In an ideal mixed flow reactor we can write-
Using the mass conservation of A
Rate of mass coming in - rate of mass going out- rate of
disappearance by the reaction = rate of accumulation of mass
Rate of mass coming in = flow rate coming in*concentration of the
incoming reactant
=q*Cao
=5 L/he * 100 mole/L
=500 mole/hr
Assuming that the reactor is well mixed and thus the concentration
in the reactor as well as in the outgoing steam is C.
Rate of mass going out = q*C
=5C mole/hr
Rate of disappearance due to the reaction=
Net rate of reaction*Volume of the reactor
(Vm1*C/km1+C )+ Vm2*C/km2+C
Values are given in the question-
Net rate of reaction
=3.3C/(0.05+C ) +5C/(200+C)
Rate of accumulation of mass = d(V*C)/dt
V= 100 L.
d(100*C)/dt
Keeping all the values in the conservation equation.
500-5C- 100*(3.3C/(0.05+C ) +5C/(200+C)=100dC/dt
This is the mass balance equation for the given case.
Hope this helps
Thanks
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