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Parents Pricing Home A-Level AQA Physics Thermodynamics & Engines Figure 4 shows the results of a test on an internal combustion engine which uses purified biogas
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 3
View full question 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
View marking scheme 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
Determine, for the speed at which the engine develops its maximum brake power:
- the overall efficiency Only available for registered users.
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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:
e x t O v e r a l l E f f i c i e n c y = 40 50 × 100 = 80 % ext{Overall Efficiency} = \frac{40}{50} \times 100 = 80\% e x t O v er a llE ff i c i e n cy = 50 40 × 100 = 80%
- the thermal efficiency Only available for registered users.
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Thermal efficiency can be calculated as follows:
e x t T h e r m a l E f f i c i e n c y = e x t O v e r a l l E f f i c i e n c y Calorific Value of Fuel × 100 ext{Thermal Efficiency} = \frac{ ext{Overall Efficiency}}{\text{Calorific Value of Fuel}} \times 100 e x t T h er ma lE ff i c i e n cy = Calorific Value of Fuel e x t O v er a llE ff i c i e n cy × 100
Using the calorific value of biogas (32.3 x 10^6 J m³) and assuming ideal conditions, the calculation leads to:
e x t T h e r m a l E f f i c i e n c y = 80 32.3 ≈ 2.47 % ext{Thermal Efficiency} = \frac{80}{32.3} \approx 2.47\% e x t T h er ma lE ff i c i e n cy = 32.3 80 ≈ 2.47%
- the mechanical efficiency Only available for registered users.
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Mechanical efficiency is determined by assessing the ratio of brake power to indicated power:
e x t M e c h a n i c a l E f f i c i e n c y = e x t B r a k e P o w e r e x t I n d i c a t e d P o w e r × 100 ext{Mechanical Efficiency} = \frac{ ext{Brake Power}}{ ext{Indicated Power}} \times 100 e x t M ec hani c a lE ff i c i e n cy = e x t I n d i c a t e d P o w er e x t B r ak e P o w er × 100
Using the same indicated power:
e x t M e c h a n i c a l E f f i c i e n c y = 40 k W 50 k W × 100 = 80 % ext{Mechanical Efficiency} = \frac{40 kW}{50 kW} \times 100 = 80\% e x t M ec hani c a lE ff i c i e n cy = 50 kW 40 kW × 100 = 80%
Go on to explain how knowledge of these efficiencies can be useful to an engineer. Only available for registered users.
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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.
Explain why it is not advisable to run this engine at speeds above 7000 rev min⁻¹. Only available for registered users.
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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.
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