In previous lesson, we have obtained molar balance equations for different reactors. Here we are going to apply that molar balance equation for a CSTR reactor to get familiar with that equation.
CSTR means continuous stirred tank reactor
There is a CSTR which has inflow and outflow operates at steady state. A and B are reactants which enters to the
reactor at a rate of 10.01 mol/min. A and B is react and produce C and D.
A + B → 2C + D.
at steady state A is reacted at a rate of 8.94 mol/min
a) producing rate of products
b) rate of A and B in inflow
According to the reaction, one A molecule and one B molecule react together and produce two C molecules and
one D molecule. Therfore producing rate of C is twice as reducing rate of A.
producing rate of C = 8.94 mol/min * 2 = 17.88 mol/min
producing rate of D = 8.94 mol/min
This reaction occurs in steady state. Therefore no accumulation in this reactor. Here we are going to apply our quantities in moles. Our time basis is per minute.
A is reacting and reducing in the reactor. So generation is negative.
flow rate of A in inflow, FA|in = 10.01 mol/min
generation of A, rA= -8.94 mol/min
accumulation = 0
substitute those values to the molar balance equation,
10.01 + ( - 8.94 ) = FA|out + 0
FA|out = 1.07 mol/min