1. CHAPTER 2: Flow through single &combined Pipelines
University of Palestine
Engineering Hydraulics
2nd semester
CHAPTER 2: Flow through single &combined Pipelines

2. Flow through single &combined Pipelines
Content
Pipelines in series & parallel
Pipelines with negative pressure
Branching pipe systems
Power in pipelines

3. Pipelines in series & parallel Pipelines with negative pressure
Flow through single &combined Pipelines
Part A
Pipelines in series & parallel
Pipelines with negative pressure

4. Any water conveying system may include the following elements:
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.

5. How to solve flow problems
Introduction
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.

6. Flow Through A Single Pipe (simple pipe flow)
Introduction
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.

7. Introduction
Simple pipe flow
(1)
(2)

8. To solve such system: Introduction Apply Bernoulli’s equation where
(1)
(2)
To solve such system:
Apply Bernoulli’s equation
where
For the same material and constant diameter (same f , same V) we can write:

9. Introduction
Example 1
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 = Also, draw the HGL and EGL. Z1 Z

10. The system is called compound pipe flow
Introduction
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

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

12. Flow Through Pipes in Series
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:

13. Flow Through Pipes in Series
Where

14. Flow Through Pipes in Series
Pipelines in Series

15. Flow Through Pipes in Series
Example 2
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.

16. 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.

17. Flow Through Parallel Pipes
Q1, L1, D1, f1
Q2, L2, D2, f2
Q3, L3, D3, f3
Discharge:
Head loss: the head loss for each branch is the same

18. Flow through single &combined Pipelines
Example 3:
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

19. Flow through single &combined Pipelines
Example 4
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.

20. Flow through single &combined Pipelines
Solution

21. Flow through single &combined Pipelines

22. Flow through single &combined Pipelines
Example 5
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.

23. Flow through single &combined Pipelines

24. Flow through single &combined Pipelines

25. Flow through single &combined Pipelines
4/21/2017
Flow through single &combined Pipelines
Group work Example
Four pipes connected in parallel as shown. The following details are given:
Pipe
L (m)
D (mm) f 1
200
0.020 2
300
250
0.018 3
150
0.015 4
100
If ZA = 150 m , ZB = 144m, determine the discharge in each pipe ( assume PA=PB = Patm)

26. Flow through single &combined Pipelines
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 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

27. 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:

28. Pipe line with negative Pressure
(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.

29. Siphon happened in the following cases:
Pipe line with negative Pressure
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.

30. Characteristics of this system
Pipe line with negative Pressure
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 = 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 )

31. Pipe line with negative Pressure
-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)

32. Pipe line with negative Pressure
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 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)

33. Pipe line with negative Pressure
Example 1
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

34. Pipe line with negative Pressure
Solution

35. Pipe line with negative Pressure
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.

36. Pipe line with negative Pressure
Pumps design will be discussed in details in next chapters

37. Branching pipe systems
Flow through single &combined Pipelines
Part B
Branching pipe systems

38. 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.
2 – all pipes that meet at the junction must share the same pressure at the junction. Pressure at point J = P

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

40. Branching pipe systems
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.

41. Branching pipe systems There are two types of Branching pipe problem:
Consider a case where three reservoirs are connected by a branched-pipe system.
There are two types of Branching pipe 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 .
This types of problems are most conveniently solved by trail and error

42. Branching pipe systems
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:
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.

43. Branching pipe systems
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.

44. Branching pipe systems
Type 1:
The problem is to determine the discharge in each pipe and the pressure head at the junction (point J). Four unknowns:
Q in each pipe and P at J
Given:
All pipes parameters
All Tanks elevation
The solution based on:
1- Assuming the pressure at J and then estimate the value of hf in each pipe
2- Calculate the flow in each pipe and check
3- try the last procedure until

45. Branching pipe systems
Example 1
In the following figure determine the flow in each pipe AJ BJ CJ
Pipe
1000
4000
2000
Length m 30 50 40
Diameter cm
0.024
0.021
0.022 f

46. Branching pipe systems
Example 1.cont.
Trial 1
ZP= 110m
Applying Bernoulli Equation between A , J :
V1 = 1.57 m/s , Q1 = m3/s
Applying Bernoulli Equation between B , J :
V2 = 1.08 m/s , Q2 = m3/s

47. Branching pipe systems
Example 1.cont.
Applying Bernoulli Equation between C , J :
V3 = m/s , Q2 = m3/s

48. Branching pipe systems
Example 1.cont.
Trial 2
ZP= 100m
Trial 3
ZP= 90m

49. Branching pipe systems
Example 1.cont.
Draw the relationship between and P

50. Branching pipe systems
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.

51. Branching pipe systems
Type 2:
The problem is to determine the flow in two pipes and the pressure head at the Junction J. Three unknown:
Q in two pipe and P at J
Given:
All pipes parameters
Q in a pipe
two Tanks elevation
The solution based on:
1- Determine the pressure at J by calculate hf in pipe with known Q
2- Estimate the value of hf in the other pipe
2- Calculate the flow in the other pipes

52. Branching pipe systems
Example 2
In the following figure determine the flow in pipe BJ & pipe CJ. Also, determine the water elevation in tank C

53. Branching pipe systems
Example 2 cont.
Solution
Applying Bernoulli Equation between A , J :
Applying Bernoulli Equation between B , J :

54. Branching pipe systems
Example 2 cont.
Applying Bernoulli Equation between C , J :

55. Flow through single &combined Pipelines
Group Work 1
Find the flow in each pipe

56. Flow through single &combined Pipelines
Group Work 2
Determine the flow if
The valve is closed
The valve is open and the flow through the small pipe = 100L/S

57. Power Transmission Through Pipes
Flow through single &combined Pipelines
Part C
Power Transmission Through Pipes

58. 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.

59. Power Transmission Through Pipes
What is the power available at the end B of the pipe?
What is the condition for maximum transmission of power?

60. Power Transmission Through Pipes
Total head (energy per unit weight) H of fluid is given by:
Therefore:
Units of power:
N . m/s = Watt
745.7 Watt = 1 HP (horse power)

61. For the system shown in the figure, the following can be stated:
Power Transmission Through Pipes
For the system shown in the figure, the following can be stated:

62. Power Transmission Through Pipes
Calculate the max transported power through pipe line
The max transported power through pipe line at
 Power transmitted through a pipe is maximum when the loss of head due
of the total head at the inlet
to friction equal

63. Maximum Efficiency of Transmission of Power:
Power Transmission Through Pipes
Maximum Efficiency of Transmission of Power:
Efficiency of power transmission is defined as or
(If we neglect minor losses)
Maximum efficiency of power transmission occurs when 

64. Power Transmission Through Pipes
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

65. Power Transmission Through Pipes