Answer / kangchuentat

PROCESS DESIGN - ANSWER 21.1 : (a) a = (k - 1) / k = (1.4 - 1) / 1.4 = 0.286. Ti = 25 degree Celsius to degree Rankine = (25 + 273.15) (9 / 5) = 536.67 degree Rankine. THp = Q [ Ti / (8130a) ] [ 1 - (Po / Pi) ^ a ] = 230 [ 536.67 / (8130 x 0.286) ] [ 1 - (14.7 / 90) ^ 0.286 ] = 21.469 horsepowers which is more than 20 horsepowers for expander to be used. (b) To = 536.67 (14.7 / 90) ^ 0.286 = 319.629 degree Rankine. 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.

BIOPROCESS ENGINEERING - EXAMPLE 14.2 : An aqueous solution with 2.5 g of a protein dissolved in 600 cubic centimeters of a solution at 20 degree Celsius was placed in a container that has a water-permeable membrane. Water permeated through the membrane until the h - level of the solution was 0.9 cm above the pure water. (a) Calculate the absolute temperature of the solution, T in Kelvin, where T (Kelvin) = T (degree Celsius) + 273.15. (b) Calculate the osmotic pressure, P of the solution by using the formula P = hrg where h is level of the solution, r is density of water with 1000 kg per cubic meter, g = 9.81 N / kg as gravitational acceleration. (c) Calculate the concentration of the protein solution, C in kg / cubic meter. (d) Calculate the molecular weight of the protein, (MW) = CRT / P where R = 8.314 Pa cubic meter / (mol K) as ideal gas constant.

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