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Set3Solutions

# Set3Solutions - ChE 101 2012 Problem Set 3 Solutions 1(a We...

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ChE 101 2012 Problem Set 3 Solutions 1. (a) We will assume that all of the CSTRs have the same residence time. Thus, for n CSTRs in series, we have τ CSTR = τ PFR n Writing a mass balance on the n th CSTR, we see that c A,n - 1 - c A,n = k τ CSTR c A,n Solving this balance, we find that c A,n = c A,n - 1 1 + k τ CSTR = c A,n - 1 1 + k τ PFR /n Since this recursion holds for all n CSTRs, we conclude that c A,n = c A, 0 (1 + k τ PFR /n ) n The resultant conversion is given by x = 1 - c A,n c A, 0 = 1 - 1 (1 + k τ PFR /n ) n In a PFR, we know that c A = c A, 0 e - k τ PFR Thus, we see that x PFR = 1 - e - k τ PFR For the problem at hand, we require that x 0 . 95 x PFR . Since PFR = 1, we require that 1 - 1 (1 + 1 /n ) n 0 . 95 ( 1 - e - 1 ) Solving this equation numerically, we require that n > 5 . 38, so a minimum of 6 CSTRs are needed to achieve the desired conversion. (b) This time, a balance on the n th CSTR tells us that c A,n - 1 - c A,n = τ CSTR k c 2 A,n 1

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Applying the quadratic formula, we find that c A,n = - 1 + p 1 + 4 k τ PFR c A,n - 1 /n 2 k τ PFR /n Hence, we obtain the relation c A,n c A, 0 = - 1 + p 1 + 4 k τ PFR c A,n - 1 /n 2 k τ PFR c A, 0 /n Since k τ PFR c A, 0 = 1, we arrive at the following recursion relation: c A,n c A, 0 = - n + p n 2 + 4 n c A,n - 1 /c A, 0 2 In a PFR, we can use the n th order batch reactor formula from last week’s set with n = 2 by simply replacing t with τ PFR . This yields c A = k τ PFR + c - 1 A, 0 - 1 = c A, 0 1 + k τ PFR c A, 0 = 0 . 5 c A, 0 We then require that 1 - c A,n c A, 0 0 . 95 (1 - 0 . 5) = 0 . 4750 , or equivalently, c A,n c A, 0 0 . 5250 Since we do not have a closed-form solution for c A,n like for the first order case, we can find the required value of n using the following Matlab script: n = 0 ; ratio = 1 ; while ratio > 0.5250 n = n + 1 ; ratio = 1 ; f o r i = 1 : n ratio = ( - n + sq rt ( n ˆ2 + 4 * n * ratio ) ) / 2 ; end end s p r i n t f ( num2str ( n ) ) This calculation yields a requirement of 7 CSTRs in series to achieve the desired con- version. 2. (a) Writing out the reaction stoichiometry in terms of the conversion of A, we have c A = c A, 0 (1 - x ) c Q = c Q, 0 + c A, 0 x 2
The reaction rate is then given by r ( x ) = kc A c Q = k c A, 0 (1 - x ) ( c Q, 0 + c A, 0 x ) Thus we find that 1 r = 1 k c A, 0 (1 - x ) ( c Q, 0 + c A, 0 x ) Using the given feed concentrations, a 1 /r plot for this system is shown in Figure 1. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 5 10 15 20 25 x k/r (M - 2 ) Figure 1: 1 /r plot for autocatalytic reaction system. (b) We know from the CSTR mass balance that τ CSTR = c A, 0 - c A r = c A, 0 x r Thus, the required τ for a CSTR is given by the area of the rectangle subtended by the 1 /r plot at the desired conversion x . In contrast, the PFR mass balance tells us that τ PFR = - Z c A c A, 0 dc A r = c A, 0 Z x 0 dx r 3

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Thus, the required τ for a PFR is given by the integral under the 1 /r plot up to the conversion x . From Figure 1, we see that at x = 0 . 5, the CSTR rectangle has a smaller area than the PFR integral, so a CSTR should be able to achieve the desired conversion with a smaller volume. Physically, this is the case because up to x = 0 . 5, the system is still in an autocatalytic regime where each turnover of the reaction increases the rate of subsequent turnovers. As such, the reaction rate actually increases with increasing
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Set3Solutions - ChE 101 2012 Problem Set 3 Solutions 1(a We...

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