EXAMPLE $9\mathrm{.}5$ OPTIMUM MIXED FLOW

REACTOR PERFORMANCE A concentrated aqueous A$-$solution of the previous examples $($CAo$4$ mol$/$liter$,$ FA$0-1000$ mol$/$min$)$ is to be $80\%$ converted in a mixed flow reactor. $($a$)$ What size of reactor is needed? $($b$)$ What is the heat duty if feed enters at $25\backslash$deg C and product is to be withdrawn at this temperature? Note that $1000$ cal $1$kg $1$ liter kg K $1$ iter $4$mol Amol A K cal $-=250$ SOLUTION $($a$)$ Reactor Volume. For Co $4$ moliter we may use the XA versus T chart of Fig. E$9\mathrm{.}3$ as long as we multiply all rate values on this chart by $4$ Following Fig. $9\mathrm{.}9$ the mixed flow operating point should be located where the locus of optima intersects the $80\%$ conversion line $($point C on Fig. E$9$Sa$).$ Here the reaction rate has the value $-?=0\mathrm{.}4$ mol A converted$/$min$.$ liter From the performance equation for mixed flow reactors, Eq$.5\mathrm{.}11.$ the volume required is given by FAO $000$ mol$/$min$)(0\mathrm{.}80)($rA $0\mathrm{.}4$ momin liter $-2000$ liters

EXAMPLE $9\mathrm{.}5$ OPTIMUM MIXED FLOW REACTOR PERFORMANCE

A concentrated aqueous A$-$solution of the previous examples $,F_{A0}=1000mo\frac{l}{m}in)$ is to be $80\%$ converted in a mixed flow reactor.

$($a$)$ What size of reactor is needed?

$($b$)$ What is the heat duty if feed enters at $25C$ and product is to be withdrawn at this temperature?

Note that

$C_{pA}=\frac{1000\text{c}\text{a}\text{l}}{kg*K}*\frac{1(kg)}{1\text{l}\text{i}\text{t}\text{e}\text{r}}*\frac{1\text{l}\text{i}\text{t}\text{e}\text{r}}{4(mol)(A)}=250\frac{\text{c}\text{a}\text{l}}{\text{m}\text{o}\text{l}\text{A}*K}$

SOLUTION

$($a$)$ Reactor Volume. For $C_{A0}=4mo\frac{l}{?}$ liter we may use the $x_A$ versus $T$ chart of Fig. E$9\mathrm{.}3$ as long as we multiply all rate values on this chart by $4.$

Following Fig. $9\mathrm{.}9$ the mixed flow operating point should be located where the locus of optima intersects the $80\%$ conversion line $($point $C$ on Fig. E$9\mathrm{.}5$a$).$ Here the reaction rate has the value

$-r_A=0\mathrm{.}4$molA converted $\frac{?}{\text{m}\text{i}\text{n}}*$ liter

From the performance equation for mixed flow reactors, Eq$.5\mathrm{.}11,$ the volume required is given by

$V_{{?}_{?}}=\frac{F_{A0}{x}_A}{(-r_A)}=\frac{(1000\frac{\text{m}\text{o}\text{l}}{m}in)(0\mathrm{.}80)}{0\mathrm{.}4\frac{\text{m}\text{o}\text{l}}{m}in*\text{l}\text{i}\text{t}\text{e}\text{r}}=2000$ liters ${?}_{(}{)}_{?}$