GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)
Original practice questions written to the published syllabus · not past papersEnergy, work, heat and the cycles behind engines, refrigerators and dryers.
In the exams: GATE AG Section 3 (thermodynamic principles of IC engines); ICAR AIEEA-PG Unit III; NCEES FE thermodynamics and heat transfer 9–14 questions
GATE — Agricultural Engineering (AG)
Thermodynamic principles of IC engines under Farm Power: Otto and Diesel efficiency, compression and cut-off ratio, MEP, and isentropic compression temperatures, usually as numerical-answer questions. Processing sections draw on psychrometry (humidity ratio, dryer air balances, evaporative cooling) and refrigeration (COP, TR, refrigerant flow). Answers must be computed to the stated precision, so carry kelvin and kJ consistently.
How our questions for GATE spread across the units of Engineering Thermodynamics. Difficulty: 9 easy, 18 medium, 9 hard.
| Air-standard cycles: Otto, Diesel, dual and Brayton | 10 | |
| Introduction to psychrometry | 10 | |
| Vapour-compression cycle basics | 6 | |
| Steady-flow energy equation | 3 | |
| Properties of steam, steam tables and the Mollier chart | 3 | |
| Ideal gas processes | 2 | |
| Second law, entropy and the Carnot cycle | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Air at 300 K and 100 kPa is compressed reversibly and adiabatically in a diesel engine cylinder with a compression ratio of 16 (γ = 1.4). The temperature at the end of compression is about
D. 909 K
For an isentropic process T2/T1 = (V1/V2)^(γ−1) = 16^0.4, so T2 = 300 × 3.031 = 909 K. 4800 K multiplies by the volume ratio itself, 2174 K uses exponent 1/γ, and 662 K uses the pressure-ratio exponent (γ−1)/γ with a volume ratio.
Further reading: Çengel, Boles and Kanoğlu — Thermodynamics: An Engineering Approach
Steam enters an adiabatic nozzle at negligible velocity and its enthalpy falls by 200 kJ/kg. The exit velocity is about
C. 632 m/s
V2 = √(2Δh) with Δh in J/kg: √(2 × 200 000) = 632 m/s. 447 m/s drops the factor 2, 20.0 m/s forgets to convert kJ to J, and 14.1 m/s makes both mistakes.
Further reading: Çengel, Boles and Kanoğlu — Thermodynamics: An Engineering Approach
The air-standard efficiency of an Otto cycle with a compression ratio of 8 (γ = 1.4) is
C. 56.5 %
η = 1 − 8^(−0.4) = 1 − 0.4353 = 56.5 %. 87.5 % uses 1 − 1/r without the exponent, 43.5 % is the term 1/r^(γ−1) itself, and 64.6 % uses exponent 0.5 instead of γ − 1 = 0.4.
Further reading: Çengel, Boles and Kanoğlu — Thermodynamics: An Engineering Approach
In the Diesel cycle the cut-off ratio is defined as
D. the volume at the end of heat addition divided by the volume at its start
Cut-off ratio rc = V3/V2, the expansion during constant-pressure heat addition. Total volume over clearance volume is the compression ratio, the pressure ratio describes the constant-volume part of a dual cycle, and the last ratio has no standard name.
Further reading: Çengel, Boles and Kanoğlu — Thermodynamics: An Engineering Approach
At a fixed compression ratio, what happens to the air-standard Diesel-cycle efficiency as the load (and so the cut-off ratio) increases?
C. It decreases, because heat is added over a longer part of the expansion stroke
The factor (rc^γ − 1)/(γ(rc − 1)) grows with rc, so efficiency falls as cut-off ratio rises; heat added later in the stroke is converted less effectively. Dependence on r alone holds only for the Otto cycle, and there is no maximum at rc = 2.
Further reading: Çengel, Boles and Kanoğlu — Thermodynamics: An Engineering Approach
In an ideal vapour-compression cycle, the enthalpies are 395 kJ/kg at evaporator exit, 425 kJ/kg at compressor exit and 250 kJ/kg at condenser exit. The COP is
D. 4.83
Refrigerating effect = 395 − 250 = 145 kJ/kg; compressor work = 425 − 395 = 30 kJ/kg; COP = 145/30 = 4.83. 5.83 is the heat-pump COP (condenser heat 175/30), 0.207 is the inverse, and 8.33 uses the condenser-exit enthalpy as the refrigerating effect.
Further reading: Stoecker and Jones — Refrigeration and Air Conditioning
A cold store needs 10 TR of refrigeration (1 TR = 3.517 kW). The refrigerating effect of the cycle is 145 kJ/kg. The refrigerant mass flow rate is about
C. 0.243 kg/s
ṁ = Q̇/RE = 35.17/145 = 0.243 kg/s. 14.6 is the flow per minute labelled per second, 0.069 divides TR by kJ/kg without converting to kW, and 1.17 divides by a compressor work of 30 kJ/kg instead of the refrigerating effect.
Further reading: Stoecker and Jones — Refrigeration and Air Conditioning
If the evaporator temperature of a cold store is lowered while the condensing temperature stays the same, the COP of the plant
B. decreases, because compressor work per kg rises and refrigerating effect falls
A lower evaporator pressure means a larger pressure ratio (more work per kg) and, usually, a smaller refrigerating effect, so COP falls — as the Carnot COP TL/(TH − TL) also shows. Suction vapour density actually falls, reducing capacity.
Further reading: Stoecker and Jones — Refrigeration and Air Conditioning
The dew-point temperature of moist air is
C. the temperature at which the air becomes saturated when cooled at constant humidity ratio
Dew point is the saturation temperature at the actual vapour pressure: cool the air at constant W until condensation begins. The wetted-wick reading is the wet-bulb temperature, adiabatic saturation defines the adiabatic-saturation temperature, and 50 % RH has no special name.
Further reading: ASHRAE Handbook — Fundamentals, Psychrometrics
Air at 30 °C has a vapour pressure of 2.0 kPa. The saturation pressure of water at 30 °C is 4.247 kPa. The relative humidity is about
C. 47.1 %
φ = pv/psat = 2.0/4.247 = 47.1 %. 2.0 % divides by barometric pressure, 212 % inverts the ratio, and 46.0 % is the degree of saturation W/Ws, which is close to but not the same as relative humidity.
Further reading: ASHRAE Handbook — Fundamentals, Psychrometrics
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