APPLICATION I: STEADY PIPE FLOW IN SIMPLE
PIPELINES
M.S. Ghidaoui (CIVL 252, Spring 2001)
Pipeline systems are used to:
1. Transport water from where it is available to where it is needed: water supply;
seawater for flushing etc.
2. Dispose treated water into assimilative water bodies such as the ocean.
3. Convey sewage from sources to treatment plants.
4. Drain rain water (storm-water) from urban areas.
5. Transport oil and gas.
6. Conveyance of various chemicals in chemical plants.
7. Transport blood in living bodies (veins are pipes; blood is the fluid; heart is the
pump!).
8. Supply fuel from reservoirs to engines in cars, planes etc.
The above cited examples are by no means exhaustive. Try to think of more examples
yourself.
A modern water supply system often comprises:
1. A water source: In Hong Kong, about 80% of our water comes from
Guangdong, China, and about 20% from local natural rainfall catchment. We
expect to pay for 840 million cubic metres to Guangdong in the year 2000.
Assuming 7 million inhabitants, then we are buying about 330 litres per capita
(person) per day!
2. Pipes: In Hong Kong, the freshwater pipes add up to a total length of  about
5000km (i.e., roughly the earth’s radius!)
3. Pumps: The number of pumping stations for freshwater in Hong Kong is  145.
Pumps provide flexibility to vary flowrates and pressures; overcome energy losses
through the pipeline system and deliver water against gravity.
4. Impounding reservoirs: main water supply reservoirs (e.g., High Island
reservoir; Plover Cove reservoir etc.). There are 17 such reservoirs in Hong Kong.
5. Service reservoirs: located within the distribution system. There are 167 fresh
water service reservoirs in Hong Kong. They provide quick response to water
needs and storage capacity for low demand periods during the day and at night.
6. Control devices: to control the magnitudes of high and low pressures which
are generated either during steady or unsteady (surge or waterhammer events)
flow conditions. Example of control devices: surge tanks, air chambers, pressure
relief valves, air valves etc.
7. Check valves: Allows flow in one direction only. Mainly used in the vicinity of
pumps.
8. Pressure relief valves (PRV): maintains a constant pressure downstream of
it regardless of how large the upstream pressure becomes.
Function of valves: control flow and pressure through the pipeline system. Some
typical control valves are given the figure below.
For data related to freshwater and flushing water supply systems in Hong Kong and
for a figures of typical distribution systems in Hong Kong you may consult the web
page of the water supply department (WSD): http://www.info.gov.hk/wsd/index.htm.
The design, analysis, planning and management of pipeline systems are based on
hydraulics principles. The hydraulic principles of pipeline systems are presented in
this part of the course.
In the design, analysis, planning and management engineers are often faced by
questions such as:
1.
2.
3.
4.
5.
Should one assume steady or unsteady flow?
What flowrate should one design for?
How large and how strong should the pipe be?
How big and how many pumps does one need?
Where, how many and what kind of control devices are needed? Etc.
The answers to these questions require good understanding of hydraulic principles. In
this course, emphasis is placed on steady state analysis and only a brief introduction
to unsteady pipe flow (waterhammer) will be given.
Steady State Flow In Pipes



Pipe sizes and pumps are often selected on the basis of steady state analysis.
Reservoir size is often selected on the basis of quasi-steady analysis.
Pipe strengths and control devices are selected on the basis of unsteady
(waterhammer, surge, transient) analysis.
Review of the Fundamental Equations:
We have up to now derived the fundamental equations of mass, momentum and
energy for steady pipe flows; went through detailed discussions of frictional head
losses in both laminar and turbulent flows; and showed how the friction factor can be
obtained for different flow regimes and pipe characteristics. The student is highly
encouraged to go back to those notes so as to understand the principles and all the
assumptions involved in arriving at the basic equations for steady pipe flows.
Continuity Equation:
V1 A1  V2 A2
where V is the cross-sectional average of the velocity distribution V (r ) . Note, density
is assumed constant.
Momentum Equation:


Fx  Q  2 V2  1V1
 1 and  2 are determined from the velocity profiles.  1 and  2 are much larger than
1 in laminar pipe flow but very close to 1 in turbulent pipe flows. Hence,  1 and  2
are usually taken as being 1.0 in turbulent steady pipe flows.
Energy Equation:
In Civil 151, you derived the following energy equation along a streamline by starting
from the energy equation for a control volume:
V12
P2
V22
 z1 
 H Pump 
 z2 
 H turbine  hl12

2g

2g
This is the energy equation along a streamline. You are required to review the
derivation of this equation an pay special attention to the assumptions made in the
course of the derivation.
P1
Taking two elemental areas dA1 and dA2 such the continuity equation gives V1 dA1 =
V2 dA2 . Therefore, multiplying the left hand side of the above energy equation by V1
dA1 and the right hand side by V2 dA2 and integrating over the areas gives:
 P1

 P2

V12
V22




z


H
V
dA


z

A   1 2 g Pump  1 1 A   2 2 g  H turbine  hl12 V2 dA2
1
2
Carrying out the integration gives:
2
2
P

P

V1
V2
1
2



Q
 z1   1
 H Pump  Q
 z2   2
 H turbine  hl12 




2g
2g




Or
P1
2
2
V
P
V
 z1  1 1  H Pump  2  z 2   2 2  H turbine  hl12

2g

2g
The energy correction factors  1 and  2  1 for steady turbulent flows and often
much larger than 1 for steady laminar flows. The above energy equation is averaged
over the streamlines.
The Head Loss Equation of Darcy-Weisbach:
The Darcy-Weisbach equation states that the head loss in pipe segment of length x
is:
x V 2
hl  f
D 2g
f  friction factor
D  pipe diameter
V  average velocity  Q A
Written in term of Q , the Darcy-Weisbach equation is as follows:
x Q 2
D 2gA 2
where f is often determined from the Moody diagram or the equations derived in the
last chapter.
hl  f
Examples:
Problem 1 : Crude oil ( = 860 kg/m3,  = 0.008 Pa.s) flows through a 100mm
diameter, 5km long pipeline at a rate of 5 m3/h. Find the head loss due to friction.
Solution:
Q 5 / 3600
V 
 0.1768 m / s
A 
2
0.1
4
 0.008
 
 9.302 10 6 m 2 / s
 860
V D 0.1768  0.1
Re 

 1900.7  2000

9.302  10 6
 Laminar Flow
64
64
f

 0.03367
Re 1900.7
2
f L V 2 f L Q2
f L
8 fL Q 2
Q


 2
 2.683m
hL 
2
D 2g
D 2 g A2 D
g D5

2
 D 
2g 
4 


Problem 2 : Water flows through a 300mm diameter, 1km long new cast iron pipe ( 
= 0.26 mm). If the discharge is 0.1m3/s, determine the head loss in the pipe by the
Colebrook-white equation and the Churchill equation.
Solution:
e 0.26

8.667 10 4
D 300
  1  10 6 m 2 / s
water at T  20  C
V D QD
4Q
Re 


 424,413

A D
Colebrook  white :


 9.35
1
e
 1.14  2 log 10 
 
 Re f
D 
f

Trial and error 
f  0.01974
Churchill :
  7  0.9
1
e 
  2 log 10  


  Re 
3.7 D 
f

f  0.019878
(note : no trial and error )
hl  6.7 m.
Problem 3: Determine the head-loss for flow of 140 L/s of oil, = 0.00001 m2/s,
through a 400m of 200mm diameter cast-iron pipe.
2
L
L Q2
Solution: h L1 2  f V  f
D 2g
D 2 gA 2
Note: f is the only unknown in this case.
Solution:
V D V D



Re 
 V
Q
2
D
4

0.14  4
3.14   0.2 
2
 4.459 m / s
4.459  0.2
 89127
0.00001
Cast iron    0.2mm
Re 


D


0.25
 0.00125
200
 0.00125 and Re  90,000
D
f  0.023 ( From Moody Diagram )
For
Hence , h L1 2  h f 1 2  0.023
400 4.4592
 46.6 m
0.2 2  9.81
Problem 4 : Water at 15C flows through a 300mm riveted steel pipe (  =3mm) with
a head loss of 6m in 300m. Find Q.
Solution:
2
2
L
300
2
hL12  f D V2 g  f 0.3 2 V9.81  50.9 f V  6
Note, for simplicity the over-bar has been dropped. The head loss relationship
provides one equation but two unknowns f and V. How do we get the second

equation? For a given
, the Moody diagram relates f to R e . Thus, the second
D
equation is given by the Moody diagram. The solution in this case is not direct and
require trial and error!
V  2m / s  f 
6
50.9  2 
2
 0.029
VD
2  0.3
2
5

m / s  5  10
ν
1.13  10 6
 0.003

 0.01
D
0.3
 From Moody diagram f  0.038  0.029
 Guess is not accurate enough
Check : R e 
Assume V  1.7m/s
Check : R e 

f 
6
50.9  1.7 
2
 0.04
VD 1.7  0.3 2
6

m / s  4.5  10
ν
1.13  10 6

 0.01 and e  4.5  106
D
 f  0.038 close enough to 0.04
 Accept
From Moody diagram with
Thus
,V  1.7 m / s
Q  VA  1.7 
 0.32
4
 0.12 m3 / s
Hazen-Williams Equation: Although the Darcy-Weisbach formula along with the
Moody diagram are considered to be the most accurate model for estimating frictional
head losses in steady pipe flow, this approach often requires trial and error solution.
Thus, the Darcy-Weisbach formula along with the Moody diagram are not efficient to
use in calculations especially in the old days when computers were not readily
available. Therefore, engineers have sought to find other empirical head loss
equations that do not require trial and error. One the most commonly used in practice
is the Hazen-Williams equation:
Q 1.852
1.852
C HW
D 4.87
where C HW =Hazen-Williams coefficient. These coefficients for clean new pipes of
different material are provided in the table below.
hL12  10.68
L
Typical Values of The Hazen-Williams Coefficient
Important: The factor of (10.68) in the Hazen-Williams equation is only for the case
when L and D are in meters and the flowrate (discharge) Q is in m 3 / s . If one
changes the units, the coefficient must also be will changed. For example, if one
wants to use L in meters and Q in m 3 / s but D in centimeters then the coefficient
becomes as follows
L
Q 1.852
L Q 1.852
10

10
.
68

5
.
869

10
hL12
1.852
1.852
C HW
(0.01D) 4.87
C HW
D 4.87
The Hazen-Williams formula is empirical and lacks physical basis. The HazenWilliams coefficients given in the table above are from tests which were carried out
for: (1) velocities around 0.9 m/s, (2) slope of energy grade line around 1/1000, (3)
pipe diameters around 0.6 m and (4) new and clean pipes. When actual flow
conditions depart significantly from the above test condition, corrections to the
Hazen-Williams coefficients may be warranted. However, most design engineers
continue to use the Hazen-Williams coefficients even for applications which depart
significantly from the test condition! One major reason for the reluctance of engineers
to adopt an improved version of the Hazen-Williams formula is that most corrections
to this formula defeat the purpose for which the Hazen-Williams equation was
established, namely, simplicity. Another reason is that many existing models are
based on the original version of the Hazen-Williams formula. The error which result
from using the Hazen-Williams formula in problems that depart from the test
conditions stipulated earlier in this paragraph, can be in the range of  30 %. Not only
is the Hazen-Williams equation used widely in practice, but it is also used extensively
in the third year ``municipal hydraulics course.’’
Of course, the Hazen-Williams formula is not the only empirical formula available.
For example, although Manning’s formula is mostly used in open channel (i.e.,
gravity driven flows), some engineers and agencies have and still use this formula for
pressurized pipe flows. The Manning’s formula will be covered in the open channel
part of this course.