Chapter 5: Isothermal Reactor Design: Conversion

Example 5.2: The Reaction From the Movie

The Laboratory Experiment

\((\text{CH}_3\text{CO})_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{CH}_3\text{COOH}\)

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

Diagram of a continuous stirred-tank reactor (CSTR) with input concentrations CA0 = 1M and CB0 = 51.2M, volumetric flow rate v0 = 0.0033 dm^3/s, and reactor volume V = 1 dm^3. The output variable X is indicated. ^{\(
\begin{aligned}
&C_{A0} = 1M,\\
&C_{B0} = 51.2M,\\
&v_0 = 0.0033 \, \text{dm}^3/\text{s}.
\end{aligned}
\)}

\(\tau = \frac{1 \, \text{dm}^3}{0.0033 \, \text{dm}^3/\text{s}} = 303 \, \text{s}.\)

A. CSTR

A.1. Single CSTR

\(C_{A0} = 1 \, \text{molar A}\)

1. Mole Balance

\(\text{In - Out + Gen = Accum}\)

\(F_{A0} - F_A + r_A V = 0\)

\(V = \frac{F_{A0} - F_A}{-r_A} = \frac{F_{A0} X}{-r_A} = \frac{v_0 C_{A0} X}{-r_A}\)

2. Rate Law

\(-r_A = k' C_A C_B, \quad k' = 1.97 \times 10^{-4} \, \frac{\text{dm}^3}{\text{mol} \cdot \text{s}}\)

3. Stoichiometry Liquid v = v

\(\text{CH}_3 - \text{C} - \text{O} - \text{C} - \text{CH}_3\)

Species	Symbol	Entering	Change	Leaving
\(\text{CH}_3 - \text{C} - \text{O} - \text{C} - \text{CH}_3\)	A	\( F_{A0} \)	\( -F_{A0}X = F_{A0}(1-X) \)	\( F_{A0}(1-X) \)
\( \text{H}_2\text{O} \)	B	\( \Theta_B F_{A0} \)	\( -F_{A0}X = F_{A0}(\Theta_B - X) \)	\( F_{A0}(\Theta_B - X) \)
\( \text{CH}_3\text{COOH} \)	C	0	\( 2F_{A0}XF_{C} \)	\( 2F_{A0}X \)

\( C_A = \frac{F_A}{\nu} = \frac{F_A}{\nu_0} = \frac{F_{A0}(1-X)}{\nu_0} = C_{A0}(1-X) \)

\( C_B = \frac{F_B}{\nu_0} = C_{A0}(\Theta_B - X) \)

\( \Theta_B = \frac{C_{B0}}{C_{A0}}, \quad C_{B0} \approx 51.2 \, \text{mol/dm}^3 \)

\( \Theta_B = 51.2 \)

\( \Theta_B = 51.2 \gg X \)

\( C_B = \Theta_B C_{A0} \equiv C_{B0} \)

4. Combine

\(-r_A = k'C_A C_B = k'C_{B0}C_A = kC_A\)

\(-r_A = kC_{A0}(1-X)\)

\(V = \frac{\nu_0 C_{A0} X}{kC_{A0}(1-X)}, \quad \tau = \frac{V}{\nu_0}\)

\(\tau k = \frac{X}{1-X}\)

\[ X = \frac{\tau k}{1+\tau k} \]

\(k = k'C_{B0} = (1.97 \times 10^{-4} \, \text{dm}^3/\text{mol} \cdot \text{s})(51.2 \, \text{mol}/\text{dm}^3)\)

\(k = 0.01 \, \text{s}^{-1}\)

\(\tau k = 3.03\)

\[ X = \frac{3.03}{1+3.03} = 0.75 \]

A.2. CSTRs in Series

Diagram of two continuous stirred-tank reactors (CSTRs) in series. The first reactor has a volume of 1 dm^3, input flow rate FA0, and output flow rate FA1 with conversion X1 = 0.75. The second reactor also has a volume of 1 dm^3 and output flow rate FA2 with unknown conversion X2.

\(F_{A1} = F_{A0}(1-X_1)\)

\(F_{A2} = F_{A0}(1-X_2)\)

\(\text{In} - \text{Out} + \text{Gen} = \)

\(F_{A1} - F_{A2} + r_{A2}V = 0\)

Balance on Second Reactor

\(V_2 = \frac{F_{A0}(X_2 - X_1)}{-r_A}\)

\(-r_{A2} = kC_{A2} = k\frac{F_{A2}}{v_0} = k\frac{F_{A0}(1-X_2)}{v_0} = kC_{A0}(1-X_2)\)

\(V = \frac{v_0 C_{A0}(X_2 - X_1)}{kC_{A0}(1-X_2)}\)

\(\tau k = \frac{X_2 - X_1}{1 - X_2}\)

\(\tau k = X_2(1 + \tau k) - X_1\)

\(X_2 = \frac{X_1 + \tau k}{1 + \tau k} = \frac{(0.75) + 3}{4.03} = \frac{3.75}{4.03}\)

B. PFR

Diagram of a plug flow reactor (PFR) with volume V = 0.311 dm^3. Two input streams labeled A and B combine and flow into the reactor, with a single outlet stream at the end.

1. Mole Balance/Design Equation

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

2. Rate Law

\(-r_A = k'C_A C_B\)

3. Stoichiometry v = v₀

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

\(C_B = C_{A0}(\Theta_B - X) \approx C_{B0}\)

4. Combine

\(-r_A = k'C_A C_B = k'C_{B0}C_A = kC_{A0}(1 - X)\)

\(\frac{dX}{dV} = \frac{-r_A}{F_{A0}} = \frac{-r_A}{v_0C_{A0}} = \frac{kC_{A0}(1 - X)}{v_0C_{A0}}\)

\(\frac{dX}{dV} = \frac{k}{v_0}(1 - X)\)

\(\frac{dX}{1 - X} = \frac{k}{v_0} dV\)

\(V = 0, X = 0\)

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

\(X = 1 - e^{-k\tau}\)

\(\tau = \frac{0.311 \, \text{dm}^3}{0.0033 \, \text{dm}^3/\text{s}} = 94.2 \, \text{s}\)

\(k\tau = (94.2)(0.01) = 0.94\)

\(X = 1 - e^{-0.94} = 0.61\)

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