1. smooth circular pipe - university of...
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058:0160 Chapter 6-part4
Professor Fred Stern Fall 2020 1
Chapter 6: Viscous Flow in Ducts
6.4 Turbulent Flow in Pipes and Channels using mean-
velocity correlations
1. Smooth circular pipe
Recall laminar flow exact solution
d
ave
w
uf Re/64
82
2000Re
duaved
A turbulent-flow “approximate” solution can be
obtained simply by computing uave based on log law.
Bv
yu
u
u
*
*ln
1
Where
rRyuByuu w ;/;5;41.0);( *
32ln
2
2
1
2ln11
**
0
**
2
Bv
Ruu
drrBv
yuu
RA
QuV
R
ave
Or:
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34.1ln44.2*
*
v
Ru
u
V
8.0]log[Re2
02.1]log[Re99.1
2/1
2/12/1
f
ff
d
d
Since f equation is implicit, it is not easy to see
dependency on ρ, μ, V, and D
4/1Re316.0)( Dpipef
g
v
D
Lf
ph f
2
2
Turbulent Flow: 4/74/54/14/3158.0 VDLp
75.175.44/14/3241.0 QDL
Laminar flow: 4/8 RLQp
EFD Adjusted constants
f only drops by a factor of 5 over 104 < Re < 108
4000 < ReD < 105
Blasius (1911) power law
curve fit to data
Nearly linear Only slightly
with μ
Drops weakly with
pipe size
Near quadratic
(as expected)
V2
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p (turbulent) decreases more sharply with D than
p (laminar) for same Q; therefore, increase D for smaller
p . 2D decreases p by 27 for same Q.
BRu
u
ru
u
u
*
**
max ln1)0(
Combine with
2
3ln
1 *
* B
Ru
u
V
V
u
V
uuuV
u
u
u
V
2
31
2
3
2
3 *
max
*
max*
max
*
Also
8/8/18/1
*
2
2*
2
2* fV
u
V
uf
Vfandu w
w
ffV
u
V
u3.118/
2
31
2
31
*
max
Or:
For Turbulent Flow: 1
max
3.11
fu
V
Recall laminar flow: 5.0/ max uV
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2. Turbulent Flow in Rough circular pipe
),( kyfU )/,(Re dkff d
)(ln1 kBByU
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)
11.1
2/1
2/1
7.3
/
Re
9.6log8.1~
Re
51.2
7.3
/log2
dk
f
dkf
d
d
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
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
4. Determine the pipe length, given Q, d, hf, ks, , g
Log law shifts downward
Moody diagram
Approximate explicit
formula
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1. Determine the head loss
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)
2
2f
L Vh f h
D g
2121
ppzzh
=
z
p
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
2/12/1
f fD/L
gh2V
known from note sign
given data
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2) using V compute Re
3) obtain a new value for f = f(Re, ks/D) and reapeat as
above until convergence
Or can use Re
2/12/3
2/12
L
ghDf
f
scale on Moody Diagram
1) compute 2/1Re f and ks/D
2) read f
3) solve V from g2
V
D
Lfh
2
f
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
5/1
5/1
f2
2
fgh
LQ8D
known from
given data
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2) using D compute Re and ks/D
3) obtain a new value of f = f(Re, ks/D) and repeat as above
until convergence
4. Determine the pipe length
The four problem of pipe length is solved by obtaining f from
the Moody diagram, using values of Re and ks/D computed
from the given data. Then using given hf ,V, D, and
calculated f to solve L from f
Dh
V
gL
f
2
2 .
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3. Concept of hydraulic diameter for noncircular
ducts
For noncircular ducts, τw= f(perimeter); thus, new
definitions of 2
8
Vf w
and
2
2
VC w
f
are required.
Define average wall shear stress
dsP
P
ww 0
1 ds = arc length, P = perimeter
Momentum:
0
L
z
W
ALPLpA w
PA
Lzph
w
//
A/P =Rh= Hydraulic radius (=R/2 for circular pipe and 2/R
Lh w
)
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Energy:
PA
Lhh
w
L/
dx
pd
P
A
dx
zpd
P
A
dx
dh
P
A
L
h
P
Aw
^
non-circular duct
Recall for circular pipe:
dx
pdD
dx
pdRw
ˆ
4
ˆ
2
In analogy to circular pipe:
P
AD
D
P
A
dx
pdD
dx
pd
P
Ah
hhw
4
4
^
4
^
For multiple surfaces such as concentric annulus P and A
based on wetted perimeter and area
h
DhD
w VDDf
Vf
hh Re)/,(Re
82
g
V
D
Lf
R
LfV
R
Lhh
hhh
w
L28
22
However, accuracy not good for laminar flow (40%) and
marginal turbulent flow (15%).
Hydraulic
diameter
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a. Accuracy for laminar flow (smooth non-circular
pipe)
Recall for pipe flow:
64Re
16Re)(#
0
0
0fP
CPPPoiseuille
f
fc f
Recall for channel flow:
hD
hhVhf
Re
42 Re
96
Re
4824
hD
hh
f
f
VhC
fC
Re
42 Re
24
Re
126
4/
96Re
24Re)(#
0
0
0
h
hf
Df
Dfc
fP
CPPPoiseuille
Therefore:
3
2
96
64
24
16
0
0
0
0
hf
f
hf
f
Donbasedchannel
pipe
Donbasedchannelc
pipec
P
P
P
P
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Thus, if we could not work out the laminar theory and
chose to use the approximation 64Re hDf or 16Re
hDfC ,
we would be 33 percent low for channel flow.
For laminar flow, 0P
varies greatly,
therefore it is better to
use the exact solution
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b. Accuracy for turbulent flow(smooth non-circular
pipe)
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. fCP 0/16 or fP0/64 .
First recall turbulent circular pipe solution and compare
with turbulent channel flow solution using log-law in both
cases
Channel Flow
dYBuyh
uh
Vh
*
0
* )(ln
11
1ln
1 ** B
huu
hhB
hB
P
AD
Bh 4
42
)2(4lim
4
h= half width
Define
hVVDh
Dh
4Re
Y=h-y wall coordinate
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19.1Relog2 2/12/1 ffhD
Therefore error in Dh concept relatively smaller for
turbulent flow.
Note 8.0Re64.0log2)( 2/12/1 fchannelfhD
Define Deffective h
f
f
h DchannelP
circlePD
24)(
16)(~64.0
0
0
(therefore, improvement on Dh is)
eff
effD
VDRe
h
C
C
h
f
f
eff DcircularnonP
circlePD
circularnonP
circlePD
f
f
)(
)(
)(
)(
0
0
0
0
Or
h
C
h
f
eff DcircularnonP
DcircularnonP
D
f)(
16
)(
64
00
Very nearly the same as circular pipe
7% to large at Re = 105
4% to large at Re = 108
From exact laminar solution
Laminar solution
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