Elementary Principles 77

# Elementary Principles 77 - 4.46 a A B = C D nA = nA 0 nB =...

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4- 36 4.46 a. A B ===== C + D Total + =− = =+ = = == + = ∑ + nn y n n y n n y n n y n n AA BB A A T CC B B T DD C C T II D D T Ti 0 00 ξ ξξ e j e j ej e j At equilibrium: yy CD AB Cc Dc Ac Bc = ++ −− = 487 bg . ( n T ’s cancel) 387 487 487 0 0 2 0 0 0 0 2 .. . [] cCD cC D A B cc n n n n abc −+ + + = = ch ∴= −± − = + + + D A B a bb a c a bnn cn n n n 1 2 4 387 2 where . . . b. Basis : 1 mol A feed n A 0 1 = n B 0 1 = 0 0 CDI nnn = Constants: 3.87 9.74 4.87 ab c = () ()() 2 1 2 1 9.74 9.74 4 3.87 4.87 0.688 23 .87 1.83 is also a solution but leads to a negative conversion e e e = = Fractional conversion: 01 0.688 AAe XX = = c. 0 0 80, 0 BC D J n n = = n n n n c C c cA c c c A A = = + =======> = = = = = = ==⇒ =⇒ = 70 0 70 80 70 10 70 70 70 70 70 10 1706 0 0 0 0 0 0 0 mol mol methanol fed . b g
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## This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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