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

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

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