NCEES FE Other Disciplines · USA · NCEES
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 NCEES FE Other Disciplines spread across the units of Engineering Thermodynamics. Difficulty: 20 easy, 47 medium, 23 hard.
| Properties of steam, steam tables and the Mollier chart | 13 | |
| Air-standard cycles: Otto, Diesel, dual and Brayton | 12 | |
| Introduction to psychrometry | 12 | |
| Ideal gas processes | 10 | |
| Steady-flow energy equation | 10 | |
| Vapour-compression cycle basics | 10 | |
| Second law, entropy and the Carnot cycle | 9 | |
| First law of thermodynamics | 8 | |
| Systems, properties and the zeroth law | 6 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A closed system receives 150 kJ of heat and does 60 kJ of work on its surroundings. The change in its internal energy is
C. +90 kJ
ΔU = Q − W = 150 − 60 = +90 kJ (heat in positive, work by the system positive). +210 kJ adds the work instead of subtracting it, −90 kJ reverses both signs, and +60 kJ confuses work with energy change.
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
Steam enters an adiabatic nozzle at negligible velocity and its enthalpy falls by 200 kJ/kg. The exit velocity is about
C. 632 m/s
V2 = √(2Δh) with Δh in J/kg: √(2 × 200 000) = 632 m/s. 447 m/s drops the factor 2, 20.0 m/s forgets to convert kJ to J, and 14.1 m/s makes both mistakes.
Further reading: Çengel, Boles and Kanoğlu — Thermodynamics: An Engineering Approach
A pump raises the pressure of 20 kg/s of water (v = 0.001 m³/kg) by 300 kPa, with negligible changes in velocity, height and temperature. The ideal power required is
B. 6 kW
For an incompressible liquid w = vΔP = 0.001 × 300 = 0.3 kJ/kg; power = 20 × 0.3 = 6 kW. 6000 kW adds a spurious factor of 1000 (kPa × m³/kg already gives kJ/kg), 0.3 kW is the work per kg without the mass flow, and 0.6 kW is a misplaced decimal.
Further reading: Çengel, Boles and Kanoğlu — Thermodynamics: An Engineering Approach
For a reversible process, the area under the process curve on a T–s diagram represents
D. the heat transferred during the process
For a reversible process δQ = T ds, so ∫T ds is the heat. Work is the area under a P–v curve; enthalpy and internal energy changes are not, in general, areas under a T–s line.
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
On a Mollier (h–s) chart, constant-pressure lines inside the wet region are straight because
C. their slope ∂h/∂s at constant pressure equals the saturation temperature, which is constant there
From T ds = dh − v dP, at constant pressure dh/ds = T; in the wet region T = Tsat is constant, so the lines are straight. Enthalpy and entropy both change with quality during boiling, and the Mollier chart uses linear axes.
Further reading: IAPWS-IF97 steam properties; any standard steam table
In the Diesel cycle the cut-off ratio is defined as
D. the volume at the end of heat addition divided by the volume at its start
Cut-off ratio rc = V3/V2, the expansion during constant-pressure heat addition. Total volume over clearance volume is the compression ratio, the pressure ratio describes the constant-volume part of a dual cycle, and the last ratio has no standard name.
Further reading: Çengel, Boles and Kanoğlu — Thermodynamics: An Engineering Approach
If the evaporator temperature of a cold store is lowered while the condensing temperature stays the same, the COP of the plant
B. decreases, because compressor work per kg rises and refrigerating effect falls
A lower evaporator pressure means a larger pressure ratio (more work per kg) and, usually, a smaller refrigerating effect, so COP falls — as the Carnot COP TL/(TH − TL) also shows. Suction vapour density actually falls, reducing capacity.
Further reading: Stoecker and Jones — Refrigeration and Air Conditioning
Air at 30 °C has a vapour pressure of 2.0 kPa. The saturation pressure of water at 30 °C is 4.247 kPa. The relative humidity is about
C. 47.1 %
φ = pv/psat = 2.0/4.247 = 47.1 %. 2.0 % divides by barometric pressure, 212 % inverts the ratio, and 46.0 % is the degree of saturation W/Ws, which is close to but not the same as relative humidity.
Further reading: ASHRAE Handbook — Fundamentals, Psychrometrics
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