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 10P

Chemical/Bio Engineering

An irreversible, first-order reaction takes place in four well mixed reactors (Fig. P12.10),

A k B

Thus, the rate at which A is transformed to B can be represented as

R a b = k V c

The reactors have different volumes, and because they are operated at different temperatures, each has a different reaction rate:

Reactor V, L k , h - 1
1 25 0.05
2 75 0.1
3 100 0.5
4 25 0.1

Determine the concentration of A and B in each of the reactors at steady state.

Chapter 12, Problem 10P, Chemical/Bio Engineering
12.10	An irreversible, first-order reaction takes place in four well mixed

FIGURE P12.10

Expert Solution & Answer
Check Mark
To determine

To calculate:

Answer to Problem 10P

Solution:

Explanation of Solution

Given Information:

Write the provided values of the volume and rate of reaction. reactorV(L)k(h-1)1250.052750.131000.54250.1

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 diagram for an irreversible first order reaction takes place in four will-mixed reactors.

Numerical Methods for Engineers, Chapter 12, Problem 10P

Balance the mass for A in reactor 1 at steady-state.

QincA,inQincA,1k1V1cA,1=0

Substitute 10 for Qin, 1 for cA,in, 25 for V1 and 0.05 for k1.

10(1)10cA,1(0.05)(25)cA,1=01010cA,11.25cA,1=01011.25cA,1=0

Further solve.

11.25cA,1=10 …… (1)

Balance the mass for A in reactor 2 at steady-state.

QincA,1+Q32cA,3(Qin+Q32)cA,2k2V2cA,2=0

Substitute 10 for Qin, 5 for Q32, 75 for V2 and 0.1 for k2.

10cA,1+5cA,3(10+5)cA,2(0.10)(75)cA,2=010cA,1+5cA,315cA,27.5cA,2=010cA,1+5cA,322.5cA,2=0

Further solve.

10cA,122.5cA,2+5cA,3=0 …… (2)

Balance the mass for A in reactor 3 at steady-state.

(Qin+Q32)cA,2+Q43cA,4(Qin+Q43+Q32)cA,3k3V3cA,3=0

Substitute 10 for Qin, 5 for Q32, 3 for Q43, 100 for V3 and 0.5 for k3.

(10+5)cA,2+3cA,4(10+3+5)cA,3(0.5)(100)cA,3=015cA,2+3cA,418cA,350cA,3=0

Further solve.

15cA,268cA,3+3cA,4=0 …… (3)

Balance the mass for A in reactor 4 at steady-state.

(Qin+Q43)cA,3(Qin+Q43)cA,4k4V4cA,4=0

Substitute 10 for Qin, 3 for Q43, 25 for V4 and 0.1 for k4.

(10+3)cA,3(10+3)cA,4(0.1)(25)cA,4=013cA,313cA,42.5cA,4=0

Further solve. 13cA,315.5cA,4=0 …… (4)

Balance the mass for B in reactor 1 at steady-state.

QincB,1+k1V1cA,1=0

Substitute 10 for Qin, 25 for V1 and 0.05 for k1.

10cB,1+(0.05)(25)cA,1=0

Further solve.

10cB,1+1.25cA,1=0 …… (5)

Balance the mass for B in reactor 2 at steady-state.

QincB,1+Q32cB,3(Qin+Q32)cB,2+k2V2cA,2=0

Substitute 10 for Qin, 5 for Q32, 75 for V2 and 0.1 for k2.

10cB,1+5cB,3(10+5)cB,2+(0.1)(75)cA,2=0

Further solve.

10cB,1+5cB,315cB,2+7.5cA,2=0 …… (6)

Balance the mass for B in reactor 3 at steady-state.

(Qin+Q32)cB,2+Q43cB,4(Qin+Q43+Q32)cB,3+k3V3cA,3=0

Substitute 10 for Qin, 5 for Q32, 3 for Q43, 100 for V3 and 0.5 for k3.

(10+5)cB,2+3cB,4(10+3+5)cB,3+(0.5)(100)cA,3=0

Further solve.

15cB,2+3cB,418cB,3+50cA,3=0 …… (7)

Balance the mass for B in reactor 4 at steady-state.

(Qin+Q43)cB,3(Qin+Q43)cB,4+k4V4cA,4=0

Substitute 10 for Qin, 3 for Q43, 25 for V4 and 0.1 for k4.

(10+3)cB,3(10+3)cB,4+(0.1)(25)cA,4=0

Further solve.

13cB,313cB,4+2.5cA,4=0 …… (8)

Now, write all the equations, to find linear system of equations.

11.25cA,1=1010cA,122.5cA,2+5cA,3=015cA,268cA,3+3cA,4=013cA,315.5cA,4=0

10cB,1+1.25cA,1=010cB,1+5cB,315cB,2+7.5cA,2=015cB,2+3cB,418cB,3+50cA,3=013cB,313cB,4+2.5cA,4=0

Write the above linear equations in matrix form as written in symbolized form.

AC=B 

Here, coefficient matrix A is,

A=[ 11.2500000001022.55000000106830000001315.500001.250001000007.500101550005000151830002.5001313 ]

Column matrix C is,

C=[ cA,1cA,2cA,3cA,4cB,1cB,2cB,3cB,4]

Column matrix B is,

B=[ 100000000 ]

Substitute the values in the matrix equation form.

AC=B[ 11.2500000001022.55000000106830000001315.500001.250001000007.500101550005000151830002.5001313 ][ cA,1cA,2cA,3cA,4cB,1cB,2cB,3cB,4]=[ 100000000 ]

Solve for C=A1B with the help of MATLAB to obtain the value of the corresponding concentration for each reactor for the column matrix X.

Code:

>> A=[-11.25 0 0 0 0 0 0 0;

10 -22.5 5 0 0 0 0 0;

0 10 -68 3 0 0 0 0;

0 0 130 -15.5 0 0 0 0;

1.25 0 0 0 -10 0 0 0;

0 7.5 0 0 10 -15 5 0;

0 0 50 0 0 15 -18 3;

0 0 0 2.5 0 0 13 -13];

B=[-10;0;0;0;0;0;0;0];

C=A\B

Type the above code into MATLAB command window and press enter to find the result.

Result is obtained as follows:

C =

0.8889

0.4167

0.0973

0.8158

0.1111

0.6014

0.9570

1.1139

Hence,

C=[ cA,1=cA,2cA,3cA,4cB,1cB,2cB,3cB,4]

Hence, the concentration of A and B is,

cA,1=0.8421cB,1=0.1578cA,2=0.3400cB,2=0.9932cA,3=0.1010cB,3=1.8989cA,4=0.0847cB,4=1.9152

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