P$10-18$ The elementary irreversible gas$-$phase catalytic reaction

$A+BkC+D$

is to be carried out in a moving$-$bed reactor at constant temperature. The reactor contains $5kg$ of cat$-$

alyst. The feed is stoichiometric in A and B$.$ The entering concentration of A is $0\mathrm{.}2mo\frac{l}{d}{m}^3.$ The cat$-$

alyst decay law is zero order with $k_D=0\mathrm{.}2s^{-1}$ and $-$cat $-s$ and the volumetric

flow rate is $v_0=1d\frac{m^3}{s}.$

$($a$)$ What conversion will be achieved for a catalyst feed rate of $0\mathrm{.}5k\frac{g}{s}?($Ans$.$: $x=0\mathrm{.}2)$

$($b$)$ Sketch the catalyst activity as a function of catalyst weight $($i$.$e$.,$ distance$)$ down the reactor length

for a catalyst feed rate of $0\mathrm{.}5k\frac{g}{s}.$

$($c$)$ What is the maximum conversion that could be achieved $($i$.$e$.,$ at an infinite catalyst loading rate$)?$

$($d$)$ What catalyst loading rate is necessary to achieve $40\%$ conversion? $($Ans$.$: $U_S=1\mathrm{.}5k\frac{g}{s})$

$($e$)$ At what catalyst loading rate $(k\frac{g}{s})$ will the catalyst activity be exactly zero at the exit of the

reactor?

$($f$)$ What does an activity of zero mean? Can catalyst activity be less than zero?

$($g$)$ How would your answer in part $($a$)$ change if the catalyst and reactant were fed at opposite ends?

Compare with part $($a$).$

$($h$)$ Now consider the reaction to be zero order with

$k=0\mathrm{.}2mo\frac{l}{k}g-$cat $*min.$

The economics:

The product sells for $$160$ per gram mole.

The cost of operating the bed is $$10$ per kilogram of catalyst exiting the bed.

What is the feed rate of solids $(k\frac{g}{m}in)$ that will give the maximum profit? $($Ans$.$: $U_s=4k\frac{g}{m}in)$

$($Note: For the purpose of this calculation, ignore all other costs, such as the cost of the reactant,

the cost to the company of providing free lunches to workers, etc.$)$