Chapter 5: Isothermal Reactor Design: Conversion

Elementary gas phase reaction in different reactor types.

CSTR

The elementary gas phase reaction

\( 2A + B \rightarrow C \)

takes place in a CSTR at constant temperature (500 K) and constant pressure (16.4 atm). The feed is equal molar in A and B.

Mole Balance

\( V = \frac{F_{A0}X}{-r_A} \)


Rate Law

\( -r_A = kC_A^2C_B \)


Stoichiometry

gas phase, isothermal (T = T0), no pressure drop (P = P0)

\( C_A = C_{A0} \frac{(1-X)}{(1-0.5X)} \)

\( C_B = C_{A0} \)

Derive

Why do you suppose CB is a constant, when B is being consumed?

\( C_{A0} = \frac{y_{A0} P_0}{R T_0} = \frac{(0.5)(16.4 \, \text{atm})}{\left(0.082 \, \text{atm} \cdot \text{m}^3 / \text{kmol} \cdot \text{K}\right)(500 \, \text{K})} = 0.2 \, \text{kmol}/\text{m}^3 = 0.2 \, \text{mol}/\text{dm}^3 \)


Combine

\( -r_A = kC_A^2C_B = kC_{A0}^3 \frac{(1 - X)^2}{(1 - 0.5X)^2} \)

\( -r_A = \left( 10 \, \frac{\text{dm}^6}{\text{mol}^2 \cdot \text{s}} \right) \left( 0.2 \, \frac{\text{mol}}{\text{dm}^3} \right)^3 \frac{(1 - X)^2}{(1 - 0.5X)^2} \)

\( -r_A = \left( 0.08 \, \frac{\text{mol}}{\text{dm}^3 \cdot \text{s}} \right) \frac{(1 - X)^2}{(1 - 0.5X)^2} \)

Evaluate

\( V = \frac{F_{A0}X}{-r_A} = \frac{\left( 5 \, \frac{\text{mol}}{\text{s}} \right)(0.9)\left[1 - 0.5(0.9)\right]^2}{\left( 0.8 \, \frac{\text{mol}}{\text{dm}^3 \cdot \text{s}} \right)(1 - 0.9)^2} \)

\( V = 1701 \, \text{dm}^3 \)


PFR and Batch Reactors

Elementary Gas Phase Reaction:

\( 2A + B \rightarrow C \)

PFR

Mole Balance

\( V = F_{A0} \int_{0}^{X} \frac{dX}{-r_A} \)


Rate Law

\( -r_A = kC_A^2C_B \)

Stoichiometry

gas phase, isothermal (T = T0), no pressure drop (P = P0), CAo=CBo (Q=1), v=vo(1+eX)

\( A + \frac{B}{2} \rightarrow \frac{C}{2} \)

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

\( \epsilon = y_{A0}\delta = \frac{1}{2}\left(\frac{1}{2} - \frac{1}{2} - 1\right) = -\frac{1}{2} \)

\( v = v_0\left(1 - \frac{1}{2}X\right) \)

\( C_B = \frac{F_{A0}}{v} = \frac{F_{A0}\left(1 - \frac{1}{2}X\right)}{v_0\left(1 - \frac{1}{2}X\right)} = C_{A0} \)

\( C_B = C_{A0} \)


Combine

\(-r_A = kC_{A0}^3 \frac{(1-X)^2}{(1+\epsilon X)^2} \)

\(V = \frac{F_{A0}}{kC_{A0}^3} \int \frac{(1+\epsilon X)^2}{(1-X)^2} \, dX \)

\(V = \frac{F_{A0}}{kC_{A0}^3} \left[ 2\epsilon(1+\epsilon)\ln(1-X) + \epsilon^2 X + (\epsilon+1)^2 \frac{X}{1-X} \right] \)


Parameter Evaluation

CAo=0.2, v=vo=25 dm3/s, k=10 dm6/mol2 s, e=-0.5, X=0.9

\( V = \frac{\left( \frac{5 \, \text{mol}}{\text{s}} \right)}{\left( 10 \, \frac{\text{dm}^6}{\text{mol}^2 \cdot \text{s}} \right) \left( 0.2 \, \frac{\text{mol}}{\text{dm}^3} \right)^3} \cdot \left[ 2 \left( -\frac{1}{2} \right) \left( 1 - \frac{1}{2} \right) \ln(1-0.9) + \left( -\frac{1}{2} \right)^2 (0.9) + \left( -\frac{1}{2} + 1 \right)^2 \frac{0.9}{1-0.9} \right] \)

\( V = 6.25 \, \text{dm}^3 \cdot [1.15 + 0.23 + 2.25] = 226.9 \, \text{dm}^3 \)

\( V = 227 \, \text{dm}^3 \)



Batch Reactor Constant Volume, V=Voand the pressure changes.

Mole Balance

\( t = N_{A0} \int_{0}^{X} \frac{dX}{-r_A V} = C_{A0} \int_{0}^{X} \frac{dX}{-r_A} \)

Rate Law

\( -r_A = kC_A^2 C_B \)

Stoichiometry

\( C_A = \frac{N_A}{V_0} = \frac{N_{A0}}{V_0} (1 - X) \)

\( C_B = \frac{N_A}{V_0} \left( \Theta_B - \frac{1}{2} X \right) = C_{A0} \left( 1 - \frac{X}{2} \right) \)

\( P = \frac{N_T}{N_{T0}} P_o = \frac{C_T}{C_{T0}} P_o \)

Combine

\(-r_A = kC_{A0}^3(1-X)^2\left(1-\frac{X}{2}\right)\)

\(t = \frac{1}{kC_{A0}^2} \int_0^X \frac{dX}{(1-X)^2\left(1-\frac{X}{2}\right)}\)

Parameter Evaluation

CAo=0.2, k=10 dm6/mol2 s,

Graph of the function 1 / [(1-X)^2 * (1-X/2)] plotted against X, highlighting two intervals: the first interval (ΔX1 = 0.3) between X = 0.3 and X = 0.6, showing a moderate rise, and the second interval (ΔX1m = 0.15) between X = 0.75 and X = 0.9, illustrating a steeper increase in the curve.

\(t = \frac{1}{kC_{A0}^2} \int_0^{0.9} \frac{dX}{(1-X)^2\left(1-\frac{X}{2}\right)}\)

\(t = \frac{1}{\left(10 \, \text{dm}^6/\text{mol}^2 \cdot \text{s}\right)\left(0.2 \, \text{mol}/\text{dm}^3\right)^2} \int_0^{0.9} \frac{dX}{(1-X)^2\left(1-\frac{X}{2}\right)}\)

\(t = 2.5 \, \text{s} \int_0^{0.9} \frac{dX}{(1-X)^2\left(1-\frac{X}{2}\right)}\)

\(t = 2.5 \, \text{s} \left[\int_0^{0.6} \frac{dX}{(1-X)^2\left(1-\frac{X}{2}\right)} + \int_{0.6}^{0.9} \frac{dX}{(1-X)^2\left(1-\frac{X}{2}\right)}\right]\)

\(t = \frac{1}{kC_{A0}^2} \left[\frac{\Delta X_1}{3}\left(f(X=0) + 4f(X=0.3) + f(X=0.6)\right) + \frac{\Delta X_2}{3}\left(f(X=0.6) + 4f(X=0.75) + f(X=0.9)\right)\right]\)

\(t = 2.5s \left[\frac{0.3}{3}[1 + 4(2.4) + 7.81] + \frac{0.15}{3}[7.81 + 4(25.6) + 181]\right]\)

\(t = 2.5s \left[\frac{0.3}{3}(18.4) + \frac{0.15}{3}(291)\right]\)

\(t = 2.5s [1.84 + 14.55]\)

\(t = 41s\)

\(P = \frac{N_r}{N_{r0}}P_0 = \frac{(N_{A0} + N_{B0} + \epsilon N_{A0}X)P_0}{N_{r0}}\)

\(P = (1 + y_{A0}\epsilon X)P_0\)

\(P = \left(1 + \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)(0.9)\right)16.4\)

\(P = 12.7 \, \text{atm}\)



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