sanitary engineering solve by hardy Cross method k=8*f*L/pi*g*D^5 f=0.02 g=9.81 pi=e.14
sanitary engineering solve by hardy Cross method k=8*f*L/pi*g*D^5 f=0.02 g=9.81 pi=e.14
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9b041149-b704-41dd-bc69-68301fd9c56e%2Facdafa29-7d02-497d-95b6-fc8bb8a99c49%2Fsw35lkq_processed.png&w=3840&q=75)
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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