Efficiency Physics Example 4

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Example 4

hard
A car engine has an efficiency of 25%25\% and a petrol pump has an efficiency of 90%90\%. The pump delivers 40 litres40 \text{ litres} of fuel containing 1.4×109 J1.4 \times 10^9 \text{ J} of chemical energy. How much useful kinetic energy does the car ultimately produce from this fuel?

Solution

  1. 1
    Energy delivered by pump: Epump=0.90×1.4×109=1.26×109 JE_{\text{pump}} = 0.90 \times 1.4 \times 10^9 = 1.26 \times 10^9 \text{ J}.
  2. 2
    Useful energy from engine: Euseful=0.25×1.26×109=3.15×108 JE_{\text{useful}} = 0.25 \times 1.26 \times 10^9 = 3.15 \times 10^8 \text{ J}.
  3. 3
    Overall efficiency: 0.90×0.25=0.225=22.5%0.90 \times 0.25 = 0.225 = 22.5\%.

Answer

Euseful=3.15×108 J(overall efficiency 22.5%)E_{\text{useful}} = 3.15 \times 10^8 \text{ J} \quad (\text{overall efficiency } 22.5\%)
When energy passes through multiple stages, the overall efficiency is the product of each stage's efficiency. Each stage loses some energy, so the overall efficiency is always lower than any individual stage.

About Efficiency

The ratio of useful output energy (or power) to total input energy, expressed as a percentage — always less than 100% due to energy losses.

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