PIPE NETWORK ANALYSIS

Presented by-Azaz Ahmed.CIB-09-015.

Department of Civil EngineeringSchool of Engineering, Tezpur University, Napaam 784028, Tezpur,

Assam, India

INTRODUCTION

Pipe Network

An interconnected system of pipes forming several loops or circuits.E.g.- Municipal water distribution systems in cities.

F E

HG

CB

A

D

Flow in

Flow out

Fig.: Pipe Network

Pipe network

Necessary conditions for any network of pipes

The flow into each junction must be equal to the flow out of the junction. This is due to the continuity equation.

The algebraic sum of head losses round each loop must be zero.

The head loss in each pipe is expressed as hf = kQn. For Turbulent flow, n = 2.

Pipe network problems are difficult to solve analytically.

As such HARDY CROSS METHOD which uses successive approximations is used.

Hardy cross method

•A trial distribution of discharges is made arbitrary in such a way that continuity equation is satisfied at each junction.

•With the assumed values of Q, the head loss in each pipe is calculated according the following equation-

•Head loss around each loop is calculated considering the head loss to be positive in CW-flow and negative in CCW-flow.

hf = kQ2Where,

K =4f x L

2g x D5 x (∏∕4)2

If the net head loss due to assumed values of Q round the loop is zero, then the assumed values of Q are correct.

But if the net head loss due to assumed values of Q is not zero, then the assumed values of Q are corrected by introducing a correction delta Q for the flows, till circuit is balanced.

The correction factor is obtained by-

ΔQ =- ∑ (kQ0

2)∑ (2kQ0)

If the Correction factor comes out to be positive, then it should be added to the flows in the CW direction and subtracted from the flows in the CCW direction.

After the corrections have been applied to each pipe in a loop and to all loops, a second trial calculation is made for all loops. The procedure is repeated till Delta Q becomes negligible.

Let us consider a problem.

A

C

B

DK= 2

K= 4K= 1

K= 1

K= 2

We have to calculate discharge in each pipe of the network.

20

30

40

90

A

C

B

DK= 2

K= 4K= 1

K= 1

K= 2

20

30

40

90

30

20

60

10 20

First Trial

Discharges are assumed as in the above figure

Loop ADBPipe k Q Hf= kQ2 2kQAD 4 30 3600 240

DB 1 10 -100 20

AB 2 60 -7200 240

Total -3700 500

Q1 = 7.4

Loop DCBPipe k Q Hf= kQ2 2kQDC 2 20 800 80

CB 1 20 -400 40

BD 1 10 100 20

Total 500 140

Q2 = -3.6

Corrected flow for second trial.

Pipe Correction Flow Direction

AD 30 + 7.4 37.4 CW

AB 60 – 7.4 52.6 CCW

BD 10 – 7.4 2.6 CCW

DC 20 – 3.6 16.4 CW

BC 20 + 3.6 23.6 CCW

BD 2.6 – 3.6 -1 CW

A

C

B

DK= 2

K= 4K= 1

K= 1

K= 2

20

30

40

90

37.4

16.4

52.6

1 23.6

Second Trial

Discharges for the second trial

Loop ADBPipe k Q Hf= kQ2 2kQAD 4 37.4 5595 299.2

DB 1 1 1 2

AB 2 52.6 -5533.5 210.4

Total 62.54 511.6

Q1 = -0.1

Loop DCBPipe k Q Hf= kQ2 2kQDC 2 16.4 537.9 65.6

CB 1 23.6 -556.9 47.2

BD 1 1 -1 2

Total -20 114.8

Q2 = 0.2

Since Q1 and Q2 are very small, the correction is applied and

furthur trials are discontinued.

Corrected flow for second trial.

Pipe Correction Flow Direction

AD 37.4 – 0.1 37.3 CW

AB 52.6 + 0.1 52.7 CCW

BD 1 – 0.1 0.9 CW

DC 16.4 + 0.2 16.6 CW

BC 23.6 – 0.2 23.4 CCW

BD 0.9 – 0.2 0.7 CCW

A

C

B

DK= 2

K= 4K= 1

K= 1

K= 2

20

30

40

90

37.3

16.6

52.7

0.7 23.4

Final Discharge

Final Distribution of discharges.

Hereby we conclude Hardy Cross method.

Thank you for your time

and patience.