The analysis of the performance of the CSTR is identical with that of the first CSTR considered in Illustration 10.3. The space time for the CSTR is
\tau=\frac{V_R}{ V _0}=\frac{1000}{40.0}=25 h
The fraction conversion at the CSTR exit for this first-order reaction is given by
f_{ A 1}=\frac{k \tau}{1+k \tau}=\frac{25 k}{1+25 k} (A)
The rate constant is given by
k=2.61 \times 10^{14} e^{-14,570 / T} (B)
The energy balance equation for adiabatic operation is
0=-\frac{F_{ A 0} f_{ A 1}}{\nu_{ A }} \Delta H_{R \text { at } T_0}+F_{ A 0} \int_{T_0}^{T_{\text {out }}} C_p d T
or, assuming a constant heat capacity for the liquid,
f_{ A 1}=\frac{v_{ A } 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) can now be solved simultaneously using a trial-and-error iterative procedure. One finds that f_{A1} = 0.705 and T = 410 K. These properties are those of the stream entering the plug flow reactor. The design equation for this reactor is
\tau=C_{ A 0} \int_{0.705}^{0.97} \frac{d f_A}{k C_{A 0}\left(1-f_A\right)}=\int_{0.705}^{0.97} \frac{d f_A}{k\left(1-f_A\right)} (D)
where k is again given by equation (B).
An energy balance on the PFR operating at steady state is given by equation (10.4.6). For adiabatic operation this equation becomes
\begin{aligned}U\left(T_m-T\right) \frac{4}{D} F_{ A 0} \frac{d f_{ A }}{\left(-r_{ A }\right)}= & \sum\left(F_i \int_{T_0}^{T \text { out of element }} \bar{C}_{p i} d T\right) \\& -\sum\left(F_i \int_{ T _0}^{T_{\text {into element }}} \bar{C}_{p i} d T\right) \\& -\frac{F_{ A 0} \Delta H_{R \text { at } T_0}}{\nu_A} d f_{ A }\end{aligned} (10.4.6)
\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_{\text {leaving PFR }}} C_p d T-F_{A 0} \int_{T_0}^{T_{\text {entering PFR }}} C_p d T\end{aligned} (E)
Because the heat capacity of the reaction mixture is described as being independent of temperature and composition, equation (E) simplifies to
\begin{aligned}& 0=\frac{\left(f_{ A 2}-f_{ A 1}\right) \Delta H_{R \text { at } T_0}}{\nu_{ A }} \\& +C_p\left(T_{\text {leaving PFR }}-T_{\text {entering PFR}}\right) \\&\end{aligned} (F)
The relationship between the temperature and the fraction conversion at a particular point in the plug flow reactor can be obtained by setting f_{A2} = f_{A} and Tleaving PFR equal to T. Hence
T=T_{\text {entering PFR }}+\frac{\Delta H_{R \text { at } T_0}}{\nu_{ A }C_p}+\left(f_{ A }-f_{ A 1}\right) (G)
In numerical terms,
T=410+\frac{-83}{-1(0.5)}\left(f_{ A }-0.705\right)=410+166\left(f_{ A}-0.705\right) (H)
At the exit from the PFR, the temperature will be 454 K or 181°C.
Equations (B), (D), and (H) may now be solved simultaneously to determine the space time required for the plug flow reactor. For the present situation τ = 3.72 h so the required PFR volume is 3.72(40) = 148.8 gal or 19.8 ft³. By operating in this mode, one eliminates or minimizes the necessity for using heat exchangers. If one regards the exit temperature of 181°C as excessive, a heat exchanger could be used between the CSTR and the PFR, or arrangements could be made for internal cooling within the PFR.