A pump is proposed as the supply unit for a Mall fountain Th

A pump is proposed as the supply unit for a Mall fountain. The following requirements have been provided by the architectural firm: The pump outlet is to be located 3 feet below ground level. The water flow of 2,000 kpph is to reach a peak height of 30 feet above ground level. The pump characteristics are given in the following plot. What head must be supplied by the pump? Report your answer in ft. What flow rate must be supplied by the pump? Report your answer in gal min (gpm). What pump impeller diameter should be used? (either 15.00, 16.00, 17.00, or 18.00 inch diameter) What is the pump efficiency? Report your answer in terms of a percentage. What power is required to derive the pump? Report your answer in horsepower (hp). What range of NPSH is acceptable at the pump inlet? Report your answer in ft.

Solution

1) The head to be supplied is (3+30) ft =33 ft minimum

2)Flow rate of 2000 kpph or 2000 kilopounds/hour equals 4000 gallon/minute ( 1 kpph = 2 gal /minute)

3) what pump dia impeller must be used: Can see ther curves from the graph: For a heat of 33 ft and flow of 4000 gpm, the curve nearest the vertical line going from 4000 upwards and meeting the horizontal line drawn from 33 on the top graphs is the D4 curve --corresponding to a dia of 15.00 inch

4) Pump efficiency can also be read off from the top graph, it is between 75% and 80 percent or about 78%

5) Power Reqd To Drive The Pump In hp :The mass flow as calculated is 4000 gal/min

6) From the graph the NPSH is 10 ft approx. If this minimum is not met at the inlet, the pump will cavitate.

Since one gallon is 8.33 lb, the flow is 4000*8.33 lp/min or 555.33 lb/sec

Since efficiency is approx 80%, the theoretical flow must be discounted by 0.8, yielding 555.33/.8= 694.2 lb/sec

The work done to raise this flow thru a head of 33 ft is 694.2*33 ft pound sec or22907.5 ft lb sec

Now one HP is 550 ft lb sec, hence the power of the pump is 22907.5/550 = 41.6 or approx 42 hp