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 papersEnergy, work, heat and the cycles behind engines, refrigerators and dryers.
In the exams: GATE AG Section 3 (thermodynamic principles of IC engines); ICAR AIEEA-PG Unit III; NCEES FE thermodynamics and heat transfer 9–14 questions
How our questions for Agricultural and Biosystems Engineers Licensure Examination spread across the units of Engineering Thermodynamics. Difficulty: 0 easy, 15 medium, 7 hard.
| Steady-flow energy equation | 5 | |
| Properties of steam, steam tables and the Mollier chart | 4 | |
| Introduction to psychrometry | 4 | |
| First law of thermodynamics | 2 | |
| Ideal gas processes | 2 | |
| Air-standard cycles: Otto, Diesel, dual and Brayton | 2 | |
| Vapour-compression cycle basics | 2 | |
| Systems, properties and the zeroth law | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
2 kg of air is heated at constant pressure from 20 °C to 80 °C (cp = 1.005, cv = 0.718 kJ/kg·K). The heat supplied is
C. 120.6 kJ
At constant pressure Q = m cp ΔT = 2 × 1.005 × 60 = 120.6 kJ. 86.2 kJ uses cv (that is ΔU only), 34.4 kJ is only the boundary work m R ΔT, and 160.8 kJ uses the final temperature instead of the rise.
Further reading: Çengel, Boles and Kanoğlu — Thermodynamics: An Engineering Approach
An insulated tank holds 50 kg of water that a 3 kW electric heater must warm from 20 °C to 90 °C for washing dairy equipment (c = 4.18 kJ/kg·K; neglect losses). The heating time is about
B. 81.3 min
Q = m c ΔT = 50 × 4.18 × 70 = 14630 kJ; t = Q/P = 14630/3 = 4877 s = 81.3 min. 421.7 min uses the absolute final temperature (363.15 K) instead of the rise, 1.35 min treats hours as minutes, and 19.4 min uses c = 1 (a kcal habit) instead of 4.18 kJ/kg·K.
Further reading: Çengel, Boles and Kanoğlu — Thermodynamics: An Engineering Approach
A pasteuriser heats 1000 kg/h of milk from 4 °C to 72 °C. Taking the specific heat of the milk as 3.9 kJ/kg·K, the heating rate needed is
C. 73.7 kW
Q̇ = ṁ c ΔT = (1000/3600) × 3.9 × 68 = 73.7 kW. 79.0 kW uses water's 4.18 instead of the stated 3.9, 4420 kW converts per hour as though it were per minute, and 78.0 kW uses the outlet temperature instead of the rise.
Further reading: Singh and Heldman — Introduction to Food Engineering
In an insulated blancher feed tank, 2 kg/s of water at 80 °C mixes steadily with 3 kg/s of water at 20 °C. The outlet temperature is
D. 44 °C
Energy balance with constant c: T = (2 × 80 + 3 × 20)/5 = 44 °C. 50 °C is the simple average that ignores flow rates, 56 °C swaps the flow rates, and 110 °C divides by one stream's flow instead of the total.
Further reading: Çengel, Boles and Kanoğlu — Thermodynamics: An Engineering Approach
In an ideal Rankine cycle, steam enters the turbine at 3 MPa and 400 °C (h = 3231.6 kJ/kg) and leaves at 10 kPa with h = 2193.0 kJ/kg. Saturated liquid at 10 kPa has h = 191.8 kJ/kg and v = 0.00101 m³/kg. The cycle efficiency is about
D. 34.1 %
Pump work = vΔP = 0.00101 × (3000 − 10) = 3.02 kJ/kg; h after pump = 194.8; q_in = 3036.7; turbine work = 1038.6; η = (1038.6 − 3.02)/3036.7 = 34.1 %. 32.0 % divides by h1 instead of the heat added, 52.6 % is the Carnot limit between the extreme temperatures, and the negative value comes from multiplying the pump work by 1000.
Further reading: IAPWS-IF97 steam properties; any standard steam table
In an ideal vapour-compression cycle, the enthalpies are 395 kJ/kg at evaporator exit, 425 kJ/kg at compressor exit and 250 kJ/kg at condenser exit. The COP is
D. 4.83
Refrigerating effect = 395 − 250 = 145 kJ/kg; compressor work = 425 − 395 = 30 kJ/kg; COP = 145/30 = 4.83. 5.83 is the heat-pump COP (condenser heat 175/30), 0.207 is the inverse, and 8.33 uses the condenser-exit enthalpy as the refrigerating effect.
Further reading: Stoecker and Jones — Refrigeration and Air Conditioning
A grain dryer heats 1.5 kg/s of dry air (humidity ratio 0.012 kg/kg) from 30 °C to 60 °C. Taking cp of dry air 1.005 and of vapour 1.88 kJ/kg·K, the heater duty is about
D. 46.2 kW
Sensible heating at constant W: Q = ṁa (1.005 + 1.88W)ΔT = 1.5 × 1.0276 × 30 = 46.2 kW. 45.2 kW ignores the vapour, 92.5 kW uses the final temperature instead of the rise, and 91.3 kW wrongly adds latent heat (1.5 × 0.012 × 2501 = 45.0 kW), although no water evaporates in a heater.
Further reading: Brooker, Bakker-Arkema and Hall — Drying and Storage of Grains and Oilseeds
Air (1.5 kg/s of dry air) enters a dryer's grain bed with a humidity ratio of 0.012 kg/kg and leaves with 0.020 kg/kg. The water removed from the grain is about
B. 43.2 kg/h
Water removed = ṁa (W2 − W1) = 1.5 × 0.008 = 0.012 kg/s = 43.2 kg/h. 108 kg/h counts all the exit moisture including what the air brought in, 0.72 kg/h converts per minute instead of per hour, and 172.8 kg/h adds the humidity ratios.
Further reading: Brooker, Bakker-Arkema and Hall — Drying and Storage of Grains and Oilseeds
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