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

Answer:

62.1 ft

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