u pdixx

vu Vττ ∂= −

∂∂∂ Φ: dissipation function

From continuity eq. 0D divV

Dtρ ρ+ =

p D D p DppdivVDt Dt Dt

ρ ρρ ρ

⎛ ⎞= − = −⎜ ⎟

⎝ ⎠

ph eρ

= +introducing enthalpy

Energy eq. ( )Dh Dp div k TDt Dt

ρ = + ∇ + Φdissipation function(viscous dissipation)

For Newtonian fluid2 2 22 2 2

2

2 2 2u v w v u w v u wx y z x y y z z x

u v wx y z

µ

λ

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛ ⎞Φ = + + + + + + + +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

⎛ ⎞∂ ∂ ∂+ + +⎜ ⎟∂ ∂ ∂⎝ ⎠

0 , 0 ; 3 2 0µ λ µΦ ≥ ≥ + ≥ Hookes’ hypothesis 2 / 3λ µ= −

Incompressible flow, ρ = const.

( )pDT Dpc T k TDt Dt

ρ β= + ∇ ∇ + Φ

Constant Properties, k=const.

perfect gas relationp

v

dh c dT

de c dT

=

=

(1 )pdpdh c dT Tβρ

= − −

1/ ( )T perfect gasβ =1

PTρβ

ρ∂⎛ ⎞= − ⎜ ⎟∂⎝ ⎠

cp, cv : specific heats

2p

DT Dpc k TDt Dt

ρ = + ∇ + Φ ideal gas

2. . 0 pneglected

DTif const V c k TDt

ρ ρ= ∇ = → = ∇ + Φ

VALID for either gas (low velocity) or liquid.

Low velocity or incompressible flow Φ → 0

2DT TDt

α= ∇ 2/ : /pk c thermal diffusivity m sα ρ ⎡ ⎤= ⎣ ⎦

( )

2

. :V T convective terms

T T T Tu v w Tt x y z

α

∇

∂ ∂ ∂ ∂+ + + = ∇

∂ ∂ ∂ ∂

Example: Fully developed laminar flow down and inclined plane surface

Given: ρ, µwidth b=1 m

h=1mmθ=15º

Find: velocity profile shear stress distributionvolume flow rate (per unit depth)average flow velocityfilm thickness in terms of volume flow rate

θh=1mm

x

y

liquid

g

Basic eq.s for ρ=const.

4 3

0u v wx y z

∂ ∂ ∂+ + =

∂ ∂ ∂

2 2 2

2 2 25 3

1 4 4 4 3

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

ρ ρ µ⎛ ⎞⎛ ⎞

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

2 2 2

2 2 25

1 4 3 4 35

yv v v v p v v vu v w gt x y z y x y z

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

2 2 2

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

ρ ρ µ⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

+ + + = − + + +⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠1. steady flow (given)2. ρ=const. (incomp. flow)3. No flow or variation of properties in the z-dir. 4. Fully developed flow, so no properties vary in the x-dir. ,

0, 0wz

∂= =

∂0

x∂

=∂

Cont. 0 .v v const cy

∂= ⇒ = =

∂ 0 0 05.

v at y v everywhere= = ⇒ =

2

20

0

x

y

ugy

pgy

ρ µ

ρ

∂= +

∂∂

= −∂

Continuity u=u(y) only.

2 2

2 2

u d uy dy

∂→

∂ 2

2

2

1 2

1

sin

si

sin(

n

)2

xgd u gdydu g y C

yu y g C C

y

y

d

ρ θρµ µ

θ

θρµ

ρµ

= − = −

=

= +

−

+

+

−

B.C.s No slip

20 0 0

0

u at y Cdu at y hdy

= = → =

= = (zero shear stress on the liquid free surface)

1 1sin sin0 g h C C g hθ θρ ρ

µ µ= − + → =

( )

2

2sin

sin sin( )2

sin

( )2

yx

yu y g g hy or

du g h

yu y g

d

hy

yy

θρµ

θ θρ ρµ µ

τ µ ρ θ

⎛ ⎞= −⎜ ⎟

− +

= =

⎠

=

−

⎝

Volume flow rate

0

2 3

0

sin sin2 3

h

A

h

Q udA ubdy

g y g bhhy bdyρ θ ρ θµ µ

= =

⎛ ⎞= − =⎜ ⎟

⎝ ⎠

∫ ∫

∫

The average velocity / QV Q Abh

= =

2sin3

g hV ρ θµ

=

Solving film thickness

1/31/33

sin .Qh h Q

g bµ

ρ θ⎡ ⎤

= →⎢ ⎥⎣ ⎦

∼ non-linear relation

h=1 mm, b=1 m, θ=15º Q = 0.846 lt/sec

Summary of Basic Equations:

• Newtonien fluid• Stokes hypothesis• gravity is the only body force• Fourier ‘s law, no internal heat sources

2

2

.

. 0

1

1

p

const

V

DV g p VDtDT TDt c

ρ

υρ

αρ

=

∇ =

= − ∇ + ∇

= ∇ + Φ

( )

( )

( )

2

var

. 0

1 1 .3

iable

Vt

DV g p V VDt

Dh Dp div k TDt Dt

ρρ ρ

υ υρ

ρ

=∂

+ ∇ =∂

= − ∇ + ∇ + ∇ ∇

= + ∇ + Φ

perfect gasin general ( , )

( , )( , )

pdh c dT

h h p Tp Tp T

µ µρ ρ

=

===

unknowns u,v,p,(T)# of eq.s 4,(5)

exact solution is possible for “simple” problems

MATHEMATICAL CHARACTER OF THE BASIC EQS.difficulties• equations are coupled• “ “ non-linear

, & ( . . )V p T temp dep property

QUASI-LINEAR 2nd order PDE

2 2 2

2 2A B C Dx x y yφ φ φ∂ ∂ ∂

+ + =∂ ∂ ∂ ∂

where coef. A,B,C,D may be non-linear functions of , , , ,x yx yφ φφ ∂ ∂

∂ ∂but not of the second derivatives of φ.

2

discriminant

0 the eq. is elliptic B.V.Pif B -4AC 0 the eq. is parabolic diffusion type (mixed BVP & IVP)

0 the eq. is hyperbolic IVP

< →⎧⎪= →⎨⎪> →⎩

2

2 2

2 2

2 2

2 2

Laplace eq.: 0 EllipticT T THeat conduction parabolict

Wave equation 0 hyperbolic

x y

x y

φ

φ φ

∇ =

∂ ∂ ∂= +

∂ ∂ ∂

∂ ∂− =

∂ ∂

Navier-Stokes eqs. are too complicated to fit into this model. can be any or mixturesof all three depending upon specific flow and geometry.

Incompressible flow with zero convective derivativesLet us assume a Newtonian fluid with constant ρ, µ, k. further assume convective derivatives vanish.

( ) ( ). 0 ; . 0V V V T∇ = ∇ =realistic assumption for flows with gradients of flow properties are normal to the flowdirection, duct flows.

• Φ = 0 (viscous dissipation is negligible when the flow velocity is much smallerthan the speed of sound of the fluid)

2

2

.V=0 (1)

1 (2)tT (3)t

V g p V

T

υρ

α

∇

∂= − ∇ + ∇

∂∂

= ∇∂

(kinematic viscosity)µυρ

=

• Temperature effects are confined to the energy eq. continuity eq. & momentum eqs. are uncoupled from T.

• energy eq. is heat conduction eq. (parabolic)temp. mixed behaviour.i.e BVP in space (x, y, z). IVP in time

Taking divergence of (2) and making use of (1),

2 '.(2) 0 (2 ) Laplace eq. (Elliptic) pure B.V.P

p∇ ∇ = →

Finally, by taking the curl (∇x) of eq. (2), can eliminate p

( ) ( )2

fluid vorticity

1 ; t

V g p V V

curlV

υ ωρ

ω

⎛ ⎞∂∇× = ∇× − ∇× ∇ + ∇× ∇ = ∇×⎜ ⎟

∂⎝ ⎠

=0 0

2 (2'')tω υ ω∂

= ∇∂

eq. (2’’) also heat-conduction eq.∴vorticity, like temp. has parabolic behaviour

∴Both vorticity & temp. have diffusion coef. or diffusivities

2

2

:viscous diffusivity (m /s)

:thermal diffusivity (m /s) p

kc

µυρ

αρ

=

=

• They are entirely fluid properties, not geometric or flow parameters

viscous diffusion ratePrthermal diffusion rate

υα

= =

Liquid metals → low Pr (Pr<<1) ,e.g. PrHg = 0.024air → Pr = 0.72 (gases of the order of the unity)water → Pr = 7.0oils → high Pr. (Pr>>1) ,e.g. Prglycerin = 12 000

Low speed viscous flow (laminar flow) past a hot wall:

n

0 1.0

/T T∞/V V∞

Tδ

/T T∞

/V V∞

0 1

viscous diffusion rate >> thermal dif. ratethermal effects are confined near the wall

Liquid metals: Pr << 1 Oils : Pr >> 1

ν: shows the effect of viscosity of a fluid: momentum diffusion

, the region effected by viscosity is narrower known as when is very small.

(B.L. thickness) (for laminar flow) viscous sprea

Boundary Layer

d

as υυ

δ δ υ∼

T T

T

ing length

(thermal B.L. thickness) (for laminar flow) thermal spreading length

good approximation for all B.L. flows,

r

P

δ δ α

δδ

→

→

∼

∼

even at high speeds

gases: Pr (1) both effects are equally important liquids: Pr 1, may neglect the effect of heat conduction

. . . Pr .Re p

O

U L cPe

kρ

• ≈• >>

• = =

DIMENSIONLESS PARAMETERS IN VISCOUS FLOW

Basic flow eqs are extremely difficult to analyzeTherefore, we need to get most efficient possible form.

Buckingham Pi theorem:

0 0 0

dependent parameters , , ( , ,15 flow parameters)

9 fluid properties: , , , , , , , ,

4 reference quantities: , , ,1 wall heat flux, 1 acceleration of gravity,

i

p v

w

V p T f x t

k c c l

V p T Lq

g

ρ µ λ β σ

=

Primary dimensions, M (mass), L (length), t (time), T (temp.)15 – 4 = 11 dimensionless numbers → which governs the viscous flows

with heat transfer

p v

1. Re 7. Nu2. Pr 8. Kn3. Fr 9. Cavitation#4. Ec 10. We (Weber number)5. =c /c 11. viscosity ratio /

6. Gr

γ λ µ

impossible to consider all of themat one time

Few of them are important for a given problemdetermined by the non-dimensionalizing the basic eqs & B.C.s

1. From the eqs: Re, Pr, Ec, Gr, (λ/µ neglected)2. Wall heat transfer conditions: Nu3. Slip-flow conditions: Kn, γ4. Free-surface conditions: Fr, We, Cavitation #.

Non-dimensionalizing the Basic Equations:Reference properties appropriate to the flow

L

τw

, ,p TUρ∞ ∞ ∞

∞

x3

x2

x1

**

* *2

** *

V ; V

; T

g ; ; g

ii

w

xxL Up p T Tp

U T T

tUtL g

ρ

ρρρ

∞

∞ ∞

∞ ∞ ∞

∞

∞ ∞

= =

− −= =

−

= = =

Star denotes dimensionless variables.

( )

( )( )

* ** **

* ** **

. V 0t

1 V 0t

. V 0t

U UL L

ρ ρ

ρ ρ ρ ρ

ρ ρ

∞∞ ∞ ∞

∂+ ∇ =

∂∂

+ ∇ =∂

∂+ ∇ =

∂no dimensionless parameters: dimensional & dimensionless cont. eq. are the same

0 0 0

0

: 1. U / if there is no free stream (as in free convection) 2. L/V (steady flows have no characteristic time of their oResidence t )me ni wNote Lµ ρ=

Momentum Equation

( )2

.

1 . 3

const

DVg p V VDt

ρ ρ

ρ µ µ ρ

∞= =

− ∇ + ∇ + ∇ ∇ =

Continuity Equation

( )( )

( )( )

( )( )2

2

2

2

*2

*

2*

*

*

**

2 * 2 *

**

2 *2 2 *2

2 2 *

*

*2

* 2

2 2** * * * * *2

*2

V

t /

p = L

U=Lx

- p + V = L L L

p= pL

U=L

D U

D L U

g g g

g U p Up px xx L

U uu uV

UDV DVDt L Dt

U

V

x xL

U U U Dg g

V

ρ ρ ρ

ρ

ρ µ ρρ ρ ρ

ρ

∞

∞ ∞

∞

∞

∞

∞ ∞

∞ ∞ ∞ ∞

∞ ∞

∞ ∞ ∞ ∞ ∞∞ ∞

= =

=

∂∂ ∂∇ ⇒ →

∂ ∂∂

∂∂ ∂∇ → →

∂

∇

∇ ∇

∂

∇

∂

∇ ∇*

* VDt

2*

* ** * * * *2 *- p + =

Lg DVg VU LU Dtµρ ρ

ρ∞

∞ ∞∞

∇ ∇

2

1Fr

1Re

** 22

f. force force force

1 1Repressure

inertia gravity viscous

DV p g VDt Fr

ρ ρ= − ∇ + + ∇

Froude number (Fr):: determines the importance of buoyancy (important for freesurface flows) Dynamically similar flows: dimensionless parameters & dimensionless B.C.s should be identical.

⇒ kinematic similarity

Re U L

UFrg L

ρµ

∞ ∞

∞

∞

∞

=

=

g

θh

x

y

liquid

Ratio of Forces Dimensionless numbers

3 2 2dV dV ds forces ma dt ds dt

I V Lnertia L ρρ ρ→ → ∀ = ∼

V

2 2

2 2

2 2 22

3

forces .Adu dy

inertia forcesRe forces f. f.

ViscousVL LL

V L VLviscous VLinertia V L VFrgravity g

V

L gL

L

τ

µ µ

ρ ρµ µ

ρρ

µ

→

= =

= = =

= =∼

2 2 2

p2

2 2 22

3

2 2 2

forces .inertia forces

coef. c 12

inertia forcesgravity forcesinertia forces

surface tension f.

p

pressure p A pEuV L V

pc pressureV

V L VFrg L gL

V L V LWeL

ρ ρ

ρ

ρρρ ρ

σ σ

∆ ∆= = =

∆= =

= = =

= = =

dynamic pressure

flows with the free surfaceeffects

VMc

=flow speed

local sonicspeed

inertia forcesforces due to compressibility

M =

2 2 22

2

compressibilitymodulus [Pa]

V L VME L Eυ υ

ρ ρ= = Edp pa E

dυ

υδδρ ρ

= = = −∀∀2

22

VMa

=

For truly incompressible flow , a= M=0Eν→ = ∞ ∞ ⇒

• inertia force• viscous force• pressure force• gravity force• surface tension force• compressibility force

Compare compressibility modulus of water and air.

What is the pressure rise to reduce the volume of water by 4% ?

Example 1: Lid-driven cavity flow with periodic boundary condition

sinbcu V ttVtL

ω∞

∗ ∞

=

= streamlines

ubc

L

H Aspect ratio: H/L

Re V Lρµ

∞=

non-dimensional boundary condition

sin sinbcbc

u Lu t tV V

ωω∗ ∗

∞ ∞

⎛ ⎞= = = ⎜ ⎟

⎝ ⎠

duplication of the b.c. requires that the parameter ωL/V∞ be the same betweentwo flows.

L number, St=V

Strouhal ω

∞

• Need to begin with the correct equations

• Non-dimensional governing diff.eqs. are useful for numerical solution.- scaling is simplified- unit conversion problems are reduced- solutions can be presented in generalized form.

Example 2: Natural (free) convection: flow is due to the variation of density

( ) 11 =-P

TTρρ ρ ρ ρ β β

ρ∞ ∞

∂⎛ ⎞= + ∆ ≈ − ∆ ⎜ ⎟∂⎝ ⎠

( )3

2-speed Gr= , & PrwgL T TLow

βυ

∞ ∞

∞

−→

qw

Tw

Tkh

∞

∞

∞

n

( ) ww w

dTq h T T kdx∞ ∞ ∞= − = ∓

convective heat transfer coef.

( )w

w

q Lh LNuk k T T∞

∞ ∞ ∞

= =−

w

w

TNu kn

T TTT T

∗∗

∗

∗ ∞

∞

⎛ ⎞∂= − ⎜ ⎟∂⎝ ⎠

−=

−

Nu is the driving parameter which effects the solution

( ).p

DhDt

DT Dpc k TDt Dt

ρ = + ∇ ∇ + Φp

p

Maxwell's relation

dh=c (1 )

gas dh=c

dpdT T

perfect dT

βρ

+ −

→

substitute the new variables,

* * *, k , , , , Tpp

p w

c T Tkc Lc k T T

µ ρµ ρµ ρ

∞

∗ ∗ ∗ ∞

∞ ∞ ∞ ∞

−= = ∇ = ∇ = = =

−

Energy Equation

( )1 .Re Pr Rep

DT Dp Ecc Ec k TDt Dt

ρ∗ ∗

∗ ∗ ∗ ∗ ∗ ∗ ∗∗ ∗= + ∇ ∇ + Φ

( )2

p

inertia forcesRe ; forces

dif. ratePr dif. rate

VEckert number , Ec=c

p

w

V Lviscous

c viscousk thermal

T T

ρµ

µ υα

∞

∞

∞ ∞

∞

∞ ∞

∞ ∞

∞

∞

=

= = =

−

( ) all three are important for heat transfer analyses. V < 30% of s

High-speed flowsLow-speed flows

incompressible fl pressure term & dissipation term is peed of s

neglectedoound

RePr Pecl

wet n

sor∞

→

→

→→

2

umber DT 1constant properties

Re.Pr

, , , ( , )p

TDt

k c f p T

α

ρ µ

∗∗ ∗ ∗

∗

∗ ∗ ∗ ∗ ∗ ∗

→ = ∇

=

a RTγ ∞=

VORTICITY CONSIDERATIONS IN INCOMPRESSIBLE VISCOUS FLOW: VORTICITY TRANSPORT EQ.

vector, V corticity V u lVrω = ∇× =-is a measure of rotational effects

2ω = Ωlocal angular velocity of a fluid element

eqs. constant , Let gNavier Stokes gkρ µ− → = −

x

y

z

ij

k g2 (1)DV p gk V

Dtρ ρ µ= −∇ − + ∇

Use the following vector identities;

( ) ( ) ( )

( ) ( ) ( )

2

2

. 22

. 3

VV V V V

V V Vω

ω

⎛ ⎞∇ = ∇ − × ∇×⎜ ⎟

⎝ ⎠

∇ = ∇ ∇ − ∇× ∇×

2

2

that .V 0 for an incomp. flow Substitute (2) &(3) into (1)

(4)2

- (5) 2

Note

V V V p gkt

V Vp gz Vt

ρ ω ρ µ ω

ρ ρ ρ ρ ω µ ω

∇ =

⎡ ⎤⎛ ⎞∂+ ∇ − × = −∇ − + ∇×⎢ ⎥⎜ ⎟∂ ⎝ ⎠⎣ ⎦

⎛ ⎞∂+ ∇ + + = × ∇×⎜ ⎟∂ ⎝ ⎠

( )

2

2

2

0, irrotational flow

02

02

tan2

Vi

Vp gz

Vp g

f

z cons t

t

Vp gt

t

z

φρ ρ ρ

φρ ρ ρ

φ

ω φ

ρ ρ ρ

∂ ∇ ⎛ ⎞+ ∇ + + =⎜ ⎟∂ ⎝ ⎠

⎛ ⎞∂∇ + + + =⎜ ⎟∂

∂

= =

+ + + =

∇

∂

⎝ ⎠

Bernoulli eq. for unsteady incompressible flow

• Bernoulli eq. is valid even forviscous fluids if flow is irrotational• difficulty- potential flows do not satisfyno-slip condition at a solid wall.

VORTICITY TRANSPORT EQUATIONWhen dealing with a real fluid, we need an equation or eqs. to determine thebehaviour of vorticity.

vorticity ⇒ rotational behaviour of fluid ~ angular momentum

2 velocity = V rotation of fluidangular localω = × ∇× →

vorticity → creation, transport, destruction, stretching etc.

To obtain an eqn. for vorticity transport, we take the curl of the N-S eqn.

( )2 2

2

2

1

since =0 where scalar & 0

,

= V

g h

V

DV p B VDt

p hThus

DVDt

ω

υρ

φ φ

υ

− ∆

∇ =∇ ∇×

⎛ ⎞∇× = − ∇ + + ∇⎜ ⎟

⎝ ⎠∇×∇ ⇒

∴ ∇×∇ ∇×∇ =

∇× ∇×∇

( )

( ) ( )

( ) ( ) ( )

2

2

0

= .

1 =2

1 = +2

= = V . V+ .

DV V V VDt t

V V V Vt

VV V

V

Vt

t t

ω

ω ωω ω ω

⎡ ⎤∂∇× ∇× + ∇⎢ ⎥∂⎣ ⎦

⎡ ⎤∂∇× + ∇ − × ∇×⎢ ⎥∂⎣ ⎦

∂ ∇×∇×∇ − ∇× × ∇×

∂

∇× × ∇∂

∂−− ∇

∂∂

Thus, vorticity transport equations becomes

( ) ( ) 2

viscconvection of vorticitytime rate localvorticity by productionof change changevelocity field termof vorticity

-by stretchingand tilting ofexistingvorticity

. .D V VDt t

ω ω ω ω υ ω∂= + ∇ = ∇ + ∇

∂ ousdiffusion ofvorticity

z

2 case . 0 flow in xy plane

. 0

D V V

V Vk i jx y

ω ω

ω

− ∇ ⇒ ⊥ ∇

⎛ ⎞∂ ∂+ =⎜ ⎟

∂ ∂⎝ ⎠

• In all viscous flows vorticity is generally present and is generated by relativemotion near solid walls.

• If Re is large, vorticity is swept downstream and remains close to the wall:

0ω =B.L.

• If flow is between two walls, e.g. duct flow

potential-flow model is generally notvalid in duct flow.0ω ≠core

TWO – DIMENSIONAL CONSIDERATIONS :THE STREAM FUNCTION , ψ

2 2

2 2

2 2

2 2

2 , . =const. V=u(x,y,t)i +v(x,y,t) ju 0x

1

1

x

y

D constvy

u u u p u uu v gt x y x x y

v v v p v vu v gt x y y x y

ρ µ

υρ

υρ

− =∂ ∂

+ =∂ ∂

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

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

3 eqs. u, v, p u=u(x,y,t) , v=v(x,y,t) , p=p(x,y,t)

2p k=const. c

momentum equation is uncoupled from the energy eq.

DTLet k TDt

ρ = ∇

Eliminate pressure & gravity by cross-differentiation, i.e.taking the curl of the 2-D vector momentum equation

( ) ( ) ( )

( )

2

0 0

2

2

1.

.

V V g p Vt

Vt

DDt

ω υρ

ω ω υ ω

ω υ ω

⎛ ⎞∂∇× + ∇× ∇ = ∇× − ∇× ∇ + ∇× ∇⎜ ⎟

∂⎝ ⎠

∂+ ∇ = ∇

∂

= ∇

VORTICITYTRANSPORT EQ.2-D

2 2

2 2

x , 0

z z z z z

z z y

u vt x y x y

v u kx y

ω ω ω ω ωυ

ω ω ω ω ω

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

= − = = =⎜ ⎟∂ ∂⎝ ⎠

2zz

DDtω υ ω= ∇

substantive variation of vorticity

rate of dissipation ofvorticity through friction

2DT TDt

α≡ = ∇

z

2

2 , incomp. flow, =const.

0

DDDtu vx y

µ ω ωω υ ω

− =

= ∇

∂ ∂+ =

∂ ∂

2 eqs.2 unknownsu, v

GOAL: Reduce the governing eq. (for incomp., 2-D, µ=const.)to just one!

2 22

2 2

u= ; v=y

the stream function (x,y,t)

x

=

Introduce

v ux y x y

ψ

ψω

ψψ

ψ ψ

∂ ∂−

∂ ∂ ∂ ∂− = − − = −∇

∂ ∂ ∂ ∂

∂ ∂(Note that ψ satisfies the continuity eq.)

2

2

, formulationDDtω υ ω

ω ψψ ω

⎫= ∇ ⎪⎬⎪∇ = − ⎭

( ) ( ) ( )2 2 2 4

term accel. term

. . (A)y viscous

local convective

t x x yψ ψψ ψ ψ υ ψ∂ ∂ ∂ ∂ ∂

∇ + ∇ − ∇ = ∇∂ ∂ ∂ ∂ ∂

1-eq. 1 unknown (ψ)4th order PDE (non-linear) → very complexvalid for 2-D, incompressible flow, µ=const.• B.C.s would be in terms of the first derivatives of ψ

Example:

U∞

infinity u=U , v=0 0 , x

the solid surface, 0x

At Uy

Aty

ψ ψ

ψ ψ

∞ ∞

∂ ∂→ = =

∂ ∂∂ ∂

= =∂ ∂

- only few simple analytic solutions are known.- numerical solutions are required

Non-dimensionalization

0/ ReV Lψ ψ∗ = →

( ) ( ) ( )2 2 2 4 *1. .y Ret x x yψ ψψ ψ ψ ψ

∗ ∗∗ ∗ ∗ ∗ ∗ ∗

∗ ∗ ∗ ∗ ∗

∂ ∂ ∂ ∂ ∂∇ + ∇ − ∇ = ∇

∂ ∂ ∂ ∂ ∂

Study of Viscous Flows :

1) Exact solutions of N-S.a) Analytical solutions (limited success

because of the non-linearity of the eqs.)b) Numerical solutions

2) Very slow motions of viscous flow → CREEPING FLOW : Re<<1

Re 0→

2 forces VRe forces<<viscous forces forces V

inertia Inertiaviscous

= ⇒∼∼

23 3 2 2

dVVdS

dV VF ma L L V Ldt L

ρ ρ ρ= ∼ ∼ ∼

2. du VA A L VLdy L

τ µ µ µ∼ ∼ ∼

4

0 Fourth order eq.

biharmonic eq. in tw

linear

o dimensions

ψ∗ ∗∇ = → −

- governing relation in the theory of 2-D elasticity- large number of practical solutions exists from solid mechanics.

See Timoshenko & Goodier (1970)- can be applied to creeping viscous flows

2-D, µ=const. , ρ=const.

3) Limit as Re→∞Boundary layer theory

inertia forces>>viscous forcesbut cannot neglect viscous termscompletely but order of eqs. reduces

∴ cannot satisfy all B.C.s simultaneously.

Boundary-Layer Flow: Re>>1

( ) 21.Re

V V V p Vt

∂+ ∇ = −∇ + ∇

∂2

2

1The term is never negligible near a solid boundary becauseRe

the no-slip condition forces to be very large, of order Re, nearthe wall B-L.

V

V

∇

∇⇒

Singular-perturbation problem Van Dyke (1964) “Perturbation Methods in F.M.”

1/Re contains the highest-order derivative in the system,i.e. solution changes mathematical character as 1/Re → ∞

Example: Flow of a uniform stream parallel to an infinite flat plate with uniform suction:

Re1 ; ; 0wv yw wu e v v v= − = <

Exact solution for arbitrary Re.

Behaviour of solution as Re → ∞

i) Re finite (no matter how large) u → 0 as y → 0ii) Re → ∞ ⇒ u=1 everywhere

⇒ frictionless solutioni.e. frictionless flow with very small viscosity is not a potential flow since

a layer of finite thickness always exists where viscous effects are important.

x

y

y=0vw

u

1small Re

medium Re

Large ReRe → ∞

Low Reynolds Number : Creeping flow

Exact solutions N-S. → valid for arbitrary Re, at least until instability sets in andturbulence ensues.Lack generality & limited

e.g. Viscous flow past an immersed body at arbitrary Re- Direct experiment- Numerical solution- B-L approach- Creeping flow approx.

Limiting case of very large viscosity: Re<<1

Basic Assumption : Re<<1, i.e inertia forces<<viscous f.pressure cannot scale with the ‘dynamic’ or inertia term ρU2 but rather must dependupon a ‘viscous’ scale µU/LNon-dimensionalize the N-S eqs. with variables

2

, V , t , p/

VRe V

p px V tUxL U L U L

D pDt

µ∗∗ ∗ ∗ ∞

∗∗∗ ∗ ∗

∗

−= = = =

⇒ = −∇ + ∇

Neglect inertia if Re<<1i.e. Creeping Flow (or Stokes flow) assumption

( )Note: inertia V. is also negligible if there is no convective acceleration

e.g., fully developed duct flow (no restriction on Re).

V• ∇

Full N-S. for ρ=const. , µ=const. (steady flow)

( )

( )

2

0

2

.

. 0 (inertial force)

p=

. 0

V V V p Vt

V V

V

V

ρ µ

µ

⎡ ⎤∂⎢ ⎥+ ∇ = −∇ + ∇⎢ ⎥∂

⎢ ⎥⎣ ⎦

∇ →

∇ ∇

∇ =

Momentum eq. inertia forcesRe 1 forcesviscous

= <<

in extended form,2 2 2

2 2 2

2 2 2

2 2 2

2 2 2

2 2 2

0

p u u ux x y z

p v v vy x y z

p w w wz x y zu v wx y z

µ

µ

µ

⎛ ⎞∂ ∂ ∂ ∂= + +⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

⎛ ⎞∂ ∂ ∂ ∂= + +⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

⎛ ⎞∂ ∂ ∂ ∂= + +⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

∂ ∂ ∂+ + =

∂ ∂ ∂

GOVERNING EQS forCREEPING FLOWB.C.s are the same as N-S eqs.

Characteristics of CREEPING flowsa) Solutions are independent of densityb) Take div (∇.) of the momentum eq.

2 2

2

2

. 0

grad p = =const.

=

=

V

div p div V

div V

divV

µ µ

µ

µ

∇ =

⎡ ⎤∇ = ∇⎣ ⎦⎡ ⎤∇⎣ ⎦⎡ ⎤∇ ⎣ ⎦

2 2 2

2 2 2

: operator

=

Laplace

x y z

∇

∂ ∂ ∂∇ + +

∂ ∂ ∂

Pressure satisfies the Laplace (potential) eq. & p(x,y,z) is a potential (harmonic)function.

c) Take the curl of the momentum2(3) , (vor0 ticity)Vωω = ∇×∇ =

ABOVE valid for any 3-D problem

2 0p∇ =

FOR 2-D CREEPING FLOWchoice #1

work with primitive variables u, v, p

choice #2 work in terms of ψ (stream function)

has only one non-zero componentω

2

4

22 24

2 2

(4) Note: u= ; v=y

(3) & (4) 0

v ux y x

x y

ψ ψω ψ

ψ

ψ ψ

∂ ∂ ∂ ∂= − = −∇ −

∂ ∂ ∂ ∂

∇ =

⎛ ⎞∂ ∂∇ = +⎜ ⎟∂ ∂⎝ ⎠

2

2 24

2 2 2

's operator in cylindrical coord.

cylindrical coord. (r, )

1 1

Laplace

In

r r r r

θ

ψ ψθ

⎛ ⎞⎜ ⎟∂ ∂ ∂

∇ = + +⎜ ⎟∂ ∂ ∂⎜ ⎟⎝ ⎠

See plane elasticity problems

Examples: low-velocity, small-scale, highly viscous flows.a) Flow past immersed bodies:b) Lubrication theory:c) Biofluid dynamics:d) Hele-Shaw flows:e) Fully dev. duct flow:f) Small particle dynamics

Linear P.D.E’s → many solutions in closed form.Happel & Brenner, Low Reynolds number hydrodynamics (1965)

Example: Stokes’ solution for an immersed sphere:

no exact solutions-numerical sol.-B.L. viscous/inviscid patching schemes-creeping flow approx.

aU

r

θ

uθ ru motion of colloidal particles underinfluence of an electric field theoryof sedimentationstudy of movement of aerosol particles.

creeping motion of a stream of speed U about a solid sphere of radius a

2

22 2

2 2 2 2

( , ) Stokes stream function relation1 1 , u (1)sin sin

note: continuity eq. identically satisfied.

1 cotMom. eq. 0 (2)

BC 's

r=a

r

r

ur r r

r r r

At

θ

θψ ψ

θ θ θ

θ ψθ θ

ψ

∂ ∂= = −

∂ ∂

⎛ ⎞∂ ∂ ∂+ − =⎜ ⎟∂ ∂ ∂⎝ ⎠

∂

2 2

0 (u 0)

1 r sin . :2

(r, )=f (4r)g( ) )(

rur

As Ur const

θψ

θ

ψ

ψ

θ

θ θ

∂= = = =

∂ ∂

→ ∞ → +cos

sinru U

u Uθ

θθ

→→ −

(3)

Substitute (4) into (2) and satisfy (3). Find the solution of Stokes (1851) forfor a creeping motion past a sphere

22 2

2

1 3 2sin4

a r rUar a a

ψ θ⎛ ⎞

= − +⎜ ⎟⎝ ⎠

3

3

3

3

3cos 12 2

3sin 14 4

ra au Ur r

a au Ur rθ

θ

θ

⎛ ⎞= + −⎜ ⎟

⎝ ⎠⎛ ⎞

= − + −⎜ ⎟⎝ ⎠

Properties1) Streamlines & velocities are entirely independent of fluid viscosity. True for all

all creeping flows.2) The streamlines posses perfect fore-and-aft. symmetry.

No wake predicted.∴convective accel. terms (neglected here) are responsible for strong flowasymmetry typical of higher Re number flows.

3) The local velocity is everywhere retarded from its freestream value. There is no faster region such as occurs in potential flow. (uθ = 1,5U at sphere shoulder)

4) The effect of the sphere extends to enermous distances: At r=10a, the velocities are still about 10% below their free stream values.

With ur & uθ known, pressure can be found2p Vµ∇ = ∇

2

3Result p=p cos p : uniform freestream pressure2

aUr

µ θ∞ ∞−

• Pressure deviation is proportional to µ & antisymmetric•

Dp

aθ

3 cos2s

Up pa

µ θ∞− = −

Dp : pressure drag

Drag forcepressure (form)

surface shear stress (friction drag)

Shear stress distribution in fluid3

3

1 sin 32

rr

u uu U ar r r r r

θ θθ

µ θτ µθ

⎛ ⎞∂∂⎛ ⎞= + − = − ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

Total drag is found by integrating pressure & shear around the surface

0 02

sin cos

2 sin

formula, Stokes (1851)2/3 viscous force + 1/3 pressur

4 2 6

e forceF=6 Ua

r r a r aF dA p dA

dA a d

Sphere dF Ua Ua Ua

rag

π π

θτ θ θ

π θ θπµ πµ πµ

πµ

= =

= + = ←

←

= − −

=

−

∫ ∫

Strictly valid only for Re<<1 but agrees with experiment up to about Re=1.(predicts 10% low at Re=1)

Drag Coef. F/ Ua=6 =const.µ π

But customary definition

22

2 A: cross-sectional area = aDFC

U Aπ

ρ=

2 2

24R

2 6 12

2Re 2a=D

eDx UaCU a Ua

a U

πµ µρ π ρ

ρµ

= = =

=

Notes:• Formula introduces a Re effect where none exists• underpredicts the actual drag when Re>1 due to a symmetrical wake forms

For Re>20- Flow seperates from the rear surface, causing markedly increased pressure

drag

Comparison with inviscid solution

32 2

3

1 sin 12

aUrr

ψ θ⎛ ⎞

= −⎜ ⎟⎝ ⎠

See handout

Streamlines are close to the body!

Recirculation is absentsphere drags the entire surrounding fluid with it circulation streamlines

sphere pushes fluid out of the way

Other three-dimensional body shapes

Happel & Brenner, Low Reynolds number hydrodynamics (1965)

Disk normal to the freestream:

F=(32/3) UaµDisk parallel to the freestream:

F=16 Uaµ

See White (1991) Viscous Fluid Flow

EXACT SOLUTIONS OF NAVIER-STOKES EQS

A) Parallel Flows: i.e. velocity is unidirectionalB) Similarity solutionsC) Generalized Beltrami Flows-• difficulties

- non-linear PDEs- no general solution is known

0 , v=w=0u ≠

0

0 3D Flow

V

V

ω

ω=

⎛ ⎞∇× × =⎜ ⎟

⎝ ⎠

Reference C.Y. Wang (1991) “Exact solutions of the steady-state N-S eqs.”Ann. Rev. of Fluid Mech. Vol.23, pp. 159-177.R.Berker (1963)

Almost all of the particular solutions are for the case of incompressible Newtonienflow with constant transport properties,

2

2

. 0

p

V

DV p VDt

DTc k TDt

ρ µ

ρ

∇ =

= −∇ + ∇

= ∇ + Φ

P: total hydrostatic pressure. i.e. it includes the gravity term for convenience.

or P=p+ gzP p gρ ρ∇ = ∇ −

y

x

z

gg gk= −( )

( )1) Convective accel. V. .

2) Non-linear solutions V. does not vanish.

vanishes∇ →

∇ →

GROUP A: PARALLEL FLOWS0 , v=w=0u ≠

parallel flow: only one velocity component is different from zero.i.e. all fluid particles moving in one direction.

00

0 0

u=u(y,z,t) ; v 0 , w 0

u v w ux y z x

∂ ∂ ∂ ∂+ + = ⇒ =

∂ ∂ ∂ ∂

⇒ ≡ ≡y-comp. of momentum eq.

2 2 2

2 2 2

v v v v p v v vu v wt x y z y x y z

ρ µ⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

+ + + = − + + +⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠0p

y∂

=∂

z- comp. of mom. eq. ( )0 p=p x,tpz

∂= ⇒

∂

x- comp. of mom. eq.2 2 2

2

2 2

2

2 2

2 (A)

u u u u p u u uu v wt x y z x

u p u ut x

x

y z

y zρ µ

ρ µ

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

⎛ ⎞∂ ∂ ∂ ∂= − + +⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

⎝ ⎠ ⎝ ⎠

Linear dif. eq. for u(y,z,t)( ).u 0 0 (In general)

t tsteady

∂∂→ = =

∂ ∂

1) Parallel flow through a straight channelCouette Flows

a) both plates stationary

x

y

2

2

dp d udx dy

µ=BC's y=0 u=0 y=a u=0

→→

2 11 2

2 12 1

22

dp0 .dx

dp 1 dp dx 2 dx

1 dp 1 dp0, 0= c2

dp2 dx

dx 2 dx

velocity profile

p consty

cu c u y y cy

cc a a

a y yua a

a

µµ

µ

µ

µ

µ

∂= ⇒ =

∂

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

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢⎣

←⎥⎦

⎠

Shear stress distribution

1 12 2yx

dp dp dp yy a adx dx dx a

τ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎡ ⎤= − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦

Volume flow rate3

0

2

. ( )12

1/12

a

A

l dpQ V d A u ldy adx

dpV Q A adx

µ

µ

⎛ ⎞= = = − ⎜ ⎟⎝ ⎠

⎛ ⎞= = − ⎜ ⎟⎝ ⎠

∫ ∫

average vel.

2max

duPoint of maximum vel. 0 & solve for ydy

y=a/2 du/dy=01 3y=a/2 u=u

8 2dp a Vdxµ

=

⇒

⎛ ⎞→ = − =⎜ ⎟⎝ ⎠

b) upper plate moving with constant speed, U:

a

U

22 1

22

1 u=2

cdp u dp y y cdx y dx

µµ µ

∂ ⎛ ⎞= → + +⎜ ⎟∂ ⎝ ⎠

2

1

0 @ y=0 c 01 @ y=a c2

uU dpu U aa dxµ

= → =

⎛ ⎞= → = − ⎜ ⎟⎝ ⎠

0 at y=0dudy

⎛ ⎞=⎜ ⎟

⎝ ⎠

( ), ,dp f a Udx

µ=

After rearrangement,

22

( )2

Uy a dp y yu ya dx a aµ

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

vel. distribution depends on

both & Udpdx

y pressure gradient 0 u=Ua

dpzerodx

→ = →

simple shear (Couette) flow

2

21 0dp Updx a

µ⎛ ⎞= − = >⎜ ⎟⎝ ⎠

1.0

0 1 2 3

y/a

u/U

0dpdx

=

0dpdx

<

0dpdx

>

U

x

ya

0dpdx

<

0dpdx

>

2

2a dpP

U dxµ⎛ ⎞= −⎜ ⎟⎝ ⎠

dimensionless pressure gradient

-Results are valid for laminar flow only

-Experiments show

-Theory of lubrication

C0 Re 1500dp Uafordx

ρµ

⎛ ⎞= = ≈⎜ ⎟⎝ ⎠

journal bearing

Temperature considerations.constµ = flow eqs are uncoupled from the temp.

Energy eq.

2

2 2

2 2

p

p

DTc k TDt

T T T T Tc u v kt x y x y

ρ

ρ

= ∇ + Φ

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

T=T1

2h

T=T0

x

x

y

y'

U0dP

dx=

2yu Uh

=

Coordinate transformation y=y'+h→

[ ]

( ) ( )22

2

' ' 12 2

0=k

U U yu y hh h

u y T y

d T dudy dy

µ

⎛ ⎞= + = + ←⎜ ⎟⎝ ⎠

→

⎛ ⎞+ ⎜ ⎟

⎝ ⎠

T=T1

T=T0

2h xy

( ) 12

/ 2

U yu yh

du U hdy

⎛ ⎞= +⎜ ⎟⎝ ⎠

=

( ) ( )2 2

1 0 1 02

linear temp. dist. temp. rise due topure conduction viscous dissipation in

the fluid

2 2

3 42

0 1

4 2 &

2

2 8

T

1

h

U yT c y c

T T T T y U yTh k

khT h T

h

T

µ

µ= − + +

−

⎛ ⎞+ −⎡ ⎤= + + −⎜ ⎟⎢ ⎥⎣ ⎦ ⎝

=

⎠

=

Non-dimensionalizing T, by (T1-T0)dimensionless dissipation parameter, Brinkman number, Br

( ) ( )2 2

1 0 1 0

Pr .

.

p

p

cU UBr Eck T T k c T T

µµ= = =

− −

→Br represents the ratio dissipation effects to fluid conduction effects

-1

0

+1

T0

T1

Br=0

Br=16

Br=8

Temperature profile

For low-speed flows, only the most viscous fluids (oils) have significant Br.

Example: U=10 m/s T1-T0=10 °C

Fluid µ (kg/m-s) k (W/m-°C) Br

Air 1.8x10-5 0.26 0.0007

Water 1.0x10-3 0.6 0.017

Mercury 1.54x10-3 8.7 0.0018

SAE 30 oil 0.29 0.145 20.0

Except for heavy oils, we neglect dissipation effects in low-speed temperatureanalyses.

Rate of heat transfer at the wall

( )2

1 02 4wh

T k Uq k T Ty h h

µ

±

∂= = − ±

∂

lower surface

upper surface

0: Wkm C

⎡ ⎤⎢ ⎥⎣ ⎦

Convective heat transfer coef. ζ W/m2°C

( )1 0

Re Pr

w

L h L

q T T

LNu Ck

ζ

ζ

= −

= = L: characteristic length of theflow geometry, L=2h

2

convection heat transfer1 pure conduction Nuconduction heat tr

2 12

ansfer

hh BrNu

N

k

u

ζ=

= ≡

= ±

FLOW BETWEEN ROTATING CONCENTRIC CYLINDERS

2ω 1ω

r1

r2

steady rotational speed (angular vel.)parallel flow: 0, 0

, T & p must be functions only of r.r zv v v

vθ

θ

= = ≠

Equations of motion of incompressible Newtonian Fluids in cylindrical and spherical coordinates: See appendix B, Viscous Fluid Flow, F. White

Continuity equation:

( ) ( ) ( ) ( )1 1 0 0 6r zVrv v v

r r r zθ

θθ θ∂∂ ∂ ∂

+ + = =∂ ∂ ∂ ∂

( )

steady flow, 0

axisymmetric flow 0

cylinder infinitely long 0

unidirectional motion v

t

zv rθ θ

θ

∂=

∂∂

=∂

∂=

∂=

Momentum eq. in r - direction

( )2

1vdpdr r

θρ=

balance between centrifugal force & forcewhich is produced by the induced pressurefield.

0 p as rdpdr

>p

streamline

( )2

2 0 2d v vddirectiondr dr r

θ θθ ⎛ ⎞− → + =⎜ ⎟⎝ ⎠

( )2

viscous dissipation term

k equation 0= 3r

dv vd dTEnergy rdr dr dr r

θ θµ ⎛ ⎞⎛ ⎞ + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

BC’s

1 1 1 1 1

2 2 2 2

r=r v & T=T p=p r=r v & T=T

At rAt r

θ

θ

ωω

=

=

Integrate eq. (2)

( ) ( )

1

2

1 1 1 2

2

0

1 2

/

dv v dv vd cdr dr r dr r

d d rrv c rv c r rv c cr dr dr

v cr c r

θ θ θ θ

θ θ θ

θ

⎡ ⎤+ = ⇒ + =⎢ ⎥⎣ ⎦

= = → = ⇒ = +

= +

( )2 2

2 2 2 2

1 11 2

rr z

rr

V V V V V V Vv vt r r r z

V VVr r z rr

V p gr r

θ θ θ θ θ θ

θ θθθ

ρθ

µ ρθ θ θ

∂ ∂⎛ ⎞⎜ ⎟

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

= + + + − +⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦∂ ∂⎝ ⎠

1. steady2. ρ = const.3. Fully developed in z-dir.4. axisymmetric in θ -dir.5. vr = vz = 0

( )

( ) ( )1 11

1

0

rV c rV c rr r r

rVr r r θ

θ θ∂ ∂

= → =∂

∂ ∂⎛ ⎞ =⎜⎝ ⎠

∂

⎟∂ ∂

Momentum eq. in θ direction

1 2

2

2 V /2rrV c cr c rc θθ = += + →

Energy eq.

2

0 (3)dV Vk d dTrr dr dr dr r

θ θµ ⎛ ⎞⎛ ⎞= + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

BC’s

1 1 1 1 1

2 2 2 2

r=r v & T=T p=p r=r v & T=T

At rAt r

θ

θ

ωω

==

using the B.C’s

( )2 22 22 1 2 12 2 1 1

12 2 2 22 1 2 1

, cr rr rc

r r r rω ωω ω −−

= =− −

( ) ( ) ( )2 2

2 2 2 12 2 1 1 2 12 2

2 1

1( ) 4r rV r r r rr r rθ ω ω ω ω

⎡ ⎤= − − −⎢ ⎥− ⎣ ⎦

Eq. (1) determines the radial pressure distribution resulting from the motion.p=p(r) ← obtain this distribution!

Having found vθ(r) , it is substituted into eq. (3) to find temperature distribution

( ) ( )( )

( )( )

24 22 2 1 1 11 1

4 4 22 1 2 1 2 1 2 1

1 / ln / ln /Pr 1 1

ln / ln /r r r r rT T rEc

T T r r r r r r rω ω ⎡ ⎤− ⎛ ⎞−

= − − +⎢ ⎥⎜ ⎟− − ⎝ ⎠ ⎣ ⎦

where

( )2 2

1 1

2 1

Pr rEck T T

µ ω=

−Brinkman numberexpressing the temp. rise due to dissipation

( )( )

11

2 1 2 1

ln / PrEc=0 heat conduction solution

ln /r rT Tif

T T r r−

=−

SPECIAL CASES :

i) case when r1 → 0 i.e. in the limit as the inner cylinder vanishes(α=0)=r1/r2

1 2.(4) V ( 0) rigid-body rotationEq r rθ ω→ =

2 2rω

i.e. fluid rotates inside the outer cylinder as a rigid body

2Vω ω= ∇× =

ii) single cylinder rotating in an infinite fluid 2 2( , 0)r ω→ ∞ =

1 1rω

21 1

2( ) rV rrθω

→ ∞ = flow

v . .0

circularr const

r vθ

θ

=

∞ →

vortex 0free ω→ =

vortex 0forced ω→ ≠

1

11 1 1

1

1 12 22

2

=-2

rr r

r

rr

M r rV VV

r

F

r r

r L

θθ

θ

θθ

τ π

τ µε µθ

µω

=

= =

∂∂⎛ ⎞= = + −⎜ ⎟∂ ∂⎝ ⎠

iii) very small clearance between the cylinders,

2 12

1

1, Let 0r rr

ω−<< =

Eq. (4) becomes, in the limit,

1

1 1 2 1

1V r rr r r

θ

ω−

≈ −−

represents linear Couette flow between parallel plates0

r1

r2

1 1rω

21 1 14M Lrπµ ω=

Torque transmitted by the fluid to the cylinder of length L

Viscometry: to determine the viscosity of a fluid calculate the moment (or torque)exerted by the cylinders upon each other.

Moment = Stress x Area x Moment arm

moments on inner and outer walls are equal since the system is in equilibrium, i.esteady flow with no heat loss.

∴ The moment on the outer wall with the cylinder of height L.

( )

( )

2

2

2 2 2

2 21 2 1

2 21 2

2 2 12 2

22 2

2

2

1

21

2

1 12 22

2

4

lateral area

r r r

rr rr r

M r L r

V VVr r r

r rr r r

r rM Lr r

θ

θ θθ θ

τ π

τ µε µ

πµ ω ω

θ

ω ωµ

=

=

=

∂∂⎛ ⎞= = + −⎜ ⎟∂ ∂⎝ ⎠⎡ ⎤−

= − ⎢ ⎥−⎣ ⎦

= −−

0

can be measured

• if inner cylinder is at rest, ω1 = 0torque transmitted by the outer cylinder to the fluid, M2

2 21 2

2 22 22 1

4 r rM Lr r

πµ ω=−

Work out

CASE I : inner cylinder rotates with the outer one at rest.ω2 = 0

CASE II : ω1 = 0

Both case I & II called Couette Flows

1 2

2 1

2

Let the ratio of radii by =r / , width of the annulus by s=r

relative radius by x=r/r

rr

current

α−

r1

r2

x=r/r2

I

II

I

II

2

2

2

1 1

2 2

V 1 I (inner rotating, outer at rest)V 1

VCasse II (outer rotating, inner at rest)V 1

V (peripheral velocity of the inner cylinder)

V (

xCasex

xx

r

r

θ

θ

θ

θ

θ

θ

αα

α αα α

ω

ω

−→ =

−

⎛ ⎞→ = −⎜ ⎟− ⎝ ⎠

=

= " " outer " )

Let 1' r rxs s

−=

0 1

1.0

I

VV

θ

θ

1' r rxs s

−=

α= 0.10.20.40.60.81.0

comment on plots

Example : Axially annular Couette flow between concentric moving cylinders

a) inner cylinder moving

b) Outer cylinder moving

See White (1991) pg. 109

Stability of Couette FlowsAll solutions up to now are exact steady flow solutions of N-S. equations.

Laminar flows → smooth streamline character

ALL Laminar flows become unstable at a finite value of some critical parameter,usually the Re.

unstable → a different type of flow sets in- sometimes still laminar (different steady solution, symmetry-asymmetry

steady → time-periodic- more often, turbulent flow: fluctuating flow regime, random

h

-h

0

laminar flow velocityprofile

turbulent flow (approx.)Re 1500h

Uhυ

= =

U

Time-averaged S-shaped profile-varies slightly with Re-τw (and heat transfer rate) increases by two orders of magnitude.

Taylor (1923) → rotating inner-cylinderlaminar profile is valid until a critical rotation rate.For small clearence (r2-r1)<<r1, critical value of instability is given byTaylor number,

23 1

1 2 1 2( ) 1700Ta r r r ωυ

= − ≈

If Ta>1700 → 3-D laminar flow consisting of counter rotating pairs of vortices

Taylor vortices

POISEUILLE FLOW:Coutte flows → flow is driven by moving wallsPoiseuille flows → “ “ “ pressure gradients.

application to duct flows.

Consider steady flow of a viscous fluid in a straight duct of arbitrary but constant shape

entrance length, Le

U0

potentialcoreU0

viscouslayers

transitionL.B.L. T.B.L.core vanishes,layer coalesce

fully developed flow

Flow in the entry region of a tube

• wall friction causes viscous layer

depending on Re # both developingand developed regions remain laminar

Fully developed flow: slightly further downstream of the coalescence, the flowprofile ceases to change with axial position

constant shape cross-section

h.d p const

dx=

x

y

z

(v=0)

(w=0)

u=u(y,z)

: total hydrostatic pressure

p

p

p gzρ= +

Fully developed duct flow:

Entrance effect : i.e. thin initial shear layer and core acceleration

Shah & London (1978) regardless of duct shape for laminar flow

1 2 1 2Re c 0.5 , c 0.05h

eD

h

L c cD

≈ + ≈ ≈

Dh : suitable “diameter” scale for the duct

eRe 2000 L 100 (pipes of circular cross-section)c D≈ ⇒ ≤

0 =0 u v w uContinuityx y z x

∂ ∂ ∂ ∂+ + = ⇒

∂ ∂ ∂ ∂

Governing eqs. for incompressible flow

2 2 2

2 2 2

2 2

2 2

ˆdirection

ˆ0

ˆ ˆ ˆ ˆ- and -direction 0 ( ) only

u u u p u u ux u v wx y z x x y z

p u ux y z

p py z p p xy z

µ

µ

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

⎛ ⎞∂ ∂ ∂= − + +⎜ ⎟∂ ∂ ∂⎝ ⎠

∂ ∂= − = − ⇒ =

∂ ∂

(3)

x > Le velocity becomes purely axial, v = w = 0

u varies only with the lateral coord. u = u(y,z)

Momentum eq.

2 2

2 2

2 2

2 2

.( )

1 .

p u ux y z

d p const negativedx

u u p consty z x

µ

µ

⎛ ⎞∂ ∂ ∂= +⎜ ⎟∂ ∂ ∂⎝ ⎠

⇓

=

∂ ∂ ∂+ = =

∂ ∂ ∂Basic eq. for fully-dev. duct flow.(A)

B.C. → no-slip condition on duct surface uw = 02 equation Poisson u c∇ = if c = 0

Laplace eq.

• no general solution for arbitrary cross-section• but many known solutions for different duct shapes.

See R. Berker (1963), and Viscous Fluid Flow F. M. White (1991)

To non-dimensionalize eq. (A) no characteristic vel. & axial length scaleh: characteristic duct width

* *

2

, z , uy z uyh h d ph

dx

µ∗ = = =⎛ ⎞

−⎜ ⎟⎜ ⎟⎝ ⎠

eq. (A) becomes

( )2 *1 u 0 on duct boundaryu∗ ∗∇ = − =

The circular pipe: Hagen-Poiseuille Flow

r0 r θ

(y,z) plane → (r,θ) circle

Cylindrical coord. system is preferred

2

2

2

1

1

2 1

1 1 .

1

1 u(r)= ln

2

4

d drr dr dr

d du d pr constr dr dr dx

du d pr r cd

d p r cd

r

r

dx

cx

µ

µ

µ

⎛ ⎞∇ = ⎜ ⎟⎝ ⎠

⎛ ⎞ = =⎜ ⎟⎝ ⎠

+

+

=

+

z0, vrv v uθ= = =

0

1

1

20 2 0

( ) 0( 0) c 0 (ln0)

or u(r) is even function u(-r)=u(r) c 0

1( ) 0 c4

u r ru r finite

d pu r rdxµ

= == = → =

→ =

= → = −

( )2 20

1( ) note: 0 u>04

d p d pu r r rdx dxµ

= − − <

paraboloid about centerline

Volume flow rate, Q0 4

0

0

28

r

A r

r d pQ udA u rdrdx

ππµ=

⎛ ⎞= = = ⎜ ⎟⎜ ⎟

⎝ ⎠∫ ∫

mean velocity

0

20

max20

0w

0

1 /8 2

wall shear stress,

4 2

r

r r

rQ d pu Q A ur dx

The

rvu d p ur x dx r

π µ

µτ µ=

⎛ ⎞= = = − =⎜ ⎟⎜ ⎟

⎝ ⎠

⎛ ⎞∂∂⎡ ⎤= − + = − =⎜ ⎟⎢ ⎥ ⎜ ⎟∂ ∂⎣ ⎦ ⎝ ⎠

Introduce the pipe-friction factor λ

2

0

0D

Darcy-Weisbach eq.2 2

2 64Re =Re

d p udx r

u r

λ ρ

ρ λµ

− =

≡ ⇒ VALID FOR Laminar flowReC ≈ 2300

f2

2 16 CRe 4

wf

D

Cu

τ λ

ρ= = → =

fanning friction factor(skin friction factor)

Non-circular ducts:

( )2* * *1 ; u 0 Dirichlet problemu∇ = − =

Berker (1963) , White, Viscous Fluid Flow

rectangle, ellipse, concentric annulus, circular sector, equilateral triangle, etc.

a

a az

y

( )2 2

h

/ 1( , ) 3 322 3

4 4 D perimeter 3

Re

h

h

d p dxu y z z a y za

A area aDP wetted

uD

µ

ρµ

− ⎛ ⎞= − −⎜ ⎟⎝ ⎠

×= = =

=

Example: equilateral triangle

Temperature distribution in fully developed duct flow:

u(y,z) is knownenergy eq. can be solved for TT(x,y,z) if B.C. change with x.

Tw=const.

r0

Ex: Pipe Flow

( )2 20

1( )4

d pu r r rdxµ

= − −

T(r) only

Energy equation: Appendix B, White (1991)22

41 2

2

40

( ) l

6

n

1k d dT du u rrr dr dr d

T r Ar r

r r

c c

µµ⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝

= +

⎠

+ A is known

at r=0 T=finite → c1 = 0

at r=r0 T=Tw →2 4

40

2

max ; T1 wwuTu rT

r kT

kµ µ⎛ ⎞

= + −⎜ ⎟⎝ ⎠

= +

max. temp. rise due to dissipation2 4

*4

0

2

T 1 1

u 100 ft/sec

max. temp. rise 1 F (air) " " 3 F (water)

w

u rBr BrkT r

uAirk

µ

µ

⎛ ⎞= = + −⎜ ⎟

⎝ ⎠

→ =

≈ °≈ °

oils → µ is large dissipation importantor gas dynamics where velocities are high.

The wall heat transfer, qw

0

w 0

0 0 0w

0 0

or q ( )

4 ( ) h(2r ) (2r ) q , Nu= 8( )

w wr r

w w

w

dTq k h T Tdr

k T T qorr k k T T

=

= = −

−= = =

−

for viscous dissipation

Homework Find u(y,z)

y

z(a,b)(-a,b)

(-a,-b) (a,-b)

Contours of u(y,z)can use method of eigen functionexpansion

Let a=1/2b=1

VISCOUS FLOW NEAR A STAGNATION POINT

y

0 rigid boundarystagnation point (u=v=0)

δ

B-L thickness(region of non-zero vorticity, )ω

• At large distances from stagnation point, the flow essentially the same as that of thecorresponding potential flow problem.

∴ can use potential flow to solve the viscous flow problem.

2-D Flow: Plane Stagnation Flow:

( )

0 (1)

Introduce stream function x,y

u= , v= (2) satisfies cont. eq. identically

u vContinuityx y

y x

ψψ ψ ψ

∂ ∂+ =

∂ ∂

∂ ∂−

∂ ∂

it must also satisfy the 2-D eq. of motions

2 2

2 2

2 2

2 2

1

1

u u p u uu vx y x x y

v v p v vu vx y y x y

υρ

υρ

⎛ ⎞∂ ∂ ∂ ∂ ∂+ = − + +⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠

⎛ ⎞∂ ∂ ∂ ∂ ∂+ = − + +⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠

(3)

Invıscid Flow near a stagnation point of a body

0

, u=Bx , v= By V

positive constant proportional to U /

Bxy Bxi By j

B L

ψ = − → = −

( )2 22

0 2 21 0

01 1

0

2 2

: pressure2BP VP V

P stagnation

P P x yρ ρ

ρ+ = + − +⇒ =

Inviscid flow is described above slips at the wall, i.e.It must be modified to account viscous effects.

0 at y=0u ≠

22

0

( ) u=xf '(y) , v= f(y) (4a)and

( ) (4b)2

viscous xf y

Bp p x F y

ψ

ρ

= → −

⎡ ⎤= − +⎣ ⎦

Note: Continuity eq. is automatically satisfied

Substituting eqs. (4a-b) into (3)

0UB is L

∼stream vel.

charact. body length

( )

( )2 2

21'. ' ( ) " ( ) 0 "'

(5' a" "' )f ff

xf f f xf B x xf

B fυ

υρ

+ − = − −ρ +

− = +

+

ODEdimensional

( )( ) ( )

( )

2

2

10+ ' '( ) 0 "2

1' 2

5b" '

Bf f F y f

ff B F f

υ

υ

⎛ ⎞− − = − −ρ + −⎜ ⎟ρ ⎝

=

⎠

−

BCs for the unknown functions f & F

u=v=0 on the wall ' 0 at y=0f f⇒ = =

( ) ( )

( ) ( )( )

0 x=0 F 0 0 at y=0 y=0y u=Bx inviscid flow Bx=xf '

f '

p p

B

= ⇒ =

→ ∞ ⇒ ⇒ ∞

∞ =

Can first solve eq. (5a) for f then substitute the result into eq. (5b) to get F.• Note :- instead of solving u,v & p(x,y) or ψ(x,y) via PDEwe solve ODE only.eliminated x, leaving only y a single similarity variable.

To eliminate dimensional constants B&υ , eq. (5a) needs to be non-dimensionalizedas follows.

2

2

since no body-length scale "L" for this flow

m use the proper length scale as B s.1/s

1 " " velocity " B /

m

m m ss s

υ

υ

•

=

= =

Appropriate dimensionless variables then

( ) ( ) ( )( )( )

2

, or =x 6a-b)

substituting eqs. 6a

"' " ' 1 0

-b) into 5ad (7) '= , etcd

f yBy BB

η φ η ψ φ η υ

ηφ + φφ −

= =ν ν

φφ + = φ

BC’s for φ0 =0, '=0

= '=1η = ⇒ φ φη ∞ ⇒ φ

No analytic solution existsNumerical integration.

Solution of eq. (7) is given first by Hiemenz (1911).Can consider thickness of B.L. y=δ where u=0.99 U

corresponding value of thepotential flow

( )

( ) ( ) ( )

( ) ( )

2.4

2.44' 2.4 0.99

' ' '

;

BBy

B

du xf y x B xB udy

f

F

yByB

δ

δ

ν≈ ν

η =ν

→ ≅ν

η= = νφ η = φ η =

η = φ η =ν

=

ν

∼

Ui → inviscid vel. profile

( ) ( )v f y B= − = − νφ η u, v satisfy the no-slip condition

( )'i

uU

= φ η

1

i

uU

0 1 2 32.4

Byη =ν

ψφ ∝

' uφ ∝

" τφ ∝

Fig. 1

2"' " 1 ' 0 φ + φφ + − φ =Runge-Kutta numerical integration3rd order ODE ⇒ 3 1st order ODE ‘s

( )( )( )

Y 1 "

Y 2 '

Y 3

Let = φ

= φ

= φ

( ) ( ) ( ) ( ) ( )

( ) ( )

( ) ( )

( ) ( ) ( )

12 2 1 3 1

21

32

for Y 1 , Y 2 & Y 3

dYY Y Y Y

ddY

Yd

dYY

dSolve

= ∗ − − ∗η

=η

=η

( ) ( ) ( ) ( ) ( ) ( )( )

0 0

Initial conditions:0 ' 0 0 Y 2 Y 3 0 i.e. Y 2 Y 3 0 at =0

' 1 (2) 1 as Y

φ = φ = ⇒ = = = = η

φ ∞ = ⇒ → η → ∞

Numerical Solution

how large is “infinity” η → ∞-5 when " becomes very small, say <10answer → φ

asymptotic analysis

( ) ( )

2

i2

- / 2

5

uas becomes large ' 1 .U

"' " 1 ' 0

"' or ""

'' 10 if > 4.8 = "infinity" 4th order R-K method " 1 to 1.5 gue

vanishes

const

e η

−

η = φ η → ⇒ φ η ≅ η +

∴ φ + φφ + − φ =

φ∴ ≅ −η φ ≈

φ

φ < η

φ ≈ ← ( )

( )0

i

|

ss Y 1 1 1.5 at =0

'

" 0

=u

1.2325

/U

9yw

τ=

φ = =

= η

φ

∼

0.10.2 . .2.42.83.0

η '

0.118260.22661

0.990550.997050.99843

φ

Numerical Solution

δ0.99u

U=

x

y

B-L behaviour : no-slip condition creates a low-vel. region which merges smoothlywith the outer inviscid flow

( )( )

( )

( )

i

outer flow called "free stream" velocity , U x

flow: U=u x,u 0.99 ' 2.4

U

=2.4= y=

Stagnation Bx

Bδ

→

∞ =

= = φ η =

ην

accelerating freestream pressuredecreases in the flow direction⇒ favorable pressure gradient

2.4 .B

constδ ν≈ =

because thinning due to stream acceleration, U=Bx,exactly balances the thickening due to viscous diffusion.

e.g. if U=cxm

the B-L will grow with x if m<1 will become thinner with x if m>1

Pressure distribution: exhibits B.L. behaviour

( ) ( )

( ) ( )

( )

22

0

2

2

2

p=p x,y2

' " 2' , F' y2 1

= f f '+ "

Bp p x F y

p dUB x Ux dxp B ff fF yy B

f

ρ

ρ ρ

ρ

ρ

⎡ ⎤= − +⎣ ⎦

∂= − = −

∂∂ + ν

= − =∂

− ν

U=BxBernoulli eq. gradient parallel to the

wall satisfies Bern. eq.

small if υ is smalltypical to laminar boundary layers

( )

( ) ( ) ( ) ( )

( )

0

i

'

=

, ( ) 0

= =U

B, =y

wy

u xf y

v f y B

u v vv v yy x x

u u By

f yB

y

τ µ

ηη ν

=

=

= − − νφ η

⎛ ⎞∂ ∂ ∂= + = → =⎜ ⎟∂ ∂ ∂⎝ ⎠

∂ ∂ ∂ ′′

φ η = ην

φ η

ν

∂ ∂ ∂

"i 0

"i 0

x2

U

=U wall shear stress is proportional to freestream vel.)

2 2 ReRe

w

w i

wf

x

BU

UxCU

τ µ

µρτ

τρ

Β= φ

ν

φ

′′φ (0)= = =

ν

∼

Friction factor varies inversly with thesquare root of the local Reynolds numberCommon in L.B. Layers.

Wall shear stress

Stagnation-point problem: Temperature distribution

( )

( ) ( ) ( )

B =y

'u xf y

v f yf y

B

=

= −

ην

φ η =ν

• velocities are known

A similarity solution exists if the wall & stream temperatures Tw and T∞are constant.* A realistic approximation in typical heat transfer problems

Goldstein (1938)

2 2

2 2pT T T Tc u v kx y x y

ρ⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂

+ = +⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

( )

( )

( ) ( ) ( ) ( ) ( )

( )

w

2 2

2 2

B , =y

with T & T being constant, the fluid temp T=T y only.

B.

B

w

w

w w

y

w

p

T TT T

T Tv f y B T T B T Ty y

T T Ty

c Bρ

∞

∞

∞ ∞

∂η∂

∞

− θ η = η

− ν

∂ ∂ ∂θ ∂θ= − = − νφ η − = − − φ η

∂ ∂ ∂η ν ∂η

∂ ∂ θ= −

∂ ∂η ν

− ( )wT T∞ − ( )wk T T∞

∂θφ = −

∂ηB

( )

( ) ( )

2

2

2

2

p

k =c

Pr= .

' 0 0 & 1

d Pr 0d

pcconst

k

B

d

C

d

s

αρ

µα

∂ θν ∂η

ν= =

θ

θ θ+ φ η =

η η

= θ ∞ =

( ) stream function is known from flow problem Fig.1 Tabulated values are knownφ η →

IVP → RK4 subroutineBut has an exact solution. (Linear eq.)

( ) ( )0 0

0 0

T

exp -Pr Pr

exp -Pr

thermal boundary layer, , is the point where 0.99

d ds

d ds

φ

φ

δ

η η

η∞

⎡ ⎤η ⎢ ⎥

⎣ ⎦θ η = = θ = θ⎡ ⎤

η ⎢ ⎥⎣ ⎦

θ ≈

∫ ∫

∫ ∫

1.0

01 2 8

10 1 0.3 0.1

0.01

Pr=0.001

B=yην

w

w

T TT T∞

−θ =

−

0.4Pr (Power law curve fit)

Pr=

Pr 1 >

u

T

T

δδ

αδ δ

α

≈

ν

> → >ν

velocity B.L. is thicker than thermal B.L. because viscousdiff. exceeds conduction effects.

thermal diff.

viscous diff.

Heat transfer at the wall is computed from Fourier’s Law

( ) ( ) ( )

( )

00

1

0 0

Pr

G Pr exp -Pr

w w wy

T d B Bq k k T T k T T Gy d

where d ds

θ

φ

∞ ∞η==

η∞−

∂= − = − − = − −

∂ η ν ν

⎡ ⎤= η ⎢ ⎥

⎣ ⎦∫ ∫

( )( ) ( ) ( )

( ) ( )( )

( ) ( )

4 Formulation: Y 3

43 Y 4

5exp Pr 4

5 Pr

RK

dYY d

ddY

Yd

Y G

= φ

= = φ ηη

= − ∗η

≡

∫

0.1

1

G(Pr)

10

0.01 1 10 100 1000 Pr

Pr (plane flow)

0.01 0.0760.1 0.221 0.57

10 1.339100 2.9861000 6.529

G

( ) ( )

( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

0.4

2

2

0.57 Pr

d Pr 0 IId

Y 4

5 Y 5 Pr 3 . 5 4

4 5 5

G

dORd

Let

dYd Y Yd d

dYY

d

≈

θ θ+ φ η =

η η

= θ η

θ= ⇒ = − ∗

η η

=η

( ) ( )

( ) ( )0

0

' 0 0 Y 4 0

1 Y 5 1

BC s θ = ⇒ =

θ ∞ = =

( )( )( )

( )

( )

( ) ( )

i

22 2

22 2 2

1

1 "

2 ' solve for Y i=1,...5 RK4 Routine depends on Pr

3

1B p=- B2

1 1 B ' K' y - B2

Y

Y

Y

xpx K yx

ypB y y K y Ky

= φ

= φ → η

= φ

∂= − ρ +

ρ ∂

∂+ = − = − = ρ +

ρ ∂ ρ

( )

( )

( )

2 2 20

2 2 20 0

2 20

x=0 , y=0 p=p stagnation pressure2

c=p2

Bernoulli Eq.2

p B x y C

p p B x y

p p u v

ρ= − + +

ρ= − +

ρ= − +

( ) ( ) ( ) ( )( )

2

1/s mm/s

22

0

1u = B x v=-By =Bxy s

y=0 u 0 slips on the wall

=xf y u=xf ' y v=-f y 2

. 0

m

Bp p x F y

u vC Ex y

ψ

ψ

⎡ ⎤⎢ ⎥⎣ ⎦≠

= − ρ +

∂ ∂+ =

∂ ∂( ) ( )

( ) ( ) ( )2 2

2 2

f ' y f ' y 0

1' " 0 "'

' " "' ODE

xf f xf B x xf

f ff B f

− =

+ − = − −ρ + ν +ρ

− = + ν

x

y

( ) ( )

( ) ( )( ) ( )

2

2 2

2

2

' " 0 "'

' " "'

1' .0 ' ' 0 "2

1 ff '= ' "2

xf f xf xf

f ff B f

Bxf y f f F f

B F f

+ − = +ν +

− = + ν

⎛ ⎞+ − − = − −ρ + ν −⎜ ⎟ρ ⎝ ⎠

− ν

( )( )

( )

( )

[ ]

0

y f '

y=0 u=0, f ' 0 0

v=0, f 0 0

0 p=p F 0 0

0

length scale : B

u Bx B

xy

proper m

= → ∞ ∞ =

=

=

= ⎞=⎟= ⎠

ν

[ ] ( )

( )

2 2 2

2

y =

B

scale B : / B

B' " =B + "'

' - "=1+ "'

velocity m sx

f B

B B

ψ

2

ην

ν = φ ην

= νφ η

Βφ − νφΒφ ν φ

ν ν

φ φφ φ

( )( )

( )2

2

3

3

' /

' "

" "'

Bdf B Bdy y

f d BB By dy

f d B B BB By dy

∂ νφ η ∂η= = νφ ν

∂η ∂

∂= φ = φ

∂ ν

⎛ ⎞∂= φ = φ⎜ ⎟⎜ ⎟∂ ν ν ν⎝ ⎠

( )( )

( ) ( )( )

( ) ( )

( )

0 u=0=xB ' ' 0 0

v=0=- B 0 0

u=xB ' ' 1

Fig s.11 Schlichting

u=xf ' y '

v=-f yiU

xB

xB

B

η = φ φ =

νφ φ =

η → ∞ φ η → ∞ = φ η → ∞ =

= φ η

= − νφ( )η

inviscid profile

Unsteady Motions of a Plate :

0

y

x

u=u(y,t) v=0

Parallel flow, v=w=0

• initially both the plate & fluid are at rest u(y,0)=0 for y>0• the plate is jerked into motion in its own plane• no-slip at the plate : u(y=0,t)=U(t) for t>0

Two-cases1- U=constant (Stoke’s First Problem)2- U(t)=U0cosωt (Stoke’s Second Problem)

Steady oscillation of the plane at Ucosωt.

zero pressure gradient

governing P.D.E. reduces

0 p=const.dpdx

→ = →

( )2

2 1-D heat conductionu ut y

∂ ∂= ν

∂ ∂

CASE I: U=const.

( )

( )

0 00,

0

,

for tu y t

U for t

u y t finite

=⎧= = ⎨ >⎩

=

Unsteady heat conduction equation

Carslaw and Jaeger (1959) – Conduction of heat in solids

Solution methods-Laplace transforms-Similarity methods

2 2

2 2 u u ut x y

⎛ ⎞∂ ∂ ∂= ν +⎜ ⎟∂ ∂ ∂⎝ ⎠

12 2

u y yerf erfcU t t

⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟ν ν⎝ ⎠ ⎝ ⎠

( ) 2

0

2 xerf e dxβ

βπ

−= ∫Similarity Solutions :• applicable for non-linear problems

similarity solutions exist for:parabolic P.D.E with two independent variables when there is no geometric lengthscale in the problem.

⇒ give ODE

y

Uu

t=0 t1 t2

increasing timeerfx

erfcx

1

1 2x

1 1

2 2

Eg: At t=t u=0.5U at y=y

At t=t u=0.5U at y=y

→

→

Expect solution exists in the form

( ) ( )

( )

( ) ( )( )

n

n

, y where =t

, : similarity variable

: constant of proportionality: will be found later to make dimensionless.u ,

=const. u=const. y t

u y tf

Uy t

y t Uf

when

α

α

= η η

η

η

= η

η ⇒ ∼

IF similarity solution exists, it will result in an ODE with f as the dependent variableand η as the independent variable.

PDE ( ) ( )

( )

2

2

n

n+1

n n

2

2 n

1 , Let u=Uf

y =t

yt

' 't t

'

''t

u ut y

u u dfU nt t du u Uf U fy y

u u u U fy y y y y

Un ft

α

α

α α

α α

∂ ∂= ν η

∂ ∂

η

∂ ∂ ∂η= = − =

∂ ∂η ∂ η∂ ∂ ∂η

= = =∂ ∂η ∂

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

η

⎝ ⎠

−

2

n 2n 't

't

U fα=

Substituting above expressions to (1)

U− 'n f Utη

= ν2

2n ''t

fα

To eliminate t (hence obtain an ODE for f) Let n=1/2

∴ for n=1/2 a similarity solution is obtained

2

1" ' 0 f "+ ' 02 2

f f fαα

2 ην + η = → =

ν constant [-]

( ) ( )

( )

( )

1/2

2

2

to make dimensionless; u=Uf use & U if necessaryy= : constant of proportionality

t

y m 1 y y , = = or [ ]t t 2 t

1 12 = f2 2

f "+2

f "+2 f '=

f '=0

0

mss

A

α α

α

α α

η η ν

η

ν ⇒ → η η = −ν ν ν

ν = → → η

η

ην

∼ ∼

( ) ( ) ( )( ) ( ) ( )

y ODE 2 t

BCs u 0,t t>0 f =0 1

u y,t=0 0 y 0 f 0

U

η =ν

= → η =

= ≥ → η → ∞ η →

( )

( )

( )

( )

( ) ( )

( )

2

2

2

2

2

0

0

0

" d f '2 ln ' 2 ln' d A

' f

0 1 B=1

f 0 0=A 1 12

2

y =2 t

y

, 21

Erfc =2 t

u

f ff

f Ae A e d B

f

e d A

A

Erfc

C

y tf

om

e dU

ξ

ξ

ξ

ξ

πξ

π

ξπ

η

η−η −

∞−

−

= − η → = − η → = −ηη

= → η = +

= ⇒

η → ∞ = ⇒ + = +

= −

ην

⎛ ⎞= ⎜ ⎟ν⎝ ⎠

= η = −

η

∫

∫

∫

( ) 2

0

2 error function Erfc 1plementary e dξ ξπ

η−η = − ∫

( )

( )( )( )( )( )

( ) ( )

2

0

2 function Erf

Numerical values of the Erfc is given in Tables. (pg. 103)

Erfc 0 1.0

Erfc 0.1 0.88754

Erfc 4.0 0.00000001

Erfc

u y=f =Erfc =ErfcU 2

0

t

Error e dξ ξπ

η−

=

=

=

∞ =

η =

η

η ην

∫

⎛ ⎞⎜ ⎟⎝ ⎠

y=2 t

ην

1.0

1.8

0.5 1.0u/U

( )Erfc η 1.82 u/U=0.01At η →

Viscous effects are confined to a region next to the plate where 1.82. Outside

the region, the vel. gradient is too small to cause a significant shear stress.

In other words, vorticity dissipate

uy

η ≤∂∂

sup to the point 1.82.Beyond this point vorticity is negligible.This region called Boundary Layer.

η ≈

It is customary to define the shear layer (thickness) (B-L thickness) as the pointwhere the wall effect on fluid drops to 1%.

( )

-5 2

-6 2 -6

y= =2 t 1.82 3.64 t

: Air at 20 C 10 m / 11 cm after 1 minute

Water at 20 C

0.01

10 m / 3.64 10 60 2.8 cm

1.82

Ex s

s

u fU

δ

δ

δ

∴

ν ≅ ν

° ν =1.5× ≈

° ν = ≈ ×

= = η ≅

STOKES’ SECOND PROBLEM:

x

y

U

u(0,t)

t2π

2T πω

=

( )( ) ( )( )

0, cos

,0 0 at rest

, 0

u t U t

u y

u t

ω=

=

∞ =

• differs from Stokes’ first problem by only in B.C. at y=0 oscillatory vel.

( )0 cos

daily or seasonal variations in surface temp.

determine temp. variati

E

o

x:

ns in ground

T T tcyclic

ω=

( )2

2 T y,tT Tt y

α∂ ∂=

∂ ∂

Fluid is also to oscillate with the same frequency, so let,

( ) ( ) ( )( )

( )2

2

2 i= -1 cos sin

2 into the gov. eq.

,

i

ii t t

i t

e t i t

substitute

u u f y et y

u y t f y e ωω

ω

ω ω

ω

= +

∂ ∂

=

= ν ⇒∂ ∂

( )" i tf y e ω= ν

( ) ( )

( )( )

21,2

ODE

Characteristic eq.

0 Roots:

sol.

f"

y 0

f y

, 0 B=0

i iy y

ii

Ge

i f y

Ae

ner

B

al

u t

e

α α

ω ω−

ν ν

ω− =

ν

=

ω ω− = ⇒ = ±

ν ν

∞ → ⇒

+

( )( )

( ) ( )( )

( ) ( )

1

1i2

=

=

f y

,

cos sin

i y y

ky ik

i y

y i ti

i t kyky

t

ky

i

k

Ae e e

A

Ae Ae e

u y t f y e

Ae t ky i t ky

e e

ω ω

ω ω ω− + − −

2ν 2ν 2ν

−

− − −

ω −−

+=

ω=

2ν=

= =

=

ω − + ω −⎡ ⎤⎣ ⎦

( ) ( )

( )

( ) ( ) shift

, is real need to take real part of f yTo evaluate A use B.C.u 0,t cos A=U Thus, vel. distribution

whereu y,t co k2

s =ky

amplitudephase

i tu y t e

U t

Ue t ky

ω

−

⇒

= ω → ←

ων

= ω −

( ) velocity of fluid decreases exponentially as the distance from the plate y increases.

rate of decrease k= , will be faster for higher frequency and smaller viscosity.2

fluid oscillates in time

•

ων

•

( )

( )

-ky

with the same frequency as the input freq. in boundary. amplitude of oscillation Ue

max. amp. at y=0 U phase shift cos t-ky

g y• =

→

• ω

specific timeinstantaneous time

U

y

Can define thickness of oscillating shear layer, δ as again where velocity amplitudedrops to 1% of U. ∴u/U=0.01.

4.6 4.60.01 =k

6.5 2

ky k

y

u e e eU

k

δ

δδ

δ δ

− − −

== = = = ⇒

ω ν= ⇒ ≅ ν

ν ω∼

remember characteristic of laminar flows

( )

w 00

: For air at 20 , with a plate frequency of 1 Hz =2 rd/sec10

The wall shear stress at the oscillating plate

sin4

lags the max. vel.

yy

Ex Cmm

u U ty

shear stress

πδ

πτ τ µ µ==

ω

≈

⎛ ⎞∂ ⎛ ⎞= = = ρω ω −⎜ ⎟ ⎜ ⎟∂ ⎝ ⎠⎝ ⎠

by 135 2 4π π⎛ ⎞+⎜ ⎟

⎝ ⎠Since governing eq. is linear, the method of superposition is applicable. Hence, superposing several oscill. of diff. freq. & ampl. the sol. for arbitrary periodic motionof plate can be obtained.

UNIFORM FLOW OVER A POROUS WALL( ) non-linear inertia terms are not zero V. 0

but linearized to permit a closed form solution

V• ∇ ≠

•

Example: steady, fully-dev. flow over a plate with suction

y

fully dev.

u=u(y)p=const.

δ

• suction on surface is sometimes used to prevent boundary layers from separating.

I. solution with primitive variables, u, v, p

ux

∂∂

2

2

0 v=const.

: u

vy

u u ux vx y x

∂+ = ⇒

∂

∂ ∂ ∂+ = ν

∂ ∂ ∂

( )

( )

2

2

0

2

2

22

2

1

v= V uniform

: 0

0 0 =0 &

V y

u py x

pyy

du d uVdy dy

d u V du V Vdy d

A

y

u y Be

α α α α

⎛ ⎞−⎜ ⎟ν⎝ ⎠

⎛ ⎞∂ ∂+ −⎜ ⎟⎜ ⎟∂ ρ ∂⎝ ⎠

−

∂=

∂

− = ν

+ = + = ⇒ = −ν

+

ν ν

=

( )( )

. . u y=0 0

u y

B C s

U

=

→ ∞ =

( )V

note if blowing instead of suction v=V u y y u

yA Be ν⇒ = +

→ ∞ →

•

→ ∞ not physically possible

u would be unbounded at large y( )

( )

( )

0 0 A+B=0

u y B.L. thicknes

1

s

V y

u

U AV

u y U e

δ

−ν

⎡ ⎤=

= →

ν⎛ ⎞→ ∞ = =

−⎢ ⎥⎣

⎜ ⎟⎝ ⎠

⎦

∼

II. solution using Vorticity transport eq.

Vorticity Transport in 2-D

( ) 2

2 2

2 2

.

z z z z z

Vt

u vt x y x y

ω ω υ ω

ω ω ω ω ωυ

∂+ ∇ = ∇

∂⎛ ⎞∂ ∂ ∂ ∂ ∂

+ + = +⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠

( )

( )

2

z2

2

2

of diffusionvorticity toward away frplate

0 & 0 fully-dev. flow

& v= V=const. from continuity

z

z zz

z z

convection viscous

steadyt x

v yy y

dy dy

d dVdy dy

ω ωυ

ω ωυ

∂ω ∂→ = ⇒

∂ ∂∂ ∂

= ⇒ ω = ω∂ ∂

∂→ −

∂

− =

om plate

2

2

1

z

1

Integrate once,

0 at y. . 0

0 at y

z z

zz

z

d dVdy dy

dV cdy

B C cddy

ω ω

ωω

ω

− =ν

− = +ν

⎫ω = → ∞⎪ ⇒ =⎬= → ∞⎪⎭

z 2 2

z

z 2 yields,

0 at y , 0=c c ??, let

zV y

z

cdV dy

but eNo

e

w

vx

ωω

−∞

−ν− =

ν

ω = → ∞ ⇒

∂ω =

=

∂

ω

( )

( )

z 2

2 3

2 3

2

z

3

2

u=U 1

V

U=c .0

0 0 0=c cV

&

V

V y

V y

y

V y

u duy dy

du c edy

u c e c

u y U c c UVu

U

y U

e

e

U

V

−

−

ν

−

ν

ν

−ν

⎡ ⎤

∂− = −

∂

ω = − =

ν= +

→ ∞ = → + → =

ν= = ⇒ + → = −

ν

−⎢ ⎥⎣ ⎦

ω = −ν

y

UV−

νzω

is max at y=0zω

terms in eq.3

2

2 3

2 2

V yz

V yz

VV U e convectiony

VU e diffusiony

−ν

−ν

∂ω− = − →

∂ ν

∂ ων = − →

∂ ν

if V vorticity moves toward wallV

if V vorticity moves away from walldif. term temdency of shear layer to grow due to viscous di

Note lengthν⇒ ⇒

⇒

ν →

∼

( )( )

ffusion

term toward the wallAs usual, define the B.L. thickness to be the point where

u=0.99U

. Air at 20 , if V=1 cm/s, 7 mV

m

=4.6V

V convective

Eg C

δ ϑ

δ

ν ν⎛ ⎞= ⎜ ⎟⎝

→

→

≈⎠

• For a plate with a leading edge (x=0), a Laminar shear layer would grow andapproach this constant value.it is estimated by Iglisch (1944)

2

4 U=10 m/s, V=1 cm/s x 6mUxVν

≅ ≅

Ex2: Flow through & between porous plates

h Flow .dp constdx

= y

x

. . u=0, v= V at y=0, h2-D , N-S eqs.

B C

ut

−

∂∂

uux

∂+

∂

2

2

1u p uvy x x

∂ ∂ ∂+ = − + ν

∂ ρ ∂ ∂

2

2

uy

vt

⎛ ⎞∂+⎜ ⎟⎜ ⎟∂⎝ ⎠

∂∂

vux

∂+

∂vvy

∂+

∂

2

2

1 p vy x

∂ ∂= − + ν

ρ ∂ ∂

2

2

vy

∂+

∂

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

-V

Assume1) Steady2) Fully-dev. flow ( )0 0 P=f xp

x y∂ ∂⎛ ⎞→ = ⇒⎜ ⎟∂ ∂⎝ ⎠

2

. uContx

∂∂

2 2

2 2

part hom

part part

hom 1

1

0 v=const.= V

1 1V .

1

solution =

1 ,

,

1

V y

V y

vy

du dp d u d u V du dp constdy dx dy dy dy dxdu d V dpLetdy dy dx

DE

dp AV dx

c e

du dpc edy V dx

µαα α

µα α α

α α

α

α

−ν

−ν

∂+ = −

∂

− = − + ν → + = =ρ ν

= ⇒ + =ν

+

= =ρ

=

= = + ⇒ρ

( )

( )

( )

1 2

2V 0

0

1

. . u=0 at y=0 & y=h

no suc

1h

tion V=0, then1u

2

1 22

u=V 1

1

1

V y

V

V y

V h

V y

V h

dpu y c e y cV V dx

B C s

ifdp y hyd

edp ydx h e

V edu h dpdy V dx h

x

dp

e

y hdx

µτ µ

µ

τ

−ν

→

−ν

−ν

−ν

−ν

=

ν= − + +

ρ

⎡ ⎤−⎢ ⎥−⎢ ⎥ρ−⎢ ⎥⎣ ⎦

⎡ ⎤−⎢ ⎥ν= = +⎢

ρ ⎢ −⎣

⇒

= −

⎢

=

⎦

−

⎥⎥⎥

parabolic vel. profile

0

0

1

1

1

1

y

y h

w V hy

V h

w V h

Vdu h dpdy V dx h

e

V eh dpV dx h

e

µτ µ

µτ

=

=

−= ν

−ν

−ν

⎡ ⎤−⎢ ⎥ν= = +⎢ ⎥ρ ⎢ ⎥−⎣ ⎦⎡ ⎤

−⎢ ⎥ν= +⎢ ⎥ρ ⎢ ⎥−⎢ ⎥⎣ ⎦

dp we have fully-dev. flow, shear stress & is relateddx

If

L

h P1P2

0ywτ=

y hwτ=

( ) ( )( )

( )

0

0

1 2

1 2

1 2

1 2

0 for fully-dev. flow

=P . . 0y h y

y y h

w w

w w

F

h P h L L

P P h L

dpP P Ldx

P Pdpdx L

τ τ

τ τ

= =

= =

=

− + − =

⇒ − = −

= −

−

−

− =

∑

L

P1

P2