A grain sample of \( 250 \) g contains \( 35 \) g of water. The moisture content on a wet basis is
Answer: A
Wet basis moisture is the water divided by the total wet mass, \( 35/250 = 0.14 \). Fourteen per cent is the standard safe storage moisture for many cereals, and trade standards are written on this basis rather than the dry one.
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For the same sample, the moisture content on a dry basis is
Answer: B
Correct answer: 16.3 per cent
The sample is 250 g holding 35 g of water, so the dry matter is
\[ 250 - 35 = 215\ \mathrm{g} \]
On the dry basis the denominator is that dry matter alone:
\[ M_{db} = \frac{35}{215} \times 100 = 16.28\% \approx 16.3\% \]
Compare the wet basis figure of \( 35/250 = 14.0\% \) for the same sample. The dry basis value is always the larger of the two, because its denominator excludes the water. The conversion is \( M_{db} = M_{wb}/(1 - M_{wb}) \), which confirms \( 0.14/0.86 = 0.163 \).
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A grain has a true density of \( 1350 \) kg/m\(^3\) and a bulk density of \( 780 \) kg/m\(^3\). Its porosity is
Answer: C
Correct answer: 42.2 per cent
Porosity is one minus the solid fraction:
\[ \varepsilon = 1 - \frac{\rho_b}{\rho_t} = 1 - \frac{780}{1350} \]
\[ \frac{780}{1350} = 0.5778 \qquad \varepsilon = 0.4222 = 42.2\% \]
The distractor 57.8 per cent is the solid fraction itself, and 26.9 per cent comes from mistakenly dividing the density difference by the true density in the wrong sense. A value near 42 per cent is entirely typical of a cereal lot, and it is that void space which carries the aeration air.
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A seed has principal dimensions \( 8.0 \), \( 4.0 \) and \( 2.0 \) mm. Its geometric mean diameter is
Answer: D
The geometric mean is \( (8 \times 4 \times 2)^{1/3} = 64^{1/3} = 4.00 \) mm. Note that the arithmetic mean of the same dimensions would be 4.67 mm, which is a different quantity altogether and should never be substituted.
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For the seed above, the sphericity is
Answer: A
Sphericity is the geometric mean diameter divided by the largest dimension, \( 4.0/8.0 = 0.50 \). An elongated seed of this kind rolls poorly, and its angle of repose is correspondingly higher than that of a near-spherical grain.
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A bin of \( 40 \) m\(^3\) is filled with grain of bulk density \( 760 \) kg/m\(^3\). The mass stored is
Answer: B
Correct answer: 30.4 tonnes
Bulk density converts a bin volume straight into a mass, which is exactly what it is for:
\[ m = \rho_b V = 760 \times 40 = 30\,400\ \mathrm{kg} = 30.4\ \mathrm{t} \]
Note that no allowance for voids is needed: the air spaces are already inside the bulk density figure, which is what distinguishes it from true density. Had true density been used the answer would have been far too large. The tenfold distractor at 3.04 tonnes is simply a decimal slip, and is worth guarding against by sanity-checking the order of magnitude - a 40 m\(^3\) bin holds tens of tonnes, not a few.
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Grain at \( 20 \) per cent wet basis is dried to \( 14 \) per cent wet basis. Starting from \( 1000 \) kg, the mass of water removed is
Answer: C
Correct answer: 69.8 kg
Dry matter is conserved during drying, so work with it rather than with the water. At 20 per cent wet basis:
\[ m_{\text{dry}} = 1000 \times 0.80 = 800\ \mathrm{kg} \]
At 14 per cent wet basis that same dry matter is 86 per cent of the new total:
\[ m_{\text{final}} = \frac{800}{0.86} = 930.2\ \mathrm{kg} \]
\[ \text{water removed} = 1000 - 930.2 = 69.8\ \mathrm{kg} \]
The tempting wrong answer is 60 kg, from treating the six-point moisture drop as six per cent of the original mass. That fails because the total mass shrinks as water leaves, so the percentages are taken on different denominators.
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A particle of mass \( 0.045 \) g reaches terminal velocity where drag equals weight. If the drag force is \( F_D = 1.2\times10^{-6} v^2 \) N with \( v \) in m/s, the terminal velocity is
Answer: D
The weight is \( 4.5\times10^{-5} \times 9.81 = 4.41\times10^{-4} \) N, so \( v^2 = 4.41\times10^{-4}/1.2\times10^{-6} = 368 \) and \( v = 19.2 \) m/s. Cereal grains sit near this figure while chaff settles far lower, which is exactly what an aspirator exploits.
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The specific heat of a grain is estimated as \( c = 1.0 + 3.2M \) kJ/kg K with \( M \) the dry basis moisture as a fraction. At \( M = 0.18 \) the specific heat is
Answer: A
Correct answer: 1.58 kJ/kg K
Substitute the moisture fraction directly into the correlation:
\[ c = 1.0 + 3.2M = 1.0 + 3.2 \times 0.18 \]
\[ c = 1.0 + 0.576 = 1.576 \approx 1.58\ \mathrm{kJ/kg\,K} \]
Check that \( M \) is used as a fraction, not as 18: entering the percentage would give an absurd 58.6 kJ/kg K. The value sits sensibly between dry grain solids near 1.0 and pure water at 4.18, which is the reassurance that the substitution was made correctly.
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Heating \( 500 \) kg of grain of specific heat \( 1.6 \) kJ/kg K from \( 15 \) to \( 45 \) degrees Celsius requires
Answer: B
Correct answer: 24.0 MJ
Sensible heating is mass times specific heat times temperature rise:
\[ Q = mc\Delta T = 500 \times 1.6 \times (45 - 15) \]
\[ Q = 500 \times 1.6 \times 30 = 24\,000\ \mathrm{kJ} = 24.0\ \mathrm{MJ} \]
Because only a temperature difference appears, degrees Celsius and kelvin are interchangeable here and no conversion is needed - a 30 degree rise is a 30 kelvin rise. This figure is the sensible heat alone; a real drier must also supply the latent heat of the water evaporated, which is much the larger term.
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