Answer / kangchuentat

REACTION ENGINEERING - ANSWER 13.1 : (a) Theoretical oxygen gas = 100 kmol C (1/1) = 100 kmol of oxygen gas. (b) Theoretical air = (100 kmol) (1 / 0.21) = 476.2 kmol. (c) Amount of air required 50 % excess = 476.2 kmol (1.5) = 714.3 kmol. (d) Carbon consumption to produce carbon dioxide gas = 0.8 x 100 = 80 kmol = oxygen consumption to produce carbon dioxide gas. Carbon consumption to produce carbon monoxide gas = 0.2 x 100 = 20 kmol = 2 times oxygen consumption to produce carbon monoxide gas, then oxygen consumption to produce carbon monoxide gas = 20 / 2 = 10 kmol. Total oxygen gas required = 80 + 10 = 90 kmol. 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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UNIT OPERATION - EXAMPLE 9.3 : In the distillation of binary systems by Mc Cabe Thiele method, the equation for the line of top section is given by y = [ R / (R + 1) ] x + XD / (R + 1). 2 points on the line are (0.99, 0.99) and (0, 0.36). Find the reflux ratio of R and XD.

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Question 43 - In a non-dilute absorber, the inlet gas stream consists of 8 mol % carbon dioxide in nitrogen. By contact with room temperature water at atmospheric pressure, 65 % of the carbon dioxide from a gas stream has been removed. (a) Find the mole ratio of carbon dioxide and nitrogen gases at inlet and outlet gas streams. (b) The Henry’s Law provides y = 1640 x for carbon dioxide in water. Find the mole ratio when x = 0.0000427. Mole ratio is y / (1 - y) for y.

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