NCEES FE Other Disciplines · USA · NCEES
Original practice questions written to the published syllabus · not past papersConduction, convection, radiation, heat exchangers and diffusion — the physics of drying, cooling, freezing and cold-store walls.
In the exams: GATE AG Section 7 (heat exchangers and composite walls each asked in 7 of 8 papers 2019–26); ICAR AICE subject 41 Unit 2; NCEES PE processing
How our questions for NCEES FE Other Disciplines spread across the units of Heat & Mass Transfer. Difficulty: 14 easy, 28 medium, 0 hard.
| Radiation: Stefan-Boltzmann law and view factors | 8 | |
| Convection and dimensionless numbers (Nu, Re, Pr) | 7 | |
| Heat exchangers: LMTD method | 5 | |
| Steady conduction through plane walls and cylinders | 4 | |
| Extended surfaces (fins) | 4 | |
| Heat exchangers: effectiveness-NTU method | 4 | |
| Composite walls and critical radius of insulation | 3 | |
| Transient conduction: lumped capacity and Heisler charts | 2 | |
| Condensation and boiling | 2 | |
| Diffusive mass transfer and Fick's law | 2 | |
| Convective mass transfer | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A steam pipe of outer radius 50 mm is covered with insulation (k = 0.05 W/m·K) to an outer radius of 100 mm. The inner and outer surfaces of the insulation are at 150 °C and 30 °C. The heat loss per metre length is closest to
B. 54.4 W/m
For a cylinder Q/L = 2πk(T1 − T2)/ln(r2/r1) = 2π × 0.05 × 120 / ln 2 = 54.4 W/m. Treating the layer as a plane wall of thickness 0.05 m with the inner area gives 37.7, with the outer area 75.4, and with the arithmetic-mean area 56.5 W/m; only the logarithmic-mean area makes the plane-wall formula exact.
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
Water (ν = 1.0 × 10⁻⁶ m²/s) flows at 0.5 m/s through a tube of 25 mm internal diameter. The Reynolds number and flow regime are
D. 12,500; turbulent
Re = VD/ν = 0.5 × 0.025/1.0 × 10⁻⁶ = 12,500. Pipe flow is laminar below about 2,300, so this is turbulent. Entering D as 2.5 mm gives 1,250; using the radius gives 6,250; calling Re = 12,500 laminar confuses the pipe limit with the flat-plate limit of about 5 × 10⁵.
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
Adding fins to a heat-exchange surface is most worthwhile when the surface transfers heat to
C. a gas with a low convection coefficient
Fin effectiveness √(kP/(hA_c)) is large when h is small. Fins are therefore used on the air side of radiators, condensers and engine cylinders. With boiling, condensation or liquid metals, h is so high that the fin's own conduction resistance limits it and effectiveness is poor.
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
For a straight fin with an insulated tip, mL = 2.0. Its fin efficiency is closest to
A. 0.482
η_f = tanh(mL)/(mL) = 0.9640/2 = 0.482. 0.964 is tanh(mL) alone, the ratio of actual heat to that of an infinitely long fin; e^(−mL) = 0.135 is the tip excess-temperature ratio of an infinite fin; 1/(1 + mL) = 0.333 is not a fin relation at all.
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
A 1-shell-pass, 2-tube-pass exchanger has U = 500 W/m²·K and A = 10 m². The counterflow LMTD for its terminal temperatures is 40 °C and the correction factor from the chart is F = 0.90. Its duty is
A. 180 kW
Q = UAF·ΔT_lm,cf = 500 × 10 × 0.90 × 40 = 180000 W = 180 kW. Omitting F gives 200 kW; dividing by F gives 222 kW. F ≤ 1 because multipass flow is less effective than pure counterflow. 162 kW applies F twice.
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
The maximum possible heat-transfer rate used in defining heat-exchanger effectiveness is
A. C_min(T_h,in − T_c,in)
Only the fluid with the smaller heat-capacity rate can undergo the full temperature difference between the two inlets, which it would do in an infinitely long counterflow exchanger. Using C_max would require the other fluid to change temperature by more than the inlet difference, which violates the second law. UA × LMTD is the actual rate, not the maximum.
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
As the excess temperature of a heated surface in a pool of saturated water is increased, the boiling regimes follow the order
A. natural convection → nucleate → transition → film boiling
At small ΔT_e heat moves by natural convection; bubbles then form at nucleation sites (nucleate boiling) up to the critical heat flux; beyond it, unstable vapour patches (transition boiling) reduce the flux to a minimum at the Leidenfrost point; above that, a stable vapour film covers the surface (film boiling).
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
For laminar film condensation on a vertical plate, Nusselt's analysis gives h ∝ L^(−1/4). If the plate height is doubled with all else the same, the average coefficient becomes
D. 0.841 times the original
h ∝ L^(−1/4), so h_new/h_old = 2^(−0.25) = 0.841. The thicker film on the lower part of a taller plate lowers the average coefficient, but only weakly; the total heat flow still rises, because the area doubles (1.68 times).
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
A solute diffuses steadily through a stagnant liquid layer 2 mm thick with D = 2 × 10⁻⁹ m²/s. Its concentrations on the two faces differ by 0.5 kmol/m³. The molar flux is
B. 5.00 × 10⁻⁷ kmol/m²·s
N = DΔC/L = 2 × 10⁻⁹ × 0.5/0.002 = 5.0 × 10⁻⁷ kmol/m²·s, the mass analogue of Q = kAΔT/L. Entering L as 2 (mm) gives 5 × 10⁻¹⁰; multiplying by L gives 2 × 10⁻¹²; omitting ΔC gives 1 × 10⁻⁶ with the wrong units.
Further reading: Incropera et al., Fundamentals of Heat and Mass Transfer
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