Example 3.4. The pipe network of two loops as shown in Fig. 3.11 has to be analyzec by the Hardy Cross method for pipe flows for given pipe lengths L and pipe diameters D The nodal inflow at node 1 and nodal outflow at node 3 are shown in the figure. Assume a constant friction factor f= 0.02.

Structural Analysis
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Author:KASSIMALI, Aslam.
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Chapter2: Loads On Structures
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hardy Cross method 

Example 3.4. The pipe network of two loops as shown in Fig. 3.11 has to be analyzed
by the Hardy Cross method for pipe flows for given pipe lengths L and pipe diameters D.
The nodal inflow at node 1 and nodal outflow at node 3 are shown in the figure. Assume
a constant friction factor f= 0.02.
Transcribed Image Text:Example 3.4. The pipe network of two loops as shown in Fig. 3.11 has to be analyzed by the Hardy Cross method for pipe flows for given pipe lengths L and pipe diameters D. The nodal inflow at node 1 and nodal outflow at node 3 are shown in the figure. Assume a constant friction factor f= 0.02.
L1 = 300 m
D; = 150 mm
K, = 6628 s²/m5
%3D
0.6 m3/s
(1 Pipe no.
AQ
Node no.
L2 = 200 m
(2. D2 = 100 mm
K2 = 33050 s/m³ [1] Loop no.
L5 = 360 m D5 = 100 mm
L4 = 200 m
D4 = 150 mm
KA = 4352 s?/m5
5
[2] +
Flow direction
(4
K5 = 59491 s?/m5
AQ
[1]
3
0.6 m3/s
L3 = 300 m
D3 = 100 mm
K3 = 49575 s?/m5
3
4
Figure 3.11. Looped network.
2.
Transcribed Image Text:L1 = 300 m D; = 150 mm K, = 6628 s²/m5 %3D 0.6 m3/s (1 Pipe no. AQ Node no. L2 = 200 m (2. D2 = 100 mm K2 = 33050 s/m³ [1] Loop no. L5 = 360 m D5 = 100 mm L4 = 200 m D4 = 150 mm KA = 4352 s?/m5 5 [2] + Flow direction (4 K5 = 59491 s?/m5 AQ [1] 3 0.6 m3/s L3 = 300 m D3 = 100 mm K3 = 49575 s?/m5 3 4 Figure 3.11. Looped network. 2.
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