Slides 4 Water Distribution in Pipeline

7/29/2019 Slides 4 Water Distribution in Pipeline

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4.1 : Series, Parallel and Branched Pipes

4.2 : Water Distribution System

4.2.1 : Hardy-Cross (Loop) Method

4.2.2 : Nodal method

Chapter 4 :Water Distribution in

Pipelines


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4.1 : Series, Parallel and Branched Pipes

When pipes of different diameters are connected end-to-end to form a

pipeline, they are said to be in series. The total loss of energy (head) will be

the sum of the losses in each pipe plus local losses at connections (and often

assumed to be negligible)


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4.1(a) : Pipes in series

For the case of new installation, the problem would be given Hst an pipescharacteristics (D,L, roughness) to find the Q and hLosses. The governing

equations are:


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Method 1


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Method 2


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Example : Pipes in series

As shown in figure below, reservoir A delivers water to reservoir B through

uniform pipelines AJ and JB of diameters 300 mm and 200 mm respectively.

Length of AJ=3 km, length of JB = 4 km, the effective roughness size of both

pipes is 0.015 mm; static head between reservoirs is 25 m. assume kinematic

viscosity 1.13x10-6 m/s2. neglect all minor losses, determine the discharge in

B


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Solutions

Guess Q

Calculate v1 and v2

Calculate Re, 1 and 2

Calculate

H using bernuolli

Compare Hst with H

Q V 1 V 2 Re 1 Re 2 1 2 Hf 1 Hf 2 H tot H

st

35 0.5 1.11 1.31x10 5 1.97x10 5 2.04 18.92 20.96 25

40 0.57 1.27 1.5x10 5 2.25x10 5 2.59 24.04 26.62 25


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4.1(b): Pipes in parallel

When two or more pipes in parallel connect two reservoirs, then the fluid may

flow down any of the available pipes at possible different rates. But the head

difference over each pipe is always the same. The total flow rate is the sum of

the flow in each pipe.

For such a system, analysis can be carried out by simply treating each pipeindividually and summing flow rates at the end. Assuming two pipes in parallel,

we have the following governing equations:


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Example : Pipes in parallel

As shown in figure, 2 pipes connect two reservoirs which have a height

difference of 10 m. Pipe 1 has a diameter of 50 mm and length of 100 m.

pipe 2 has a diameter of 100 mm and length of 100 m. Both have entry loss

coefficient kL=0.5 and exit loss coefficient kL=1.0 and friction factor =0.032.

Calculate:

Rate of flow for each pipe

Diameter D of a pipe 100 m long that could replace the two pipes to provide

the same flow.


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4.1(c) : Branched pipes


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Cont: Branched pipes

If pipes connect three reservoirs, then the problems become more complex. Infigure below, flow in pipe 1 and pipe 3 are obviously from A to D and D to C

respectively. If one assumes that the flow in pipe 2 is from B to D then the

following relationships could be written;


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Trial and error method for Branched pipes

The trial and error are recommended to solve this problem with the following

steps:

1. Estimate a value of the head at the junction, HJ

2. Substitute this into energy equations (4.9) or (4.11) to get an estimate Q

for each pipes.

3. Check to see if continuity equation in (4.8) or (4.10) are satisfied.4. If flow into junction is too high, choose larger Hj or vice versa and return to

step (2)


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Example : Branched Pipes

Given the following data for a branched system. Calculate the discharge and

the pressure head at the junction J. Assume v=1.13x10-6 m/s2

pipe Length

(km)

Diameter

(mm)

Roughness

(mm)

1 5 300 0.03

2 2 150 0.03

3 4 350 0.03

item Elevation

(m ad)

Reservoir 1 800

Reservoir 2 780

Reservoir 3 700

Junctin J 720


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Solutions

Assume Hj

Calculate hf1, hf2, hf3

Calculate Sf

Calculate v1,v2,v3 using Colbrooke White-Darcy Weisbach

Calculate Q1,Q2, Q3 (Q=Av)

Produce table Hj = 750 Pipe 1 Pipe 2 Pipe 3

hf 50 30 50

Sf 0.01 0.015 0.013

V 2.028 1.614 2.516

A 0.071 0.018 0.096

Q 143 29 242

143+29 242 thus, proceed

with 2nd

trial


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solutions

Hj = 740 Pipe 1 Pipe 2 Pipe 3

hf 60 40 40

Sf

V

A

Q

Hj = 735 Pipe 1 Pipe 2 Pipe 3

hf 65 45 35

Sf

V

A

Q


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4.2 : Water Distribution System


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4.2.1 : Hardy-Cross method/Loop method


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Example : Loop method

For a square loop shown below, find:

The discharge in the loop

The pressure heads at points B, C and D if the pressure head at A is 70 m and

A, B C and D have the same elevation

All pipes are 1 km long and 300 mm in diameter, with roughness 0.03 mm. thekinematic viscosity of water is 1.13x10-6 m/s2.

100 l/s20 l/s

40 l/s 40 l/s

A

D C

B


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solutions

pipe Q V Re hf hf/Q

AB 60 0.848 2.25x10 5 0.016 1.98 33

BC 40 0.566 1.5x10 5 0.017 0.94 23.5

CD 0 0 0 - 0 0

DA -40 -0.566 1.5x10 5 0.017 -0.94 23.5

1.98 23.5

pipe Q V Re hf hf/Q

AB 47.6 0.674 1.79x10 5 0.017 1.293 27.2

BC 27.6 0.391 1.04x10 5 0.018 0.479 17.4

CD -12.4 -0.1750 4.65x10 4 0.022 -0.113 9.1

DA -52.4 -0.741 2.97x10 5 0.017 -1.539 29.4

0.12 83


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4.2.2 : Node method

Consists of eliminating the discharge in equation (4.12) and (4.14)-(4.15) togive a set of equations in head losses only. It may be applied to loops or

branches where the external heads are known and the heads within networks

are required. The basis of the method is as follows:

1. Assume values for the head HJ at each junction

2. Calculate Qi from HJ using combined Colebrook-White and Darcy-Weisbach equation

3. If , then the solution is correct else apply correction factor H and

return to step (2)

where

Appropriate sign convention must be used for Qi from Hlossi (+ve entering a

node)

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