Pertemuan 15 - 16
CLOSED CONDUIT FLOW 2
Types of Engineering Problems
• How big does the pipe have to be to carry
a flow of x m3/s?
• What will the pressure in the water
distribution system be when a fire hydrant
is open?
• Can we increase the flow in this old pipe
by adding a smooth liner?
Viscous Flow in Pipes: Overview
• Boundary Layer Development
• Turbulence
• Velocity Distributions
• Energy Losses
– Major
– Minor
• Solution Techniques
Laminar and Turbulent Flows
• Reynolds apparatus
VD inertia
Re 

damping

Transition at Re of 2000
Boundary layer growth:
Transition length
What does the water near the pipeline wall experience?
Drag or shear
_________________________
Why does the water in the center of the pipeline speed
Conservation of mass
up? _________________________
Non-Uniform Flow
v
v
v
Entrance Region Length
le
 f  Re 
D
le
 0.06 Re
D
le
1/ 6
 4.4  Re 
D
100
10
Re
100
000
0
00
0
000
0
100
000
0
100
000
100
00
100
100
Shear in the
entrance region vs laminar
shear in long pipes?
0
1
100
l e /D
10
Distance
for
velocity
profile to
develop
turbulent
Velocity Distributions
• Turbulence causes transfer of momentum
from center of pipe to fluid closer to the
pipe wall.
• Mixing of fluid (transfer of momentum)
causes the central region of the pipe to
uniform
have relatively
_______velocity
(compared to laminar flow)
• Close to the pipe wall, eddies are smaller
(size proportional to distance to the
Log Law for Turbulent,
Established Flow, Velocity
Profiles
yu*
u
 2.5ln
 5.0 Dimensional analysis and measurements
u*

yu*
 20
Valid for
u* 
0 

0
Turbulence produced by shear!

Shear velocity
 hf d
4l
ghf d
u* 
4l
f
u*  V
8
Velocity of large eddies
rough
Force balance
smooth
y
u/umax
Pipe Flow: The Problem
• We have the control volume energy
equation for pipe flow
• We need to be able to predict the head
loss term.
• We will use the results we obtained using
dimensional analysis
Viscous Flow: Dimensional
Analysis
• Remember dimensional analysis?
D
e

C p  f  , Re 
l
D

 2p
VD
and C p 
Where Re 
V 2

• Two important parameters!
– Re - Laminar or Turbulent
– e/D - Rough or Smooth
• Flow geometry
– internal in a bounded region (pipes, rivers): find Cp
_______________________________
flow around an immersed object : find Cd
Pipe Flow Energy Losses
 D
e

f   C p   f  , Re 
L

D

Cp 
 2p
V 2
2 ghf D
f 2
V L
L V2
hf  f
D 2g
u*2
f=8 2
V
Dimensional Analysis
 ghl  p
 ghl  p   g z
2 ghl
More general
Cp 
V2
Assume horizontal flow
Always true (laminar or turbulent)
Darcy-Weisbach equation
L u*2
hf  8
D 2g
Friction Factor : Major losses
• Laminar flow
• Turbulent (Smooth, Transition, Rough)
• Colebrook Formula
• Moody diagram
• Swamee-Jain
Laminar Flow Friction Factor
 gD 2 hl
V
32 L
32 LV
hf 
 gD 2
L V2
hf  f
D 2g
32LV
LV2
f
2
D 2g
gD
64 64
f

VD Re
Hagen-Poiseuille
hf  V
 D 4  ghl
Q
128 l
Darcy-Weisbach
f independent of roughness!
Slope of ___
-1 on log-log plot
Turbulent Flow:
Smooth, Rough, Transition
• Hydraulically smooth
•
•
pipe law (von Karman,
1930)
Rough pipe law (von
Karman, 1930)
Transition function for
both smooth and
rough pipe laws
(Colebrook)
f
u*  V
8
L V2
hf  f
D 2g
 Re f 
1
 2 log 

2.51
f


1
 3.7 D 
 2 log 

f
 e 
1
2.51 
e D
 2 log 


3.7
f
Re f 

(used to draw the Moody diagram)
Moody Diagram
0.10
0.08
 D
f  Cp 
l  0.06

0.05
0.04
0.03
friction factor
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.004
0.03
laminar
0.002
0.02
0.001
0.0008
0.0004
0.0002
0.0001
0.00005
0.01
1E+03
smooth
1E+04
1E+05
1E+06
Re
1E+07
1E+08
e
D
Swamee-Jain
• 1976
• limitations
f
– e/D < 2 x 10-2
– Re >3 x 103
– less than 3% deviation
from results obtained Q   D 5 / 2
2
with Moody diagram
• easy to program for
computer or
calculator use
0.25
  e
5.74  
log  3.7 D  Re0.9  

 
 e
ghf
log 
 2.51
L
3.7
D

Colebrook

1.25  LQ 
D  0.66 e 

gh

 f 
2
4.75
2
no f

L

2 ghf D 3 
5.2 0.04
 L  
 Q 
 
 ghf  
9.4
Each equation has two terms. Why?
Pipe roughness
pipe material
glass, drawn brass, copper
commercial steel or wrought iron
asphalted cast iron
galvanized iron
cast iron
concrete
rivet steel
corrugated metal
PVC
pipe roughness e (mm)
0.0015
0.045
e
0.12
d Must be
0.15 dimensionless!
0.26
0.18-0.6
0.9-9.0
45
0.12
Solution Techniques
find
head loss given (D, type of pipe, Q)
0.25
2
8
LQ
f
4Q
2
hf  f 2
Re 
5
  e
5.74  

g
D
D
log  3.7 D  Re0.9  

 
find flow rate given (head, D, L, type of pipe)

Q
D5 / 2
2
find
 e
ghf
log 
 2.51
L
 3.7 D

L

2 ghf D 3 
pipe size given (head, type of pipe,L, Q)
2



LQ
1.25
D  0.66 e 

gh

 f 
4.75
 L  
 Q 
 
 ghf  
5.2
9.4
0.04
Example: Find a pipe diameter
• The marine pipeline for the Lake Source Cooling
project will be 3.1 km in length, carry a maximum
flow of 2 m3/s, and can withstand a maximum
pressure differential between the inside and
outside of the pipe of 28 kPa. The pipe roughness
is 2 mm. What diameter pipe should be used?
Minor Losses: Expansions!
• We previously obtained losses through
an expansion using conservation of
energy, momentum, and mass
• Most minor losses can not be obtained
analytically, so they must be measured
• Minor losses are often expressed as a
V2
loss coefficient, K, times the velocityhex  K
2g
head. High Re  2p
V2
2 ghex
C p  f  geometry, Re  C p 
hex  C p
Cp 
2
2
V
V
2g
Venturi
hex
2
in
V

2g
Sudden Contraction
EGL
Cc 
2
 1
 V2
hc  
 1 2
C
 2g
 c

HGL
Ac
A2

Ain 
1 

Aout 

1
0.95
0.9
0.85
Cc 0.8
0.75
0.7
0.65
0.6
0
0.2
0.4
0.6
A2/A1
0.8
1
c
2
V1
•
•
2
vena contracta
Losses are reduced with a gradual contraction
Equation has same form as expansion equation!
V2
Entrance Losses
V2
Losses can be
he  K e
2g
reduced by
K e  1.0
accelerating the flow
Estimate based on
gradually and
contraction equations!
K e  0.5
eliminating the
vena contracta
K e  0.04
Head Loss in Bends
High pressure
• Head loss is a function
of the ratio of the
bend radius to the
pipe diameter (R/D)
• Velocity distribution
returns to normal
several pipe diameters
downstream
high
R
Possible
separation
from wall
n
V
p    dn   z  C
 R
2
D
Low pressure
hb  K b
Kb varies from 0.6 - 0.9
V2
2g
Head Loss in Valves
• Function of valve type and
valve position
• The complex flow path
through valves can result in
high head loss (of course,
one of the purposes of a
valve is to create head loss
when it is not fully open)
Yes!
Can Kvbe greater than 1? ______
What is V?
hv  K v
V2
2g
8Q 2
hv  K v
g 2 D 4
Solution Techniques
• Neglect minor losses
• Equivalent pipe lengths
• Iterative Techniques
– Using Swamee-Jain equations for D and Q
– Using Swamee-Jain equations for head loss
– Assume a friction factor
• Pipe Network Software
Solution Technique: Head Loss
• Can be solved explicitly
hminor
Re 
V2
 K
2g
4Q
D
f 
hminor
8Q 2

g 2
K
 D4
0.25

 e
5.74 



log

 3.7 D Re 0.9 

hl   hf   hminor
2
hf  f
8
g 2
LQ 2
D5
Find D or Q
Solution Technique 1
• Assume all head loss is major head loss
• Calculate D or Q using Swamee-Jain
equations
8Q 2
• Calculate minor losses
hex  K
g 2 D 4
• Find new major losses by subtracting
minor losses from total head hloss
f  hl   hex

 LQ 
D  0.66 e 1.25 


 ghf 
2
Q

D5 / 2
2
 e
ghf
log 
 2.51
L
 3.7 D

L

2 ghf D 3 
4.75
5.2 0.04
 L  
 Q 
 
 ghf  
9.4
Find D or Q
Solution Technique 2: Solver
• Iterative technique
• Solve these equations
Re 
4Q
D
hminor  K
f 
0.25
5.74

 e
log


0.9

 3.7 D Re

8Q 2
g 2 D 4
hl   hf   hminor



2
hf  f
8
g 2
LQ 2
D5
Use goal seek or Solver to
find discharge that makes the
calculated head loss equal
the given head loss.
Spreadsheet
Find D or Q
Solution Technique 3: assume f
• The friction factor doesn’t vary greatly
• If Q is known assume f is 0.02, if D is
known assume rough pipe law 1f  2 log  3.7e D 
• Use Darcy Weisbach and minor loss
equations
• Solve for Q or D
• Calculate Re and e/D
• Find new f on Moody diagram
• Iterate
Example: Minor and Major
Losses
• Find the maximum dependable flow between the
reservoirs for a water temperature range of 4ºC
to 20ºC.
Water
25 m elevation difference in reservoir water levels
Reentrant pipes at reservoirs
Standard elbows
2500 m of 8” PVC pipe
Sudden contraction
1500 m of 6” PVC pipe
Spreadsheet
Gate valve wide open
Directions
Example (Continued)
• What are the Reynolds numbers in the
two pipes?
90,000 & 125,000 e/D= 0.0006, 0.0008
• Where are we on the Moody Diagram?
• What is the effect of temperature?
• Why is the effect of temperature so small?
• What value of K would the valve have to
produce to reduce the discharge by 50%?
140
friction factor
0.1
0.02
0.015
0.01
0.008
0.006
0.004
laminar
0.01
1E+03
Spreadsheet
0.05
0.04
0.03
1E+04
1E+05
1E+06
Re
1E+07
1E+08
0.002
0.001
0.0008
0.0004
0.0002
0.0001
0.00005
smooth
Example (Continued)
• Were the minor losses negligible? Yes
• Accuracy of head loss calculations? 5%
• What happens if the roughness increases
f goes from 0.02 to 0.035
by a factor of 10?
• If you needed to increase the flow by 30%
what could you do? Increase small pipe diameter
0.1
0.05
0.04
0.03
friction factor
0.02
0.015
0.01
0.008
0.006
0.004
laminar
0.002
0.001
0.0008
0.0004
0.0002
0.0001
0.00005
0.01
1E+03
smooth
1E+04
1E+05
1E+06
Re
1E+07
1E+08
Pipe Flow Summary (1)
• Shear increases linearly
_________ with
distance from the center of the pipe (for
both laminar and turbulent flow)
• Laminar flow losses and velocity
distributions can be derived based on
momentum (Navier Stokes) and energy
conservation
• Turbulent flow losses and velocity
distributions requireexperimental
___________
Pipe Flow Summary (2)
• Energy equation left us with the elusive head
•
•
loss term
Dimensional analysis gave us the form of the
head loss term (pressure coefficient)
Experiments gave us the relationship between
the pressure coefficient and the geometric
parameters and the Reynolds number (results
summarized on Moody diagram)
Pipe Flow Summary (3)
• Dimensionally correct equations fit to
the empirical results can be
incorporated into computer or calculator
solution techniques
• Minor losses are obtained from the
pressure coefficient based
on the fact
constant
that the pressure coefficient is _______
at high Reynolds numbers
• Solutions for discharge or pipe diameter
Pressure Coefficient for a
Venturi Meter
Cp
10
 2p
Cp 
V 2
1
1E+01
1E+02
1E+03
1E+04
1E+05
1E+06
Re
Vl
Re 

0.1
0.05
0.04
0.03
0.02
0.015
friction factor
1E+00
0.01
0.008
0.006
0.004
laminar
0.002
0.001
0.0008
0.0004
0.0002
0.0001
0.00005
0.01
1E+03
smooth
1E+04
1E+05
1E+06
Re
1E+07
1E+08
Moody Diagram
0.1
 D
f  Cp 
l 

0.05
0.04
0.03
friction factor
0.02
0.015
0.01
0.008
0.006
0.004
e
D
laminar
0.002
0.001
0.0008
0.0004
0.0002
0.0001
0.00005
0.01
1E+03
smooth
1E+04
1E+05
1E+06
Re
1E+07
1E+08
Minor Losses
LSC Pipeline
z=0
cs1
cs2
0
Ignore minor losses
V12 p2
V22
 z1  1

 z2   2
 hl

2g 
2g
p1
KE will be small
Q  2m 3 / s
-2.85 m
28 kPa is equivalent to 2.85 m of water
5.2 0.04


2 



LQ
L

D  0.66 e 1.25 
 Q9.4 


 gh 
 gh  

f 

 f  

V22
D  154
. m V  1.07m / s  2
 0.06 m
2g
4.75
  106 m 2 / s
L  3100m
e  0.002m
h f  2.85m
Directions
• Assume fully turbulent (rough pipe law)
– find f from Moody (or from von Karman)
• Find total head loss (draw control volume)
• Solve for Q using symbols (must include
minor losses) (no iteration required)
hl   hf   hminor
Solution
0.1
0.05
0.04
0.03
friction factor
0.02
0.015
0.01
0.008
0.006
0.004
laminar
0.002
0.001
0.0008
0.0004
0.0002
0.0001
0.00005
0.01
1E+03
smooth
1E+04
1E+05
1E+06
Re
1E+07
1E+08
Pipe roughness
Find Q given pipe system
hminor  K
8Q 2
g D
2
4
hf  f
8
g 2
hl   hf   hminor
8Q 2 
 L 
 K
hl 
f


 4


5 
g 2 
D


D




ghl
L 
K



8   f 5     4
 D 
D




Q 
LQ 2
D5