Chapter 12, Problem 20P

(a)

To determine

To calculate: The concentration of carbon mono-oxide at steady state in each room, for the system provided as shown below:

Answer to Problem 20P

Solution: The concentrations of carbon mono-oxide $c_1,c_2,c_3,c_4$ in milligram cubic per meter are $8.4415,23.3333,19.9733$

and $31.8667$ respectively.

Explanation of Solution

Given Information:

The air flow through the room with proper description is provided as follows:

Also, for the smoking section (room 1) the balance can be written as follows:

$0=W_{\text{smoker}}+Q_a{c}_a-Q_a{c}_1+E_{13}\left(c_3-c_1\right)$

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 figure:

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

Taking Smoking section into consideration the mass balance equation for room 1 is as follows:

$0=W_{\text{smoker}}+Q_a{c}_a-Q_a{c}_1+E_{13}(c_3-c_1)$

Substitute the known values:

$\begin{array}{c}0=1000+200\times 2-200c_1+25(c_3-c_1)\\ =1000+400-200c_1+25c_3-25c_1\\ =1400-225c_1+25c_3\end{array}$

Taking Kid’s section into consideration the mass balance equation for room 2 is as follows:

$0=Q_b{c}_b+(Q_a-Q_d)c_4-Q_c{c}_2+E_{24}(c_4-c_2)$

Substitute the known values and solve:

$\begin{array}{c}0=50\times 2+\left(200-100\right)c_4-150\times c_2+25\left(c_4-c_2\right)\\ =100+100c_4-150c_2+25c_4-25c_2\\ =100-175c_2+125c_4\end{array}$

The mass balance equation for room 3 is as follows:

$0=W_{\text{grill}}+Q_a{c}_1+E_{13}(c_1-c_3)+E_{34}(c_4-c_3)-Q_a{c}_3$

Substitute the known values and solve:

$\begin{array}{c}0=2000+200c_1+25(c_1-c_3)+50(c_4-c_3)-200c_3\\ =2000+200c_1+25c_1-25c_3+50c_4-50c_3-200c_3\\ =2000+225c_1-275c_3+50c_4\end{array}$

The mass balance equation for room 4 is as follows:

$0=Q_a{c}_3+E_{34}(c_3-c_4)+E_{24}(c_2-c_4)-Q_d{c}_4$

Substitute the known values and solve;

$\begin{array}{c}0=200c_3+50(c_3-c_4)+25(c_2-c_4)-100c_4\\ =200c_3+50c_3-50c_4+25c_2-25c_4-100c_4\end{array}$

Consider all the linear equation obtained above.

$\begin{array}{c}1400=225c_1-25c_3\\ 100=175c_2-125c_4\\ 2000=-225c_1+275c_3-50c_4\\ 0=-25c_2-250c_3+175c_4\end{array}$

The system of equations is written in augmented form as shown below:

$AX=D\text{ or }X=A^{-1}D$

Here, the coefficient matrix A is:

$A=\left[\begin{array}{cccc}225& 0& -25& 0\\ 0& 175& 0& -125\\ -225& 0& 275& -50\\ 0& -25& -250& 175\end{array}\right]$

The column matrix X is,

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

And

The column matrix D is:

$D=\left[\begin{array}{c}1400\\ 100\\ 2000\\ 0\end{array}\right]$

Substitute the values in the augmented form of the equation:

$\begin{array}{c}AX=D\\ \left[\begin{array}{cccc}225& 0& -25& 0\\ 0& 175& 0& -125\\ -225& 0& 275& -50\\ 0& -25& -250& 175\end{array}\right]\left[\begin{array}{c}{c}_1\\ c_2\\ c_3\\ c_4\end{array}\right]=\left[\begin{array}{c}1400\\ 100\\ 2000\\ 0\end{array}\right]\end{array}$

To solve the equation use MATLAB, follow the following seeps to obtained the result:

Once these details are added in the MATLAB “cmd” press enter and the result is obtained as follows:

Hence, the concentration matrix for carbon mono-oxide is as follows:

$\left[\begin{array}{c}{c}_1\\ c_2\\ c_3\\ c_4\end{array}\right]=\left[\begin{array}{c}8.4415\\ 23.3333\\ 19.9733\\ 31.8667\end{array}\right]$

Therefore, the concentrations $c_1,c_2,c_3,c_4$ in milligram cubic per meter are $8.4415,23.3333,19.9733$ and $31.8667$ respectively. $$

(b)

To determine

To calculate: The percentage of carbon mono-oxide due to (i) the smokers, (ii) the grill and the air in the intake vents.

Answer to Problem 20P

Solution: The percent of the carbon mon-oxide in the kids’ section due to smokers, grills, and air intake vents is $28.71\%,50.57\%$

and $14.35\%$ respectively.

Explanation of Solution

Given Information:

The air flow through the room with proper description is provided as follows:

Also, for the smoking section (room 1) the balance can be written as follows:

$0=W_{\text{smoker}}+Q_a{c}_a-Q_a{c}_1+E_{13}\left(c_3-c_1\right)$

Calculation:

Rewrite all the linear equation obtained in part (a).

$\begin{array}{c}1400=225c_1-25c_3\\ 100=175c_2-125c_4\\ 2000=-225c_1+275c_3-50c_4\\ 0=-25c_2-250c_3+175c_4\end{array}$

Open MATLAB and enter values as shown below,

Once, you press enter after adding these all in the matlab “cmd” the result is obtained as:

(i)

Follow the following procedure to obtain the value of the carbon mono-oxide concentration in kid’s section.

The concentration of the carbon mono-oxide in the kid’s section is:

$c_{2,\text{smokers}}=a_{21}^{-1}{W}_{\text{smokers}}$

Substitute $a_{21}^{-1}=0.0067\text{ and }{W}_{\text{smokers}}=1000$.

$\begin{array}{c}{a}_{21}^{-1}{W}_{\text{smokers}}=0.0067(1000)\\ =6.7\end{array}$

Carbon monoxide percentage in the kids’ section due to smokers is as follows,

$\begin{array}{c}{\%}_{\text{smokers}}=\frac{6.7}{23.3333}\times 100\%\\ =28.71\%\end{array}$ (ii)

The concentration of the carbon mono-oxide in the kid’s section due to grill is:

$c_{2,\text{grill}}=a_{31}^{-1}{W}_{\text{grill}}$

Substitute $a_{31}^{-1}=0.0059\text{ and }{W}_{\text{grill}}=2000$.

$\begin{array}{c}{a}_{31}^{-1}{W}_{\text{grill}}=0.0059(2000)\\ =11.8\end{array}$

Carbon monoxide percentage in the kids’ section due to grill is as follows,

$\begin{array}{c}{\%}_{\text{grills}}=\frac{11.8}{23.3333}\times 100\%\\ =50.57\%\end{array}$

(iii)

The concentration of the carbon mono-oxide in the kid’s section due to air intake vents is:

$c_{2,\text{intakes}}=a_{21}^{-1}{Q}_a{c}_a+a_{22}^{-1}{Q}_b{c}_b$

Substitute the known values:

$\begin{array}{c}{a}_{21}^{-1}{Q}_a{c}_a+a_{22}^{-1}{Q}_b{c}_b=0.0067(200)2+0.0067(50)2\\ =2.68+0.67\\ =3.35\end{array}$

Carbon monoxide percentage in the kids’ section due to air intake vents is:

$\begin{array}{c}{\%}_{\text{air intakes}}=\frac{3.35}{23.333}\times 100\%\\ =14.35\%\end{array}$

Hence, the percent of the carbon monoxide in the kids’ section due to smokers, grills, and air intake vents is $28.71\%,50.57\%$ and $14.35\%$ respectively.

(c)

To determine

To calculate: The increase in concentration in the kid’s section by using matrix inverse if the smokers and grill loads are increased to $2000$ and $5000$ $\text{mg/hr}$.

Answer to Problem 20P

Solution: The increase in the kid’s section is $24.4\text{ mg/hr}$.

Explanation of Solution

Given Information:

The air flow through the room with proper description is provided as follows:

Also, for the smoking section (room 1) the balance can be written as follows:

$0=W_{\text{smoker}}+Q_a{c}_a-Q_a{c}_1+E_{13}\left(c_3-c_1\right)$

Calculation:

The smoker and grill loads are increased to 2000 and 5000 $\text{mg/hr}$ respectively.

Therefore, increase in concentration of carbon mono-oxide after increase of smoker loads is:

$\Delta c_{2,\text{smokers}}=a_{21}^{-1}\Delta W_{\text{smokers}}$

Substitute $a_{21}^{-1}=0.0067\text{ and }\Delta W_{\text{smokers}}=2000-1000$.

$\begin{array}{c}{a}_{21}^{-1}{W}_{\text{smokers}}=0.0067(2000-1000)\\ =0.0067(1000)\\ =6.7\end{array}$

Therefore, increase in concentration of carbon mono-oxide after increase of grill load is:

$\Delta c_{2,\text{grill}}=a_{31}^{-1}\Delta W_{\text{grill}}$

Substitute $a_{31}^{-1}=0.0059\text{ and }\Delta W_{\text{grill}}=5000-2000$.

$\begin{array}{c}{a}_{31}^{-1}{W}_{\text{grill}}=0.0059(5000-2000)\\ =0.0059(3000)\\ =17.7\end{array}$

Therefore, increase in the concentration of the carbon mon-oxide in the kid’s section is:

$\begin{array}{c}\Delta c_2=\Delta c_{2,\text{smokers}}+\Delta c_{2,\text{grill}}\\ =6.7+17.7\\ =24.4\end{array}$

$$

The incereases in concetration is $24.4$.

(d)

To determine

To calculate: The change in the concentration in the kid’s section if the mixing between area 2 and 4 is decreases to $5{\text{ m}}^3\text{/hr}$.

Answer to Problem 20P

Solution: There won’t be any changes in concertation in kid’s section.

Explanation of Solution

Given Information:

The air flow through the room with proper description is provided as follows:

Also, for the smoking section (room 1) the balance can be written as follows:

$0=W_{\text{smoker}}+Q_a{c}_a-Q_a{c}_1+E_{13}\left(c_3-c_1\right)$

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:

The mixing between areas 2 and 4 is decreased to $5{\text{ m}}^3/\text{hr}$.

Therefore, the value of $E_{24}=5{\text{ m}}^3/\text{hr}$.

The balance equation for room 1 is written as follows:

$0=W_{\text{smoker}}+Q_a{c}_a-Q_a{c}_1+E_{13}(c_3-c_1)$

Substitute the known values;

$\begin{array}{c}0=1000+200\times 2-200c_1+25(c_3-c_1)\\ =1000+400-200c_1+25c_3-25c_1\\ =1400-225c_1+25c_3\end{array}$

The balance equation for room 2 is written as follows:

$0=Q_b{c}_b+(Q_a-Q_d)c_4-Q_c{c}_2+E_{24}(c_4-c_2)$

Substitute the known values and solve:

$\begin{array}{c}0=50\times 2+\left(200-100\right)c_4-150\times c_2+5\left(c_4-c_2\right)\\ =100+100c_4-150c_2+5c_4-5c_2\\ =100-155c_2+105c_4\end{array}$

The balance equation for room 3 is written as follows:

$0=W_{\text{grill}}+Q_a{c}_1+E_{13}(c_1-c_3)+E_{34}(c_4-c_3)-Q_a{c}_3$

Substitute the known values and solve:

$\begin{array}{c}0=2000+200c_1+25(c_1-c_3)+50(c_4-c_3)-200c_3\\ =2000+200c_1+25c_1-25c_3+50c_4-50c_3-200c_3\\ =2000+225c_1-275c_3+50c_4\end{array}$

The balance equation for room 4 is written as follows:

$0=Q_a{c}_3+E_{34}(c_3-c_4)+E_{24}(c_2-c_4)-Q_d{c}_4$

Substitute the known values and solve.

$\begin{array}{c}0=200c_3+50(c_3-c_4)+5(c_2-c_4)-100c_4\\ =200c_3+50c_3-50c_4+5c_2-5c_4-100c_4\end{array}$

Rewrite all the equations obtained by applying mass balance equation.

$\begin{array}{c}1400=225c_1-25c_3\\ 100=155c_2-105c_4\\ 2000=-225c_1+275c_3-50c_4\\ 0=-5c_2-250c_3+155c_4\end{array}$

Rewrite the linear equations system in augmented form as shown below.

$AX=D\text{ or }X=A^{-1}D$

Here, the coefficient matrix A is,

$A=\left[\begin{array}{cccc}225& 0& -25& 0\\ 0& 155& 0& -105\\ -225& 0& 275& -50\\ 0& -5& -250& 155\end{array}\right]$

The column matrix X is:

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

The column matrix D is;

$D=\left[\begin{array}{c}1400\\ 100\\ 2000\\ 0\end{array}\right]$

Substitute the value of the matrices.

$\begin{array}{c}AX=D\\ \left[\begin{array}{cccc}225& 0& -25& 0\\ 0& 155& 0& -105\\ -225& 0& 275& -50\\ 0& -5& -250& 155\end{array}\right]\left[\begin{array}{c}{c}_1\\ c_2\\ c_3\\ c_4\end{array}\right]=\left[\begin{array}{c}1400\\ 100\\ 2000\\ 0\end{array}\right]\end{array}$

Add the following code into MATLAB “cmd”.

Press enter to get the result.

Hence, the equivalent concentration matrix is:

$\left[\begin{array}{c}{c}_1\\ c_2\\ c_3\\ c_4\end{array}\right]=\left[\begin{array}{c}8.4776\\ 23.3333\\ 20.2984\\ 33.4921\end{array}\right]$

Therefore, the concentration in the kids’ area is:

$23.333-23.333=0{\text{ mg/m}}^{\text{3}}$

Therefore, there won’t be any changes in the concentration of the kid’s room area.