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Efficiency Formula (DP IB Physics)

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Ashika

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DP Physics: HLDP Physics: SL

Efficiency Formula

  • The efficiency of a system is a measure of how successfully energy is transferred in a system

  • Efficiency is defined as:

    The ratio of the useful power or energy transfer output from a system to its total power or energy transfer 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

  • Determining which type of energy is useful or wasted depends on the system

    • When energy is transferred from the thermal store of a kettle's heating element to the thermal store of the water, this is useful energy

    • When energy is transferred to the plastic or metal casing of the kettle and to the surrounding air, this energy is wasted

  • Efficiency is represented as a fraction, and can be calculated using the equation:

eta space equals fraction numerator space E space open parentheses o u t p u t close parentheses over denominator E space open parentheses i n p u t close parentheses end fraction space equals fraction numerator space P space open parentheses o u t p u t close parentheses over denominator P space open parentheses i n p u t close parentheses end fraction

  • Where:

    • η = efficiency (the greek letter "eta") 

    • E = energy (J)

    • P = power (W)

  • To turn this equation into a percentage, just × 100 %

  • It can also be written in words as:

eta space equals space fraction numerator useful space work space out space over denominator total space work space in end fraction space equals space fraction numerator useful space power space out space over denominator total space power space in end fraction

Worked Example

An electric motor has an efficiency of 35 %. It lifts a 7.2 kg load through a height of 5 m in 3 s.

Calculate the power of the motor.

Answer:

Step 1: Write down the efficiency equation (as a percentage)

eta space equals space fraction numerator u s e f u l space p o w e r space o u t space over denominator u s e f u l space p o w e r space i n end fraction space cross times space 100 percent sign

Step 2: Rearrange equation for the useful power in

u s e f u l space p o w e r space i n space space equals space fraction numerator u s e r f u l space p o w e r space o u t space cross times space 100 percent sign space over denominator eta end fraction

Step 3: Calculate the power output

  • The power output is equal to energy ÷ time

  • The electric motor transferred electric energy into gravitational potential energy to lift the load

Gravitational potential energy = mgh = 7.2 × 9.81 × 5 = 353.16 J

Power = fraction numerator 353.16 over denominator 3 end fraction = 117.72 W

Step 4: Substitute values into power input equation 

u s e f u l space p o w e r space i n space equals space fraction numerator 117.72 space cross times space 100 space over denominator 35 end fraction space equals space 336 space W

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 and the kinetic energy of this water is used to lift a small quantity of water by 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

  • The pump transfers energy from the kinetic store to the gravitational potential store of the water

  • Since its efficiency is 20%, the kinetic energy can be multiplied by 0.2 since only 20% of the kinetic energy will be transferred (not 20% of the gravitational potential energy)

Examiner Tips and Tricks

Efficiency can be in a ratio (between 0 and 1) or percentage format (between 0% and 100%). 

If the question asks for 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: if the ratio = 0.25, percentage = 0.25 × 100 = 25 %.

Remember that efficiency has no units. It is a ratio with both the numerator and denominator with the same units.

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    Ashika

    Author: Ashika

    Expertise: Physics Project Lead

    Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.