Motor Sizing Calculator
Motor Selection Details
Torque Analysis
Derating Factors
Choosing a motor means matching its rating to both the steady load and the torque needed to accelerate the driven inertia, then derating for the operating environment. This calculator works from mechanical inputs—load power, speed, acceleration time, and inertia—to find the required torque and power, applies an application service factor and duty-cycle and environmental derating, and recommends a standard IEC power rating, NEMA frame, and VFD current.
Formula
Tload = (P × 9550) / n; Tacc = J × (2π·n / 60) / t; Ttotal = (Tload + Tacc) × Kapp
- P
- Mechanical load power (kW)
- n
- Required shaft speed (RPM)
- J
- Load moment of inertia (kg·m²)
- t
- Acceleration time to reach speed (s)
- Kapp
- Application service factor (1.15 fan … 1.75 hoist)
How it works
- Enter the application (conveyor, pump, fan, hoist, or compressor), the load power in kW, required speed in RPM, acceleration time, load inertia, duty cycle (S1/S3/S6), ambient temperature, and altitude.
- Load torque is found from power and speed, acceleration torque from inertia and the speed ramp, and the two are summed and multiplied by an application factor (1.15 for a fan up to 1.75 for a hoist) to get total torque.
- Required power is computed back from total torque, then divided by the combined temperature, altitude, and duty-cycle derating before being rounded up to the next standard IEC motor size with its frame, service factor, and VFD current rating.
Worked example
A pump needing 15 kW at 1450 RPM, accelerating in 5 s against 2.5 kg·m² of inertia, S1 continuous duty, 30 °C ambient, 500 m altitude.
- Load torque: (15 × 9550) / 1450 ≈ 98.79 N·m.
- Acceleration torque: 2.5 × (2π·1450 / 60) / 5 ≈ 75.92 N·m.
- Total torque with pump factor 1.25: (98.79 + 75.92) × 1.25 ≈ 218.39 N·m → required power ≈ 33.16 kW (no derating at 30 °C / 500 m / S1).
- Round up to the next standard IEC size: 37 kW.
Recommended 37 kW motor, NEMA frame 364T, service factor 1.25, VFD current ≈ 59.4 A at 460 V; rated speed ≈ 1455 RPM.
Frequently asked questions
- Why include acceleration torque on top of the running load?
- During start-up the motor must not only drive the load but also spin up the rotating inertia from rest to speed within the acceleration time. That extra torque can be significant for high-inertia loads and is added to the steady load torque when sizing the motor.
- What does the application service factor account for?
- Different loads impose different shock, starting, and overload conditions. A smooth fan barely stresses the motor (factor 1.15), while a hoist must handle high starting torque and is safety-critical (factor 1.75). The factor pads the required torque so the motor is not run at its absolute limit.
- Why does the motor get derated for temperature and altitude?
- Motors are rated for cooling at up to 40 °C ambient and 1000 m altitude. Hotter air and thinner air both reduce heat removal, so the usable output falls (about 3% per 5 °C and 1% per 100 m here), meaning a larger nameplate motor is needed for the same shaft load.
- How is the VFD current rating determined?
- The VFD must supply the motor's full-load current, computed from the rated power using I = P / (√3 × V × PF × η) with typical assumptions of 460 V, 0.85 power factor, and 0.92 efficiency. Drives are selected at or above this current rating.