GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)
Original practice questions written to the published syllabus · not past papersConduction, convection, radiation, heat exchangers and diffusion — the physics of drying, cooling, freezing and cold-store walls.
In the exams: GATE AG Section 7 (heat exchangers and composite walls each asked in 7 of 8 papers 2019–26); ICAR AICE subject 41 Unit 2; NCEES PE processing
GATE — Agricultural Engineering (AG)
Heat and Mass Transfer sits in Section 7 (Dairy and Food Engineering). Composite-wall conduction and heat-exchanger calculations (LMTD, and increasingly effectiveness–NTU) appear almost every year as one- or two-mark numerical-answer questions. Also expected: critical radius of insulation, lumped-capacitance cooling time, fin heat flow and efficiency, Stefan–Boltzmann exchange with view factors, Nusselt/Reynolds/Prandtl definitions and the Dittus–Boelter equation, Fick's law, and drying-rate and moisture-basis conversions that overlap with Section 6. Watch units (kJ/h against W, °C against K in radiation) and the choice of characteristic length.
How our questions for GATE spread across the units of Heat & Mass Transfer. Difficulty: 21 easy, 55 medium, 14 hard.
| Convection and dimensionless numbers (Nu, Re, Pr) | 11 | |
| Steady conduction through plane walls and cylinders | 8 | |
| Composite walls and critical radius of insulation | 8 | |
| Transient conduction: lumped capacity and Heisler charts | 8 | |
| Extended surfaces (fins) | 8 | |
| Simultaneous heat and mass transfer in drying | 8 | |
| Radiation: Stefan-Boltzmann law and view factors | 7 | |
| Heat exchangers: LMTD method | 7 | |
| Heat exchangers: effectiveness-NTU method | 7 | |
| Condensation and boiling | 7 | |
| Diffusive mass transfer and Fick's law | 6 | |
| Convective mass transfer | 5 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A spherical storage vessel has an inner radius of 0.5 m and is covered with 0.1 m of insulation (k = 0.04 W/m·K). The insulation inner surface is at −5 °C and its outer surface at 25 °C. The steady heat gain through the insulation is
D. 45.2 W
Spherical shell resistance R = (r2 − r1)/(4πk r1 r2) = 0.1/(4π × 0.04 × 0.5 × 0.6) = 0.6631 K/W, so Q = 30/0.6631 = 45.2 W. Using the plane-wall formula with the inner area gives 37.7 W and with the outer area 54.3 W; dropping r1 r2 from the denominator gives 151 W.
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
A cold-store wall of 150 m² consists of 230 mm brick (k = 0.7 W/m·K) and 100 mm insulation (k = 0.025 W/m·K). The outside air is at 40 °C with h = 20 W/m²·K and the inside air at 2 °C with h = 10 W/m²·K. The steady heat gain through the wall is closest to
B. 1,270 W
Resistance per m²: 1/20 + 0.23/0.7 + 0.10/0.025 + 1/10 = 0.05 + 0.3286 + 4.0 + 0.1 = 4.4786 m²·K/W. Flux q = 38/4.4786 = 8.485 W/m², and Q = 8.485 × 150 = 1,270 W. Ignoring the air films gives 1,320 W; leaving out the insulation gives 11,900 W; 8.5 W is the flux per square metre, not the total.
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
Water (ν = 1.0 × 10⁻⁶ m²/s) flows at 0.5 m/s through a tube of 25 mm internal diameter. The Reynolds number and flow regime are
D. 12,500; turbulent
Re = VD/ν = 0.5 × 0.025/1.0 × 10⁻⁶ = 12,500. Pipe flow is laminar below about 2,300, so this is turbulent. Entering D as 2.5 mm gives 1,250; using the radius gives 6,250; calling Re = 12,500 laminar confuses the pipe limit with the flat-plate limit of about 5 × 10⁵.
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
A small body obeying the lumped-capacitance model is plunged into a fluid. After a time equal to one time constant, its excess temperature (T − T∞) has fallen to about what fraction of the initial excess?
B. 0.368
θ/θ_i = exp(−t/τ) = e⁻¹ = 0.368 at t = τ. 0.632 is the fraction of the change already completed; 0.5 is the half-time (t = 0.693τ); 0.135 is reached after two time constants.
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
Soil initially at a uniform 20 °C (α = 5 × 10⁻⁷ m²/s) has its surface suddenly brought to 0 °C and held there. Treat the soil as a semi-infinite solid, with (T − T_s)/(T_i − T_s) = erf[x/(2√(αt))]. Given erf(0.24) = 0.266, erf(0.48) = 0.503 and erf(0.96) = 0.825, the temperature at 0.1 m depth after 6 hours is closest to
D. 10.1 °C
√(αt) = √(5 × 10⁻⁷ × 21,600) = 0.1039 m, so x/(2√(αt)) = 0.1/0.2078 = 0.48; erf(0.48) = 0.503. T = 0 + 0.503 × 20 = 10.1 °C. Forgetting the 2 gives the argument 0.96 and 16.5 °C; using 4 gives 0.24 and 5.3 °C; using 1 − erf gives 9.94 °C, which would be the temperature for heating from 0 °C towards 20 °C.
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
Oil (2.0 kg/s, c = 2.1 kJ/kg·K) is cooled by water (1.0 kg/s, c = 4.18 kJ/kg·K). The capacity ratio C_r and the fluid that has C_min are
D. C_r = 0.995; water has C_min
C_oil = 2.0 × 2.1 = 4.20 kW/K; C_water = 1.0 × 4.18 = 4.18 kW/K. Water has the smaller C, so C_r = 4.18/4.20 = 0.995. Comparing mass flows alone (0.50) or specific heats alone (0.502) is wrong: C is the product m·c.
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
As the excess temperature of a heated surface in a pool of saturated water is increased, the boiling regimes follow the order
A. natural convection → nucleate → transition → film boiling
At small ΔT_e heat moves by natural convection; bubbles then form at nucleation sites (nucleate boiling) up to the critical heat flux; beyond it, unstable vapour patches (transition boiling) reduce the flux to a minimum at the Leidenfrost point; above that, a stable vapour film covers the surface (film boiling).
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
Saturated steam condenses on a vertical plate 0.5 m high and 1 m wide whose surface is 10 °C below saturation. Liquid properties: ρ_l = 960 kg/m³, k_l = 0.68 W/m·K, μ_l = 2.8 × 10⁻⁴ Pa·s; h_fg = 2,257 kJ/kg; vapour density 0.6 kg/m³. Using h = 0.943[ρ_l(ρ_l − ρ_v)g h_fg k_l³/(μ_l L ΔT)]^(1/4), the condensate produced is closest to
A. 61.9 kg/h
h = 0.943 × [960 × 959.4 × 9.81 × 2.257 × 10⁶ × 0.68³/(2.8 × 10⁻⁴ × 0.5 × 10)]^0.25 = 7758 W/m²·K. Q = hAΔT = 7758 × 0.5 × 10 = 38788 W. Condensate = Q/h_fg = 38788/2.257 × 10⁶ = 0.01719 kg/s = 61.9 kg/h. Using the horizontal-tube constant gives 47.6; leaving ΔT out of Q = hAΔT gives 6.19; dividing by h_fg in kJ/kg gives a value 1,000 times too large. That value is 61,869 kg/h.
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
Water at 25 °C evaporates from the bottom of a narrow tube into dry air flowing across the top, 0.15 m above the water surface. Total pressure is 101.325 kPa, the vapour pressure of water is 3.17 kPa, D = 2.6 × 10⁻⁵ m²/s and R = 8.314 kJ/kmol·K. Treating the air as stagnant, the evaporation flux is closest to
C. 14.60 g/m²·h
Diffusion of A through stagnant B: N_A = [DP/(RTL)] ln[(P − p_A2)/(P − p_A1)] = [2.6 × 10⁻⁵ × 101.325/(8.314 × 298 × 0.15)] × ln(101.325/98.155) = 2.25 × 10⁻⁷ kmol/m²·s. × 18 kg/kmol × 1,000 g/kg × 3,600 s/h = 14.60 g/m²·h. The simple Fick form without the bulk-flow factor gives 14.37 (about 1.6% lower here, because the vapour is dilute); 0.0146 is a kg-to-g slip; using ln(P/p_A1) mixes up the partial pressures of the stagnant gas and gives 1591.
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
Grain with a moisture content of 20% (wet basis) has a dry-basis moisture content of
A. 25%
M_db = M_wb/(1 − M_wb) = 0.20/0.80 = 0.25 = 25%. 16.7% is the reverse conversion, M_wb = M_db/(1 + M_db), applied to 20%; 80% is the dry-matter fraction.
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
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