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 12P

Chemical/Bio Engineering

Figure P12.12 depicts a chemical exchange process consisting of a series of reactors in which a gas fl owing from left to right is passed over a liquid fl owing from right to left. The transfer of a chemical from the gas into the liquid occurs at a rate that is proportional to the difference between the gas and liquid concentrations in each reactor. At steady state, a mass balance for the first reactor can be written for the gas as

Q G C G O Q G C G I + D ( C L 1 C G I ) = 0

and for the liquid as

Q L C 12 Q L C L 1 + D ( C G I C L 1 ) = 0

Where Q G and Q L are the gas and liquid flow rates, respectively, and D = the gas liquid exchange rate . Similar balances can be written for the other reactors. Solve for the concentrations given the following values:

Q G = 2 , Q L = 1 , D = 0.8 , C G O = 100 , C L 6 = 20 .

Chapter 12, Problem 12P, Chemical/Bio Engineering
12.12	Figure P12.12 depicts a chemical exchange process consisting of a

FIGURE P12.12

Expert Solution & Answer
Check Mark
To determine

To calculate: The flow rate of the flow through every pipe if the flow equation through he reactors is provided by following system of equations:

For gas it is provided as:

QGcG0QGcG1+D(cL1cG1)=0

And for liquid it is provided as:

QLcL2QLcL1+D(cG1cL1)=0

Answer to Problem 12P

Solution: The concentration of liquid and gases in the reactor is:

LiquidGas0100.0000196.207486.7258291.331179.1405385.061769.3880477.000956.8491566.637140.7276620.0000

Explanation of Solution

Given Information:

The system of reactors is provided as follows:

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

Flow rate and the concentration are provided as:

QG=2, QL=1, cG0=100 and cL6=20.

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 below,

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

Here, QG is the gas flow rate and QL is the liquid flow rates.

For reactor 1, the system is in steady state. Therefore, themass balance equation for the gas reactor 1.

QGcG0QGcG1+D(cL1cG1)=0

Substitute 2 for QG, 0.8 for D and 100 for cG0.

2×1002cG1+0.8(cL1cG1)=02002cG1+0.8cL10.8cG1=0200=2.8cG10.8cL1

For reactor 2, the system is in steady state. Therefore, the mass balance for the gas reactor 2 is,

QGcG1QGcG2+D(cL2cG2)=0

Substitute 2 for QG and 0.8 for D.

2cG12cG2+0.8(cL2cG2)=02cG12cG2+0.8cL20.8cG2=02cG12.8cG2+0.8cL2=02cG1+2.8cG20.8cL2=0

For reactor 3, the system is in steady state, therefore, the mass balance equation for the gas reactor 3 is,

QGcG2QGcG3+D(cL3cG3)=0

Substitute 2 for QG, 0.8 for D and 100 for cG0.

QGcG2QGcG3+D(cL3cG3)=02cG22cG3+0.8cL30.8cG3=02cG22.8cG3+0.8cL3=02cG2+2.8cG30.8cL3=0

For reactor 4, the system is in steady state. Therefore, the mass balance for the gas reactor 4 is,

QGcG3QGcG4+D(cL4cG4)=0

Substitute 2 for QG, 0.8 for D and 100 for cG0.

QGcG3QGcG4+D(cL4cG4)=02cG32cG4+0.8cL40.8cG4=02cG32.8cG4+0.8cL4=02cG3+2.8cG40.8cL4=0

For reactor 5, the system is in steady state. Therefore, the mass balance for the gas reactor 5 is,

QGcG4QGcG5+D(cL5cG5)=0

Substitute 2 for QG, 0.8 for D and 100 for cG0.

QGcG4QGcG5+D(cL5cG5)=02cG42cG5+0.8cL50.8cG5=02cG42.8cG5+0.8cL5=02cG4+2.8cG50.8cL5=0

For reactor 1, the system is in steady state. Therefore, the mass balance for the liquid reactor 1 is,

QLcL2QLcL1+D(cG1cL1)=0

Substitute 1 for QL and 0.8 for D.

QLcL2QLcL1+D(cG1cL1)=01cL21cL1+0.8cG10.8cL1=01cL21.8cL1+0.8cG1=0

1cL2+1.8cL10.8cG1=0 …… (6)

For reactor 2, the system is in steady state. Therefore, the mass balance for the liquid reactor 2 is,

QLcL3QLcL2+D(cG2cL2)=0

Substitute 1 for QL and 0.8 for D.

QLcL3QLcL2+D(cG2cL2)=01cL31cL2+0.8cG20.8cL2=01cL31.8cL2+0.8cG2=0

1cL3+1.8cL20.8cG2=0 …… (7)

For reactor 3, the system is in steady state. Therefore, the mass balance for the liquid reactor 3 is,

QLcL4QLcL3+D(cG3cL3)=0

Substitute 1 for QL and 0.8 for D.

QLcL4QLcL3+D(cG3cL3)=01cL41cL3+0.8cG30.8cL3=01cL41.8cL3+0.8cG3=01cL4+1.8cL30.8cG3=0

For reactor 4, the system is in steady state. Therefore, the mass balance for the liquid reactor 4 is,

QLcL5QLcL4+D(cG4cL4)=0

Substitute 1 for QL and 0.8 for D.

QLcL5QLcL4+D(cG4cL4)=01cL51cL4+0.8cG40.8cL4=01cL51.8cL4+0.8cG4=01cL5+1.8cL40.8cG4=0

For reactor 5, the system is in steady state. Therefore, the mass balance for the liquid reactor 5 is,

QLcL6QLcL5+D(cG5cL5)=0

Substitute 1 for QL, 20 for cL6, and 0.8 for D.

QLcL6QLcL5+D(cG5cL5)=01×201cL5+0.8cG50.8cL5=01.8cL5+0.8cG5=201.8cL50.8cG5=20

Now, recollect all the linear equations for mass balance equation of gases and liquids.

200=2.8cG10.8cL12cG1+2.8cG20.8cL2=02cG2+2.8cG30.8cL3=02cG3+2.8cG40.8cL4=0

2cG4+2.8cG50.8cL5=01cL2+1.8cL10.8cG1=01cL3+1.8cL20.8cG2=01cL4+1.8cL30.8cG3=0

1cL5+1.8cL40.8cG4=01.8cL50.8cG5=20

There are too many linear equations which is complex to solving manually. So, write the equation in the form ofmatrices that is augmented form as shown below:

AX=D

With the help of the linear system of equations provided above, the coefficient matrix A is,

A=[2.800000.8000022.800000.8000022.800000.8000022.800000.8000022.800000.80.800001.8100000.800001.8100000.800001.8100000.800001.8100000.800001.8]

With the help of the linear system of equations provided above, the column matrix X is:

X=[cG1cG2cG3cG4cG5cL1cL2cL3cL4cL5]

And With the help of the linear system of equations provided above, the column matrix D is,

D=[2000000000020]

Write the system of equation in the augmented form.,

AX=D[2.800000.8000022.800000.8000022.800000.8000022.800000.8000022.800000.80.800001.8100000.800001.8100000.800001.8100000.800001.8100000.800001.8][cG1cG2cG3cG4cG5cL1cL2cL3cL4cL5]=[2000000000020]

Solve the matrix X=A1D in order to find the solutions:

MATLAB is used to perform the calculation, type the following code into MATLAB cmd.

A=[2.8 0 0 0 0 -0.8 0 0 0 0;

-2 2.8 0 0 0 0 -0.8 0 0 0;

0 -2 2.8 0 0 0 0 -0.8 0 0;

0 0 -2 2.8 0 0 0 0 -0.8 0;

0 0 0 -2 2.8 0 0 0 0 -0.8;

-0.8 0 0 0 0 1.8 -1 0 0 0;

0 -0.8 0 0 0 0 1.8 -1 0 0;

0 0 -0.8 0 0 0 0 1.8 -1 0;

0 0 0 -0.8 0 0 0 0 1.8 -1;

0 0 0 0 -0.8 0 0 0 0 1.8];

D=[200; 0; 0; 0; 0; 0; 0; 0; 0; 20];

X=A\D

Once you press eneter, the resut is obtained as follows:

X =

96.2074

91.3311

85.0617

77.0009

66.6371

86.7258

79.1405

69.3880

56.8491

40.7276

Hence, the values of concentrations passing through the reactors is shown in the below table:

LiquidGas0100.0000196.207486.7258291.331179.1405385.061769.3880477.000956.8491566.637140.7276620.0000

The plot of concentration of liquid and gas is provided as:

Numerical Methods for Engineers, Chapter 12, Problem 12P , additional homework tip  3

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