Question

- Vimy a -page IISL 01 cquauOLID WIum ile HUL S SiginalUIE WIII UE IUweu. Problem 1 The conversion of So, into SO3 using a V205 catalyst is an essential part of the formation of SO; with the final objective of manufacturing H2SO4, SO2(g)+ '/2 O2 (g) A SO2 (g); SO2(g)=A; O2(g)=B; SO3(g)=C; -0 a) It is proposed to develop this reaction in a fluidized bed unit (laboratory scale) with the CSTR model bcing adequate. Derive the equation allowing calculation of the space time. E b) Calculate the space time in minutes for a SO2 conversion of 80%. Data • Isothermal operation and xa,eq=l LE: SO2 and oxygen are feed under equimolar conditions (Fso2,in=Fo2,in) • Volume of the reactor = 500 cm3, Weight of the Catalyst= 200 g. • 20 oxygen and 80% nitrogen in air fed. • r= -k Cso2 (adequate approximation) with k= 1 cm3/gcat min at 700K. Note: The space time equation has to be derived step-by-step starting from basic principles. Units have to be given for the variables involved. A short answer will not do.

Answer 1

a.

Bass balance

[Flow in] -[ Flow out] + [rate of accumulation by chemical reaction] = [rate of accumulation inside reactor]

Assuming that the tank is well-mixed and the reaction rate is constant throughout the reactor we get

Since Steady state is an assumption of this derication that means all temperature dependent terms are 0

So dN/dt = 0

$V =\frac{ (F_{ao} - F_{a}) }{-r_{a}}$

$V =\frac{ (F_{ao} * X) }{-r_{a}}$

X= conversion

We also know that space tiem is given by

$\tau = V/ \vartheta$

v = velocity

So the $V =\frac{ (F_{ao} * X) }{-r_{a}}$ eqn becomes

$V/ \vartheta =\frac{ (C_{ao} * X) }{-r_{a}}$

$\tau =\frac{ (C_{ao} * X) }{-r_{a}}$

X = 0.8

r= - k [SO2]

$\tau =\frac{ (C_{ao} * X) }{ 1 * C_{ao} * (1-x)}$

$\tau =\frac{ X }{1-x}$

$\tau =\frac{ 0.8}{ 0.2}$

$\tau =4$