IC Engines - ISRO Mechanical & Aerospace Subject-Wise PYQ Practice
Practice topic-wise IC Engines previous year questions for ISRO Mechanical & Aerospace with instant step-by-step solutions and formula breakdowns.
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4 questions.
Question 1
Year: 2015 | Topic: IC Engines
A large diesel engine runs on a stroke cycle at 2000 rpm. The engine has a displacement of 25 litre and a brake mean effective pressure of 0.6 $MN/m²$. It consumes 0.018 kg/s of fuel (calorific value = 42000 kJ/kg). Determine the brake power.
- A. 250 KW
- B. 225 KW
- C. 275 KW
- D. none of the above
Answer: A. 250 KW
Explanation: The formula for brake power is $BP = P_{mep} \times V_s \times n$, where $n$ is the number of power cycles per second. For a 4-stroke cycle, $n = N / (2 \times 60) = 2000 / 120$. The brake mean effective pressure is $P_{mep} = 0.6 \times 10^6$ N/m$^2$ and stroke volume $V_s = 25 \times 10^{-3}$ m$^3$. Multiplying these gives $BP = 0.6 \times 10^6 \times 25 \times 10^{-3} \times \frac{2000}{120} = 250,000$ W, which equals 250 kW.
Question 2
Year: 2018 | Topic: IC Engines
The compression ratio for diesel engines is close to
- A. 3 to 6
- B. 5 to 8
- C. 15 to 20
- D. 20 to 30
Answer: C. 15 to 20
Explanation: For diesel engine compression ratio rc = 15-22. For Otto engine rc = 7-12. For Dual cycle engine rc = 13-17.
Question 3
Year: 2024 | Topic: IC Engines
An air standard Otto cycle operates with a compression ratio of 9 and the ambient temperature is 300 K. If the adiabatic index of the medium is 1.5, what is the efficiency?
- A. 75%
- B. 66.67%
- C. 50%
- D. 33%
Answer: B. 66.67%
Explanation: The efficiency of an Otto cycle is given by eta = 1 - (1 / r^(gamma - 1)). Substituting r = 9 and gamma = 1.5 gives eta = 1 - (1 / 9^(1.5 - 1)) = 1 - (1 / 9^0.5) = 1 - (1/3) = 2/3 = 66.67%.
Question 4
Year: 2025 | Topic: IC Engines
Calculate the ideal air standard thermal efficiency for a petrol engine with a cylinder bore of 70 mm and stroke of 100 mm and clearance volume of 42.78 cm³? Consider ratio of specific heats, gamma = 1.4, pi = 22/7 and log10 2.5 = 0.4.
Answer: D. 60%
Explanation: Swept volume V_s = (pi/4) * D² * L = (22/28) * 7^2 * 10 = (22/28) * 490 = 385 cm³. Compression ratio r = (V_s + V_c) / V_c = (385 + 42.78) / 42.78 = 427.78 / 42.78 = 10. Otto cycle efficiency = 1 - 1 / (r^(gamma - 1)) = 1 - 1 / (10^(1.4 - 1)) = 1 - 1 / 10^0.4. Given log10 2.5 = 0.4, it implies 10^0.4 = 2.5. Efficiency = 1 - 1/2.5 = 1 - 0.4 = 0.6 = 60%.