ICAR AICE-JRF/SRF (PhD) — Agricultural Engineering subjects 38–43 · 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 AICE-JRF/SRF (PhD) — Processing and Food Engineering (41) and Dairy Engineering (38)
PG-level depth: Heisler charts and product solutions, fin effectiveness, radiation networks and shields, ε–NTU relations and regeneration efficiency in pasteurisers, condensation and boiling regimes in evaporators, the heat–mass analogy and Lewis relation, Fick's second law and effective diffusivity, and thin-layer drying models.
How our questions for ICAR AICE-JRF/SRF spread across the units of Heat & Mass Transfer. Difficulty: 3 easy, 41 medium, 27 hard.
| Diffusive mass transfer and Fick's law | 10 | |
| Simultaneous heat and mass transfer in drying | 9 | |
| Condensation and boiling | 8 | |
| Convective mass transfer | 7 | |
| Transient conduction: lumped capacity and Heisler charts | 6 | |
| Extended surfaces (fins) | 6 | |
| Heat exchangers: effectiveness-NTU method | 6 | |
| Convection and dimensionless numbers (Nu, Re, Pr) | 5 | |
| Steady conduction through plane walls and cylinders | 4 | |
| Radiation: Stefan-Boltzmann law and view factors | 4 | |
| Heat exchangers: LMTD method | 4 | |
| Composite walls and critical radius of insulation | 2 |
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
In laminar flow of a heavy oil (Pr = 1,000) along a heated flat plate, with heating starting at the leading edge, the thermal boundary-layer thickness is approximately
B. one-tenth of the velocity boundary-layer thickness
For laminar flat-plate flow δ/δ_t ≈ Pr^(1/3). With Pr = 1,000, Pr^(1/3) = 10, so δ_t ≈ δ/10. High-Pr fluids diffuse momentum much faster than heat. Ten times thicker would apply to a very low-Pr fluid such as a liquid metal; one-thousandth wrongly uses Pr to the first power.
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
An electrically heated element submerged in water is controlled by its power input. If the power is raised just above the critical heat flux, the element
C. jumps to the film-boiling branch at a much higher surface temperature, risking burnout
With heat flux imposed, the operating point cannot follow the falling transition branch; it jumps at the same flux to the film-boiling curve, where the surface temperature can be hundreds of degrees higher and may exceed the element's melting point. Transition boiling can be followed only when surface temperature is controlled, not heat flux.
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
For laminar film condensation on a vertical plate, Nusselt's analysis gives h ∝ L^(−1/4). If the plate height is doubled with all else the same, the average coefficient becomes
D. 0.841 times the original
h ∝ L^(−1/4), so h_new/h_old = 2^(−0.25) = 0.841. The thicker film on the lower part of a taller plate lowers the average coefficient, but only weakly; the total heat flow still rises, because the area doubles (1.68 times).
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
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
A solute diffuses steadily through a stagnant liquid layer 2 mm thick with D = 2 × 10⁻⁹ m²/s. Its concentrations on the two faces differ by 0.5 kmol/m³. The molar flux is
B. 5.00 × 10⁻⁷ kmol/m²·s
N = DΔC/L = 2 × 10⁻⁹ × 0.5/0.002 = 5.0 × 10⁻⁷ kmol/m²·s, the mass analogue of Q = kAΔT/L. Entering L as 2 (mm) gives 5 × 10⁻¹⁰; multiplying by L gives 2 × 10⁻¹²; omitting ΔC gives 1 × 10⁻⁶ with the wrong units.
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
Over a moist product in a dryer the convective heat-transfer coefficient is 30 W/m²·K. Air has ρ = 1.2 kg/m³ and c_p = 1,007 J/kg·K, and Le = 0.88. By the Chilton–Colburn analogy, h/h_m = ρ c_p Le^(2/3), the mass-transfer coefficient is closest to
D. 27.0 × 10⁻³ m/s
h_m = h/(ρ c_p Le^(2/3)) = 30/(1.2 × 1007 × 0.88^(2/3)) = 30/(1.2 × 1007 × 0.9183) = 0.02703 m/s. Taking Le^(2/3) = 1 (the Lewis relation) gives 24.8 × 10⁻³, a small error for air–water; using c_p in kJ/kg·K gives an impossible 27.0 m/s; using Le^(1/3) gives 25.9 × 10⁻³.
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
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