chapter 2: flow through single &combined pipelines university of palestine engineering...

CHAPTER 2: Flow through single &combined Pipelines

CHAPTER 2: Flow through single &combined Pipelines

University of PalestineEngineering Hydraulics2nd semester 2010-2011

1

ContentPipelines in series &

parallel

Pipelines with negative

pressure

Branching pipe systems

Power in pipelines

Flow through single &combined Pipelines

2

Part APipelines in series &

parallel Pipelines with

negative pressure


3

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.

Introduction

4

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.

Introduction

5

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.

Introduction

6

Simple pipe flow

(1)

(2)

Introduction

7

To solve such system:

• Apply Bernoulli’s equation

• where

pL hhzg

VPz

g

VP 2

222

1

211

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

TotalmfL K

D

fL

g

Vhhh

2

2

Introduction

8

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 = 0.032. Also, draw the HGL and EGL.

Z1 Z

Introduction

9

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

Introduction

10

• pipes of different lengths and different diameters connected end to end (in series) to form a pipeline

Flow Through Pipes in Series

11

• 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


12

13

LBBB

AAA hz

g

VPz

g

VP

22

22

Hhzz LBA

Where

4

1

3

1 jmj

ifiL hhh

g

VK

g

VK

g

VK

g

VK

g

V

D

Lfh exitenlcent

i

i

i

iiL 22222

23

22

22

21

3

1

2


Pipelines in Series

nQQQQ 21

LnLLL ....hhhh 21


14

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.


15

• 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.

Flow Through Parallel Pipes

n

iiQQ

1

LnLLLL ....hhhhh 321

16

• Discharge:

• Head loss: the head loss for each branch is the same

3

1321

iiQQQQQ

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

23

3

33

22

2

22

21

1

11

Flow Through Parallel Pipes

17

18

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

/sm...π

AVQ

m/s.V.

V

..

g

V.

D

Lf

332111

1

21

21

1

1

1015350620404

5062

28192040

25010

22

221 ff hh

/sm.QQQ

/sm...π

Q

m/s.V.

V

..

g

V.

D

Lf

3321

3322

2

22

22

2

2

10178

1002555720504

55728192050

30010

22


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.


19

22

031402

20 m.

.πAa

22

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 2

1502

22

g

V

g

V.

g

V

D

Lfh bbb

b

bbb 2

102

19022

222

Solution


20

g

V.

.

.f

g

V.

.f b

ba

a 210380

120

46

2150

20

4 22

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


21

Example 5Determine 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.


22


23


24

Group work Example

• Four pipes connected in parallel as shown. The following details are given:

PipeL (m)D (mm)f

12002000.020

23002500.018

31503000.015

41002000.020

• If ZA = 150 m , ZB = 144m, determine the discharge in each pipe ( assume PA=PB = Patm)


25

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.

=180mQN QN1

QN2


26

• 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:

(siphon phenomena)

Pipe line with negative Pressure

27

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)


28

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.


29

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 )


30

A S

LSS

SA

LSSS

AAA

hP

g

VZZ

hZP

g

VZ

P

g

V

2

222

22

-ve valueMust 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)


31

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)


32

Example 1Siphon 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


33

Solution


34

• 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


35

LsRP hHHH

Pumps design will be discussed in details in next chapters


36

Part B



37

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

0Q


38

Three-reservoirs problem

)Branching System(

How we can demonstrate the hydraulics of branching pipe System??

by the classical three-reservoirs problem


39

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????


40

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

Consider a case where three reservoirs are connected by a branched-pipe system .

There are two types of Branching pipe problem:


41

• 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.

Q Q Q Q 1 2 3 0


42

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.


43

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 JGiven:All pipes parametersAll Tanks elevation

The solution based on:1- Assuming the pressure at J and then estimate the value of hf in each pipe2- Calculate the flow in each pipe and check3- try the last procedure until

Q

0Q


Type 1:

44

Example 1

PipeCJBJAJ

Length m200040001000

Diameter cm405030

f0.0220.0210.024

In the following figure determine the flow in each pipe


45

Trial 1 ZP= 110mApplying Bernoulli Equation between A , J :

g

V

g

V

D

LfZZ PA 23.0

1000024.0110120

2.

21

21

1

11

V1 = 1.57 m/s , Q1 = 0.111 m3/s

g

V

g

V

D

LfZZ BP 25.0

4000021.0100110

2.

22

22

2

22

V2 = 1.08 m/s , Q2 = - 0.212 m3/s

Applying Bernoulli Equation between B , J :

Branching pipe systemsExample 1.cont.

46

g

V

g

V

D

LfZZ CP 24.0

2000022.080110

2.

23

23

3

33

Applying Bernoulli Equation between C , J :

V3 = 2.313 m/s , Q2 = - 0.291 m3/s

0392.0291.0212.0111.0321 QQQQ


47

Trial 2 ZP= 100m

0/ 08.0237.00157.0 3321 smQQQQ

Trial 3ZP= 90m

0/ 324.0168.03.0192.0 3321 smQQQQ

Example 1.cont.


48

Draw the relationship between and P Q

99m Pat 0 Q


49

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.


50

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 JGiven:All pipes parametersQ in a pipetwo Tanks elevationThe solution based on:1- Determine the pressure at J by calculate hf in pipe with known Q2- Estimate the value of hf in the other pipe2- Calculate the flow in the other pipes

Type 2:


51

Example 2

In the following figure determine the flow in pipe BJ & pipe CJ. Also, determine the water elevation in tank C


52

m.Z

.

.

..Z

g

V.

D

LfZZ

m/s..

π.

A

QV

P

PPA

47536

8192

8490

30

1200024040

2

849030

4

060

221

1

11

21

11

Solution

/sm 0.0203Q0.645m/sV

9.812

V

0.2

6000.02436.47538

2g

V.

D

LZZ

322

22

22

2

22PB

f

Applying Bernoulli Equation between B , J :

Applying Bernoulli Equation between A , J :

Branching pipe systemsExample 2 cont.

53

mZ

gg

V

D

LfZZ

c

CP

265.32

2

136.1

3.0

800024.0 Z- 6.4753

2.

2

c

23

3

33

Applying Bernoulli Equation between C , J :

smQQQ

QQQQ

/ 0803.00203.006.0

03

213

321

sm

A

QV / 136.1

3.04

0803.0

23

33

Branching pipe systemsExample 2 cont.

54

Group Work 1

ACAB

ACAB

BDBC

BDBCAB

VV

VV

QQ

QQQ

125.1

3.024.0

0

24

24

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


55

Group Work 2Determine the flow ifa) The valve is closedb) The valve is open and the flow through the small pipe = 100L/S


56

Part C

Power Transmission Through Pipes


57

• 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.


58

• What is the power available at the end B of the pipe?

• What is the condition for maximum transmission of power?


59

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 HUnits of power:

N . m/s = Watt745.7 Watt = 1 HP (horse

power)


60

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


61

Calculate the max transported power through pipe line

f

π

π

f

mf

hg

V

D

f LH

g

V

D

f L HDγ

dV

dP

g

V

D

Lf HVDγ P

VAQhHγ Q

hhHγ Q

32

..3

2..30at Max.

2

P

lossminor neglect

PExit At

2

22

4

32

4

The max transported power through pipe line at 3

H h f


Power transmitted through a pipe is maximum when the loss of head due

of the total head at the inlet 3

1to friction equal 62

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 when3

Hh f

%67.663

2]

3[

max

H

HH

)If we neglect minor losses(


%67.661003max

H

HH

H

hHη

H

hhH

γQH

hhHγQη

f

mfmf

63

ExamplePipe 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 f 3

500

3at Power Max.

g

V

..

g

V

D

Lfh f 230

35000240

2

22

m/s 3.417V

/s m...AVQ π 32

4 24150417330


64

HP.

tt) N.m/s (Wa

..

HgQ

HHgQ

hHγQP f

10597745

789785789785

500241508191000

3

32

32


65

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chapter 2: flow through single &combined pipelines university of palestine engineering...

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