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                                                                                  6.6 Office Studies of Pipe Networks  207


                                             3. Section c-c population   14,700:
                                                                                6
                                                (a) Demands: domestic   14,700   150/10   2.2 MGD (8.3 MLD); fire   5.4 MGD
                                                   (20.4 MLD); total   7.6 MGD (28.7 MLD).
                                                (b) Existing pipes: one 20-in. (500 mm) at 3.7 MGD; two 12-in. (300 mm) at 1.0 MGD
                                                   2.0 MGD; five 6-in. (150 mm) at 0.16 MGD   0.8 MGD; total   6.5 MGD (24.6 MLD).
                                                (c) Deficiency   7.6   6.5   1.1 MGD or (28.7   24.6   4.1 MLD).
                                                (d) Pipes added: two 10-in. (250 mm) at 0.6 MGD   1.2 MGD (4.5 MLD).
                                                   Pipes removed: one 6-in. at 0.2 MGD (one 150-mm at 0.76 MLD).
                                                   Net added capacity: 1.2   0.2   1.0 MGD (3.8 MLD).
                                                   Reinforced capacity   6.5   1.0   7.5 MGD or (24.6   3.8   28.4 MLD).
                                                (e) The reinforced system equivalent pipe at 7.5 MGD (28.4 MLD) and a hydraulic gradient
                                                   of 2‰ is 26.0-in. (650 mm)
                                                   This will carry 7.6 MGD (28.8 MLD) with a loss of head of 2.1‰.





                    6.6.2  Relaxation (Hardy Cross)
                                         A method of relaxation, or controlled trial and error, was introduced by Hardy Cross,
                                         whose procedures are followed here with only a few modifications. In applying a method
                                         of this kind, calculations become speedier if pipe-flow relationships are expressed by an
                                         exponential formula with unvarying capacity coefficient, and notation becomes simpler if
                                         the exponential formula is written
                                                                          H = kQ n                            (6.1)

                                         where, for a given pipe, k is a numerical constant depending on C, d, and L, and Q is the flow,
                                         n being a constant exponent for all pipes. In the Hazen-Williams equation, for example,
                                                           Q = 405 Cd 2.63 0.54  (U.S. Customary Units)
                                                                      s
                                         where Q   rate of discharge, gpd; d   diameter of circular conduits, in.; C   Hazen-William
                                          coefficient, dimensionless; S   H/L   hydraulic gradient, dimensionless; H   loss of head, ft;
                                          L   conduit length, ft.

                                                             Q = 0.278 CD 2.63 0.54   (SI Units)
                                                                            S
                                                                  3
                                          where Q   rate of discharge, m /s; D   diameter of circular conduits, m; C   Hagen-William
                                         coefficient, dimensionless; S   H/L   hydraulic gradient, dimensionless; H   loss of head,
                                         m; L   conduit length, m. For the Hazen William using either the U.S. customary units or the
                                          SI units, the following relationship hold true.
                                                                       s = k¿Q 1>0.54  = k¿Q 1.85             (6.2)
                                                                       H = sL
                                                                       H = kQ 1.85

                                             Two procedures may be involved, depending on whether (a) the quantities of water en-
                                          tering and leaving the network or (b) the piezometric levels, pressures, or water table ele-
                                          vations at inlets and outlets are known.
                                             In balancing heads by correcting assumed flows, necessary formulations are made alge-
                                          braically consistent by arbitrarily assigning positive signs to clockwise flows and associated
                                          head losses, and negative signs to counterclockwise flows and associated head losses. For the
                                          simple network shown in Fig. 6.11a, inflow Q and outflow Q are equal and known, inflow
                                                                                          o
                                                                              i
                                          being split between two branches in such a manner that the sum of the balanced head losses
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