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                                          Chapter 5
                                      EXAMPLE 5.50 CAPACITY AND SIZE OF PARALLEL PIPES
                                      A town is receiving its water supply from a groundwater tank located at the treatment plant through a gravity main that terminates at
                                      an elevated water tank adjacent to the town. The difference between the levels of the tanks is 100 m, and the distance between them
                                      is 50 km. The tanks were originally connected by a single pipeline designed to carry 13,000 m ∕day. It was later found necessary to
                                                             3
                                      increase the supply to 19,500 m /day and consequently the decision was made to lay another pipeline of the same diameter alongside
                                      part of the original line and cross-connected to it.
                                          Calculate (given C for all pipes = 100)
                                          1. The diameter of the pipes
                                          2. The length of the second pipe, which was necessary to install
                                      Solution:    Water Hydraulics, Transmission, and Appurtenances      3
                                          1. The diameter of the pipes:
                                                                                                                3
                                                               3
                                                                            3
                                            At present: Q = 13,000 m ∕d = (13,000 m ∕d)∕ (24 h∕d) (60 min ∕h) (60 s∕min) = 0.150 m ∕s (150 L∕s).
                                                                              S = h ∕L = 100 m∕50 km = 2 .
                                                                                  f
                                                                                     3
                                            The nomogram (C = 100) of Fig. 5.17 for Q = 0.150 m ∕sand s = 2  will give a pipe diameter = 500 mm.
                                            Diameter of existing pipe no. 1 = 500 mm.
                                                                      L 2  = 50 – L 1      L 1
                                                        3
                                            Q = (19,500 m ∕d)∕(24 h∕d) (60 min ∕h) (60 s∕min)
                                              2
                                                       3
                                               = 0.226 m ∕s = (226 L∕s) .
                                                              3
                                            Q = 0.226∕2 = 0.113 m ∕s = 113 L∕s.
                                             1
                                            Lines are in parallel, their flows are equal.
                                          2. The length of the second pipe, which had to be installed:
                                            h = (h ) + (h ) .
                                                        f 2
                                             f
                                                  f 1
                                            Lines are in series, their head losses are additive.
                                            100 = L s + L s .
                                                       2 2
                                                  1 1
                                                      (      )
                                            100 = L s + 50 − L 1  s .
                                                               2
                                                  1 1
                                                                                      3
                                            The nomogram (C = 100) of Fig. 5.17 for Q = 0.113 m ∕sand D = 500 mm will give s = 1.0 .
                                                                                                             1
                                                                             1
                                                                                      3
                                            The nomogram (C = 100) of Fig. 5.17 for Q = 0.226 m ∕sand D = 500 mm will give s = 4.3 .
                                                                             2
                                                                                                             2
                                                                         100 = L km × 1.0  + (50 − L ) × 4.3 .
                                                                               1
                                                                                                  1
                                                                         100 = L + 215 − 4.3 L .
                                                                                           1
                                                                               1
                                                                         3.3 L = 115.
                                                                            1
                                                                         L = 115∕3.3 = 35 km.
                                                                          1
                                            The length of the second pipe, which had to be installed, is 35 km.
                                    5.4 MULTIPLE LINES
                                                                                       for example; (b) when possible failure would put the line
                                    Although masonry aqueducts and tunnels of all kinds are  out of commission for a long time; and (c) when pipe loca-
                                    best designed to the full projected capacity of the sys-  tion presents special hazards—floods, ice, and ships’ anchors
                                    tem, this is not necessarily so for pipe lines. Parallel lines  endangering river crossings or submarine pipes and cave-ins
                                    (Fig. 5.22) built a number of years apart may prove to be  rupturing pipe lines in mining areas, for example.
                                    more economical. Cost, furthermore, is not the only consid-  Twin lines generally cost 30–50% more than a single
                                    eration. It may be expedient to lay more than one line (a)  line of equal capacity. If they are close enough to be inter-
                                    when the maximum pipe size of manufacture is exceeded;  connected at frequent intervals, gates should be installed in
                                    36 in. (900 mm) in the case of centrifugal cast iron pipe,  the bridging pipes to keep most of the system in operation
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