Pump power, affinity laws and pipe friction
Estimate hydraulic, shaft and motor power, predict the effect of a speed change and check friction loss in the pipework.
From fluid power to electrical power
A pump adds energy to a liquid. The useful rate of energy transfer is hydraulic power. The pump shaft must supply more than that because of the pump efficiency, and the motor must draw more electrical power than the shaft receives because of motor losses. The pump power calculator follows that chain. These are estimates at one duty point; motor selection and NPSH need separate checks against the pump curve and applicable standard.
Density enters the hydraulic power directly, so a liquid heavier or lighter than water needs the correct value. Head is used rather than pressure because the same pump produces the same head for liquids of different density, while the pressure rise changes.
Power formulas
- Hydraulic power (kW) = ρ × g × Q × H ÷ 1000, with Q in m³/s (m³/h ÷ 3600) and g = 9.80665 m/s².
- Shaft power = hydraulic ÷ pump efficiency.
- Motor input = shaft ÷ motor efficiency.
Total dynamic head includes static lift, pressure difference and friction at the operating flow. Use the pump efficiency from the curve at the duty point and the motor efficiency at its actual load. Drive and coupling losses are not included.
Overall efficiency is the product of the two efficiencies. Because shaft power is divided by two efficiencies in turn, a modest drop in either one raises the electrical power noticeably, which is why the efficiency values should come from the curve and the motor data rather than a typical figure.
Worked example 1: water duty
Flow 120 m³/h, head 35 m, density 1000 kg/m³, pump efficiency 72 %, motor efficiency 92 %.
- Q = 120 ÷ 3600 = 0.03333 m³/s.
- Hydraulic = 1000 × 9.80665 × 0.03333 × 35 ÷ 1000 = 11.44 kW.
- Shaft = 11.44 ÷ 0.72 = 15.89 kW.
- Motor input = 15.89 ÷ 0.92 = 17.27 kW.
Overall efficiency is 0.72 × 0.92 = 66.24 %. A lighter liquid of 850 kg/m³ at the same flow, head and efficiencies gives 9.72 kW hydraulic, 13.51 kW shaft and 14.68 kW input, because power is proportional to density.
Affinity laws and a worked speed change
For the same pump and impeller, with speed ratio r = N2 ÷ N1: Q2 = Q1 × r, H2 = H1 × r² and P2 = P1 × r³. The pump affinity laws calculator applies these directly and assumes efficiency stays about the same.
The pump above runs at 1450 RPM with 15.89 kW shaft power. Slowing it to 1160 RPM gives r = 0.8:
- Flow = 120 × 0.8 = 96 m³/h.
- Head = 35 × 0.64 = 22.4 m.
- Shaft power = 15.89 × 0.512 = 8.136 kW, a 48.8 % reduction.
At 1305 RPM (r = 0.9) the values are 108 m³/h, 28.35 m and 11.58 kW, a 27.1 % reduction. Using the energy and running cost calculator, 17.27 kW for 20 h/day over 300 days is 103,620 kWh, or 12,434.40 at a tariff of 0.12 per kWh. At the reduced speed the motor input is about 8.136 ÷ 0.92 = 8.84 kW, giving 53,040 kWh and 6,364.80. That is only valid if the system really delivers the reduced duty, efficiencies hold, and drive losses (not included) are small.
Pipe friction and the system curve
The head a pump must supply is static head plus friction. The pipe pressure drop calculator uses Darcy–Weisbach for straight, full liquid pipe: v = Q ÷ A, Re = ρ × v × D ÷ μ, then ΔP = f × (L ÷ D) × ρ × v² ÷ 2. The friction factor is 64 ÷ Re for laminar flow (Re below 2000) and the Swamee–Jain expression for turbulent flow. Between Re 2000 and 4000 the result is indicative only.
Example: 120 m³/h through 200 m of 107 mm internal-diameter pipe, roughness 0.045 mm, density 1000 kg/m³, viscosity 1 mPa·s. Velocity is 3.707 m/s, Re = 396,648, f = 0.01750, ΔP = 2.248 bar (224.8 kPa) and head loss 22.92 m. At 60 m³/h the velocity is 1.853 m/s, f = 0.01852, ΔP = 0.595 bar and head loss 6.06 m, so halving flow cuts friction to about 26 %, not exactly 25 %, because f changes. Adding an assumed 12 m static lift gives 12 + 22.9 = 34.9 m, close to the 35 m used in example 1. With a large static part the system curve does not pass through the origin, so flow will not fall in direct proportion to speed.
Common mistakes
- Using the best-efficiency value rather than the efficiency at the duty point.
- Entering head in bar or flow in L/s without converting.
- Treating motor nameplate power as the actual draw.
- Applying the cube law to a system dominated by static head.
- Using nominal pipe size instead of the actual internal diameter, or water viscosity for a hot or viscous liquid.
- Forgetting fittings, valves and elevation in the head.
Checks before you trust the result
- Compare the calculated shaft power with the pump curve at the same flow and head.
- Verify flow and head by measurement where possible.
- Check speed limits for pump and motor, minimum flow and NPSH.
- For savings, use the real duty profile rather than a single point.
A quick plausibility check on the power chain: for water, hydraulic kW is roughly flow in m³/h times head in m divided by 367. For example 1, 120 × 35 ÷ 367 = 11.44 kW, which agrees with the calculator.
Frequently asked questions
Why does power fall faster than flow?
Power scales with the cube of speed, while flow scales linearly, provided the system is friction-dominated.
Does the calculator size the motor?
No. It gives power at one duty point; motor selection needs margin and checks.
Can I use the pipe calculator for steam or air?
No. It assumes an incompressible liquid in a full pipe.
Record the data
Log flow, head, density, measured motor input power, speed, efficiencies used and their source, pipe diameter and length, and the date. A baseline measured before any speed change is the only fair basis for later energy comparison.