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Jianming Yang
Fall 2012
Chapter 6: Viscous Flow in Ducts
1 Laminar Flow
1.1 Entrance, developing, and fully developed flow
Chapter 6
1
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Jianming Yang
Chapter 6
2
Fall 2012
Entry length
𝐿𝑒 = 𝑓(𝐷, 𝑉, , 𝜇)
From Π theorem
𝐿𝑒 /𝐷 = 𝑓(𝑅𝑒)
Recrit ~ 2300
Re < Recrit laminar
Re > Recrit unstable
Re > Retrans turbulent
Laminar Flow:
𝐿𝑒 /𝐷 ≅ 0.06𝑅𝑒
Maximum entry length for laminear flow: 𝐿𝑒 = 0.06𝑅𝑒crit 𝐷 ~ 138𝐷
Turbulent flow:
𝐿𝑒 /𝐷~4.4𝑅𝑒 1/6
Recent CFD results indicate
𝐿𝑒 /𝐷~1.6𝑅𝑒 1/6 for 𝑅𝑒 < 107
Relatively shorter than for laminar flow
Some computed turbulent entry length estimates
𝑅𝑒
4000
104
105
106
𝐿𝑒 /𝐷
13
16
28
51
107
90
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Jianming Yang
Fall 2012
Chapter 6
3
Laminar vs. Turbulent Flow
Left: Hagen (1839) noted difference in Δ𝑝 = Δ𝑝(𝑉) but could not explain two regimes
Reynolds 1883 showed that the difference depends on 𝑅𝑒 = 𝑉𝐷/𝜐
Right: Reynolds’ sketches of pipe flow transition: (a) low-speed, laminar flow; (b) highspeed, turbulent flow; (c) spark photograph of condition (b)
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Jianming Yang
Chapter 6
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Fall 2012
1.2 Laminar Pipe Flow
1.2.1 CV Analysis
Continuity:
0   V  dA  Q1  Q2  const.
CS
i.e. V1  V2
sin ce
A1  A2 ,   const., and V  Vave
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Chapter 6
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Fall 2012
Momentum:
F
x

 ( p1  p2 ) R   w 2 RL   R L sin   m(  2V2  1V1 )
W z / L
0
p
2
2
pR   2RL  R z  0
2
2
w
p  z 
2 L
R
w
h  h1  h2  ( p /   z ) 
2 w L
 R
or
w 
R h
R dh
Rd


( p  z )
2 L
2 dx
2 dx
For fluid particle control volume:
r d
 
( p  z )
2 dx
i.e. shear stress varies linearly in 𝑟 across pipe for either laminar or turbulent flow
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Chapter 6
6
Fall 2012
Energy:
p1


1
2g
p2
V1 2  z1 

h  h 
L

2
2g
V22  z 2  hL
2 L
 R
w
 once τw is known, we can determine pressure drop
In general,
   (  ,V ,  , D,  )
w
Πi Theorem
8 w
V
2
w
where 𝜀 is wall roughness height
 f  friction factor  f (Re D ,  / D )
where
Re D 
VD

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.
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Fall 2012
Chapter 6
7
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Jianming Yang
Fall 2012
Chapter 6
8
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Chapter 6
9
Fall 2012
1.2.2 Differential Analysis
Continuity:
 V  0
Use cylindrical coordinates (r, θ, z) where z replaces x in previous CV analysis
1
1 
vz
(rv r ) 
(v ) 
0
r r
r 
z
where V  vr eˆr  v eˆ  vz eˆz
vz
Assume v = 0 i.e. no swirl and fully developed flow
 0 , which shows v r =
z
constant = 0 since vr (R ) =0
V  vz eˆz  u (r ) eˆz
Momentum:
DV

 ( p  z )   2 V
Dt
z equation:
u

   V  u    ( p  z )   2 u
z
 t

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Chapter 6
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Fall 2012

1   u 
0   ( p  z )  
r 
z  r
r  

r


f (z)
f (r )
 both terms must be constant
 
u
p
ˆ
)
r r r
z
u
1 p
ˆ 2
r

r A
r 2  z
u
1 p
1
ˆ


rA
r 2  z
r
1 p
ˆ 2
u 
r  A ln r  B
4  z
u ( r  0)
(r
 A=0
R 2 dpˆ
 B
4 dz
finite
u (r=R) = 0
r 2  R 2 dpˆ
u (r ) 
4 dz
umax
R 2 dpˆ
 u (0)  
4 dz
pˆ  p  z
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Chapter 6
11
Fall 2012
v
u
u
y r R
r pˆ

2 z
u
R pˆ
 

r r R
2 z
u
    r    
fluid shear stress
r
 z r 
w  
As per CV analysis
y=R-r,
𝑅
−𝜋𝑅4 𝑑𝑝̂ 1
𝑄 = ∫ 𝑢(𝑟)2𝜋𝑟𝑑𝑟 =
= 𝑢max 𝜋𝑅2
8𝜇 𝑑𝑧 2
0
𝑄
1
−𝑅2 𝑑𝑝̂
𝑉𝑎𝑣𝑒 =
= 𝑢
=
𝜋𝑅2 2 max
8𝜇 𝑑𝑧
Substituting V = Vave
8
f  w2
V
R 8Vave 4Vave 8V


2
2 R
R
D
w   
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Chapter 6
12
Fall 2012
f 
64
64

DV Re D
or
Cf 
w
1
V 2
2
L V 2 64 L V 2 32LV
h  hL  f

 

D 2 g DV D 2 g
gD 2
for z  0

f
16

4 Re D
V
 p  V
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:
P0 c f  C f Re  16
Poiseuille # ( P0 )
P0 f  f Re  64
for pipe flow
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Chapter 6
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Fall 2012
Compare with previous solution for flow between parallel plates with p̂x
  y
u  u 1   
 h
max
2



4
2h 3
 pˆ x 
q  humax 
3
3
q h2
 pˆ x   2 umax
v

2h 3
3
 w  3V h
umax
 h2

pˆ x
2
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Chapter 6
14
Fall 2012
f 
24
48
96


Vh Re 2 h Re
4h

Re Dh
Cf  f / 4 
Cf 
6
12
24


Vh Re 2 h Re
4h

Re Dh
 P0 c  C f Re D  24
Poiseuille# ( P0 )
 P0 f  f Re D  96
h
f
h
Same as pipe other than constants!
P0 c f
P0 c f
pipe
channel based on Dh

P0 f
P0 f
pipe
channel based on Dh

16 64 2


24 96 3
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Chapter 6
15
Fall 2012
1.2.3 Noncircular Ducts
Exact laminar solutions are available for any “arbitrary” cross section for laminar steady
fully developed duct flow
BVP
𝜕𝑢
=0
𝜕𝑥
𝜕𝑝̂
𝜕2𝑢 𝜕2𝑢
0=−
+ 𝜇 ( 2 + 2)
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝑢(𝑤𝑎𝑙𝑙 ) = 0
𝑦
𝑧
∗
𝑦 =
𝑦𝑧 =
ℎ
ℎ
∗
∇2 𝑢∗ = −1
𝑢∗ (𝑤𝑎𝑙𝑙 ) = 0
𝑢
𝑢 =
𝑈
∗
ℎ2
𝜕𝑝̂
𝑈 = (− )
𝜇
𝜕𝑥
Poisson equation
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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Fall 2012
Chapter 6
16
2 Stability and Transition
Stability: a physical state can withstand a disturbance and still return to its original state.
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Chapter 6
17
Fall 2012
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.
2.1 Transition from one to another laminar flow
2.1.1 Thermal instability: Bernard Problem
A layer of fluid heated from below is top heavy, but only undergoes convective
“cellular” motion for Raleigh number larger than the critical value
Raleigh number: 𝑅𝑎 =
1 𝜕𝜌
𝑔𝛼𝛤𝑑
𝜐𝑤/𝑑 2
=
𝑔𝛼𝛤𝑑 4
𝑘𝜐
> 𝑅𝑎𝑐𝑟
bouyancy force
viscous force
α=− ( )
coefficient of thermal expansion
𝛤=
vertical temperature gradient
𝜌 𝜕𝑇 𝑃
∆𝑇
𝑑𝑇
𝑑
=−
𝑑𝑧
𝜌 = 𝜌0 (1 − 𝛼∆𝑇)
𝑑
𝑘, 𝜈
𝑤 = 𝑘/𝑑
density
depth of layer
thermal, viscous diffusivities
velocity scale: convection (𝑤𝛤) ~ diffusion (𝑘𝛤/𝑑)
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Chapter 6
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Solution for two rigid plates:
Racr = 1708
for progressive wave disturbance
http://www.youtube.com/watch?v=xb_pHQzEFJg
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Jianming Yang
2.1.2
Chapter 6
19
Fall 2012
Double-diffusive instability:
e.g., hot/salty over cold/fresh water and vise versa.
𝜌 = 𝜌0 [1 − 𝛼(𝑇 − 𝑇0 ) + 𝛽(𝑆 − 𝑆0 )]
𝑅𝑎 =
𝑆
𝑔𝛼(𝑑𝑇/𝑑𝑧)𝑑 4
𝑘𝜐
𝑅𝑠 =
𝑔𝛽(𝑑𝑆/𝑑𝑧)𝑑 4
𝑘𝜐
salt content
1 𝜕𝜌
𝛽=− ( )
𝜌 𝜕𝑆 𝑃
determines how fast the density increases with salinity
(𝑅𝑠 – 𝑅𝑎)𝑐𝑟 = 657
http://salty.oce.orst.edu/SC07_Web/index.html
http://www.youtube.com/watch?v=1d1vPyDQHbA
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Chapter 6
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Fall 2012
2.1.3 Centrifugal instability: Taylor Problem
Taylor Instability: Couette flow between two rotating cylinders where centrifugal
force (outward from center opposed to centripetal force) > viscous force.
Taylor number : 𝑇𝑎 =
𝑇𝑎𝑐𝑟 =
Ω1 𝑅12 −Ω2 𝑅22 Ω1 𝑑 4
4 ( 2 2 ) 2 (𝑑 = 𝑅2 − 𝑅1 ≪ 𝑅1 )
𝑅2 −𝑅1
𝜐
centrifugal force
viscous force
1708
0.5(1+Ω2 /Ω1 )
http://atoc.colorado.edu/TeachingandLearning/Demonstrations/TaylorCouette/Taylorcouette.htm
http://www.youtube.com/watch?v=cEqvx0N_txI
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Chapter 6
21
Fall 2012
2.1.4 Gortler Vortices
Longitudinal vortices in concave curved wall boundary layer induced by centrifugal
force and related to swirling flow in curved pipe or channel induced by radial
pressure gradient and discussed later with regard to minor losses.
For δ/R > .02~.1
and
Reδ = Uδ/υ > 5
http://media.efluids.com/galleries/boundary?medium=221
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Jianming Yang
2.1.5
Chapter 6
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Fall 2012
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.
http://en.wikipedia.org/wiki/Kelvin%E2%80%93Helmholtz_instability
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2.2 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.
𝑅𝑒𝑐𝑟 =
𝑈𝛿
𝜐
~1000
𝑅𝑒trans > 𝑅𝑒𝑐𝑟
𝛿 = transverse viscous thickness
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.
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Chapter 6
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Fall 2012
Outline linearized stability theory for parallel viscous flows:
a) Select basic solution of interest;
b) Add disturbance;
c) Derive disturbance equation;
d) Linearize and simplify;
e) Solve for eigenvalues;
f) Interpret stability conditions and draw thumb curves.
𝑢 = 𝑢̅ + 𝑢̂
𝑣 = 𝑣̅ + 𝑣̂
𝑝 = 𝑝̅ + 𝑝̂
u̅, v̅: mean flow, which is solution of steady NS
𝑢̂, 𝑣̂: small 2D disturbance oscillating in time, which is solution of unsteady NS
̅
𝜕𝑢
𝜕𝑡
𝜕𝑣̅
𝜕𝑡
̂
𝜕𝑢
𝜕𝑡
𝜕𝑣̂
𝜕𝑡
̅
𝜕𝑢
𝜕𝑥
+ 𝑢̅
+ 𝑢̅
+ 𝑢̅
+ 𝑢̅
+
̅
𝜕𝑢
𝜕𝑥
𝜕𝑣̅
𝜕𝑥
̂
𝜕𝑢
𝜕𝑥
𝜕𝑣̂
𝜕𝑥
𝜕𝑣̅
𝜕𝑦
+ 𝑣̅
+ 𝑣̅
+ 𝑢̂
+ 𝑢̂
̅
𝜕𝑢
𝜕𝑦
𝜕𝑣̅
𝜕𝑦
̅
𝜕𝑢
𝜕𝑥
𝜕𝑣̅
𝜕𝑥
=0 →
=−
=−
+ 𝑣̅
+ 𝑣̅
1 𝜕𝑝̅
𝜌 𝜕𝑥
1 𝜕𝑝̅
𝜌 𝜕𝑦
̂
𝜕𝑢
𝜕𝑦
𝜕𝑣̂
+ 𝜐∇2 𝑢̅
+ 𝜐∇2 𝑣̅
+ 𝑣̂
+ 𝑣̂
̅
𝜕𝑢
𝜕𝑦
𝜕𝑣̅
𝜕𝑦
𝜕𝑦
̂
𝜕𝑢
𝜕𝑣̂
𝜕𝑥
+
𝜕𝑦
=−
=−
1 𝜕𝑝̂
𝜌 𝜕𝑥
1 𝜕𝑝̂
𝜌 𝜕𝑦
+ 𝜐∇2 𝑢̂
+ 𝜐∇2 𝑣̂
=0
Linear PDE for 𝑢̂, 𝑣̂, 𝑝̂ , for (u̅, v̅, 𝑝̅ ) known.
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Chapter 6
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Fall 2012
Assume disturbance is sinusoidal waves propagating in 𝑥 direction at speed 𝑐: TollmienSchlicting (T-S) waves.
Stream function
̂ (𝑥, 𝑦, 𝑡) = 𝜙(𝑦)𝑒 𝑖𝛼(𝑥−𝑐𝑡)
𝛹
𝑦 is distance across shear layer
𝑢̂ =
̂
𝜕𝛹
𝜕𝑦
̂
𝜕𝛹
𝑣̂ = −
̂
𝜕𝑢
𝜕𝑥
+
= 𝜙 ′ (𝑦)𝑒 𝑖𝛼(𝑥−𝑐𝑡) ,
𝜕𝑥
𝜕𝑣̂
𝜕𝑦
= −𝑖𝛼𝜙(𝑦)𝑒 𝑖𝛼(𝑥−𝑐𝑡) :
=0
Identically!
Wave number:
𝛼 = 𝛼𝑟 + 𝑖𝛼𝑖 = wave number 2𝜋/𝜆
𝑐 = 𝑐𝑟 + 𝑖𝑐𝑖 = wave speed 𝜔/𝛼
Where λ = wave length and ω =wave frequency
Temporal stability:
Disturbance (𝛼 = 𝛼𝑟 only and 𝛼𝑟 real)
𝑐𝑖 > 0
unstable
=0
neutral
<0
stable
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Fall 2012
Chapter 6
26
Spatial stability:
Disturbance (𝑐 = real only)
𝛼𝑖 < 0
unstable
=0
neutral
>0
stable
Inserting 𝑢̂, 𝑣̂ into small disturbance equations and eliminating 𝑝̂ results in OrrSommerfeld equation:
𝑖
′′
2
′′
(𝑢 − 𝑐 )(𝜙 − 𝛼 𝜙) − 𝑢 𝜙 = −
(𝜙 ′′′′ − 2𝛼 2 𝜙 ′′ + 𝛼 4 𝜙)
𝛼𝑅𝑒
𝑢 = 𝑢̅/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 (u̅, v̅, 𝑝̅ ) solution to steady NS.
Although difficult, methods are now available for the solution of the O-S equation.
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Chapter 6
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Fall 2012
3 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 (root mean square) 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.
3.1 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.
𝑢 = 𝑢̅ + 𝑢′
mean motion:
1
𝑡 +𝑇
𝑢̅ = ∫𝑡 0
𝑇
0
𝑢𝑑𝑇
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Chapter 6
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Fall 2012
superimposed random fluctuation: 𝑢′
𝑢̅′ = 0
1 𝑡 +𝑇 2
̅̅̅̅
Reynolds stresses:
𝑢′ 2 = ∫𝑡 0 𝑢′ 𝑑𝑇 (RMS = √̅̅̅̅
𝑢′ 2 )
𝑇
0
(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.
(3) Diffusion
Large scale mixing of fluid particles greatly enhances diffusion of momentum/heat.
(4) Vorticity/eddies/energy cascade
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. Largest eddies contain most of energy, which break up into successively
smaller eddies with energy transfer to yet smaller eddies until LK is reached and energy is
dissipated by molecular viscosity (i.e. viscous diffusion).
(5) Dissipation
Energy comes from largest scales and fed by mean motion. Dissipation occurs at
smallest scales.
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Fall 2012
Chapter 6
29
3.2 Averages
For turbulent flow 𝐕(𝐱, 𝑡) and 𝑝(𝐱, 𝑡) are random functions of time and must be
evaluated statistically using averaging techniques: time, ensemble, phase, or conditional.
3.2.1
Time Averaging
For stationary flow, the mean is not a function of time and we
can use time averaging.
1
𝑡 +𝑡
𝑢̅ = ∫𝑡 0
𝑇
0
𝑢(𝑡)𝑑𝑡
𝑇 > any significant period of 𝑢′ = 𝑢 − 𝑢̅
(e.g. 1 sec. for wind tunnel and 20 min. for ocean)
3.2.2
Ensemble Averaging
For non-stationary flow, the mean is a function of time and
ensemble averaging is used
1
𝑖
𝑢̅(𝑡) = ∑𝑁
𝑖=1 𝑢 (𝑡)
𝑁
𝑁 is large enough that 𝑢̅ is independent
𝑢𝑖 (𝑡) = collection of experiments performed under identical
conditions (also can be phase aligned for same t=0).
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3.2.3
Chapter 6
30
Fall 2012
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 = 𝑢̅ + 𝑢′′ + 𝑢′ where 𝑢′′
is called organized oscillation. Phase/conditional
averaging extracts all three components.
3.2.4
Averaging Rules
𝑓 = 𝑓̅ + 𝑓 ′
𝑔 = 𝑔̅ + 𝑔′
𝑓 = 𝑓(𝑠) and 𝑔 = 𝑔(𝑠) with 𝑠 = 𝐱 or 𝑡
𝑓̅′ = 0
̅̅̅̅
𝜕𝑓
𝜕𝑠
=
𝜕𝑓̅
𝜕𝑠
𝑓̿ = 𝑓̅
̅̅̅̅
𝑓𝑔̅ = 𝑓 ̅𝑔̅
̅̅̅̅
𝑓𝑔 = 𝑓 ̅𝑔̅ + ̅̅̅̅̅̅
𝑓 ′ 𝑔′
̅̅̅̅̅
𝑓 ′ 𝑔̅ = 0
̅̅̅̅̅̅̅
𝑓𝑑𝑠 = ∫ 𝑓 ̅𝑑𝑠
∫
̅̅̅̅̅̅̅
𝑓 + 𝑔 = 𝑓 ̅ + 𝑔̅
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Chapter 6
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Fall 2012
3.3 Reynolds-Averaged Navier-Stokes Equations
For convenience of notation use uppercase for mean and lowercase for fluctuation in
Reynolds decomposition.
𝑢̃𝑖 = 𝑈𝑖 + 𝑢𝑖
𝑝̃ = 𝑃 + 𝑝
Navier-Stokes equations for incompressible flow with constant fluid properties
̃𝑗
𝜕𝑢
𝜕𝑥𝑗
̃𝑖
𝜕𝑢
𝜕𝑡
3.3.1
̃𝑗
𝜕𝑢
𝜕𝑥𝑗
̅̅̅̅
̃𝑗
𝜕𝑢
𝜕𝑥𝑗
̃𝑗
𝜕𝑢
𝜕𝑥𝑗
+ 𝑢̃𝑗
̃𝑖
𝜕𝑢
𝜕𝑥𝑗
=−
=0
1 𝜕𝑝̃
𝜌 𝜕𝑥𝑖
+𝜐
̃𝑖
𝜕2 𝑢
𝜕𝑥𝑗 𝜕𝑥𝑗
− 𝑔𝛿𝑖3
Mean Continuity Equation
=0
=
=
̅̅̅̅̅̅̅̅̅̅̅̅
𝜕(𝑈
𝑗 +𝑢𝑗 )
𝜕𝑥𝑗
𝜕𝑈𝑗
𝜕𝑥𝑗
+
=
𝜕𝑢𝑗
𝜕𝑥𝑗
̅̅̅̅𝑗 +𝑢
̅̅̅)
𝜕(𝑈
𝑗
𝜕𝑥𝑗
=0
=
̅̅̅̅𝑗
𝜕𝑈
𝜕𝑥𝑗
⇒
+
̅̅̅𝑗
𝜕𝑢
𝜕𝑥𝑗
=
𝜕𝑈𝑗
𝜕𝑥𝑗
𝜕𝑢𝑗
𝜕𝑥𝑗
=0
=0
Both mean and fluctuation satisfy divergence = 0 condition.
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3.3.2
Chapter 6
32
Fall 2012
Mean Momentum Equation
̃𝑖
𝜕𝑢
𝜕𝑡
+ 𝑢̃𝑗
𝜕(𝑈𝑖 +𝑢𝑖 )
𝜕𝑡
̃𝑖
𝜕𝑢
=−
𝜕𝑥𝑗
1 𝜕𝑝̃
𝜌 𝜕𝑥𝑖
+ (𝑈𝑗 + 𝑢𝑗 )
̅̅̅̅̅̅̅̅̅
(𝑈𝑖 +𝑢𝑖 )
𝜕
+𝜐
− 𝑔𝛿𝑖3
𝜕𝑥𝑗 𝜕𝑥𝑗
𝜕(𝑈𝑖 +𝑢𝑖 )
𝜕𝑥𝑗
̅
𝜕𝑢
𝜕𝑈
̃𝑖
𝜕2 𝑢
=−
1 𝜕(𝑃+𝑝)
𝜌
𝜕𝑥𝑖
+𝜐
𝜕2 (𝑈𝑖 +𝑢𝑖 )
𝜕𝑥𝑗 𝜕𝑥𝑗
− 𝑔𝛿𝑖3
𝜕𝑈
= 𝑖+ 𝑖= 𝑖
𝜕𝑡
𝜕𝑡
𝜕𝑡
𝜕𝑡
̅̅̅̅̅̅̅
𝜕𝑢
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
̅̅̅̅̅̅̅
̅
𝜕(𝑈 +𝑢 )
𝜕𝑈
𝜕𝑢
𝜕𝑈
𝜕𝑢
𝜕𝑈
𝑖 𝑢𝑗
(𝑈𝑗 + 𝑢𝑗 ) 𝑖 𝑖 = 𝑈𝑗 𝑖 + 𝑈𝑗 𝑖 + 𝑢̅𝑗 𝑖 + 𝑢𝑗 𝑖 = 𝑈𝑗 𝑖 +
𝜕𝑥𝑗
̅̅̅̅̅̅̅
𝜕𝑢
𝑖 𝑢𝑗
Since
̅̅̅̅̅̅̅̅
𝜕(𝑃+𝑝)
𝜕𝑥𝑖
=
𝜕𝑥𝑗
𝜕𝑥𝑗
𝜕𝑃
𝜕𝑥𝑖
+
𝜕𝑝̅
𝜕𝑥𝑖
𝜕𝑈𝑖
𝜕𝑡
Or
+ 𝑈𝑗
𝜕𝑈𝑖
𝜕𝑥𝑗
=
𝜕𝑥𝑗
𝜕𝑥𝑗 𝜕𝑥𝑗
+
̅̅̅̅̅̅̅
𝜕𝑢
𝑖 𝑢𝑗
𝜕𝑥𝑗
𝐷𝑈𝑖
𝐷𝑡
=−
=−
𝜕𝑥𝑗
𝜕𝑥𝑗
̅̅̅̅̅̅̅
𝜕𝑢𝑗
̅̅̅̅̅̅̅
̅̅̅̅̅̅̅
𝜕𝑢
𝜕𝑢
= 𝑢𝑖
+ 𝑢𝑗 𝑖 = 𝑢𝑗 𝑖
̅̅̅̅̅̅
𝑔𝛿𝑖3 = 𝑔𝛿𝑖3
2 (𝑈 +𝑢 )
̅̅̅̅̅̅̅̅̅̅̅
𝜕
𝜕 2 𝑈𝑖
𝑖
𝑖
=
+
𝜕𝑥𝑗 𝜕𝑥𝑗
𝜕𝑥𝑗
𝜕𝑥𝑗
𝜕𝑥𝑗
𝜕𝑃
𝜕𝑥𝑖
𝜕 2 𝑢𝑖
𝜕𝑥𝑗 𝜕𝑥𝑗
1 𝜕𝑃
𝜌 𝜕𝑥𝑖
1 𝜕𝑃
𝜌 𝜕𝑥𝑖
=
+𝜐
𝜕 2 𝑈𝑖
𝜕𝑥𝑗 𝜕𝑥𝑗
𝜕 2 𝑈𝑖
𝜕𝑥𝑗 𝜕𝑥𝑗
− 𝑔𝛿𝑖3 +
𝜕
𝜕𝑥𝑗
− 𝑔𝛿𝑖3
(𝜐
𝜕𝑈𝑖
𝜕𝑥𝑗
− ̅̅̅̅̅)
𝑢𝑖 𝑢𝑗
𝜕𝑥𝑗
𝜕𝑥𝑗
058:0160
Jianming Yang
Or
𝐷𝑈𝑖
𝐷𝑡
with
Chapter 6
33
Fall 2012
= −𝑔𝛿𝑖3 +
𝜕𝑈𝑗
𝜕𝑥𝑗
1 𝜕
𝜌 𝜕𝑥𝑗
[−𝑃𝛿𝑖𝑗 + 𝜇 (
𝜕𝑈𝑖
𝜕𝑥𝑗
+
𝜕𝑈𝑗
𝜕𝑥𝑖
) − 𝜌𝑢
̅̅̅̅̅]
𝑖 𝑢𝑗 = −𝑔𝛿𝑖3 +
1 𝜕
𝜎̅
𝜌 𝜕𝑥𝑗 𝑖𝑗
=0
RANS equation
The difference between the NS and RANS equations is the Reynolds stresses −𝜌𝑢
̅̅̅̅̅,
𝑖 𝑢𝑗
which acts like additional stress.
−𝜌𝑢
̅̅̅̅̅
̅̅̅̅̅
(i.e. Reynolds stresses are symmetric)
𝑖 𝑢𝑗 = −𝜌𝑢
𝑗 𝑢𝑖
−𝜌𝑢𝑢
̅̅̅̅ −𝜌𝑢𝑣
̅̅̅̅ −𝜌𝑢𝑤
̅̅̅̅
̅̅̅̅ −𝜌𝑣𝑣
̅̅̅ −𝜌𝑣𝑤
̅̅̅̅ ]
= [ −𝜌𝑢𝑣
−𝜌𝑢𝑤
̅̅̅̅ −𝜌𝑣𝑤
̅̅̅̅ −𝜌𝑤𝑤
̅̅̅̅̅
−𝜌𝑢
̅̅̅̅̅
𝑖 𝑢𝑖 are normal stresses
−𝜌𝑢
̅̅̅̅̅
𝑖 𝑢𝑗 (𝑖 ≠ 𝑗) are shear stresses
For isotropic turbulence
̅̅̅̅̅
𝑢
𝑖 𝑢𝑗 = 0 (𝑖 ≠ 𝑗)
𝑢𝑢
̅̅̅̅ = 𝑣𝑣
̅̅̅ = 𝑤𝑤
̅̅̅̅̅ = const.
however, turbulence is generally non-isotropic.
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3.3.3
Chapter 6
34
Fall 2012
Turbulent Kinetic Energy Equation
Turbulent kinetic energy
1
1
2
2
𝑘 = ̅̅̅̅̅
𝑢𝑖 𝑢𝑖 = (𝑢𝑢
̅̅̅̅ + 𝑣𝑣
̅̅̅ + 𝑤𝑤
̅̅̅̅̅ )
Subtracting NS equation for 𝑢̃𝑖 and RANS equation for 𝑈𝑖 results in equation for 𝑢𝑖 :
̃𝑖
𝜕𝑢
𝜕𝑡
𝜕𝑈𝑖
𝜕𝑡
̃𝑖
𝜕𝑢
+ 𝑢̃𝑗
+ 𝑈𝑗
𝜕𝑈𝑖
𝜕𝑥𝑗
(𝑈𝑗 + 𝑢𝑗 )
𝜕𝑢𝑖
𝜕𝑡
=−
𝜕𝑥𝑗
+ 𝑈𝑗
+
1 𝜕𝑝̃
𝜌 𝜕𝑥𝑖
̅̅̅̅̅̅̅
𝜕𝑢
𝑖 𝑢𝑗
𝜕𝑥𝑗
𝜕(𝑈𝑖 +𝑢𝑖 )
𝜕𝑢𝑖
𝜕𝑥𝑗
𝜕𝑥𝑗
+ 𝑢𝑗
+𝜐
=−
= 𝑈𝑗
𝜕𝑈𝑖
𝜕𝑥𝑗
̃𝑖
𝜕2 𝑢
𝜕𝑥𝑗 𝜕𝑥𝑗
1 𝜕𝑃
𝜌 𝜕𝑥𝑖
𝜕𝑈𝑖
𝜕𝑥𝑗
+ 𝑢𝑗
+𝜐
+ 𝑈𝑗
𝜕𝑢𝑖
𝜕𝑥𝑗
− 𝑔𝛿𝑖3
−
𝜕 2 𝑈𝑖
𝜕𝑥𝑗 𝜕𝑥𝑗
𝜕𝑢𝑖
𝜕𝑥𝑗
+ 𝑢𝑗
̅̅̅̅̅̅̅
𝜕𝑢
𝑖 𝑢𝑗
𝜕𝑥𝑗
− 𝑔𝛿𝑖3
𝜕𝑈𝑖
𝜕𝑥𝑗
=−
+ 𝑢𝑗
1 𝜕𝑝
𝜌 𝜕𝑥𝑖
𝜕𝑢𝑖
𝜕𝑥𝑗
+𝜐
𝜕 2 𝑢𝑖
𝜕𝑥𝑗 𝜕𝑥𝑗
Multiply by 𝑢𝑖 and average
̅̅̅̅̅̅̅̅̅
̅̅̅̅̅̅̅̅̅̅̅
̅̅̅̅̅̅̅
𝜕𝑢
̅̅̅̅̅̅̅
̅̅̅̅̅̅̅̅̅̅
̅̅̅̅̅̅̅̅̅̅
̅̅̅̅̅̅̅̅̅̅
𝜕𝑢𝑖
𝜕𝑢𝑖
𝜕𝑈𝑖
𝜕𝑢𝑖
1 ̅̅̅̅̅̅̅
𝜕𝑝
𝜕 2 𝑢𝑖
𝑖 𝑢𝑗
𝑢𝑖
+ 𝑢𝑖 𝑈𝑗
+ 𝑢𝑖 𝑢𝑗
+ 𝑢𝑖 𝑢𝑗
− 𝑢𝑖
= − 𝑢𝑖
+ 𝜐𝑢𝑖
𝜕𝑡
𝜕𝑥𝑗
1
̅̅̅̅̅̅̅̅̅
𝜕( 𝑢𝑖 𝑢𝑖 )
2
𝜕𝑡
𝜕𝑡
𝜕𝑥𝑗
𝜕𝑥𝑗
𝜌
𝜕𝑥𝑖
𝜕𝑥𝑗 𝜕𝑥𝑗
1
1
̅̅̅̅̅̅̅̅̅̅̅̅
̅̅̅̅̅̅̅̅̅̅̅̅
𝜕( 𝑢𝑖 𝑢𝑖 )
𝜕(
𝑢𝑢)
̅̅̅̅̅̅̅̅̅
̅̅̅̅̅̅̅̅̅̅̅
̅̅̅̅̅̅̅
𝜕𝑢
̅̅̅̅̅̅̅̅̅̅
𝜕𝑈
1 ̅̅̅̅̅̅̅
𝜕𝑝
𝜕 2 𝑢𝑖
𝑖 𝑢𝑗
𝑖
2
2 𝑖 𝑖
+ 𝑈𝑗
+ 𝑢𝑖 𝑢𝑗
+ 𝑢𝑗
− 𝑢𝑖
= − 𝑢𝑖
+ 𝜐𝑢𝑖
𝜕𝑥𝑗
1
̅̅̅̅̅̅)
𝜕( 𝑢
𝑢
2 𝑖 𝑖
𝜕𝑥𝑗
1
+ 𝑈𝑗
̅̅̅̅̅̅)
𝜕( 𝑢
𝑢
2 𝑖 𝑖
𝜕𝑥𝑗
𝜕𝑥𝑗
𝜕𝑈𝑖
̅̅̅̅̅̅̅̅̅)
1 𝜕(𝑢
𝑖 𝑢𝑖 𝑢𝑗
+ ̅̅̅̅̅
𝑢𝑖 𝑢𝑗 𝜕𝑥 + 2
𝑗
𝜕𝑥𝑗
𝜕𝑥𝑗
− 𝑢̅𝑖
𝜕𝑥𝑗
̅̅̅̅̅̅
𝜕𝑢
𝑖 𝑢𝑗
𝜕𝑥𝑗
𝜌
̅̅̅̅̅)
1 𝜕 (𝑢
𝑖𝑝
= −𝜌
𝜕𝑥𝑖
+𝜐
𝜕𝑥𝑖
𝜕𝑥𝑗 𝜕𝑥𝑗
1
2
𝜕 2 ( ̅̅̅̅̅̅)
𝑢𝑖 𝑢𝑖
𝜕𝑥𝑗 𝜕𝑥𝑗
̅̅̅̅̅̅̅̅
𝜕𝑢 𝜕𝑢
− 𝜐 𝜕𝑥 𝑖 𝜕𝑥 𝑖
𝑗
𝑗
058:0160
Jianming Yang
Chapter 6
35
Fall 2012
Since
1
2
𝜕( 𝑢𝑖 𝑢𝑖 𝑢𝑗 )
𝜕𝑥𝑗
= 𝑢𝑗
1
𝜕2 ( 𝑢𝑖 𝑢𝑖 )
2
𝜕𝑥𝑗 𝜕𝑥𝑗
𝜕𝑘
𝜕𝑡
𝑈𝑗
+ 𝑈𝑗
𝜕𝑘
𝜕𝑥𝑗
̅̅̅̅̅̅̅̅̅̅̅̅̅̅
1
𝜕𝑢𝑗 ( 𝑢𝑖 𝑢𝑖 )
2
𝜕𝑥𝑗
𝜐
̅̅̅̅̅̅̅̅
𝜕𝑢𝑖 𝜕𝑢𝑖
𝜕𝑥𝑗 𝜕𝑥𝑗
𝑢𝑖 𝑢𝑗
̅̅̅̅̅
𝜕𝑈𝑖
𝜕𝑥𝑗
𝜕𝑘
𝜕𝑥𝑗
=−
=
̅̅̅̅̅)
1 𝜕(𝑢
𝑖𝑝
𝜌
𝜕𝑥𝑖
1
2
𝜕( 𝑢𝑖 𝑢𝑖 )
𝜕𝑥𝑗
𝜕
𝜕𝑥𝑗
−
(𝑢𝑖
1
𝜕𝑢𝑗
𝜕(𝑢𝑖 𝑝)
2
𝜕𝑥𝑗
𝜕𝑥𝑖
+ 𝑢𝑖 𝑢𝑖
𝜕𝑢𝑖
𝜕𝑥𝑗
) = 𝑢𝑖
̅̅̅̅̅̅̅̅̅̅̅̅̅̅
1
𝜕𝑢𝑗 ( 𝑢𝑖 𝑢𝑖 )
2
𝜕𝑥𝑗
= convection
= turbulent diffusion
+𝜐
𝜕 2 𝑢𝑖
𝜕𝑥𝑗 𝜕𝑥𝑗
𝜕2 𝑘
𝜕𝑥𝑗 𝜕𝑥𝑗
+
𝜐
𝜕𝑝
𝜕𝑥𝑖
+𝑝
𝜕𝑢𝑖
𝜕𝑥𝑖
𝜕𝑢𝑖 𝜕𝑢𝑖
𝜕𝑥𝑗 𝜕𝑥𝑗
− ̅̅̅̅̅
𝑢𝑖 𝑢𝑗
̅̅̅̅̅)
1 𝜕 (𝑢
𝑖𝑝
𝜌
= 𝑢𝑖
𝜕𝑥𝑖
𝜕2 𝑘
𝜕𝑥𝑗 𝜕𝑥𝑗
𝜕𝑈𝑖
𝜕𝑥𝑗
−𝜐
̅̅̅̅̅̅̅̅
𝜕𝑢𝑖 𝜕𝑢𝑖
𝜕𝑥𝑗 𝜕𝑥𝑗
= pressure work
= viscous diffusion
= isotropic dissipation rate
= rate of turbulent kinetic energy production, represents loss of mean
kinetic energy and gain of turbulent kinetic energy due to interactions of ̅̅̅̅̅
𝑢𝑖 𝑢𝑗 and
𝜕𝑈𝑖
𝜕𝑥𝑗
.
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
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Fall 2012
Chapter 6
36
3.4 Velocity Profiles in Turbulent Wall Flow: Inner, Outer, and Overlap Layers
Mean velocity profile of a smooth-flat-plate turbulent boundary layer plotted in log-linear
coordinates with law-of-the-wall normalizations
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Detailed examination of turbulent-flow velocity profiles indicates the existence of a
three-layer structure:
(1) a thin inner layer close to the wall where turbulence is inhibited by the presence of
the solid boundary, i.e. ̅̅̅̅̅
𝑢𝑖 𝑢𝑗 are negligibly small (= 0 at the wall) and the flow is
controlled by molecular viscosity.
(2) An outer layer where turbulent shear dominates.
(3) An overlap layer where both types of shear are important.
More information is obtained from dimensional analysis and confirmed by experiment.
Inner law:
𝑢 = 𝑓(𝜏𝑤 , 𝜌, 𝜇, 𝑦)
+
𝑢 =
𝑢
𝑢∗
= 𝑓(
𝑦𝑢∗
𝜐
) = 𝑓 (𝑦 + )
Wall shear velocity (friction velocity): 𝑢∗ = √𝜏𝑤 ⁄𝜌
𝑢+ , 𝑦 + are called inner-wall variables
Note that the inner law is independent of geometries, i.e., flat plate boundary
layer or pipe flow.
Outer Law: the velocity defect
𝑈𝑒 − 𝑢 = 𝑔(𝜏𝑤 , 𝜌, 𝑦, 𝛿 ) for 𝑝𝑥 = 0
𝑈𝑒 −𝑢
𝑢∗
𝑦
= 𝑔 ( ) = 𝑔(𝜂 )
𝛿
Note that the outer wall is independent of 𝜇.
=
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Chapter 6
38
Fall 2012
Overlap law: both laws are valid
It is not that difficult to show that for both laws to overlap, 𝑓 and 𝑔 are logarithmic
functions:
∗
𝑦𝑢∗
𝑑𝑢
Inner region:
𝑢 = 𝑢 𝑓(
Outer region:
𝑈𝑒 − 𝑢 = 𝑢∗ 𝑔 ( )
+
At large 𝑦 and small 𝜂:
𝜐
𝑓 (𝑦
)
+)
𝑦
𝑑𝑦
𝑑𝑢
𝛿
𝑑𝑦
=
𝑦 𝑢∗
2
𝑑𝑓
𝑢∗ 𝜐 𝑑𝑦 +
=
𝑢∗
𝑑𝑓
𝜐 𝑑𝑦 +
𝑢∗ 𝑑𝑔
=−
=
2
𝛿 𝑑𝜂
𝑦 𝑢∗ 𝑑𝑔
𝑢∗ 𝛿 𝑑𝜂
= 𝑔(𝜂 )
Therefore, both sides must equal universal constant, 𝜅 −1
𝑢
𝑓 (𝑦 + ) = 𝜅 −1 ln 𝑦 + + 𝐵 = ∗
(inner variables)
𝑔(𝜂 ) = −𝜅 −1 ln 𝜂 + 𝐴 =
𝑢
𝑈𝑒 −𝑢
𝑢∗
(outer variables)
𝜅, 𝐴, and 𝐵 are pure dimensionless constants
𝜅 = 0.41
Von Karman constant
𝐵 = 5.5
𝐴 = 2.35
boundary layer flow
= 0.65
pipe flow
Values vary somewhat depending on different experimental arrangements
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3.4.1
Chapter 6
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Fall 2012
Details of Inner Law
Very near the wall, 𝜏𝑤 = 𝜇
𝑢=
+
𝜕𝑢
𝜕𝑦
𝜏𝑤
𝜇
𝑢 =
Or
3.4.2
~𝑐𝑜𝑛𝑠𝑡.
𝑦
𝑢
𝑢∗
i.e., varies linearly
+
=𝑦 =
𝑦𝑢∗
𝜐
𝑦+ ≤ 5
Details of the Outer Law
With pressure gradient included, the outer law becomes:
𝑈𝑒 −𝑢
𝑢∗
= 𝑔(𝜂, 𝛽)
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
solution for simple geometries: pipe and channel flow 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
boundary layer equations for arbitrary 𝑝𝑥 .
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3.4.3 Summary of Inner, Outer, and Overlap Layers
Mean velocity correlations
Inner layer: 𝑈 + = 𝑈⁄𝑢∗
𝑦 + = 𝑦 ⁄𝑢 ∗
𝑢∗ = √𝜏𝑤 ⁄𝜌
Sub-layer: 0 ≤ 𝑦 + ≤ 5
𝑈+ = 𝑦+
Buffer layer: 5 ≤ 𝑦 + ≤ 30
where sub-layer merges smoothly with log law
Outer Layer:
𝑈𝑒 −𝑢
𝑢∗
= 𝑔(𝜂, 𝛽)
𝑦
𝜂= ,
Overlap layer (log region):
𝑈 + = 𝜅 −1 ln 𝑦 + + 𝐵
𝑈𝑒 −𝑢
𝑢∗
𝛽=
𝛿
𝛿∗
𝑝
𝜏𝑤 𝑥
inner variables
= −𝜅 −1 ln 𝜂 + 𝐴
outer variables
Composite Inner/Overlap layer correlation
+
1
1
2
6
𝑦 + = 𝑈 + + 𝑒 −𝜅𝐵 [𝑒 𝜅𝑈 − 1 − 𝜅𝑈 + − (𝜅𝑈 + )2 − (𝜅𝑈 + )3 ]
Composite Overlap/Outer layer correlation
Π
𝑈 + = 𝜅 −1 ln 𝑦 + + 𝐵 + 2 𝑊 (𝜂 )
𝜅
Π and 𝑊 (𝜂 ) are the wake strength parameter and a wake function, respectively, both
introduced by Coles (1956).
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Solution: This is accomplished by straight substitution:
 turb
du  2 2 du  du
du u*
  w  u*  
  y
,
solve
for


dy 
dy  dy
dy  y
Integrate:
2
u*
 du   
dy
u*
, or: u 
ln(y)  constant
y

Ans.
To convert this to the exact form of Eq. (6.28) requires fitting to experimental data.
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Chapter 6
42
Fall 2012
4 Turbulent Flow in Ducts Using Mean-Velocity Correlations
4.1 Smooth Circular Pipe
Recall laminar flow exact solution
8𝜏
64
𝑓 = 2𝑤 =
𝜌𝑢𝑎𝑣𝑒
𝑅𝑒𝑑 =
𝑅𝑒𝑑
𝑢𝑎𝑣𝑒 𝑑
𝜐
≤ 2300
A turbulent-flow “approximate” solution can be obtained simply by computing 𝑢𝑎𝑣𝑒
based on log law.
𝑢
𝑢∗
1
𝑦𝑢∗
𝜅
𝜐
= ln
+𝐵
Where 𝑢 = 𝑢(𝑦); 𝜅 = 0.41; 𝐵 = 5; 𝑢∗ = √𝜏𝑤 ⁄𝜌 ; 𝑦 = 𝑅 − 𝑟
𝑅 ∗ 1
𝑦𝑢∗
𝑉 = 𝑢𝑎𝑣𝑒 = 𝐴 = 𝜋𝑅2 ∫0 𝑢 (𝜅 ln 𝜐 + 𝐵) 2𝜋𝑟𝑑𝑟
(𝑅−𝑟)𝑢∗
2𝜋𝑢∗ 𝑅 1
= 𝜋𝑅2 ∫0 [𝜅 ln 𝜐 + 𝐵] 𝑟𝑑𝑟
2𝜋𝑢∗ 𝑅 1
1
𝑢∗
= 𝜋𝑅2 ∫0 {[𝜅 ln(𝑅 − 𝑟) + 𝜅 ln 𝜐 + 𝐵] (𝑅 − 𝑟) −
2𝑢∗ 1
1
1
𝑢∗
2 1
𝑄
=
{ (𝑅 − 𝑟) [𝜅 ln(𝑅 − 𝑟) − 2𝜅 + 𝜅 ln
𝑅2 2
2𝑢∗ 1
=−
=
=
1
𝑅2
1
1
1
{2 𝑅2 [𝜅 ln 𝑅 − 2 + 𝜅 ln
𝑢∗
𝜐
1
1
𝑢∗
∗ 1
−𝑢 {𝜅 ln 𝑅 − 2𝜅 + 𝜅 ln 𝜐 + 𝐵
1
1
𝑢∗
3
𝑢∗ {𝜅 ln 𝑅 + 𝜅 ln 𝜐 + 𝐵 − 2𝜅}
𝜐
1
1
𝑅 [𝜅 ln(𝑅 − 𝑟) + 𝜅 ln
1
𝜐
+ 𝐵]} 𝑑(𝑅 − 𝑟)
1
1
+ 𝐵] − 𝑅(𝑅 − 𝑟) [𝜅 ln(𝑅 − 𝑟) − 𝜅 + 𝜅 ln
1
1
1
+ 𝐵] − 𝑅2 [𝜅 ln 𝑅 − 𝜅 + 𝜅 ln
1
𝑢∗
2
1
− 2 𝜅 ln 𝑅 + 𝜅 − 2 𝜅 ln
𝑢∗
𝜐
𝑢∗
𝜐
− 2𝐵}
+ 𝐵]}
𝑢∗
𝜐
+ 𝐵]}
𝑅
0
058:0160
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Fall 2012
1
2
= 2 𝑢∗ (𝜅 ln
Where
Or:
Chapter 6
43
𝑅𝑢∗
𝜐
3
+ 2𝐵 − 𝜅)
∫(𝑅 − 𝑟)𝑚 ln(𝑅 − 𝑟) 𝑑 (𝑅 − 𝑟) = (𝑅 − 𝑟)𝑚+1 (
𝑉
𝑢∗
≈ 2.44 ln
8𝜏𝑤
𝑓=
𝑉
𝑢∗
=
𝑅𝑢∗
𝜐
2
𝜌𝑢𝑎𝑣𝑒
𝑉
√𝜏𝑤 ⁄𝜌
1 𝑑𝑉 𝑢∗
=2
8 1⁄2
( )
𝑓
1
𝑓1⁄2
1
𝑓1⁄2
=
𝜐 𝑉
𝑅𝑢∗
𝜐
ln(𝑅−𝑟 )
𝑚+1
−(
1
𝑚+1)2
)
+ 1.34
8𝜏𝑤
𝜌𝑉 2
=(
𝜌𝑉 2
𝜏𝑤
1⁄2
)
8 1⁄2
=( )
𝑓
1
8 1⁄2
𝑅𝑒𝑑 (𝑓)
2
8 1⁄2
2
𝑓
1
= 2 𝑅𝑒𝑑
𝑢∗
𝑉
=
1
= 2.44 ln [ 𝑅𝑒𝑑 ( )
] + 1.34
= 1.99 log(𝑅𝑒𝑑 𝑓 1⁄2 ) − 1.02
= 2.0 log(𝑅𝑒𝑑 𝑓 1⁄2 ) − 0.8
Fit friction data better
𝑓 only drops by a factor of 5 over 104 < Re < 108
Since 𝑓 equation is implicit, it is not easy to see dependency on 𝜌, 𝜇, 𝑉, and 𝐷
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Chapter 6
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Fall 2012
Explicit correlation of 𝑓 and 𝑅𝑒𝑑 by Blasius (1911), a student of Prandtl:
⁄
4
𝑓 = 0.316𝑅𝑒−1
,
𝑑
Hagen’s 1839 pressure-drop data
ℎ𝑓 =
∆𝑝
𝜌𝑔
𝐿 𝑉2
=𝑓
𝑑 2𝑔
4000 < 𝑅𝑒𝑑 < 105
⁄
𝑑𝑉 −1 4 𝐿 𝑉 2
= 0.316 ( )
𝑑 2𝑔
𝜐
For turbulent Flow:
1
𝑑𝑉 −4 𝐿 𝑉 2
∆𝑝 ≈ 0.316𝜌𝑔 ( )
𝜐
𝑑 2𝑔
= 0.158𝐿𝜌
1
𝑄 = 𝜋𝑑 2 𝑉
𝑉=
4
∆𝑝 ≈ 0.158𝐿𝜌
3⁄4
1
4
5
4
−
3⁄4
7
4
𝜇 𝑑 𝑉 ≈ 0.241𝐿𝜌
1
5
−
𝜇 4𝑑 4𝑉
7
4
4𝑄
𝜋𝑑 2
3⁄4
1
4
𝜇 𝑑 −4.75 𝑄1.75
For laminar flow
∆𝑝 =
128𝜇𝐿𝑄
𝜋𝑑 4
In turbulent pipe flow ∆𝑝 decreases more sharply than
that in laminar pipe flow for same 𝑄; therefore, one can
increase the pipe diameter 𝑑 for smaller ∆𝑝.
For example, 2𝑑 decreases ∆𝑝 by 27 ((2𝑑 )−4.75 ≈ 26.91𝑑 −4.75 )for same 𝑄.
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Fall 2012
The maximum velocity in turbulent pipe flow:
𝑢max
𝑢(𝑟=0)
=
𝑢∗
𝑢∗
1
𝑅𝑢∗
𝜅
𝜐
= ln
+𝐵
Combine with
𝑉
𝑢∗
1
𝑅𝑢∗
𝜅
𝑉
𝜐
= ln
𝑢∗
𝑢
3
2𝜅
3
= 𝑢max
∗ − 2𝜅
𝑢max
𝑉
+𝐵−
=1+
3𝑢 ∗
2𝜅𝑉
Also
∗
𝑢 = √𝜏𝑤 ⁄𝜌
𝑓=
8𝜏𝑤
2
𝜌𝑢𝑎𝑣𝑒
𝑢max
𝑉
=
=1+
8𝜏𝑤
𝜌𝑉 2
3𝑢 ∗
2𝜅𝑉
=1+
𝑓=
3
2𝜅
8𝜌𝑢∗
2
𝜌𝑉 2
𝑢∗
𝑉
= √𝑓⁄8
√𝑓⁄8 ≈ 1 + 1.3√𝑓
Or:
𝑉
𝑢max
Recall laminar flow:
𝑉
𝑢max
≈ (1 + 1.3√𝑓)
= 0.5
−1
for turbulent flow
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Chapter 6
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Fall 2012
4.2 Rough Circular Pipe
Surface roughness has a neglibible effect for laminar
pipe flow. But turbulent flow is strongly affected by
roughness.
Roughness height 𝜖
𝜖
𝑓 = 𝑓 (𝑅𝑒𝑑 , )
𝑑
∗
𝜖𝑢
𝑢+ = 𝑔(𝑦 + , 𝜖 + ) 𝜖 + =
𝜐
1
𝑢+ = ln 𝑦 + + 𝐵 − ∆𝐵(𝜖 + )
𝜅
which leads to three roughness regimes:
1. 𝜖 + < 5
hydraulically smooth
2. 5 ≤ 𝜖 + ≤ 70 transitional roughness (Re dependence)
3. 𝜖 + > 70
full rough (no Re dependence)
Moody chart:
Approximate explicit formula:
1
𝑓1⁄2
1
𝑓1⁄2
= −2.0 log (
𝜖⁄𝑑
2.51
)
𝑅𝑒𝑑 𝑓 1⁄2
𝜖⁄𝑑 1.11
6.9
3.7
= −1.8 log [(
+
Log law shifts downward
3.7
)
+
𝑅𝑒𝑑
]
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Fall 2012
Chapter 6
47
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Jianming Yang
Chapter 6
48
Fall 2012
4.3 Four Types of Pipe Flow Problems
Moody chart:
1
𝑓1⁄2
= −2.0 log (
𝜖⁄𝑑
3.7
+
2.51
𝑅𝑒𝑑 𝑓 1⁄2
)
There are basically four types of problems involved with uniform flow in a single pipe:
1. Determine the head loss, given the kind and size of pipe and the flow rate, 𝑄 = 𝐴𝑉
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
4. Determine the pipe length, given 𝑄, 𝑑, ℎ𝑓 , 𝜖, 𝜇, 𝑔
4.3.1
Head Loss Problem
Determine the head loss ℎ𝑓 , given 𝑑, 𝐿, and 𝑉 or 𝑄, 𝜌, 𝜇, and 𝑔
The first problem of head loss is solved readily by obtaining 𝑓 from the Moody
diagram, using values of 𝑅𝑒 and 𝜖 ⁄𝑑 computed from the given data. The head loss ℎ𝑓
is then computed from the Darcy-Weisbach equation.
𝜖
𝑓 = 𝑓 (𝑅𝑒𝑑 , )
𝑑
𝑅𝑒𝑑 = 𝑅𝑒𝑑 (𝑉, 𝑑 )
ℎ𝑓 = 𝑓
𝐿 𝑉2
𝑑 2𝑔
= ∆ℎ
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4.3.2
Chapter 6
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Fall 2012
Flow Rate Problem
Determine the velocity 𝑉 or flow rate 𝑄, given head loss, given 𝑑, 𝐿, ℎ𝑓 , 𝜌, 𝜇, and 𝑔
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
𝑉=(
2𝑔ℎ𝑓 1⁄2
𝐿/𝑑
)
𝑓 −1⁄2
2) using 𝑉 compute 𝑅𝑒
𝜖
3) obtain a new value for 𝑓 = 𝑓 (𝑅𝑒𝑑 , ) and repeat as above until convergence
𝑑
Or can use 𝑅𝑒𝑑 𝑓 1⁄2 = (
2𝑔𝑑 3 ℎ𝑓
𝐿𝜐2
1⁄2
)
scale proposed by Prof. Hunter Rouse
1) compute 𝑅𝑒𝑑 𝑓 1⁄2 and 𝜖 ⁄𝑑
2) read 𝑓 from Moody Chart of 𝑅𝑒𝑑 𝑓 1⁄2 scale, or
3) solve 𝑉 from ℎ𝑓 = 𝑓
4) flow rate 𝑄 = 𝐴𝑉
𝐿 𝑉2
𝑑 2𝑔
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4.3.3 Sizing Problem
Determine the size of the pipe 𝑑, given 𝑄, 𝐿, ℎ𝑓 , 𝜌, 𝜇, and 𝑔
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
𝑑=(
1⁄5
8𝐿𝑄2
𝜋2 𝑔ℎ𝑓
)
𝑓 1⁄5
2) using 𝑑 compute 𝑅𝑒 and 𝜖 ⁄𝑑
𝜖
3) obtain a new value of 𝑓 = 𝑓 (𝑅𝑒𝑑 , ) and repeat as above until convergence
𝑑
4.3.4
Pipe Length Problem
The four problem of pipe length is solved by obtaining 𝑓 from the Moody diagram,
using values of 𝑅𝑒 and ϵ⁄d computed from the given data. Then using given ℎ𝑓 ,𝑉, 𝑑,
and calculated 𝑓 to solve 𝐿 from
𝐿=
2𝑔 𝑑ℎ𝑓
𝑉2 𝑓
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4.4 Noncircular Ducts
For noncircular ducts, 𝜏𝑤 = 𝑓 (perimeter); thus, new definition of 𝑓 =
8𝜏𝑤
𝜌𝑉 2
is required.
Define average wall shear stress
1
𝑃
𝜏̅𝑤 = ∫0 𝜏𝑤 𝑑𝑠
𝑃
𝑃 = perimeter
𝑑𝑠 = arc length
Control volume analysis for momentum equation:
∆𝑧
∆𝑝𝐴 − 𝜏̅𝑤 𝑃𝐿 + 𝛾𝐴𝐿 ( ) = 0
𝑝
∆ℎ = ∆ ( + 𝑧) =
𝛾
hydraulic radius
𝜏̅𝑤 =
𝐴𝛾 ∆ℎ
𝑃 𝐿
𝑅ℎ =
=−
𝐿
𝜏̅𝑤 𝐿
𝛾𝐴⁄𝑃
𝐴
cross−sectional area
=
𝑃
𝐴𝛾 𝑑ℎ
𝑃 𝑑𝑥
wetted perimeter
𝐴 𝑑 (𝑝+𝛾𝑧)
𝐴 𝑑𝑝̂
=−
Recall for circular pipe:
𝜏𝑤 = −
In analogy to circular pipe:
𝜏̅𝑤 = −
hydraulic diameter
𝐷ℎ =
4𝐴
𝑃
=−
𝑃
𝑑𝑥
𝑅 𝑑𝑝̂
𝐷 𝑑𝑝̂
2 𝑑𝑥
𝐴 𝑑𝑝̂
4 𝑑𝑥
𝐷ℎ 𝑑𝑝̂
𝑃 𝑑𝑥
=−
=−
𝑃 𝑑𝑥
4 𝑑𝑥
= 4𝑅ℎ
For multiple surfaces such as concentric annulus, 𝑃 and 𝐴 are based on wetted perimeter
and area
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Δℎ = ℎ𝑓 = 𝑓
𝑓={
64⁄𝑅𝑒𝐷ℎ
𝐿 𝑉2
𝐷ℎ 2𝑔
± 40%
𝑓Moody (𝑅𝑒𝐷ℎ , 𝜖 ⁄𝐷ℎ )
± 15%
laminar flow
turbulent flow
𝑅𝑒𝐷ℎ =
𝑉𝐷ℎ
𝜐
Recall for laminar channel flow between two parallel plates of 2ℎ apart, 𝐷ℎ = 4ℎ:
24𝜇
48
𝑓 = 𝜌𝑉ℎ = 𝑅𝑒
96
2ℎ
= 𝑅𝑒
4ℎ
96
= 𝑅𝑒
𝐷ℎ
For laminar flows, it is better to use the exact solutions. If we chose to use the
approximation 𝑓 = 64⁄𝑅𝑒𝐷ℎ , we would be 33 percent low for channel flow.
For turbulent flow, 𝐷ℎ works much better especially if combined with “effective
diameter” concept.
Turbulent channel flow
𝑢(𝑌)
𝑢∗
𝑅𝑒𝐷ℎ =
1
𝑓1⁄2
1
𝑓1⁄2
𝑉 = 𝑢∗ (𝜅 ln
𝑉𝐷ℎ
𝑢∗
𝜐
𝑉
𝑌𝑢∗
𝜅
𝜐
≈ ln
1
Mean velocity
1
ℎ𝑢∗
𝜐
+𝐵
1
+ 𝐵 − 𝜅)
= √𝑓 ⁄8
≈ 2.0 log(𝑅𝑒𝐷ℎ 𝑓 1⁄2 ) − 1.19
≈ 2.0 log(0.64𝑅𝑒𝐷ℎ 𝑓 1⁄2 ) − 0.8
Exactly the pipe law
Therefore error in 𝐷ℎ concept relatively smaller for turbulent flow.
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For laminar flow, if we define
64
2
𝐷eff = 96 𝐷ℎ = 3 𝐷ℎ ≈ 0.667𝐷ℎ ~0.64𝐷ℎ
(𝑓𝑅𝑒𝐷ℎ )
laminar
= 96
For turbulent flow,
𝐷eff =
64
(𝑓𝑅𝑒𝐷 )
𝐷ℎ
ℎ laminar
4.5 Minor or Local Losses in Pipe Systems
For real pipe systems, in addition to friction head loss, there are minor losses due to
1.
2.
3.
4.
Entrane or exit
Expansions and contractions
Bends, elbows, tees, and other fittings
Valves, open or partially closed
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For such complex geometries we rely on experimental data to obtain a loss coefficient
𝐾=
ℎ𝑚
𝑉 2 ⁄(2𝑔)
∆𝑝
=1
2𝜌𝑉
2
𝜖
In general, 𝐾 = 𝐾 (geometry, 𝑅𝑒, )
𝐷
Loss coefficient data is supplied by manufacturers and also listed in handbooks.
Modified energy equation to include minor losses
∆ℎ𝑡𝑜𝑡 = ℎ𝑓 + ∑ ℎ𝑚 =
𝑉 2 𝑓𝐿
2𝑔
(
𝑑
+ ∑ 𝐾)
4.6 Multiple-Pipe Systems
4.6.1 Pipes in Series
Flow rate is the same in all pipes:
𝑄1 = 𝑄2 = 𝑄3 = ⋯ = 𝑄𝑛
𝑉1 𝑑12 = 𝑉2 𝑑22 = 𝑉3 𝑑32 = ⋯ = 𝑉𝑛 𝑑𝑛2
Total head loss equals the sum of head loss in each pipe:
∆ℎ𝐴→𝐵 = ∆ℎ1 + ∆ℎ2 + ∆ℎ3 + ⋯ + ∆ℎ𝑛
𝑉 2 𝑓1 𝐿1
1
∆ℎ𝐴→𝐵 = 2𝑔
(
𝑉12
𝑑1
𝑉 2 𝑓2 𝐿21
2
+ ∑ 𝐾1 ) + 2𝑔
(
𝑑2
+ ∑ 𝐾2 ) + ⋯
∆ℎ𝐴→𝐵 = 2𝑔 (𝛼0 + 𝛼1 𝑓1 + 𝛼2 𝑓2 + 𝛼3 𝑓3 + ⋯ + 𝛼𝑛 𝑓𝑛 )
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4.6.2 Pipes in Parallel
Pressure drop is the same in each pipe
∆ℎ𝐴→𝐵 = ∆ℎ1 = ∆ℎ2 = ∆ℎ3
𝑄 = 𝑄1 + 𝑄2 + 𝑄3
If ∆ℎ𝐴→𝐵 is known, solve for 𝑄𝑖 in each pipe directly
𝑅𝑒𝑑 = −(8𝜁
) 1 ⁄2
𝜖 ⁄𝑑
log ( 3.7 +
1.775
√𝜁
)
𝜁=
For the inverse problem, iterate: ℎ𝑓 =
𝑔𝑑3 ℎ𝑓
𝐿𝜐2
𝑄2
(∑ √𝐶𝑖 ⁄𝑓𝑖 )
2
𝐶𝑖 =
𝜋2 𝑔𝑑𝑖5
8𝐿𝑖
4.6.3 Three-Reservoir Junction
If all flows are considered positive toward the junction, then
𝑄1 + 𝑄2 + 𝑄3 = 0
ℎ𝐽 = 𝑧𝐽 +
∆ℎ1 =
∆ℎ2 =
∆ℎ3 =
𝑝𝐽
𝜌𝑔
𝑉21 𝑓1 𝐿1
2𝑔 𝑑1
𝑉22 𝑓2 𝐿2
2𝑔 𝑑2
𝑉23 𝑓3 𝐿3
2𝑔 𝑑3
= 𝑧1 − ℎ𝐽
= 𝑧2 − ℎ𝐽
= 𝑧3 − ℎ𝐽
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Solution: With no driving pressure and negligible tank surface velocity, the energy
equation can be combined with a control-volume mass conservation:
h (t ) 
V2
L V2

2gh
dh
 fav
, or: Qout  ApipeV  D2
 WY
2g
D 2g
4
1  fav L/D
dt
We can separate the variables and integrate for time to drain:

4
D2
2g
1  fav L/D
𝑡drain ≈
4𝑊𝑌
t

0
√
𝜋𝐷2
0
dt  WY 
ho
dh
h
2ℎ0 (1+𝑓𝑎𝑣 𝐿/𝐷)
𝑔

 WY 0  2 ho

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Solution: For air at 20C and 1 atm, take  = 1.20 kg/m3 and  = 1.8E-5 kg/m-s. The pressure
drop is related to the hydraulic diameter of the duct. Convert L = 200 ft = 60.96 m. For sheet
metal, from Table 6.1, the roughness = 0.05 mm.
p  f
L  2
V ,
Dh 2
Solve for V 
where Dh 
2 p Dh

f L
2BH
4A


, f related to Re Dh and
P
BH
Dh
2 (80 Pa) Dh
Dh
m
 1.479
which gives V in
f (1.20)(60.96)
f
s
The duct area A = 2BH increases with B for a fixed H, and so does the hydraulic diameter. The
Reynolds number (VDh/) also increases, hence the friction factor f decreases. All of these factors
make the flow rate Q increase with H. Therefore, without even making calculations, we conclude
that the widest H (36 inches) produces the most flow rate. Ans.
30
We can calculate the actual flow rate for H = 36 inches = 0.9144 m:
Q, m3/s
H  36 in , Dh  0.261 m , Re Dh  95, 200 ,
Giving V  5.46

Dh
20
 0.000191 , f  0.01914,
3
m
m
, Q  V B H  0.761
s
s
 27
10
3
ft
s
Ans.
Here is a plot of flow rate Q vs. width H. It is almost exactly linear.
0
0
5
Duct width
B, inches
10 15 20 25
30 35 40
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/d 
(a) L 
d  12 L/d
1000 ft, in,
 1000,
0.00333
(b) 1500 ft
8 in
2250
0.00500
800 ft
12 in
800
0.00333
(d) 1200 ft
15 in
960
0.00267
(c)
With the flow rate known, we can find everything in pipe (a):
Va 
Qa
20
ft
1.94(25.5)(1)

 25.5 , Rea 
 2.36E6, fa  0.0270
2
Aa ( /4)(1 ft)
s
2.09E5
Then pipes (b,c,d) are in parallel, each having the same head loss and with flow rates which must
add up to the total of 20 ft3/s:
hfb 
8fb L bQ2b
 2 g d5b
 hfc 
8fc Lc Qc2
 2 g d5c
 h fd 
8fd L d Qd2
 2 g d5d
ft 3
, and Q b  Qc  Qd  20
s
Introduce Lb, db, etc. to find that Qc  3.77Qb(fb/fc)1/2 and Qd  5.38Qb(fb/fd)1/2
Then the flow rates are iterated from the relation
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 Q  20
ft 3
 Q b [1  3.77(fb /fc )1/2  5.38(fb /fd )1/2 ]
s
First guess: fb  fc  fd : Qb  1.97 ft 3 /s; Qc  7.43 ft 3 /s; Qd  10.6 ft 3 /s
Improve by computing Reb  349000, fb  0.0306, Rec  878000, fc  0.0271, Red  1002000, fd 
0.0255. Repeat to find Qb  1.835 ft3/s, Qc  7.351 ft3/s, Qd  10.814 ft3/s. Repeat once more and
quit: Qb  1.833 ft3/s, Qc  7.349 ft3/s, Qd  10.819 ft3/s, from which Vb  5.25 ft/s, Vc  9.36 ft/s, Vd
 8.82 ft/s. The pressure drop is
p1  p2  pa  p b  fa
La  Va2
L  Vb2
 fb b
da 2
db 2
 17000  1800  18800 psf  131
lbf
in 2
Ans.