sanitary engineering  solve by hardy Cross method  k=8*f*L/pi*g*D^5 f=0.02 g=9.81 pi=e.14

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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sanitary engineering 

solve by

hardy Cross method 

k=8*f*L/pi*g*D^5

f=0.02

g=9.81

pi=e.14

L1 = 200 m
D1 = 150 mm
L2 = 300 m
D2 = 150 mm
%3D
0.5 m/s -
L7= 200 m
D7 = 150 mm
L6 = 200 m
[2]
D6 = 150 mm
L3 = 200 m
%3D
%3D
[1]
3
D3 = 150 mm
→0.3 m3/s
L5 = 200 m
D5 = 150 mm 0.2 m³/s
L4 = 200 m
D4 = 150 mm
%3D
%3D
Figure 3.19. A pipe network with two loops.
3.4. Analyze a looped pipe network as shown in Fig. 3.19 for pipe discharges using
Hardy Cross, Newton-Raphson, and linear theory methods. Assume a constant
friction factor f=0.02 for all pipes in the network.
Transcribed Image Text:L1 = 200 m D1 = 150 mm L2 = 300 m D2 = 150 mm %3D 0.5 m/s - L7= 200 m D7 = 150 mm L6 = 200 m [2] D6 = 150 mm L3 = 200 m %3D %3D [1] 3 D3 = 150 mm →0.3 m3/s L5 = 200 m D5 = 150 mm 0.2 m³/s L4 = 200 m D4 = 150 mm %3D %3D Figure 3.19. A pipe network with two loops. 3.4. Analyze a looped pipe network as shown in Fig. 3.19 for pipe discharges using Hardy Cross, Newton-Raphson, and linear theory methods. Assume a constant friction factor f=0.02 for all pipes in the network.
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