Ind. Eng. Chem. Res. 1988,27, 214
214
Response to Comments on “Optimization of Consecutive Reactions with Recovery and Reuse of Unconverted Reactant” Sir: Kambitsis et al. (1988) have written a comment about our paper (Liu et al., 1987). The purpose of this note is the following: to correct an error in eq 13 of our paper, to extend the analysis to the more practical case where not only is the separation of A from R not perfect but the separation of R from S is not perfect, and to show the significance of the optimum equation in industry. First, in eq 13 of our original paper (Liu et al., 1987), there is a typographical error. This equation should read + k (“*AY-’ -
m ( 1 - k ) (“*A)’ PFR
- 1= 0
(13)
PFR
Second, the more practical situation for consecutive reaction should be that not only is the separation of A from R not perfect, but the separation of R from S is not perfect. Figure 1 of Kambitsis et al. (1988) gives the simplified process. According to Figure 1 of Kambitsis et al., the yield of the system considered here should be
where m = the recovery ratio of A, n = recycle ratio of R, 1 = the loss ratio of R, and Z = the one-pass nonconversion of A through the reactor. The optimum equation for consecutive reactions reads
optimum conversion in the development and design of the process. Naturally, if the raw material is very expensive, and the operating cost and depreciation charge are very low and can be neglected, the conversion given in eq 3 is optimum for the process. On the other hand, if the raw material is very cheap and the costs for the operation and depreciation charge are very high, then unconverted reactant need not be recovered and reused; i.e., m = n = 0. In this case, eq 2 gives the optimum conversion for the process. But since the operating cost and depreciation charge generally can not be neglected in contrast to the costs of raw materials, the conversion for the development and design of the process must be between the optimum determined by eq 2 under m = n = 0 (upper boundary) and the optimum determined by the same equation under the practical m and n (lower boundary). Thus, using an optimum algorithm derivative free, we can find the optimum conversion of the process according to eq 7 in this interval. In addition, the costs for the raw material take a great proportion of the total cost in a practical chemical process, so the optimum conversion for the development and design is generally near the lower boundary of the interval.
Example If the degree of chlorination (mole of chlorine/mole of 1-methyl-4-chlorobenzene) is less than 1, the following reaction is a first-order consecutive reaction: C
H
3
e
C
l t C12
If the first-order consecutive reaction is carried out in a PFR (piston flow reactor), then we obtain (1 - n - 1)(Zk - 2 ) YPFR= h # 1 (3) (1 - k ) ( l - n Z k ) ( l - mZ) ( 1 - n - 1)Z I n 2
YPFR= (1 - mZ)(nZ - 1)
k = l
(4)
If 1 = 0, eq 3 and 4 reduce to eq 1 and 2 of Kambisis et al. If the first-order consecutive reaction is carried out in a CSTR (continuous stirred tank reactor), then Z ( l - Z)(1 - n - 1) YCsTR = (1 - mZ)(Z(l - h - n) + k ) Z(1 - h - n) + k # 0 ( 5 ) and -k + ( k ( 1 - m ) ( l - n))1/2 ( Z ~ ~ ~ ) C=S T R(1 - m ) ( l - n) - k (1 - m)(l - n) - k # 0 ( 6 )
If 1 = 0 and n = 0, eq 5 and 6 reduce to eq 17 and 11 in our paper (Liu et al., 1987). All the equations in our previous paper, in the comment of Kambitsis et al., and in this note are obtained under the condition that only the utilization of the reactant is considered. In industry, the operating cost and equipment depreciation charge should also be considered. So the proper objective function should be j = cost for material + cost for operation equipment depreciation (7)
+
The conversion that is obtained under aJ/aZ = 0 is the
light 7 CHpCl
e.1 t HCI
(S)
In this system, the reactor corresponds to two CSTR in series; that is, N = 2. k is between 0.161 and 0.179, the recovery ratio of A is 0.99, and n is 0. What is the optimum degree of chlorination?
Solution 1. Combining eq 10 (Liu et al., 1987) for m = 0.99 and n = 0 with the material balance equation, we can obtain the degree of chlorination Dlopt= 0.2-0.22 (lower boundary). 2. Combining eq 10 for m = n = 0 with the material balance equation, we can obtain the degree of chlorination Dzopt= 0.9 (upper boundary). 3. Between 0.2 and 0.9, we find the optimum degree of chlorination Do, = 0.3 using an optimum algorithm of one variation. This value is what is used in industry. Literature Cited Kambitsis, C.; Turton, R.; Levenspiel, 0. “Comments on ‘Optimization of Consecutive Reactions with Recovery and Reuse of Unconverted Reactant”’. Ind. Eng. Chem. Res. 1988, 2 preceding paper in this issue. Liu, D.-Z.; Xu, H A ;Wang, A.-Z. Ind. Eng. Chem. Res. 1987,26, 376.
Liu Da-Zhuang,* Xu Hai-Sheng, Wang An-Zhong Chemical Engineering Department Zhengzhou Institute of Technology Zhengzhou, The People’s Republic of China
