Handbook of Civil Engineering Calculations

(singke) #1

the tank and system is 60 lb/in
2
(gage) (413.6 kPa) and the allowable low pressure is 30
lb/in
2
(gage) (206.8 kPa). How many starts per hour will the pump make if the system
draws 3000 gal/min (189.3 L/s) from the tank?


Calculation Procedure:



  1. Compute the required tank capacity
    If the usual hydropneumatic system, a storage-tank capacity in gal of 10 times the pump
    capacity in gal/min is used, if this capacity produces a moderate running time for the
    pump. Thus, this system would have a tank capacity of (10)(200) = 2000 gal (7570.8 L).

  2. Compute the quantity of liquid withdrawn per cycle
    For any hydropneumatic tank the withdrawal, expressed as the number of gallons (liters)
    withdrawn per cycle, is given by W = (VL - %)/C, where V 1 = air volume in tank at the
    lower pressure, ft^3 (m^3 ); % = volume of air in tank at higher pressure, ft^3 (m^3 ); C = con-
    version factor to convert ft^3 (m^3 ) to gallons (liters), as given below.
    Compute VL and Vn using the gas law for VH and either the gas law or the reserve per-
    centage for VL. Thus, for vffi the gas law gives vH=pLvL/pH, where pL = lower air pressure
    in tank, lb/in^2 (abs) (kPa); pH = higher air pressure in tank lb/in^2 (abs) (kPa); other sym-
    bols as before.
    In most hydropneumatic tanks a liquid reserve of 10 to 20 percent of the total tank vol-
    ume is kept in the tank to prevent the tank from running dry and damaging the pump. As-
    suming a 10 percent reserve for this tank, VL = 0.1 V, where V= tank volume in ft^3 (m^3 ).
    Since a 2000-gal (7570-L) tank is being used, the volume of the tank is 2000/7.481 fWgal
    = 267.3 ft^3 (7.6 m^3 ). With the 10 percent reserve at the 44.7 lb/in^2 (abs) (308.2-kPa) low-
    er pressure, VL = 0.9 (267.3) = 240.6 ft^3 (6.3 m^3 ), where 0.9 = V- 0.1 V.
    At the higher pressure in the tank, 74.7 lb/in^2 (abs) (514.9 kPa), the volume of the air
    will be, from the gas law, vH = pLvL/pH = 44.1 (24Q.6)114.1 = 143.9 ft^3 (4.1 m^3 ). Hence,
    during withdrawal, the volume of liquid removed from the tank will be W 8 = (240.6 -
    143.9)70.1337 = 723.3 gal (2738 L). In this relation of the constant converts from cubic
    feet to gallons and is 0.1337. To convert from cubic meters to liters, use the constant 1000
    in the denominator.

  3. Compute the pump running time
    The pump has a capacity of 200 gal/min (12.6 L/s). Therefore, it will take 723/200 = 3.6
    min to replace the withdrawn liquid. To supply 3000 gal/h (11,355 L/h) to the system, the
    pump must start 3000/723 = 4.1, or 5 times per hour. This is acceptable because a system
    in which the pump starts six or fewer times per hour is generally thought satisfactory.
    Where the pump capacity is insufficient to supply the system demand for short peri-
    ods, use a smaller reserve. Compute the running time using the equations in steps 2 and 3.
    Where a larger reserve is used—say 20 percent—use the value 0.8 in the equations in step

  4. For a 30 percent reserve, the value would be 0.70, and so on.
    Related Calculations. Use this procedure for any liquid system having a hydropneu-
    matic tank—well drinking water, marine, industrial, or process.


USING CENTRIFUGAL PUMPS


AS HYDRAULIC TURBINES


Select a centrifugal pump to serve as a hydraulic turbine power source for a 1500-gal/min
(5677.5-L/min) flow rate with 1290 ft (393.1 m) of head. The power application requires

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