# Chapter 6: Viscous Flow in Ducts

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```					058:0160                                                            Chapters 6
Professor Fred Stern          Fall 2007                                      1

Chapter 6: Viscous Flow in Ducts

Entrance, developing, and fully developed flow

Le = f (D, V, , μ)
L
 theorem       f (Re) f(Re) from AFD and EFD
e
i
D

Laminar Flow: Recrit ~ 2000               Re < Recrit          laminar
Re > Recrit          turbulent
L / D  .06 Re
e

L e max
 .06 Re D ~ 138D
crit               Max Le for laminar flow
058:0160                                                                  Chapters 6
Professor Fred Stern   Fall 2007                                                   2

Turbulent flow:

Re                 Le/D
4000                 18                          L / D ~ 4.4 Re     1/ 6

104
e
20
105                 30
106                 44               (Relatively shorter than for
107                 65                      laminar flow)
108                 95

Laminar vs. Turbulent Flow

Hagen 1839 noted
difference in
Δp=Δp(u) but could
not explain two
regimes

Laminar                        Turbulent                      Spark photo
Reynolds 1883 showed that the difference depends on Re = VD/ν
058:0160                                                                                Chapters 6
Professor Fred Stern        Fall 2007                                                            3

Laminar pipe flow:

1. CV Analysis

Continuity:
0   V  dA  Q1  Q2  const.
CS

i.e. V1  V2   sin ce   A1  A2 ,   const., and V  Vave

Momentum:

 Fx  ( p1  p2 ) R2  w 2 RL   R2 L sin   m(2V2  1V1 )
p                         W z / L           0
pR   2RL  R z  0
2
w
2

2 L
p  z              w

R
058:0160                                                                                Chapters 6
Professor Fred Stern       Fall 2007                                                             4

2 w L
h  h1  h2   ( p /   z ) 
 R
or
R h    R dh
 
w

2 L      2 dx
R d
       ( p  z )
2 dx

r d
              ( p  z )
2 dx

i.e. shear stress varies linearly in r across pipe for either
laminar or turbulent flow

Energy:
p1       1                        p2          2
           V1  z1                          V2  z 2  hL
        2g                                   2g

2 L
h  h            w
L
 R

 once τw is known, we can determine pressure drop

In general,                                                                 roughness
   (  ,V ,  , D,  )
w      w

Πi Theorem
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Professor Fred Stern       Fall 2007                                  5

8 w
 f  friction factor  f (Re D ,  / D)
V      2

VD
where              Re D 


LV2
h  hL  f                       Darcy-Weisbach Equation
D 2g

f (ReD, ε/D) still needs to be determined. For laminar
flow, there is an exact solution for f since laminar pipe
flow has an exact solution. For turbulent flow,
approximate solution for f using log-law as per Moody
diagram and discussed late.

2. Differential Analysis

Continuity:
 V  0

Use cylindrical coordinates (r, θ, z) where z replaces
x in previous CV analysis
1             1          
(r r )       ( )  z  0
r r           r          z
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Professor Fred Stern         Fall 2007                                               6

^         ^       ^
where V   r er   e   z e z

Assume  = 0 i.e. no swirl and fully developed
 z
flow        0 , which shows  r = constant = 0 since
z
r (R ) =0
^             ^
V  z e z  u ( r ) e z
Momentum:
DV
       ( p  z )   2 V
Dt

z equation:
u                



 V  u    ( p  z )   2u
 t              z

               1   u 
0   ( p  z )       r 
z  r   
               r r 
 
f ( z)          f (r )

 both terms must be constant

^
2
r dp                                              ^
u         A ln r  B                              p  p  z
4  dz

u (r  0) finite                            A=0
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Professor Fred Stern    Fall 2007                                                             7

^
2
R dp
u (r=R) = 0                            B
4  dz

^                                                      ^
r R d p
2      2                                                     2
R dp
u (r )                                         u           u (0)  
4  dz                                                      4  dz
m ax

 u     u
   r    
 z r 
          r                                                      As per CV analysis
^
rp

2 z
^
u            u          Rp
w                                
y r  R      r r  R    2 z
y=R-r,
^
 R 4 d p 1
R
Q   u (r )2 r dr               umax R 2
0
8 dz 2
^
Q       1       R2 d p
Vave         u 
 R2 2 max 8 dz

Substituting V = Vave

8 w
f 
V 2
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Professor Fred Stern           Fall 2007                                        8

R 8Vave 4Vave 8V
w                     
2 R  2
R     D

64    64
f          
DV Re D

L V 2 64  L V 2 32 LV
h  hL  f                                                         V
D 2 g DV D 2 g   gD 2

for z  0                     p  V

w                       16
or     Cf                          
1                         Re D
V 2
2

Both f and Cf based on V2 normalization, which is
appropriate for turbulent but not laminar flow. The more
appropriate case for laminar flow is:

Poiseuille #  P  C Re  16  0       f
for pipe flow

Compare with previous solution for flow between parallel
^
plates with p              x
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Professor Fred Stern        Fall 2007                                                   9

  y                                                h ^
2                                  2

u  u 1   
 h                     
               u              p
2
max                                         m ax            x

                         

 ^p 
3
4                   2h
q  hu                              
3            
m ax                   x
3

q   h  ^  2     2

v       p   u
2h 3      3
x        max

24 
 w  3V h
48
f               
Vh Re 2h

C  12                            Po =12
f
Re   2h

Same as pipe other than constants!
Exact laminar solutions are available for any “arbitrary”
cross section for laminar steady fully developed duct flow
058:0160                                                           Chapters 6
Professor Fred Stern        Fall 2007                                     10

BVP

u 0
x

^
0   p   (u  u )
x          YY       ZZ

u ( h)  0

2
Re only                                                                 h         ^
enters       y  y/h
*
z  z/h
*
u  u /U
*
U           ( p )
through                                                                              x

stability
and                                                                  Related umax
transition           u  1
2
Poisson equation

u(1) = 0                  Dirichlet boundary condition

Can be solved by many methods such as complex
variables and conformed mapping, transformation into
Laplace equation by redefinition of dependent variables,
and numerical methods.
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058:0160                           Chapters 6
Professor Fred Stern   Fall 2007          12
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Professor Fred Stern   Fall 2007          13
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Professor Fred Stern   Fall 2007          14
058:0160                                                 Chapters 6
Professor Fred Stern   Fall 2007                                15

Stability and Transition

Stability: can a physical state withstand a disturbance and

In fluid mechanics, there are two problems of particular
interest: change in flow conditions resulting in (1)
transition from one to another laminar flow; and (2)
transition from laminar to turbulent flow.

(1) Transition from one to another laminar flow

(a) Thermal instability: Bernard Problem

A layer of fluid heated from below is top heavy,
but only undergoes convective “cellular” motion
for
gd      gd 4bouyancy force
Raleigh #:             Ra 
w / d 2

k
 Racr
viscous force
1 
α = coefficient of thermal expansion =          
  T    P

  T / d   dT
dz
   1  T 
0

d = depth of layer
k, ν =thermal, viscous diffusivities
w=velocity scale: convection (w) = diffusion (k/d)
from energy equation, i.e., w=k/d
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Professor Fred Stern   Fall 2007                                          16

Solution for two rigid plates:

Racr = 1708          for progressive wave disturbance
^ i ( x ct )
αcr/d = 3.12         w  we
ct
 e i [cos( x  ct )  sin( x  ct )]
^ i ( x ct )
λcr = 2π/α = 2d      T T e
α = αr               c=cr + ici                  For temporal stability
αr = 2π/λ=wavenumber
cr = wave speed
Ra > 5 x 104 transition
ci : > 0             unstable
to turbulent flow
=0              neutral
<0              stable

Thumb curve: stable for low Ra < 1708 and very long or
short λ.
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Professor Fred Stern   Fall 2007                           17

(b) finger/oscillatory instability: hot/salty over
cold/fresh water and vise versa.

Rs  gd ds dz  /k
4

(Rs – Ra)cr = 657                                   S

   (1  T  S )
0
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Professor Fred Stern   Fall 2007                                      18

(c) Centrifugal instability: Taylor Problem

Bernard Instability: buoyant force > viscous force
Taylor Instability: Couette flow between two
rotating cylinders where centrifugal force (outward
from center opposed to centripetal force) > viscous
force.

r c  
2       2

Ta      i      i       o
c  r  r  r
   2              0   i     i

 centrifugal force / viscous force

Tacr            = 1708          αcrc = 3.12  λcr = 2c
Tatrans         = 160,000                        Square counter rotating vortex
pairs with helix streamlines
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058:0160                                          Chapters 6
Professor Fred Stern   Fall 2007                         20

(d) Gortler Vortices

Longitudinal vortices in concave curved wall
boundary layer induced by centrifugal force and
related to swirling flow in curved pipe or channel
later with regard to minor losses.

For δ/R > .02~.1       and   Reδ = Uδ/υ > 5
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Professor Fred Stern   Fall 2007                                                     21

(e) Kelvin-Helmholtz instability

Instability at interface between two horizontal
parallel streams of different density and velocity with
heavier fluid on bottom, or more generally
ρ=constant and U = continuous (i.e. shear layer
instability e.g. as per flow separation). Former case,
viscous force overcomes stabilizing density
stratification.

         
g 2  12  12 U1  U 2 
2                             2
 ci  0 (unstable)

U U
1      2
large α i.e. short λ always unstable

Vortex Sheet
1
1  2  ci  (U1  U 2 )
2
>0
Therefore always unstable
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058:0160                                                    Chapters 6
Professor Fred Stern   Fall 2007                                   23

(2) Transition from laminar to turbulent flow

Not all laminar flows have different equilibrium states,
but all laminar flows for sufficiently large Re become
unstable and undergo transition to turbulence.

Transition: change over space and time and Re range of
laminar flow into a turbulent flow.

U
Re             ~ 1000             δ = transverse viscous thickness

cr

Retrans > Recr              with       xtrans ~ 10-20 xcr

Small-disturbance (linear) stability theory can predict Recr
with some success for parallel viscous flow such as plane
Couette flow, plane or pipe Poiseuille flow, boundary
layers without or with pressure gradient, and free shear
flows (jets, wakes, and mixing layers).

Note: No theory for transition, but recent DNS helpful.
058:0160                                                                       Chapters 6
Professor Fred Stern         Fall 2007                                                24

Outline linearized stability theory for parallel viscous
flows: select basic solution of interest; add disturbance;
derive disturbance equation; linearize and simplify; solve
for eigenvalues; interpret stability conditions and draw
thumb curves.
^
u  u u                        u, v  mean flow, which is solution steady NS
^
v vv
^                       ^ ^
p  p p                        u,v  small 2D oscillating in time disturbance is

^       ^ ^             ^ ^            1^          ^
u t  u u x  u u x  v u y  v u y   p x   2 u

^       ^ ^             ^ ^            1^          ^
v t  u v x  u v x  v v y  v v y   p x   2 u

^ ^
ux  v x  0

^     ^    ^
Linear PDE for    u, v, p            for ( u , v ,   p ) known.

Assume disturbance is sinusoidal waves propagating in x
direction at speed c: Tollmien-Schlicting waves.
^                         i ( xct )
 ( x, y , t )   ( y ) e
^
Stream function
^                i ( xct )
u            e                                         y =distance across
y
shear layer
^
^                        i ( xct )
v             i e
x
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Professor Fred Stern       Fall 2007                         25

^     ^
u v 0
x     y
Identically!

    i = wave number = 2 
r       i

c  c  ic = wave speed = 
r        i


Where λ = wave length and ω =wave frequency

Temporal stability:
Disturbance (α = αr only and cr real)

ci     >0                unstable
=0                neutral
<0                stable

Spatial stability:
Disturbance (cα = real only)

αi     <0                unstable
=0                neutral
>0                stable
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Professor Fred Stern       Fall 2007                                                 26

^   ^
Inserting u , v into small disturbance equations and
^
eliminating p results in Orr-Sommerfeld equation:

inviscid Raleigh equation
i
(u  c)( ''  2 )  u ''          ( IV  2 2 ''  4 )
 Re

u  u /U                 Re=UL/υ                               y=y/L

4th order linear homogeneous equation with
homogenous boundary conditions (not discussed here) i.e.
eigen-value problem, which can be solved albeit not
easily for specified geometry and ( u , v , p ) solution to
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Professor Fred Stern   Fall 2007                         27

Although difficult, methods are now available for the
solution of the O-S equation. Typical results as follows

(1) Flat Plate BL:
U          

Re          520

crit

(2) αδ* = 0.35
 λmin = 18 δ* = 6 δ (smallest unstable λ)
 unstable T-S waves are quite large

(3) ci = constant represent constant rates of damping
(ci < 0) or amplification (ci > 0). ci max = .0196 is
small compared with inviscid rates indicating a
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Professor Fred Stern   Fall 2007                             28

(4) (cr/U0)max = 0.4  unstable wave travel at average
velocity.

(5) Reδ*crit = 520  Rex crit ~ 91,000

Exp: Rex crit ~ 2.8 x 106 (Reδ*crit = 2,400) if care taken,
i.e., low free stream turbulence
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Professor Fred Stern   Fall 2007                               29

Falkner-Skan Profiles:
Reδ*crit :
67 sep bl
(1) strong influence of                     520 fp bl
12,490 stag point bl
Recrit   > 0                 fpg

Recrit   < 0                 apg
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Professor Fred Stern   Fall 2007          31
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Professor Fred Stern   Fall 2007                      32

Extent and details of these processes depends on Re and
many other factors (geometry, pg, free-stream, turbulence,
roughness, etc).
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Professor Fred Stern   Fall 2007            33

Rapid development of span-
wise flow, and initiation of
nonlinear processes

- stretched vortices disintigrate
families of smaller and smaller
vortices
- onset of turbulence

Note: apg may undergo
much more abrupt
transition. However, in
general, pg effects less
on transition than on
stability
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Professor Fred Stern   Fall 2007          34
058:0160                                            Chapters 6
Professor Fred Stern   Fall 2007                           35

Some recent work concerns recovery distance:
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Professor Fred Stern   Fall 2007                         36

Turbulent Flow

Most flows in engineering are turbulent: flows over
vehicles (airplane, ship, train, car), internal flows (heating
and ventilation, turbo-machinery), and geophysical flows
(atmosphere, ocean).

V (x, t) and p(x, t) are random functions of space and
time, but statistically stationary flows such as steady and
forced or dominant frequency unsteady flows display
coherent features and are amendable to statistical analysis,
i.e. time and place (conditional) averaging. RMS and
other low-order statistical quantities can be modeled and
used in conjunction with averaged equations for solving
practical engineering problems.

Turbulent motions range in size from the width in the
flow δ to much smaller scales, which become
progressively smaller as the Re = Uδ/υ increases.
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Professor Fred Stern   Fall 2007          37
058:0160                                                             Chapters 6
Professor Fred Stern   Fall 2007                                            38

Physical description:

(1) Randomness and fluctuations:
Turbulence is irregular, chaotic, and unpredictable.
However, for statistically stationary flows, such as steady
flows, can be analyzed using Reynolds decomposition.

1   t0 T
1t 0 T

u  u  u'             u   u dT     u' 0       u '   u ' dT
2             2
etc.
T    t0                      T t0

u = mean motion
u ' = superimposed random fluctuation
u ' = Reynolds stresses; RMS = u '
2                                           2

Triple decomposition is used for forced or dominant
frequency flows

u  u  u' 'u'

Where u ' ' = organized oscillation

(2) Nonlinearity
Reynolds stresses and 3D vortex stretching are direct
results of nonlinear nature of turbulence. In fact,
Reynolds stresses arise from nonlinear convection term
after substitution of Reynolds decomposition into NS
equations and time averaging.
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Professor Fred Stern   Fall 2007                                           39

(3) Diffusion
Large scale mixing of fluid particles greatly enhances
diffusion of momentum (and heat), i.e.,
viscous stress
 

Reynolds Stresses:                   u ' u '    
i    j      ij        ij

2
Isotropic eddy viscosity:           u 'i u ' j  t  ij   ij k
3

Turbulence is characterized by flow visualization as
eddies, which vary in size from the largest Lδ (width of
flow) to the smallest. The largest eddies have velocity
scale U and time scale Lδ/U. The orders of magnitude of
the smallest eddies (Kolmogorov scale or micro scale)
are:
1
 3  4
LK = Kolmogorov micro-scale =  3 
U 
LK = O(mm) >> Lmean free path = 6 x 10-8 m
Velocity scale = (νε)1/4= O(10-2m/s)
Time scale = (ν/ε)1/2= O(10-2s)

Largest eddies contain most of energy, which break up
into successively smaller eddies with energy transfer to
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Professor Fred Stern          Fall 2007                                     40

yet smaller eddies until LK is reached and energy is
dissipated by molecular viscosity (i.e. viscous diffusion).
Richardson (1922):
Lδ Big whorls have little whorls
Which feed on their velocity;
And little whorls have lesser whorls,
LK And so on to viscosity (in the molecular sense).

(5) Dissipation
0    L
u0  k           k  u '2  v '2  w '2                      Energy comes from
 0 (U )                                                   largest scales and
Re  u0 0 /   l arg e
fed by mean motion

ε = rate of dissipation = energy/time

2                                                       Dissipation
u0                           0

o               o         u0
occurs at
smallest
scales
1
3
u0                                          3 4
=               independent υ              LK   
 
l0
058:0160                                           Chapters 6
Professor Fred Stern   Fall 2007                          41

Fig. below shows measurements of turbulence for
Rex=107.

Note the following mean-flow features:

(1) Fluctuations are large ~ 11% U∞

(2) Presence of wall cause anisotropy, i.e., the
fluctuations differ in magnitude due to geometric and
physical reasons. u ' is largest, v ' is smallest and reaches
2   2

its maximum much further out than u ' or w' . w' is
2   2     2

intermediate in value.

(3) u ' v'  0 and, as will be discussed, plays a very
important role in the analysis of turbulent shear flows.

(4) Although u u  0 at the wall, it maintains large values
i   j

right up to the wall

(5) Turbulence extends to y > δ due to intermittency.
The interface at the edge of the boundary layer is called
the superlayer. This interface undulates randomly
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Professor Fred Stern   Fall 2007                       42

between fully turbulent and non-turbulent flow regions.
The mean position is at y ~ 0.78 δ.

(6) Near wall turbulent wave number spectra have more
energy, i.e. small λ, whereas near δ large eddies dominate.
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Professor Fred Stern   Fall 2007          43
058:0160                                             Chapters 6
Professor Fred Stern   Fall 2007                            44

Averages:

For turbulent flow V (x, t), p(x, t) are random functions of
time and must be evaluated statistically using averaging
techniques: time, ensemble, phase, or conditional.

Time Averaging

For stationary flow, the mean is not a function of time and
we can use time averaging.

1 t0  t
u        u (t ) dt T > any significant period of u '  u  u
T t0
(e.g. 1 sec. for wind tunnel and 20 min. for ocean)

Ensemble Averaging

For non-stationary flow, the mean is a function of time
and ensemble averaging is used

1 N i
u (t )   u (t ) N is large enough that u independent
N i 1
ui(t) = collection of experiments performed under
identical conditions ( phase aligned for same
t=tref).
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Professor Fred Stern   Fall 2007          45
058:0160                                                         Chapters 6
Professor Fred Stern   Fall 2007                                        46

Phase and Conditional Averaging

Similar to ensemble averaging, but for flows with
dominant frequency content or other condition, which is
used to align time series for some phase/condition. In this
case triple velocity decomposition is used: u  u  u' 'u'
where u΄΄ is called organized oscillation.
Phase/conditional averaging extracts all three
components.

Averaging Rules:

f  f  f'                 g  g  g'            s = x or t

f ' 0               f  f              fg fg         f 'g  0

f  f
f g f g                                fg  f g  f ' g '
s s

 f ds   f ds
058:0160                                                   Chapters 6
Professor Fred Stern          Fall 2007                           47

Reynolds-Averaged Navier-Stokes Equations

For convenience of notation use uppercase for mean and
lowercase for fluctuation in Reynolds decomposition.

~
u i  U i  ui
~
p  P p
~
 ui
0
xi
~        ~        ~          ~                                   NS
 ui ~  ui      1 p      2 u i
 ui                        g i 3                       equation
t       xi     xi    x j x j

Mean Continuity Equation

             U u U
(U  u )              0
i
i                  i

x             x   x    x
i            i
i                            i         i        i

~
 u U u                u
         0 
i       i
0                i

x i
x x  i
x
i                    i

Both mean and fluctuation satisfy divergence = 0
condition.
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Professor Fred Stern                 Fall 2007                                                                                          48

Mean Momentum Equation

                                            1 
U i  ui   (U j  u j ) (U i  ui )         ( P  p) 
t                           x j              xi
2
(U i  ui )  g i 3
x j x j

             U u U
(U  u )          
i
i                               i

t             t   t   t
i       i

              U      u     U    u
(U  u )               (U  u )  U    U       u    u
i
i                                           i          i

x              x      x     x    x
j        j                i               i               j                       j                       j           j
j                                                j                           j               j              j

U    
U          uu                              i

x   x
j                               i       j
j                   j

        u    u    u
Since                      uu  u    u    u                    j                   i                       i

x        x    x    x
i       j           i                       j                   j
j                                   j                   j                       j

            P  p P
( P  p)       
x i
x x x               i               i               i

 g   gi3               i3
058:0160                                                                         Chapters 6
Professor Fred Stern        Fall 2007                                                   49

2                                  2U i      2 ui     2U i
              (U i  ui )                            
x j   2
x 2
j        x j
2
x 2
j

U i      U i  (ui u j )    1 P     2
U j                          2 U i  g i 3
t       x j    x j         xi    x j

DU i    1 P                                              U i          
Or              g i 3                                         ui u j 
Dt      xi            x j                              x j          

DU i              1 
Or         g i3         ij
Dt                xi                                                       RANS
 U U                                                       Equations
   P                  
 x  x    u u
ij
i        j
i   j

                    j        i

U
with             0             i

x               i

The difference between the NS and RANS equations is
the Reynolds stresses   u u , which acts like additional
i       j

stress: (1) This leads to the closure problem: 4 equations
10 unknowns for RANS versus 4 equations 4 unknowns
for NS, (2) NS and RANS paradox: NS deterministic with
non-deterministic solutions while RANS non-
deterministic with deterministic solutions.
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Professor Fred Stern          Fall 2007                                      50

 u u =  u u
i   j           j    i
(i.e. Reynolds stresses are symmetric)

   u 2   uv   uw
                            
    uv   v   2
  vw 
  uw   vw   w 2 
                            
u are normal stresses
2
i

uu i  j
i   j
are shear stresses

For isotropic turbulence u u i  j = 0 and u  v  w 
i   j
2     2      2

constant; however, turbulence is generally non-isotropic.

dU
For example, consider shear flow with                           0,
dy
V  (U  u, v, w) :                                  x-momentum tends
towards decreasing y as
v0  u0                                            turbulence diffuses
uv  0
v0  u0                                            gradients and decreases
dU
dy
x-momentum transport across y = constant AA per unit
area

 (U  u )v  U v   uv   uv

i.e  u u         i   j
=        average flux of j-momentum in
058:0160                                                           Chapters 6
Professor Fred Stern       Fall 2007                                      51

i-direction = average flux of
i-momentum in j-direction

Turbulent Kinetic Energy Equation

k  u  u  v  w  = turbulent kinetic energy
1   1  2            2       2       2
i
2   2

~
Subtracting NS equation for u i and RANS equation for Ui
results in equation for ui:

ui      ui      U i     ui                    1 p     2 ui
U j      uj      uj          (ui u j )          2
t      x j     x j     x j x j                xi   x j

Multiply by ui and average

Dk    1            1  2                                 U i
        pu j         ui u j  2      ui eij  uiu j       2 eij e ji
Dt     x j        2 x j             x j                x j

PT                         T         D             P

Dk k       k         1  ui u j           
   U j         e                       
Dt t      x j and ij 2  x                 
 j xi
Where

C
058:0160                                                    Chapters 6
Professor Fred Stern   Fall 2007                                   52

T= turbulent transport illustrates the rate at which TKE is
transported through the fluid by turbulent fluctuations.

PT =pressure transport is another form of turbulent
transport resulting from correlation of pressure and
velocity fluctuations.

D=viscous diffusion represents the diffusion of TKE
caused by the fluid’s natural molecular transport process.

P = shear production (usually > 0) represents loss of mean
kinetic energy and gain of turbulent kinetic energy due to
U
interactions of u u and                   .    i
i   j
x  j

ε = viscous dissipation represents the rate at which TKE is
converted into thermal internal energy.
 2 ui u j
                    
xi x j
0
where  is called pseudo-dissipation and defined by
ui ui
 
x j x j

C= turbulent convection represents the rate of TKE
convected by the mean flow
058:0160                                                      Chapters 6
Professor Fred Stern   Fall 2007                                     53

Recall previous discussions of energy cascade and
dissipation:
Energy fed from mean flow to largest eddies and cascades
to smallest eddies where dissipation takes place.
(1) Kolmogorov Hypothesis 1:
local isotropy: for large Re, micro-scale l l0 and
turbulence is isotropic.

Where eddies in the largest size range are characterized
by the length scale l0 , velocity scale u 0 , and turn-over
time scale  0 . The Reynolds number of these eddies is
Re0  u0l0 

All of the experimental evidence confirms that most
eddies break-up on a timescale of  0 . So the rate at which
energy per unit mass is passed down the energy cascade
from the largest eddies with size l0 is:
2  2      3
u0 u0     u0
             
 0  l0 u0  l0
For statistically steady conditions, this must match exactly
the rate of dissipation of energy at the smallest scale
  Sij Sij (from TKE equation), where the turbulent rate
1  u u j              
Sij   i                     Since Sij   u  , so:
of strain      2  x j xi

.

058:0160                                                              Chapters 6
Professor Fred Stern   Fall 2007                                             54

   u  2 
2

Apply             
3
  u  2 
u0         2

l0
It is also known that Re based on u and  is of order
unity: u   1 we get:

  3 / 
1/ 4

length        / l0  Re0 3/4


u   1 / 4

velocity       u / u0  Re01/4

   u   /  
1/2                    /  0  Re01/2

time
Micro-scale<<large scale
As Re increases, the range of scales increases. It is also
known as
(2) Kolmogorov’s first similarity hypothesis: for large
Re, micro-scale has universal form uniquely determined
by υ and ε.

(3) Kolmogorov’s second similarity hypothesis: for
large Re, intermediate scale ( l0 l  ) has a universal
form for u l and  l that are uniquely determined by  and
l and independent of υ, i.e., dimensional reasoning results
in:
ul   l   u  l     u0  l l0 
13           13            13
058:0160                                                            Chapters 6
Professor Fred Stern   Fall 2007                                           55

 l   l      l           0  l l0 
2     13           23                  23

Show that the velocity scales and time scales decrease as
l decreases.

(2) and (3) are called universal equilibrium range in
distinction from non-isotropic energy-containing range.
(2) is the dissipation range and (3) is the inertial subrange.

Spectrum of turbulence in the Inertial subrange

u   S (k ) dk
2
k = wave number in inertial subrange.
0
S  A 2 / 3k  5 / 3
For l01  k   1 (based on dimensional analysis)
A ~ 1.5          Called Kolmogorov k-5/3 law
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Professor Fred Stern   Fall 2007          56
058:0160                                                                  Chapters 6
Professor Fred Stern   Fall 2007                                                 57

Velocity Profiles: Inner, Outer, and Overlap Layers

Detailed examination of turbulent-flow velocity
profiles indicates the existence of a three-layer structure:

(1) a thin inner layer close to the wall ( y /   .02) where
turbulence is inhibited by the presence of the solid
boundary, i.e. u u are negligibly small (= 0 at the
i    j

wall) and the flow is controlled by molecular
viscosity.

(2) An outer layer                  (.2  y /   1)   where turbulent shear
dominates.

(3) An overlap layer (.02  y /   .2) where both types of
shear are important.

dimensional analysis and confirmed by experiment.

Inner law:             u  f ( ,  ,  , y)
w

           yu* 
u  u /u  f 
  
*
                          u*   w /    Wall shear
velocity
     

 f ( y )
058:0160                                                         Chapters 6
Professor Fred Stern   Fall 2007                                        58

u+, y+ are called inner-wall variables

Note that the inner law is independent of δ or r0, for
boundary layer and pipe flow, respectively.

Outer Law:                 Ue  u       g ( w ,  , y,  )   for px = 0

 
velocity defect

Ue  u
*
 g ( )        where   y / 
u

Note that the outer wall is independent of μ.

Overlap law: both laws are valid

It is not that difficult to show (also see AMF chapter VII)
that for both laws to overlap, f and g are logarithmic
functions:
Inner region:
2
dU u       df

dy      dy 
Outer region:
dU u dg

dy  d
058:0160                                                      Chapters 6
Professor Fred Stern   Fall 2007                                     59

2
y u         y u dg
df
     
        ; valid at large y+ and small η.
u     dy    u  d
f(y+)                g(η)

Therefore, both sides must equal universal constant,  1
1
f ( y )           ln y   B  u / u (inner variables)

1           U u
g ( )        ln   A  e                  (outer variables)
            u

 , A, and B are pure dimensionless constants
 =             0.41        Von Karman constant
Values vary
somewhat
depending on       B      =        5.5         (or 5.0)
different exp.
arrangements
A      =        2.35        BL flow The difference is due to
=        0.65        pipe flow loss of intermittency in
duct flow. A = 0 means
small outer layer

The validity of these laws has been established
experimentally as shown in Fig. 6-9, which shows the
profiles of Fig 6-8 in inner-law variable format. All the
profiles, with the exception of the one for separated flow,
058:0160                                                            Chapters 6
Professor Fred Stern    Fall 2007                                          60

are seen to follow the expected behavior. In the case of
separated flow, scaling the profile with u* is inappropriate
since u* ~ 0.

Details of Inner Law

u
Very near the wall,                      ~ constant
y
w

w
 u               y                  i.e. varies linearly


Or u+ = y+                    y+ ≤ 5

This is the linear sub-layer which must merge smoothly
with the logarithmic overlap law. This takes place in the
blending zone/buffer layer (5  y  30) . Evaluating the


RANS equation near the wall using an eddy-viscosity μt
turbulence model shows that:

μt ~ y3               y  0

Several expressions that satisfy this requirement have
been derived and are commonly used in turbulent-flow
analysis. That is:
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Professor Fred Stern     Fall 2007                                  61

 
t  e B eu  1  u  
u  2
 
                   2 
                        
Assuming the total shear is constant very near the wall the
equation for μt can be integrated to obtain a composite
formula which is valid in the sub-layer, blending layer,
and logarithmic-overlap regions.
 
y   u   e B eu  1  u  
u  2

   
 3
u 
                   2         6 
                                

Fig. 6-11 shows a comparison of this equation with
experimental data obtained very close to the wall. The
agreement is excellent. It should be recognized that
obtaining data this close to the wall is very difficult.

Details of the Outer Law

With pressure gradient included, the outer law
becomes (Fig 6-10):

Ue  u
 g ( ,  )                                (*)
*
u

 * dpc Clauser’s equilibrium
  y /                     
 w dx = parameter
058:0160                                                         Chapters 6
Professor Fred Stern       Fall 2007                                    62

Clauser (1954, 1956):

BL’s with different px but constant  are in equilibrium,
i.e., can be scaled with a single parameter:

Ue  u
*       vs. y / 
u

u              U
  defect thickness   e * dy   *
0 u

  2/C         f

Also, G = Clauser Shape parameter

1 U u 
2

   e *  dy                      6.1   1.81  1.7
 0 u 
Curve fit by Nash

Which is related to the usual shape parameter by

H  1  G /  1  const. due to    ( x)

Finally, Clauser showed that the outer layer has a wake-
like structure such that
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Professor Fred Stern   Fall 2007                                         63

t  0.016 U e *                                     (**)

Mellor and Gibson (1966) combined (*) and (**) into a
theory for equilibrium outer law profiles with excellent
agreement with experimental data: Fig. 6-12

Coles (1956):

A weakness of the Clauser approach is that the
equilibrium profiles do not have any recognizable shape.
This was resolved by Coles who showed that:
Deviations above log-overlap layer

u   2.5 ln y   5.5  1
 W(y / )                     (***)
U e   2.5 ln    5.5 2

Max deviation at δ            Single wake-like function of y/δ

 y 
W  wake function  2 sin       3  2 ,   y / 
2     2    3

  2  

curve fit
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Professor Fred Stern   Fall 2007                                         64

Thus, from (***), it is possible to derive a composite
function, which covers both the overlap and outer layers,
as shown in Fig 6-13.

1               
u             ln y   B  W ( y /  )
                   

  wake parameter = π(β)
 0.8(   0.5) 0.75                           (curve fit to data)

Note that the agreement of Coles’ wake law even for β 
constant Bl’s is quite good.

We see that the behavior in the outer layer is more
complex that that of the inner layer due to pressure
gradient effects. In general, the above velocity profile
correlations are extremely valuable both in providing
physical insight and, as we shall see, in providing
approximate solutions for simple wall bounded
geometries: pipe, channel, and flat plate boundary layer.
Furthermore, such correlations have been extended
through the use of additional parameters to provide
velocity formulas for use with integral methods for
solving the b.l. equations for arbitrary px. Also used for
CFD turbulence modeling.
058:0160                                                                    Chapters 6
Professor Fred Stern    Fall 2007                                                  65

6-4 Summary of Inner, Outer, and Overlap Layers: Mean
velocity correlations


Inner layer: U  U / u y   y / u  u*   w / 
*

Sub-layer: 0  y  5         U+ = y+


Buffer layer: 5  y  30        
region where sub-layer
merges smoothly
with log law
Outer Layer:

Ue U                                                 *
u*
 g ( ,  )             y / ,              px
w

Overlap layer (log region):

1
U             ln y   B                 inner variables


Ue  U             1
  ln   A               outer variables
u   *           

Composite Inner/Overlap layer correlation
                (U  )2 (U  )3 
U   y   e B eU  1  U                    
                    2       6 

Composite Overlap/Outer layer correlation
058:0160                                                                  Chapters 6
Professor Fred Stern   Fall 2007                                                 66

1                2     

U  ln y  B     
W ( ) W  sin 2     3 2  2 3
                             2 
  0.8(   0.5) 0.75

Equilibrium Boundary Layers                            (  = constant)

Ue U
*
vs. y /  ,        ,*
  2/C           f
u
(shows similarity)

Reynolds Number Dependence of Mean-Velocity
correlations and Reynolds stresses

1. inner/overlap U+ scaling shows similarity; extent of
overlap region (i.e. similarity) increases with Reθ.
Fig 10, 11, 12

2. outer layer may asymptotically approach similarity
for large Re as shown by U ( 2 /  ) vs. Reθ, but


controversial due to lack of data for Reθ > 5 x 104.
Fig 13, 14

2
3. streamwise turbulence intensity                   u  u            vs. y+
u*
shows similarity for 0  y  15 (i.e., just beyond the


point of kmax, y+ = 12). u+max vs. Reθ increases with
Reθ. Fig 18, 27
058:0160                                      Chapters 6
Professor Fred Stern   Fall 2007                     67

4. v+ and uv do not show similarity: peak values


increase with Rand move away from the wall. Fig
25, 31, 34
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058:0160                                                                    Chapters 6
Professor Fred Stern          Fall 2007                                            74

6-5 Turbulent Flow in Pipes and Channels using mean-
velocity correlations

Circular pipe
Recall laminar flow exact solution
8 w
f            64 / Re D           Re D 
uave D
 2000
 uave
2

A turbulent-flow “approximate” solution can be obtained
simply by computing uave based on log law.

1        yu*                   u  u ( y )   .41,
u                  ln         B              B  5 u*   w / 
u   *                                    y  Rr       dy   dr

1 R *  1 ( R  r )u *     
V  uave  Q / A  2  u  ln              B  2r dr
R 0                    

1 *  2 Ru*
 ln          3
 u           2R  
2             


V             Ru *
 2.44 ln       1.34
Or                   u *

f 1/ 2  1.99 log(Re D f 1/ 2 )  1.02
f 1 / 2  2 log(ReD f 1 / 2 )  0.8                  constants

f only drops by a factor of 5 over 104 < Re < 108
058:0160                                                            Chapters 6
Professor Fred Stern     Fall 2007                                         75

Turbulent Flow in Rough Pipes

U   f ( y , k  )                     f  f (Re d , k / d )

1
U             ln y   B  B(k  )                 Log law shifts downward


which leads to three roughness regimes:

1. k+ < 4                     hydraulically smooth
2. 4 < k+ < 60                transitional roughness (Re dependence)
3. k+ > 60                    full rough (no Re dependence)

1 / 2          k / d     2.51            Moody diagram
f           2 log                     

  3.7 Re d f 1 / 2 
approximate explicit
formula
 6.9  k / d 1.11 
~ 1.8 log             

 Re d  3.7      

There are basically three types of problems involved with
uniform flow in a single pipe:

1. Determine the head loss, given the kind and size of pipe
along with the flow rate, Q = A*V
2. Determine the flow rate, given the head, kind, and size of
pipe
3. Determine the pipe diameter, given the type of pipe, head,
and flow rate
058:0160                                              Chapters 6
Professor Fred Stern   Fall 2007                             76

The first problem of head loss is solved readily by obtaining f
from the Moody diagram, using values of Re and ks/D
computed from the given data. The head loss hf is then
computed from the Darcy-Weisbach equation.

f = f(Red, ks/D)

L V2                   p         p         
hf  f       h        h   2  z 2    1  z1 
D 2g                                     
p 
=    z 
    
Red = Red(V, D)

2. Determine the flow rate
The second problem of flow rate is solved by trial, using a
successive approximation procedure. This is because both
Re and f(Re) depend on the unknown velocity, V. The
solution is as follows:
1) solve for V using an assumed value for f and the Darcy-
Weisbach equation
1/ 2
 2gh f 
V                  f 1/ 2
L/D   

known from           note sign
given data
2) using V compute Re
058:0160                                                Chapters 6
Professor Fred Stern   Fall 2007                               77

3) obtain a new value for f = f(Re, ks/D) and reapeat as
above until convergence
1/ 2
D 3 / 2  2gh f 
Or can use Re  f 1/ 2                 
  L 
scale on Moody Diagram

1) compute Re f 1/ 2 and ks/D
L V2
3) solve V from h f  f
D 2g
4) Q = VA

3. Determine the size of the pipe
The third problem of pipe size is solved by trial, using a
successive approximation procedure. This is because hf, f,
and Q all depend on the unknown diameter D. The solution
procedure is as follows
1) solve for D using an assumed value for f and the Darcy-
Weisbach equation along with the definition of Q
1/ 5
 8LQ 2 
D 2                 f 1/ 5
  gh f 

known from
given data
2) using D compute Re and ks/D

3) obtain a new value of f = f(Re, ks/D) and reapeat as
above until convergence
058:0160                           Chapters 6
Professor Fred Stern   Fall 2007          78
058:0160                                                    Chapters 6
Professor Fred Stern     Fall 2007                                 79

Concept of hydraulic diameter for noncircular ducts

For noncircular ducts, τw= f(perimeter); thus, new
8
definitions of f  w and C f  2 w are required.
V 2            V 2

Define average wall shear stress

1P
 w    w ds            ds = arc length, P =
P0
perimeter of duct

Momentum:

 z 
pA   w PL  AL    0
 L 
W

 wL
h    p /   z              A/P=Rh=hydraulic radius
 A/ P

circle: A/P = R2/2R=R/2=D/4

Energy:

 wL
h  hL 
 A/ P
058:0160                                                                     Chapters 6
Professor Fred Stern     Fall 2007                                                  80

   ^
A h  A dh  A d ( p   z ) A  d p 
w 
P L

P dx

P       dx
 
P  dx 
non-circular duct

     
A/P = length= Dh/4                                Dh = 4A/P           hydraulic
diameter

For multiple surfaces such as concentric annulus P and A
based on wetted perimeter and area

8 w                                                  VDh
f          f (Re Dh ,  / Dh )                Re Dh 
V 2                                                   

L V 2 f     L V2      L V2
h  hL               f         f
 Rh 8        4Rh 2 g    Dh 2 g

However, accuracy not good for laminar flow ( 40 %) and
marginal turbulent flow ( 15 %)

8 w
fVDh /          VD /   64      using         8V / Dh i.e.   exact solution
V 2 h
w

f
Cf        64 / Re Dh  16 / Re D
4
P 0  C f Re Dh  64  16                                          Circular pipe
058:0160                                          Chapters 6
Professor Fred Stern   Fall 2007                         81

For laminar flow, P     0

varies greatly,
therefore it is better to
use the exact solution

For turbulent flow, Dh works much better especially if
combined with “effective diameter” concept based on
ratio of exact laminar circular and noncircular duct P0
numbers, i.e. 16 / P = exact laminar solution for
0

noncircular duct.

First recall turbulent circular pipe solution and compare
with turbulent channel flow solution using log-law in both
cases
058:0160                                                                 Chapters 6
Professor Fred Stern          Fall 2007                                         82

1        yu*                   u  u ( y )   .41,
u                  ln         B              B  5 u*   w / 
u   *                                    y  Rr     dy   dr

1 R *  1 ( R  r )u *     
V  uave  Q / A  2  u  ln              B  2r dr
R 0                    

1 *  2 Ru*
 ln          3
 u           2R  
2             


V             Ru *
 2.44 ln       1.34
Or                   u *

f 1/ 2  1.99 log(Re D f 1/ 2 )  1.02
f 1 / 2  2 log(ReD f 1 / 2 )  0.8               constants

f only drops by a factor of 5 over 104 < Re < 108
058:0160                                                                Chapters 6
Professor Fred Stern   Fall 2007                                               83

Since f equation is implicit, it is not easy to see
dependency on ρ, μ, V, and D

f   ( pipe)  0.316 Re1 / 4
D
4000 < ReD < 105
Blasius (1911) power law
curve fit to data
p   L V2
hf     f
     D 2g

p  0.158 L 3 / 4 1 / 4 D 5 / 4v 7 / 4

Drops weakly with
Only slightly       pipe size
Nearly linear       with μ

 0.241L 3 / 4  1/ 4 D 4.75Q1.75

laminar flow: p  8LQ / R 4

p (turbulent) increases more sharply than p (laminar)
for same Q; therefore, increase D for smaller p . 2D
decreases p by 27 for same Q.

Ru*
 u (r  0) *  ln
umax                                1
B
u *
u       
V  2.44 ln  Ru *
Combine with                         1.34
u*          
058:0160                                                                Chapters 6
Professor Fred Stern       Fall 2007                                           84

V
u max  1  1.3 f                1

Recall laminar flow: V / u max  0.5

Channel Flow

Laminar Solution C f  Re h  24 Re D P0 = 6 based h=
6
h

half width (=24 based on Dh=4h)

1 h *  1 (h  y )u*     
V   u  ln             B  dY Y=h-y wall coordinate
h 0                  


* 1
       1     hu*
u    ln   B 
        
           

VDh                             A=2h*w
Define Re Dh                           Dh 
4A
 lim
4(2hw)
 4h
P=4h+2w
          P    w 2 w  4h

           
f 1/ 2  2log Re Dh f 1/ 2  1.19

Very nearly the same as circular pipe
7% to large at Re = 105
4% to large at Re = 108
058:0160                                                                    Chapters 6
Professor Fred Stern         Fall 2007                                             85

Therefore error in Dh concept relatively smaller for
turbulent flow.

Note           f   1 / 2
(channel )  2 log0.64 Re f
Dh
1/ 2
  0.8
P0 (circle)  16
Define Deffective  0.64 Dh ~                                  Dh
P0 (channel)  24

Laminar solution

(therefore, improvement on Dh is)

VDeff                     16
Re Deff                            Deff             Dh
                   P0 (lam)

From exact laminar solution
058:0160                           Chapters 6
Professor Fred Stern   Fall 2007          86
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Professor Fred Stern   Fall 2007          87
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Professor Fred Stern   Fall 2007         102
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