Application of Numerical Methods in Chemical Engineering

General Ideas Regarding the Use of the Interpolation in the Chemical Engineering There are certain cases when approximation solutions are useful within the numerical methods field itself. Some of these cases are presented in the next sections. Approximation of the Initial Solution for Iterative Methods Let us consider a theoretical problem where the formulation of the problem depends on some parameters and an iterative solution is possible. It is well known that the final solution depends on the initial approximation in terms of accuracy and of computer runtime.

Approximation is used to evaluate an initial solution which is close to the final solution If some solutions can be found using an initial solution which is not accurate, the next initial solutions can be evaluated by the use of an interpolation method. In this way the computer time decreases and the accuracy is increased. Calculus of the Integrals Let us consider that a set of points are known, that is. . Usually these points result from experimental studies. There must be computed the integral of the function approximated by this set of points.

Two ways to compute the integral of a function defined by a set of points

ENGINEERING APPLICATIONS OF SYSTEMS OF LINEAR EQUATIONS

Many “real life” situations are governed by a system of differential equations. Physical problems usually have more than one dependent variable to be considered. Models of such problems can give rise to system of differential equations in which there are two or more dependent variable and one independent variable. Example: A Yogurt Company makes three yogurt blends: Lime Orange, using 2 quarts of lime yogurt and 2 quarts of orange yogurt per gallon; Lime Lemon, using 3 quarts of lime yogurt and 1 quart of lemon yogurt per gallon; and Orange Lemon, using 3 quarts of orange yogurt and 1 quart of lemon yogurt per gallon. Each day the company has 800 quarts of lime yogurt, 650 quarts of orange yogurt, and 350 quarts of lemon yogurt available. How many gallons of each blend should it make each day if it wants to use up all of the supplies? The unknowns in the problem are related to what? The number of gallons of each yogurt blends How many unknowns in the problem? The three unknowns can be specified as follows: Let x = the number of gallons of Lime Orange the company should make Let y = the number of gallons of Lime Lemon the company should make Let z = the number of gallons of Orange Lemon the company should make You can also list these in some other order, for instance: Let x = the number of gallons of Lime Lemon the company should make Let y = the number of gallons of Lime Orange the company should make Let z = the number of gallons of Orange Lemon the company should make. Note For the rest of the problem, we shall assume the first ordering. We can organize the given information in a table. To set up the table, do the following:  

Place the categories corresponding to the unknowns along the top. Add an extra column for the "Total Available"

Now read across the first row of the table: it gives the amounts of lime yogurt needed for the three blends, and also the total available. If the company makes x quarts of Lime Orange, y quarts of Lime Lemon, and z quarts of Orange Lemon, it will need a total of 2x + 3y quarts of lime yoghurt. Since Soft flow has a total of 800 quarts of lime yogurt on hand, and it wants nothing left over, we must have Amount used=Amount Available 2x + 3y=800 Similarly, we get two more equations for orange and lemon yogurt:

Q Equation for Orange Yogurt: 2x + 3z = 650 Q Equation for Lemon Yogurt: y + z = 350

Application of Gaussian Elimination method in chemical engineering

This method reduces the system of linear equations to a simpler form.

Example A liquid-liquid extraction process conducted in the Electrochemical Materials Laboratory involved the extraction of nickel from the aqueous phase into an organic phase. A typical set of experimental data from the laboratory is given below. Ni aqueous phase, a g l 

2

2.5

3

Ni organic phase, g g l 

8.57

10

12

Assuming g is the amount of Ni in the organic phase and a is the amount of Ni in the aqueous phase, the quadratic interpolant that estimates g is given by

g  x1 a 2  x 2 a  x3 , 2  a  3 The solution for the unknowns x1 , x2 and x3 is given by

2 1  x1  8.57   4 6.25 2.5 1  x    10     2    9 3 1  x3   12  Find the values of x1 , x 2 and x3 using naïve Gauss elimination. Estimate the amount of nickel in the organic phase when 2.3 g l is in the aqueous phase using quadratic interpolation.

Solution Forward Elimination of Unknowns Since there are three equations, there will be two steps of forward elimination of unknowns.

First step

Divide Row 1 by 4 and then multiply it by 6.25, that is, multiply Row 1 by 6.25 4  1.5625 .

Row 1  1.5625   6.25 3.125 1.5625

13.391

Subtract the result from Row 2 to get

2 1 4   x1   8.57  0  0.625  0.5625  x    3.3906     2   9   x3   12  3 1 Divide Row 1 by 4 and then multiply it by 9 , that is, multiply Row 1 by 9 4  2.25 .

Row 1  2.25   9 4.5 2.25

19.283

Subtract the result from Row 3 to get

2 1 4   x1   8.57  0  0.625  0.5625  x    3.3906     2   0  1 .5  1.25   x3   7.2825

Second step

We now divide Row 2 by −0.625 and then multiply it by −1.5, that is, multiply Row 2 by

 1.5  0.625  2.4 . Row 2  2.4   0  1.5  1.35

 8.1375

Subtract the result from Row 3 to get

2 1 4   x1   8.57  0  0.625  0.5625  x    3.3906     2   0 0 0.1   x3   0.855 

Back Substitution From the third equation,

0.1x3  0.855

x3 

0.855 0.1

 8.55 Substituting the value of x3 in the second equation,

 0.625x2   0.5625x3  3.3906  3.3906   0.5625 x3  0.625

x2 



 3.3906   0.5625  8.55  0.625

 2.27 Substituting the values of x 2 and x3 in the first equation,

4 x1  2 x 2  x3  8.57 8.57  2 x 2  x3 4

x1 



8.57  2   2.27   8.55 4

 1.14 Hence the solution vector is

 x1   1.14   x     2.27   2    x3   8.55  The polynomial that passes through the three data points is then

g a   x1 a 2  x 2 a  x3  1.14a 2   2.27 a  8.55 Where g is the amount of nickel in the organic phase and a is the amount of nickel in the aqueous phase. When 2.3 g l is in the aqueous phase, using quadratic interpolation, the estimated amount of nickel in the organic phase is

g 2.3  1.14  2.3   2.27   2.3  8.55 2

 9.3596 g/l Application of Euler method in chemical engineering

Euler method is used to check the stability analysis of system of linear equations

Example 1 The concentration of salt x in a homemade soap maker is given as a function of time by

dx  37.5  3.5 x dt At the initial time, t  0 , the salt concentration in the tank is 50 g/L. Using Euler’s method and a step size of h  1.5 min , what is the salt concentration after 3 minutes?

Solution

dx  37.5  3.5 x dt f t , x   37.5  3.5 x The Euler’s method reduces to

xi 1  xi  f t i , xi h For i  0 , t 0  0 , x0  50

x1  x 0  f t 0 , x 0 h  50  f 0,50 1.5  50  37.5  3.5(50) 1.5  50   137.51.5

 156.25 g/L x1 Is the approximate concentration of salt at

t  t1  t 0  h  0  1.5  1.5 min x 1.5  x1  156 .25 g/L For i  1 , t1  1.5 , x1  156 .25

x 2  x1  f t1 , x1 h  156.25  f 1.5,156.251.5  156.25  37.5  3.5(156.25) 1.5  156.25  584.381.5

 720.31 g/L

x2 is the approximate concentration of salt at t  t 2  t1  h  1 .5  1 .5  3 min x 3  x 2  720.31 g/L

Figure 1 compares the exact solution with the numerical solution from Euler’s method for the step size of h  1.5 .

Figure 1 Comparing exact and Euler’s method.

The problem was solved again using smaller step sizes. The results are given below in Table 1.

Table 1 Concentration of salt at 3 minutes as a function of step size, h . step size, h

x 3

Et

|t | %

3

362.5

373.22

3483.0

1.5

720.31

709.60

6622.2

0.75

284.65

273.93

2556.5

0.375

10.718

0.0024912

0.023249

0.1875

10.714

0.0010803

0.010082

Figure 2 shows how the concentration of salt varies as a function of time for different step sizes.

Figure 2 Comparison of Euler’s method with exact solution for different step sizes.

While the values of the calculated concentration of salt at t  3 min as a function of step size are plotted in Figure 3.

Figure 3 Effect of step size in Euler’s method.

The exact solution of the ordinary differential equation is given by

x(t )  10.714  39.286e 3.5t The solution to this nonlinear equation at t  3 min is

x(3)  10.715 g/L Application of r-k method in chemical engineering

The Runge-Kutta Method is a commonly used numerical method for solving 1st-order ordinary differential equations (ODEs) with a known initial condition. The method starts at a known point and develops the solution to the ODE by proceeding stepwise in small increments.  

It is used to solve Variable Volume/Concentration Tank Problem Used to check the set of equations for absorber calculation

Application of Iterative Methods in chemical engineering

These were particularly important in the early days of computing, when memory was limited. In fact, many can be implemented without reference to matrices. MANY areas of chemical engineering analysis require the use of efficient techniques for determining the roots of algebraic or transcendental equations. In many cases, the equation is not explicit, but is represented by an algorithm of a mathematical mode1 of elements of a process, with which an output variable is calculated for a certain input variable, and it is required to find the value of the input variable for which the value of the output variable is minimal or zero. EXAMPLES

Two cases of typical problems in chemical engineering, that result in a variety of forms Case I. A chemical equilibrium problem Case II. AJash distillation problem

Application of Taylor Series method in chemical engineering Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point. In Taylor Series the polynomial functions are easy and that if we could find a way of representing complicated functions as series ("infinite polynomials") then maybe some properties of functions would be easy to study too.  

Taylor series in used by engineers in analyzing circuits Taylor series also used in chemical sensors and vibration sensors