Question

For all positive reaction orders and for a given reactor duty:

• A. Mixed reactor is always larger than the plug-flow reactor
• B. The ratio of the volume of the mixed reactor to that of the plug-flow reactor decreases with order
• C. Reactor size is independent of the type of flow
• D. Density variation during reaction affects design

Hint:

Reactor Size Comparison Rules: - For positive reaction orders (\(n \gt 0\)): \(V_{\text{MFR}} \gt V_{\text{PFR}}\) (MFR requires a larger volume because it operates entirely at the lowest reactant concentration). - For zero-order reactions (\(n = 0\)): \(V_{\text{MFR}} = V_{\text{PFR}}\) (the reaction rate is independent of concentration). - For negative reaction orders (\(n \lt 0\)): \(V_{\text{MFR}} \lt V_{\text{PFR}}\).

Answer 1

The Correct Option is A

Concept: The performance and required size of continuous flow reactors depend on the flow pattern inside the vessel. We can compare the required volumes of a Mixed Flow Reactor (MFR / CSTR) and a Plug Flow Reactor (PFR) for a given duty (the same feed rate, initial concentration, and target fractional conversion $X_A$) by examining their respective performance equations. For a chemical reaction of positive order ($n \gt 0$), the rate of reaction is a monotonically increasing function of reactant concentration: \[ -r_A = k \cdot C_A^n = k \cdot C_{A0}^n \cdot (1 - X_A)^n \] As the fractional conversion $X_A$ increases from 0 to its target value, the reactant concentration drops, which causes the reaction rate ($-r_A$) to decrease continuously.

Step 1: Analyzing the performance equations.
Let us examine the performance equations that determine the required volumes for MFR and PFR systems:

• For a Plug Flow Reactor (PFR): \[ \frac{V_{\text{PFR}}}{F_{A0}} = \int_{0}^{X_A} \frac{dX_A}{-r_A} \] The PFR volume is proportional to the area under the curve on a Levenspiel plot ($1/{-r_A}$ versus $X_A$). Inside a PFR, the concentration drops gradually along the length of the reactor, meaning the reaction rate starts high at the inlet and drops continuously toward the outlet.

• For a Mixed Flow Reactor (MFR / CSTR): \[ \frac{V_{\text{MFR}}}{F_{A0}} = \frac{X_A}{(-r_A)_{\text{exit}}} \] The MFR volume is proportional to the area of a rectangle on a Levenspiel plot evaluated entirely at the exit conversion conditions ($X_A$). Because an MFR is perfectly stirred, the reactant concentration drops immediately to the low exit concentration upon entering the vessel. Consequently, the entire reaction occurs at this minimum rate ($(-r_A)_{\text{exit}}$).

Step 2: Comparing the reaction rates graphically via a Levenspiel plot.
For any positive reaction order ($n \gt 0$), the reaction rate at any intermediate point inside a PFR is strictly greater than or equal to the exit reaction rate: \[ (-r_A)_{\text{internal, PFR}} \ge (-r_A)_{\text{exit, MFR}} \] Taking the reciprocal of these rates invert the inequality: \[ \frac{1}{(-r_A)_{\text{internal, PFR}}} \le \frac{1}{(-r_A)_{\text{exit, MFR}}} \] On a Levenspiel plot, the area under the curve (integrated for the PFR) will always be strictly smaller than the area of the bounding rectangle (calculated for the MFR) for any conversion greater than zero ($X_A \gt 0$). Mathematically, this confirms that: \[ V_{\text{MFR}} \gt V_{\text{PFR}} \] Therefore, for all positive reaction orders and a given reactor duty, a mixed flow reactor is always larger than a plug-flow reactor. This makes statement (1) the correct choice.