Physics 3.11 - 3.12 - Efficiency

Calculating efficiency

Efficiency measures the fraction of the supplied energy that is transferred usefully by a device. First decide the device's intended output; only then can the useful energy be identified.

Efficiency has no unit because it is one energy divided by another energy. It can be written as a decimal between $0$ and $1$, or as a percentage between $0\%$ and $100\%$:

An efficiency above $1$ or $100\%$ would mean the claimed useful output exceeds the total input, so it is not physically possible for this definition.

Worked example 1: finding efficiency

A motor receives $1500\,\mathrm{J}$ of energy and transfers $960\,\mathrm{J}$ usefully to the gravitational store of a lifted load. Calculate its efficiency as a decimal and a percentage, then account for the remaining energy.

1. The quantities already use the same unit:
   • Total energy supplied is $1500\,\mathrm{J}$
   • Useful energy transferred is $960\,\mathrm{J}$
2. Select the equation:

$\text{efficiency}=\frac{E_{\mathrm{useful}}}{E_{\mathrm{input}}}$
3. Substitute and calculate:

$\text{efficiency}=\frac{960}{1500}=0.64$
4. Convert to a percentage:

$\text{percentage efficiency}=0.64\times100\%=64\%$
5. Account for the rest:

$E_{\mathrm{dissipated}}=1500\,\mathrm{J}-960\,\mathrm{J}=540\,\mathrm{J}$

Sense-check: $960\,\mathrm{J}$ is a little under two-thirds of $1500\,\mathrm{J}$, so $64\%$ is reasonable. The useful $960\,\mathrm{J}$ plus the dissipated $540\,\mathrm{J}$ equals the supplied $1500\,\mathrm{J}$, so energy is conserved.

Worked example 2: finding useful energy

A lifting device is $72\%$ efficient and receives $2.50\,\mathrm{kJ}$ of energy. Calculate the useful energy transferred.

1. Convert the percentage to a decimal: $72\% = 0.72$.
2. Rearrange the efficiency equation:

$E_{\mathrm{useful}}=\text{efficiency}\times E_{\mathrm{input}}$
3. Substitute:

$E_{\mathrm{useful}}=0.72\times2.50\,\mathrm{kJ}=1.80\,\mathrm{kJ}$
4. If joules are required: $1.80\,\mathrm{kJ} = 1800\,\mathrm{J}$.