Example 3.3. A single looped network as shown in Fig. 3.10 has to be analyzed by the Hardy Cross method for given inflow and outflow discharges. The pipe diameters D and lengths L are shown in the figure. Use Darcy-Weisbach head loss-discharge relation- ship assuming a constant friction factor f= 0.02.

Example 3.3. A single looped network as shown in Fig. 3.10 has to be analyzed by the Hardy Cross method for given inflow and outflow discharges. The pipe diameters D and lengths L are shown in the figure. Use Darcy-Weisbach head loss-discharge relation- ship assuming a constant friction factor f= 0.02.

Structural Analysis

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ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Open Channel Flow

Open channel flow is a branch of fluid mechanics. It deals with the flow of fluid in an open conduit or free surface. In an open channel flow, gravity induces the flow of water whose surface is open to the air. This flow is free-falling rather than confined in rigid body conduits. The open channel flow depends upon two parameters that are the cross-section of the conduit and the velocity profile. Practical examples of open channel flow are streams, rivers, canals, and other drainage systems.In open channel flow, the cross-sectional area of at least one of the sides is exposed to air, and the surface of water faces atmospheric pressure. This flow is also known as free gravity flow. The cross-sectional area of the conduit can be either man-made or a natural channel. For instance, the flow in rivers and streams are natural conduits, and the flow in a canal and sanitary systems are man-made or engineered channels.

Question

Example 3.3. A single looped network as shown in Fig. 3.10 has to be analyzed by the
Hardy Cross method for given inflow and outflow discharges. The pipe diameters D and
lengths L are shown in the figure. Use Darcy-Weisbach head loss-discharge relation-
ship assuming a constant friction factor f= 0.02.

L = 300 m
D, = 150 mm
K, = 6528 s²/m5
2
0.6 m³/s
Pipe no.
Node no.
L4 = 200 m
D4 = 150 mm
K4 = 4352 s2/m5
L, = 200 m
D, = 150 mm
K = 4352 s?/m5 [1] Loop no.
Flow direction
[1]
(3
0.6 m/s
3
L3 = 300 m
D3 = 150 mm
K3 = 6528 s²/m5
Figure 3.10. sSingle looped network.

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