Ch4 part 1

1
The Islamic University of Gaza
Faculty of Engineering
Civil Engineering Department
Hydraulics - ECIV 3322
Chapter 4
Part 1
Pipelines and Pipe Networks

2
Introduction
Any water conveying system may include
the following elements:
• pipes (in series, pipes in parallel)
• elbows
• valves
• other devices.
• If all elements are connected in series,
The arrangement is known as a pipeline.
• Otherwise, it is known as a pipe network.

3
How to solve flow problems
• Calculate the total head loss (major and
minor) using the methods of chapter 3
• Apply the energy equation (Bernoulli’s
equation)
This technique can be applied for
differentsystems.

4
Flow Through A Single Pipe
(simple pipe flow)
• A simple pipe flow: It is a
• flow takes place in one pipe
• having a constant diameter
• with no branches.
• This system may include bends, valves,
pumps and so on.

6
To solve such system:
• Apply Bernoulli’s equation
• where
pL hhz
g
VP
z
g
VP
 2
2
22
1
2
11
22 
(1)
(2)
 
g
V
K
g
V
D
fL
hhh LmfL
22
22
For the same material and constant diameter (same f , same V) we can write:






  L
Total
mfL K
D
fL
g
V
hhh
2
2

7
Example
Determine the difference in the elevations between
the water surfaces in the two tanks which are
connected by a horizontal pipe of diameter 30 cm and
length 400 m. The rate of flow of water through the
pipe is 300 liters/sec. Assume sharp-edged entrance
and exit for the pipe. Take the value of f = 0.032.
Also, draw the HGL and EGL.
Z1 Z

8
Compound Pipe flow
• When two or more pipes with different
diameters are connected together head to
tail (in series) or connected to two common
nodes (in parallel)
The system is called compound pipe flow

9
Flow Through Pipes in Series
• pipes of different lengths and different
diameters connected end to end (in series) to
form a pipeline

10
• Discharge:The discharge through each pipe is the same
• Head loss: The difference in liquid surface levels is equal to the sum
of the total head loss in the pipes:
332211 VAVAVAQ 
LB
BB
A
AA
hz
g
VP
z
g
VP

22
22

332211 VAVAVAQ 

11
LB
BB
A
AA
hz
g
VP
z
g
VP

22
22

Hhzz LBA 
Where
 

4
1
3
1 j
mj
i
fiL hhh
g
V
K
g
VV
K
g
V
K
g
V
K
g
V
D
L
fh exitenlcent
i
i
i
i
iL
22
)(
222
2
3
2
32
2
2
2
1
3
1
2


 

12
Flow Through Parallel Pipes
• If a main pipe divides into two
or more branches and again
join together downstream to
form a single pipe, then the
branched pipes are said to be
connected in parallel
(compound pipes).
• Points A and B are called
nodes.
Q1, L1, D1, f1
Q2, L2, D2, f2
Q3, L3, D3, f3

13
• Discharge:
• Head loss: the head loss for each branch is the same


3
1
321
i
iQQQQQ
Q1, L1, D1, f1
Q2, L2, D2, f2
Q3, L3, D3, f3
321 fffL hhhh 
g
V
D
L
f
g
V
D
L
f
g
V
D
L
f
222
2
3
3
3
3
2
2
2
2
2
2
1
1
1
1 

14
Example
Determine the flow in each pipe and the main pipe if the head
loss between nodes A and B is 2 m and f=0.01.
Solution
  /sm...
π
AVQ
m/s.V
.
V
.
.
g
V
.
D
L
f
332
111
1
2
1
2
1
1
1
101535062040
4
5062
2
8192040
25
010
2
2







221  ff hh
 
/sm.QQQ
/sm...
π
Q
m/s.V
.
V
.
.
g
V
.
D
L
f
33
21
332
2
2
2
2
2
2
2
2
10178
100255572050
4
5572
8192050
30
010
2
2









15
Example
The following figure shows pipe system from cast iron steel.
The main pipe diameter is 0.2 m with length 4m at the end
of this pipe a Gate Valve is fixed as shown. The second pipe
has diameter 0.12 m with length 6.4m, this pipe connected
to two bends R/D = 2.0 and a globe valve. Total Q in the
system = 0.26 m3/s at T=10oC. Determine Q in each pipe at
fully open valves.

16
2
2
03140
2
20
m.
.
πAa 






2
2
01130
2
120
m.
.
πAb 






ba
babbaa
hh
V.V.VAVAm.
QQQ



0113003140260 3
21
g
V
.
g
V
D
L
fh aa
a
a
aa
2
150
2
22
  
g
V
g
V
.
g
V
D
L
fh bbb
b
b
bb
2
10
2
1902
2
222

Solution

17
g
V
.
.
.
f
g
V
.
.
f b
b
a
a
2
10380
120
46
2
150
20
4
22
























    22
3810335315020 bbaa V.f.V.f  0255.0
0185.0
b
a


f
f
      22
38100255033531500185020 ba V...V.. 
ba V.V 7194
m/s.V
m/s.V
b
a
6301
6937


V.V.VAVAm. bbbbaa 01130)719.4(03140260 3

 
  /sm...VAQ
/sm...VAQ
bbb
aaa
3
3
0180630101130
2420693703140


by trial and error

18
Example
Determine the flow rate in each pipe (f=0.032).
Also, if the two pipes are replaced with one pipe of the
same length determine the diameter which give the same
flow.

21
Example
Four pipes connected in parallel as shown. The following
details are given:
Pipe L (m) D (mm) f
1 200 200 0.020
2 300 250 0.018
3 150 300 0.015
4 100 200 0.020
• If ZA = 150 m , ZB = 144m, determine the
discharge in each pipe ( assume PA=PB = Patm)

22
Example
Two reservoirs with a difference in water levels of 180 m
and are connected by a 64 km long pipe of 600 mm
diameter and f = 0.015. Determine the discharge through
the pipe. In order to increase this discharge by 50%, another
pipe of the same diameter is to be laid from the lower
reservoir for part of the length and connected to the first
pipe (see figure below). Determine the length of additional
pipe required.
=180m
QN QN1
QN2

23
Pipeline with negative Pressure
(Siphon phenomena)
• Long pipelines laid to transport water from one reservoir
to another over a large distance usually follow the natural
contour of the land.
• A section of the pipeline may be raised to an elevation
that is above the local hydraulic gradient line (siphon
phenomena) as shown:

24
Definition:
It is a long bent pipe which is used to transfer liquid
from a reservoir at a higher elevation to another
reservoir at a lower level when the two reservoirs are
separated by a hill or high ground
Occasionally, a section of the pipeline may be
raised to an elevation that is above the local HGL.
(siphon phenomena)

25
Siphon happened in the following cases:
• To carry water from one reservoir to another
reservoir separated by a hill or high ground
level.
• To take out the liquid from a tank which is not
having outlet
• To empty a channel not provided with any
outlet sluice.

26
Characteristics of this system
• Point “S” is known as the summit.
• All Points above the HGL have pressure less
than atmospheric (negative value)
• If the absolute pressure is used then the
atmospheric absolute pressure = 10.33 m
• It is important to maintain pressure at all
points (above HGL) in a pipeline above the
vapor pressure of water (not be less than
zero Absolute )

27
L
S
Sp
LS
SS
p
pp
h
P
ZZ
hZ
P
g
V
Z
P
g
V



 22
22
A S
-ve value Must be -ve value ( below the atmospheric pressure)
Negative pressure exists in the pipelines wherever the pipe line is raised above the
hydraulic gradient line (between P & Q)
Sp VV 

28
The negative pressure at the summit point can reach
theoretically to -10.33 m water head (gauge pressure) and
zero (absolute pressure). But in the practice water
contains dissolved gasses that will vaporize before -10.33
m water head which reduces the pipe flow cross section.
Generally, this pressure reach to -7.6 m water head (gauge
pressure) and 2.7 m (absolute pressure)
In practice…

29
Example
Siphon pipe between two tanks and pipe has diameter of
20 cm and length 500 m as shown. The difference
between reservoir levels is 20 m. The distance between
reservoir A and summit point S is 100 m. Calculate the
flow in the system and the pressure head at summit.
f=0.02
3m
20m

31
• Pumps may be needed in a pipeline to lift water
from a lower elevation or simply to boost the rate
of flow. Pump operation adds energy to water in
the pipeline by boosting the pressure head
• The computation of pump installation in a
pipeline is usually carried out by separating the
pipeline system into two sequential parts, the
suction side and discharge side.
Pumps

32
LsRP hHHH 
See example 4.5
Pumps selection will
be discussed in details
in next chapters

33
Branching in pipes occur when water is brought by pipes to a
junction when more than two pipes meet.
This system must simultaneously satisfy two basic conditions:
1 – The total amount of water brought by pipes to a junction must
equal to that carried away from the junction by other pipes.
2 – All pipes that meet at the junction must share the same pressure
at the junction. Pressure at point J = P
Branching pipe systems
  0Q

34
Three-reservoirs problem
(Branching System)
How we can demonstrate the
hydraulics of branching pipe System??
by the classical three-reservoirs problem

35
This system must satisfy:
Q3 = Q1 + Q2
2) All pipes that meet at junction “J” must
share the same pressure at the junction.
1) The quantity of water brought to junction “J” is equal
to the quantity of water taken away from the junction:
Flow Direction????

36
Types of three-reservoirs problem:
Type 1:
• given the lengths, diameters, and materials of all pipes
involved
D1 , D2 , D3 , L1 , L2 , L3 , and e or f
• given the water elevation in each of the three reservoirs
Z1 , Z2 , Z3
• determine the discharges to or from each reservoir,
Q1 , Q2 and Q3
Two types
This types of problems are most conveniently
solved by trial and error

37
• First assume a piezometric surface elevation, P , at the junction.
• This assumed elevation gives the head losses hf1, hf2, and hf3
• From this set of head losses and the given pipe diameters, lengths,
and material, the trial computation gives a set of values for
discharges Q1 , Q2 ,and Q3 .
• If the assumed elevation P is correct, the computed Q’s should
satisfy:
• Otherwise, a new elevation P is assumed for the second trial.
• The computation of another set of Q’s is performed until the above
condition is satisfied.
Q Q Q Q    1 2 3 0

38
Note:
• It is helpful to plot the computed trial values of P
against ΣQ.
• The resulting difference may be either plus or minus
for each trial.
• However, with values obtained from three trials, a
curve may be plotted as shown in the next example.
The correct discharge is indicated by the
intersection of the curve with the vertical axis.

39
Example
AJBJCJPipe
100040002000Length m
305040Diameter cm
0.0240.0210.022f
In the following figure determine the flow in each pipe

40
Trial 1
ZP= 110m
Applying Bernoulli Equation between A , J :
g
V
g
V
D
L
fZZ PA
23.0
1000
024.0110120
2
.
2
1
2
1
1
1
1 
V1 = 1.57 m/s , Q1 = 0.111 m3/s
g
V
g
V
D
L
fZZ BP
25.0
4000
021.0100110
2
.
2
2
2
2
2
2
2 
V2 = 1.08 m/s , Q2 = - 0.212 m3/s
Applying Bernoulli Equation between B , J :

41
g
V
g
V
D
L
fZZ CP
24.0
2000
022.080110
2
.
2
3
2
3
3
3
3 
Applying Bernoulli Equation between C , J :
V3 = 2.313 m/s , Q2 = - 0.291 m3/s
0392.0291.0212.0111.0321  QQQQ

42
Trial 2
ZP= 100m
0/08.0237.00157.0 3
321  smQQQQ
Trial 3
ZP= 90m
0/324.0168.03.0192.0 3
321  smQQQQ

43
Draw the relationship between and PQ
99mPat0 Q

44
Type 2:
• Given the lengths , diameters, and materials of all pipes involved;
D1 , D2 , D3 , L1 , L2 , L3 , and e or f
• Given the water elevation in any two reservoirs,
Z1 and Z2 (for example)
• Given the flow rate from any one of the reservoirs,
Q1 or Q2 or Q3
• Determine the elevation of the third reservoir Z3 (for example) and
the rest of Q’s
This types of problems can be solved by simply using:
• Bernoulli’s equation for each pipe
• Continuity equation at the junction.

45
Example
In the following figure determine the flow in pipe BJ & pipe
CJ. Also, determine the water elevation in tank C. take f =
0.024

46
 
m.Z
.
.
.
.Z
g
V
.
D
L
fZZ
m/s.
.
π
.
A
Q
V
P
PPA
47536
8192
8490
30
1200
024040
2
8490
30
4
060
22
1
1
1
1
2
1
1
1




Solution
/sm0.0203Q0.645m/sV
9.812
V
0.2
600
0.02436.47538
2g
V
.
D
L
ZZ
3
22
2
2
2
2
2
2
2PB


 f
Applying Bernoulli Equation between B , J :
Applying Bernoulli Equation between A , J :

47
mZ
gg
V
D
L
fZZ
c
CP
265.32
2
136.1
3.0
800
024.0Z-6.4753
2
.
2
c
2
3
3
3
3


Applying Bernoulli Equation between C , J :
smQQQ
QQQQ
/0803.00203.006.0
0
3
213
321


 
sm
A
Q
V /136.1
3.0
4
0803.0
23
3
3 


48
Group Work
ACAB
ACAB
BDBC
BDBCAB
VV
VV
QQ
QQQ
125.1
3.024.0
0
2
4
2
4





smQsmV
smQQsmV
VV
VV
g
V
g
V
hh
ABAB
BDBCBC
BCBC
BCAB
ABAB
BCAB
/31.0/5.2
/155.0/2.2
10816.0)125.1(55.2
107.155.2
10
23.0
1000
01.0
24.0
2000
01.0
10
3
3
22
22






01.0f
Find the flow in each pipe
VBC
2

49
Power Transmission Through Pipes
• Power is transmitted through pipes by the
water (or other liquids) flowing through them.
• The power transmitted depends upon:
(a) the weight of the liquid flowing through the pipe
(b) the total head available at the end of the pipe.

50
• What is the power available at the end B of
the pipe?
• What is the condition for maximum
transmission of power?

51
Total head (energy per unit weight) H of fluid is
given by:
time
Weight
x
weight
Energy
time
Energy
Power
Z
P
g
V
H


2
2
QQg
time
Weight
 
Therefore:
Power Q H 
Units of power:
N . m/s = Watt
745.7 Watt = 1 HP (horse power)

52
For the system shown in figure, the following can be stated:
 mf
m
f
hhHγ Q
γ Q h
γ Q h
γ Q H




PowerExitAt
lossminortoduedissipatedPower
frictiontoduedissipatedPower
PowerEntranceAt

53
Condition for Maximum Transmission of Power:
The condition for maximum transmission of power occurs when : 0
dV
dP
][ mf hhHQP 
Neglect minor losses and use VDAVQ ]
4
[ 2

So ]
2
[
4
3
2
g
V
D
L
fHVDP 


0]
2
3
[
4
22
 V
Dg
fL
HD
dV
dP 

fh
g
V
D
fL
H 3
2
3
2

3
H
hf 
 Power transmitted through a pipe is maximum when the loss of head due
of the total head at the inlet
3
1

to friction equal

54
Maximum Efficiency of Transmission of Power:
Efficiency of power transmission is defined as
inletat thesuppliedPower
outletat theavailablePower

H
hhH
QH
hhHQ mfmf ][][ 






or
H
hH f ][ 

Maximum efficiency of power transmission occurs when
3
H
hf 
%67.66
3
2
]
3
[
max 


H
H
H

(If we neglect minor losses)


55
Example
Pipe line has length 3500m and Diameter 0.3m is used to
transport Power Energy using water. Total head at entrance
= 500m. Determine the maximum power at the Exit. f =
0.024
 fout hHγ QP 
m
H
hf
3
500
3
atPowerMax. 
g
V
.
.
g
V
D
L
fhf
230
3500
0240
2
22

m/s3.417V 
    /sm...AVQ π 32
4 24150417330 

56
 
    
HP
.
tt)N.m/s (Wa
..
HgQ
H
HgQ
hHγQP f
1059
7745
789785
789785
500241508191000
3
3
2
3
2














Ch4 part 1

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