The document discusses analyzing pipe networks through various methods. It describes the Hardy Cross method which involves iteratively solving for pipe flows (Q) until head losses (hf) around loops equal zero. The key steps are: 1) Assume initial Q values; 2) Calculate hf from Q; 3) Apply correction factor ΔQ to Q if hf ≠ 0 and repeat; 4) Terminate when hf < 0.01m or ΔQ < 1 L/s. It provides an example application analyzing a sample network loop and presenting initial and final pipe discharge values. The document also discusses using computer programs like Epanet and WaterGEM to analyze pipe networks.

1. Pipe Network Distribution
Md. Mehedi Hassan Masum
Lecturer
Department of Civil Engineering
Port City International University
Department of CEN, PCIU

2. Department of CE, CUET2
Introduction Result & Discussion Conclusion
Analysis of pipe Network
The following are the most common methods used to analyze the
Grid-system networks:
1.Hardy Cross method.
2.Computer programs (WaterCAD, Epanet, WaterGEMs etc.)

3. Department of CE, CUET3
Introduction Result & Discussion Conclusion
Hardy Cross Method
Assumptions
1. Water is withdrawn from nodes only, not directly from
pipes.
2. The discharge entering the system will have (+) value, and
the discharge leaving the system will have (-) value.
3. Usually neglect minor losses.
4. Assume flows for each individual pipe in the network.
5. At any junction (node) discharge input is equal to output.
6. Around any loop in the grid, the sum of head losses must
equal to zero

4. Department of CE, CUET4
Introduction Result & Discussion Conclusion
Hardy Cross Method
Steps
1. Assume values of Q to satisfy 𝑄 = 0.
2. Calculate 𝐻𝑓 from Q using ℎ 𝑓 = 𝐾1 𝑄2
.
3. If ℎ 𝑓 = 0, then the solution is correct.
4. If ℎ 𝑓 ≠ 0, then apply a correction factor, ∆𝑄, to all
Q and repeat from step (2).
5. For practical purposes, the calculation is usually
terminated when ℎ 𝑓< 0.01 m or ∆𝑄 < 1 L/s.

5. Department of CE, CUET5
Introduction Result & Discussion Conclusion
Hardy Cross Method
Steps
A reasonably efficient value of ∆𝑄 for rapid convergence is
given by;
∆𝑄 = −
𝐻𝐿
2
𝐻𝐿
𝑄
N.B. Head loss can be calculated using the Darcy-Weisbach
equation
ℎ 𝑓 = 𝜆
𝐿
𝐷
𝑉2
2𝑔
= 𝜆
𝐿
𝐷
𝑄2
2𝑔𝐴2
= 𝐾1 𝑄2

6. Department of CE, CUET6
Introduction Result & Discussion Conclusion
Computer Programs
𝐸𝑃𝐴𝑁𝐸𝑇 𝑊𝑎𝑡𝑒𝑟𝐺𝐸𝑀

7. Department of CE, CUET7
Introduction Result & Discussion Conclusion
Pipe Network System
𝐶𝑉
𝑆𝐶
𝐵
𝑆𝑀
𝑀
M : Mains
SM : Sub Main
B : Branch
SC :Service Connection
CV :CUTOFF Valve

8. Department of CE, CUET8
Introduction Result & Discussion Conclusion
𝐹𝑖𝑔 01: 𝑃𝑙𝑎𝑛 𝑜𝑓 𝑎 𝑝𝑜𝑤𝑒𝑟 𝑝𝑙𝑎𝑛𝑡

9. Department of CE, CUET9
Introduction Result & Discussion Conclusion
𝐹𝑖𝑔 02: 𝑁𝑒𝑡𝑤𝑜𝑟𝑘 𝑑𝑖𝑎𝑔𝑟𝑎𝑚

10. Department of CE, CUET10
Introduction Result & Discussion Conclusion
𝐹𝑖𝑔 03: 𝑁𝑒𝑡𝑤𝑜𝑟𝑘 𝑑𝑖𝑎𝑔𝑟𝑎𝑚
𝑊𝑇 − 1𝑊𝑇 − 2

11. Department of CE, CUET11
Introduction Result & Discussion Conclusion
𝐹𝑖𝑔 04: 𝑁𝑒𝑡𝑤𝑜𝑟𝑘 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 (𝑊𝑇 − 1)

12. Department of CE, CUET12
Introduction Result & Discussion Conclusion
𝐹𝑖𝑔 05: 𝑁𝑒𝑡𝑤𝑜𝑟𝑘 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 (𝑊𝑇 − 2)

13. Department of CE, CUET13
Introduction Result & Discussion Conclusion
𝐹𝑖𝑔 06: 𝐿𝑜𝑜𝑝 𝑠𝑒𝑙𝑒𝑐𝑡𝑒𝑑 𝑓𝑜𝑟 𝑎𝑛𝑎𝑙𝑦𝑠𝑖𝑠

14. Department of CE, CUET14
Introduction Result & Discussion Conclusion
Initial Assumptions
47.12 𝑙/𝑠𝑒𝑐
20𝑙/𝑠𝑒𝑐
10 𝑙/𝑠𝑒𝑐
10𝑙/𝑠𝑒𝑐
10 𝑙/𝑠𝑒𝑐
20 𝑙/𝑠𝑒𝑐
𝐴 𝐵
𝐶
𝐷

15. Department of CE, CUET15
Pipe name Dia (m)
Length
(m)
hf (m) Q (L/sec) hf/Q
AB 0.15 219.512 0.305773 -10 -30.5773
BC 0.2 60.98 0.08063 -20 -4.03149
CD 0.2 219.51 0.533681 27.12 19.67849
DA 0.15 115.85 0.161375 10 16.13754
Sum 1.081459 1.207208
First Trial
= -0.44792
Introduction Result & Discussion Conclusion

16. Department of CE, CUET16
Pipe name Dia (m)
Length
(m)
hf (m) Q (L/sec) hf/Q
AB 0.15 219.512 0.23807 -8.82374 -26.9806
BC 0.2 60.98 0.070918 -18.7568 -3.78089
CD 0.2 219.51 0.58373 28.36318 20.58055
DA 0.15 115.85 0.203993 11.24318 18.14373
Sum 1.096711 7.962748
First Trial
= -0.06887
Introduction Result & Discussion Conclusion

17. Department of CE, CUET17
Introduction Result & Discussion Conclusion
Final Discharge
47.12 𝑙/𝑠𝑒𝑐
18.76𝑙/𝑠𝑒𝑐
8.82 𝑙/𝑠𝑒𝑐
11.24𝑙/𝑠𝑒𝑐
9.94 𝑙/𝑠𝑒𝑐
20.06 𝑙/𝑠𝑒𝑐
𝐴 𝐵
𝐶
𝐷

18. Sewer Network Distribution
Department of CE, CUET
Md. Mehedi Hassan Masum
Lecturer
Department of Civil Engineering
Port City International University

19. Department of CE, CUET19
Introduction Result & Discussion Conclusion
𝐹𝑖𝑔 01: 𝑃𝑙𝑎𝑛 𝑜𝑓 𝑎 𝑝𝑜𝑤𝑒𝑟 𝑝𝑙𝑎𝑛𝑡

20. Department of CE, CUET20
Introduction Result & Discussion Conclusion
𝐹𝑖𝑔 03: 𝑁𝑒𝑡𝑤𝑜𝑟𝑘 𝑑𝑖𝑎𝑔𝑟𝑎𝑚

21. Department of CE, CUET21
Introduction Result & Discussion Conclusion
𝐹𝑖𝑔 04: 𝑁𝑒𝑡𝑤𝑜𝑟𝑘 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 (𝑃 − 1)

22. Department of CE, CUET22
Introduction Result & Discussion Conclusion
𝐹𝑖𝑔 05: 𝑁𝑒𝑡𝑤𝑜𝑟𝑘 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 (𝑃 − 2)

23. Department of CE, CUET23
Introduction Result & Discussion Conclusion
𝐹𝑖𝑔 06: 𝐿𝑜𝑜𝑝 𝑠𝑒𝑙𝑒𝑐𝑡𝑒𝑑 𝑓𝑜𝑟 𝑎𝑛𝑎𝑙𝑦𝑠𝑖𝑠

24. Department of CE, CUET24
Introduction Result & Discussion Conclusion
Pipe
Sub-
Area
Population
Per capita
consumpti
on (lpcd)
Water
Demand
(m3/day)
Peak
factor
% of
wastewater
Total
wastewater
(m3/day)
P1
A1 432 135 58.32 2.50 70 102.06
A2 768 135 103.68 2.50 70 181.44
A9 384 135 51.84 3.00 90 139.97
P2
A3 768 135 103.68 2.50 70 181.44
A4 768 135 103.68 2.50 70 181.44
P3
A5 1152 135 155.52 2.50 70 272.16
A6 1152 135 155.52 2.50 70 272.16
A7 1152 135 155.52 2.50 70 272.16
A8 300 100 30.00 3.00 90 81.00
Total 917.76 23.50 1683.83
Table: Wastewater Generation from household

25. Department of CE, CUET25
Introduction Result & Discussion Conclusion
Table: Wastewater Generation from household
Pipe Sub-Area Area (sq.m) Area (ha)
Peak
infiltration
factor
(m3/ha-day)
Infiltration , I
(m3/day)
P1
A1 7432.2400 0.7432
8.7500 22.6962
A2 7618.0460 0.7618
A9 10888.2316 1.0888
P2
A3 7543.7236 0.7544
8.7500 13.1690
A4 7506.5624 0.7507
P3
A5 7580.8848 0.7581
8.7500 24.8097
A6 7618.0460 0.7618
A7 7432.2400 0.7432
A8 5722.8248 0.5723
Total 69342.7992 6.9343 60.6749

26. Department of CE, CUET26
Introduction Result & Discussion Conclusion
Total Wastewater Generation
Total wastewater generation =
1683.73+60.6749 = 1744.59 𝑚3/𝑑𝑎𝑦

27. Department of CE, CUET27

28. Department of CE, CUET28