FOOD ENGINEERING - QUESTION 23.2 : (a) A dryer reduces the moisture content of 100 kg of a potato product from 80 % to 10 % moisture. Find the mass of the water removed in such drying process. (b) During the drying process, the air is cooled from 80 °C to 71 °C in passing through the dryer. If the latent heat of vaporization corresponding to a saturation temperature of 71 °C is 2331 kJ / kg for water, find the heat energy required to evaporate the water only. (c) Assume potato enters at 24 °C, which is also the ambient air temperature, and leaves at the same temperature as the exit air. The specific heat of potato is 3.43 kJ / (kg °C). Find the minimum heat energy required to raise the temperature of the potatoes. (d) 250 kg of steam at 70 kPa gauge is used to heat 49,800 cubic metre of air to 80 °C, and the air is cooled to 71 °C in passing through the dryer. If the latent heat of steam at 70 kPa gauge is 2283 kJ / kg, find the heat energy in steam. (e) Calculate the efficiency of the dryer based heat input and output, in drying air. Use the formula (Ti - To) / (Ti - Ta) where Ti is the inlet (high) air temperature into the dryer, To is the outlet air temperature from the dryer, and Ta is the ambient air temperature.
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FOOD ENGINEERING - ANSWER 23.2 : (a) The mass of dried material is constant throughout the drying process, or 100 kg x 0.2 = 20 kg. Mass of water = mass of dried material (% of water / % of dried material). Initial mass of water = 20 (80 / 20) = 80 kg. Final mass of water = 20 (10 / 90) = 2.222 kg. Mass of the water removed = Final mass - Initial mass = 80 - 2.222 = 77.778 kg. (b) Heat energy required to evaporate the water only = Mass of the water removed x Latent heat of vaporization = 77.778 kg x 2331 kJ / kg = 181300.518 kJ. (c) Minimum heat energy required to raise the temperature of the potatoes = Mass of potatoes of 80 % moisture x specific heat of potato x temperature changes = 100 kg x 3.43 kJ / (kg °C) x (71 - 24) °C = 16121 kJ. (d) Heat energy in steam = mass of steam x latent heat of steam = 250 kg x 2283 kJ / kg = 570750 kJ. (e) Efficiency of the dryer based heat input and output = (Ti - To) / (Ti - Ta) = (80 - 71) / (80 - 24) = 0.1607. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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