Pump hydraulic power:
formula and worked example

The first number in pump sizing

Hydraulic power is the rate at which a pump adds energy to the liquid: lifting it against gravity and pushing it through the pipe at the required flow. It is the theoretical minimum power for the duty. Real pumps and motors have losses, so the power you must supply is larger.

Engineers calculate it first because it is determined purely by the duty, meaning how much liquid and how much head, and not by the machine chosen. Every later step, from shaft power to motor rating to energy cost, builds on it.

It is also a handy sanity check. If a vendor quotes a motor that is many times larger than the hydraulic power for your duty, there is either a large allowance, a very inefficient pump or a mistake in the stated head or flow, and it is worth asking why.

The formula and units

• Density: kg per cubic metre. Clean water is close to 1,000.
• Gravity: 9.81 metres per second squared.
• Flow: volume per second in cubic metres. A flow of 10 litres per second is 0.01.
• Head: the total height, in metres, the pump must develop, including static lift, friction losses in pipes and fittings, and any required delivery pressure.
• Division by 1,000: converts watts to kilowatts.

Unit consistency is the main risk. Flow quoted in litres per minute or cubic metres per hour must be converted to cubic metres per second before use.

The product of density, gravity and head is simply the pressure rise across the pump in pascals. Multiply by flow in cubic metres per second to get watts. Seen that way, the formula is pressure times flow, which is a good way to remember it.

Worked example: 10 L/s of water at 15 m head

A pump moves water at 0.01 cubic metres per second against a total head of 15 metres.

1. 1Density: 1,000 kg/m³.
2. 2Gravity: 9.81 m/s².
3. 3Flow: 0.01 m³/s.
4. 4Head: 15 m.
5. 5Multiply: 1,000 × 9.81 × 0.01 × 15 = 1,471.5 W.
6. 6Divide by 1,000: 1.4715, shown as 1.472 kW.

A flow of 0.01 m³/s is 36 cubic metres per hour, a sanity check that is useful when a datasheet lists the flow in another unit.

For scale, 1.47 kW is roughly the power of a small domestic water heater element. It is a modest duty, which is why a small pump at around 2 to 3 kW of input would cover it once losses and a margin are included.

From hydraulic power to motor size

To find the power at the pump shaft, divide by pump efficiency. If the pump is 65% efficient, shaft power is 1.4715 ÷ 0.65, about 2.26 kW. The electrical input is higher again, by the motor efficiency, and a margin is added when selecting a standard motor.

• Hydraulic power: the duty, from the formula.
• Shaft power: hydraulic divided by pump efficiency.
• Motor input: shaft divided by motor efficiency.
• Motor rating: the next standard size above the shaft power plus a margin appropriate to the duty.

Efficiencies vary with pump type, size and operating point, and a pump running far from its best-efficiency flow can waste a considerable share of its input. For energy cost, the electrical input is the figure that matters, since that is what the meter records.

How a heavier liquid changes the duty

A 10% denser liquid, such as a brine or a light slurry, needs 10% more power: 1.472 rises to 1.619 kW. Flow and head behave the same way in the formula, but head is the one most often underestimated. A system with long pipes or throttled valves can need far more head than the simple lift suggests.

Mistakes that undersize pumps

• Using only the static lift as the head, and leaving out pipe friction and fittings.
• Entering flow in litres per second or cubic metres per hour without converting.
• Treating hydraulic power as the motor rating.
• Using the density of water for a slurry or a hot liquid.
• Forgetting that the pump's efficiency varies with the operating point.

Head is the most common reason for a pump that disappoints. Friction rises roughly with the square of flow, so a pipe that looked fine at one flow can add a lot of head at a higher one.

Checklist before you pick a pump

1. 1Define the flow you need in m³/s.
2. 2Calculate total head including friction.
3. 3Compute hydraulic power.
4. 4Divide by pump and motor efficiency.
5. 5Add a margin and choose the next standard size.

If the duty varies through the day, repeat the calculation at the highest and lowest flow. A pump chosen only for the peak may run inefficiently for much of the time, and a variable speed drive can recover some of that.