108ff7f5-3c54-4030-b788-44e3b1f06c36 65 Javier Ochoa 4 5 false false 0 0 decimal

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Example I. Bubble Point calculation with several nonlinear solvers in SMath Studio installed on an Arch Linux based distro. Source: Chemical Engineering, July 23, 1984, Nonlinear Models.

U Component xj aj bj cj dj K, 350 K C2H6 0.03 20.178 0.15613 - 3.7218 104 -^* 1.749 107 -^* - 3.5267 C3H8 0.2 16.363 0.12471 - 2.9544 104 -^* 1.767 107 -^* - 1.3299 C4H10 0.37 5.466 - 0.04088 1.1174 104 -^* - 1.283 107 -^* 0.6547 C5H12 0.35 2.131 - 0.01780 0.5787 104 -^* - 0.763 107 -^* 0.2813 C6H14 0.05 3.1465 0.02175 - 0.4070 104 -^* 0.927 108 -^* - 0.1223 67 mat :

U U

U
Component	xj	aj	bj	cj	dj	K, 350 K
C2H6	0.03	20.18	-0.16	0	0	3.53
C3H8	0.2	16.36	-0.12	0	0	1.33
C4H10	0.37	-5.47	0.04	0	0	0.65
C5H12	0.35	-2.13	0.02	0	0	0.28
C6H14	0.05	3.15	-0.02	0	0	0.12
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coef U 26 range 36 range el : 20.178 0.1561 - 0.0004 1.749 107 -^* - 16.363 0.1247 - 0.0003 1.767 107 -^* - 5.466 - 0.0409 0.0001 - 1.283 107 -^* 2.131 - 0.0178 5.787 105 -^* - 7.63 108 -^* 3.1465 0.0217 - 4.07 105 -^* 9.27 109 -^* - 54 mat

x U 26 range 2 el : 0.03 0.2 0.37 0.35 0.05 51 mat

T temp 1 T T 2^T 3^41 mat :

Check the equilibrium coefficients at 350 K, see table above.

T0 350 :

K coef T0 temp * : 3.6257 1.3299 0.6547 0.2813 0.1223 51 mat

Let's find the absolute temperature T in Kelvin from the convergence equation of the article. Chemical Engineering, July 23, 1984 as the example 1. First, the definition of the convergence equation, in order to solve the absolute temperature is the following.

T f coef T temp * x * ln :

King, C . J . Separation Process, McGraw Hill Book Co. NY, 1980.

Using different nonlinear solvers to find the solution.

epsilon 106 -^:

iterations 400 :

res T f T 285450 solve :

res 374.3678

res f 8.8527 1012 -^*

res1 T f T 350 roots :

res1 374.3678

res1 f 2.2204 1016 -^*

res2 T f 285450 epsilon epsilon iterations 110 Secant :

res2 374.3678 411 mat i x1 x2 X3 f(x1) f(x2) f(X3) 0285450 390.9783 1.4475 - 0.8062 0.2007 1450 390.9783 371.417 0.8062 0.2007 0.0374 - 2 390.9783 371.417 374.4884 0.2007 0.0374 - 0.0015 3 371.417 374.4884 374.3687 0.0374 - 0.0015 1.1103 105 -^* 4 374.4884 374.3687 374.3678 0.0015 1.1103 105 -^* 3.3173 109 -^* - 67 mat 31 mat

res2 1 el f 3.3173 109 -^* -

res3 T f 350 epsilon Broyden :

res3 f 9.1786 1010 -^* -

res3 374.3678

Bisected Direct Quadratic Regula Falsi root- finding method.

res4 T f 300 epsilon BDQRF :

res4 f 9.1041 106 -^*

res4 29.6255

res5 T f 350 epsilon NewtonRaphson :

res5 f 9.8144 1010 -^* -

res5 374.3678

res6 T f 350450 epsilon epsilon iterations 110 Secant :

res6 374.3679 311 mat i x1 x2 X3 f(x1) f(x2) f(X3) 0350450 378.8161 0.3263 - 0.8062 0.0553 1450 378.8161 373.5711 0.8062 0.0553 0.01 - 2 378.8161 373.5711 374.3766 0.0553 0.01 - 0.0001 3 373.5711 374.3766 374.3679 0.01 - 0.0001 2.1773 107 -^* 57 mat 31 mat

The article compared the number of iterations at each method, used to find the solution. As you can see with minimun parameters the solvers found excellent results. The NewtonRaphson, Broyden and BDQFR methods could not be solved with the option which shows the iterations matrix