GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)
Original practice questions written to the published syllabus · not past papersRefrigeration cycles, equipment and cold-store design, from COP to the heat load of a room full of potatoes.
In the exams: GATE AG Section 7 (refrigeration COP/TR in 4 of 8 papers 2019–26; composite cold-store walls recur); ICAR AIEEA-PG Unit VI; ICAR AICE subject 38 Unit 6
GATE — Agricultural Engineering (AG)
Refrigeration sits in the Dairy and Food Engineering and Storage Systems parts of the syllabus. Expect one- or two-mark numerical-answer questions: COP from given enthalpies or from Carnot temperatures, tonnes of refrigeration, mass flow and compressor power; conduction through a composite cold-store wall (the most repeated calculation); product and latent heat loads; freezing time by Plank's equation; and psychrometric properties feeding a load. Read units carefully: kJ/min against kW and °C against K are the usual traps.
How our questions for GATE spread across the units of Refrigeration, Air-Conditioning & Cold-Chain Engineering. Difficulty: 9 easy, 19 medium, 12 hard.
| Vapour-compression refrigeration cycle | 6 | |
| Cold-store heat load: transmission, product and respiration | 6 | |
| Pressure-enthalpy charts | 5 | |
| COP and tonnes of refrigeration | 4 | |
| Psychrometric air-conditioning processes | 4 | |
| Condensers and evaporators | 3 | |
| Vapour-absorption refrigeration | 2 | |
| Refrigerants: ODP and GWP | 2 | |
| Compressors | 2 | |
| Insulation and vapour barriers | 2 | |
| Expansion devices | 1 | |
| Cold-store heat load: infiltration, occupancy and equipment | 1 | |
| Ice-bank systems | 1 | |
| Refrigerated transport and cold-chain logistics | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A two-stage ammonia system has suction pressure 2.0 bar and condensing pressure 12.5 bar. Using the usual first choice for intermediate pressure, the intercooler pressure is about
D. 5.0 bar
The geometric mean √(p_s p_d) = √(2.0 × 12.5) = 5.0 bar roughly equalises the pressure ratios (2.5 each) and the work of the stages. The arithmetic mean 7.25 bar gives unequal ratios; 6.25 is the overall pressure ratio, not a pressure, and 10.5 bar is the pressure difference.
Further reading: Stoecker and Jones, Refrigeration and Air Conditioning
One tonne of refrigeration (TR) is equal to about
C. 3.517 kW
1 TR = 12,000 Btu/h = 211 kJ/min ≈ 3.517 kW. 12 kW confuses it with 12,000 Btu/h; 4.187 is the specific heat of water in kJ/kg K.
Further reading: Arora, Refrigeration and Air Conditioning
From a p–h chart an ammonia cycle has h1 = 1420 kJ/kg (suction), h2 = 1650 kJ/kg (discharge) and h3 = 320 kJ/kg (condenser outlet). For a capacity of 50 kW, the refrigerant mass flow rate is
A. 0.0455 kg/s
Refrigerating effect = h1 − h4 = 1420 − 320 = 1100 kJ/kg (h4 = h3). m = 50/1100 = 0.0455 kg/s. Using the condenser enthalpy change (1330) or the work (230) as divisor are the usual slips (0.0376 and 0.2174 kg/s); dividing by the suction enthalpy alone gives 0.0352 kg/s. COP here = 1100/230 = 4.78.
Further reading: Arora, Refrigeration and Air Conditioning
At the evaporator pressure, saturated liquid enthalpy is 180 kJ/kg and saturated vapour enthalpy 390 kJ/kg. Liquid at 250 kJ/kg is throttled into the evaporator. The dryness fraction at evaporator inlet is
B. 0.333
x = (h4 − h_f)/(h_g − h_f) = (250 − 180)/(390 − 180) = 70/210 = 0.333. 0.667 is the liquid fraction; h4/h_g ignores the liquid enthalpy datum. 0.641 is h4/h_g (250/390), and 0.359 is (h_g − h4)/h_g (140/390).
Further reading: Arora, Refrigeration and Air Conditioning
A water-cooled condenser rejects 50 kW with refrigerant condensing at 40 °C. Water enters at 30 °C and leaves at 35 °C. With U = 800 W/m² K, the required area is about
B. 8.66 m²
ΔT1 = 10 K, ΔT2 = 5 K. LMTD = (10 − 5)/ln(10/5) = 7.21 K. A = Q/(U·LMTD) = 50,000/(800 × 7.21) = 8.66 m². The arithmetic mean (7.5 K) gives 8.33 m²; using the water rise (5 K) gives 12.5 m²; 0.0087 m² forgets kW → W.
Further reading: Stoecker and Jones, Refrigeration and Air Conditioning
A thermostatic expansion valve is designed to maintain a nearly constant
B. Superheat of vapour at the evaporator outlet
The bulb on the suction line senses outlet temperature, and the diaphragm balances bulb pressure against evaporator pressure plus spring: the controlled variable is superheat. Constant evaporator pressure is the automatic (constant-pressure) valve; liquid level is a float valve.
Further reading: Arora, Refrigeration and Air Conditioning
A cold-store wall is 230 mm brick (k = 0.7), 100 mm polyurethane (k = 0.025) and 15 mm plaster (k = 0.7 W/m K). Outside and inside film coefficients are 23 and 8 W/m² K. For 100 m² of wall and a 35 K temperature difference, the heat gain is about
A. 775 W
R = 1/23 + 0.23/0.7 + 0.10/0.025 + 0.015/0.7 + 1/8 = 0.0435 + 0.3286 + 4.000 + 0.0214 + 0.125 = 4.518 m² K/W; U = 0.2213 W/m² K. Q = 0.2213 × 100 × 35 = 775 W. Ignoring films gives 805 W; insulation alone gives 875 W; multiplying by R instead of dividing gives a huge wrong number. 15815 W multiplies by R (4.518 × 100 × 35).
Further reading: Singh and Heldman, Introduction to Food Engineering
100 t of potatoes (c_p = 3.6 kJ/kg K) enter a store at 30 °C and must be cooled to 4 °C in 24 h. The average product cooling load is about
C. 30.8 TR
Q = m c_p Δt = 100,000 × 3.6 × 26 = 9,360,000 kJ. ÷ 86,400 s = 108.3 kW = 30.8 TR. 108.3 is the kW value mislabelled; using 30 K instead of 26 K gives 35.5 TR; dividing by 3,600 treats the day as one hour. 739 TR divides by 3,600 instead of 86,400 (2,600 kW/3.517).
Further reading: ASHRAE Handbook — Refrigeration
By Siebel's relation, the specific heat above freezing of a fruit with 80% water is about
D. 3.52 kJ/kg K
c_p = 3.35 × 0.8 + 0.84 = 3.52 kJ/kg K. 1.85 is the below-freezing relation; omitting the 0.84 solids term gives 2.68.
Further reading: Singh and Heldman, Introduction to Food Engineering
In a cold-store wall, the vapour barrier should be placed
D. On the warm (outer) side of the insulation
Vapour moves from warm to cold. A barrier on the warm side stops it before it reaches layers cold enough to condense or freeze it. A tight barrier on the cold side traps any moisture that enters.
Further reading: ASHRAE Handbook — Refrigeration
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