= 200 mL2 = 300 m%3D= 150 mmD2= 150 mm0.5 m/sL7= 200 mD7 = 150 mmL6 = 200 m[2]D6 = 150 mmL3 = 200 mD3 = 150 mm[1]%3D(5-0.3 m/sL5 = 200 mD5 = 150 mm 0.2 m3/sL4 = 200 mD4 = 150 mm%3DFigure 3.19. A pipe network with two loops.3.4. Analyze a looped pipe network as shown in Fig. 3.19 for pipe discharges usingHardy Cross, Newton-Raphson, and linear theory methods. Assume a constantfriction factor f= 0.02 for all pipes in the network.%3D

The Hardy Cross method is to determine the flow in pipe network systems and the inputs and outputs…

L = 200 m D, = 150 mm L2 = 300 m D2 = 150 mm %3D 0.5 m/s L7= 200 m (7 [1] 6 L6 = 200 m [2] 3 L3 = 200 m D7 = 150 mm D6 = 150 mm D3 = 150 mm 5 4) ►0.3 m³/s L4 = 200 m D4 = 150 mm L5 = 200 m %3D %3D D5 = 150 mm 0.2 m³/s

L = 200 m D, = 150 mm L2 = 300 m D2 = 150 mm %3D 0.5 m/s L7= 200 m (7 [1] 6 L6 = 200 m [2] 3 L3 = 200 m D7 = 150 mm D6 = 150 mm D3 = 150 mm 5 4) ►0.3 m³/s L4 = 200 m D4 = 150 mm L5 = 200 m %3D %3D D5 = 150 mm 0.2 m³/s

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Storm Sewers and Detention

The storm drains system transports rainwater or stormwater and other runoff from roads, parking lots, and roof downpipes directly to nearby creeks, rivers, or the sea without any treatment.

Question

= 200 m
L2 = 300 m
%3D
= 150 mm
D2
= 150 mm
0.5 m/s
L7= 200 m
D7 = 150 mm
L6 = 200 m
[2]
D6 = 150 mm
L3 = 200 m
D3 = 150 mm
[1]
%3D
(5
-0.3 m/s
L5 = 200 m
D5 = 150 mm 0.2 m3/s
L4 = 200 m
D4 = 150 mm
%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.
%3D

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