PE Civil Construction · Complete public lesson

From pressure and elevation to electrical pump power

Build an SI energy balance, distinguish hydraulic output from electrical input, and test how a changed boundary affects the result.

Suggested 18 minutes · No account required

The situation and givens

A pump transfers water at 0.035 m³/s from an open sump to a receiving vessel. The vessel water surface is 12 m above the sump surface and its gauge pressure is 19.62 kPa. Pipe and fitting losses total 4 m of water head at this flow. Velocities at both water surfaces are negligible. The combined pump-and-motor efficiency is 70%. Use water density 1,000 kg/m³ and g = 9.81 m/s². What is most nearly the required electrical input power at this operating point?

Flow Q: 0.035 m³/s
Surface elevation rise: 12 m
Discharge vessel gauge pressure: 19.62 kPa; sump gauge pressure: 0
Pipe and fitting loss: 4 m of water
Combined pump-and-motor efficiency: 0.70
Water density: 1,000 kg/m³; g: 9.81 m/s²

Water moves from an open sump to a pressurized vessel. The vessel surface is 12 m above the sump, its gauge pressure is 19.62 kPa, flow is 0.035 m³/s, and pipe and fitting loss is 4 m. Combined efficiency is 70%. — Open full-size technical exhibit

Complete worked method

1. Convert pressure to water head

19.62 kPa is 19,620 Pa. Dividing pressure by density times gravity gives metres of water, so it can be combined with elevation and loss head.

H_p=\frac{p_2-p_1}{\rho g}=\frac{19{,}620}{1{,}000(9.81)}=2.00\ \mathrm{m}

2. Assemble the required pump head

Both water surfaces have negligible velocity. The receiving vessel's pressure remains part of the boundary condition; an open sump does not make the other vessel open.

H=\Delta z+H_p+h_L=12+2+4=18\ \mathrm{m}

3. Find hydraulic output

Density, gravity, flow and total head produce watts in base SI units. This is the rate of useful energy delivered to the water.

P_{\mathrm{hyd}}=\rho gQH=1{,}000(9.81)(0.035)(18)=6{,}180.3\ \mathrm{W}

4. Account for combined efficiency

Electrical input must exceed hydraulic output at an efficiency below 100%. Divide once by the combined pump-and-motor efficiency, then convert W to kW.

P_{\mathrm{electrical}}=P_{\mathrm{hyd}}/\eta=6{,}180.3/0.70=8{,}829\ \mathrm{W}\approx8.83\ \mathrm{kW}

Common errors and why they fail

6.18 kW is the hydraulic power delivered to the water. Electrical input must be larger because combined efficiency is below 100%.
7.85 kW includes the 12 m elevation rise and 4 m loss but omits the 2 m pressure head required at the receiving vessel.
12.6 kW divides by 0.70 twice. The given 70% already combines pump and motor efficiencies.

Apply the method

Which value is the required electrical input at the stated operating point?

Choose one response6.18 kW7.85 kW12.6 kW8.83 kW

Change one condition

The receiving vessel is vented to atmosphere, with the same flow, elevation, losses and efficiency. What changes? Work it out before revealing the answer.

Reveal the worked transfer

The pressure head falls from 2 m to zero. Total head is 16 m, so input is 1,000 × 9.81 × 0.035 × 16 / 0.70 = 7,848 W = 7.85 kW. Elevation and friction remain. This is a stated fixed-flow exercise; an actual pump's operating point would require its curve.

Put the method to work

Try different questions to check whether you can identify the governing condition and explain the next action. The free sampler includes answers and worked explanations.

Try 5 free questions →See program and price

The result is operating input power. Motor sizing, starting demand, pump curves and a specified design margin would require additional information. This public application also appears in the supplementary SI practice collection; it is instructional practice, not an unseen scored assessment.

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