The energy, which must be supplied to vaporize the water at any temperature, depends upon this temperature. The quantity of energy required per kg of water is called the latent heat of vaporization, if it is from a liquid, or latent heat of sublimation if it is from a solid. The heat energy required to vaporize water under any given set of conditions can be calculated from the latent heats given in the steam table in Appendix 8, as steam and water vapour are the same thing.
EXAMPLE 7.1. Heat energy in air drying A food containing 80% water is to be dried at 100°C down to moisture content of 10%. If the initial temperature of the food is 21°C, calculate the quantity of heat energy required per unit weight of the original material, for drying under atmospheric pressure. The latent heat of vaporization of water at 100°C and at standard atmospheric pressure is 2257 kJ kg-1. The specific heat capacity of the food is 3.8 kJ kg-1 °C-1 and of water is 4.186 kJ kg-1 °C-1. Find also the energy requirement/kg water removed.
Calculating for 1 kg foodHeat energy required for 1kg original material
Initial moisture = 80%
800 g moisture are associated with 200 g dry matter.
Final moisture = 10 %,
100 g moisture are associated with 900 g dry matter,
Therefore (100 x 200)/900 g = 22.2 g moisture are associated with 200 g dry matter.
1kg of original matter must lose (800 - 22) g moisture = 778 g = 0.778 kg moisture.
= heat energy to raise temperature to 100°C + latent heat to remove water
= (100 - 21) x 3.8 + 0.778 x 2257
= 300.2 + 1755.9
= 2056 kJ.
Energy/kg water removed, as 2056 kJ are required to remove 0.778 kg of water,
= 2056/0.778
= 2643 kJ.
Steam is often used to supply heat to air or to surfaces used for drying. In condensing, steam gives up its latent heat of vaporization; in drying, the substance being dried must take up latent heat of vaporization to convert its liquid into vapour, so it might be reasoned that 1 kg of steam condensing will produce 1 kg vapour. This is not exactly true, as the steam and the food will in general be under different pressures with the food at the lower pressure. Latent heats of vaporization are slightly higher at lower pressures, as shown in Table 7.1. In practice, there are also heat losses and sensible heat changes which may require to be considered.
TABLE 7.1
LATENT HEAT AND SATURATION TEMPERATURE OF WATER
| Absolute pressure | Latent heat of vaporization | Saturation temperature |
| (kPa) | (kJ kg-1) | (°C) |
| 1 | 2485 | 7 |
| 2 | 2460 | 18 |
| 5 | 2424 | 33 |
| 10 | 2393 | 46 |
| 20 | 2358 | 60 |
| 50 | 2305 | 81 |
| 100 | 2258 | 99.6 |
| 101.35 (1 atm) | 2257 | 100 |
| 110 | 2251 | 102 |
| 120 | 2244 | 105 |
| 200 | 2202 | 120 |
| 500 | 2109 | 152 |
EXAMPLE 7.2. Heat energy in vacuum drying Using the same material as in Example 7.1, if vacuum drying is to be carried out at 60°C under the corresponding saturation pressure of 20 kPa abs. (or a vacuum of 81.4 kPa), calculate the heat energy required to remove the moisture per unit weight of raw material.
Heat energy required per kg raw material
= heat energy to raise temperature to 60°C + latent heat of vaporization at 20 kPa abs.
= (60 - 21) x 3.8 + 0.778 x 2358
= 148.2 + 1834.5
= 1983 kJ.
In freeze drying the latent heat of sublimation must be supplied. Pressure has little effect on the latent heat of sublimation, which can be taken as 2838 kJ kg-1.
EXAMPLE 7.3. Heat energy in freeze drying If the foodstuff in the two previous examples were to be freeze dried at 0°C, how much energy would be required per kg of raw material, starting from frozen food at 0°C?
Heat energy required per kilogram of raw material= latent heat of sublimation
= 0.778 x 2838
= 2208 kJ.
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