Photo AI

Figure 4 shows the results of a test on an internal combustion engine which uses purified biogas - AQA - A-Level Physics - Question 3 - 2021 - Paper 6

Question icon

Question 3

Figure-4-shows-the-results-of-a-test-on-an-internal-combustion-engine-which-uses-purified-biogas-AQA-A-Level Physics-Question 3-2021-Paper 6.png

Figure 4 shows the results of a test on an internal combustion engine which uses purified biogas. Figure 4 shows how the indicated power, brake power (or output) p... show full transcript

Worked Solution & Example Answer:Figure 4 shows the results of a test on an internal combustion engine which uses purified biogas - AQA - A-Level Physics - Question 3 - 2021 - Paper 6

Step 1

Determine, for the speed at which the engine develops its maximum brake power: - the overall efficiency

96%

114 rated

Answer

To find the speed at maximum brake power, we refer to Figure 4. The maximum brake power occurs around 4000 rev min⁻¹ at approximately 40 kW.

Overall efficiency can be calculated using the formula:

ext{Overall Efficiency} = rac{ ext{Brake Power}}{ ext{Indicated Power}} \times 100

From the figure, if the indicated power at this speed is about 50 kW:

extOverallEfficiency=4050×100=80% ext{Overall Efficiency} = \frac{40}{50} \times 100 = 80\%

Step 2

- the thermal efficiency

99%

104 rated

Answer

Thermal efficiency can be calculated as follows:

extThermalEfficiency=extOverallEfficiencyCalorific Value of Fuel×100 ext{Thermal Efficiency} = \frac{ ext{Overall Efficiency}}{\text{Calorific Value of Fuel}} \times 100

Using the calorific value of biogas (32.3 x 10^6 J m³) and assuming ideal conditions, the calculation leads to:

extThermalEfficiency=8032.32.47% ext{Thermal Efficiency} = \frac{80}{32.3} \approx 2.47\%

Step 3

- the mechanical efficiency

96%

101 rated

Answer

Mechanical efficiency is determined by assessing the ratio of brake power to indicated power:

extMechanicalEfficiency=extBrakePowerextIndicatedPower×100 ext{Mechanical Efficiency} = \frac{ ext{Brake Power}}{ ext{Indicated Power}} \times 100

Using the same indicated power:

extMechanicalEfficiency=40kW50kW×100=80% ext{Mechanical Efficiency} = \frac{40 kW}{50 kW} \times 100 = 80\%

Step 4

Go on to explain how knowledge of these efficiencies can be useful to an engineer.

98%

120 rated

Answer

Understanding overall, thermal, and mechanical efficiencies is crucial for engineers because:

  • It allows them to optimize engine performance and minimize fuel consumption.
  • Engineers can make design adjustments based on efficiency data, enhancing the sustainability of the engine systems.
  • Knowledge of these efficiencies can guide maintenance practices to improve the longevity and reliability of the engine.

Step 5

Explain why it is not advisable to run this engine at speeds above 7000 rev min⁻¹.

97%

117 rated

Answer

Running the engine at speeds above 7000 rev min⁻¹ may lead to:

  • A significant reduction in brake power as shown in Figure 4, where power begins to decline after the maximum speed.
  • Increased fuel consumption without proportional power output, leading to inefficiencies.
  • Potential wear and failure of engine components due to excessive operational stress, risking safety and reliability.

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;