Page 217 - Water Engineering Hydraulics, Distribution and Treatment

EXAMPLE 6.5 ANALYSIS OF A WATER NETWORK USING THE EQUIVALENT PIPE METHOD
Find an equivalent pipe for the network of Fig. 6.14. Express Q in MGD or MLD; s in ‰; H in ft or m; and assume a Hazen–Williams
coefficient C of 100.
D
4000´−16˝
3000´−12˝ B Equivalent pipe 5000´−14˝ 6.6 Office Studies of Pipe Networks 195
3000´−8˝

A 4000´−10˝ C
Figure 6.14 Plan of network analyzed by the method of equivalent pipes (Example 6.5).

Solution:
1. Line ABD. Assume Q = 1 MGD (3.785 MLD).
(a) Pipe AB, 3,000 ft, 12 in.; s = 2.1; H = 2.1 × 3 = 6.3 ft (1.92 m).
(b) Pipe BD, 4,000 ft, 16 in.; s = 0.52; H = 0.52 × 4 = 2.1 ft (0.64 m).
(c) Total H = 6.3 + 2.1 = 8.4 ft (2.56 m).
(d) Equivalent length of 12 in. (300 mm) pipe: 1,000 × 8.4∕2.1 = 4,000 ft (1,219 m).
2. Line ACD. Assume Q = 0.5 MGD (1.89 MLD).
(a) Pipe AC, 4,000 ft, l0 in.; s = 1.42; H = 1.42 × 4 = 5.7 ft (1.74 m).
(b) Pipe CD, 3,000 ft, 8 in.; s = 4.2; H = 4.2 × 3 = 12.6 ft (3.84 m).
(c) Total H = 5.7 + 12.6 = 18.3 ft (5.58 m).
(d) Equivalent length of 8 in. (200 mm) pipe: 1,000 × 18.3∕4.2 = 4,360 ft (1,329 m).
3. Equivalent line AD. Assume H = 8.4ft(2.56 m).
(a) Line ABD, 4,000 ft, 12 in.; s = 8.4∕4.00 = 2.1; Q = 1.00 MGD (3.785 MLD).
(b) Line ACD, 4360 ft, 8 in.; s = 8.4∕4.36 = 1.92; Q = 0.33 MGD (1.25 MLD).
(c) Total Q = 1 + 0.33 = 1.33 MGD (5.03 MLD).
(d) Equivalent length of 14 in. (350 mm) pipe: Q = 1.33, s = 1.68, 1,000 × 8.4∕1.68 = 5,000 ft (1,524 m).
(e) Result: 5,000 ft of 14 in. (1,524 m of 350 mm) pipe.

Necessary calculations are as follows:

1. Because line ABD consists of two pipes in series, the losses of head created by a given flow of water are additive. Find,
therefore, from the Hazen–Williams diagram (Chapter 5 and/or Appendix 14) the frictional resistance s for some reasonable
flow (1 MGD or 3.78 MLD) (a) in pipe AB and (b) in pipe BD. Multiply these resistances by the length of pipe to obtain
the loss of head H. Add the two losses to find the total loss H = 8.4 ft (2.56 m). Line ABD, therefore, must carry 1 MGD
(3.78 MLD) with a total loss of head of 8.4 ft (2.56 m). Any pipe that will do this is an equivalent pipe. Because a 12 in.
(300 mm) pipe has a resistance s of 2.1‰ when it carries 1 MGD (3.78 MLD) of water, a 12 in. (300 mm) pipe, to be an
equivalent pipe, must be 1,000 × 8.4∕2.1 = 400 ft (122 m) long.
2. Proceed in the same general way with line ACD to find a length of 4360 ft (1329 m) for the equivalent 8 in. (200 mm) pipe.