How To Calculate Motor Efficiency

The goal of a motor is to get something to move. Often, that something is an axle, the rotational motion of which can be converted into translational motion, as in a car, or otherwise be put to use to do mechanical ​work​ (which has units of energy).

The ​power​ (energy per unit time) for the motor usually comes from electricity, the ultimate source of which could be a coal-powered plant, a windmill or a bank of solar cells.

Applied physics can be used to determine ​motor efficiency,​ which is a measure of the fraction of energy put into a mechanical system that results in useful work . The more efficient the motor, the less energy wasted as heat, friction and so on, and the more ultimate cost savings to a business owner in a manufacturing scenario.

​Energy​ is physics takes many forms: kinetic, potential, heat, mechanical, electrical and more. Work is defined as the amount of energy expended in moving a mass ​m​ through a distance ​x​ by applying a force ​F​. Work in the SI (metric) system has units of Newton-meters, or Joules (J).

​Power​ is energy per unit time. You might expend a given number of joules crossing a parking lot, but if you sprint and cover the distance in 20 seconds rather than amble and take two minutes, your power output is correspondingly higher in the sprinting example. The SI unit is Watts (W), or J/s.

Efficiency is simply output (useful) power divided by input power, with the difference being losses due to imperfections in design and other inevitabilities. Efficiency in this context is a decimal varying from 0 to 1.0, or sometimes a percentage.

Usually, the more powerful the motor, the more efficient it is expected to be. An efficiency of 0.80 is good for a 1 to 4 hp motor, but it is normal to aim for above 0.90 for 5-hp and more powerful motors.

Efficiency is often denoted by the Greek letter eta (​η​), and is calculated using the following formula:

\(η = \frac{0.7457 × \text{hp} × \text{load}}{P_i}\)

Here, ​hp​ = motor horsepower, ​load​ = Output power as a percentage of rated power, and ​**​P​_i**​ = input power in kW.

•The constant factor 0.7457 is used to convert horsepower to kilowatts. This is because 1 hp = 745.7 W, or 0.7457 kW. 

​Example​: Given a 75-hp motor, a measured load of 0.50 and input power of 70 kW, what is the motor efficiency?

\(\begin{aligned}
η &= \frac{0.7457 \;\text{kW/hp} × 75 \;\text{hp} × 0.50}{70 \;\text{kW}} \
&= 0.40
\end{aligned}\)

Sometimes you are given the efficiency in a problem and asked to solve for a different variable, such as the input power. In this case you rearrange the equation as needed.

​Example:​ Given a motor efficiency of 0.85, a load of 0.70 and a 150-hp motor, what is the input power?

\(\begin{aligned}
η &= \frac{0.7457 × \text{hp} × \text{load}}{P_i} \
\text{Therefore} \;P_i &= \frac{0.7457 × \text{hp} × \text{load}}{η}\)
\(&= \frac{0.7457 \;\text{kW/hp}×150 \;\text{hp}×0.70}{0.85}\)
\(&= 92.1 \;\text{kW}
\end{aligned}\)

Sometimes you are given the parameters of a motor, such as its torque (force applied about an axis of rotation) and its revolutions per minute (rpm). You can use the relationship ​**​η​ = ​P​_o/​P​_i**​, where ​P​_o is output power, to determine efficiency in such cases, because ​P​_i is given by ​I​ × ​V​, or current times voltage, whereas ​P​_o is equal to torque ​**​τ​ ​times rotational velocity ​​ω​**​. Rotational velocity in radians per second is given in turn by ​ω​ = (2π)(rpm)/60.

Thus​:​

\(\begin{aligned}
η &= P_o/P_i \
&=\frac{τ×2π×\text{rpm}/60}{I×V} \
&= \frac{(π /30)(τ × \text{rpm})}{I×V} \
\end{aligned}\)