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PFR Volume for the Same Duty

The same liquid-phase first-order reaction ( $k = 0.10\ \text{min}^{-1}$ ) is instead carried out in a plug flow reactor.

Again target $90\%$ conversion at $v_0 = 10\ \text{L/min}$ , constant density.

Find the required PFR volume.

Given

k	0.1 1/min	Rate constant
X	0.9 -	Fractional conversion
v0	10 L/min	Volumetric feed rate

Hint 1

Concentration falls continuously along a PFR, so the design equation is an integral.

Hint 2

For first order the integral gives a logarithm.

Hint 3

V = (v₀/k)·ln(1/(1−X)).

Worked solution — try the problem first

PFR design equation — concentration varies along the reactor, so this is an integral, not an algebraic ratio:

$V = F_{A0}\int_0^X\frac{dX}{-r_A}$

For first order this integrates to

$V = \frac{v_0}{k}\ln\!\left(\frac{1}{1-X}\right)$

$= \frac{10}{0.10}\ln\!\left(\frac{1}{0.10}\right) = 100\ln(10) = 100(2.3026)$

$\boxed{V = 230\ \text{L}}$

Against the CSTR's 900 L, the PFR needs 3.9× less volume for identical duty. The reason is that a PFR passes through the full concentration profile from $C_{A0}$ down to $0.1C_{A0}$ , spending much of its length at high concentration where the rate is fast. A CSTR operates entirely at the slowest condition.

This advantage holds for any reaction with positive reaction order. It reverses for autocatalytic or strongly product-inhibited kinetics, where a CSTR can be smaller.

Concepts:PFR designReactor comparisonFirst-order kinetics

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