Pump Work Calculator

Inputs

Formula used

The pump adds energy to the fluid as a pressure rise or an equivalent head. The core relationships are:

• Pressure–head: ΔP = ρ g H
• Hydraulic power: Pₕ = ΔP Q = ρ g Q H
• Shaft power: Pₛ = Pₕ / η
• Work (energy): E = P × t
• Specific work: w = g H (J/kg)

Units check: ΔP is Pa, Q is m³/s, so ΔP·Q is W. Multiply by time in seconds to get joules.

How to use this calculator

1. Enter fluid density and keep gravity at the default value.
2. Select whether you know head or pressure rise.
3. Enter flow rate directly, or compute it from volume and time.
4. Provide pump efficiency as a percent or decimal.
5. Enter operating time to get total energy and work.
6. Press Calculate to show results above the form.

Example data table

Pump work fundamentals for practical sizing

1) What “pump work” means

Pump work is the energy the pump must add to a fluid to overcome elevation changes, pressure requirements, and losses. In steady flow, the useful hydraulic energy is often expressed through head (meters or feet) or pressure rise (kPa, bar, psi). This calculator reports work, energy, and power so you can connect system requirements to motor selection.

2) Core relationship used in industry

The hydraulic power delivered to the fluid is P_hyd = ρ g Q H, where ρ is density, g is gravitational acceleration, Q is volumetric flow rate, and H is total head. For water at room temperature, ρ is about 1000 kg/m³ and g is 9.80665 m/s². If efficiency is η, the shaft power is P_shaft = P_hyd/η.

3) Head, pressure, and why units matter

Head converts to pressure by Δp = ρ g H. For water, 10 m of head is roughly 98 kPa (about 0.98 bar). Many specifications are given in mixed units—gpm, psi, feet—so the calculator includes conversions to keep calculations consistent and to reduce rounding mistakes.

4) Efficiency ranges you should expect

Efficiency depends on pump type and operating point. Small centrifugal pumps may run around 50–70%, while larger well-selected centrifugal units can reach 75–90%. Positive displacement pumps can be high, but leakage and viscosity effects matter. Using a realistic efficiency prevents underestimating motor power and overheating risk.

5) From power to total work over time

Energy is power multiplied by time: E = P × t. If the pump runs intermittently, use the actual duty time. For example, 2 kW over 3 hours consumes 6 kWh. The calculator also reports Joules for engineering reports and kWh for utility-style energy tracking.

6) Interpreting results for pump selection

Compare calculated shaft power to the motor nameplate. A margin is common because real systems include uncertainty in losses, density changes, fouling, and control valves. If your calculated shaft power is close to the motor rating, consider a higher frame size or verify the pump curve at your Q–H operating point.

7) Common sources of error

Typical mistakes include mixing gauge and absolute pressures, forgetting elevation head, using the wrong fluid density, and entering efficiency as “80” when the input expects “0.80” (or vice versa). This calculator accepts both percent and decimal formats and shows intermediate values so you can validate inputs quickly.

FAQs

1) What is the difference between head and pressure?

Head is energy per unit weight, expressed as a fluid column height. Pressure is force per unit area. They relate by Δp = ρ g H, so the same head gives different pressure for different fluids.

2) Which efficiency should I use?

Use pump efficiency if you want shaft power at the pump. If you need electrical input power, include motor and drive efficiencies too. For quick estimates, many designs start with 0.6–0.8 for a centrifugal pump.

3) Why does density affect pump work?

Hydraulic power is proportional to ρ. A heavier fluid needs more power for the same flow and head. If you pump glycol, seawater, or oils, enter a density close to your operating temperature.

4) Can I compute work from pressure rise instead of head?

Yes. Work to raise pressure is W = Δp × V (pressure times volume). The calculator can convert head to pressure internally, so you can enter whichever value you have and still get consistent energy and power outputs.

5) What time value should I enter for energy?

Enter the pump’s actual run time, not the calendar time. If a pump cycles 15 minutes each hour for 8 hours, the duty time is 2 hours. That produces a more accurate kWh estimate.

6) Why is my calculated power lower than the motor rating?

Motors are selected with margin for startup, uncertainties, wear, and future expansion. Also, pump curves vary with speed and impeller trim. Verify the operating point and consider system losses that may not be included yet.

7) How do I include pipe friction losses?

Add friction losses to total head. Calculate static head plus dynamic losses from pipes, fittings, and valves, then enter that total. Using total head helps align the result with real-world pump curve selection.