Answer / kangchuentat

CHEMICAL MATERIAL BALANCE – ANSWER 2.6 : (a) For component 1, ln (PSAT) = 15 - 3010 / (150 + 250) = 7.475, PSAT = 1763 kPa. For component 2, ln (PSAT) = 14 - 2700 / (150 + 205) = 6.394, PSAT = 598 kPa. Let x(1) PSAT(1) = y(1) P, x(2) PSAT(2) = y(2) P, then x(1) PSAT(1) + x(2) PSAT(2) = [ y(1) + y(2) ] P = P, x(1) (1763) + [ 01 - x(1) ] (598) = 1210, x(1) (1763 - 598) = 1210 - 598, x(1) = 0.525, x(2) = 1 - x(1) = 0.475. (b) y(1) = x(1) PSAT(1) / P = 0.525 (1763) / 1210 = 0.765, y(2) = 1 - y(1) = 0.235. Overall mass balance is assumed 1 mole for F = V + L where F is incoming mole, V is flashed vapor in mole and L is outgoing liquid in mole. For component 1, zF = xL + yV then 0.6(1) = 0.525 L + 0.765 (1 - L), then L = 0.6875, V = 1 - L = 0.3125. Let z to be 60 mole %. 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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