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                                          Chapter 8
                                      EXAMPLE 8.8 PUMPING AND STORAGE SYSTEM
                                      Draw a sketch, showing a system (including the raw water reservoir, pump station, water transmission lines, elevated water storage
                                      tank, hydraulic grade line, and elevation in ft or m) based on the following information: the raw water reservoir of a city is at an
                                      elevation of 500 ft (152.4 m). An automatic booster pumping station is proposed, having for its control point a 550,000 gal (2,082 m )
                                      water storage tank. The design tank water level is at an elevation of 490 ft (149.4 m).
                                          Measurement of the water storage tank level is to be transmitted to the pumping station and is to be the only variable used for
                                      the control of the pumps in the pumping station. The head losses of the water transmission line between the raw water reservoir and
                                      the elevated water storage tank (including the miscellaneous head losses in the proposed pumping station) computed for the rate of
                                      flow of 1,600 gpm (101 L/s) are 40 ft (12.2 m), which is to be overcome by the pumps. A chlorination system is used for treating the
                                      water entering the elevated water storage tank. The chlorinated water from the tank is transmitted to the city’s water filtration plant
                                      for further treatment prior to domestic consumption.
                                      Solution:    Pumping, Storage, and Dual Water Systems                                        3
                                      Figure 8.11 shows a sketch of the required water supply system.
                                                     El. 500 ft
                                                                                                                 El. 490'
                                                                               Hydraulic
                                                                               grade line
                                                                                                    40 ft
                                                                                                 at 1600 gpm
                                      Figure 8.11 Sketch of water supply system for Example 8.8.
                                                                                               3
                                                     ′
                                      Conversion factors: 1 = ft = 0.3048 m; 1 gpm = 3.785 L∕min = 0.0000631 m ∕s = 0.0631 L∕s.
                                      EXAMPLE 8.9 PUMPS AND WATER STORAGE TANKS
                                      The water supply system shown in Example 8.8 provides water to a city at the rates of 2.0 MGD (88 L/s) for average daily water
                                      demand, 3.4 MGD (150 L/s) for maximum daily water demand, and 4.8 MGD (210 L/s) for peak hourly water demand. The total
                                      head losses are 40 ft (12.2 m) at a flow of 1,600 gpm (101 L/s).
                                          What are the head losses of the transmission line between the raw water reservoir and the elevated water storage tank (including
                                      all miscellaneous head losses), which must be overcome by the pumps at 2, 3.4, and 4.8 MGD (88, 150, and 210 L/s; or 7.6, 12.9, and
                                      18.2 MLD)? Explain which water demand is used for the design of a pumping station between the raw water reservoir and the water
                                      storage tank, and why. Is the size of the existing water storage tank sufficient for the city in accordance with the current Ten-States
                                      Standards for Water Works?
                                          The following equation shows the relationship between the head loss and the water flow:
                                                                               H = H (Q ∕Q ) 2
                                                                                          1
                                                                                       2
                                                                                    1
                                                                                2
                                      where
                                                                 3
                                         Q = the first water flow, gpm (m /s)
                                          1
                                                                   3
                                         Q = the second water flow, gpm (m /s)
                                          2
                                         H = head loss at flow of Q ,ft(m)
                                          1
                                                            1
                                         H = head loss at flow of Q ,ft(m)
                                                            2
                                          2
                                      Solution 1 (US Customary System):
                                      Head loss at average daily water demand of 2 MGD (or 1388 gpm):
                                                                                           2
                                                                         H = (40)(1,388∕1,600) = 30 ft.
                                                                           2
                                      Head loss at maximum daily water demand of 3.4 MGD (or 2,360 gpm):
                                                                                           2
                                                                         H = (40)(2,360∕1,600) = 87 ft.
                                                                           2
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