Grain at \( 22 \) per cent wet basis is dried to \( 13 \) per cent. From \( 1000 \) kg, the water removed is
Answer: A
Dry matter is \( 780 \) kg, so the final wet mass is \( 780/0.87 = 896.6 \) kg and the water removed is \( 1000 - 896.6 = 103.4 \) kg. Subtracting the percentages to get 90 kg is the classic error and understates the duty.
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Removing \( 103.4 \) kg of water at a latent heat of \( 2.3 \) MJ/kg needs a theoretical heat input of
Answer: B
Correct answer: 238 MJ
The theoretical heat is the mass of water multiplied by its latent heat of vaporisation:
\[ Q = m_w h_{fg} = 103.4 \times 2.3 = 237.8 \approx 238\ \mathrm{MJ} \]
This is the irreducible minimum - the energy needed to turn that water to vapour and nothing more. A real dryer must supply considerably more, because the exhaust air leaves hot, the product itself is warmed, and heat escapes through the structure; the ratio of the two is the drying efficiency.
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A dryer supplies \( 480 \) MJ and evaporates water requiring \( 238 \) MJ. Its drying efficiency is
Answer: C
Correct answer: 49.6 per cent
Drying efficiency is the heat usefully spent on evaporation divided by the heat supplied:
\[ \eta = \frac{238}{480} \times 100 = 49.6\ \mathrm{per\ cent} \]
The distractor \( 201.7 \) inverts the ratio, and \( 50.4 \) is the complement - the fraction lost rather than used. A figure near 50 per cent is typical of a hot air dryer without heat recovery, the losses being chiefly sensible heat carried away in the exhaust.
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Air enters a dryer at a humidity ratio of \( 0.010 \) and leaves at \( 0.022 \) kg per kg dry air. To evaporate \( 60 \) kg of water the dry air required is
Answer: D
Correct answer: 5000 kg
Each kilogram of dry air picks up the difference between the outlet and inlet humidity ratios:
\[ \Delta W = 0.022 - 0.010 = 0.012\ \mathrm{kg\ water\ per\ kg\ dry\ air} \]
\[ m_{air} = \frac{60}{0.012} = 5000\ \mathrm{kg\ dry\ air} \]
The distractor \( 2727 \) divides by the outlet humidity alone, ignoring the moisture the air already carried in. The calculation shows how large the air flow must be: five tonnes of air to remove sixty kilograms of water, which is why fan power is a significant part of a dryer's energy budget.
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Air at \( 30 \) degrees Celsius has a saturation vapour pressure of \( 4.24 \) kPa and an actual vapour pressure of \( 1.70 \) kPa. Its relative humidity is
Answer: A
Correct answer: 40.1 per cent
Relative humidity is the actual vapour pressure as a fraction of the saturation value at the same temperature:
\[ \phi = \frac{1.70}{4.24} \times 100 = 40.1\ \mathrm{per\ cent} \]
The distractor \( 2.54 \) is the difference of the two pressures rather than their ratio. Note what this air can do: at 40 per cent relative humidity it is well short of saturation and therefore has substantial drying potential, and heating it further would lower the relative humidity again by raising \( p_{sat} \).
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A product at \( 4.0 \) kg water per kg dry matter has an equilibrium moisture of \( 0.12 \) on the same basis. Its free moisture content is
Answer: B
Correct answer: 3.88 kg per kg
Free moisture is the moisture present above the equilibrium value:
\[ X_{free} = X - X_{eq} = 4.00 - 0.12 = 3.88\ \mathrm{kg\ water\ per\ kg\ dry\ matter} \]
The distractor \( 4.12 \) adds instead of subtracting. Only the free moisture can be removed in air of this condition: drying rate curves fall to zero as the free moisture approaches zero, leaving the \( 0.12 \) behind however long the dryer runs.
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A tray of product loses \( 2.4 \) kg of water in \( 30 \) minutes from a surface of \( 0.8 \) m\(^2\). The constant rate drying flux is
Answer: C
The rate is \( 2.4/0.5 = 4.8 \) kg per hour and the flux is \( 4.8/0.8 = 6.0 \) kg per square metre per hour. During the constant rate period this flux depends only on the air's temperature, humidity and velocity, not on the material.
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A dryer handles \( 3.0 \) tonnes of grain per hour and removes \( 8 \) per cent of its mass as water. The evaporation rate is
Answer: D
Correct answer: 240 kg per hour
The water removed is the stated fraction of the mass throughput:
\[ \dot{m}_w = 3000\ \mathrm{kg/h} \times 0.08 = 240\ \mathrm{kg/h} \]
The distractor \( 24 \) misplaces the decimal and \( 2400 \) treats the 8 per cent as 80. At \( 2.3 \) MJ/kg this evaporation alone needs about \( 552 \) MJ per hour of useful heat, so at 50 per cent efficiency the dryer must be supplied with roughly \( 1.1 \) GJ per hour.
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A thin layer follows \( MR = \exp(-kt) \) with \( k = 0.25 \) per hour. The moisture ratio after \( 4 \) hours is
Answer: A
Correct answer: 0.368
Substituting into the thin layer model:
\[ MR = \exp(-kt) = \exp(-0.25 \times 4) = \exp(-1) = 0.368 \]
So after four hours the free moisture has fallen to about 37 per cent of its initial value - \( 1/k \) is the time constant, and one time constant removes about 63 per cent of the free moisture. The distractor \( 0.632 \) is that complement, the fraction removed rather than the fraction remaining.
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For the same layer, the time to reach a moisture ratio of \( 0.10 \) is
Answer: B
Correct answer: 9.2 hours
Rearranging \( MR = \exp(-kt) \) for the time:
\[ t = -\frac{\ln MR}{k} = -\frac{\ln 0.10}{0.25} = \frac{2.303}{0.25} = 9.2\ \mathrm{hours} \]
Compare this with the previous item: four hours took the moisture ratio to \( 0.368 \), but reaching \( 0.10 \) takes more than twice as long again. The exponential form means each further reduction costs progressively more time, which is why the last few points of moisture are the most expensive to remove.
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