NCEES PE Agricultural and Biological Engineering · USA · NCEES (ASABE is the content partner)
Original practice questions written to the published syllabus · not past papersNCEES PE Agricultural and Biological Engineering is a licensing and registration exam run by NCEES (ASABE is the content partner), for candidates in USA. Computer-based, 85 questions, 9.5-hour appointment (8.5 hours of exam). Once a year (next: 28 Oct 2026). Closed book with the electronic NCEES handbook and supplied searchable ASABE standards; SI and US customary units.
Source: ncees.org.
Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.
Conduction, convection, radiation, heat exchangers and diffusion — the physics of drying, cooling, freezing and cold-store walls.
Our 26 NCEES PE Agricultural and Biological Engineering questions in this subject: 2 easy, 14 medium, 10 hard, across 12 units.
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 NCEES PE Agricultural and Biological Engineering and are not shown on any other exam's page.
A 1 m² section of an animal-house wall is 100 mm thick. Timber studs (k = 0.12 W/m·K) occupy 10% of the area and insulation (k = 0.04 W/m·K) the remaining 90%. Using the parallel-path model and ignoring surface films, the heat flow for a 20 °C difference across the wall is
B. 9.6 W
Parallel paths: conductances add. UA = (0.1 × 0.12 + 0.9 × 0.04)/0.1 = 0.48 W/K, so Q = 0.48 × 20 = 9.6 W. Ignoring the studs gives 8.0 W, a 17% underestimate of the thermal bridging; averaging the two conductivities equally gives 16.0 W; an all-timber wall would give 24.0 W.
A cold-room panel has 150 mm of foam insulation (k = 0.03 W/m·K), with an inside film coefficient of 10 W/m²·K and an outside coefficient of 25 W/m²·K. Neglecting the thin metal skins, the overall heat-transfer coefficient U is
D. 0.195 W/m²·K
1/U = 1/10 + 0.15/0.03 + 1/25 = 0.1 + 5.0 + 0.04 = 5.14 m²·K/W, so U = 0.195 W/m²·K. 0.200 ignores the films (a small error here because the insulation dominates); 5.14 is the total resistance reported as if it were U; adding the coefficients directly treats series resistances as if they were in parallel. Adding the terms that way gives 35.2 W/m²·K.
An apple (radius 40 mm, α = 1.4 × 10⁻⁷ m²/s) at 25 °C is hydro-cooled in water at 1 °C. For its Biot number, the one-term sphere solution has ζ1 = 1.5708 and C1 = 1.2732, so that θ_centre/θ_i = C1 exp(−ζ1² Fo). The centre temperature after 1 hour is closest to
B. 15.0 °C
Fo = αt/r² = 1.4 × 10⁻⁷ × 3600/0.04² = 0.315 (> 0.2, so one term is enough). θ/θ_i = 1.2732 × exp(−1.5708² × 0.315) = 0.585. T = 1 + 0.585 × 24 = 15.0 °C. Omitting C1 gives 12.0 °C; not squaring ζ1 gives 19.6 °C; multiplying the ratio by the initial temperature (25 °C) instead of using the excess temperature gives 14.6 °C.
On a finned evaporator coil, fins make up 90% of the total air-side area and have a fin efficiency of 0.80. The overall surface efficiency η_o = 1 − (A_f/A)(1 − η_f) is
D. 0.82
η_o = 1 − 0.9 × (1 − 0.80) = 0.82. The unfinned base (10%) works at full efficiency, so η_o is slightly above η_f. 0.80 ignores the base area; 0.72 multiplies the fractions; 0.98 swaps the fin and base fractions.
Milk (2.0 kg/s, c = 3.9 kJ/kg·K) is cooled from 35 °C to 10 °C in a counterflow plate cooler using chilled water (3.0 kg/s, c = 4.18 kJ/kg·K) entering at 2 °C. The overall coefficient is 1,500 W/m²·K. The heat-transfer area required is closest to
D. 10.7 m²
Duty Q = 2.0 × 3.9 × 25 = 195 kW. Water outlet = 2 + 195/(3.0 × 4.18) = 17.55 °C. Counterflow: ΔT1 = 35 − 17.55 = 17.45 °C, ΔT2 = 10 − 2 = 8 °C; LMTD = 12.12 °C. A = Q/(U·LMTD) = 195000/(1500 × 12.12) = 10.7 m². The arithmetic mean (12.72 °C) gives 10.2 m²; 7.53 m² follows from no consistent terminal pairing (the parallel-flow pairing is impossible here, because the water would leave hotter than the milk); leaving Q in kW with U in W/m²·K gives 0.0107 m².
The regeneration section of a milk pasteuriser is a counterflow plate exchanger in which raw milk (entering at 5 °C) and pasteurised milk (entering at 72 °C) have equal heat-capacity rates. The section has NTU = 4. The raw milk leaves the section at about
B. 58.6 °C
Counterflow with C_r = 1: ε = NTU/(1 + NTU) = 4/5 = 0.80. Raw milk outlet = 5 + 0.80 × (72 − 5) = 58.6 °C; the regeneration efficiency is 80%. Parallel flow with C_r = 1 could never exceed ε = 0.5 (38.5 °C); 1 − e^(−NTU) applies only when C_r = 0 (70.8 °C); applying ε to 72 °C instead of to the inlet difference (72 − 5) gives 62.6 °C.
A balanced counterflow heat-recovery exchanger (C_r = 1) must reach an effectiveness of 0.75. C_min = 5,000 W/K and U = 1,000 W/m²·K. The area needed is
A. 15 m²
For counterflow with C_r = 1, ε = NTU/(1 + NTU), so NTU = ε/(1 − ε) = 0.75/0.25 = 3. A = NTU × C_min/U = 3 × 5,000/1,000 = 15 m². The C_r = 0 relation gives NTU = −ln(0.25) = 1.386 and 6.93 m²; treating ε as NTU gives 3.75 m²; inverting the ratio gives 0.6 m².
A packaging film of a given polymer has a water-vapour transmission rate of 10 g/m²·day at 25 µm thickness under stated test conditions. At the same temperature and humidity difference, a 50 µm film of the same polymer would transmit about
A. 5 g/m²·day
For steady diffusion through a homogeneous film, flux = permeability × Δp/thickness, so doubling thickness halves the transmission rate. 2.5 would follow a thickness-squared law, which applies to transient diffusion time, not steady flux. Real films can deviate (pinholes, multilayer structures), so measured values are preferred. 20 g/m²·day doubles instead of halving; 10 g/m²·day ignores the thickness change.
In the constant-rate period, air at 60 °C (wet-bulb 30 °C) flows over a wet product with h = 40 W/m²·K. Taking h_fg = 2,430 kJ/kg at the wet-bulb temperature, the drying rate is closest to
A. 1.78 kg/m²·h
R_c = h(T_a − T_wb)/h_fg = 40 × 30/2.43 × 10⁶ = 4.94 × 10⁻⁴ kg/m²·s = 1.78 kg/m²·h. Using T_a instead of the difference gives 3.56; using h_fg in kJ/kg gives a value 1,000 times too large; the kg/s figure labelled per hour is a time-unit slip. That value is 1778 kg/m²·h.
Air at 30 °C and 3 m/s flows across a 0.4 m long flat plate at 50 °C (film properties: ν = 1.7 × 10⁻⁵ m²/s, k = 0.027 W/m·K, Pr = 0.71). The flow is laminar throughout. The heat lost from one side of a 0.4 m × 1 m plate is closest to
A. 85.0 W
Re = 3 × 0.4/1.7 × 10⁻⁵ = 70,600 (laminar). Nu = 0.664 × 265.7 × 0.71^(1/3) = 157.4; h = 157.4 × 0.027/0.4 = 10.62 W/m²·K. Q = hAΔT = 10.62 × 0.4 × 20 = 85.0 W. The local coefficient (0.332) gives half; using the plate temperature (50) instead of the difference gives 212 W; using Nu in place of h gives 1259 W. Half is 42.5 W.
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.
Read the manual chapters for these first and do every worked question you can find on them.
Aim for one sitting per unit, then mixed questions across them.
Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.
Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.
Do these early in revision so they are fast on the day.
Last updated · How Laisha writes and checks its pages · Report a mistake
US$10 a month · one price worldwide · cancel any time
| Free | Smarter | |
|---|---|---|
| Grow guides, pest pages, articles | Read all | Read all |
| Ask Laisha | 5 a day | Up to 1,000 a month |
| Exam practice | The sample on each page | Every question, timed papers |
| Crop calendar and pest alerts | — | For your own fields |
| Calculators and tools | — | All |
| Uploads and reports | — | 100 a month, real files |
An AI operating system for agriculture: free guides, manuals and tools written for every country.