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
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