Engineering Properties, Moisture and Psychrometry, Drying and Evaporation of Agricultural Produce
2. Moisture content, and psychrometry — the properties of moist air
Moisture content is stated on a wet basis, M_wb = W_w/(W_w + W_d), as traders and graders do, or on a dry basis, M_db = W_w/W_d, as drying calculations do because the dry matter stays constant. They convert by M_db = M_wb/(1 − M_wb) and M_wb = M_db/(1 + M_db): 20 per cent wet basis is 25 per cent dry basis. The water removed in drying a mass W_i from M_i to M_f (both wet basis, per cent) is W_i(M_i − M_f)/(100 − M_f): 1000 kg of paddy dried from 25 to 14 per cent loses 1000 × 11/86 = 127.9 kg — not 110 kg, because the percentage is of a shrinking total.
Psychrometry treats moist air as a mixture of dry air and water vapour, each an ideal gas, at a total pressure p. The humidity ratio (specific humidity) is W = 0.622 p_v/(p − p_v) kg of water per kg of dry air, where p_v is the partial pressure of the vapour; the relative humidity is φ = p_v/p_s, with p_s the saturation pressure at the dry-bulb temperature; the dew point is the temperature at which p_v would saturate the air; and the enthalpy per kg of dry air is h = 1.005t + W(2501 + 1.88t) kJ/kg, t in °C. With p_v = 2.0 kPa at p = 101.325 kPa, W = 0.622 × 2/99.325 = 0.0125; if the saturation pressure at the 30 °C dry-bulb temperature is given as 4.246 kPa, φ = 47.1 per cent; and at W = 0.0125, h = 30.15 + 0.0125 × 2557.4 = 62.1 kJ/kg of dry air. The wet-bulb temperature from a psychrometer is, for air–water mixtures, very close to the adiabatic-saturation temperature.
| Process | What stays constant | What changes |
|---|---|---|
| Sensible heating (air heater) | humidity ratio W and dew point | dry-bulb rises, relative humidity falls — the air’s drying capacity rises |
| Sensible cooling (above the dew point) | humidity ratio W | dry-bulb falls, relative humidity rises |
| Cooling below the dew point | — (water condenses) | both dry-bulb and W fall; dehumidification |
| Drying air passing through a bed (adiabatic) | wet-bulb temperature, nearly (enthalpy almost constant) | dry-bulb falls, W and relative humidity rise as the air picks up water |
3. Drying kinetics, dryers, and the drying and milling of cereals, pulses and oilseeds
A moist solid dries first at a constant rate, while its surface stays wet and evaporation is limited by heat and mass transfer from the air, and then — below the critical moisture content X_c — at a falling rate, as moisture must diffuse from inside. It approaches the equilibrium moisture content (EMC), the moisture at which it neither gains nor loses water in air of a given temperature and relative humidity (described by sorption isotherms and by equations such as Henderson’s). Most cereal grains dry almost entirely in the falling-rate period. The drying times, with W_s the dry-solid mass, A the drying area and N_c the constant rate (kg/m²·h), are t_c = W_s(X₀ − X_c)/(AN_c) for the constant-rate period and, when the falling rate is linear in X down to X_e, t_f = [W_s(X_c − X_e)/(AN_c)] ln[(X_c − X_e)/(X − X_e)]. For 50 kg of dry solid on 2 m² at 2.5 kg/m²·h from X₀ = 0.6 to X_c = 0.3, t_c = 3 h; falling on to X = 0.1 with X_e = 0.05 takes a further 2.5 ln 5 = 4.02 h.
Thin-layer drying of grain is described by the moisture ratio MR = (M − M_e)/(M₀ − M_e). The Lewis (Newton) model MR = e^(−kt) treats drying like Newtonian cooling, and the Page model MR = e^(−ktⁿ) adds an exponent for a better fit. With k = 0.2 h⁻¹, after 5 h MR = e⁻¹ = 0.368, and a grain from 30 per cent dry basis with M_e = 10 per cent reaches 10 + 20 × 0.368 = 17.4 per cent. Deep-bed drying is a stack of thin layers: a drying front moves up the bed, the bottom over-dries while the top may stay wet, which is why deep beds are stirred, reversed or kept shallow.
| Dryer | How it works | Suits |
|---|---|---|
| Flat-bed / tray (batch) | heated air through a static layer, or across trays of product | small lots of grain; fruits and vegetables |
| Continuous-flow (mixing, e.g. LSU; non-mixing columnar) | grain flows down by gravity through heated-air zones | large grain throughputs; paddy with tempering |
| Recirculating batch | grain is cycled through the drying zone repeatedly | uniform drying of paddy |
| Drum | a thin film on a steam-heated rotating drum, scraped off by a knife | purées, flakes, pastes |
| Spray | a liquid atomised into hot air (rotary wheel or nozzle) dries in seconds; powder separated in a cyclone | milk, coffee, egg powder — heat-sensitive liquids |
| Freeze | the frozen product’s ice sublimes under vacuum, below the triple point of water (about 611 Pa, 0.01 °C) | highest quality and rehydration; costliest |
Osmotic dehydration soaks fruit or vegetable pieces in a concentrated sugar or salt solution, removing water by osmosis with some gain of solute, usually as a pre-treatment before air or freeze drying. Hydrothermal treatment — soaking, steaming and drying — is parboiling of paddy: it gelatinises the starch, seals cracks, drives nutrients from the bran into the endosperm and raises the head-rice yield. Milling of paddy is cleaning, dehusking (rubber-roll or under-runner disc shellers; husk is roughly a fifth of paddy mass), paddy separation, polishing (whitening) and grading; head rice (whole kernels) is the measure of milling quality. Pulse milling loosens the husk by pre-treatment (oil and water, or drying) and then dehulls and splits the grain (dal). Oilseeds are cleaned, dehulled, conditioned by heat and moisture, and pressed in screw expellers, with the cake then solvent-extracted (hexane) to recover the residual oil.
4. Concentration of liquid foods: evaporators
Evaporation concentrates a liquid food — juice, milk, sugar solution — by boiling off water, usually under vacuum so that it boils at a low temperature and keeps its colour, flavour and nutrients. The types are the open pan (simple, slow, heat-damaging), the short-tube vertical (calandria) evaporator, long-tube rising-film and falling-film evaporators (short residence time, the usual choice for heat-sensitive liquids like milk and juice), forced-circulation evaporators for viscous or scaling liquids, plate evaporators, and agitated (wiped) thin-film evaporators for very viscous products. A heat balance on one effect is m_s λ_s = F c_p(T₁ − T_F) + Vλ, and the heat transferred is q = UA ΔT. The steam economy — kg of water evaporated per kg of steam — is a little under 1 for a single effect; in multiple-effect evaporation the vapour from one effect heats the next at a lower pressure, raising the economy roughly in proportion to the number of effects (evaporating 3600 kg/h with 2000 kg/h of steam is an economy of 1.8). Boiling-point elevation by dissolved solids reduces the available temperature difference as the liquid concentrates.
Key takeaways
- Sphericity = (abc)^(1/3)/a; porosity of a bulk = 1 − ρ_b/ρ_t.
- M_db = M_wb/(1 − M_wb); water removed = W_i(M_i − M_f)/(100 − M_f) on wet-basis percentages.
- W = 0.622p_v/(p − p_v); φ = p_v/p_s; h = 1.005t + W(2501 + 1.88t) — computed from given pressures, not a remembered chart.
- Sensible heating keeps W constant and lowers RH; drying air through a bed follows a line of nearly constant wet-bulb temperature.
- t_c = W_s(X₀ − X_c)/(AN_c); linear falling rate adds a logarithm; thin-layer MR = e^(−kt).
- Steam economy = water evaporated/steam used, raised by multiple effects; freeze drying sublimes ice below the triple point.
Practice questions (14)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
A grain sample has a moisture content of 20 per cent on a wet basis. What is its moisture content on a dry basis, in per cent?
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Answer: 25
M_db = M_wb/(1 − M_wb) = 0.20/0.80 = 0.25 = 25 per cent: 20 g of water with 80 g of dry matter. Going the other way, 20/(1 + 0.20) = 16.7 per cent, converts dry basis to wet basis and answers the reverse question.1000 kg of paddy at 25 per cent moisture (wet basis) is dried to 14 per cent (wet basis). How much water is removed, in kg (to one decimal place)?
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Answer: 127.9
Dry matter = 1000 × 0.75 = 750 kg; final mass = 750/0.86 = 872.1 kg; water removed = 1000 − 872.1 = 127.9 kg, the same as 1000 × (25 − 14)/(100 − 14). Subtracting the percentages (110 kg) ignores that 14 per cent is of the smaller final mass.Moist air at a total pressure of 101.325 kPa has a water-vapour partial pressure of 2.0 kPa. What is its humidity ratio, in kg water per kg dry air (to four decimal places)?
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Answer: 0.0125
W = 0.622 p_v/(p − p_v) = 0.622 × 2.0/(101.325 − 2.0) = 1.244/99.325 = 0.0125. Dividing by the total pressure instead of the dry-air partial pressure gives 0.01228, and dropping the 0.622 (the ratio of the molar masses of water and air) gives 0.0201.Air at a dry-bulb temperature of 30 °C has a vapour partial pressure of 2.0 kPa. The saturation vapour pressure at 30 °C is given as 4.246 kPa. What is its relative humidity, in per cent (to one decimal place)?
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Answer: 47.1
φ = p_v/p_s = 2.0/4.246 = 0.471 = 47.1 per cent. The ratio of humidity ratios (the degree of saturation) is close but not equal; dividing by the total pressure gives 2 per cent, which is not a relative humidity at all.Moist air at 30 °C has a humidity ratio of 0.0125 kg/kg dry air. Using h = 1.005t + W(2501 + 1.88t), what is its enthalpy, in kJ per kg of dry air (to one decimal place)?
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Answer: 62.1
h = 1.005 × 30 + 0.0125 × (2501 + 1.88 × 30) = 30.15 + 0.0125 × 2557.4 = 30.15 + 31.97 = 62.1 kJ/kg dry air. Leaving out the latent term 2501 gives only 30.9 kJ/kg, missing most of the energy the vapour carries.Grain at an initial moisture content of 30 per cent (dry basis) is dried in a thin layer in air for which its equilibrium moisture content is 10 per cent (dry basis). The drying follows MR = e^(−kt) with k = 0.2 h⁻¹. What is its moisture content after 5 h, in per cent dry basis (to one decimal place)?
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Answer: 17.4
MR = e^(−0.2 × 5) = e^(−1) = 0.368. M = M_e + MR(M₀ − M_e) = 10 + 0.368 × 20 = 17.4 per cent. Applying MR to the whole 30 per cent (30 × 0.368 = 11.0) forgets that grain dries towards M_e, not towards zero.A tray dryer holds 50 kg of dry solid on a drying area of 2 m². The constant drying rate is 2.5 kg/(m²·h). How long does it take to dry the solid from 0.6 to the critical moisture content of 0.3 (kg water per kg dry solid), in hours?
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Answer: 3
t_c = W_s(X₀ − X_c)/(AN_c) = 50 × 0.3/(2 × 2.5) = 15/5 = 3 h. Using the total wet mass instead of the dry solid overstates the water to be removed; forgetting the area gives 6 h.The same solid (50 kg dry solid, 2 m², constant rate 2.5 kg/(m²·h), critical moisture 0.3) dries in the falling-rate period with a rate linear in moisture content down to an equilibrium moisture of 0.05. How long does the falling-rate period take to reach 0.1, in hours (to two decimal places)?
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Answer: 4.02
t_f = [W_s(X_c − X_e)/(AN_c)] ln[(X_c − X_e)/(X − X_e)] = [50 × 0.25/5] ln(0.25/0.05) = 2.5 × ln 5 = 2.5 × 1.609 = 4.02 h. Treating the falling period at the constant rate gives 50 × 0.2/5 = 2 h — too short, because the rate keeps falling.A grain kernel has principal dimensions of 6 mm, 4 mm and 3 mm. What is its sphericity (to three decimal places)?
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Answer: 0.693
Geometric-mean diameter = (6 × 4 × 3)^(1/3) = 72^(1/3) = 4.160 mm; sphericity = 4.160/6 = 0.693. Using the arithmetic mean (4.33 mm) gives 0.722, and dividing by the smallest dimension gives a value above 1, which no sphericity can have.Wheat in a bin has a bulk density of 800 kg/m³ and a true (kernel) density of 1250 kg/m³. What is the porosity of the bulk, in per cent?
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Answer: 36
ε = 1 − ρ_b/ρ_t = 1 − 800/1250 = 1 − 0.64 = 0.36 = 36 per cent. The ratio 0.64 itself is the solid fraction, the complement of the answer.A double-effect evaporator removes 3600 kg/h of water from juice while using 2000 kg/h of steam. What is its steam economy?
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Answer: 1.8
Economy = water evaporated/steam used = 3600/2000 = 1.8, a little under the ideal 2 for two effects because of heat losses and the sensible heat of the feed. Inverting the ratio (0.56) gives steam per kg of water, the steam consumption.In freeze drying, water is removed from the frozen product mainly by:
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Answer: A — sublimation of ice under vacuum, below the triple point of water
Below the triple point (about 611 Pa and 0.01 °C) liquid water cannot exist, so ice passes directly to vapour; the product keeps its structure and rehydrates well. Osmosis describes osmotic dehydration and the hot drum describes drum drying — both liquid-phase removal.Which of the following statements about the flow behaviour of food materials are correct?
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Answer: B — A power-law fluid with n < 1 is shear-thinning (pseudoplastic); C — A Bingham plastic does not flow until a yield stress is exceeded; D — A Newtonian fluid is a power-law fluid with n = 1
For τ = Kγ̇ⁿ the apparent viscosity is Kγ̇ⁿ⁻¹: it falls with shear rate when n < 1 (pseudoplastic) and rises when n > 1 (dilatant), so the first statement is false. A Bingham plastic needs τ > τ₀ to flow, and n = 1 recovers the Newtonian τ = μγ̇.Which of the following statements about psychrometric processes are correct?
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Answer: A — Sensible heating leaves the humidity ratio unchanged; C — Air passing adiabatically through a wet grain bed follows a line of nearly constant wet-bulb temperature; D — The dew point does not change during sensible heating
Heating without adding water keeps W — and therefore the vapour pressure and the dew point — fixed, while p_s rises, so relative humidity falls; that is why heated air dries. Adiabatic evaporative drying trades sensible for latent heat at almost constant enthalpy, along a wet-bulb line. The second statement has the direction reversed.