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For example, initially this reactor is having a 2nd order reaction of $-ra=kaCa^2$ and it has a fractional conversion of 50% in a CSTR.

I replace it with something 1/6 as large. What happens to the conversion assuming everything else stays the same

Through theory, I know the conversion will decrease as its a smaller CSTR, but is there a way to calculate it using the formula - $fa = Fao - Fa1 / Fao$?

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  • $\begingroup$ Are the volumetric flow rates in and out to the reactor the same? $\endgroup$ – Jeffrey J Weimer Dec 12 '18 at 4:52
  • $\begingroup$ @JeffreyJWeimer yes $\endgroup$ – mutu mumu Dec 12 '18 at 5:03
  • $\begingroup$ What do you mean by "fa"? fa has one meaning I am clear on... $\endgroup$ – Solar Mike Dec 12 '18 at 7:05
  • $\begingroup$ @SolarMike fa is the fractional conversion, Fao is the Initial Molar flowrate of reactant A $\endgroup$ – mutu mumu Dec 12 '18 at 7:07
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The mole balance on component A through the reactor balances flow rates out and in with rate of change.

$$ \dot{n}_{Ao} - \dot{n}_{Ai} = \frac{dn_A}{dt} $$

The left side can be written in terms of overall conversion for a consumption reaction.

$$ f_A \equiv \frac{\dot{n}_{Ai} - \dot{n}_{Ao}}{\dot{n}_{Ai}} $$

The right side can be rewritten as a rate expression on concentration $dn_A / dt = V dC_A/dt$. Use an empirical rate law expression for the rate. Assume the reaction consumes component A. In a CSTR, the outlet concentration is the reactor concentration. Combined we obtain

$$ f_A\ \dot{n}_{Ai} = V\ k\ C_{Ao}^n $$

When you want to decrease the volume of the reactor $V$, you must change these factors with these consequences:

  • Decrease the volumetric flow rate of A in to keep the overall conversion $f_A$ constant.

  • Decrease the overall conversion $f_A$ to keep the same volumetric flow rate of A in.

  • Increase the concentration of A in the outlet for the same overall conversion and volumetric flow rate of A in.

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