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Unformatted text preview: ChE 101 2012 Problem Set 4 Solutions 1. (a) We start with the balance on A in a CSTR, which tells us that = c A, c A k 1 c 2 A + k 2 c A Rearranging terms, we obtain the equation k 1 c 2 A + (1 + k 2 ) c A c A, = 0 Applying the quadratic formula and taking the positive root, we find that c A = (1 + k 2 ) + q (1 + k 2 ) 2 + 4 k 1 c A, 2 k 1 Now we use the CSTR balance for B, c B = ( k 2 c A k 3 ) = (1 + k 2 ) + q (1 + k 2 ) 2 + 4 k 1 c A, 2 k 1 /k 2 k 3 Rearranging terms, we conclude that Y CSTR = k 2 2 k 1 c A, q (1 + k 2 ) 2 + 4 k 1 c A, k 2 2 k 1 c A, (1 + k 2 ) k 3 c A, Next, we note that c A, c A = 2 k 1 c A, + (1 + k 2 ) q (1 + k 2 ) 2 + 4 k 1 c A, 2 k 1 Thus we conclude that S CSTR = k 2 q (1 + k 2 ) 2 + 4 k 1 c A, k 2 (1 + k 2 ) 2 k 1 k 3 2 2 k 1 c A, + 1 + k 2  q (1 + k 2 ) 2 + 4 k 1 c A, (b) In a PFR, the mass balance for A becomes = Z c A c A, dc A k 1 c 2 A + k 2 c A 1 Integrating this in Mathematica, we find that = 1 k 2 ln c A ( c A, k 1 + k 2 ) c A, ( c A k 1 + k 2 ) Then solving for c A in Mathematica, we obtain the result c A = k 2 c A, ( k 1 c A, + k 2 ) e k 2  k 1 c A, The differential balance for B is then given by dc B d = k 2 c A k 3 Integrating this balance, we find that c B = Z k 2 2 c A, ( k 1 c A, + k 2 ) e k 2  k 1 c A, k 3 d Evaluating the integral in Mathematica, we conclude that Y PFR = k 2 k 1 c A, ln 1 + k 1 c A, k 2 e k 2  k 1 c A, k 2 k 2 2 k 1 c A, + k 3 c A, To compute the selectivity for B, we use Mathematica to find c A, c A = c A, ( e k 2  1 ) ( k 1 c A, + k 2 ) ( k 1 c A, + k 2 ) e k 2  k 1 c A, Thus we conclude that S PFR = k 2 k 1 c A, ln 1 + k 1 c A, k 2 e k 2  k 1 c A, k 2 k 2 2 k 1 c A, + k 3 c A, ( k 1 c A, + k 2 ) e k 2  k 1 c A, ( e k 2  1)( k 1 c A, + k 2 ) (c) To maximize the yield of B in a CSTR we require that 0 = dY CSTR d = k 2 [2 k 2 (1 + k 2 ) + 4 k 1 c A, ] 4 k 1 c A, q (1 + k 2 ) 2 + 4 k 1 c A, k 2 2 2 k 1 c A, k 3 c A, For convenience, we multiply both sides by 2 k 1 c A, : 0 = k 2 2 (1 + k 2 ) + 2 k 1 k 2 c A, q (1 + k 2 ) 2 + 4 k 1 c A, k 2 2 2 k 1 k 3 We then separate out the square root as follows: q (1 + k 2 ) 2 + 4 k 1 c A, = k 2 2 (1 + k 2 ) + 2 k 1 k 2 c A, k 2 2 + 2 k 1 k 3 This is now a quadratic equation that can be solved in Mathematica to yield the solution max = (2 c A, k 1 + k 2 ) k 3 ( k 2 2 + k 1 k 3 ) + q c A, ( k 1 c A, + k 2 ) k 3 ( k 2 2 + k 1 k 3 )( k 2 2 + 2 k 1 k 3 ) 2 k 2 2 k 3 ( k 2 2 + k 1 k 3 ) 2 (d) The analogous optimization in a PFR requires that 0 = dc B d But we know from the PFR mass balance that the right hand side above is given by k 2 c A k 3 , so the relevant equation to solve is k 3 k 2 = k 2 c A, ( k 1 c A, + k 2 ) e k 2  k 1 c A, Solving in Mathematica, we find that max = 1 k 2 ln " c A, ( k 2 2 + k 1 k 3 ) k 3 ( k 1 c A, + k 2 ) # 2. (a) A mass balance on species...
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 Winter '11
 ARNOLD

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