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NCEES PE Agricultural and Biological Engineering Post-Harvest Engineering of Cereals, Pulses & Oilseeds: practice questions

NCEES PE Agricultural and Biological Engineering · USA · NCEES (ASABE is the content partner)

Original practice questions written to the published syllabus · not past papers
10worked here, free
22written to this exam in this subject
7syllabus units

About NCEES PE Agricultural and Biological Engineering

NCEES 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.

Post-Harvest Engineering of Cereals, Pulses & Oilseeds in NCEES PE Agricultural and Biological Engineering

Processing Systems (12–18 questions) covers mass transfer and drying, psychrometrics, bulk solids and properties of biological materials, with ASAE D271 (psychrometric data) and D272 (airflow resistance of grains) supplied. Common System Applications covers materials handling and energy and mass balances. Expect applied problems in SI or US customary units: water removal and shrink, dryer energy, fan airflow and static pressure, conveyor capacity and power.

Moisture, drying, parboiling, milling, cleaning, size reduction and conveying of grains, pulses and oilseeds.

Our 22 NCEES PE Agricultural and Biological Engineering questions in this subject: 0 easy, 12 medium, 10 hard, across 7 units.

10 practice questions, worked

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.

Question 1 · medium

An elevator receives 50.0 t of maize at 18% wet-basis moisture and dries it to 15% wet basis. Ignoring handling losses, the dried mass is

  1. 48.50 t
  2. 51.83 t
  3. 48.17 t
  4. 48.24 t
Answer and working

D. 48.24 t

Dry matter = 50 × 0.82 = 41.0 t; dried mass = 41.0/0.85 = 48.24 t. 48.50 t subtracts 3% of the initial mass; 48.17 t divides the moisture drop by (1 − M_initial) instead of (1 − M_final); 51.83 t inverts the ratio and gives a mass larger than the wet lot.

Question 2 · medium

Air at 101.325 kPa has a water-vapour partial pressure of 2.0 kPa. Its humidity ratio is closest to

  1. 0.0200 kg/kg dry air
  2. 0.0125 kg/kg dry air
  3. 0.0123 kg/kg dry air
  4. 0.0201 kg/kg dry air
Answer and working

B. 0.0125 kg/kg dry air

W = 0.622 p_v/(P − p_v) = 0.622 × 2.0/99.325 = 0.01252. Dividing by total pressure instead of dry-air pressure gives 0.0123; omitting the 0.622 molar-mass ratio gives 0.0201.

Question 3 · hard

The humid volume of air at 30 °C, humidity ratio 0.015 kg/kg dry air and 101.325 kPa is closest to

  1. 0.872 m³/kg dry air
  2. 0.087 m³/kg dry air
  3. 0.879 m³/kg dry air
  4. 0.859 m³/kg dry air
Answer and working

C. 0.879 m³/kg dry air

v = 0.287 T(1 + 1.608 W)/P = 0.287 × 303.15 × 1.0241/101.325 = 0.8794 m³/kg dry air. Ignoring the vapour term gives the dry-air value 0.859; using (1 + W) instead of (1 + 1.608 W) gives 0.872; using °C instead of K gives an absurdly small volume.

Question 4 · hard

Recirculated air (2 kg/s dry air, 35 °C, W = 0.018 kg/kg) mixes adiabatically with ambient air (3 kg/s dry air, 20 °C, W = 0.008 kg/kg). The humidity ratio and temperature of the mixture are closest to

  1. W = 0.0120 kg/kg, t = 27.5 °C
  2. W = 0.0130 kg/kg, t = 27.5 °C
  3. W = 0.0140 kg/kg, t = 29.0 °C
  4. W = 0.0120 kg/kg, t = 26.1 °C
Answer and working

D. W = 0.0120 kg/kg, t = 26.1 °C

Mass-weighted: W₃ = (2×0.018 + 3×0.008)/5 = 0.0120. Enthalpies: h₁ = 81.40, h₂ = 40.43 kJ/kg; h₃ = (2h₁ + 3h₂)/5 = 56.82; t₃ = (h₃ − 2501W₃)/(1.006 + 1.86W₃) = 26.07 °C. A simple average (27.5 °C) ignores the unequal flows; swapping the weights gives the wrong W.

Question 5 · hard

A dryer fan delivers 2.0 kg/s of dry air with W = 0.012 kg/kg, heated from 25 °C to 60 °C. The heat input required is closest to

  1. 70.4 kW
  2. 72.0 kW
  3. 36.0 kW
  4. 123.4 kW
Answer and working

B. 72.0 kW

Q = m_a·c_s·Δt with humid heat c_s = 1.006 + 1.86 × 0.012 = 1.0283 kJ/kg K, so Q = 2.0 × 1.0283 × 35 = 71.98 kW. Using the dry-air specific heat alone gives 70.4 kW; using 60 °C rather than the 35 K rise overstates it. 36.0 kW omits the 2.0 kg/s air flow (1.0283 × 35); 123.4 kW uses 60 K.

Question 6 · hard

A dryer uses 1.5 kg/s of dry air. Air enters the grain at humidity ratio 0.010 kg/kg and leaves at 0.018 kg/kg. The moisture removed from the grain is

  1. 54.0 kg/h
  2. 43.2 kg/h
  3. 97.2 kg/h
  4. 28.8 kg/h
Answer and working

B. 43.2 kg/h

Water picked up = m_a(W₂ − W₁) = 1.5 × 0.008 = 0.012 kg/s = 43.2 kg/h. Using the outlet W alone counts water the air already carried; 28.8 kg/h forgets the air mass flow. 54.0 kg/h uses the inlet W alone (1.5 × 0.010 × 3600); 97.2 kg/h uses the outlet W alone.

Question 7 · hard

A dryer evaporates 120 kg of water per hour while the burner supplies 600 MJ/h. Taking latent heat as 2.4 MJ/kg, the thermal efficiency and specific heat consumption are

  1. 48% and 0.2 MJ/kg water
  2. 208% and 5.0 MJ/kg water
  3. 20% and 5.0 MJ/kg water
  4. 48% and 5.0 MJ/kg water
Answer and working

D. 48% and 5.0 MJ/kg water

Heat used for evaporation = 120 × 2.4 = 288 MJ/h; efficiency = 288/600 = 48%. Specific heat consumption = 600/120 = 5.0 MJ/kg water, about twice the latent-heat minimum. Dividing kg by MJ (20%) ignores the latent heat; the inverted ratio exceeds 100%.

Question 8 · medium

In ASABE D272, grain airflow resistance is expressed in the Hukill–Ives form ΔP/L = aQ²/ln(1 + bQ). For a bin fan design, this means that

  1. static pressure is independent of grain depth
  2. fines in the grain lower the static pressure by opening up more space between the kernels
  3. static pressure rises more than proportionally with airflow, and in proportion to depth
  4. doubling airflow halves the static pressure
Answer and working

C. static pressure rises more than proportionally with airflow, and in proportion to depth

ΔP/L grows faster than linearly with superficial velocity Q, and ΔP scales with depth L. Fines fill voids and raise resistance, which is why cleaning before drying helps.

Question 9 · hard

A bucket elevator has 2.0 L buckets filled to 75%, spaced 0.30 m apart on a belt running at 2.0 m/s, handling wheat of bulk density 750 kg/m³. Its capacity is

  1. 36.0 t/h
  2. 7.50 t/h
  3. 27.0 t/h
  4. 4.05 t/h
Answer and working

C. 27.0 t/h

Buckets per second = 2.0/0.30 = 6.667; mass per bucket = 0.002 × 0.75 × 750 = 1.125 kg; capacity = 7.50 kg/s = 27.0 t/h. Ignoring the fill fraction gives 36.0 t/h; 7.50 is in kg/s mislabelled as t/h. 4.05 t/h assumes one bucket per second (1.125 kg × 3.6).

Question 10 · medium

A troughed belt conveyor carries a grain load of cross-section 0.020 m² at 2.5 m/s. With bulk density 780 kg/m³, its capacity is

  1. 39.0 t/h
  2. 2.34 t/h
  3. 140.4 t/h
  4. 22.5 t/h
Answer and working

C. 140.4 t/h

Q = ρAv = 780 × 0.020 × 2.5 = 39.0 kg/s = 140.4 t/h. 39.0 is kg/s, not t/h; dividing by the speed is dimensionally wrong. 2.34 t/h treats 39.0 kg/s as kg/min (39 × 60/1000); 22.5 t/h divides by the speed.

A study plan for Post-Harvest Engineering of Cereals, Pulses & Oilseeds in NCEES PE Agricultural and Biological Engineering

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.

  1. First, the heavy units (73% of the 22 questions): Psychrometry and properties of air (7, 32%); Conveying: belt, bucket, screw and pneumatic (6, 27%); Grain dryers: bin, LSU, recirculating, fluidised-bed and solar (3, 14%).

    Read the manual chapters for these first and do every worked question you can find on them.

  2. Then the middle band: Moisture content: wet and dry basis (2, 9%); Size reduction: Rittinger, Kick and Bond laws (2, 9%); Drying theory: constant and falling rate periods (1, 5%).

    Aim for one sitting per unit, then mixed questions across them.

  3. Then the lighter units: Thin-layer drying models (1, 5%).

    Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.

  4. Leave time for the hard ones: Drying theory: constant and falling rate periods (1 hard of 1), Thin-layer drying models (1 hard of 1), Psychrometry and properties of air (4 hard of 7).

    Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.

  5. Finish timed. Sit the subject as a timed set without notes, mark it, and send each wrong answer back to its unit above.

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