Pumps & pump power: practice problems with step-by-step solutions

Free Pumps & pump power practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: hydraulic power, pump efficiency, system head and power.

Free practice problemEasy
PQ = 15 L/sH = ?

A pump delivers 15 L/s of water (ρ = 1000 kg/m³) and gives it a hydraulic power of 11 kW. Use g = 9.81 m/s². Find the head H the pump adds.

Ph = 11 kWQ = 15 L/sρ = 1000 kg/m³g = 9.81 m/s²
m
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What you practise

  1. Easyhydraulic power
  2. Mediumpump efficiency
  3. Hardsystem head and power

Key formulas

Hydraulic power
Ph = ρ·g·Q·H = Q·Δp
Shaft power
Pshaft = Ph/ηpump
Electrical power
Pel = Pshaft/ηmotor
Pump head
H = Δz + Δp/(ρ·g) + hL
System losses
hL = (f·L/D + ΣK)·V²/(2g)

Learn Pumps & pump power from scratch

Before this: Pipe friction & head loss

1What a pump does

A pump gives energy to a liquid. It can lift water to a higher tank, push it through a long pipe against friction, or raise its pressure.

Engineers describe that energy as a height, the pump head H in metres. A head of 40 m means each kilogram of water gets as much energy as lifting it 40 m.

A head can also be a pressure rise. The two are linked by the weight of the liquid:

Δp = ρ · g · H

For water (ρ = 1000 kg/m³) every metre of head is 9.81 kPa. So a pump that adds 98.1 kPa adds a head of 10 m.

Self-check A pump raises the pressure of water by 196.2 kPa. What head does it add? (ρ = 1000 kg/m³, g = 9.81 m/s²)
  1. 0.2 m
  2. 196.2 m
  3. 2 m
  4. 20 m

Answer: 20 m. H = Δp/(ρ·g) = 196 200/(1000 · 9.81) = 20 m. The pressure must be in Pa, not kPa, before you divide.

2Hydraulic power and efficiency

The power the water actually receives is the hydraulic power (sometimes called water power). It is the weight of water moved per second times the head:

Ph = ρ · g · Q · H = Q · Δp

With ρ in kg/m³, g in m/s², Q in m³/s and H in m, the power comes out in watts. Remember: 1 L/s = 0.001 m³/s.

A pump is not perfect. Friction and leaks inside it waste part of the power on its shaft. The pump efficiency is the share that reaches the water. The electric motor has its own efficiency:

MPPel​Pshaft​Ph​ηmotor​ηpump​
Pshaft = Ph / ηpump Pel = Pshaft / ηmotor

Each step loses some power, so going back towards the supply you always divide by the efficiency. The input is always larger than the output.

Self-check Water receives 8 kW of hydraulic power from a pump with an efficiency of 80%. What shaft power does the pump need?
  1. 1.6 kW
  2. 10 kW
  3. 6.4 kW
  4. 8 kW

Answer: 10 kW. Pshaft = Ph/η = 8/0.8 = 10 kW. Multiplying by the efficiency (6.4 kW) would make the input smaller than the output, which is impossible.

3The head a system needs

When a pump moves water from one reservoir to another, it must supply three things:

ΔzP
  • The static lift Δz: the height between the two water surfaces.
  • The losses hL in the pipe: friction plus fittings, hL = (f·L/D + ΣK)·V²/(2g).
  • A pressure head Δp/(ρ·g) if the upper tank is closed and pressurised.
H = Δz + Δp / (ρ · g) + hL

With open tanks on both sides the pressure term is zero. Then the pump only has to lift the water and overcome the losses.

Self-check A pump lifts water 15 m into a closed tank where the air is at 50 kPa gauge. The pipe losses are 3 m. What head must the pump add?
  1. 18 m
  2. 20.1 m
  3. 23.1 m
  4. 68 m

Answer: 23.1 m. H = 15 + 50 000/(1000 · 9.81) + 3 = 15 + 5.1 + 3 = 23.1 m. Forgetting the tank pressure gives 18 m, and adding 50 as if it were metres gives 68 m.

4The method step by step

  1. Convert the flow rate to m³/s (L/s ÷ 1000) and any pressure to Pa.
  2. Find the head H. If it is not given, add the static lift, the pressure head and the pipe losses.
  3. Hydraulic power: Ph = ρ·g·Q·H, in W. Divide by 1000 for kW.
  4. Shaft power: divide by the pump efficiency. Electrical power: divide again by the motor efficiency.
  5. To find an efficiency from a test, divide the output by the input: η = Ph / Pshaft.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A pump with an efficiency of 80% delivers 20 L/s of water (ρ = 1000 kg/m³) against a total head of 40 m. It is driven by an electric motor with an efficiency of 95%. Use g = 9.81 m/s². Find the shaft power and the electrical power.

1

Convert the flow rate

Q = 20 L/s = 0.020 m³/s

The head is given directly: H = 40 m.

2

Hydraulic power

Ph = 1000 · 9.81 · 0.020 · 40 = 7848 W = 7.85 kW

This is the power that ends up in the water.

3

Shaft power

Only 80% of the shaft power reaches the water, so the shaft must deliver more:

Pshaft = 7.848 / 0.80 = 9.81 kW
4

Electrical power

Pel = 9.81 / 0.95 = 10.33 kW
5

Check the answer

The overall efficiency is 0.80 · 0.95 = 0.76, and 7.848 / 0.76 = 10.33 kW. Same result.

The powers grow towards the supply: 7.85 → 9.81 → 10.33 kW. That is the right order.

Answer

Pshaft ≈ 9.81 kW and Pel ≈ 10.3 kW.

6Common mistakes

  • Using the flow rate in L/s as if it were m³/s. That makes the power 1000 times too large.
  • Multiplying by the efficiency instead of dividing when you go from the water back to the shaft.
  • Mixing W and kW in the final answer.
  • Forgetting the pipe losses or the tank pressure when you build up the system head.
  • Writing the efficiency as 80 instead of 0.80.
Self-check A pump delivers 5 L/s of water against a head of 20 m. What is the hydraulic power? (ρ = 1000 kg/m³, g = 9.81 m/s²)
  1. 0.1 kW
  2. 0.981 kW
  3. 98.1 kW
  4. 981 kW

Answer: 0.981 kW. Ph = 1000 · 9.81 · 0.005 · 20 = 981 W = 0.981 kW. Using Q = 5 instead of 0.005 gives 981 kW, and forgetting g gives 0.1 kW.

7How exam questions look

  • Easy: hydraulic power. From Q and H find Ph, or from Ph and Q find the head. Some questions give a pressure rise instead of a head: then Ph = Q·Δp.
  • Medium: pump efficiency. Find the shaft power from the efficiency, the efficiency from a test, or the electrical power when the motor efficiency is also given.
  • Hard: system head and power. A pump moves water between two reservoirs through a pipe with friction and fittings. Find the pipe losses, the pump head and then the shaft power.

In the hard questions, find V = Q/A first, then hL = (f·L/D + ΣK)·V²/(2g), then H = Δz + Δp/(ρ·g) + hL, and only then the power.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
PQ = 60 L/sH = ?

A pump delivers 60 L/s of water (ρ = 1000 kg/m³) and gives it a hydraulic power of 15 kW. Use g = 9.81 m/s². Find the head H the pump adds.

Ph = 15 kWQ = 60 L/sρ = 1000 kg/m³g = 9.81 m/s²

Find H (m)

Show hints
  1. Hydraulic power P = ρ·g·Q·H, so H = P/(ρ·g·Q).
  2. Use watts and m³/s.
  3. ρ·g·Q = 1000·9.81·0.06 = 588.6 W/m.
Show the answer and the step-by-step solution

Pump power: the power given to the water is the hydraulic power Ph = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: Pshaft = Ph/ηpump, and the motor draws Pel = Pshaft/ηmotor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.

Head

H = P/(ρ·g·Q) = 15000/(1000·9.81·0.06) = 25.48 m

Answer

H = 25.48 m
Example 2Medium
PQ = 80 L/sH = 64 m

A pump with an efficiency of 60% delivers 80 L/s of water (ρ = 1000 kg/m³) against a head of 64 m. Use g = 9.81 m/s². Find the shaft power Pshaft needed, in kW.

Q = 80 L/sH = 64 mη = 60%g = 9.81 m/s²

Find Pshaft (kW)

Show hints
  1. First find the hydraulic power ρ·g·Q·H.
  2. Then Pshaft = Ph/η.
  3. Ph = 50.23 kW.
Show the answer and the step-by-step solution

Pump power: the power given to the water is the hydraulic power Ph = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: Pshaft = Ph/ηpump, and the motor draws Pel = Pshaft/ηmotor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.

Hydraulic power

Ph = ρ·g·Q·H = 1000·9.81·0.08·64 = 50.23 kW

Shaft power

Pshaft = Ph/η = 50.23/0.6 = 83.71 kW

Answer

Pshaft = 83.71 kW
Example 3Hard
PQ = 60 L/sΔz = 5 mL = 200 m, D = 200 mm, f = 0.025, ΣK = 5

A pump moves 60 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 5 m higher. The pipe is 200 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.025; the minor-loss coefficients add up to ΣK = 5 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.

Q = 60 L/sΔz = 5 mL = 200 mD = 200 mmf = 0.025ΣK = 5

Find H (m)

Show hints
  1. Energy equation from the lower surface to the upper surface: H = Δz + hL.
  2. hL = (f·L/D + ΣK)·V²/(2g).
  3. V = 1.91 m/s, V²/(2g) = 0.1859 m.
Show the answer and the step-by-step solution

Pump power: the power given to the water is the hydraulic power Ph = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: Pshaft = Ph/ηpump, and the motor draws Pel = Pshaft/ηmotor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.

Velocity

V = Q/A = 0.06/0.03142 = 1.91 m/s
V²/(2g) = 0.1859 m

Head losses

hL = (f·L/D + ΣK)·V²/(2g) = (25 + 5)·0.1859 = 5.577 m

Pump head

Both surfaces are open and practically at rest, so pressure and velocity terms drop out.

H = Δz + hL = 5 + 5.577 = 10.58 m

Answer

H = 10.58 m

Questions and answers

What is Pumps & pump power?

A pump gives energy to a liquid. It can lift water to a higher tank, push it through a long pipe against friction, or raise its pressure.

Engineers describe that energy as a height, the pump head H in metres. A head of 40 m means each kilogram of water gets as much energy as lifting it 40 m.

How do I solve Pumps & pump power problems step by step?

Pump power: the power given to the water is the hydraulic power Ph = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: Pshaft = Ph/ηpump, and the motor draws Pel = Pshaft/ηmotor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.

What are common mistakes in Pumps & pump power?
  • Divide by g as well: P = ρ·g·Q·H.
  • Convert kW to W and L/s to m³/s, then H = P/(ρ·g·Q).
  • The shaft must supply more than the water receives: divide by the efficiency, don't multiply.
What do the Pumps & pump power exercises cover?
  • Easy: hydraulic power
  • Medium: pump efficiency
  • Hard: system head and power

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

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Printable worksheet

Pumps & pump power worksheet with answers
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