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Unformatted text preview: 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 CSTR = PFR n Writing a mass balance on the n th CSTR, we see that c A,n 1 c A,n = k CSTR c A,n Solving this balance, we find that c A,n = c A,n 1 1 + k CSTR = c A,n 1 1 + k PFR /n Since this recursion holds for all n CSTRs, we conclude that c A,n = c A, (1 + k PFR /n ) n The resultant conversion is given by x = 1 c A,n c A, = 1 1 (1 + k PFR /n ) n In a PFR, we know that c A = c A, e k PFR Thus, we see that x PFR = 1 e k PFR For the problem at hand, we require that x . 95 x PFR . Since k PFR = 1, we require that 1 1 (1 + 1 /n ) n . 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 = CSTR k c 2 A,n 1 Applying the quadratic formula, we find that c A,n = 1 + p 1 + 4 k PFR c A,n 1 /n 2 k PFR /n Hence, we obtain the relation c A,n c A, = 1 + p 1 + 4 k PFR c A,n 1 /n 2 k PFR c A, /n Since k PFR c A, = 1, we arrive at the following recursion relation: c A,n c A, = n + p n 2 + 4 nc A,n 1 /c A, 2 In a PFR, we can use the n th order batch reactor formula from last weeks set with n = 2 by simply replacing t with PFR . This yields c A = k PFR + c 1 A, 1 = c A, 1 + k PFR c A, = 0 . 5 c A, We then require that 1 c A,n c A, . 95(1 . 5) = 0 . 4750 , or equivalently, c A,n c A, . 5250 Since we do not have a closedform 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; for i = 1: n ratio = ( n + sqrt ( n 2 + 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, (1 x ) c Q = c Q, + c A, x 2 The reaction rate is then given by r ( x ) = kc A c Q = k c A, (1 x )( c Q, + c A, x ) Thus we find that 1 r = 1 k c A, (1 x )( c Q, + c A, x ) Using the given feed concentrations, a 1 /r plot for this system is shown in Figure 1. 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 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 CSTR = c A, c A r = c A, x r Thus, the required 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 PFR = Z c A c A, dc A r = c A, Z x dx r 3 Thus, the required 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...
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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.
 Winter '11
 ARNOLD
 Mass Balance

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