ICAR AIEEA-PG — Agricultural Engineering & Technology (Major Subject Group 10) · India · National Testing Agency for ICAR
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
ICAR AIEEA-PG — Agricultural Engineering & Technology
Unit III (thermodynamics) and Unit VI (processing) ask single-answer MCQs: definitions of the dimensionless numbers, units of k, h, α and D, the physical meaning of the Biot and Fourier numbers, critical radius, counterflow versus parallel flow, Kirchhoff's law, black and grey bodies, and simple one-step calculations. Negative marking rewards knowing exact definitions.
How our questions for ICAR AIEEA-PG spread across the units of Heat & Mass Transfer. Difficulty: 31 easy, 32 medium, 0 hard.
| Convection and dimensionless numbers (Nu, Re, Pr) | 9 | |
| Steady conduction through plane walls and cylinders | 6 | |
| Composite walls and critical radius of insulation | 6 | |
| Radiation: Stefan-Boltzmann law and view factors | 6 | |
| Extended surfaces (fins) | 6 | |
| Transient conduction: lumped capacity and Heisler charts | 5 | |
| Heat exchangers: LMTD method | 5 | |
| Diffusive mass transfer and Fick's law | 5 | |
| Simultaneous heat and mass transfer in drying | 5 | |
| Heat exchangers: effectiveness-NTU method | 4 | |
| Condensation and boiling | 3 | |
| Convective mass transfer | 3 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
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
The maximum possible heat-transfer rate used in defining heat-exchanger effectiveness is
A. C_min(T_h,in − T_c,in)
Only the fluid with the smaller heat-capacity rate can undergo the full temperature difference between the two inlets, which it would do in an infinitely long counterflow exchanger. Using C_max would require the other fluid to change temperature by more than the inlet difference, which violates the second law. UA × LMTD is the actual rate, not the maximum.
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
On the pool-boiling curve, the Leidenfrost point is
B. the point of minimum heat flux between transition and film boiling
The Leidenfrost point marks the minimum heat flux, where a stable vapour film first covers the surface; water drops skittering on a very hot plate show it. The maximum is the critical heat flux at the end of nucleate boiling.
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
The typical order of magnitude of binary diffusion coefficients at ordinary temperature is
A. gases about 10⁻⁵ m²/s, liquids about 10⁻⁹ m²/s, solids much smaller
Gas-phase diffusivities are around 10⁻⁵ m²/s (water vapour in air about 2.5 × 10⁻⁵), liquid-phase around 10⁻⁹ m²/s, and diffusion in solids and food matrices is slower still. This four-orders-of-magnitude gap is why internal diffusion controls drying in the falling-rate period.
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
The Sherwood number, Sh = h_m L/D_AB, is the mass-transfer analogue of the
B. Nusselt number
Sh is the dimensionless convective mass-transfer coefficient, as Nu is for heat. The Schmidt number Sc = ν/D is the analogue of Pr; the mass-transfer Biot number h_m L/D_solid uses the diffusivity inside the solid.
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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