Efficiency (Cambridge International AS Physics)

Efficiency of a System

Spec point — 5.1.3

• The efficiency of a system is the ratio of the useful energy output from the system to the total energy input

  • If a system has high efficiency, this means most of the energy transferred is useful
  • If a system has low efficiency, this means most of the energy transferred is wasted

• Multiplying this ratio by 100 gives the efficiency as a percentage

• The efficiency is calculated using the equation:

$\text{efficiency} = \frac{ \text{useful energy} output}{total \text{energy} input} \times 100 \%$

• Efficiency can also be written in terms of power (the energy transferred per second):

$\mathrm{efficiency}=\frac{\mathrm{useful} \mathrm{power} \mathrm{output}}{\mathrm{total} \mathrm{power} \mathrm{input}}$

Worked Example An electric motor has an input power, P_in, useful output power, P_out, power lost P_lost, and an efficiency η.

What is the output power of the motor?

A.$\frac{η}{P_{in}}$

B.$\frac{-ηP_{lost}}{η-1}$

C.$ηP_{lost}$

D.$- \\eta P_{lost} ( \\eta - 1 )$

Answer: B

Step 1: State the efficiency equation

$\mathrm{efficiency}=\frac{\mathrm{useful} \mathrm{power} \mathrm{output}}{\mathrm{total} \mathrm{power} \mathrm{input}}$

Step 2: Substitute the terms given in the question

$η=\frac{P_{out}}{P_{in}}=\frac{P_{out}}{P_{out}+P_{lost}}$

Step 3: Manipulate the equation

• Multiply by P_out + P_lost on both sides

$\\eta ( p_{out} + p_{lost} ) = P_{out}$

• Expand the brackets

$ηP_{out}+ηP_{lost}=P_{out}$

• Minus P_out from both sides

  • It is helpful to first minus ηP_lost

$ηP_{out}=P_{out}-ηP_{lost}$

$ηP_{out}-P_{out}=-ηP_{lost}$

• Factor out P_out

$P_{out} ( \\eta - 1 ) = - \\eta P_{lost}$

• Divide by η − 1

$P_{out} = \frac{- \\eta P_{lost} }{( \\eta - 1 )}$

Exam Hint Efficiency can be in a ratio or percentage format. If the question asks for an efficiency as a ratio, give your answer as a fraction or decimal. If the answer is required as a percentage, remember to multiply the ratio by 100 to convert it, e.g. Ratio = 0.25, Percentage = 0.25 × 100 = 25 %

Solving Problems Involving Efficiency

• Efficiency calculations are often part of a larger multi-step problem
• Since efficiency deals with energy and power, questions will often involve energy or power calculations

Worked Example The diagram shows a pump called a hydraulic ram.

In one such pump, the long approach pipe holds 700 kg of water. A valve shuts when the speed of this water reaches 3.5 m s^-1. The kinetic energy of this water is used to lift a small quantity of water by a height of 12m. The efficiency of the pump is 20%.

Which mass of water could be lifted 12 m?

A. 6.2 kg

B. 4.6 kg

C. 7.3 kg

D. 0.24 kg

Answer: C

Step 1: List the known quantities

• Mass of water in approach pipe = 700 kg
• Speed of water in approach pipe, v = 3.5 m s^-1
• Height of lifted water, h = 12 m

Step 2: Consider the energy transfer taking place

• Energy is transferred from the kinetic store of the water to its gravitational potential store

Step 3: Consider the efficiency of the energy transfer

• The transfer is 20% efficient
• Therefore, 20% of the kinetic input energy is output as gravitational potential energy

$0.2E_{k}=E_{p}$

$0.2\times\frac{1}{2}mv^{2}=mgh$

Step 4: Calculate the mass of water lifted water

$0.2 \times 0.5 \times 700 \times ( 3.5 )^{2} = m \times 9.81 \times 12$

$857.5=m\times117.72$

$m=\frac{857.5}{117.72}$

$m=7.3\mathrm{kg}$

Exam Hint Equations for kinetic and potential energies are important for these types of questions. Also, familiarise yourself with the different equations for power depending on what quantities are given.