Answer / kang chuen tat (malaysia - pen

Answer 73 - (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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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.20 : Assume an engineer buys a $1 bond in period T while the nominal interest rate is R. The inflation rate at T + 1 is anticipated to be I. (a) If the bond is redeemed in period T + 1, how much money will the buyer engineer receive, in term of R, which is not affected by inflation? (b) Find the present value, PV of the proceeds from the bond, in term of R and I. (c) If the bond is redeemed in period T + 1, calculate the real growth or real value of the money that the buyer engineer will receive, in term of r = real interest rate, which is affected by inflation. (d) From the answers in (b) and (c), find the values of x, y and z in the following Fisher equation : (1 + x) = (1 + y) / (1 + z), in term of r, R and I.

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Question 72 - (a) According to United States Department of Agriculture (USDA) (http://ndb.nal.usda.gov/ndb/search/list, accessed 12 August 2016), 100 g of potatoes generate 77 kcal of energy. For raw tomatoes, 111 g have 18 kcal of energy. Question : How much energy will one gain if 150 g of heated potatoes are eaten with 200 g of raw tomatoes? (b) If 1 Calorie = 1 food Calorie = 1 kilocalorie and 1000 calories = 1 food Calorie, then how many Calories are there in 9600 calories? (c) According to a food package of potato chips, 210 Calories are produced per serving size of 34 g. In actual experiment of food calorimetry lab, 1.75 g of potato chips, when burnt, will produce 9.6 Calories. For each serving size of potato chip, find the difference of Calories between the actual experimental value and the value stated on the food package. (d) The specific heat of water is c = 1 cal / (g.K) where cal is calory, g is gram and K is Kelvin. Then what is the temperature rise of water, in degree Celsius, when 150 g of water is heated by 9600 calories of burning food?

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Question 73 - (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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