Pipe Network Analysis using Hardy Cross method

PIPE NETWORK ANALYSIS

Presented byAzaz Ahmed. CIB-09-015. Department of Civil Engineering School 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. Fl o

w

in

F

A

H

G

B

E

C Fig.: Pipe Network

D Flo

ou w t

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•

hf = kQ2

•

Wher e,

4f x K L 2g x (∏∕4) = D5 x considering 2 Head loss around each loop is calculated

the head loss to be positive in CW-flow and negative in CCW-flow.

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 (kQ02) ∑ = (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. D

K= 4

2 0

C

K= 2

K= 1

4 0 K= 1

A K= 2

9 0

B

3 0

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

First Trial D

K= 4

A

2 0

3 0

C

K= 2

4 0

2 0 1 0 6 0

K= 1

2 0

K= 2

9 0 Discharges are assumed as in the above figure

K= 1

B

3 0

Loop ADB Pipe

k

Q

Hf= kQ2

2kQ

AD DB AB Total

4 1 2

30 10 60

3600 -100 -7200 -3700

240 20 240 500

Q1 = 7.4

Loop DCB Pipe

k

Q

Hf= kQ2

2kQ

DC CB BD Total

2 1 1

20 20 10

800 -400 100 500

80 40 20 140

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

Second Trial D

K= 4

A

2 0

C

K= 2

4 0

16. 4

37. 4

1 52. 6

K= 1

23. 6

K= 2

9 0 Discharges for the second trial

K= 1

B

3 0

Loop ADB Pipe

k

Q

AD DB AB

4 1 2

37.4 1 52.6

Total

Hf= kQ2

2kQ

5595 299.2 1 2 -5533.5 210.4 62.54

Q1 = -0.1

511.6

Loop DCB Pipe

k

Q

Hf= kQ2

2kQ

DC CB BD Total

2 1 1

16.4 23.6 1

537.9 -556.9 -1 -20

65.6 47.2 2 114.8

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

Final Discharge D

K= 4

A

2 0

C

K= 2

4 0

16. 6

37. 3

0. 7 52. 7

K= 1

23. 4

K= 2

9 0 Final Distribution of discharges.

K= 1

B

3 0

Hereby we conclude Hardy Cross method.

Thank you for your time and patience.