Oneeq16

Name: Oneeq16- Adiabatic Flame Temperature for Natural Gas Combustion
Source: Cutlip, M. B. and Shacham, M, Problem Solving in Chemical
  Engineering with Numerical Methods, Prentice Hall Inc., 1999
Reference/s      
       
Model: 1 implicit equation, indep. variable name: T
  Lower difficulty level
  Constraints: T>0
  Discontinuities: none in the range of interest
  Initial range: Tmin=1000, Tmax=3000
       
Solved by Shacham, M., POLYMATH 5.1, build 19, Dec. 12, 2001
       
Model Eqs. Adiabatic Flame Temperature for Natural Gas Combustion |POLVER05_1

EXCEL FILE  

f(T) = 212798*y*xx+372820*z*xx+H0-Hf #

TEXT FILE 

y = 0.7 #

POLYMATH FILE 

x = 1.7 #
  z=1-y-0.02 #
  CH4=if (x<1) then (y*(1-x)) else (0) #
  C2H6=if (x<1) then (z*(1-x)) else (0) #
  COtwo = if (x<1) then ((y+2*z)*x) else (y+2*z) #
  H2O=if (x<1) then ((2*y+3*z)*x) else (2*y+3*z) #
  N2=0.02+3.76*(2*y+7*z/2)*x #
  alp = 3.381*CH4+2.247*C2H6+6.214*COtwo+7.256*H2O+6.524*N2+6.148*O2 #
  bet = 18.044*CH4+38.201*C2H6+10.396*COtwo+2.298*H2O+1.25*N2+3.102*O2 #
  gam = -4.3*CH4-11.049*C2H6-3.545*COtwo+0.283*H2O-0.001*N2-0.923*O2 #
  H0 = alp*298+bet*0.001*298*298/2+gam*1e-6*298^3/3 #
  Hf = alp*T+bet*0.001*T^2/2+gam*1e-6*T^3/3 #
  O2 = if (x<1) then (0) else ((2*y+7*z/2)*(x-1)) #
  xx = if (x<=1) then (x) else (1) #
  T(min)=1000, T(max)=3000
       
Variable/function values Variable Value f(x)
  T 2000 -4.2930E+04
  y 0.7  
  x 1.7  
  z 0.28  
  CH4 0  
  C2H6 0  
  COtwo 1.26  
  H2O 2.24  
  N2 15.23296  
  O2 1.666  
  alp 133.705479  
  bet 42.455612  
  gam -5.38573096  
  H0 41681.83824  
  Hf 337960.2329  
  xx 1  
       
Root Variable Value f(x)
  V 1779.48406483697 0.0000E+00
  y 0.7  
  x 1.7  
  z 0.28  
  CH4 0  
  C2H6 0  
  COtwo 1.26  
  H2O 2.24  
  N2 15.23296  
  O2 1.666  
  alp 133.705479  
  bet 42.455612  
  gam -5.38573096  
  H0 41681.83824  
  Hf 295030.0382  
  xx 1  
       
Function plot