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.
Energy sources and waste treatment applications: manure digester sizing by loading rate and HRT, methane and energy production, combustion air and efficiency, and life-cycle energy comparisons.
Turning crop residues, dung and oils into fuel: briquettes, gasifiers, biogas plants and liquid biofuels, and auditing farm energy.
Our 21 NCEES PE Agricultural and Biological Engineering questions in this subject: 1 easy, 10 medium, 10 hard, across 10 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.
Wood with a dry-basis HHV of 18.5 MJ/kg and dry-basis hydrogen of 6% is burned at 30% moisture (wet basis). Using 2.442 MJ/kg for latent heat, its lower heating value on the as-fired (wet) basis is closest to:
C. 11.29 MJ/kg
Per kg of wet fuel: dry matter = 0.70 kg, so HHV_wet = 18.5 × 0.70 = 12.95 MJ. Water in products = 9H × dry fraction + moisture = 9 × 0.06 × 0.70 + 0.30 = 0.678 kg. LHV_wet = 12.95 − 2.442 × 0.678 = 11.29 MJ/kg. Scaling the dry LHV by (1 − M) (12.03) forgets to evaporate the fuel moisture itself. 17.18 MJ/kg is the dry-basis LHV, with the moisture ignored altogether.
A dry biomass contains, by mass, C 0.48, H 0.06, O 0.42 (neglect N, S and ash). Taking air as 23.2% oxygen by mass, the stoichiometric air–fuel ratio is:
A. 5.78 kg air per kg fuel
O2 needed = (32/12)C + 8H − O = 2.667 × 0.48 + 8 × 0.06 − 0.42 = 1.280 + 0.48 − 0.42 = 1.340 kg/kg. Air = 1.340 ÷ 0.232 = 5.78 kg/kg. Ignoring the oxygen already in the fuel gives 7.59; using 0.21 (the volume fraction) instead of 0.232 (the mass fraction) gives 6.38. Biomass needs roughly half the air per kg that coal does, because it carries much of its own oxygen.
A rice-husk furnace for a grain dryer burns 50 kg/h of husk (heating value 13 MJ/kg) and heats 5.0 kg/s of air by 20 °C (c_p = 1.005 kJ/kg·K). The furnace efficiency is:
C. 55.7%
Heat input = 50 × 13 MJ/h = 650 MJ/h = 180.6 kW. Heat output = 5.0 × 1.005 × 20 = 100.5 kW. Efficiency = 100.5 ÷ 180.6 = 55.7%. Dividing kW by kJ/h without converting (÷3600) gives a meaningless tiny number; inverting the ratio gives 180%; 15.5% treats 650 MJ/h as if it were 650 kW.
A digester treating agro-industrial wastewater removes 100 kg of COD per day. Taking the theoretical methane yield as 0.35 m³ CH4 per kg COD removed (at STP) and a biogas methane content of 65%, the biogas production is about:
D. 53.8 m³/day
Methane = 100 × 0.35 = 35 m³/day. Biogas = methane ÷ methane fraction = 35 ÷ 0.65 = 53.8 m³/day. Reporting 35 m³ as biogas ignores the CO2; multiplying by 0.65 applies the fraction the wrong way. The 0.35 figure comes from the fact that 1 mol CH4 (22.4 L) needs 64 g of oxygen demand to oxidise. 22.8 m³/day multiplies by 0.65 (35 × 0.65); 285.7 m³/day divides the COD by the yield (100 ÷ 0.35).
In a biogas pipeline from a plant to a kitchen 30 m away, a water trap (condensate drain) should be installed:
C. At the lowest point of the line
Biogas leaves the digester saturated with water vapour, which condenses as the pipe cools. Water collects in low points and blocks gas flow, causing fluctuating flames. Sloping the pipe toward a drain at the lowest point, and emptying it regularly, prevents this.
For medium to large farms digesting manure slurry, a continuously stirred tank reactor (CSTR) is often chosen over an unmixed plug-flow digester because:
A. Mixing keeps solids suspended and spreads heat and feed, suiting slurries of 3–10% solids
In a CSTR the whole contents are mixed, so the digester behaves as one uniform zone; this suits dilute slurries and avoids scum and sediment. Plug-flow digesters suit thicker manures (roughly 11–13% solids) that move through as a plug. Very dry feeds need dry (solid-state) digestion systems. Decoupling solids and liquid retention is the idea of UASB and similar high-rate reactors, used for soluble wastewater, not for solid-rich manure.
A dairy unit produces 200 kg of fresh manure per day at 18% total solids; volatile solids are 80% of total solids. With a biogas yield of 0.25 m³ per kg of VS added, daily biogas production is:
D. 7.2 m³
TS = 200 × 0.18 = 36 kg; VS = 36 × 0.8 = 28.8 kg; biogas = 28.8 × 0.25 = 7.2 m³/day. Applying the yield to total solids (9.0 m³) forgets that ash does not make gas; applying it to fresh manure (50 m³) counts water as feed. 40 m³ skips the total-solids step (200 × 0.8 × 0.25).
The first-order kinetic model of methane production, B(t) = B0 [1 − exp(−k t)], is used in biogas plant design. In this model, B0 represents:
D. The ultimate methane yield of the substrate at very long retention time
B0 (for example in m³ CH4 per kg VS) is the maximum yield the substrate can give, measured in a long batch biochemical methane potential test; k is the first-order rate constant (per day). For a chosen HRT, the fraction 1 − exp(−k·HRT) shows how much of the potential is achieved, which lets a designer trade digester size against yield.
An energy audit finds an irrigation pump set drawing 7.5 kW for 1000 h per year at an overall efficiency of 30%. Replacing it with a matched set of 50% efficiency delivering the same hydraulic output would save about:
D. 3000 kWh per year
Hydraulic output = 7.5 × 0.30 = 2.25 kW. New input = 2.25 ÷ 0.50 = 4.5 kW. Saving = (7.5 − 4.5) × 1000 = 3000 kWh/year. Subtracting efficiencies (50 − 30 = 20%) applied to input (1500) understates the saving; 4500 kWh is the new consumption, not the saving.
When a life-cycle energy analysis of a liquid biofuel reports a net energy ratio (energy delivered in the fuel ÷ fossil energy consumed in its production), the result is especially sensitive to:
D. How co-product energy, such as oilcake, distillers' grains, glycerol or bagasse, is allocated or credited
Biofuel chains yield co-products that displace other products. Whether energy is allocated by mass, energy content or economic value, or credited by system expansion, can change the reported ratio substantially. This is a main reason why published net energy figures for the same biofuel vary and are sometimes contested.
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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