Numerical Methods for Engineers
Numerical Methods for Engineers
7th Edition
ISBN: 9780073397924
Author: Steven C. Chapra Dr., Raymond P. Canale
Publisher: McGraw-Hill Education
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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 , and  Q 44 = 12 .

Expert Solution & Answer
Check Mark
To determine

To calculate: The concentration of the fluid flow rate through the remaining reactors if the value of the following flow changes to Q01=6, Q12=14, Q24=2 and Q44=12 from the value provided in figure 12.3 in text book.

Answer to Problem 1P

Solution: The concentrations c1,c2,c3,c4,c5 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:

Q01=6  Q15=3, Q12=4, Q31=1, Q03=8, Q25=1, Q23=1, Q54=2, Q55=2, Q24=2, Q34=8 and Q44=12

Also,

c01=20 and c03=6

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 figure provided in the section 12.1.

Numerical Methods for Engineers, Chapter 12, Problem 1P , additional homework tip  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,

Q01c01+Q31c3

Mass flow out rate is,

Q12c1+Q15c1

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

So,

Mass flow out=Mass flow in

Substitute the values of the flow rates.

Q12c1+Q15c1=Q01c01+Q31c3

Substitute the known values from the diagram.

4c1+3c1=6×20+1×c37c1c3=120…… (1)

For reactor 2, the mass flow rate in is,

Q12c1

The mass flow rate out is,

Q25c2+Q24c2+Q23c2

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

Mass flow out =Mass flow in

Substitute the values of mass flow rates.

Q25c2+Q24c2+Q23c2=Q12c1

Substitute the values of rates of flows from the figure.

c2+2c2+1c2=4c14c24c1=0…… (2)

For reactor 3, mass flow rate in is,

Q03c03+Q23c2

The mass flow rate out is,

Q31c3+Q34c3

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

Mass flow out=Mass flow in

Substitute the mass flow rate in the above expression.

Q31c3+Q34c3=Q03c03+Q23c2

Substitute the values of the flow rate from the figure.

1c3+8c3=8×6+1c29c3=48+c29c3c2=48…… (3)

For reactor 4, mass flow rate in is,

Q24c2+Q34c3+Q54c5

The mass flow out rate is,

Q44c4

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

Mass flow out=Mass flow in

Substitute the mass flow rate values in the above expression.

Q44c4=Q24c2+Q34c3+Q54c5

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

12c4=2c2+8c3+2c512c42c28c32c5=0…… (4)

For reactor 5, the mass flow rate in is,

Q15c1+Q25c2

The mass flow rate out is,

Q55c5+Q54c5

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

Mass flow out=Mass flow in

Substitute the mass flow rate into the expression.

Q55c5+Q54c5=Q15c1+Q25c2

Substitute the value of flow rate into the above expression.

2c5+2c5=3c1+1c24c53c1c2=0…… (5)

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

7c1c3=1204c24c1=09c3c2=4812c42c28c32c5=0

4c53c1c2=0

Write the above equations in the matrix

AX=DX=A1D

Here, coefficient matrix A is,

A=[70100440000190002812231004]

Column matrix X is,

X=[c1c2c3c4c5]

Column matrix D is,

D=[12004800]

Use MATLAB to solve the matrix system X=A1D.

>> 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:

Numerical Methods for Engineers, Chapter 12, Problem 1P , additional homework tip  2

Hence,

[c1c2c3c4c5]=[18.193518.19357.354810.967718.1935]

Hence, the concentrations c1,c2,c3,c4,c5 in milligram cubic per meter are 18.1935,18.1935,7.3548,10.9677 and 18.1935 respectively.

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