Storage Engineering Questions and Answers

Answer: A

Correct answer: 305 tonnes

Find the volume of the filled cylinder, then multiply by the bulk density:

\[ V = \pi r^2 h = \pi (3.0)^2 (14) = 395.8\ \mathrm{m}^3 \]

\[ m = 395.8 \times 770 = 304\,800\ \mathrm{kg} = 305\ \mathrm{tonnes} \]

The distractor \( 396 \) is the volume in cubic metres mistaken for tonnes, and \( 1220 \) would follow from using the diameter as the radius. Note that bulk density, not grain density, is the right figure - it already accounts for the void space between kernels.

Answer: B

Correct answer: 36 m\(^3\) per minute

The fan duty is the specific aeration rate times the tonnage:

\[ Q = 0.12\ \mathrm{m^3/min/t} \times 300\ \mathrm{t} = 36\ \mathrm{m^3/min} \]

This is a modest flow for a large mass of grain, which is the character of aeration - enough to move a cooling front through the bulk over a day or more, not enough to dry it. The fan must nonetheless develop appreciable static pressure, because the resistance of a deep packed bed of grain is high.

Answer: C

Correct answer: 200 g

Multiply the specific respiration rate by the mass, watching the units:

\[ 0.8\ \mathrm{mg/kg/day} \times 250\,000\ \mathrm{kg} = 200\,000\ \mathrm{mg} = 200\ \mathrm{g\ per\ day} \]

This is a low rate, typical of sound dry grain, and it is precisely why carbon dioxide monitoring is such a sensitive early warning: a developing wet pocket or insect focus raises the figure many times over long before anything is visible or the temperature probes respond.

Answer: D

Correct answer: 3192 MJ

The sensible heat removed is mass times specific heat times temperature drop:

\[ Q = mc\Delta T = 40\,000 \times 3.8 \times (25-4) = 40\,000 \times 3.8 \times 21 \]

\[ Q = 3\,192\,000\ \mathrm{kJ} = 3192\ \mathrm{MJ} \]

This is the product load alone. Respiration, transmission and infiltration must be added before the plant is sized, and the pull-down time matters as much as the total: 3192 MJ removed over 24 hours is about 37 kW, but over 8 hours it is 111 kW.

Answer: A

Correct answer: 1.8 kW

Multiply the specific respiration heat by the mass:

\[ Q = 45\ \mathrm{mW/kg} \times 40\,000\ \mathrm{kg} = 1\,800\,000\ \mathrm{mW} = 1.8\ \mathrm{kW} \]

Unlike the product load, this is a continuous demand for as long as the produce is in store, so it must be carried in the refrigeration duty indefinitely. It also falls as the produce cools - respiration roughly halves for each ten degree drop - which is a further reason to precool promptly.

Answer: B

Correct answer: A higher moisture content

Grain and the air around it exchange moisture until their vapour pressures match, and the moisture content at which that occurs is the equilibrium moisture content for the air's relative humidity. Raising the store's humidity from 68 to 80 per cent raises the equilibrium value, so the grain absorbs moisture from the air until it reaches the new balance.

The consequence is a real risk: grain that was safe at 14 per cent may rise above the threshold at which moulds can grow, typically around 70 per cent relative humidity in equilibrium terms. Store humidity is therefore as important as store temperature.

Answer: C

Correct answer: 9.0 tonnes

The loss is the stated percentage of the tonnage stored:

\[ 500 \times 0.018 = 9.0\ \mathrm{tonnes} \]

The distractor \( 90 \) misplaces the decimal by treating 1.8 per cent as 18. A loss of this size is unremarkable for six months of untreated tropical storage, and it is entirely avoidable - which is why the economics of drying, cooling and sealing are usually favourable when set against the value of the grain saved.

Answer: D

Correct answer: 1.36 kW

The transmission load is the overall heat transfer coefficient times the area times the temperature difference:

\[ Q = UA\Delta T = 0.28 \times 220 \times 22 = 1355\ \mathrm{W} = 1.36\ \mathrm{kW} \]

The distractor \( 13.6 \) misplaces the decimal. A \( U \) value of \( 0.28 \) implies substantial insulation, and the resulting load is modest next to the respiration load calculated for a similar store - which is a useful reminder that in a produce store the contents, not the fabric, often dominate.

Answer: A

Correct answer: 8.8 kg

Take the oxygen fraction of the free air volume, then convert to mass using the density of oxygen at ambient conditions, about \( 1.4 \) kg/m\(^3\):

\[ V_{O_2} = 0.21 \times 30 = 6.3\ \mathrm{m}^3 \qquad m = 6.3 \times 1.4 \approx 8.8\ \mathrm{kg} \]

The distractor \( 6.3 \) is the volume mistaken for a mass. The point of the calculation is how little oxygen a sealed store holds: 20 tonnes of respiring grain and insects will consume that quantity within days, which is exactly how hermetic storage kills the infestation.

Answer: B

Correct answer: 216 tonnes

Count the bags, then convert to mass:

\[ N = 90\ \mathrm{m^2} \times 4\ \mathrm{bags/m^2/layer} \times 12\ \mathrm{layers} = 4320\ \mathrm{bags} \]

\[ m = 4320 \times 50 = 216\,000\ \mathrm{kg} = 216\ \mathrm{tonnes} \]

The distractor \( 180 \) omits a factor by using 10 layers, and \( 54 \) counts one layer only. Note that the footprint given is the stack area, not the store area: aisles for access and inspection must be added on top.