# 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