Chapter 32, Problem 1P

Perform the same computation as in Sec. 32.1, but use $\Delta x=1.25$.

To determine

To calculate: The concentration of the distributed-parameter system equations for $\Delta x=1.25$. (Refer Sec. 32.1)

Answer to Problem 1P

Solution: The concentration of equations for $\Delta x=1.25$ is shown below.

Explanation of Solution

Given Information:

The distributed-parameter system depicts chemical modelling being subjected to first order decay and the tank is mixed well vertically and laterally.

Apply finite length of segment, $\Delta x$. So, the equation formed is of the form,

$V\frac{\Delta c}{\Delta t}=Qc\left(x\right)-Q\left[c\left(x\right)+\frac{\partial c\left(x\right)}{\partial x}\Delta x\right]-D{A}_c\frac{\partial c\left(x\right)}{\partial x}+D{A}_c\left[\frac{\partial c\left(x\right)}{\partial x}+\frac{\partial }{\partial x}\frac{\partial c\left(x\right)}{\partial x}\Delta x\right]-k{V}_c$

Here, $V$

is the volume, $Q$

is the flow rate, $c$

is the concentration, $D$

is the dispersion coefficient, $A_c$

is the tank’s cross-sectional area, and $k$

is the first order decay coefficient.

Calculation:

$Q{c}_{in}=Q{c}_0-D{A}_c\frac{\partial c_0}{\partial x}$ …… (1)

Substitute centred finite differences for the first and the second derivatives in equation (1) to develop steady-state solution,

$D\frac{c_{i+1}-2c_i+c_{i-1}}{\Delta x^2}-U\frac{c_{i+1`}-c_{i-1}}{2\Delta x}-k{c}_i=0$

Further simplify the equation,

$\left(\frac{2D}{U\Delta x}+\frac{k\Delta x}{U}+2+\frac{\Delta xU}{D}\right)c_0-\left(\frac{D}{U\Delta x}\right)c_1=\left(2+\frac{\Delta xU}{D}\right)c_{in}$ …… (2)

And,

$-\left(\frac{D}{U\Delta x}+\frac{1}{2}\right)c_{i-1}+\left(\frac{2D}{U\Delta x}+\frac{k\Delta x}{U}\right)c_i-\left(\frac{D}{U\Delta x}-\frac{1}{2}\right)c_{i+1}=0$ …… (3)

Also,

$-\left(\frac{D}{U\Delta x}\right)c_{n-1}+\left(\frac{2D}{U\Delta x}+\frac{k\Delta x}{U}\right)c_n=0$ …… (4)

Solve the equation (2),

$\left(\frac{2D}{U\Delta x}+\frac{k\Delta x}{U}+2+\frac{\Delta xU}{D}\right)c_0-\left(\frac{D}{U\Delta x}\right)c_1=\left(2+\frac{\Delta xU}{D}\right)c_{in}$

Substitute $D=2$, $U=1$, $c_{in}=100$, $\Delta x=1.25$, and $k=0.2$,

$\begin{array}{c}\left(\frac{2\times 2}{1\times 1.25}+\frac{0.2\times 1.25}{1}+2+\frac{1.25\times 1}{2}\right)c_0-\left(\frac{2}{1\times 1.25}\right)c_1=\left(2+\frac{1.25\times 1}{2}\right)100\\ \left(3.2+0.25+2+0.625\right)c_0-\left(1.6\right)c_1=\left(2+0.625\right)100\\ 6.075c_0-1.6c_1=262.5\end{array}$

Hence, the required first equation is $6.075c_0-1.6c_1=262.5$.

Solve the equation (3),

$-\left(\frac{D}{U\Delta x}+\frac{1}{2}\right)c_{i-1}+\left(\frac{2D}{U\Delta x}+\frac{k\Delta x}{U}\right)c_0-\left(\frac{D}{U\Delta x}-\frac{1}{2}\right)c_{i+1}=0$

Substitute $D=2$, $U=1$, $c_{in}=100$, $\Delta x=1.25$, and $k=0.2$,

$\begin{array}{c}-\left(\frac{2}{1\times 1.25}+\frac{1}{2}\right)c_{i-1}+\left(\frac{2\times 2}{1\times 1.25}+\frac{0.2\times 1.25}{1}\right)c_0-\left(\frac{2}{1\times 1.25}-\frac{1}{2}\right)c_{i+1}=0\\ -\left(2.1\right)c_{i-1}+\left(3.45\right)c_0-\left(1.1\right)c_{i+1}=0\end{array}$

Hence, the required middle equation is $-\left(2.1\right)c_{i-1}+\left(3.45\right)c_0-\left(1.1\right)c_{i+1}=0$.

Substitute $i=1$

in middle equation,

$\begin{array}{c}-\left(2.1\right)c_{i-1}+\left(3.45\right)c_0-\left(1.1\right)c_{i+1}=0\\ -\left(2.1\right)c_{1-1}+\left(3.45\right)c_1-\left(1.1\right)c_{1+1}=0\\ -2.1c_0+3.45c_1-1.1c_2=0\end{array}$

Substitute $i=2$ in middle equation,

$\begin{array}{c}-\left(2.1\right)c_{i-1}+\left(3.45\right)c_i-\left(1.1\right)c_{i+1}=0\\ -\left(2.1\right)c_{2-1}+\left(3.45\right)c_2-\left(1.1\right)c_{2+1}=0\\ -2.1c_1+3.45c_2-1.1c_3=0\end{array}$

Substitute $i=3$ in middle equation,

$\begin{array}{c}-\left(2.1\right)c_{i-1}+\left(3.45\right)c_i-\left(1.1\right)c_{i+1}=0\\ -\left(2.1\right)c_{3-1}+\left(3.45\right)c_3-\left(1.1\right)c_{3+1}=0\\ -2.1c_2+3.45c_3-1.1c_4=0\end{array}$

Substitute $i=4$ in middle equation,

$\begin{array}{c}-\left(2.1\right)c_{4-1}+\left(3.45\right)c_i-\left(1.1\right)c_{4+1}=0\\ -\left(2.1\right)c_{4-1}+\left(3.45\right)c_4-\left(1.1\right)c_{4+1}=0\\ -2.1c_3+3.45c_4-1.1c_5=0\end{array}$

Substitute $i=5$

in middle equation,

$\begin{array}{c}-\left(2.1\right)c_{i-1}+\left(3.45\right)c_i-\left(1.1\right)c_{i+1}=0\\ -\left(2.1\right)c_{5-1}+\left(3.45\right)c_5-\left(1.1\right)c_{5+1}=0\\ -2.1c_4+3.45c_5-1.1c_6=0\end{array}$

Substitute $i=6$

in middle equation,

$\begin{array}{c}-\left(2.1\right)c_{i-1}+\left(3.45\right)c_i-\left(1.1\right)c_{i+1}=0\\ -\left(2.1\right)c_{6-1}+\left(3.45\right)c_6-\left(1.1\right)c_{6+1}=0\\ -2.1c_5+3.45c_6-1.1c_7=0\end{array}$

Substitute $i=7$

in middle equation,

$\begin{array}{c}-\left(2.1\right)c_{i-1}+\left(3.45\right)c_i-\left(1.1\right)c_{i+1}=0\\ -\left(2.1\right)c_{7-1}+\left(3.45\right)c_7-\left(1.1\right)c_{7+1}=0\\ -2.1c_6+3.45c_7-1.1c_8=0\end{array}$

Substitute $i=8$

in middle equation,

$\begin{array}{c}-\left(2.1\right)c_{i-1}+\left(3.45\right)c_i-\left(1.1\right)c_{i+1}=0\\ -\left(2.1\right)c_{8-1}+\left(3.45\right)c_8-\left(1.1\right)c_{8+1}=0\\ -2.1c_7+3.45c_8-1.1c_9=0\end{array}$

Solve equation (4) for last equation,

$-\left(\frac{D}{U\Delta x}\right)c_{n-1}+\left(\frac{2D}{U\Delta x}+\frac{k\Delta x}{U}\right)c_n=0$

Substitute $D=2$, $U=1$, $c_{in}=100$, $\Delta x=1.25$, and $k=0.2$,

$\begin{array}{l}-\left(\frac{2}{1\times 1.25}\right)c_{9-1}+\left(\frac{2\times 2}{1\times 1.25}+\frac{0.2\times 1.25}{1}\right)c_9=0\\ -1.6c_8+3.45c_9=0\end{array}$

Hence, the required last equation is $-1.6c_8+3.45c_9=0$

Write all the equations in matrix form,

$\left[\begin{array}{c}6.075c_0-1.6c_1\\ -2.1c_0+3.45c_1-1.1c_2\\ -2.1c_1+3.45c_2-1.1c_3\\ -2.1c_2+3.45c_3-1.1c_4\\ -2.1c_3+3.45c_4-1.1c_5\\ -2.1c_4+3.45c_5-1.1c_6\\ -2.1c_5+3.45c_6-1.1c_7\\ -2.1c_6+3.45c_7-1.1c_8\\ -2.1c_7+3.45c_8-1.1c_9\\ -1.6c_8+3.45c_9\end{array}\right]=\left[\begin{array}{c}262.5\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right]$

Express the matrix in tri-diagonal form,

$\left[\begin{array}{cccccccccc}6.075& -1.6& 0& 0& 0& 0& 0& 0& 0& 0\\ -2.1& 3.45& -1.1& 0& 0& 0& 0& 0& 0& 0\\ 0& -2.1& 3.45& -1.1& 0& 0& 0& 0& 0& 0\\ 0& 0& -2.1& 3.45& -1.1& 0& 0& 0& 0& 0\\ 0& 0& 0& -2.1& 3.45& -1.1& 0& 0& 0& 0\\ 0& 0& 0& 0& -2.1& 3.45& -1.1& 0& 0& 0\\ 0& 0& 0& 0& 0& -2.1& 3.45& -1.1& 0& 0\\ 0& 0& 0& 0& 0& 0& -2.1& 3.45& -1.1& 0\\ 0& 0& 0& 0& 0& 0& 0& -2.1& 3.45& -1.1\\ 0& 0& 0& 0& 0& 0& 0& 0& -1.6& 3.45\end{array}\right]\left[\begin{array}{c}{c}_0\\ c_1\\ c_2\\ c_3\\ c_4\\ c_5\\ c_6\\ c_7\\ c_8\\ c_9\end{array}\right]=\left[\begin{array}{c}262.5\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right]$

Use MATLAB to solve the above matrix.

% declare the matrix size

A = zeros(10,10);

% define the values of matrix A

A(1,1) = 6.075;

A(1,2) = -3.2;

A(2,1) = -2.1;

A(2,2) = 3.45;

A(2,3) = -1.1;

A(3,2) = -2.1;

A(3,3) = 3.45;

A(3,4) = -1.1;

A(4,3) = -2.1;

A(4,4) = 3.45;

A(4,5) = -1.1;

A(5,4) = -2.1;

A(5,5) = 3.45;

A(5,6) = -1.1;

A(6,5) = -2.1;

A(6,6) = 3.45;

A(6,7) = -1.1;

A(7,6) = -2.1;

A(7,7) = 3.45;

A(7,8) = -1.1;

A(8,7) = -2.1;

A(8,8) = 3.45;

A(8,9) = -1.1;

A(9,8) = -2.1;

A(9,9) = 3.45;

A(9,10) = -1.1;

A(10,9) = -3.2;

A(10,10) = 3.45;

% define the matrix b

b = [262.5;0;0;0;0;0;0;0;0;0];

% solve the inverse of matrix

Sol = A\b

The output of the program is given below.