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