Question 8.PS.1: Concentrated Differential Distillation In this final case st...
Concentrated Differential Distillation
In this final case study, we will consider the case of differential distillation of benzene and toluene from our colleague Ed Cussler’s book Diffusion, which is illustrated schematically in Fig. 8.10. In this system, we have a saturated liquid feed of 3500 mol/h of 40% benzene that is to be distilled into a top product of D = 1400 mol/h that is x_D = 98% benzene and a bottom product of 2100 mol/h that is x_B = 2% benzene. The flow of gas in the (upper) rectifying part of the column is G = 4730 mol/h and the flow of liquid in the rectifying part of the column is L = 3330 mol/h. The balance on toluene on the upper part of the column gives the operating line for rectification
y_r =\frac{Dx_D}{G} +\frac{L}{G}x = 0.287 + 0.704x (8.6.1)
Table 8.4 Vapor–liquid equilibrium data for the mole fraction of toluene.
x 0.01 0.08 0.17 0.35 0.40 0.52 0.58 0.62 0.75 0.90 0.97
y^∗ 0.03 0.17 0.33 0.57 0.63 0.73 0.78 0.81 0.88 0.96 0.99
where x is the mole fraction of toluene in the liquid flowing downward at some point in the column and y is the corresponding mole fraction of toluene in the vapor flowing upward at that point in the column. Since the feed is saturated liquid, it all goes down into the stripping part of the column such that the liquid flow rate is L′ = 6830 mol/h and the gas flow rate is G′ = 4730 mol/h to satisfy the mass and energy balances at the feed location. The balance on toluene in the lower part of the column gives the second operating line for stripping
y_s = −\frac{Bx_B}{G′} +\frac{L′}{G′}x = −0.009 + 1.444x (8.6.2)
These two lines intersect at the feed line, which for a saturated liquid at this concentration is x = 0.4. The goal of the problem is to determine the number of theoretical units (NTU), which is a measure of the difficulty of the separation and an intermediate step to compute the height of the column (which also requires knowing the height of a theoretical unit, or HTU). The analysis of the mass transfer inside the column leads to
NTU =\int_{x_D}^{y_l(x_B)}{\frac{dy}{y − y^∗}} (8.6.3)
where y_l(x_B) is the composition of the vapor phase that would be in equilibrium with the bottoms product at x_B. For the toluene–benzene system with x_B = 0.02, the equilibrium phase data gives y_l(x_B) = 0.085. The quantity y^∗ is the vapor-phase equilibrium composition, which is given in tabulated form in Table 8.4. The goal of this problem is to use the methods in this chapter to compute the NTU for this process.
The first item of business is to determine a relationship between y^∗ and x so that we can use it in our integration. The simplest approach is to create an interpolating function to the equilibrium data, which we can then use directly in our integral. We can readily accomplish this task using Program 8.1 to create a tenth-order polynomial for y^∗(x). We can then break up the integral in Eq. (8.6.3) into two pieces,
NTU =\int_{x_D}^{y_F}{\frac{dy_r}{y_s − y^∗}} +\int_{y_F}^{y_l(x_B)}{\frac{dy_s}{y_s − y^∗}} (8.6.4)
where y_F = 0.5686 is the point at which the two operating lines meet, i.e. the intersection of Eq. (8.6.1) and Eq. (8.6.2). We can then use any of our methods to evaluate the integrals. Let’s use five-point Gauss quadrature, since it is quite accurate and this is a relatively smooth integrand. To do so, we first identify the Gauss points in y_r or y_s over their range of integration. At each Gauss point y_i, we can compute the corresponding value of x_i from either Eq. (8.6.1) or Eq. (8.6.2). We then use Program 8.2 to compute the corresponding value of y^∗ at the location x_i. We then are able to do the function evaluations for Gauss quadrature and get the result.
Let’s see how this strategy is implemented. The main challenge is implementing the function evaluations, which are handled by the following program:
1 function f = getf(y,xpts,coeff)
2 n = length(coeff)-1;
3 if y >= 0.5686
4 x = (y - 0.287)/0.704;
5 else
6 x = (y + 0.009)/1.444;
7 end
8 ys = integrate_interpolate_f(x,xpts,coeff,n);
9 f = 1/(y - ys);
The if-else block selects the appropriate operating line, and we then evaluate the integrand at that value of y. Note that line 8 calls Program 8.2.
Actually computing the integrals is handled by a master program:
1 function integration_distillation
2 clc
3 xD = 0.98;
4 yF = 0.5868;
5 yl = 0.085;
6 %equilibrium data
7 x = [0.01,0.08,0.17,0.35,0.40,0.52,0.58,0.62,0.75,0.90,0.97];
8 ys = [0.03,0.17,0.33,0.57,0.63,0.73,0.78,0.81,0.88,0.96,0.99];
9 %interpolate ys(x)
10 [n,coeff] = integrate_interpolation_tabulated(x,ys);
11 %integrate lower
12 I1 = integrate_gauss(xD,yF,5,x,coeff);
13 I2 = integrate_gauss(yF,yl,5,x,coeff);
14 I = I1 + I2;
15 fprintf('NTU = %6.2f \n',I)
This function calls a modified version of Program 8.8 for Gauss quadrature that also passes the interpolation parameters, which are required by our function getf.m
1 function I = integrate_gauss(xmin,xmax,n,xpts,coeff)
2 % xmin = lower bound for integral
3 % xmax = upper bound for integral
4 % n = order of Gauss quadrature
5 % xpts, coeff = parameters needed for getf
6
7 switch n
8 case 2
9 w = [1, 1];
10 x=[-1/sqrt(3), 1/sqrt(3)];
11 case 3
12 w = [5/9, 8/9, 5/9];
13 x=[-sqrt(3/5), 0, sqrt(3/5)];
14 case 4
15 a = sqrt(30)/36;
16 b = (2/7)*sqrt(6/5);
17 c = sqrt(3/7 - b);
18 d = sqrt(3/7 + b);
19 w = [0.5 + a, 0.5 + a, 0.5 - a, 0.5 - a];
20 x = [c, -c, d, -d];
21 case 5
22 a = 322/900;
23 b = 13*sqrt(70)/900;
24 c=2*sqrt(10/7);
25 d = sqrt(5 - c)/3;
26 e = sqrt(5 + c)/3;
27 w = [128/225, a + b, a + b, a - b, a - b];
28 x = [0, d, -d, e, -e];
29 otherwise
30 fprintf('Value of n is not valid.\n')
31 x = 0;
32 w = 0;
33 end
34
35 for i = 1:n
36 y(i) = ((xmax + xmin) + (xmax - xmin)*x(i))/2;
37 end
38
39 I = 0;
40 for i = 1:n
41 I = I + w(i)*getf(y(i),xpts,coeff);
42 end
43 I = (xmax-xmin)*I/2;
The result of this program is NTU = 8.42. If you used Professor Cussler’s book for mass transfer, you may recall that he got an answer of NTU = 13.9 for the same distillation specifications. These are substantially different answers, since the column required from Professor Cussler’s calculation is 65% larger than our result. The difference lies in the thermodynamic data. The data you are provided with here are quite similar to those used by Professor Cussler in his book, but we have added on 5% random noise to each of his data points. If we run the same program but use Professor Cussler’s data points, we indeed get the same answer, NTU = 13.9.
The sensitivity of the height of the distillation column to small fluctuations in the data is worth keeping in mind in other courses. It would seem that 5% uncertainty in the data is relatively small, but it is amplified in the final result via the interpolation and then the integration. In many instances, you may have an estimate for the error in the data. One very nice feature of understanding how numerical integration works (and, indeed, all of the other numerical methods in this book) is that you can readily do propagation of error through the numerical method to figure out exactly how the uncertainty in the data leads to uncertainty in the NTU. Since you also know something about the inherent error in the numerical methods, you can also make an informed conclusion about which factor is more important – error in your data or error in your calculation.