Set4Solutions

# Set4Solutions - ChE 101 2012 Problem Set 4 Solutions 1(a We...

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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 τ = c A,- c A k 1 c 2 A + k 2 c A Rearranging terms, we obtain the equation k 1 τ c 2 A + (1 + k 2 τ ) c A- c A, = 0 Applying the quadratic formula and taking the positive root, we find that c A =- (1 + k 2 τ ) + q (1 + k 2 τ ) 2 + 4 k 1 τ c A, 2 k 1 τ Now we use the CSTR balance for B, c B = τ ( k 2 c A- k 3 ) =- (1 + k 2 τ ) + q (1 + k 2 τ ) 2 + 4 k 1 τ c A, 2 k 1 /k 2- k 3 τ Rearranging terms, we conclude that Y CSTR = k 2 2 k 1 c A, q (1 + k 2 τ ) 2 + 4 k 1 τ c A,- k 2 2 k 1 c A, (1 + k 2 τ )- k 3 τ c A, Next, we note that c A,- c A = 2 k 1 τ c A, + (1 + k 2 τ )- q (1 + k 2 τ ) 2 + 4 k 1 τ c A, 2 k 1 τ Thus we conclude that S CSTR = k 2 τ q (1 + k 2 τ ) 2 + 4 k 1 τ c A,- k 2 τ (1 + k 2 τ )- 2 k 1 k 3 τ 2 2 k 1 τ c A, + 1 + k 2 τ- q (1 + k 2 τ ) 2 + 4 k 1 τ c A, (b) In a PFR, the mass balance for A becomes- τ = Z c A c A, dc A k 1 c 2 A + k 2 c A 1 Integrating this in Mathematica, we find that- τ = 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 τ- k 1 c A, The differential balance for B is then given by dc B dτ = k 2 c A- k 3 Integrating this balance, we find that c B = Z τ k 2 2 c A, ( k 1 c A, + k 2 ) e k 2 τ- k 1 c A,- k 3 dτ 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 τ- k 1 c A, k 2- k 2 2 k 1 c A, + k 3 c A, τ To compute the selectivity for B, we use Mathematica to find c A,- c A = c A, ( e k 2 τ- 1 ) ( k 1 c A, + k 2 ) ( k 1 c A, + k 2 ) e k 2 τ- 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 τ- k 1 c A, k 2- k 2 2 k 1 c A, + k 3 c A, τ × ( k 1 c A, + k 2 ) e k 2 τ- k 1 c A, ( e k 2 τ- 1)( k 1 c A, + k 2 ) (c) To maximize the yield of B in a CSTR we require that 0 = dY CSTR dτ = k 2 [2 k 2 (1 + k 2 τ ) + 4 k 1 c A, ] 4 k 1 c A, q (1 + k 2 τ ) 2 + 4 k 1 τ 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 τ ) + 2 k 1 k 2 c A, q (1 + k 2 τ ) 2 + 4 k 1 τ c A,- k 2 2- 2 k 1 k 3 We then separate out the square root as follows: q (1 + k 2 τ ) 2 + 4 k 1 τ c A, = k 2 2 (1 + k 2 τ ) + 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 τ 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 dτ 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 τ- k 1 c A, Solving in Mathematica, we find that τ 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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## This note was uploaded on 03/21/2012 for the course CHE 101 taught by Professor Arnold during the Winter '11 term at Caltech.

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Set4Solutions - ChE 101 2012 Problem Set 4 Solutions 1(a We...

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