Chapter 12, Problem 1P

Chemical/Bio Engineering

Perform the same computation as in Sec. 12.1, but change $c_{01}$ to 20 and $c_{03}$ to 6. Also change the following flows: $Q_{01}=6,Q_{12}=4,Q_{24}=2,\text{and }{Q}_{44}=12$.

To determine

To calculate: The concentration of the fluid flow rate through the remaining reactors if the value of the following flow changes to $Q_{01}=6$, $Q_{12}=14$, $Q_{24}=2$ and $Q_{44}=12$ from the value provided in figure 12.3 in text book.

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Answer to Problem 1P

Solution: The concentrations $c_1,c_2,c_3,c_4,c_5$ in milligram cubic per meter are $18.1935,18.1935,7.3548,10.9677$ and $18.1935$ respectively.

Explanation of Solution

Given Information:

Flow rates are provided as follows:

$Q_{01}=6$ $Q_{15}=3$, $Q_{12}=4$, $Q_{31}=1$, $Q_{03}=8$, $Q_{25}=1$, $Q_{23}=1$, $Q_{54}=2$, $Q_{55}=2$, $Q_{24}=2$, $Q_{34}=8$ and $Q_{44}=12$

Also,

$c_{01}=20$ and $c_{03}=6$

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 provided in the section 12.1.

Here, Q is the flow rate in cubic meter per minute and c is the concentration in milligrams cubic per meter.

For reactor 1, mass flow rate in is expressed as follows,

$Q_{01}{c}_{01}+Q_{31}{c}_3$

Mass flow out rate is,

$Q_{12}{c}_1+Q_{15}{c}_1$

Provided system is in the steady state. Therefore, mass flow rate will be constant.

So,

$\text{Mass flow out}=\text{Mass flow in}$

Substitute the values of the flow rates.

$Q_{12}{c}_1+Q_{15}{c}_1=Q_{01}{c}_{01}+Q_{31}{c}_3$

Substitute the known values from the diagram.

$\begin{array}{c}4c_1+3c_1=6\times 20+1\times c_3\\ 7c_1-c_3=120\end{array}$…… (1)

For reactor 2, the mass flow rate in is,

$Q_{12}{c}_1$

The mass flow rate out is,

$Q_{25}{c}_2+Q_{24}{c}_2+Q_{23}{c}_2$

The system is in steady state flow therefore, the mass flow in and out is same.

$\text{Mass flow out }=\text{Mass flow in}$

Substitute the values of mass flow rates.

$Q_{25}{c}_2+Q_{24}{c}_2+Q_{23}{c}_2=Q_{12}{c}_1$

Substitute the values of rates of flows from the figure.

$\begin{array}{c}{c}_2+2c_2+1c_2=4c_1\\ 4c_2-4c_1=0\end{array}$…… (2)

For reactor 3, mass flow rate in is,

$Q_{03}{c}_{03}+Q_{23}{c}_2$

The mass flow rate out is,

$Q_{31}{c}_3+Q_{34}{c}_3$

The system is in steady state flow therefore, the mass flow in and out is same.

$\text{Mass flow out}=\text{Mass flow in}$

Substitute the mass flow rate in the above expression.

$Q_{31}{c}_3+Q_{34}{c}_3=Q_{03}{c}_{03}+Q_{23}{c}_2$

Substitute the values of the flow rate from the figure.

$\begin{array}{c}1c_3+8c_3=8\times 6+1c_2\\ 9c_3=48+c_2\\ 9c_3-c_2=48\end{array}$…… (3)

For reactor 4, mass flow rate in is,

$Q_{24}{c}_2+Q_{34}{c}_3+Q_{54}{c}_5$

The mass flow out rate is,

$Q_{44}{c}_4$

The system is in steady state. Therefore, the mass flow rate in and out will be same.

$\text{Mass flow out}=\text{Mass flow in}$

Substitute the mass flow rate values in the above expression.

$Q_{44}{c}_4=Q_{24}{c}_2+Q_{34}{c}_3+Q_{54}{c}_5$

Substitute the value of the flow rate from the diagram in the above equation.

$\begin{array}{c}12c_4=2c_2+8c_3+2c_5\\ 12c_4-2c_2-8c_3-2c_5=0\end{array}$…… (4)

For reactor 5, the mass flow rate in is,

$Q_{15}{c}_1+Q_{25}{c}_2$

The mass flow rate out is,

$Q_{55}{c}_5+Q_{54}{c}_5$

The system is in steady state therefore, the mass flow rate in and out is same.

$\text{Mass flow out}=\text{Mass flow in}$

Substitute the mass flow rate into the expression.

$Q_{55}{c}_5+Q_{54}{c}_5=Q_{15}{c}_1+Q_{25}{c}_2$

Substitute the value of flow rate into the above expression.

$\begin{array}{c}2c_5+2c_5=3c_1+1c_2\\ 4c_5-3c_1-c_2=0\end{array}$…… (5)

Write all the system of equations to express in a linear system of equations.

$\begin{array}{c}7c_1-c_3=120\\ 4c_2-4c_1=0\\ 9c_3-c_2=48\\ 12c_4-2c_2-8c_3-2c_5=0\end{array}$

$4c_5-3c_1-c_2=0$

Write the above equations in the matrix

$\begin{array}{c}AX=D\\ X=A^{-1}D\end{array}$

Here, coefficient matrix A is,

$A=\left[\begin{array}{ccccc}7& 0& -1& 0& 0\\ -4& 4& 0& 0& 0\\ 0& -1& 9& 0& 0\\ 0& -2& -8& 12& -2\\ -3& -1& 0& 0& 4\end{array}\right]$

Column matrix X is,

$X=\left[\begin{array}{c}{c}_1\\ c_2\\ c_3\\ c_4\\ c_5\end{array}\right]$

Column matrix D is,

$D=\left[\begin{array}{c}120\\ 0\\ 48\\ 0\\ 0\end{array}\right]$

Use MATLAB to solve the matrix system $X=A^{-1}D$.

>> A=[7 0 -1 0 0; -4 4 0 0 0; 0 -1 9 0 0;0 -2 -8 12 -2; -3 -1 0 0 4];

D = [120; 0; 48; 0; 0];

X=A\D

The result is obtained as follows:

Hence,

$\left[\begin{array}{c}{c}_1\\ c_2\\ c_3\\ c_4\\ c_5\end{array}\right]=\left[\begin{array}{c}18.1935\\ 18.1935\\ 7.3548\\ 10.9677\\ 18.1935\end{array}\right]$

Hence, the concentrations $c_1,c_2,c_3,c_4,c_5$ in milligram cubic per meter are $18.1935,18.1935,7.3548,10.9677$ and $18.1935$ respectively. $$