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 28, Problem 15P
To determine

To calculate: The concentration of each reactant as the function of distance by using the finite difference approach, and apply centred finite-difference approximations with Δx=0.05 m.

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Answer to Problem 15P

Solution:

The concentration of each reactant as the function of distance is,

Numerical Methods for Engineers, Chapter 28, Problem 15P , additional homework tip  1

The below plot shows the distance versus reactant.

Explanation of Solution

Given Information:

The series of first order, liquid phase reactions is,

Ak1Bk2C

The second order ODEs by using the steady-state mass balance.

Dd2cadx2Udcadxk1ca=0Dd2cbdx2Udcbdx+k1cak2cb=0Dd2ccdx2Udccdx+k2cb=0

Here, c is the concentration (mol/L), x is the distance (m), D is the dispersion coefficient (0.1 m2/min), U is the velocity (1 m/min), k1 is the reaction rate (3/min), k2 is the reaction rate (1/min).

Refer to the Prob 28.14, the Danckwerts boundary conditions is,

Ucin=Uc(x=0)Ddcdx(x=0)

dcdx(x=L)=0

Here, ca,in is the concentration in the inflow (10 mol/L), and L is the length of the reactor (0.5 m).

Δx=0.05 m

Formula used:

The finite divided difference formula is,

f(x0)=ci12ci+ci+1Δx2

Calculation:

Recall the ordinary differential equations,

Dd2cadx2Udcadxk1ca=0Dd2cbdx2Udcbdx+k1cak2cb=0Dd2ccdx2Udccdx+k2cb=0

Substitute the finite divided difference formula in the above differential equations.

D(ca,i12ca,i+ca,i+1Δx2)U(ca,i+1ca,i12Δx)k1ca,i=0D(cb,i12cb,i+cb,i+1Δx2)U(cb,i+1cb,i12Δx)+k1ca,ik2cb,i=0D(cc,i12cc,i+cc,i+1Δx2)U(cc,i+1cc,i12Δx)+k2cb,i=0

Substitute (0.1 m2/min) for D, (1 m/min) for U, 0.05 m for Δx, (3/min) for k1, and (1/min) for k2 yield unit.

(0.1)(ca,i12ca,i+ca,i+1(0.05)2)(1)(ca,i+1ca,i12(0.05))(3)ca,i=0(0.1)(cb,i12cb,i+cb,i+1(0.05)2)(1)(cb,i+1cb,i12(0.05))+(3)ca,i(1)cb,i=0(0.1)(cc,i12cc,i+cc,i+1(0.05)2)(1)(cc,i+1cc,i12(0.05))+(1)cb,i=0

Solve further,

50ca,i1+83ca,i30ca,i+1=050cb,i1+81cb,i30cb,i+1=3ca,i50cc,i1+80cc,i30cc,i+1=cb,i

Now solve for inlet node i=1, use a finite difference approximation for the first derivative.

Here use the second order version from the Table 19.3 for the interior nodes,

Uca,in=Uca,1Dca,3+4ca,23ca,12Δx

Ucb,in=Ucb,1Dcb,3+4cb,23cb,12Δx

Ucc,in=Ucc,1Dcc,3+4cc,23cc,12Δx

Can be solved for,

(3D2Δx2+UΔx)ca,1(2DΔx2)ca,2+(D2Δx2)ca,3=UΔxca,in

(3D2Δx2+UΔx)cb,1(2DΔx2)cb,2+(D2Δx2)cb,3=UΔxcb,in

(3D2Δx2+UΔx)cc,1(2DΔx2)cc,2+(D2Δx2)cc,3=UΔxcc,in

Substitute (0.1 m2/min) for D, (1 m/min) for U, 0.05 m for Δx, and 10 mol/L for ca,in yield unit.

80ca,180ca,2+20ca,3=200

80cb,180cb,2+20cb,3=0

80cc,180cc,2+20cc,3=0

Solve for the outer node (i=10), the zero derivative condition which implies that c11=c9,

(2DΔx2)c9+(2DΔx2+k)c10=0

The similar equations can be written for the other nodes, because the condition does not include reaction rates Substitute all the parameter gives,

80ca,9+83ca,10=080cb,9+81cb,10=3ca,10

80cc,9+80cc,10=cb,10

Rearrange the all equations in matrix form for each reactant separately, because the reactions are in series.

Write for the reactant A.

[808020000000005083300000000005083300000000005083300000000005083300000000005083300000000005083300000000005083300000000005083300000000005083300000000008380][ca,1ca,2ca,3ca,4ca,5ca,6ca,7ca,8ca,9ca,10ca,11]=[2000000000000]

Write the following code in MATLAB.

A=[80 -80 20 0 0 0 0 0 0 0 0

-50 83 -30 0 0 0 0 0 0 0 0

0 -50 83 -30 0 0 0 0 0 0 0

0 0 -50 83 -30 0 0 0 0 0 0

0 0 0 -50 83 -30 0 0 0 0 0

0 0 0 0 -50 83 -30 0 0 0 0

0 0 0 0 0 -50 83 -30 0 0 0

0 0 0 0 0 0 -50 83 -30 0 0

0 0 0 0 0 0 0 -50 83 -30 0

0 0 0 0 0 0 0 0 -50 83 -30

0 0 0 0 0 0 0 0 0 83 -80];

b=[200;0;0;0;0;0;0;0;0;0;0];

c=A\b;

c'

The output is,

ans =

Columns 1 through 5

8.0655 7.1509 6.3416 5.6270 4.9988

Columns 6 through 10

4.4516 3.9847 3.6050 3.3328 3.2123

Column 11

3.3328

Write the all the above equations in matrix form for the reactant B.

[808020000000005081300000000005081300000000005081300000000005081300000000005081300000000005081300000000005081300000000005081300000000005081300000000008180][cb,1cb,2cb,3cb,4cb,5cb,6cb,7cb,8cb,9cb,10cb,11]=[021.452719.024816.881014.996413.354811.954110.81509.99849.63699.6369]

Write the following code in MATLAB.

A=[80 -80 20 0 0 0 0 0 0 0 0

-50 81 -30 0 0 0 0 0 0 0 0

0 -50 81 -30 0 0 0 0 0 0 0

0 0 -50 81 -30 0 0 0 0 0 0

0 0 0 -50 81 -30 0 0 0 0 0

0 0 0 0 -50 81 -30 0 0 0 0

0 0 0 0 0 -50 81 -30 0 0 0

0 0 0 0 0 0 -50 81 -30 0 0

0 0 0 0 0 0 0 -50 81 -30 0

0 0 0 0 0 0 0 0 -50 81 -30

0 0 0 0 0 0 0 0 0 81 -80];

b=[0;21.4527;19.0248;16.881;14.9964;13.3548;

11.9541;10.815;9.9984;9.6369;9.6369];

c=A\b;

c'

The output is,

ans =

1.6486 2.4090 3.0415 3.5629 3.9880 4.3295 4.5977 4.7997 4.9358

4.9938 4.9358

Write the all the above equations in matrix form for the reactant C.

[808020000000005080300000000005080300000000005080300000000005080300000000005080300000000005080300000000005080300000000005080300000000005080300000000008080][cc,1cc,2cc,3cc,4cc,5cc,6cc,7cc,8cc,9cc,10cc,11]=[02.40903.04153.56293.98804.32954.59774.79974.93584.99384.9358]

Write the following code in MATLAB.

A=[80 -80 20 0 0 0 0 0 0 0 0

-50 80 -30 0 0 0 0 0 0 0 0

0 -50 80 -30 0 0 0 0 0 0 0

0 0 -50 80 -30 0 0 0 0 0 0

0 0 0 -50 80 -30 0 0 0 0 0

0 0 0 0 -50 80 -30 0 0 0 0

0 0 0 0 0 -50 80 -30 0 0 0

0 0 0 0 0 0 -50 80 -30 0 0

0 0 0 0 0 0 0 -50 80 -30 0

0 0 0 0 0 0 0 0 -50 80 -30

0 0 0 0 0 0 0 0 0 80 -80];

b=[0;2.4090;3.0415;3.5629;3.9880;4.3295;

4.5977;4.7997;4.9358;4.9938;4.9358];

c=A\b;

c'

The output is,

Columns 1 through 7

0.2859 0.4402 0.6169 0.8101 1.0133 1.2191 1.4178

Columns 8 through 11

1.5957 1.7321 1.7949 1.7332

The reaction is in series, thus the system for each reactant is,

Numerical Methods for Engineers, Chapter 28, Problem 15P , additional homework tip  2

The below plot shows the distance versus reactant.

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Chapter 28 Solutions

Numerical Methods for Engineers

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