PolymathPlus Report
📈   Nonlinear Equation 2022-04-01 12:32 

# Example 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


Calculated values

    Variable Value f(x) Initial Guess Initial f(x)
1 Tbp 95.585087 6.1E-08 90 -1.4E02

    Variable Value Initial Value
1 PA 1196.2189 1020.9913
2 PB 485.67158 406.73785
3 xA 0.6 0.6
4 xB 0.4 0.4
5 yA 0.78698615 0.67170482
6 yB 0.21301385 0.17839379

Root Finding Plot

 -500 0 500 1000 40 60 80 100 120 140 Tbp Zero Line f(Tbp)



Formatted equations

  f T b p ( ) = + x A P A x B P B 7 6 0 1 . 2 = 0
  x A = 0 . 6
  P A = 1 0 ( ) ^ 6 . 9 0 5 6 5 1 2 1 1 . 0 3 3 + T b p 2 2 0 . 7 9
  P B = 1 0 ( ) ^ 6 . 9 5 4 6 4 1 3 4 4 . 8 + 2 1 9 . 4 8 2 T b p
  x B = 1 x A
  y A = x A P A 7 6 0 1 . 2
  y B = x B P B 7 6 0 1 . 2
  6 0 < T b p < 1 2 0

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)

General Settings

Total number of equations 7
Number of implicit equations 1
Number of explicit equations 6
Elapsed time 0.00 sec
Reporting digits 10
Solution method safenewt
Max iterations 150
Tolerance F 1E-07
Tolerance X 1E-07
Tolerance min 1E-07