Chapter 12, Problem 10P

Chemical/Bio Engineering

An irreversible, first-order reaction takes place in four well mixed reactors (Fig. P12.10),

$A\stackrel{k}{\to }B$

Thus, the rate at which A is transformed to B can be represented as

$R_{ab}=kVc$

The reactors have different volumes, and because they are operated at different temperatures, each has a different reaction rate:

Reactor	V, L	$k,h^{-1}$
1	25	0.05
2	75	0.1
3	100	0.5
4	25	0.1

Determine the concentration of A and B in each of the reactors at steady state.

FIGURE P12.10

To determine

To calculate:

Answer to Problem 10P

Solution:

Explanation of Solution

Given Information:

Write the provided values of the volume and rate of reaction. $\begin{array}{ccc}reactor& V\left(\text{L}\right)& k\left({\text{h}}^{\text{-1}}\right)\\ 1& 25& 0.05\\ 2& 75& 0.1\\ 3& 100& 0.5\\ 4& 25& 0.1\end{array}$

Formula used:

Write system of linear equations in matrix form.

$AX=D$

And,

$X=A^{-1}D$

The term $X$, represent the variable matrix, $A$ is the co-efficient matrix, and $D$ is the constant column matrix.

Calculation:

Consider the provided diagram for an irreversible first order reaction takes place in four will-mixed reactors.

Balance the mass for A in reactor 1 at steady-state.

$Q_{\text{in}}{c}_{A,\text{in}}-Q_{\text{in}}{c}_{A,1}-k_1{V}_1{c}_{A,1}=0$

Substitute $10$ for $Q_{\text{in}}$, $1$ for $c_{A\text{,in}}$, $25$ for $V_1$ and $0.05$ for $k_1$.

$\begin{array}{c}10\left(1\right)-10c_{A,1}-\left(0.05\right)\left(25\right)c_{A,1}=0\\ 10-10c_{A,1}-1.25c_{A,1}=0\\ 10-11.25c_{A,1}=0\end{array}$

Further solve.

$-11.25c_{A,1}=-10$ …… (1)

Balance the mass for A in reactor 2 at steady-state.

$Q_{\text{in}}{c}_{A,1}+Q_{32}{c}_{A,3}-\left(Q_{\text{in}}+Q_{32}\right)c_{A,2}-k_2{V}_2{c}_{A,2}=0$

Substitute $10$ for $Q_{\text{in}}$, $5$ for $Q_{\text{32}}$, $75$ for $V_2$ and $0.1$ for $k_2$.

$\begin{array}{c}10c_{A,1}+5c_{A,3}-(10+5)c_{A,2}-\left(0.10\right)\left(75\right)c_{A,2}=0\\ 10c_{A,1}+5c_{A,3}-15c_{A,2}-7.5c_{A,2}=0\\ 10c_{A,1}+5c_{A,3}-22.5c_{A,2}=0\end{array}$

Further solve.

$10c_{A,1}-22.5c_{A,2}+5c_{A,3}=0$ …… (2)

Balance the mass for A in reactor 3 at steady-state.

$\left(Q_{\text{in}}+Q_{32}\right)c_{A,2}+Q_{43}{c}_{A,4}-\left(Q_{\text{in}}+Q_{43}+Q_{32}\right)c_{A,3}-k_3{V}_3{c}_{A,3}=0$

Substitute $10$ for $Q_{\text{in}}$, $5$ for $Q_{\text{32}}$, $3$ for $Q_{\text{43}}$, $100$ for $V_3$ and $0.5$ for $k_3$.

$\begin{array}{c}\left(10+5\right)c_{A,2}+3c_{A,4}-\left(10+3+5\right)c_{A,3}-\left(0.5\right)\left(100\right)c_{A,3}=0\\ 15c_{A,2}+3c_{A,4}-18c_{A,3}-50c_{A,3}=0\end{array}$

Further solve.

$15c_{A,2}-68c_{A,3}+3c_{A,4}=0$ …… (3)

Balance the mass for A in reactor 4 at steady-state.

$\left(Q_{\text{in}}+Q_{43}\right)c_{A,3}-\left(Q_{\text{in}}+Q_{43}\right)c_{A,4}-k_4{V}_4{c}_{A,4}=0$

Substitute $10$ for $Q_{\text{in}}$, $3$ for $Q_{\text{43}}$, $25$ for $V_4$ and $0.1$ for $k_4$.

$\begin{array}{c}\left(10+3\right)c_{A,3}-\left(10+3\right)c_{A,4}-\left(0.1\right)\left(25\right)c_{A,4}=0\\ 13c_{A,3}-13c_{A,4}-2.5c_{A,4}=0\end{array}$

Further solve. $13c_{A,3}-15.5c_{A,4}=0$ …… (4)

Balance the mass for B in reactor 1 at steady-state.

$-Q_{\text{in}}{c}_{B,1}+k_1{V}_1{c}_{A,1}=0$

Substitute $10$ for $Q_{\text{in}}$, $25$ for $V_1$ and $0.05$ for $k_1$.

$-10c_{B,1}+\left(0.05\right)\left(25\right)c_{A,1}=0$

Further solve.

$-10c_{B,1}+1.25c_{A,1}=0$ …… (5)

Balance the mass for B in reactor 2 at steady-state.

$Q_{\text{in}}{c}_{B,1}+Q_{32}{c}_{B,3}-\left(Q_{\text{in}}+Q_{32}\right)c_{B,2}+k_2{V}_2{c}_{A,2}=0$

Substitute $10$ for $Q_{\text{in}}$, $5$ for $Q_{\text{32}}$, $75$ for $V_2$ and $0.1$ for $k_2$.

$10c_{B,1}+5c_{B,3}-\left(10+5\right)c_{B,2}+\left(0.1\right)\left(75\right)c_{A,2}=0$

Further solve.

$10c_{B,1}+5c_{B,3}-15c_{B,2}+7.5c_{A,2}=0$ …… (6)

Balance the mass for B in reactor 3 at steady-state.

$\left(Q_{\text{in}}+Q_{32}\right)c_{B,2}+Q_{43}{c}_{B,4}-\left(Q_{\text{in}}+Q_{43}+Q_{32}\right)c_{B,3}+k_3{V}_3{c}_{A,3}=0$

Substitute $10$ for $Q_{\text{in}}$, $5$ for $Q_{\text{32}}$, $3$ for $Q_{\text{43}}$, $100$ for $V_3$ and $0.5$ for $k_3$.

$\left(10+5\right)c_{B,2}+3c_{B,4}-\left(10+3+5\right)c_{B,3}+\left(0.5\right)\left(100\right)c_{A,3}=0$

Further solve.

$15c_{B,2}+3c_{B,4}-18c_{B,3}+50c_{A,3}=0$ …… (7)

Balance the mass for B in reactor 4 at steady-state.

$\left(Q_{\text{in}}+Q_{43}\right)c_{B,3}-\left(Q_{\text{in}}+Q_{43}\right)c_{B,4}+k_4{V}_4{c}_{A,4}=0$

Substitute $10$ for $Q_{\text{in}}$, $3$ for $Q_{\text{43}}$, $25$ for $V_4$ and $0.1$ for $k_4$.

$\left(10+3\right)c_{B,3}-\left(10+3\right)c_{B,4}+\left(0.1\right)\left(25\right)c_{A,4}=0$

Further solve.

$13c_{B,3}-13c_{B,4}+2.5c_{A,4}=0$ …… (8)

Now, write all the equations, to find linear system of equations.

$\begin{array}{c}-11.25c_{A,1}=-10\\ 10c_{A,1}-22.5c_{A,2}+5c_{A,3}=0\\ 15c_{A,2}-68c_{A,3}+3c_{A,4}=0\\ 13c_{A,3}-15.5c_{A,4}=0\end{array}$

$\begin{array}{c}-10c_{B,1}+1.25c_{A,1}=0\\ 10c_{B,1}+5c_{B,3}-15c_{B,2}+7.5c_{A,2}=0\\ 15c_{B,2}+3c_{B,4}-18c_{B,3}+50c_{A,3}=0\\ 13c_{B,3}-13c_{B,4}+2.5c_{A,4}=0\end{array}$

Write the above linear equations in matrix form as written in symbolized form.

$AC=B$

Here, coefficient matrix A is,

$A=\left[\begin{array}{cccccccc}-11.25& 0& 0& 0& 0& 0& 0& 0\\ 10& -22.5& 5& 0& 0& 0& 0& 0\\ 0& 10& -68& 3& 0& 0& 0& 0\\ 0& 0& 13& -15.5& 0& 0& 0& 0\\ 1.25& 0& 0& 0& -10& 0& 0& 0\\ 0& 7.5& 0& 0& 10& -15& 5& 0\\ 0& 0& 50& 0& 0& 15& -18& 3\\ 0& 0& 0& 2.5& 0& 0& 13& -13\end{array}\right]$

Column matrix $C$ is,

$C=\left[\begin{array}{l}{c}_{A,1}\\ c_{A,2}\\ c_{A,3}\\ c_{A,4}\\ c_{B,1}\\ c_{B,2}\\ c_{B,3}\\ c_{B,4}\end{array}\right]$

Column matrix B is,

$B=\left[\begin{array}{c}-10\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right]$

Substitute the values in the matrix equation form.

$\begin{array}{c}AC=B\\ \left[\begin{array}{cccccccc}-11.25& 0& 0& 0& 0& 0& 0& 0\\ 10& -22.5& 5& 0& 0& 0& 0& 0\\ 0& 10& -68& 3& 0& 0& 0& 0\\ 0& 0& 13& -15.5& 0& 0& 0& 0\\ 1.25& 0& 0& 0& -10& 0& 0& 0\\ 0& 7.5& 0& 0& 10& -15& 5& 0\\ 0& 0& 50& 0& 0& 15& -18& 3\\ 0& 0& 0& 2.5& 0& 0& 13& -13\end{array}\right]\left[\begin{array}{l}{c}_{A,1}\\ c_{A,2}\\ c_{A,3}\\ c_{A,4}\\ c_{B,1}\\ c_{B,2}\\ c_{B,3}\\ c_{B,4}\end{array}\right]=\left[\begin{array}{c}-10\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right]\end{array}$

Solve for $C=A^{-1}B$ with the help of MATLAB to obtain the value of the corresponding concentration for each reactor for the column matrix $X$.

Code:

>> A=[-11.25 0 0 0 0 0 0 0;

10 -22.5 5 0 0 0 0 0;

0 10 -68 3 0 0 0 0;

0 0 130 -15.5 0 0 0 0;

1.25 0 0 0 -10 0 0 0;

0 7.5 0 0 10 -15 5 0;

0 0 50 0 0 15 -18 3;

0 0 0 2.5 0 0 13 -13];

B=[-10;0;0;0;0;0;0;0];

C=A\B

Type the above code into MATLAB command window and press enter to find the result.

Result is obtained as follows:

C =

0.8889

0.4167

0.0973

0.8158

0.1111

0.6014

0.9570

1.1139

Hence,

$C=\left[\begin{array}{l}{c}_{A,1}=\\ c_{A,2}\\ c_{A,3}\\ c_{A,4}\\ c_{B,1}\\ c_{B,2}\\ c_{B,3}\\ c_{B,4}\end{array}\right]$

Hence, the concentration of A and B is,

$\boxed{\begin{array}{c}{c}_{A,1}=0.8421\\ c_{B,1}=0.1578\\ c_{A,2}=0.3400\\ c_{B,2}=0.9932\\ c_{A,3}=0.1010\\ c_{B,3}=1.8989\\ c_{A,4}=0.0847\\ c_{B,4}=1.9152\end{array}}$

$$