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 papersUnit operations of food and milk plants: balances, fluid flow, heat exchange, thermal processing, evaporation, separation, freezing and hygienic design.
In the exams: GATE AG Section 7 (~9 marks a year 2019–26; thermal processing in all 8 papers); ICAR AICE subject 38; ICAR AIEEA-PG Unit VI; NCEES PE processing
ICAR AICE-JRF/SRF (PhD) — Dairy Engineering (38) and Processing & Food Engineering (41)
Subject 38 units on dairy and food engineering, evaporation and drying, thermal engineering and process equipment design go to PG depth: fouling mechanisms, ε-NTU, UHT direct versus indirect heating, multiple-effect feed arrangements, MVR and TVR, spray-drier balances and powder properties, homogeniser valve theory, disc-stack Σ theory, membrane flux decline and diafiltration, Plank's assumptions and hygienic design standards.
How our questions for ICAR AICE-JRF/SRF spread across the units of Food & Dairy Engineering. Difficulty: 0 easy, 39 medium, 20 hard.
| Thermal processing kinetics: D, z and F values | 6 | |
| Spray and drum drying | 6 | |
| Centrifugation and cream separation | 6 | |
| Pasteurisation: HTST and LTLT | 5 | |
| UHT processing and sterilisation | 5 | |
| Evaporation: single and multiple effect | 5 | |
| Filtration and membrane processes: MF, UF, NF and RO | 5 | |
| Cleaning-in-place and sanitary design | 5 | |
| Homogenisation | 4 | |
| Freezing and Plank's equation | 4 | |
| Material and energy balances | 3 | |
| Heat exchangers: plate, tubular and scraped-surface | 3 | |
| Fluid flow in food plants: pumps, pipes and rheology | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
In a milk-to-milk regenerator the two heat-capacity rates are equal. If NTU = 4 in counter-current flow, the effectiveness is
D. 0.80
For counter-current flow with C_r = 1, ε = NTU/(1 + NTU) = 4/5 = 0.80. 0.98 is 1 − e^(−4), valid only when C_r = 0 (condensing or boiling on one side). 0.50 is the co-current value, (1 − e^(−8))/2; 0.20 is 1/(1 + NTU), the ineffectiveness.
Further reading: Singh and Heldman, Introduction to Food Engineering
Fouling of plate heat exchangers when milk is heated above about 70 °C is caused mainly by
B. Denatured β-lactoglobulin depositing with calcium phosphate
β-lactoglobulin unfolds and aggregates above about 70 °C and deposits with calcium phosphate, whose solubility falls as temperature rises. Lactose browning is a minor factor; fat crystallises on cold, not hot, surfaces.
Further reading: Walstra et al., Dairy Science and Technology
D is 5 min at 110 °C and 0.5 min at 120 °C. The z value is
D. 10 °C
z = (T₂ − T₁)/log(D₁/D₂) = 10/log(10) = 10 °C. 4.3 °C uses the natural log (10/2.303); 1 °C divides by the ratio 10 instead of its log; 2.2 °C divides by the difference in D (10/4.5).
Further reading: Singh and Heldman, Introduction to Food Engineering
Milk at 75 °C is heated to 140 °C by direct steam injection. If steam gives up 2200 kJ/kg and milk cp = 3.9 kJ/kg K, the steam condensed per kg of milk is about
B. 0.115 kg
m_s = cp ΔT/λ = 3.9 × 65/2200 = 0.115 kg/kg, about 10 % dilution, removed later by flash cooling. Using a 140 K rise (from 0 °C) instead of 65 K gives 0.248 kg. 1.15 kg is a factor-of-ten slip; 0.124 kg uses water's cp of 4.18 instead of milk's 3.9.
Further reading: Singh and Heldman, Introduction to Food Engineering
In a high-pressure valve homogeniser under industrial conditions, globule disruption is generally attributed mainly to
B. Turbulent eddies in and just beyond the valve gap
Turbulence, cavitation and shear are all proposed; for typical industrial pressures and gaps, inertial forces in turbulent eddies are generally considered dominant.
Further reading: Walstra et al., Dairy Science and Technology
A solute is at 5.0 % in the feed and 0.2 % in the permeate. The observed rejection is
A. 0.96
R = 1 − C_p/C_f = 1 − 0.2/5.0 = 0.96. 0.04 is the passage (C_p/C_f). 25 is the concentration ratio C_f/C_p; 24 divides by the permeate instead of the feed ((5.0 − 0.2)/0.2).
Further reading: Singh and Heldman, Introduction to Food Engineering
The van 't Hoff osmotic pressure of a 0.1 mol/L non-electrolyte solution at 25 °C is about
D. 248 kPa
π = C R T = 100 mol/m³ × 8.314 × 298.15 = 247,900 Pa = 248 kPa. Using 0.1 mol/L without converting to mol/m³ gives 248 Pa (0.248 kPa), so 2.48 and 24.8 kPa are decimal slips; 496 kPa assumes i = 2, which applies to a salt such as NaCl.
Further reading: Singh and Heldman, Introduction to Food Engineering
Using Plank's equation, a 5 cm thick slab (ρ = 1000 kg/m³, λ = 266.6 kJ/kg, T_f = −1.5 °C, frozen k = 1.5 W/m K) frozen in air at −30 °C takes 2.7 h with h = 30 W/m² K. Doubling the surface heat transfer coefficient from 30 to 60 W/m² K (all else equal) gives a freezing time of about
D. 1.6 h
Surface term halves: 0.5 × 0.05/60 = 4.17e-4; internal term unchanged at 2.08e-4; sum = 6.25e-4; t = 9,354,000 × 6.25e-4 = 5850 s = 1.6 h. It does not halve (1.35 h) because the internal conduction term is unchanged. 2.7 h assumes h has no effect; 0.54 h keeps only the internal conduction term.
Further reading: Singh and Heldman, Introduction to Food Engineering
Using Plank's equation, a 5 cm thick slab (ρ = 1000 kg/m³, λ = 266.6 kJ/kg, T_f = −1.5 °C, frozen k = 1.5 W/m K) frozen in air at −30 °C with h = 30 W/m² K takes 2.7 h. Doubling the thickness to 10 cm, all else unchanged, gives a freezing time of about
B. 6.5 h
Pa/h doubles (1.67e-3) and Ra²/k quadruples (8.33e-4); sum 2.5e-3; t = 9,354,000 × 2.5e-3 = 23,390 s = 6.5 h. It lies between doubling (5.4 h) and quadrupling (10.8 h). 4.9 h doubles only the surface term and leaves the internal term unchanged.
Further reading: Singh and Heldman, Introduction to Food Engineering
A dairy wants to lower its CIP caustic temperature to save energy. According to Sinner's principle, which change best keeps cleaning performance?
B. Extend circulation time and/or raise detergent concentration, then verify by ATP swabs and inspection
Lowering one factor (temperature) must be compensated by others (time, chemical, mechanical action), and any change must be validated. Lowering velocity cuts mechanical action further; skipping steps leaves soil or mineral.
Further reading: Tamime, Cleaning-in-Place
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