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                                          Chapter 4
                                                   Quantities of Water Demand
                                          For management:
                                                                      + Q
                                                        Q
                                                              = Q
                                                                              = 590 + 913 = 1,503 gpud.
                                                         max day
                                                                 domestic
                                                                         sprinkling
                                                        By Eq. (4.21a): Q
                                                                         = 334 + 2.02Q
                                                                                        = 334 + 2.02 × 1,503 = 3,640 gpud.
                                                                                    max day
                                                                     peak h
                                          For design based on 200 dwellings:
                                      From Table 4.11: Q
                                                         = 1,503 + 0.4(1,000) = 1,903 gpud.
                                                    max day
                                                        = 3,640 + 0.7(1,000) = 4,340 gpud.
                                      From Table 4.11: Q
                                                    peak h
                                      Solution 2 (SI System):
                                      Average maximum and peak daily domestic demands per dwelling unit:
                                                    By Eq. (4.20b): Q  = 594 + 13.1 MF = 594 + 13.1 × 125 × 1.0 = 2,232 Lpud.
                                                                 domestic
                                                    By Eq. (4.19b): A = 1.016 D −1.26  = 1.016∕(7.415) 1.26  = 1.016∕12.4836 = 0.08138 ha∕unit.
                                                    By Eq. (4.18b): Q sprinkling  = 6,003 A(E − P) = 6,003 × 0.08138 × (7.11 − 0)
                                                      = 3,473 Lpud excluding precipitation.
                                          For management:
                                                     Q max day  = Q domestic  + Q sprinkling  = 2,232 + 3,473 = 5,705 Lpud.
                                                     By Eq. (4.21b): Q peak h  = 1,264 + 2.02 Q max day  = 1,264 + 2.02 × 5,705.2 = 12,789 Lpud.
                                          For design based on 200 dwellings:
                                      From Table 4.11: Q max day  = 5,705 + 0.4 (1,000) × 3.785 = 7,219 Lpud.
                                      From Table 4.11: Q peak h  = 12,789 + 0.7 (1,000) × 3.785 = 15,438 Lpud.
                                      EXAMPLE 4.4 DETERMINATION OF FIRE FLOW AND DURATION FOR BUILDINGS
                                                                                                                 2
                                                                                                          2
                                      A four-story wooden-frame building (combustible, unprotected) has each floor of area 5,490 ft (510 m ). This wooden building is
                                                                                                                     2
                                                                                                              2
                                      adjacent to a five-story building of commercial construction (noncombustible, protected) with 9,800 ft (910 m ) per floor. Determine
                                      the fire flow and duration (a) for four-story building; (b) for five-story building; and (c) for both buildings assuming they are connected
                                      together.
                                      Solution 1 (US Customary System):
                                                                                         2
                                         a. Area occupied by the four-story building = 4 (5,490) = 21,960 ft .
                                             From Table 4.13, column Type V-B,
                                             Fire flow = 4,000 gpm and duration = 4h.
                                                                                         2
                                        b. Area occupied by the five-story building = 5 (9,800) = 49,000 ft .
                                             From Table 4.13, column Type I-A,
                                             Fire flow = 2,500 gpm and duration = 2h.
                                                                                     2
                                         c. Total area of two buildings = 21,960 + 49,000 = 70,960 ft .
                                             Fractional area of the four-story building = 21,960/70,960 = 0.3095.
                                             Fractional area of the five-story building = 49,000/70,960 = 0.6905.
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