Drying and Dehydration Questions and Answers

Answer: A

The initial moisture content on a wet basis is 62.31%.

Wet-basis moisture is water mass over total wet mass:

\[ M_{wb} = \frac{405}{650} \times 100 = 62.31\% \]

The dry-basis value, water over dry solids, is \(405/245 \times 100 = 165.31\%\), which is one of the distractors; dry-basis moisture can exceed 100 percent, wet-basis cannot. 162.31 and 70.45 do not follow from the data.

Answer: D

The drying rate is about 0.0485 kg/h.

The dry solids are 650 - 405 = 245 g. At 6.8% moisture on a dry basis the final product holds

\[ \begin{aligned} m_{w,final} &= 0.068 \times 245 = 16.66\ \text{g} \\ \text{water removed} &= 405 - 16.66 = 388.3\ \text{g} \\ \text{rate} &= \frac{388.3}{8} = 48.5\ \text{g/h} = 0.0485\ \text{kg/h} \end{aligned} \]

Since drying is entirely in the constant-rate period, the average rate equals the constant rate. The large options, such as 128.79 and 126.35 kg/h, are far too high for a 650 g sample dried over 8 h, and 77.81 does not follow from the mass balance either.

Answer: B

The most efficient solution is 20% NaCl solution.

Water moves out of the potato only if the solution’s water activity is below the potato’s 0.942, and the lower it is, the stronger the drive. By Raoult’s law, water activity is the mole fraction of water. Take 100 g of each solution; the 20% solutions contain 80 g of water, which is 80/18 = 4.44 mol:

\[ \begin{aligned} \text{NaCl} &: \tfrac{20}{58.5} = 0.342\ \text{mol},\ a_{w} = 0.929 \\ \text{sucrose} &: \tfrac{20}{342} = 0.058\ \text{mol},\ a_{w} = 0.987 \end{aligned} \]

The mixtures give 0.200 mol of solute (10% + 10%), \(a_{w} = 0.957\), and 0.129 mol (15% + 5%), \(a_{w} = 0.972\).

Only the 20% NaCl solution falls below 0.942, because salt, with its far smaller molar mass, supplies many more moles of solute per gram than sucrose. (Counting the dissociation of NaCl into two ions would lower its value further and strengthen the same conclusion.)

Answer: A

The partial pressure is equal to the partial pressure of water vapour in air.

Water activity is defined as \(a_{w} = p/p_{0}\), the vapour pressure of water over the food divided by that of pure water at the same temperature. The relative humidity of air is \(p_{v}/p_{0}\) at that temperature. If the food is in equilibrium with the air, \(a_{w}\) equals the equilibrium relative humidity, and since both ratios share \(p_{0}\), the vapour pressures themselves are equal.

Equilibrium also means no net moisture transfer: if the food's vapour pressure were lower it would take up moisture from the air, and if higher it would dry. So 'lesser' and 'greater' describe a food that is not in equilibrium, and 'not dependent' contradicts the definition itself.

Answer: D

The rate of drying is 0.0485 kg/h.

Dry solids = 650 - 405 = 245 g. A final moisture of 6.8% on a dry basis means water left = 0.068 x 245 g:

\[ \begin{aligned} \text{water left} &= 0.068 \times 245 = 16.66\ \text{g} \\ \text{removed} &= 405 - 16.66 = 388.34\ \text{g} \\ \text{rate} &= \frac{0.38834\ \text{kg}}{8\ \text{h}} = 0.0485\ \text{kg/h} \end{aligned} \]

A sample of 650 g cannot lose anywhere near 78 to 129 kg in an hour, so the large options cannot be right whatever the working. Because the whole drying is in the constant-rate period, the average rate equals the constant rate.

Answer: D

The relative humidity is about 81.80%.

First the saturation pressure at 30°C from the given equation:

\[ \begin{aligned} \ln p_{w}^{0} &= 18.6556 - \frac{5217.635}{303} \\ &= 18.6556 - 17.2199 = 1.4357 \\ p_{w}^{0} &= 4.20\ \text{kPa} \end{aligned} \]

Then the actual vapour pressure from the absolute humidity, \(H = \dfrac{18}{29}\dfrac{p_{w}}{P - p_{w}}\):

\[ \begin{aligned} p_{w} &= \frac{HP}{0.6207 + H} = \frac{0.0218 \times 101.325}{0.6425} \\ &= 3.438\ \text{kPa} \\ RH &= \frac{3.438}{4.2025} \times 100 = 81.8\% \end{aligned} \]

Using 0.622 for the molar mass ratio gives 81.6%, still closest to 81.80. 62.82, 68.22 and 86.62 do not follow from these numbers.

Answer: B

The one that is not conductive drying is Air drying.

In conductive (contact) drying, heat reaches the food through a hot solid surface in contact with it. Drum drying, where a thin film dries on a steam-heated drum, and vacuum drying on heated shelves or plates are standard examples, and drying in superheated steam is grouped with them in the classification the paper follows.

In air drying, heated air flows over or through the food and supplies the heat while carrying away the vapour. That is convective drying, not conductive drying, so it is the odd one out.

Answer: A

Both Statement I and Statement II are true.

Statement I: drying needs heat to evaporate the water (heat transfer into the food) and the vapour must move out of the food and away from its surface (mass transfer). The two happen together, so drying is a simultaneous heat and mass transfer operation.

Statement II: convective drying is defined by that double role of the hot air. It carries the heat for evaporation to the food surface and sweeps away the water vapour released, keeping the surrounding air dry enough for drying to continue. Both statements are accurate, so every option calling either one false is wrong.

Answer: C

The method is Lyophilization.

Lyophilization (freeze drying) first freezes the food, then lowers the pressure below the triple point of water (about 611 Pa) and supplies a little heat, so the ice turns directly into vapour by sublimation. Because the water never passes through the liquid state and temperatures stay low, the food keeps its structure, flavour and nutrients, and rehydrates readily.

Convective drying uses hot air, conductive drying uses a heated surface, and both evaporate liquid water. 'Kinetic drying' is not a standard drying method.

Answer: A

The process is Freeze drying.

Below the triple point of water (0.01 degrees C and about 611 Pa), ice cannot melt into liquid; when heated it sublimes directly into vapour. Freeze drying uses this by freezing the food and drying it under high vacuum, so the water leaves without ever becoming liquid.

Sun drying, chemical drying and bag drying all remove liquid water by evaporation at ordinary pressure and do not depend on sublimation.