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1
The Islamic University of Gaza
Faculty of Engineering
Civil Engineering Department
Hydraulics - ECIV 3322
Chapter 4
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 different systems.
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.
5
Simple pipe flow
(1)
(2)
6
To solve such system:
• Apply Bernoulli’s equation
• where
pL hhzg
VPz
g
VP 2
2
221
2
11
22
(1)
(2)
g
VK
g
V
D
fLhhh LmfL
22
22
For the same material and constant diameter (same f , same V) we can write:
L
Total
mfL KD
fL
g
Vhhh
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
LBBB
AAA hz
g
VPz
g
VP
22
22
332211 VAVAVAQ
11
LBBB
AAA hz
g
VPz
g
VP
22
22
Hhzz LBA
Where
4
1
3
1 j
mj
i
fiL hhh
g
VK
g
VK
g
VK
g
VK
g
V
D
Lfh exitenlcent
i
i
i
i
iL22222
2
3
2
2
2
2
2
13
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
Lf
g
V
D
Lf
g
V
D
Lf
222
2
3
3
3
3
2
2
2
22
2
1
1
11
14
Example Determine the flow in each pipe and the flow in the main pipe if Head loss
between A & B is 2m & f=0.01
Solution
/sm...π
AVQ
m/s.V
.
V
..
g
V.
D
Lf
332
111
1
2
1
2
1
1
1
1015350620404
5062
28192040
25010
22
221 ff hh
/sm.QQQ
/sm...π
Q
m/s.V
.
V
..
g
V.
D
Lf
33
21
332
2
2
2
2
2
2
2
2
10178
1002555720504
5572
8192050
30010
22
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 0.12m 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
031402
20 m.
.πAa
2
2
011302
120 m.
.πAb
ba
babbaa
hh
V.V.VAV A m.
QQQ
0113003140260 3
21
g
V.
g
V
D
Lfh aa
a
aaa
2150
2
22
g
V
g
V.
g
V
D
Lfh bbb
b
bbb
210
21902
2
222
Solution
17
g
V.
.
.f
g
V.
.f b
ba
a2
10380120
46
2150
20
422
22 38103353 15020 bbaa V.f.V.f
0255.0
0185.0
b
a
f
f
22 3810025503353 1500185020 ba V...V..
ba V.V 7194
m/s.V
m/s.V
b
a
6301
6937
V.V.VAV A m. bbbbaa 01130)719.4(03140260 3
/s m...VAQ
/s m...VAQ
bbb
aaa
3
3
0180630101130
2420693703140
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.
19
20
21
Group work 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
Group work 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 of 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
Pipe line 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 H.G.L.) in a pipeline above the vapor pressure of water (not be less than zero Absolute )
27
LSS
Sp
LSSS
p
pp
hP
g
VZZ
hZP
g
VZ
P
g
V
2
22
2
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)
28
The negative pressure at the summit point can reach theoretically
-10.3 m water head (gauge pressure) and zero (absolute pressure)
But in the practice water contains dissolved gasses that will vaporize
before -10.3 m water head which reduces the pipe flow cross
section.
Generally, this pressure reach to -7.6m water head (gauge pressure)
and 2.7m (absolute pressure)
29
Example
Siphon pipe between two pipe has diameter of 20cm and length
500m as shown. The difference between reservoir levels is 20m.
The distance between reservoir A and summit point S is 100m.
Calculate the flow in the system and the pressure head at summit.
f=0.02
30
Solution
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 design 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 trail.
• 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
AJ BJ CJ Pipe
1000 4000 2000 Length m
30 50 40 Diameter cm
0.024 0.021 0.022 f
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
LfZZ PA
23.0
1000024.0110120
2.
2
1
2
1
1
1
1
V1 = 1.57 m/s , Q1 = 0.111 m3/s
g
V
g
V
D
LfZZ BP
25.0
4000021.0100110
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
LfZZ CP
24.0
2000022.080110
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
99m Pat 0 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
46
m.Z
.
.
..Z
g
V.
D
LfZZ
m/s.
.π
.
A
QV
P
PPA
47536
8192
8490
30
1200024040
2
8490
304
060
22
1
1
11
21
11
Solution
/sm 0.0203Q0.645m/sV
9.812
V
0.2
6000.02436.47538
2g
V.
D
LZZ
3
22
2
2
2
2
2
22PB
f
Applying Bernoulli Equation between B , J :
Applying Bernoulli Equation between A , J :
47
mZ
gg
V
D
LfZZ
c
CP
265.32
2
136.1
3.0
800024.0 Z- 6.4753
2.
2
c
2
3
3
33
Applying Bernoulli Equation between C , J :
smQQQ
QQQQ
/ 0803.00203.006.0
0
3
213
321
sm
A
QV / 136.1
3.04
0803.0
23
3
3
48
Group Work
ACAB
ACAB
BDBC
BDBCAB
VV
VV
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
10 816.0)125.1(55.2
10 7.155.2
1023.0
100001.0
24.0
200001.0
10
3
3
22
22
01.0f
Find the flow in each pipe
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
Weightx
weight
Energy
time
EnergyPower
ZP
g
VH
2
2
QQgtime
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 the figure, the following can be stated:
mf
m
f
hhHγ Q
γ Q h
γ Q h
γ Q H
PowerExit At
lossminor todue dissipatedPower
friction todue dissipatedPower
Power EntranceAt
53
Condition for Maximum Transmission of Power:
The condition for maximum transmission of power occurs when : 0dV
dP
][ mf hhHQP
Neglect minor losses and use VDAVQ ]4
[ 2
So ]2
[4
32
g
V
D
LfHVDP
0]2
3[
4
22 VDg
fLHD
dV
dP
fhg
V
D
fLH 3
23
2
3
Hh f
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
inlet at the suppliedPower
outlet at the availablePower
H
hhH
QH
hhHQ mfmf ][][
or
H
hH f ][
Maximum efficiency of power transmission occurs when 3
Hh f
%67.663
2]
3[
max
H
HH
(If we neglect minor losses)
55
Example
Pipe line has length 3500m and Diameter 0.5m is used to transport
Power Energy using water. Total head at entrance = 500m. Determine
the maximum power at the Exit. F = 0.024
fout h Hγ QP
mH
h f3
500
3at Power Max.
g
V
..
g
V
D
Lfh f
230
35000240
2
22
m/s 3.417V
/s m...AVQ π 32
424150417330
56
HP.
tt) N.m/s (Wa
..
HgQ
HHgQ
hHγQP f
10597745
789785789785
500241508191000
3
32
32