Question

Measurements made on an operating centrifugal pump indicate that for a flow rate of 240 gpm, 6 HP are being consumed. The operating manual says the pump is 62% efficient at these conditions. What is the actual head increase across the pump? (61.3 ft)

Answer 1

62.1 ft

Step-by-step explanation:

We know that
`1 gpm= 0.0022 ft^(3)/s` hence 240 gpm will be

`240*0.0022=0.528 ft^(3)/s`

From the formula of obtaining efficiency

`\eta=\frac {\gamma Q h_a}{500W_(in)}` and making
`h_a` the subject we have

`h_a=\frac {\eta*550*W_(in)}{\gamma Q}` where Q is flow rate,
`\gamma` is specific weight of water,
`h_a` is actual head rise and
`W_(in)` is the power input to the compressor

Substituting 0.62 for
`\eta`, 6hp for
`W_(in)`,
`62.4 lb/ft^(3)` for
`\gamma` and
`0.528 ft^(3)/s` for Q then

`h_a=\frac {0.62*550*6}{62.4*0.528}=\frac {2046}{32.9472}=62.09936\approx 62.1 ft`

Therefore, actual head rise of the water being pumped is 62.1 ft