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