Consider two CSTRs of unequal sizes, V1 and V2, operating in parallel, as shown in Figure 17.2. The liquid-phase reaction, A → products, is first-order and both tanks operate at the same T. How should the total feed rate, qo, be split (at S) so as to maximize the rate of production in the combined

Question

Consider two CSTRs of unequal sizes, [latex]V_1 [/latex] and [latex]V_2 [/latex], operating in parallel, as shown in Figure 17.2. The liquid-phase reaction, A → products, is first-order and both tanks operate at the same T. How should the total feed rate, [latex]q_o[/latex], be split (at S) so as to maximize the rate of production in the combined exit streams (at M)?

Accepted Answer

For a given throughput ([latex]q_o[/latex]), the maximum rate of production is obtained if [latex]c_A[/latex]at M is minimized. We let the ratio of the feed to the two tanks be [latex]r = q_{o2} /q_{o1}[/latex] so that [latex]q_{o2} = rq_{o1} [/latex], where [latex]q_{o1} + q_{o2} = q_o [/latex]. A material balance around M serves to relate [latex]c_{A}[/latex] to [latex]c_{A1} [/latex] and [latex]c_{A2}[/latex] ,each of which can be related to the variable r by means of the material balance for a CSTR and the rate law. We then minimize [latex]c_{A}[/latex]with respect to r, all other quantities being fixed.Material balance for A around point M:

[latex] q_oc_A = q_{o1}c_{A1} + q_{o2}c_{A2} [/latex]

[latex] (1 + r) q_{o1} c_A = q_{o1}c_{A1} + rq_{o1}c_{A2} [/latex]

[latex] c_A = \frac {c_{A1}}{1 + r} + \frac {rc_{A2} }{1 + r} [/latex] (A)

Material balance for A around [latex] V_1 [/latex] , from equation 14.3-15 :

[latex] V_1 = (c_{Ao} - c_{A1}) q_{o1} / ( -r_A) = (c_{Ao} - c_{A1})[q_o /(1 + r)] / k_Ac_{A1} [/latex]

This results in:

[latex]\frac {c_{A1}}{c_{Ao}} = \frac {1}{1 + \frac {k_AV1}{q_o}(1 + r) } [/latex] (B)

Similarly, from a material balance for A around [latex] V_2 [/latex] :

[latex]\frac {c_{A2}}{c_{A0}} - \frac {1}{1 \frac {k_AV_2 1 + r}{q_o r }} [/latex] (C)

Substituting (B) and (C) into (A), we obtain :

[latex]\frac {c_A}{c_{Ao}} = \frac {1}{1 + r}\left[ \frac {1}{(1 + a_1) + a_1 r} + \frac {r^2}{a_2 + ( 1 + a_2)r} ight] [/latex] (D)

where :

[latex]a_1 = k_AV_1/q_o ; a_2 = k_AV_2 /q_o [/latex] (E)

Setting [latex] d(c_A/c_{Ao}) /dr = 0 [/latex] for minimum [latex] c_A[/latex], we solve for r in terms of [latex] a_1 [/latex] and [latex] a_2[/latex]to obtain:

[latex] r \left( = \frac {q_{o2}}{q_{o1}} ight) = \frac {a_2}{a_1} = \frac {V_2}{V_1} [/latex] (17.2.1)

(The analytical solution is rather tedious, but is aided by computer algebra software.)
The result contained in equation 17.2-1 is that the feed should be split in the same ratio as the sizes of the tanks. An implication of this result is that the space times in the two tanks should be equal :

[latex] au _1 = V_1 / q_{o1} = V_2 / q_{o2} = au _2[/latex] (17.2.2)

Furthermore, from equations 17.2-2, (A), (B), and (C)

[latex]c_{A1} = c_{A2} = c_A [/latex] (17.2.3)

That is, the two exit streams are at the same composition before being mixed at M. With the feed divided as in equation (17.2-1) the two tanks act together the same as one tank of volume [latex] V = V_1 + V_2 [/latex],. There is thus no inherent performance advantage of multiple CSTRs in parallel, but there may be increased operating flexibility.

Question 17.3: Consider two CSTRs of unequal sizes, V1 and V2, operating in...

Related Answered Questions

Verified Answer:

Verified Answer:

Verified Answer:

Compare τ for a CSTR and t for a BR, and interpret the comparison graphically. The comparison is for the same fA (in reaction 12.3-l), constant-density, isothermal BR (at same T as in CSTR), and any form of rate law for (-rA) ...

Verified Answer:

For the autocatalytic reaction described in Example 15-10 and the data given there, calculate the volume of a combined CSTR + PFR reactor arranged as in Figure 17.7. ...

Verified Answer:

Verified Answer: