This document discusses pipelines and pipe networks in hydraulics and hydrology. It begins by defining a pipeline as a system where all elements are connected in series, while a pipe network has elements connected to two common nodes. It then provides examples of calculating flow through single pipes, pipes in series, parallel pipes, and compound pipe flows. The document also discusses flow through pipelines that require negative pressure like siphons. Finally, it briefly mentions the use of pumps to boost pressure head in pipelines.

1. Civil Engineering Department
Prof. Majed Abu-Zreig
Hydraulics and Hydrology – CE 352
Chapter 4
Pipelines and Pipe Networks
10/4/2012 1

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.
10/4/2012 2

3. How to solve flow problems
• Calculate the total head loss (major and
minor) using the methods of chapter 2
• Apply the energy equation (Bernoulli’s
equation)
This technique can be applied for
different systems.
10/4/2012 3

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.
10/4/2012 4

5. (2)
(1)
Simple pipe flow
10/4/2012 5

6. (2)
To solve such system:
(1)
• Apply Bernoulli’s equation
P1 V12 P2 V22
  z1    z 2  hL  h p
 2g  2g
• where
fL V 2 V2
hL  h f  hm     KL
D 2g 2g
For the same material and constant diameter (same f , same V) we can write:
V 2  fLTotal 
hL  h f  hm 
2g  D   KL 

10/4/2012 6

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
10/4/2012 7

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
10/4/2012 8

9. Flow Through Pipes in Series
• pipes of different lengths and different
diameters connected end to end (in series) to
form a pipeline
10/4/2012 9

10. • Discharge:The discharge 1  A2V2  pipe is the same
Q  A1Vthrough each A3V3
Q  A1V1  A2V2  A3V3
• Head loss: The difference in liquid surface levels is equal to the sum
of the total head loss in the pipes:
PA V A2 PB VB2
  zA    z B  hL
 2g  2g
10/4/2012 10

11. PA V A2 PB VB2
  zA    z B  hL
 2g  2g
z A  z B  hL  H
Where
3 4
hL   h fi   hmj
i 1 j 1
3
Li Vi 2 V12 V22 V22 V32
hL   f i  K ent  Kc  K enl  K exit
i 1 Di 2 g 2g 2g 2g 2g
10/4/2012 11

12. Example
• Two new cast-iron pipes in series connect two reservoirs. Both pipes
are 300 m long and have diameters of 0.6 m and 0.4 m, respectively.
• The elevation of water surface in reservoir A is 80 m. The discharge of
10o C water from reservoir A to reservoir B is 0.5 m3/sec.
• Find the elevation of the surface of reservoir B.
• Assume a sudden contraction at the junction and a square-edge
entrance.
10/4/2012 12

13. Flow Through Parallel Pipes
• If a main pipe divides into two
or more branches and again
join together downstream to Q1, L1, D1, f1
form a single pipe, then the Q2, L2, D2, f2
branched pipes are said to be
connected in parallel Q3, L3, D3, f3
(compound pipes).
• Points A and B are called
nodes.
10/4/2012 13

14. Q1, L1, D1, f1
Q2, L2, D2, f2
Q3, L3, D3, f3
• Discharge:
3
Q  Q1  Q2  Q3   Qi
i 1
• Head loss: the head loss for each branch is the same
hL  h f 1  h f 2  h f 3
L1 V12 L2 V22 L3 V32
f1  f2  f3
D1 2 g D2 2 g D3 2 g
10/4/2012 14

15. 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
hf 1  hf 2  2 L2 V22
f . 2
L1 V12 D2 2 g
f . 2
D1 2 g 30 V22
0.01 
25 V12 0.05 2  9.81
0.01  2 V2  2.557 m/s
0.04 2  9.81
V1  2.506 m/s π
Q2  0.052  2.557  5.02 103 m3 /s
π 4
10/4/2012
Q1  V1 A1  0.042  2.506  3.15 103 m3 /s 3 3
Q  Q1  Q2  8.17 10 m /s
15
4

16. 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.
10/4/2012 16

17. Solution
2
 0.2 
Aa  π    0.0314 m
2
 2 
2
 0.12 
Ab  π    0.0113 m
2
 2 
Q  Q1  Q2
0.26 m3  AaVa  AbVb  0.0314Va  0.0113Vb
ha  hb
2 2 2
 20.19
2 2 Lb Vb Vb Vb
La Va Va hb  f b  10
ha  f a  0.15 Db 2 g 2g 2g
Da 2 g 2g
10/4/2012 17

18.   4  V   6.4  V
2 2
 fa    0.15 a   f b    0.38  10 b
  0.2   2 g   0.12   2g
f a  0.0185
20 f a  0.15Va 2
 53.33 f b  10.38Vb
2
f b  0.0255
200.0185  0.15Va 2  53.330.0255  10.38Vb 2
Va  4.719 Vb
0.26 m3  AaVa  AbVb  0.0314(4.719Vb )  0.0113Vb
Va  7.693 m/s Qa  AaVa  0.03147.693  0.242 m3 /s
Vb  1.630 m/s Qb  AbVb  0.01131.630  0.018 m3 /s
10/4/2012 18

19. Example
Determine the flow rate in each pipe (f=0.015)
Also, if the two pipes are replaced with one pipe of the same length
determine the diameter which give the same flow.
10/4/2012 19

20. 10/4/2012 20

21. 10/4/2012 21

22. 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)
10/4/2012 22

23. 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
10/4/2012 23

24. 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:
10/4/2012 24

25. (siphon phenomena)
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.
10/4/2012 25

26. 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.
10/4/2012 26

27. 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 )
10/4/2012 27

28. A S
2 2
Vp Pp VS P
  Zp   S  Z S  hL
2g  2g 
2
VS PS
Z p  ZS    hL
2g 
-ve value Must be -ve value ( below the atmospheric pressure)
Negative pressure exists in the pipelines wherever the pipe line is raised above the
10/4/2012 hydraulic gradient line (between P & Q) 28

29. 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)
10/4/2012 29

30. 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
10/4/2012 30

31. Solution
10/4/2012 31

32. Pumps
• 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.
10/4/2012 32

33. H P  H R  H s  hL
See example 4.5
Pumps design will be
discussed in details in
next chapters
10/4/2012 33

34. Branching pipe systems
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.
Q  0
2 – All pipes that meet at the junction must share the same pressure at the
junction. Pressure at point J = P
10/4/2012 34

35. How we can demonstrate the hydraulics of branching
pipe System??
by the classical three-reservoirs problem
Three-reservoirs problem
(Branching System)
10/4/2012 35

36. This system must satisfy:
1) The quantity of water brought to junction “J” is equal
to the quantity of water taken away from the junction:
Q3 = Q1 + Q2 Flow Direction????
2) All pipes that meet at junction “J” must share the
same pressure at the junction.
10/4/2012 36

37. Types of three-reservoirs problem:
Two types
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 .
This types of problems are most conveniently
solved by trail and error
10/4/2012 37

38. • 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 trail 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:
 Q  Q1  Q2  Q3  0
• Otherwise, a new elevation P is assumed for the second trail.
• The computation of another set of Q’s is performed until the above
10/4/2012 38
condition is satisfied.

39. Note:
• It is helpful to plot the computed trail values of P against .
• The resulting difference may be either plus or minus for each
trail.
• However, with values obtained from three trails, 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.
10/4/2012 39

40. Example
In the following figure determine the flow in each pipe
Pipe CJ BJ AJ
Length m 2000 4000 1000
Diameter cm 40 50 30
f 0.022 0.021 0.024
10/4/2012 40

41. Trial 1
ZP= 110m
Applying Bernoulli Equation between A , J :
2
L1 V1 1000 V12
Z A  Z P  f1 . 
 120  110  0.024  
D1 2 g 0.3 2 g
V1 = 1.57 m/s , Q1 = 0.111 m3/s
Applying Bernoulli Equation between B , J :
2
L V 4000 V22
ZP  ZB  f2 2 . 2 
 110  100  0.021  
D2 2 g 0.5 2 g
V2 = 1.08 m/s , Q2 = - 0.212 m3/s
10/4/2012 41

42. Applying Bernoulli Equation between C , J :
2
L V 2000 V32
Z P  ZC  f3 3 . 3 
 110  80  0.022  
D3 2 g 0.4 2 g
V3 = 2.313 m/s , Q2 = - 0.291 m3/s
Q  Q  Q 1 2  Q3  0.111  0.212  0.291  0.392  0
10/4/2012 42

43. Trial 2
ZP= 100m
 Q  Q1  Q2  Q3  0.157  0  0.237  0.08 m 3 / s  0
Trial 3
ZP= 90m
 Q  Q1  Q2  Q3  0.192  0.3  0.168  0.324 m 3 / s  0
10/4/2012 43

44. Draw the relationship between  Q and P
  Q  0  at P  99m
10/4/2012 44

45. 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.
10/4/2012 45

46. Example
In the following figure determine the flow in pipe BJ & pipe CJ. Also,
determine the water elevation in tank C
10/4/2012 46

47. Solution
Applying Bernoulli Equation between A , J :
Q1 0.06
V1    0.849 m/s
π
A1 0.32
4
2
L1 V1 1200 0.849 2
Z A  Z P  f1 .  40  Z P  0.024  
D1 2 g 0.3 2  9.81
Z P  36.475 m
Applying Bernoulli Equation between B , J :
2 2
L V 600 V2
ZB  ZP  f 2 2 . 2 
 38  36.475  0.024  
D 2 2g 0.2 2  9.81
V2  0.645m/s  Q 2  0.0203 m 3 /s

10/4/2012 47

48. Applying Bernoulli Equation between C , J :
Q  Q 1  Q2  Q3  0
Q3  Q1  Q2  0.06  0.0203  0.0803 m 3 / s
Q3 0.0803
V3    1.136 m / s
A3 
0.32
4
2 2
L3 V3 800 1.136
Z P  ZC  f3 . 
 36.475 - Zc  0.024  
D3 2 g 0.3 2g
Z c  32.265 m
10/4/2012 48

49. Group Work
Find the flow in each pipe
f  0.01
hAB  hBC  10
QAB  QBC  QBD  0
2 2
QBC  QBD 0.01
2000 VAB
  0.01
1000 VAB
  10

0.4 2 VAB  2  0.32 VAC 0.4 2g 0.3 2g
4 4
2.55VAB  1.7 VBC  10
2 2
VAB  1.125 VAC
2.55  (1.125VBC )  0.816 VBC  10
VBC  2.2m / s  QBC  QBD  0.155m 3 / s
VAB  2.5m / s  QAB  0.31m 3 / s
10/4/2012 49

50. 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.
10/4/2012 50

51. • What is the power available at the end B of
the pipe?
• What is the condition for maximum
transmission of power?
10/4/2012 51

52. Total head (energy per unit weight) H of fluid is
given by:
V2 P
H  Z
2g 
Energy Energy Weight
Power   x
time weight time
Weight
 gQ Q
time
Therefore:
Power   Q H
Units of power:
N . m/s = Watt
10/4/2012
745.7 Watt = 1 HP (horse power) 52

53. For the system shown in the figure, the following can be stated:
At Entrance  Power  γ Q H
Power dissipated due to friction  γ Q hf
Power dissipated due to minor loss  γ Q hm
At Exit  Power  γ Q H  h f  hm 
10/4/2012 53

54. Condition for Maximum Transmission of Power:
dP
The condition for maximum transmission of power occurs when : 0
dV
P  Q[ H  h f  hm ]

Neglect minor losses and use Q  AV  [ D 2 ]V
4
 2 L V3
So P  D [ HV  f ]
4 D 2g
dP  2 3 fL 2
 D [ H  V ]0
dV 4 2 Dg
H
H 3
fL V 2
 3h f  hf 
D 2g 3
 Power transmitted through a pipe is maximum when the loss of head due
1
to friction equal of the total head at the inlet
3
10/4/2012 54

55. Maximum Efficiency of Transmission of Power:
Efficiency of power transmission  is defined as
Power available at the outlet

Power supplied at the inlet
Q[ H  h f  hm ] [ H  h f  hm ]
 
QH H
or
[H  h f ]
 (If we neglect minor losses)
H
H
Maximum efficiency of power transmission occurs when hf 
3
H
[H  ]
  max  3  2  66.67%
H 3
10/4/2012 55

56. 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
Pout  γ Q H  h f 
H 500
Max. Power at  h f   m
3 3
V  3.417m/s
L V2 3500 V 2
hf  f   0.024
D 2g 0.3 2 g
Q  AV  π 0.3 3.417  0.2415 m3 /s
2
4
10/4/2012 56

57. P  γQH  h f 
 H
 gQ  H  
 3
 gQ 2 H
3
 10009.810.2415 2 500
3
789785
 789785 N.m/s (Watt)   1059 HP
745.7
10/4/2012 57