Chapter 5: Isothermal Reactor Design: Conversion

Note: Pressure drop does NOT affect liquid phase reactions

Sample Question:

Analyze the following second order gas phase reaction that occurs isothermally in a PBR: 

\( A \rightarrow B \)

\( F_{A0} \frac{dX}{dW} = -r'_A \)

Must use the differential form of the mole balance to separate variables


Rate Law

Second order in A and irreversible:

\(-r'_A = kC_A^2\)

\(C_A = \frac{F_A}{\nu} = C_{A0} \frac{(1-X)}{(1+\epsilon X)} \frac{P \, T_0}{P_0 \, T}\)

Isothermal, T = T₀

\(C_A = C_{A0} \frac{(1-X)}{(1+\epsilon X)} \frac{P}{P_0}\)

\(\frac{dX}{dW} = \frac{kC_{A0}^2}{F_{A0}} \frac{(1-X)^2}{(1+\epsilon X)^2} \left(\frac{P}{P_0}\right)^2\)

Need to find (P/P₀) as a function of W (or V if you have a PFR).

Pressure Drop in Packed Bed Reactors

Ergun Equation

\( \frac{dP}{dz} = - \frac{G}{\rho g_c D_p} \left(\frac{1-\phi}{\phi^3}\right) \left[ \underbrace{\frac{150 (1-\phi) \mu}{D_p}}_{\text{laminar}} + \underbrace{1.75G}_{\text{turbulent}} \right] \)

Variable Gas Density

\( \rho = \rho_0 \frac{P}{P_0} \frac{T_0}{T} \frac{F_{T_0}}{F_T} \)

\( \frac{dP}{dz} = - \frac{G}{\rho_0 g_c D_p} \left( \frac{1-\phi}{\phi^3} \right) \left[ \frac{150 (1-\phi) \mu}{D_p} + 1.75G \right] \frac{P_0}{P} \frac{T}{T_0} \frac{F_T}{F_{T_0}} \)

let \( \beta_0 = \frac{G}{\rho_0 g_c D_p} \left( \frac{1-\phi}{\phi^3} \right) \left[ \frac{150 (1-\phi) \mu}{D_p} + 1.75G \right] \)

Catalyst Weight

\( W = zA_c \rho_b = zA_c (1-\phi) \rho_c \)

where \( \rho_b = \text{bulk density}, \quad \rho_c = \text{solid catalyst density}, \quad \phi = \text{porosity (void fraction)} \)

\( \frac{dP}{dW} = \frac{-\beta_0}{A_c (1-\phi) \rho_c} \frac{P_0}{P} \frac{T}{T_0} \frac{F_T}{F_{T_0}} \)

let \( \alpha = \frac{2\beta_0}{A_c (1-\phi) \rho_c P_0} \)

then \( \frac{dP}{dW} = -\frac{\alpha}{2} \frac{P_0}{P} \frac{T}{T_0} \frac{F_T}{F_{T_0}} \)

\( \frac{d\left(\frac{P}{P_0}\right)}{dW} = -\frac{\alpha}{2} \frac{T}{T_0} \frac{F_T}{F_{T_0}} \left(\frac{P}{P_0}\right) \)

We will use this form for multiple reactions:

\(\frac{\frac{dp}{dW}}{\alpha} = -\frac{\frac{T}{T_o} \frac{F_T}{F_{T_o}}}{2p}\)

\(p = \frac{P}{P_o}\)

We will use this form for single reactions:

\(\frac{d(P/P_o)}{dW} = -\frac{\alpha}{2} \frac{1}{(P/P_o)} \frac{T}{T_o}(1+\epsilon X)\)

\(\frac{dp}{dW} = -\frac{\alpha}{2p} \frac{T}{T_o}(1+\epsilon X)\)

Isothermal Operation:

\(\frac{dp}{dW} = -\frac{\alpha}{2p}(1+\epsilon X)\)

Recall that:

\(\frac{dX}{dW} = \frac{kC_{A0}^2 (1-X)^2}{F_{A0} (1+\epsilon X)^2} p^2\)

Notice that:

\(\frac{dX}{dW} = f(X, P) \text{ and } \frac{dP}{dW} = f(X, P) \text{ or } \frac{dp}{dW} = f(p, X)\)

The two expressions are coupled ordinary differential equations. We can solve them simultaneously using an ODE solver such as Polymath. For the special case of isothermal operation and epsilon = 0, we can obtain an analytical solution.

Polymath will combine the mole balance, rate law and stoichiometry.

Analytical Solution \( A \to B \), [e], PFR with \( -r_A = kC_A^2 \)

\(\frac{dp}{dW} = -\frac{\alpha}{2p}(1+\epsilon X)\)

IF \( T = T_0 \) AND \( \epsilon = 0 \)

\( y = (1 - \alpha W)^{1/2} \)

CAUTION: Never use this form if \( \epsilon \neq 0 \)

For \( \epsilon = 0 \):

\( C_A = C_{A0}(1 - X)p = C_{A0}(1 - X)(1 - \alpha W)^{\frac{1}{2}} \)

Combine:

\( -r_A' = kC_A^2 = kC_{A0}^2(1 - X)^2(1 - \alpha W) \)

\( F_{A0} \frac{dX}{dW} = -r_A' \)

\( \frac{dX}{dW} = \frac{kC_{A0}^2}{F_{A0}}(1 - X)^2(1 - \alpha W) \)

Solve:

\( \frac{X}{1 - X} = \frac{kC_{A0}^2}{F_{A0}} \left[ W - \frac{\alpha W^2}{2} \right] \)

Could now solve for X given W, or for W given X.

For gas phase reactions, as the pressure drop increases, the concentration decreases, resulting in a decreased rate of reaction, hence a lower conversion when compared to a reactor without a pressure drop. 

Here are some links to example problems dealing with packed bed reactors. You could also use these problems as self tests.

Analysis using Simulation Tools

Consider the following gas phase reaction carried out isothermally in a packed bed reactor containing 100 kg of catalyst. Pure A is fed at a rate of 2.5 mol/s and with $ {C_{A0} = 0.2 \frac{mol}{dm^3}} $, and $ {\alpha = 0.0162kg^{-1}} $.

\( 2A \rightarrow B \)

Mole Balance

$ {\frac{dX}{dW} = -r_A^{'} / F_{A0}} $

Rate Law

Elementary: $ {r_A^{'} = -kC_A^2} $

Stoichiometry

Gas with ${T = T_0}$

${A \rightarrow \frac{1}{2}B \\ C_A = \frac{F_A}{v} = \frac{C_{A0}(1-X)p}{1+\epsilon X} \\ \frac{dp}{dW} = -\alpha \frac{1+\epsilon X}{2p} \\ \epsilon = -1/2 \\ F_{A0} = 2.5 \\ C_{A0} = 0.2 \\ k = 5 \\ \alpha = 0.0162 \\ W_{final} = 100 \\ X(W = 0) = 0, p(W = 0) = 1 }$

The above equations can be easily combined with the help of simulation tools such as Polymath, Wolfram or Python.

Below is the result from Polymath

Here are links to Polymath, Wolfram, and Python code for the above example

Optimum Particle Diameter

\( A \rightarrow B \)

\( -r_A = kC_{A0}(1 - X)p, \, p = \frac{P}{P_0} \)

\( \frac{dp}{dW} = -\frac{\alpha}{2p}(1 + \epsilon X) \)

as \( \alpha \) increases the pressure drop increases

\( \alpha = \frac{2G}{A_c \rho_c P_0 \rho_0 D_P \phi^3} \left[ \frac{150(1 - \phi)\mu}{D_P} + 1.75G \right] \)

Laminar Flow, Fix \( P_0, \rho_0, \phi \)

\( \rho_0 = \frac{P_0 (\text{MW})}{RT_0} \)

\( \rho_0 P_0 \sim P_0^2 \)

\( \alpha \sim \frac{G}{A_C D_P^2 P_0^2} \)

\( \alpha_2 = \alpha_1 \left( \frac{G_2}{G_1} \right) \left( \frac{D_{P1}}{D_{P2}} \right)^2 \left( \frac{A_{C1}}{A_{C2}} \right) \left( \frac{P_{01}}{P_{02}} \right)^2 \)

If \( D_{P2} = 2D_{P1}, \, G_1 = G_2, \, A_{C1} = A_{C2} \)

\( \alpha_2 = \alpha_1 \left( \frac{1}{2} \right)^2 = \frac{\alpha_1}{4} \)

Increasing the particle diameter descreases the pressure drop and increases the rate and conversion.

However, there is a competing effect. The specific reaction rate decreases as the particle size increases, therefore so deos the conversion.

k ∼ 1/D_p

D_P1 > D_P2
k₁ > k₂

Higher k, higher conversion

The larger the particle, the more time it takes the reactant to get in and out of the catalyst particle. For a given catalyst weight, there is a greater external surgace area for smaller particles than larger particles. Therefore, there are more entry ways into the catalyst particle.

Later on in the course, we will learn that effectiveness factor decreases as the particle size increases