Chapter 12, Problem 12P

Chemical/Bio Engineering

Figure P12.12 depicts a chemical exchange process consisting of a series of reactors in which a gas fl owing from left to right is passed over a liquid fl owing from right to left. The transfer of a chemical from the gas into the liquid occurs at a rate that is proportional to the difference between the gas and liquid concentrations in each reactor. At steady state, a mass balance for the first reactor can be written for the gas as

$Q_{G^{C}GO}-Q_{G^{C}GI}+D\left(C_{L1-C_{GI}}\right)=0$

and for the liquid as

$Q_{L^{C}12}-Q_{L^{C}L1}+D\left(C_{GI}-C_{L1}\right)=0$

Where $Q_G$ and $Q_L$ are the gas and liquid flow rates, respectively, and $D=\text{the gas liquid exchange rate}$. Similar balances can be written for the other reactors. Solve for the concentrations given the following values:

$Q_G=2,Q_L=1,D=0.8,C_{GO}=100,C_{L6}=20$.

FIGURE P12.12

To determine

To calculate: The flow rate of the flow through every pipe if the flow equation through he reactors is provided by following system of equations:

For gas it is provided as:

$Q_G{c}_{G0}-Q_G{c}_{G1}+D\left(c_{L1}-c_{G1}\right)=0$

And for liquid it is provided as:

$Q_L{c}_{L2}-Q_L{c}_{L1}+D\left(c_{G1}-c_{L1}\right)=0$

Answer to Problem 12P

Solution: The concentration of liquid and gases in the reactor is:

$\begin{array}{lll}\hfill & \text{Liquid}\hfill & \text{Gas}\hfill \\ 0\hfill & 100.0000\hfill & \hfill \\ 1\hfill & 96.2074\hfill & 86.7258\hfill \\ 2\hfill & 91.3311\hfill & 79.1405\hfill \\ 3\hfill & 85.0617\hfill & 69.3880\hfill \\ 4\hfill & 77.0009\hfill & 56.8491\hfill \\ 5\hfill & 66.6371\hfill & 40.7276\hfill \\ 6\hfill & \hfill & 20.0000\hfill \end{array}$

Explanation of Solution

Given Information:

The system of reactors is provided as follows:

Flow rate and the concentration are provided as:

$Q_G=2$, $Q_L=1$, $c_{G0}=100$ and $c_{L6}=20$.

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 figure below,

Here, $Q_G$ is the gas flow rate and $Q_L$ is the liquid flow rates.

For reactor 1, the system is in steady state. Therefore, themass balance equation for the gas reactor 1.

$Q_G{c}_{G0}-Q_G{c}_{G1}+D\left(c_{L1}-c_{G1}\right)=0$

Substitute $2$ for $Q_G$, $0.8$ for $D$ and $100$ for $c_{G0}$.

$\begin{array}{c}2\times 100-2c_{G1}+0.8\left(c_{L1}-c_{G1}\right)=0\\ 200-2c_{G1}+0.8c_{L1}-0.8c_{G1}=0\\ 200=2.8c_{G1}-0.8c_{L1}\end{array}$

For reactor 2, the system is in steady state. Therefore, the mass balance for the gas reactor 2 is,

$Q_G{c}_{G1}-Q_G{c}_{G2}+D\left(c_{L2}-c_{G2}\right)=0$

Substitute $2$ for $Q_G$ and $0.8$ for $D$.

$\begin{array}{c}2c_{G1}-2c_{G2}+0.8\left(c_{L2}-c_{G2}\right)=0\\ 2c_{G1}-2c_{G2}+0.8c_{L2}-0.8c_{G2}=0\\ 2c_{G1}-2.8c_{G2}+0.8c_{L2}=0\\ -2c_{G1}+2.8c_{G2}-0.8c_{L2}=0\end{array}$

For reactor 3, the system is in steady state, therefore, the mass balance equation for the gas reactor 3 is,

$Q_G{c}_{G2}-Q_G{c}_{G3}+D\left(c_{L3}-c_{G3}\right)=0$

Substitute $2$ for $Q_G$, $0.8$ for $D$ and $100$ for $c_{G0}$.

$\begin{array}{c}{Q}_G{c}_{G2}-Q_G{c}_{G3}+D\left(c_{L3}-c_{G3}\right)=0\\ 2c_{G2}-2c_{G3}+0.8c_{L3}-0.8c_{G3}=0\\ 2c_{G2}-2.8c_{G3}+0.8c_{L3}=0\\ -2c_{G2}+2.8c_{G3}-0.8c_{L3}=0\end{array}$

For reactor 4, the system is in steady state. Therefore, the mass balance for the gas reactor 4 is,

$Q_G{c}_{G3}-Q_G{c}_{G4}+D\left(c_{L4}-c_{G4}\right)=0$

Substitute $2$ for $Q_G$, $0.8$ for $D$ and $100$ for $c_{G0}$.

$\begin{array}{c}{Q}_G{c}_{G3}-Q_G{c}_{G4}+D\left(c_{L4}-c_{G4}\right)=0\\ 2c_{G3}-2c_{G4}+0.8c_{L4}-0.8c_{G4}=0\\ 2c_{G3}-2.8c_{G4}+0.8c_{L4}=0\\ -2c_{G3}+2.8c_{G4}-0.8c_{L4}=0\end{array}$

For reactor 5, the system is in steady state. Therefore, the mass balance for the gas reactor 5 is,

$Q_G{c}_{G4}-Q_G{c}_{G5}+D\left(c_{L5}-c_{G5}\right)=0$

Substitute $2$ for $Q_G$, $0.8$ for $D$ and $100$ for $c_{G0}$.

$\begin{array}{c}{Q}_G{c}_{G4}-Q_G{c}_{G5}+D\left(c_{L5}-c_{G5}\right)=0\\ 2c_{G4}-2c_{G5}+0.8c_{L5}-0.8c_{G5}=0\\ 2c_{G4}-2.8c_{G5}+0.8c_{L5}=0\\ -2c_{G4}+2.8c_{G5}-0.8c_{L5}=0\end{array}$

For reactor 1, the system is in steady state. Therefore, the mass balance for the liquid reactor 1 is,

$Q_L{c}_{L2}-Q_L{c}_{L1}+D\left(c_{G1}-c_{L1}\right)=0$

Substitute $1$ for $Q_L$ and $0.8$ for $D$.

$\begin{array}{c}{Q}_L{c}_{L2}-Q_L{c}_{L1}+D\left(c_{G1}-c_{L1}\right)=0\\ 1c_{L2}-1c_{L1}+0.8c_{G1}-0.8c_{L1}=0\\ 1c_{L2}-1.8c_{L1}+0.8c_{G1}=0\end{array}$

$-1c_{L2}+1.8c_{L1}-0.8c_{G1}=0$ …… (6)

For reactor 2, the system is in steady state. Therefore, the mass balance for the liquid reactor 2 is,

$Q_L{c}_{L3}-Q_L{c}_{L2}+D\left(c_{G2}-c_{L2}\right)=0$

Substitute $1$ for $Q_L$ and $0.8$ for $D$.

$\begin{array}{c}{Q}_L{c}_{L3}-Q_L{c}_{L2}+D\left(c_{G2}-c_{L2}\right)=0\\ 1c_{L3}-1c_{L2}+0.8c_{G2}-0.8c_{L2}=0\\ 1c_{L3}-1.8c_{L2}+0.8c_{G2}=0\end{array}$

$-1c_{L3}+1.8c_{L2}-0.8c_{G2}=0$ …… (7)

For reactor 3, the system is in steady state. Therefore, the mass balance for the liquid reactor 3 is,

$Q_L{c}_{L4}-Q_L{c}_{L3}+D\left(c_{G3}-c_{L3}\right)=0$

Substitute $1$ for $Q_L$ and $0.8$ for $D$.

$\begin{array}{c}{Q}_L{c}_{L4}-Q_L{c}_{L3}+D\left(c_{G3}-c_{L3}\right)=0\\ 1c_{L4}-1c_{L3}+0.8c_{G3}-0.8c_{L3}=0\\ 1c_{L4}-1.8c_{L3}+0.8c_{G3}=0\\ -1c_{L4}+1.8c_{L3}-0.8c_{G3}=0\end{array}$

For reactor 4, the system is in steady state. Therefore, the mass balance for the liquid reactor 4 is,

$Q_L{c}_{L5}-Q_L{c}_{L4}+D\left(c_{G4}-c_{L4}\right)=0$

Substitute $1$ for $Q_L$ and $0.8$ for $D$.

$\begin{array}{c}{Q}_L{c}_{L5}-Q_L{c}_{L4}+D\left(c_{G4}-c_{L4}\right)=0\\ 1c_{L5}-1c_{L4}+0.8c_{G4}-0.8c_{L4}=0\\ 1c_{L5}-1.8c_{L4}+0.8c_{G4}=0\\ -1c_{L5}+1.8c_{L4}-0.8c_{G4}=0\end{array}$

For reactor 5, the system is in steady state. Therefore, the mass balance for the liquid reactor 5 is,

$Q_L{c}_{L6}-Q_L{c}_{L5}+D\left(c_{G5}-c_{L5}\right)=0$

Substitute $1$ for $Q_L$, $20$ for $c_{L6}$, and $0.8$ for $D$.

$\begin{array}{c}{Q}_L{c}_{L6}-Q_L{c}_{L5}+D\left(c_{G5}-c_{L5}\right)=0\\ 1\times 20-1c_{L5}+0.8c_{G5}-0.8c_{L5}=0\\ -1.8c_{L5}+0.8c_{G5}=-20\\ 1.8c_{L5}-0.8c_{G5}=20\end{array}$

Now, recollect all the linear equations for mass balance equation of gases and liquids.

$\begin{array}{c}200=2.8c_{G1}-0.8c_{L1}\\ -2c_{G1}+2.8c_{G2}-0.8c_{L2}=0\\ -2c_{G2}+2.8c_{G3}-0.8c_{L3}=0\\ -2c_{G3}+2.8c_{G4}-0.8c_{L4}=0\end{array}$

$\begin{array}{c}-2c_{G4}+2.8c_{G5}-0.8c_{L5}=0\\ -1c_{L2}+1.8c_{L1}-0.8c_{G1}=0\\ -1c_{L3}+1.8c_{L2}-0.8c_{G2}=0\\ -1c_{L4}+1.8c_{L3}-0.8c_{G3}=0\end{array}$

$\begin{array}{c}-1c_{L5}+1.8c_{L4}-0.8c_{G4}=0\\ 1.8c_{L5}-0.8c_{G5}=20\end{array}$

There are too many linear equations which is complex to solving manually. So, write the equation in the form ofmatrices that is augmented form as shown below:

$AX=D$

With the help of the linear system of equations provided above, the coefficient matrix A is,

$A=\left[\begin{array}{cccccccccc}2.8& 0& 0& 0& 0& -0.8& 0& 0& 0& 0\\ -2& 2.8& 0& 0& 0& 0& -0.8& 0& 0& 0\\ 0& -2& 2.8& 0& 0& 0& 0& -0.8& 0& 0\\ 0& 0& -2& 2.8& 0& 0& 0& 0& -0.8& 0\\ 0& 0& 0& -2& 2.8& 0& 0& 0& 0& -0.8\\ -0.8& 0& 0& 0& 0& 1.8& -1& 0& 0& 0\\ 0& -0.8& 0& 0& 0& 0& 1.8& -1& 0& 0\\ 0& 0& -0.8& 0& 0& 0& 0& 1.8& -1& 0\\ 0& 0& 0& -0.8& 0& 0& 0& 0& 1.8& -1\\ 0& 0& 0& 0& -0.8& 0& 0& 0& 0& 1.8\end{array}\right]$

With the help of the linear system of equations provided above, the column matrix X is:

$X=\left[\begin{array}{c}{c}_{G1}\\ c_{G2}\\ c_{G3}\\ c_{G4}\\ \begin{array}{c}{c}_{G5}\\ c_{L1}\\ c_{L2}\\ c_{L3}\\ c_{L4}\\ c_{L5}\end{array}\end{array}\right]$

And With the help of the linear system of equations provided above, the column matrix D is,

$D=\left[\begin{array}{c}200\\ 0\\ 0\\ 0\\ \begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\\ 20\end{array}\end{array}\right]$

Write the system of equation in the augmented form.,

$\begin{array}{c}AX=D\\ \left[\begin{array}{cccccccccc}2.8& 0& 0& 0& 0& -0.8& 0& 0& 0& 0\\ -2& 2.8& 0& 0& 0& 0& -0.8& 0& 0& 0\\ 0& -2& 2.8& 0& 0& 0& 0& -0.8& 0& 0\\ 0& 0& -2& 2.8& 0& 0& 0& 0& -0.8& 0\\ 0& 0& 0& -2& 2.8& 0& 0& 0& 0& -0.8\\ -0.8& 0& 0& 0& 0& 1.8& -1& 0& 0& 0\\ 0& -0.8& 0& 0& 0& 0& 1.8& -1& 0& 0\\ 0& 0& -0.8& 0& 0& 0& 0& 1.8& -1& 0\\ 0& 0& 0& -0.8& 0& 0& 0& 0& 1.8& -1\\ 0& 0& 0& 0& -0.8& 0& 0& 0& 0& 1.8\end{array}\right]\left[\begin{array}{c}{c}_{G1}\\ c_{G2}\\ c_{G3}\\ c_{G4}\\ \begin{array}{c}{c}_{G5}\\ c_{L1}\\ c_{L2}\\ c_{L3}\\ c_{L4}\\ c_{L5}\end{array}\end{array}\right]=\left[\begin{array}{c}200\\ 0\\ 0\\ 0\\ \begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\\ 20\end{array}\end{array}\right]\end{array}$

Solve the matrix $X=A^{-1}D$ in order to find the solutions:

MATLAB is used to perform the calculation, type the following code into MATLAB cmd.

A=[2.8 0 0 0 0 -0.8 0 0 0 0;

-2 2.8 0 0 0 0 -0.8 0 0 0;

0 -2 2.8 0 0 0 0 -0.8 0 0;

0 0 -2 2.8 0 0 0 0 -0.8 0;

0 0 0 -2 2.8 0 0 0 0 -0.8;

-0.8 0 0 0 0 1.8 -1 0 0 0;

0 -0.8 0 0 0 0 1.8 -1 0 0;

0 0 -0.8 0 0 0 0 1.8 -1 0;

0 0 0 -0.8 0 0 0 0 1.8 -1;

0 0 0 0 -0.8 0 0 0 0 1.8];

D=[200; 0; 0; 0; 0; 0; 0; 0; 0; 20];

X=A\D

Once you press eneter, the resut is obtained as follows:

X =

96.2074

91.3311

85.0617

77.0009

66.6371

86.7258

79.1405

69.3880

56.8491

40.7276

Hence, the values of concentrations passing through the reactors is shown in the below table:

$\begin{array}{lll}\hfill & \text{Liquid}\hfill & \text{Gas}\hfill \\ 0\hfill & 100.0000\hfill & \hfill \\ 1\hfill & 96.2074\hfill & 86.7258\hfill \\ 2\hfill & 91.3311\hfill & 79.1405\hfill \\ 3\hfill & 85.0617\hfill & 69.3880\hfill \\ 4\hfill & 77.0009\hfill & 56.8491\hfill \\ 5\hfill & 66.6371\hfill & 40.7276\hfill \\ 6\hfill & \hfill & 20.0000\hfill \end{array}$

The plot of concentration of liquid and gas is provided as: