POLYMATH Report NLE Nonlinear Equation 2013-Dec-1

Calculated values of NLE variables
 Variable Value f(x) Initial Guess Initial f(x) 1 Tbp 95.5851 4.6E-07 90 -1.4E02

 Variable Value Initial Value 1 PA 1196.22 1020.99 2 PB 485.672 406.738 3 xA 0.6 0.6 4 xB 0.4 0.4 5 yA 0.786986 0.671705 6 yB 0.213014 0.178394

Nonlinear equations
 1 f(Tbp) = xA*PA+xB*PB-760*1.2 = 0

Explicit equations
 1 xA = 0.6 2 PA = 10^(6.90565-1211.033/(Tbp+220.79)) 3 PB = 10^(6.95464-1344.8/(219.482+Tbp)) 4 xB = 1-xA 5 yA = xA*PA/(760*1.2) 6 yB = xB*PB/(760*1.2)

Problem source text
 # S. 26(a) - Solution of DAE system # Initial Bubble Point. # Verified Solution: Tbp = 95.5851 # Ref.: Comput. Appl. Eng. Educ. 6: 176, 1998 f(Tbp)=xA*PA+xB*PB-760*1.2 xA=0.6 PA=10^(6.90565-1211.033/(Tbp+220.79)) PB=10^(6.95464-1344.8/(219.482+Tbp)) xB=1-xA yA=xA*PA/(760*1.2) yB=xB*PB/(760*1.2) Tbp(min)=60 Tbp(max)=120

Matlab formatted problem
Create m file called PolyNle.m and paste the following text into it.
 % Polymath NLE problem conversion to Matlab % NLE function PolyNle    xguess = 90 ;    x = fzero(@NLEfun,xguess);    fprintf('The NLE solution is %g\n', x); end function fTbp = NLEfun(Tbp)    xA = 0.6;    PA = 10 ^ (6.90565 - (1211.033 / (Tbp + 220.79)));    PB = 10 ^ (6.95464 - (1344.8 / (219.482 + Tbp)));    xB = 1 - xA;    yA = xA * PA / (760 * 1.2);    yB = xB * PB / (760 * 1.2);    fTbp = xA * PA + xB * PB - (760 * 1.2); end

General Settings
 Total number of equations 7 Number of implicit equations 1 Number of explicit equations 6 Elapsed time 0.01 sec Reporting digits 8 Solution method fastnewt Tolerance X 1E-07 Tolerance F 1E-07 Max trials 150

Data file: no file