CivilHydraulicseasyFE ExamCourseworkPE / Advanced

Shaft Power Required by a Pump

A pump delivers $Q = 0.050\ \text{m}^3/\text{s}$ of water against a total head of $H = 20\ \text{m}$ .

The pump efficiency is $\eta = 0.75$ . Take $\rho = 1000\ \text{kg/m}^3$ and $g = 9.81\ \text{m/s}^2$ .

Find the shaft power input required.

Given

Q	0.05 m^3/s	Flow rate
H	20 m	Total head
η	0.75 -	Pump efficiency

Hint 1

First find the useful power delivered to the water: ρgQH.

Hint 2

Then account for the pump's inefficiency.

Hint 3

Input power must exceed output power — so divide by η, don't multiply.

Worked solution — try the problem first

Hydraulic (water) power — the useful power actually delivered to the fluid:

$P_{hyd} = \rho g Q H = 1000 \times 9.81 \times 0.050 \times 20 = 9810\ \text{W}$

Shaft power. The pump is only 75% efficient, so the input must be larger than the output — divide, do not multiply:

$P_{shaft} = \frac{P_{hyd}}{\eta} = \frac{9810}{0.75} = 13{,}080\ \text{W}$

$\boxed{P_{shaft} = 13.1\ \text{kW}}$

Direction check. If you multiply by efficiency you get 7.36 kW, which is less than the useful output — a machine producing more than it consumes. Whenever an efficiency appears, ask which quantity must be larger and let that fix the operation.

Concepts:Pump powerHydraulic powerEfficiency

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