NCEES PE Agricultural and Biological Engineering · USA · NCEES (ASABE is the content partner)
Original practice questions written to the published syllabus · not past papersDesigning greenhouses and controlling the climate inside them, with the plastics and soilless systems that go with them.
In the exams: ICAR AICE subject 40 Unit 9 and subject 43 Unit 5; NCEES PE greenhouses; skill modules and electives in the ICAR 6th Deans' Committee B.Tech
NCEES PE Agricultural and Biological Engineering
Facilities and Structures lists greenhouses: heat loss from U A ΔT plus infiltration, ventilation rate from a solar heat balance, fan and pad sizing, psychrometrics of evaporative cooling, and structural loads, in SI and US customary units.
How our questions for NCEES PE Agricultural and Biological Engineering spread across the units of Protected Cultivation & Greenhouse Engineering. Difficulty: 1 easy, 22 medium, 15 hard.
| Greenhouse types and orientation | 5 | |
| Forced ventilation | 5 | |
| Glazing and cladding materials | 4 | |
| Greenhouse heat balance | 4 | |
| Natural ventilation | 4 | |
| Fan-pad and fogging cooling | 4 | |
| CO2 enrichment | 4 | |
| Greenhouse heating | 3 | |
| Hydroponics and aeroponics | 2 | |
| Automation of climate control | 2 | |
| Plastics in agriculture: mulch and lining | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A free-standing single-span greenhouse is to be built at 45° N for winter vegetable production. Which ridge orientation gives the most winter light under the classic North American guidance?
D. Ridge running east–west, so the long south roof faces the low winter sun
Above about 40° latitude the low winter sun is best intercepted by a long south-facing roof, which an east–west ridge gives a free-standing house. North–south ridges are chosen for gutter-connected houses so that gutter shadows move; wind direction governs vent placement, not light.
Further reading: Aldrich and Bartok, Greenhouse Engineering (NRAES-33); ASABE EP406
Taking air density as 1.2 kg/m³, what is the wind velocity pressure q = 0.5ρV² for a design wind speed of 35 m/s on a greenhouse?
D. 735 Pa
q = 0.5 × 1.2 × 35² = 0.5 × 1.2 × 1,225 = 735 Pa. Leaving out the 0.5 gives 1,470 Pa; forgetting to square V gives 21 Pa; 368 Pa halves twice. The code then multiplies q by pressure coefficients for each face.
Further reading: Aldrich and Bartok, Greenhouse Engineering (NRAES-33); ASABE EP406
A film house has 600 m² of cover. Replacing single film (U = 6.8 W m⁻² K⁻¹) with inflated double film (U = 4.0 W m⁻² K⁻¹) at a design difference of 15 K reduces the conduction loss by how much?
B. 25.2 kW
Saving = (U₁ − U₂) A ΔT = (6.8 − 4.0) × 600 × 15 = 25,200 W = 25.2 kW. 61.2 kW is the whole single-film loss and 36.0 kW the remaining double-film loss; 1.7 kW forgets the temperature difference.
Further reading: Aldrich and Bartok, Greenhouse Engineering (NRAES-33); ASABE EP406
Summer design: τ = 0.75, outside irradiance 900 W/m², floor area 2,000 m², fraction of absorbed energy used for evapotranspiration E = 0.4, cover area 3,000 m² with U = 5.5 W m⁻² K⁻¹, and the inside may run 5 K above outside. With ρcp = 1,200 J m⁻³ K⁻¹, what ventilation rate is needed?
B. 121 m³/s
Sensible gain = 0.75 × 900 × 2,000 × (1 − 0.4) = 810,000 W; cover loss = 5.5 × 3,000 × 5 = 82,500 W; Qv = (810,000 − 82,500) / (1,200 × 5) = 121 m³/s. Ignoring the cover loss gives 135; ignoring evapotranspiration gives 211; using 1,000 J m⁻³ K⁻¹ (cp without density) gives 146.
Further reading: Aldrich and Bartok, Greenhouse Engineering (NRAES-33); ASABE EP406
In still air, a greenhouse has 30 m² of effective opening split between side and roof vents whose centres are 4 m apart. Inside is 32 °C and outside 26 °C. Using Q = Cd A √[2gΔh(Ti − To)/Ti] with Cd = 0.65, what is the stack ventilation rate?
B. 24.2 m³/s
Ti must be absolute: 305.15 K. Q = 0.65 × 30 × √(2 × 9.81 × 4 × 6 / 305.15) = 0.65 × 30 × 1.242 = 24.2 m³/s. Using 32 °C in the denominator gives 74.8; leaving out Cd gives 37.3; dropping the 2 gives 17.1.
Further reading: ASHRAE Handbook — Fundamentals, natural ventilation
Complete combustion of propane follows C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. How much CO₂ does one kilogram of propane produce?
C. 3.0 kg
Molar mass of propane is 44 g/mol and each mole gives 3 mol CO₂ (3 × 44 = 132 g), so 132/44 = 3.0 kg CO₂ per kg. 2.75 kg is the methane figure (44/16); 1.0 kg counts one mole of CO₂ only; 4.0 kg uses the water moles.
Further reading: Aldrich and Bartok, Greenhouse Engineering (NRAES-33); ASABE EP406
What is the occupational exposure limit for CO₂, averaged over an eight-hour working day, used in many countries including the United States?
C. 5,000 ppm
The common 8-hour limit is 5,000 ppm (0.5%). Enrichment set points of roughly 700–1,000 ppm are far below it, but faulty dosing or a burst cylinder can exceed it, so alarms are fitted. 1,000 ppm is a typical enrichment set point, not a safety limit; 500 ppm is close to outdoor air. 50,000 ppm is a factor-of-ten slip (5%).
Further reading: Aldrich and Bartok, Greenhouse Engineering (NRAES-33); ASABE EP406
A recipe needs 250 mg/L potassium from potassium nitrate containing 38% K. The stock is dosed by a 1:200 injector. What concentration of potassium nitrate is needed in the stock tank?
A. 131.6 g/L
Salt in the final solution = 250 / 0.38 = 658 mg/L. The stock is 200 times stronger: 657.9 × 200 = 131,579 mg/L = 131.6 g/L. 65.8 g/L assumes a 1:100 ratio; 0.66 g/L is the final strength, not the stock; 50.0 g/L forgets that the salt is only 38% K.
Further reading: Resh, Hydroponic Food Production; Nelson, Greenhouse Operation and Management
A square farm pond is 30 m × 30 m at the top, 2.5 m deep, with side slopes of 2:1 (horizontal:vertical). What area of liner covers the bottom and the four sloping faces, before anchor trench and overlaps?
C. 959 m²
Bottom side = 30 − 2 × 2 × 2.5 = 20 m. Slant length = 2.5 × √(1 + 2²) = 5.590 m. Each face = (30 + 20)/2 × 5.590 = 139.8 m². Total = 400 + 4 × 139.8 = 959 m². 900 m² is the plan area at the top; 650 m² uses the vertical depth instead of the slant length; 1,459 m² uses the top instead of the bottom area.
Further reading: Orzolek (ed.), Plastics in Agriculture
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