Question 10.2.2: Transient Behavior of a Stirred-Tank Reactor A liquid-phase ......
Transient Behavior of a Stirred-Tank Reactor
A liquid-phase reaction with stoichiometry A → B takes place in a continuous well-mixed 10.0-liter stirred tank reactor. A schematic diagram of the process is shown below.
The reactor may be considered perfectly mixed, so that the contents are uniform and the concentration of A in the product stream equals that inside the tank. The tank is initially filled with a solution that contains 2.00 mol A/L, and the inlet and outlet flows then begin.
1. Write a balance on species A in the tank and provide an initial condition.
2. Calculate C_{AS}, the steady-state concentration of A in the tank (the value approached as t → ∞).
3. Sketch the shape expected for a plot of C_{A} versus t.
4. Solve the balance equation for C_{A}(t), check the solution, and draw the actual plot of C_{A} versus t.
Compare what follows with the general procedure given prior to this example.
1. The total moles of A in the reactor at any time equals (10.0 L)[C_{A}(mol A/L)] = 10.0C_{A} (mol A).
Therefore,
These terms are substituted into the balance equation on A (accumulation = input – output – consumption), which is then divided by 10.0 to obtain an expression for dC_{A}/dt. The result along with the initial condition for the equation [C_{A}(0) = 2.00 mol A/L] is
\boxed{ \begin{aligned}\frac{d C_{A}}{d t}=0.150\,\mathrm{mol~A/s}-(0.0200\,\mathrm{L/s})C_{A} \\ t=0, \qquad C_A = 2.00 \text{ mol A/L}\end{aligned}}2. At steady state nothing varies with time, so that the derivative of C_{A} (and of every other system variable) with respect to time must equal zero. Setting dC_{A}/dt = 0 in the balance equation and letting C_{A} = C_{AS} (steady state) in the resulting equation yields
0=0.150\,\mathrm{mol/s}-0.0200C_{\mathrm{AS}}\Longrightarrow \boxed{C_{\mathrm{AS}}=7.50\,\mathrm{mol~A}/\mathrm{L}}If C_{AS} had not had an asymptotic (steady-state) limit, the equation would not have had a finite solution.
3. We can now deduce quite a lot about the plot of C_{A} versus t, even though we have not yet solved the differential balance equation. We know a point at t = 0 (the initial condition) and the asymptotic value as t ! 1 (the steady-state solution), and we also have an expression for the slope of the plot at any time (dC_{A}=dt) as a function of the concentration (slope = 0.150 – 0.0200C_{A}). Let us summarize what we can deduce:
• The plot originates at (t = 0, C_{A} = 2.00 mol/L).
• At t = 0, the slope of the plot is [0.150 – 0.0200](2.00 mol/s) = 0.110 mol/s. Since it is positive, C_{A} must increase as t increases.
• As t continues to increase and C_{A} also increases, the slope of the curve (0.150 – 0.0200C_{A}) gets progressively less positive. The curve must therefore be concave down.
• After a long time, the plot asymptotes to C_{A} = 7.50 mol/L.
Combining all these observations leads to the following sketch:
4. We may now solve the differential balance to determine the values of C_{A} at specific values of t or vice versa. Reconsider the equation
\frac{d C_{A}}{d t}=0.150-0.0200C_{A} \\ t=0 , \qquad C_A =2.00Separating variables and integrating as in Equation 10.2-4 yields
{\frac{d y}{d x}}=f_{1}(x)f_{2}(y)\stackrel{\mathrm{separate}}{\longrightarrow} \frac{d y}{f_{2}(y)}=f_{1}(x)d x \stackrel{\mathrm{integrate}}{\longrightarrow} \int_{y(0)}^{y}{\frac{d y}{f_{2}(y)}}=\int_{0}^{x}f_{1}(x)d x (10.2-4)
\frac{d C_{A}}{0.150-0.0200C_{A}}=d t\Longrightarrow \int_{2.00}^{C_{A}}\frac{d C_{A}}{0.150-0.0200C_{A}}=\int_{0}^{t}d t=t \\ \Longrightarrow-{\frac{1}{0.0200}}\ln(0.150-0.0200C_{A}){\bigg]}_{2.00}^{C_{A}}=t\Longrightarrow \ln\left[{\frac{0.150-0.0200C_{\mathrm{A}}}{0.150-0.0200(2.00)}}\right]=-0.0200t \\ \Longrightarrow{\frac{0.150-0.0200C_{\mathrm{A}}}{0.110}}=e^{-0.0200t}\Longrightarrow\qquad \boxed{C_{\mathrm{A}}({\mathrm{mol}}/\mathrm{L})=7.50-5.50e^{-0.0200t}}(Verify each step.)
We next check the solution.
Check 1: Substitute t = 0 into the solution C_{A}(0) = 2.00 mol/L (the correct initial
value). \pmb{✓}
Check 2: Substitute t → ∞ into the solution to obtain C_{A}(∞) = 7.50 mol/L (the previously determined steady-state value). \pmb{✓}
Check 3: Differentiate the equation to obtain an expression for dC_{A}/dt, and then substitute for both dC_{A}/dt and CA(t) in the original equation [dC_{A}/dt = 1.50 – 0.200C_{A}] to show that the solution satisfies the equation. \pmb{✓}
Verify that the derived solution satisfies each of these conditions.
A plot of the solution matches the form of the plot previously sketched
