sanitary engineering solve byhardy Cross method k=8*f*L/pi*g*D^5f=0.02g=9.81pi=e.14 L1 = 200 mD1 = 150 mmL2 = 300 mD2 = 150 mm%3D0.5 m/s -L7= 200 mD7 = 150 mmL6 = 200 m[2]D6 = 150 mmL3 = 200 m%3D%3D[1]3D3 = 150 mm→0.3 m3/sL5 = 200 mD5 = 150 mm 0.2 m³/sL4 = 200 mD4 = 150 mm%3D%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.

Given data in question Length of pipe Diameter of pipe Constant value of f Acceleration due to…

Engineering

Civil Engineering

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=8fL/pigD^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

Section: Chapter Questions

Problem 1P

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Concept explainers

Dimensional homogeneity

The term dimensional homogeneity means that in an equation, the dimensions of each term on both sides are equal, which means the dimensions of the left-hand side (LHS) of an equation are similar to the dimensions of the right-hand side (RHS). For a dimensionally homogeneous equation, the powers of fundamental dimensions will be identical on both sides of the equation.

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.

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