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Electrical Engineering2D

Motor Power Calculator

Calculate electric motor shaft power from electrical power input and motor efficiency, or convert between kW and HP.

Inputs
Enter your values below

Enter as decimal (0.9 = 90%)

Result
Shaft Power Output
—
Enter values and click Calculate
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Formula
power_input × efficiency ÷ 1000
Result unit: kW

Shaft power output P_out = P_in × η. P_in converted from kW to W by engine; /1000 returns result in kW. η is efficiency as decimal (0-1).

Introduction

Calculate the actual mechanical shaft power delivered by an electric motor given its electrical input power and efficiency. Essential for motor sizing and pump/fan drive calculations.

How This Calculator Works

Shaft power output P_out = P_in × η. P_in converted from kW to W by engine; /1000 returns result in kW. η is efficiency as decimal (0-1).

Step-by-step process:

  1. Enter your input values in the calculator above
  2. The engine converts all inputs to SI base units (meters, kg, Pa)
  3. The formula is evaluated: power_input * efficiency / 1000
  4. Result is formatted with the appropriate unit and precision

Calculation Example

A 22 kW motor operating at 90% efficiency delivers 19.8 kW of shaft power.

Inputs: power_input=22, efficiency=0.9
Result: 19.8 kW

Motor Power Formula Explained

The mechanical shaft power delivered by an electric motor is the electrical input power times efficiency:

P_shaft = P_in x eta

Where P_shaft = shaft power output (kW), P_in = electrical input power (kW), eta = motor efficiency (decimal, 0-1).

Example: a 22 kW motor at 90% efficiency delivers 22 x 0.9 = 19.8 kW of shaft power. The remaining 2.2 kW is lost as heat. For kW to HP: 1 HP = 0.7457 kW, so 19.8 kW = 19.8 / 0.7457 = 26.6 HP.

Motor Efficiency Classes

| Efficiency Class | Typical Efficiency (large motors) |

|---|---|

| IE1 (Standard) | 82 - 90% |

| IE2 (High) | 86 - 93% |

| IE3 (Premium) | 88 - 96% |

| IE4 (Super Premium) | 90 - 97% |

Efficiency rises with motor size — a 200 kW IE3 motor can exceed 96% while a 1 kW motor is typically below 85%. IE3 is now the regulatory minimum in most regions. The efficiency difference between IE2 and IE3 on a continuously running motor often pays back the price premium in under a year.

Worked Example: kW to HP and Energy Cost

A pump drive needs 19.8 kW of shaft power. Select a 22 kW (30 HP) motor:

- In HP: 22 kW / 0.7457 = 29.5 HP → standard 30 HP frame

- Input power at 90% efficiency: 22 / 0.9 = 24.4 kW

- Annual energy at 8,000 h/yr: 24.4 x 8,000 = 195,200 kWh

- At $0.10/kWh: $19,520/yr

Upgrading from IE2 (88%) to IE3 (93%) cuts input power to 23.7 kW, saving 5,600 kWh and $560/yr on this duty.

Engineering Applications

  • •Motor sizing
  • •Pump drive selection
  • •Energy consumption estimation
  • •Electrical design

Frequently Asked Questions

What is typical motor efficiency?

IE3 premium efficiency motors typically achieve 88-96% depending on size. Smaller motors are less efficient than larger ones.

How do I convert kW to HP?

1 HP = 0.7457 kW. Divide kW by 0.7457 to get HP. Example: 22 kW = 22 / 0.7457 = 29.5 HP.

What is the difference between input power and shaft power?

Input power is the electrical power drawn from the supply. Shaft power is the mechanical power delivered to the load, equal to input power times efficiency. The difference is lost as heat in the motor windings and bearings.

How do I select the right motor power?

Calculate the driven load's shaft power requirement (e.g., pump power, fan power), then select the next standard motor size above it. Add margin for starting torque, service factor, and future load growth. For pumps, use the pump shaft power at the duty point.

What are IE3 and IE4 efficiency classes?

IE3 (Premium) and IE4 (Super Premium) are international efficiency classes defined by IEC 60034-30-1. IE3 is the regulatory minimum in most regions; IE4 adds 1-3% efficiency. Higher classes reduce energy cost and are required for new large motors in many jurisdictions.

How do I estimate annual motor energy cost?

Annual cost = input power (kW) x operating hours x electricity rate. A 22 kW motor at 90% efficiency running 8,000 h/yr at $0.10/kWh costs 24.4 x 8,000 x 0.10 = $19,520/yr. Improving efficiency by 5% saves roughly $1,000/yr.

Related Calculators

Pump Power CalculatorPump Efficiency CalculatorPump Energy Cost CalculatorPump Sizing CalculatorPump Head CalculatorPump Affinity Law CalculatorPump Selection CalculatorPump Flow Calculator

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Engineering Disclaimer: Calculations are for reference and educational purposes only. Always verify results independently for engineering design. See full disclaimer.
Reviewed by: Industrial Engineering Team
References: ASME B31.3, ASTM A36, Perry's Chemical Engineers' Handbook