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Agricultural and Biosystems Engineers Licensure Examination Refrigeration, Air-Conditioning & Cold-Chain Engineering: practice questions

Agricultural and Biosystems Engineers Licensure Examination (ABELE) · Philippines · Professional Regulatory Board of Agricultural and Biosystems Engineering, PRC

Original practice questions written to the published syllabus · not past papers
10worked here, free
17written to this exam in this subject
10syllabus units

About Agricultural and Biosystems Engineers Licensure Examination

Agricultural and Biosystems Engineers Licensure Examination (ABELE) is a licensing and registration exam run by Professional Regulatory Board of Agricultural and Biosystems Engineering, PRC, for candidates in Philippines. Paper-based, 2 days (Day 1 6 hours; Day 2 5 hours + 4 hours). Non-programmable calculator only; the Dec 2025 programme bans the Casio fx-991ES/ES Plus by name.

Source: prc.gov.ph.

Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.

Refrigeration, Air-Conditioning & Cold-Chain Engineering in Agricultural and Biosystems Engineers Licensure Examination

Refrigeration cycles, equipment and cold-store design, from COP to the heat load of a room full of potatoes.

Our 17 Agricultural and Biosystems Engineers Licensure Examination questions in this subject: 0 easy, 12 medium, 5 hard, across 10 units.

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

These 10 were written to Agricultural and Biosystems Engineers Licensure Examination and are not shown on any other exam's page.

Question 1 · medium

A single-effect absorption chiller delivers 100 kW of cooling with a COP of 0.7. Neglecting pump work, the heat that the cooling tower must remove from absorber and condenser is about

  1. 243 kW
  2. 143 kW
  3. 125 kW
  4. 100 kW
Answer and working

A. 243 kW

Generator heat Q_G = 100/0.7 = 142.9 kW. Heat rejected = Q_E + Q_G = 100 + 142.9 = 242.9 kW. 143 kW is only the driving heat; 125 kW is what a compression chiller of COP 4 would reject, which is why absorption plants need larger cooling towers.

Question 2 · medium

A designer wants 90–95% relative humidity in a room storing leafy vegetables. The evaporator should be selected with

  1. A large temperature difference to remove moisture quickly
  2. A small air-to-refrigerant temperature difference, and therefore a large coil surface
  3. The lowest possible evaporating temperature to save compressor size and speed pull-down
  4. Bare pipes with no fan
Answer and working

B. A small air-to-refrigerant temperature difference, and therefore a large coil surface

A large TD cools air far below its dew point and strips moisture, drying the room and the produce. A small TD (often a few kelvin) keeps humidity high at the cost of a larger coil. A low evaporating temperature also lowers COP.

Question 3 · medium

A technician finds a thermostatic expansion valve whose sensing bulb has lost its charge. The system will most likely show

  1. High condensing pressure only
  2. A flooded evaporator, with liquid refrigerant returning to the compressor and frosting the suction line
  3. A starved evaporator with high superheat and low suction pressure, as the valve closes
  4. No change, because the spring alone sets the flow
Answer and working

C. A starved evaporator with high superheat and low suction pressure, as the valve closes

Bulb pressure is the opening force; without it, evaporator pressure and spring close the valve. The evaporator is starved, suction pressure falls and superheat is high. Flooding would follow if the bulb were detached and warm, or the valve stuck open.

Question 4 · hard

2 m³/s of air (density 1.2 kg/m³, humid specific heat 1.02 kJ/kg K) is cooled sensibly from 30 °C to 18 °C. The sensible cooling load is about

  1. 29.4 kW
  2. 120.6 kW
  3. 24.5 kW
  4. 8.4 kW
Answer and working

A. 29.4 kW

Q = V ρ c_p Δt = 2 × 1.2 × 1.02 × 12 = 29.4 kW (≈ 8.4 TR). Using volume without density gives 24.5; using water's c_p gives 120.6.

Question 5 · medium

A freezer wall must limit heat flux to 8 W/m² with a 40 K temperature difference. Neglecting surface resistances, the thickness of insulation with k = 0.024 W/m K required is

  1. 208 mm
  2. 12 mm
  3. 4.8 mm
  4. 120 mm
Answer and working

D. 120 mm

q = kΔT/x → x = kΔT/q = 0.024 × 40/8 = 0.120 m = 120 mm. 12 mm is a metre-to-millimetre slip; 4.8 mm comes from the wrong rearrangement qk/ΔT, and 208 mm from ΔT/(qk) with the units mislabelled.

Question 6 · medium

A store holds 500 t of a vegetable whose tabulated heat of respiration at the storage temperature is taken as 20 mW/kg. The respiration load is

  1. 10000 kW
  2. 10 kW
  3. 240 kW
  4. 0.01 kW
Answer and working

B. 10 kW

Q = 500,000 kg × 0.02 W/kg = 10,000 W = 10 kW. Treating watts as kilowatts gives 10,000; multiplying by 24 turns a rate into a daily energy in the wrong unit. The figure is illustrative: respiration heat varies strongly with commodity, cultivar and temperature. 0.01 kW treats the 500 t as 500 kg (500 × 0.02 W).

Question 7 · medium

A cold store's calculated 24-hour load is 2,000 MJ. With a 10% safety factor and the plant planned to run 20 h a day, the refrigeration capacity to specify is about

  1. 1833 kW (521 TR)
  2. 27.8 kW (7.90 TR)
  3. 30.6 kW (8.69 TR)
  4. 25.5 kW (7.24 TR)
Answer and working

C. 30.6 kW (8.69 TR)

Capacity = 2,000,000 kJ × 1.10/(20 × 3,600 s) = 30.6 kW = 8.69 TR. Dividing by 24 h (25.5 kW) leaves no time for defrost or reserve; omitting the safety factor gives 27.8 kW. 1833 kW divides by minutes instead of seconds (2,200,000/1,200).

Question 8 · medium

Four workers, each taken as releasing 300 W in a freezer, work inside for 8 h a day. The daily occupancy load is

  1. 103,680 kJ
  2. 9,600 kJ
  3. 34,560 kJ
  4. 576.0 kJ
Answer and working

C. 34,560 kJ

E = 4 × 300 W × 8 h × 3,600 s/h = 34,560,000 J = 34,560 kJ (9.6 kWh). 9,600 is watt-hours written as kJ; 24 h assumes the workers never leave. 103,680 kJ assumes 24 h of occupancy, and 576.0 kJ uses minutes instead of seconds (4 × 0.3 kW × 8 × 60).

Question 9 · medium

Under the UN ATP Agreement, a 'heavily insulated' (IR) body for carriage of perishable foods must have an overall heat-transfer coefficient K of not more than

  1. 1.00 W/m² K
  2. 0.70 W/m² K
  3. 0.40 W/m² K
  4. 0.04 W/m² K
Answer and working

C. 0.40 W/m² K

ATP defines normal insulated equipment (IN) as K ≤ 0.70 W/m² K and heavily insulated (IR) as K ≤ 0.40 W/m² K.

Question 10 · hard

A reefer body has K = 0.4 W/m² K and mean surface area 80 m². Frozen cargo is carried at −18 °C with an ambient of 27 °C. The steady heat leak through the body is

  1. 1440 W
  2. 1.44 W
  3. 9000 W
  4. 288 W
Answer and working

A. 1440 W

Q = K A ΔT = 0.4 × 80 × 45 = 1440 W. ΔT is 27 − (−18) = 45 K; using 27 − 18 = 9 K is a sign slip. The refrigeration unit must exceed this with margin for door openings and ageing of insulation. 1.44 W is a kW-to-W slip, and 288 W uses the 9 K sign slip (0.4 × 80 × 9).

A study plan for Refrigeration, Air-Conditioning & Cold-Chain Engineering in Agricultural and Biosystems Engineers Licensure Examination

This plan follows where our questions actually fall, so the time goes where the marks are likely to be. It is our reading of the syllabus, not the exam body's weighting.

  1. First, the heavy units (59% of the 17 questions): Cold-store heat load: transmission, product and respiration (3, 18%); Refrigerated transport and cold-chain logistics (3, 18%); Cold-store heat load: infiltration, occupancy and equipment (2, 12%); COP and tonnes of refrigeration (2, 12%).

    Read the manual chapters for these first and do every worked question you can find on them.

  2. Then the middle band: Psychrometric air-conditioning processes (2, 12%); Condensers and evaporators (1, 6%); Expansion devices (1, 6%); Insulation and vapour barriers (1, 6%).

    Aim for one sitting per unit, then mixed questions across them.

  3. Then the lighter units: Refrigerants: ODP and GWP (1, 6%); Vapour-absorption refrigeration (1, 6%).

    Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.

  4. Leave time for the hard ones: Insulation and vapour barriers (1 hard of 1), Refrigerants: ODP and GWP (1 hard of 1), Refrigerated transport and cold-chain logistics (2 hard of 3).

    Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.

  5. Finish timed. Sit the subject as a timed set without notes, mark it, and send each wrong answer back to its unit above.

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