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.
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.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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sanitary engineering
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hardy Cross method

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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9b041149-b704-41dd-bc69-68301fd9c56e%2F46fc7ad4-9ed8-4a42-9d8b-f113c8dd517d%2Fpxirk9g_processed.png&w=3840&q=75)
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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