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NCEES PE Agricultural and Biological Engineering Post-Harvest Engineering of Horticultural Crops: 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
23written to this exam in this subject
8syllabus 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 Horticultural Crops in NCEES PE Agricultural and Biological Engineering

Processing Systems and Facilities: applied problems on cooling load and refrigeration capacity, forced-air and hydrocooling design, psychrometrics of cold rooms, water removal in drying, and energy balances, in SI or US customary units.

Keeping fruits and vegetables alive and saleable after harvest: cooling, handling, packhouses, controlled atmospheres and processing.

Our 23 NCEES PE Agricultural and Biological Engineering questions in this subject: 2 easy, 6 medium, 15 hard, across 8 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 · hard

A commodity has a transpiration coefficient of 42 mg per kg per second per MPa of vapour pressure deficit. It is stored at 10 degC in air at 90% relative humidity, and the product is at air temperature (saturation vapour pressure at 10 degC = 1.228 kPa). Its water loss per day is about

  1. 446 g per kg
  2. 0.45 g per kg
  3. 4.46 g per kg
  4. 0.019 g per kg
Answer and working

B. 0.45 g per kg

VPD = 1.228 x (1 - 0.90) = 0.1228 kPa = 1.228 x 10^-4 MPa. Loss = 42 x 1.228e-4 x 86 400 s = 446 mg/kg per day = 0.45 g per kg (about 0.045% per day). Using the full saturation pressure instead of the deficit gives 4.46 g; 446 g confuses mg with g. 0.019 g per kg is the loss per hour (x 3 600 s), not per day.

Question 2 · medium

Peaches at 30 degC are hydrocooled in water at 0 degC. Their half-cooling time is 20 minutes and cooling is exponential with no lag. Their temperature after 50 minutes is about

  1. 5.3 degC
  2. 3.8 degC
  3. 7.5 degC
  4. 11.2 degC
Answer and working

A. 5.3 degC

Y = 0.5^(t/Z) = 0.5^(50/20) = 0.5^2.5 = 0.1768; T = 0 + 30 x 0.1768 = 5.3 degC. 7.5 degC is the value after 40 min (two half-cooling times) and 3.8 degC after 60 min (three); 11.2 degC does not follow the exponential cooling law at all.

Question 3 · hard

Cartons of fruit at 28 degC are cooled in air at 1 degC and reach 10 degC after 90 minutes. Assuming exponential cooling with a lag factor of 1, the half-cooling time is about

  1. 45 min
  2. 131 min
  3. 57 min
  4. 30 min
Answer and working

C. 57 min

Y = (10 - 1)/(28 - 1) = 9/27 = 0.333. C = -ln(0.333)/90 = 0.01221 per min. Z = ln 2 / C = 0.693/0.01221 = 57 min. Halving 90 min assumes the product reached half-cooling at 90 min, but it has actually gone past it (Y = 1/3); 30 min divides 90 min by 3, wrongly reading Y = 1/3 as three half-cooling periods. Halving 90 min gives 45 min; 131 min uses log10 for C but ln 2 for Z.

Question 4 · hard

Twenty tonnes of table grapes (specific heat 3.6 kJ/kg K) are to be forced-air cooled from 30 degC to 5 degC in 6 hours. The average product (sensible) heat load, ignoring respiration and other gains, is about

  1. 5000 kW
  2. 83.3 kW
  3. 3.33 kW
  4. 500 kW
Answer and working

B. 83.3 kW

Q = m cp dT = 20 000 x 3.6 x 25 = 1 800 000 kJ. Over 6 h = 21 600 s: 1 800 000/21 600 = 83.3 kW. 500 kW divides by 3 600 s only (forgets the 6 hours); 5000 kW divides by 6 x 60 s. 3.33 kW forgets the 25 K temperature drop.

Question 5 · hard

An electronic cup grader has 6 lanes, each carrying 10 cups per second. On average 80% of cups are filled and fruit average 200 g. Its throughput is about

  1. 43.2 t/h
  2. 34.6 t/h
  3. 5.76 t/h
  4. 9.60 t/h
Answer and working

B. 34.6 t/h

6 x 10 x 0.8 = 48 fruit/s; 48 x 0.2 kg = 9.6 kg/s = 9.6 x 3.6 = 34.6 t/h. Ignoring the fill efficiency gives 43.2 t/h; using one lane gives 5.76 t/h. 9.60 t/h reads 9.6 kg/s as t/h.

Question 6 · hard

A packhouse packs 8 t/h into 5 kg cartons. Each packer fills 4 cartons per minute. The minimum number of packers is

  1. 6
  2. 7
  3. 34
  4. 40
Answer and working

B. 7

Cartons per hour = 8 000/5 = 1 600. One packer = 4 x 60 = 240 cartons/h. 1 600/240 = 6.67, rounded up to 7. Rounding down to 6 leaves the station a bottleneck. 34 divides 8 000 kg/h by 240 without first converting to cartons.

Question 7 · hard

A store holds 50 t of apples at 0 degC respiring at 5 mg CO2/kg h. Using 10.7 J of heat per mg of CO2, the heat of respiration to be removed is about

  1. 17778 W
  2. 0.74 W
  3. 2667 W
  4. 741 W
Answer and working

D. 741 W

CO2 = 50 000 kg x 5 mg/kg h = 250 000 mg/h. Heat = 250 000 x 10.7 J = 2.675 MJ/h; / 3 600 s = 743 W, about 741 W with the unrounded 10.67 J/mg. Multiplying by 24 gives a daily figure wrongly labelled as W. 17778 W is that daily total; 2667 W is the kJ/h figure; 0.74 W uses 50 kg instead of 50 000 kg.

Question 8 · hard

A packhouse manager wants to add a second grading line but the cold store and loading dock are unchanged. The first engineering check should be

  1. whether the new grader has more lanes and a higher rated fruit-per-second capacity than the first
  2. whether the wash tank can be moved to make floor space for the second line
  3. whether precooling, cold storage and dispatch capacity can absorb the higher peak flow
  4. whether the packhouse roof is shaded
Answer and working

C. whether precooling, cold storage and dispatch capacity can absorb the higher peak flow

A line runs at the rate of its slowest element; doubling grading only helps if precooling, storage and dispatch can handle the higher output. Otherwise produce waits warm, losing quality.

Question 9 · medium

In a fresh-cut lettuce line, the step after the sanitising wash that most affects shelf life of the sealed bag is

  1. adding more chlorine to the bag
  2. removing free surface water by centrifugal or air dewatering
  3. raising the packing hall temperature to dry the leaves faster
  4. using a blunt blade
Answer and working

B. removing free surface water by centrifugal or air dewatering

Free water in the sealed bag promotes decay and microbial growth, so dewatering is critical. Blunt blades crush cells and worsen leakage and browning.

Question 10 · medium

The single most common cause of shelf-life failure of fresh-cut products in distribution is

  1. excessive calcium, usually from the dip
  2. too sharp cutting blades
  3. temperature abuse
  4. too many antioxidants
Answer and working

C. temperature abuse

Respiration, microbial growth and browning all accelerate with temperature. Keeping fresh-cut product at or below about 5 degC throughout is the main control.

A study plan for Post-Harvest Engineering of Horticultural Crops 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 (70% of the 23 questions): Packhouse design (7, 30%); Pre-cooling: hydro, forced-air and vacuum cooling (7, 30%); Minimal processing (2, 9%).

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

  2. Then the middle band: Respiration and ripening physiology (2, 9%); Washing, sorting and waxing (2, 9%); Dehydration of fruits and vegetables (1, 4%).

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

  3. Then the lighter units: Grading, including machine vision (1, 4%); Juice extraction (1, 4%).

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

  4. Leave time for the hard ones: Washing, sorting and waxing (2 hard of 2), Dehydration of fruits and vegetables (1 hard of 1), Grading, including machine vision (1 hard of 1).

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

  5. Bank the easy marks: Packhouse design (2 easy of 7).

    Do these early in revision so they are fast on the day.

  6. 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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