pipe network analysis using hardy cross method
DESCRIPTION
A small presentation made by me. Few minor mistakes are still in the document. Will upload the revised one soon.TRANSCRIPT
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.