A free radical reaction involving nitration of decane is carried out in two sequential reactor stages, each of which operates like a continuous stirred-tank reactor (CSTR). Decane and nitrate (as nitric acid) in varying amounts (Di and Ni, respectively) are added to each reactor stage,as shown in

Question

A free radical reaction involving nitration of decane is carried out in two sequential reactor stages, each of which operates like a continuous stirred-tank reactor (CSTR). Decane and nitrate (as nitric acid) in varying amounts ([latex]D_{i}[/latex] and [latex]N_{i} [/latex], respectively) are added to each reactor stage, asshown in Fig. 19.4. The reaction of nitrate with decane is very fast and forms the following products by successive nitration [latex]DNO_{3} , D(NO_{3})_{2}, D(NO_{3})_{3}, D(NO_{3})_{4}[/latex], and so on. The desired product is [latex]DNO_{3}[/latex], whereas dinitrate, trinitrate, and so on, are undesirable products.

The flow rates of [latex]D_{1}[/latex] and [latex]D_{2}[/latex] are selected to satisfy temperature requirements in the reactors, while [latex]N_{1}[/latex] and [latex]N_{2}[/latex] are optimized to maximize the amount of [latex]DNO_{3}[/latex] produced from stage 2, subject to satisfying an overall level of nitration. In this case, we stipulate that [latex](N_{1} + N_{2})/ (D_{1} + D_{2}) = 0.4[/latex]. There is an excess of [latex]D[/latex] in each stage,and [latex]D_{1}=D_{2} = 0.5 mol/s[/latex].Asteady-state reactor model has been developed to maximize selectivity. Define [latex]r_{1} ≜ N_{1}/D_{1}[/latex] and [latex]r_{2} ≜ N_{2}/(D_{1} + D_{2})[/latex]. The amount of [latex]DNO_{3}[/latex] leaving stage 2 (as mol/s in [latex]F_{2}[/latex]) is given by

[latex]f_{DNO_{3}} = \frac{r_{1}D_{1} }{(1+r_{1} )^{2}(1+r_{2} ) } +\frac{r_{2}D_{2}}{(1+r_{1})(1+r_{2} )^{2} }[/latex] (19-9)

This equation can be derived from the steady-state equations for a continuous stirred reactor with the assumption that all reaction rate constants are equal.

Formulate a one-dimensional search problem in [latex]r_{1}[/latex] that will permit the optimum values of [latex]r_{1}[/latex] and [latex]r_{2}[/latex]to be found. Employ quadratic interpolation using an initial interval of [latex]0 ≤ r_{1} ≤ 0.8[/latex]. Use enough iterations so that the final value of [latex]f_{DNO_{3}}[/latex] is within [latex] ±0.0001[/latex] of the maximum.

Accepted Answer

The six steps described earlier are used to formulate the optimization problem.

Step 1. Identify the process variables. The process variables to be optimized are [latex]N_{1}[/latex] and [latex]N_{2}[/latex], the nitric acid molar flow rates for each stage. Because [latex]D_{1}[/latex] and [latex]D_{2}[/latex] are specified, we can just as well use [latex]r_{1}[/latex] and [latex]r_{2}[/latex], because the conversion model is stated in terms of [latex]r_{1}[/latex] and [latex]r_{2}[/latex].

Step 2. Select the objective function. The objective is to maximize production of [latex]DNO_{3}[/latex] that can be made into useful products, while the other nitrates are unwanted byproducts with a value of zero. The objective function f is given in Eq. 19-9, which is proportional to the profit function, as in Eq. 19-1.

[latex]P= \sum\limits_{s}{F_{s} V_{s} } - \sum\limits_{r}{F_{r}C_{r} } -OC[/latex] (19-1)

Step 3. Develop models for the process and constraints. The values of [latex]N_{1}[/latex] and [latex]N_{2}[/latex]are constrained by the overall nitration level:

[latex]\frac{N_{1}+N_{2} }{D_{1}+D_{2}} =0.4[/latex] (19-10)

which can be expressed in terms of [latex]r_{1}[/latex] and [latex]r_{2}[/latex] as

[latex]\frac{r_{1}D_{1}+r_{2}D_{1}+r_{2}D_{2} }{D_{1}+D_{2}} =0.4[/latex] (19-11)

Inequality constraints on [latex]r_{1}[/latex]and [latex]r_{2}[/latex] do exist, namely, [latex]r_{1}[/latex] ≥ 0 and [latex]r_{2}[/latex] ≥ 0 — because all [latex]N_{i}[/latex] and [latex]D_{i}[/latex] are positive. These constraints can be ignored except when the search method incorrectly leads to negative values of [latex]r_{1}[/latex] or [latex]r_{2}[/latex].

Step 4. Simplify the model. Given [latex]D_{1} = D_{2} = 0.5[/latex], then, from Eq. 19-11,

[latex]r_{2} = 0.4 − 0.5r_{1}[/latex] (19-12)

Because [latex] r_{1}[/latex] and [latex] r_{2}[/latex] are nonnegative, Eq. 19-12 implies that [latex] r_{1}≤ 0.8[/latex] and [latex] r_{2}≤ 0.4[/latex] . After variable substitution, there is only one independent variable [latex] (r_{1})[/latex] in the objective function in Eq. 19-9.

Step 5. Compute the optimum. Because [latex]r_{1}[/latex] lies between [latex]0[/latex] and [latex]0.8[/latex] (the interval of uncertainty), select the three interior points for the search to be [latex] r_{1}= 0.2, 0.4, and 0.6[/latex] . The corresponding values of [latex] r_{2} are 0.3, 0.2, and 0.1[/latex]. Table 19.2 shows the numerical results for three iterations, along with objective function values. After the first iteration, the worst point ([latex]r_{1} = 0.2[/latex]) is discarded and the new point [latex](r = 0.4536)[/latex] is added. After the second iteration, the point with the lowest value of [latex]f( r_{1} = 0.6[/latex]) is discarded. The tolerance on the objective function change is satisfied after only three iterations, with the value of [latex]r_{1}[/latex] that maximizes [latex]f_{DNO_{3}}[/latex] computed to be [latex] r^{opt}_{1} = 0.4439[/latex] . The converted mononitrate is [latex] 0.1348 mol/s[/latex] from stage 2; the remainder of the nitrate is consumed to make higher molecular weight byproducts.

Step 6. Perform sensitivity studies. Based on the results in Table 19.2, the yield is not significantly different from the optimum as long as [latex]0.4 ≤ r_{1} ≤ 0.6[/latex]. Practically speaking, this situation is beneficial, because it allows a reasonable range of decane flows to achieve temperature control. If either[latex]D_{1}[/latex] or [latex]D_{2}[/latex] changes by more than 10%, one should recalculate the optimum. There also might be a need to reoptimize [latex] r_{1}[/latex] and [latex] r_{2}[/latex] if ambient conditions change (e.g., summer vs. winter operation). Even a 1% change in yield can be economically significant if production rates and the selling price of the product are sufficiently high.

Table 19.2 Search Iterations for Example 19.2 (Quadratic Interpolation)

Iteration	[latex] x_{a}[/latex]	[latex] f_{a}[/latex]	[latex] x_{b}[/latex]	[latex] f_{b}[/latex]	[latex] x_{c}[/latex]	[latex] f_{c}[/latex]	[latex] x^{*}[/latex]
1	0.2	0.1273	0.4	0.1346	0.6	0.1324	0.4536
2	0.4	0.1346	0.6	0.1324	0.4536	0.1348	0.4439
3	0.4	0.1346	0.4536	0.1348	0.4439	0.1348	(not needed)
[latex] r^{opt}_{1}[/latex] = 0.4439

Question 19.2: A free radical reaction involving nitration of decane is car...

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