Governing Formula
Pump power sizing proceeds through three stages: the ideal hydraulic power (energy delivered to the water), the brake/shaft power (after pump losses), and the motor input power (after motor losses).
P_hyd = ρ × g × Q × H Where:
-
Q= Flow rate (converted internally to m³/s) [m³/day, m³/h, L/min, GPD] -
H= Total dynamic head (static lift + friction + residual pressure) [m] -
η_pump= Pump hydraulic efficiency (as a fraction of 1) [%] -
η_motor= Motor/electrical efficiency [%] -
ρ= Fluid density (water ≈ 1000 kg/m³) [kg/m³]
Derived Equations:
P_hyd = ρ × g × Q × H P_shaft = P_hyd / η_pump P_motor = P_shaft / η_motor How the Calculation Works
The flow rate is converted to m³/s and combined with the total dynamic head to compute the ideal hydraulic power in watts (÷1000 for kW):
P_hyd (kW) = ρ g Q H / 1000
Dividing by the pump and motor efficiencies (as fractions) propagates the machine losses to give the shaft power and finally the electrical draw at the motor terminals. Mechanical horsepower is reported using the 0.7457 kW/hp conversion.
Worked Engineering Example
Design Scenario: Transfer Pump, 50 m³/day @ 30 m, 70%/90%
- Flow in SI:
Q = 50 / 86400 = 5.787 × 10⁻⁴ m³/s - Hydraulic power:
P_hyd = 1000 × 9.80665 × 5.787×10⁻⁴ × 30 / 1000 = 0.170 kW - Shaft power (η = 70%):
P_shaft = 0.170 / 0.70 = 0.243 kW - Motor input (η = 90%):
P_motor = 0.243 / 0.90 = 0.270 kW ≈ 0.36 HP
Engineering Notes & Design Benchmarks
| BEP Pump Efficiency | Typical Range | Motor Class | Efficiency |
|---|---|---|---|
| Small pumps (≤ 5 kW) | 40 – 60% | IE2 (standard) | 85 – 90% |
| Centrifugal (process) | 60 – 80% | IE3 (premium) | 90 – 94% |
| Large pumps (≥ 50 kW) | 75 – 88% | IE4 (super premium) | 94 – 96% |
Standard practice is to select the next standard motor rating above the calculated input power, and to check the duty point against the pump curve's best efficiency point (BEP).
Assumptions & Limitations
- Constant density (water at ~20 °C) — brine, sludge, or elevated-temperature water increases the power requirement proportionally.
- Total dynamic head must already include static lift, all pipe friction, fittings, and any residual discharge pressure expressed in head of water.
- Efficiencies are fixed inputs; real pumps have efficiency curves that vary with flow rate — verify the duty point against the manufacturer's curve.
Frequently Asked Questions
What is "total dynamic head" and how do I get it?
Total dynamic head (TDH) is the full energy the pump must add: static suction+discharge lift, friction losses in suction and discharge piping and fittings, and the residual pressure required at the delivery point — all expressed in meters of water column. Sum them before using this calculator.
Should I size to hydraulic, shaft, or motor power?
Select the motor on the motor input power (electrical draw), adding a service/startup margin (often 10–20%, or more for frequent starts). The shaft power matters for coupling and drive selection, and hydraulic power for the water work actually done.
How does a 50 m³/day, 30 m pump become only 0.27 kW?
Convert the daily flow per second: 50 m³/day = 0.00058 m³/s. Lifting that small flow 30 m delivers roughly 0.17 kW of water power, which after pump and motor losses grows to ~0.27 kW of electrical input — the numbers are simply small because the duty is small.
Engineering Disclaimer
Engineering Note: This calculator provides sizing estimates for preliminary design and education. Full pump selection must check the operating point against the manufacturer's pump curve, NPSH requirements, and energy efficiency regulations.
Technical References
- Hydraulic Institute, Pump Systems Optimization / Pump Standards (ANSI/HI 1.1-1.6 for centrifugal pumps).
- Crane Co., Flow of Fluids Through Valves, Fittings, and Pipe (TP 410).
- Karassik, I. et al., Pump Handbook, McGraw-Hill.