Design of a Mixed Flow Reactor / Continuous Stirred Tank Reactor

Hello All, Hope everyone doing good ....!!!

Welcome back to PharmaCalculations.com! After our detailed discussion on Plug Flow Reactor (PFR) design, let’s now explore the design of another commonly used reactor type in pharmaceutical and chemical process industries — the Mixed Flow Reactor (MFR), also known as Continuous Stirred Tank Reactor (CSTR).

This post will walk you through the step-by-step design of an MFR using the same kinetic data and feed conditions used in our previous PFR design post.

If you haven't gone through the post of designing a PFR, please do it now: https://www.pharmacalculations.com/2025/05/design-of-plug-flow-reactor-pfr.html

Objective

To design a Mixed Flow Reactor (MFR) based on available kinetic data, rate constants, and conversion targets, assuming the reaction follows first-order kinetics.

Given Data (Same as PFR Post)

Parameter	Value
Type of reaction	A → Products
Order of reaction	First-order
Rate constant, k	0.0616 min⁻¹
Initial concentration, Cₐ₀	1.5 mol/L
Desired conversion, X	70% or 0.70
Volumetric flow rate, v₀	100 L/min

Step-by-Step MFR Design

Step 1: Performance Equation of MFR

For a Mixed Flow Reactor, the performance equation is:

$$V = \frac{v_0 \cdot X}{-r_A}$$

Where:

• $V$ = Volume of reactor (L)

• $v_0$ = Volumetric flow rate (L/min)

• $X$ = Fractional conversion

• $r_A$ = Rate of reaction at outlet conditions (mol/L·min)

Step 2: Calculate Outlet Concentration

$$C_A = C_{A0} \cdot (1 - X) = 1.5 \cdot (1 - 0.70) = 0.45 \text{ mol/L}$$

Step 3: Rate of Reaction at Outlet

For a first-order reaction,

$$-r_A = k \cdot C_A = 0.0616 \cdot 0.45 = 0.02772 \text{ mol/L·min}$$

Step 4: Calculate Reactor Volume

$$V = \frac{v_0 \cdot X}{-r_A} = \frac{100 \cdot 0.70}{0.02772} = 2525.27 \text{ L}$$

So, the required MFR volume is approximately 2525 L.

Step 5: Reactor Dimensions (Assume Vertical Cylindrical Tank)

Assume Height-to-Diameter (H/D) ratio = 2:1 (common for stirred reactors)

Let:

$$V = \pi \cdot \left(\frac{D}{2}\right)^2 \cdot H \Rightarrow V = \pi \cdot \left(\frac{D}{2}\right)^2 \cdot 2D = \frac{\pi}{2} D^3 \Rightarrow D = \left(\frac{2V}{\pi}\right)^{1/3}$$
$$D = \left(\frac{2 \cdot 2525}{3.1416}\right)^{1/3} = \left(1607.5\right)^{1/3} \approx 11.8\ \text{inches} \approx 0.3\ \text{m}$$

Diameter = 0.3 m, Height = 0.6 m

Step 6: Agitator Power Estimation

Use the Power number (Nₚ) method:

$$P = N_p \cdot \rho \cdot N^3 \cdot D^5$$

Assume:

• $N_p = 5$ (for Rushton turbine)

• $\rho = 1000$ kg/m³ (approx. for liquid)

• $D = 0.2$ m (agitator diameter, 2/3 of tank diameter)

• $N = 2$ rev/s (120 RPM)

$$P = 5 \cdot 1000 \cdot 2^3 \cdot 0.2^5 = 5 \cdot 1000 \cdot 8 \cdot 0.00032 = 12.8\ \text{W}$$

Required agitator power = ~13 W

Hope you understood the post well.

If any queries, pl free to write us at pharmacalc823@gmail.com

Comments are most appreciated .......!!

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About The Author

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Hi! I am Ajay Kumar Kalva, Currently serving as the CEO of this site, a tech geek by passion, and a chemical process engineer by profession, i'm interested in writing articles regarding technology, hacking and pharma technology.
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