GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)
Original practice questions written to the published syllabus · not past papersHow tractor engines work and how their performance is measured — the single most reliably examined farm-power topic.
In the exams: GATE AG Section 3 (engine performance asked in all 8 papers 2019–26); ICAR AIEEA-PG Unit V; ICAR AICE subject 39 Unit 4; NCEES PE IC engines; RPSC AAE
GATE — Agricultural Engineering (AG)
Numerical engine performance every year: IP from IMEP, BP from torque, mechanical and brake thermal efficiency, BSFC, volumetric efficiency; occasionally Otto/air-standard efficiency, fuel properties (calorific value, cetane, API gravity) and governor regulation or torque reserve. NAT questions need correct units: watch N/2 for four-stroke engines and kJ vs MJ.
How our questions for GATE spread across the units of Tractor & Automotive Engines (IC Engines). Difficulty: 5 easy, 12 medium, 22 hard.
| Otto, Diesel and dual cycles | 6 | |
| Engine performance: IP, BP, FP and efficiencies | 6 | |
| BSFC and heat balance | 5 | |
| Fuel properties | 4 | |
| Engine testing | 4 | |
| Engine classification and terminology | 3 | |
| Governing | 3 | |
| Two-stroke and four-stroke engines | 2 | |
| Combustion and knock | 2 | |
| Air intake and turbocharging | 2 | |
| Diesel fuel system: injection pump and injector | 1 | |
| Cooling system | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A diesel cylinder has a bore of 100 mm, a stroke of 125 mm and a clearance volume of 70 cm³. Its compression ratio is closest to
A. 15.0
Vs = (π/4)(10 cm)²(12.5 cm) = 981.7 cm³. r = (Vs + Vc)/Vc = (981.7 + 70)/70 = 15.02. Dividing Vs by Vc alone gives 14.0 (that is r − 1); using πD²L without the 1/4 and then also forgetting to add Vc back (Vs/Vc) gives 56.1. 16.0 counts the clearance volume twice, (Vs + 2Vc)/Vc.
The air-standard efficiency of an Otto cycle with a compression ratio of 8 (γ = 1.4) is closest to
C. 56.5%
η = 1 − 1/r^(γ−1) = 1 − 8^(−0.4) = 1 − 0.4353 = 56.5%. Using 1 − 1/r (87.5%) or exponent γ instead of γ − 1 (94.6%) are the common slips; 43.5% is the fraction rejected, not the efficiency.
Further reading: Heywood, Internal Combustion Engine Fundamentals
An air-standard Diesel cycle has a compression ratio of 16 and a cut-off ratio of 2 (γ = 1.4). Its efficiency is closest to
A. 61.4%
η = 1 − (1/r^0.4)·(ρ^γ − 1)/[γ(ρ − 1)] = 1 − (1/3.0314)·(2.6390 − 1)/(1.4 × 1) = 61.4%. 67.0% is the Otto value at r = 16; dropping γ from the denominator gives 45.9%. 93.8% is 1 − 1/r, forgetting the exponent γ − 1.
Further reading: Heywood, Internal Combustion Engine Fundamentals
Raising the compression ratio of an Otto cycle from 6 to 9 (γ = 1.4) raises its air-standard efficiency by about how many percentage points?
A. 7.3
η6 = 1 − 6^(−0.4) = 51.2%; η9 = 1 − 9^(−0.4) = 58.5%; difference = 7.3 points. 14.3 is the relative (percentage) increase, not percentage points. 5.6 drops the exponent: 1/6 − 1/9 = 0.056.
Load control in a diesel engine is called quality governing because
D. the air is not throttled and only the fuel quantity (hence mixture strength) is varied
A diesel takes a full charge of air at all loads; the governor varies fuel, so the air-fuel ratio (quality) changes. SI engines throttle the mixture quantity at near-constant ratio: quantity governing.
Taking diesel as n-dodecane (C12H26) and air as 21% O2 and 79% N2 by volume, the stoichiometric air-fuel ratio by mass is closest to
A. 14.9
C12H26 + 18.5 O2 → 12 CO2 + 13 H2O. Air = 18.5 × (32.0 + 3.76 × 28.01) = 2541 kg per kmol fuel; fuel = 170.3 kg/kmol. A/F = 2541/170.3 = 14.9. 3.48 is the oxygen-fuel ratio only; 88.1 is the molar air-fuel ratio. Burning the carbon only (12 O2) gives 9.66.
At constant speed an engine has BP = 40 kW and FP = 10 kW. If the load is reduced to BP = 20 kW with friction power unchanged, the mechanical efficiency becomes
C. 66.7%
At 40 kW: ηm = 40/50 = 80%. At 20 kW: IP = 20 + 10 = 30 kW, ηm = 20/30 = 66.7%. Mechanical efficiency falls at part load because friction stays roughly constant.
An engine uses 10 kg of diesel per hour while delivering 40 kW brake power. Its BSFC is
B. 250 g/kWh
BSFC = mf/BP = 10 kg/h ÷ 40 kW = 0.25 kg/kWh = 250 g/kWh. 400 comes from inverting the ratio (40/10 = 4, read as 400 g/kWh); fuel goes on top. 4.0 kg/kWh also inverts it.
A tractor engine has a BSFC of 250 g/kWh with fuel of lower calorific value 42 MJ/kg. Its brake thermal efficiency is
B. 34.3%
ηbt = 3600/(BSFC [kg/kWh] × CV [kJ/kg]) = 3600/(0.25 × 42,000) = 0.343 = 34.3%. Mixing MJ and kJ gives a factor-of-1000 error; inverting the formula gives 2.9, which is (1/η) × 100. Leaving out the 3600 kJ/kWh conversion gives 1/(0.25 × 42) = 9.5%.
A rope-brake dynamometer has a 0.6 m drum and a 20 mm rope. At 1,500 rev/min the dead load is 400 N and the spring balance reads 50 N. The brake power is closest to
C. 17.04 kW
BP = (W − S)π(D + d)N/60 = 350 × π × 0.62 × 25 = 17043 W = 17.04 kW. Ignoring the rope diameter gives 16.49 kW; adding W and S instead of subtracting gives 21.91 kW. Using T × N/60 without 2π gives 2.71 kW.
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