Numerical Methods For Engineers, 7 Ed
Numerical Methods For Engineers, 7 Ed
7th Edition
ISBN: 9789352602131
Author: Canale Chapra
Publisher: MCGRAW-HILL HIGHER EDUCATION
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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:

Numerical Methods For Engineers, 7 Ed, Chapter 12, Problem 20P , additional homework tip  1

(a)

Expert Solution
Check Mark

Answer to Problem 20P

Solution: The concentrations of carbon mono-oxide c1,c2,c3,c4 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:

Numerical Methods For Engineers, 7 Ed, Chapter 12, Problem 20P , additional homework tip  2

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

0=Wsmoker+QacaQac1+E13(c3c1)

Formula used:

Write system of linear equations in matrix form.

AX=D

And,

X=A1D

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:

Numerical Methods For Engineers, 7 Ed, Chapter 12, Problem 20P , additional homework tip  3

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=Wsmoker+QacaQac1+E13(c3c1)

Substitute the known values:

0=1000+200×2200c1+25(c3c1)=1000+400200c1+25c325c1 =1400225c1+25c3

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

0=Qbcb+(QaQd)c4Qcc2+E24(c4c2)

Substitute the known values and solve:

0=50×2+(200100)c4150×c2+25(c4c2)=100+100c4150c2+25c425c2 =100175c2+125c4

The mass balance equation for room 3 is as follows:

0=Wgrill+Qac1+E13(c1c3)+E34(c4c3)Qac3

Substitute the known values and solve:

0=2000+200c1+25(c1c3)+50(c4c3)200c3 =2000+200c1+25c125c3+50c450c3200c3 =2000+225c1275c3+50c4

The mass balance equation for room 4 is as follows:

0=Qac3+E34(c3c4)+E24(c2c4)Qdc4

Substitute the known values and solve;

0=200c3+50(c3c4)+25(c2c4)100c4 =200c3+50c350c4+25c225c4100c4

Consider all the linear equation obtained above.

1400=225c125c3 100=175c2125c4 2000=225c1+275c350c4 0=25c2250c3+175c4

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

AX=D or X=A1D

Here, the coefficient matrix A is:

A=[ 225025001750125225027550025250175 ]

The column matrix X is,

X=[ c1c2c3c4 ]

And

The column matrix D is:

D=[ 140010020000 ]

Substitute the values in the augmented form of the equation:

AX=D[ 225025001750125225027550025250175 ][ c1c2c3c4 ]=[ 140010020000 ]

To solve the equation use MATLAB, follow the following seeps to obtained the result: Numerical Methods For Engineers, 7 Ed, Chapter 12, Problem 20P , additional homework tip  4

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

Numerical Methods For Engineers, 7 Ed, Chapter 12, Problem 20P , additional homework tip  5

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

[ c1c2c3c4 ]=[ 8.441523.333319.973331.8667 ]

Therefore, the concentrations c1,c2,c3,c4 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.

(b)

Expert Solution
Check Mark

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:

Numerical Methods For Engineers, 7 Ed, Chapter 12, Problem 20P , additional homework tip  6

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

0=Wsmoker+QacaQac1+E13(c3c1)

Calculation:

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

1400=225c125c3 100=175c2125c4 2000=225c1+275c350c4 0=25c2250c3+175c4

Open MATLAB and enter values as shown below,

Numerical Methods For Engineers, 7 Ed, Chapter 12, Problem 20P , additional homework tip  7

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

Numerical Methods For Engineers, 7 Ed, Chapter 12, Problem 20P , additional homework tip  8

(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:

c2,smokers=a211Wsmokers

Substitute a211=0.0067 and Wsmokers=1000.

a211Wsmokers=0.0067(1000)=6.7

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

%smokers=6.723.3333×100%=28.71% (ii)

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

c2,grill=a311Wgrill

Substitute a311=0.0059 and Wgrill=2000.

a311Wgrill=0.0059(2000)=11.8

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

%grills=11.823.3333×100%=50.57%

(iii)

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

c2,intakes=a211Qaca+a221Qbcb

Substitute the known values:

a211Qaca+a221Qbcb=0.0067(200)2+0.0067(50)2=2.68+0.67=3.35

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

%air intakes=3.3523.333×100%=14.35%

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 mg/hr.

(c)

Expert Solution
Check Mark

Answer to Problem 20P

Solution: The increase in the kid’s section is 24.4 mg/hr.

Explanation of Solution

Given Information:

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

Numerical Methods For Engineers, 7 Ed, Chapter 12, Problem 20P , additional homework tip  9

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

0=Wsmoker+QacaQac1+E13(c3c1)

Calculation:

The smoker and grill loads are increased to 2000 and 5000 mg/hr respectively.

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

Δc2,smokers=a211ΔWsmokers

Substitute a211=0.0067 and ΔWsmokers=20001000.

a211Wsmokers=0.0067(20001000)=0.0067(1000)=6.7

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

Δc2,grill=a311ΔWgrill

Substitute a311=0.0059 and ΔWgrill=50002000.

a311Wgrill=0.0059(50002000)=0.0059(3000)=17.7

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

Δc2=Δc2,smokers+Δc2,grill =6.7+17.7=24.4

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 m3/hr.

(d)

Expert Solution
Check Mark

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:

Numerical Methods For Engineers, 7 Ed, Chapter 12, Problem 20P , additional homework tip  10

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

0=Wsmoker+QacaQac1+E13(c3c1)

Formula used:

Write system of linear equations in matrix form.

AX=D

And,

X=A1D

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 m3/hr.

Therefore, the value of E24=5 m3/hr.

The balance equation for room 1 is written as follows:

0=Wsmoker+QacaQac1+E13(c3c1)

Substitute the known values;

0=1000+200×2200c1+25(c3c1)=1000+400200c1+25c325c1 =1400225c1+25c3

The balance equation for room 2 is written as follows:

0=Qbcb+(QaQd)c4Qcc2+E24(c4c2)

Substitute the known values and solve:

0=50×2+(200100)c4150×c2+5(c4c2)=100+100c4150c2+5c45c2 =100155c2+105c4

The balance equation for room 3 is written as follows:

0=Wgrill+Qac1+E13(c1c3)+E34(c4c3)Qac3

Substitute the known values and solve:

0=2000+200c1+25(c1c3)+50(c4c3)200c3 =2000+200c1+25c125c3+50c450c3200c3 =2000+225c1275c3+50c4

The balance equation for room 4 is written as follows:

0=Qac3+E34(c3c4)+E24(c2c4)Qdc4

Substitute the known values and solve.

0=200c3+50(c3c4)+5(c2c4)100c4 =200c3+50c350c4+5c25c4100c4

Rewrite all the equations obtained by applying mass balance equation.

1400=225c125c3 100=155c2105c4 2000=225c1+275c350c4 0=5c2250c3+155c4

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

AX=D or X=A1D

Here, the coefficient matrix A is,

A=[ 22502500155010522502755005250155 ]

The column matrix X is:

X=[ c1c2c3c4 ]

The column matrix D is;

D=[ 140010020000 ]

Substitute the value of the matrices.

AX=D[ 22502500155010522502755005250155 ][ c1c2c3c4 ]=[ 140010020000 ]

Add the following code into MATLAB “cmd”.

Numerical Methods For Engineers, 7 Ed, Chapter 12, Problem 20P , additional homework tip  11

Press enter to get the result.

Numerical Methods For Engineers, 7 Ed, Chapter 12, Problem 20P , additional homework tip  12

Hence, the equivalent concentration matrix is:

[ c1c2c3c4 ]=[ 8.477623.333320.298433.4921 ]

Therefore, the concentration in the kids’ area is:

23.33323.333=0 mg/m3

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

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Chapter 12 Solutions

Numerical Methods For Engineers, 7 Ed

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