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.
Machinery Systems: engine and tractor power, fuel consumption and efficiency calculations applied to machinery management; using supplied ASABE standards and data rather than memorised values.
How tractor engines work and how their performance is measured — the single most reliably examined farm-power topic.
Our 17 NCEES PE Agricultural and Biological Engineering questions in this subject: 2 easy, 5 medium, 10 hard, across 8 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.
An ideal cycle produces 0.45 kJ of net work per cycle in a cylinder of 0.5 L swept volume. Its mean effective pressure is
D. 9.0 bar
mep = net work ÷ swept volume = 450 J ÷ 0.0005 m³ = 900,000 Pa = 9.0 bar. 0.9 bar and 90 bar are decimal-place slips in converting pascals to bar (1 bar = 10⁵ Pa); 4.5 bar halves the correct value.
A four-stroke diesel of 4.0 L displacement at 2,000 rev/min draws 0.068 kg/s of air. Ambient air density is 1.18 kg/m³. Its volumetric efficiency is closest to
D. 86.4%
Ideal air flow = ρ × Vd × N/2/60 = 1.18 × 0.004 × 2000/120 = 0.0787 kg/s. ηv = 0.068/0.0787 = 86.4%. Using N instead of N/2 gives 43.2%; ignoring density gives a volume ratio above 100%. Dividing N by 4 gives 172.9%, and ignoring density gives 102.0%.
A naturally aspirated tractor rated at 50 kW at 101.3 kPa and 25 °C works at a hill site where air is at 85 kPa and 15 °C. Assuming full-load power is proportional to air density (R = 287 J/kg K), the expected power is closest to
C. 43.4 kW
ρ = p/(RT): sea level 1.184 kg/m³; site 1.028 kg/m³ (T in kelvin: 298.15 K and 288.15 K). Power ≈ 50 × 1.028/1.184 = 43.4 kW. Using pressure alone gives 42.0 kW; inverting the ratio gives 57.6 kW; 25.2 kW uses Celsius temperatures and multiplies by their ratio (15/25) instead of dividing, which is doubly wrong.
An engine delivers a torque of 200 N m at 2,100 rev/min. Its brake power is
C. 44.0 kW
BP = 2πNT/60 = 2π × 2100 × 200/60 = 43982 W = 44.0 kW. Omitting 2π gives 7.0 kW; omitting /60 gives a per-minute figure; 59.0 is the value in horsepower, not kW. Omitting /60 gives 2639 kW.
Heat-balance test on a diesel: fuel 12 kg/h (CV 42 MJ/kg); BP 45 kW; cooling water 1.0 kg/s heated by 9 °C (c = 4.187 kJ/kg K); air 300 kg/h; exhaust 420 °C, ambient 30 °C, cp of exhaust 1.1 kJ/kg K. The unaccounted heat as a percentage of heat supplied is closest to
C. 14.4%
Supplied = 12 × 42,000/3600 = 140.0 kW. Coolant = 1.0 × 4.187 × 9 = 37.68 kW. Exhaust = (300 + 12)/3600 × 1.1 × (420 − 30) = 37.18 kW. Unaccounted = 140.0 − 45 − 37.68 − 37.18 = 20.14 kW = 14.4%. Leaving out the fuel mass in the exhaust gives 15.4%; using 420 °C without subtracting ambient gives 12.3%. Ignoring the exhaust loss gives (140.0 − 45 − 37.68)/140 = 40.9%.
A tractor works at 35 kW with a BSFC of 260 g/kWh; diesel density is 0.835 kg/L. Its fuel consumption is closest to
C. 10.9 L/h
Mass rate = 0.26 × 35 = 9.1 kg/h. Volume = 9.1/0.835 = 10.9 L/h. Quoting kg/h as litres, or multiplying by density instead of dividing, are the usual errors. 109 L/h is a decimal slip.
For any one tractor, which ordering of power figures is normally correct?
B. Gross engine power > net engine power > PTO power > drawbar power
Each step loses power: installed accessories (net), transmission to the PTO, and at the drawbar wheel slip and rolling resistance as well.
During a test mode an engine delivering 45 kW emits 180 g of NOx per hour. The brake-specific NOx emission is
D. 4.0 g/kWh
Brake-specific emission = mass rate ÷ power = 180 g/h ÷ 45 kW = 4.0 g/kWh, the unit in which non-road limits are written. Inverting gives 0.25; dividing by 3.6 wrongly converts to g/MJ-style units. Multiplying gives 8100 g/kWh.
An engine with DOC, DPF and SCR after-treatment must run on ultra-low-sulphur diesel mainly because
B. sulphur poisons the catalysts and forms sulphate particulate
Sulphur compounds deactivate catalyst sites and add sulphate to the particulate; high-sulphur fuel quickly degrades after-treatment. Low-ash oil is also needed to avoid clogging the DPF.
Emission limits for agricultural tractor engines in the US, EU and India are expressed and tested in which way?
C. As brake-specific limits (g/kWh) by power band, tested on steady-state multi-mode and later transient cycles
Non-road rules set g/kWh limits by power band, measured on dynamometer cycles such as the ISO 8178 C1 eight-mode cycle, with a transient cycle added in later stages. Road vehicles use g/km; smoke opacity is only one test. Limit values differ by country and stage.
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.
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