Question 10.3: Adiabatic Operation of a Cascade of Continuous Flow Stirred-...

The solution to this problem requires a trial-and-error iterative procedure. Because both the reactor volume and the initial volumetric flow rate are known, the space time per
reactor may be calculated and we may focus our attention initially on the first reactor.

The volumetric flow rate is equal to

\frac{2.1 \times 10^6}{7 \times 10^3} \frac{ lb }{ h } \times 454 \frac{ g }{ lb } \times \frac{ cm ^3}{0.9  g } \times \frac{1  gal }{3785  cm ^3} \text { or } 40.0  gal / h.

This value corresponds to a reactor space time given by

\tau=\frac{V_R}{ V _0}=\frac{1000}{40.0}=25  h

For a first-order reaction it was shown in Illustration 10.2 that

f_{ Al }=\frac{k \tau}{1+k \tau}=\frac{25  k}{1+25  k}       (A)

where the rate constant is a temperature dependent quantity given by equation (C) in Illustration 10.1.

k = 2.61 × 10^{14}e^{−14,570∕T}       (B)

The energy balance equation for adiabatic operation becomes

0=\frac{-F_{ A 0} f_{ A 1}}{v_{ A }} \Delta H_{R \text{ at }T_0}+\sum\left(F_i \int_{T_0}^{T_{\text {out }}} \bar{C}_{p i} d T\right)

where the last term is equal to F_{A0}C_{p}(T_{out} − T_{0}). Division by F_{A0} and rearrangement gives

f_{ A 1}=\nu_{ A } \frac{C_p\left(T_{\text {out }}-T_0\right)}{\Delta H_{R\text { at } T_0}}=\frac{0.5\left(T_{\text {out }}-293\right)}{83}           (C)

Equations (A) to (C) must now be solved simultaneously. Combining these equations gives

f_{ A 1}=\nu_{ A } \frac{C_p\left(T_{\text {out }}-T_0\right)}{\Delta H_{R \text{ at } T_0}}=\frac{0.5\left(T_{\text {out }}-293\right)}{83}=\frac{25  k}{1+25  k}               (D)

where k is given by equation (B) with T = T_{out}. This equation can be solved for T using a trial-and-error procedure. This exercise gives T = 410 K. Thus,

f_{ A 1}=\frac{0.5(410-293)}{83}=0.705

From Illustration 10.2,

f_{ A 2}=1-\frac{1-f_{ A 1}}{1+k \tau}=\frac{f_{ A 1}+k \tau}{1+k \tau}

or

f_{A2} = \frac{0.705  +  25  k}{1  +  25  k}             (E)

where k is given by equation (B).

The energy balance equation for adiabatic operation becomes

\begin{aligned}0= & \frac{F_{ A 0}\left(f_{ A 2}-f_{ A 1}\right)}{\nu_{ A }}\Delta H_{R \text { at } T_0} \\& +F_{ A 0} \int_{T_0}^{T_2} C_p d T-F_{A 0}\int_{T_0}^{T_1} C_p d T\end{aligned}

Division by F_{A0} and combination of the integrals gives

\left(f_{ A 2}-f_{ A 1}\right) \frac{\Delta H_{R \text { at } T_{0}}}{\nu_{ A}}=\int_{T_1}^{T_2} C_p d T

or

f_{ A 2}=f_{ A 1}+\frac{\nu_{ A } \int_{T_1}^{T_2} C_p d T}{\Delta H_{R \text{ at } T_0}}=0.705+\frac{0.5\left(T_2-410\right)}{83}            (F)

Combination of equations (B), (E), and (F) gives

\begin{aligned}0.705 & +\frac{0.5\left(T_2-410\right)}{83} \\& =\frac{0.705+25\left(2.61 \times 10^{14} e^{-14,570 / T_2}\right)}{1+25\left(2.61\times 10^{14} e^{-14,570 / T_2}\right)}\end{aligned}              (G)

A trial-and-error solution gives T_{2} = 458.5  K (approximately 186°C) and f_{A2} = 0.997.

With this reactor configuration one would be able to produce in excess of 2 million pound of B annually if side reactions do not occur at this higher temperature and if the vapor pressure of the solution still lies within a range that does not create problems.