Handbook of Civil Engineering Calculations

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horsepower (kW) and flow curves. The horsepower (kW) for—say a 200-gal/min (18.6
L/s) flow rate would be about half of a 400-gal/min (37.2 L/s) flow rate.
If a pump were suddenly given a 300-gal/min (27.9 L/s) flow-rate demand at its cross-
ing point on a larger system-head curve, the hp required might be excessive. Hence, the
pump drive motor must be chosen carefully so that the power required does not exceed
the motor's rating. The power input required by any pump can be obtained from the pump
characteristic curve for the unit being considered. Such curves are available free of charge
from the pump manufacturer.
The pump operating point is at the intersection of the pump characteristic curve and
the system-head curve in conformance with the first law of thermodynamics, which states
that the energy put into the system must exactly match the energy used by the system. The
intersection of the pump characteristic curve and the system-head curve is the only point
that fulfills this basic law.
There is no practical limit for pumps in parallel. Careful analysis of the system-head
curve versus the pump characteristic curves provided by the pump manufacturer will fre-
quently reveal cases where the system load point may be beyond the desired pump curve.
The first cost of two or three smaller pumps is frequently no greater than for one large
pump. Hence, smaller pumps in parallel may be more desirable than a single large pump,
from both the economic and reliability standpoints.
One frequently overlooked design consideration in piping for pumps is shown in Fig.


  1. This is the location of the check valve to prevent reverse-flow pumping. Figure 20
    shows the proper location for this simple valve.

  2. Compute the energy saving possible
    Since one pump can carry the fluid flow load about 90 percent of the time, and this same
    percentage holds for the design conditions, the saving in energy is 0.9 x (0.5 kW - .25 kW)



  • x 90 h per week = 20.25 kWh/week. (In this computation we used the assumption that 1 hp
    = 1 kW.) The annual savings would be 52 weeks x 20.25 kW/week = 1053 kWh/yr. If elec-
    tricity costs 5 cents per kWh, the annual saving is $0.05 x 1053 = $52.65/yr.
    While a saving of some $51 per year may seem small, such a saving can become much


FIGURE 20. Check valve locations to prevent reverse flow.

PUMP

CHECK

VALVE
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