Motor Power Calculator — Formula, Example & Step-by-Step Guide
Motor power calculation determines the electrical or mechanical power required to drive a load at a specified torque and speed. The fundamental relationship P = T × ω links power (W), torque (N·m), and angular velocity (rad/s). Including the drive train efficiency η accounts for losses in gearboxes, belts, couplings, and the motor itself. This calculation is the starting point for every motor selection process — from small servo motors in robotic arms to 10+ MW drives in mining and cement mills. Selecting the correct motor size prevents overheating (undersized) and energy waste (oversized), directly impacting operational costs and equipment lifetime.
Formula
Quick Calculation Result
Interactive Calculator:
How to Calculate Motor Power Calculator (Step-by-Step)
- 1
Determine the load torque T at the driven equipment (N·m). Include friction, gravity, and acceleration torques.
- 2
Determine the required speed n (rpm) at the driven equipment.
- 3
If a gearbox is used: T_motor = T_load / (i × η_gear). n_motor = n_load × i.
- 4
Convert speed to angular velocity: ω = 2π × n_motor / 60.
- 5
Calculate power: P = T_motor × ω.
- 6
Apply service factor (1.15–1.50) and select the next standard motor size (0.37, 0.55, 0.75, 1.1, 1.5, 2.2, 3, 4, 5.5, 7.5, 11, 15, 18.5, 22, 30 kW).
Why This Matters
Motor sizing affects energy costs, reliability, and system performance. An oversized motor operates at low load factor with poor efficiency and power factor, wasting electricity over its 20+ year lifetime. An undersized motor overheats, trips on thermal protection, and fails prematurely. In conveyor design, the motor must overcome belt friction, material lift, and acceleration loads. In pump applications, power scales with the cube of flow rate — a 10% increase in flow requires 33% more power. Variable frequency drives (VFDs) allow motors to operate at optimal efficiency across varying loads but add 2–5% losses. IE3/IE4 premium efficiency motors cost more upfront but save 2–5% energy over standard motors, typically paying back within 1–2 years in continuous duty applications.
Worked Example
Problem: A conveyor requires 450 N·m torque at 60 rpm. Gearbox ratio 25:1, efficiency 95%. Motor efficiency 92%. Solution: T_motor = 450 / (25 × 0.95) = 18.95 N·m. n_motor = 60 × 25 = 1500 rpm. ω = 2π × 1500 / 60 = 157.08 rad/s. P_shaft = 18.95 × 157.08 = 2,977 W. P_electrical = 2977 / 0.92 = 3,236 W → Select 4 kW motor (next standard size with service factor).
Standard Motor Sizes (IEC)
| kW | Frame |
|---|---|
| 0.75 | 80 |
| 1.5 | 90S |
| 4.0 | 112M |
| 7.5 | 132S |
| 15 | 160M |
✓ Design Checklist
- • Include gearbox and coupling losses
- • Apply service factor for load type
- • Verify motor frame thermal capacity
⚠ Common Pitfalls
- • Forgetting acceleration torque for cyclic loads
- • Using shaft power instead of electrical input power
Frequently Asked Questions
How do you calculate motor power?+
P = T × ω / η, where T is torque in N·m, ω is angular velocity in rad/s (= 2πn/60), and η is overall efficiency.
What is a motor service factor?+
A service factor (1.0–1.5) is a multiplier applied to the calculated power to account for overload conditions, ambient temperature, and altitude.
How do you convert HP to kW?+
1 HP = 0.7457 kW. Multiply horsepower by 0.7457 to get kilowatts.