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Adiabatic Flame Temperature calculator using Python

1. Introduction Adiabatic Flame Temperature is the final temperature of products when one mole of a particular fuel is burnt using the exact stoichiometric amount of air in an insulated box. This is the temperature obtained when the heat of combustion is used to heat up the products, and no heat is lost to the environment.…

Abdul Rehman Sadiq K
updated on 15 Feb 2020

1. Introduction

Adiabatic Flame Temperature is the final temperature of products when one mole of a particular fuel is burnt using the exact stoichiometric amount of air in an insulated box. This is the temperature obtained when the heat of combustion is used to heat up the products, and no heat is lost to the environment.

E.g., for methane, combustion using the stoichiometric amount of air is given below.

$C H_{4} + 2 (O_{2} + 3.76 N_{2}) \to C O_{2} + 2 H_{2} O + 7.52 N_{2}$ $CH_4 + 2(O_2 + 3.76N_2)-> CO_2 + 2H_2O + 7.52N_2$

AFT for methane would be the temperature of the products (1 mole of CO2, 2 moles of water vapor and 7.52 moles of nitrogen).

2. How to calculate AFT?

Since the heat loss is considered to be zero, the total enthalpy of reactants must be equal to the enthalpy of products. We can calculate the enthalpy for each participating species using NASA polynomials. NASA polynomial to calculate the enthalpy is given below.

$h = R T (a 1 + a 2 \cdot \frac{T}{2} + a 3 \cdot \frac{T^{2}}{3} + a 4 \cdot \frac{T^{3}}{4} + a 5 \cdot \frac{T^{4}}{5} + \frac{a 6}{T})$ $h = RT(a1 + a2*T/2 + a3*T^2/3 + a4*T^3/4 + a5*T^4/5 + (a6)/T)$

Since enthalpies of reactants and products are equal, we define a function $f (T) = h_{r e a c \tan t} - h_{p r o d u c t} = 0$ $f(T) = h_(reactant)-h_(pro duct) =0$

AFT is then calculated using the Newton Raphson method as shown below.

$T_{n e w} = T + α (\frac{f (T)}{f' (T)})$ $T_ ( n e w) = T+ alpha(f(T)/(f'(T)))$

where $f' (T)$ $f'(T)$ is the derivative of $f (T)$ $f(T)$ and $α$ $alpha$ is the relaxation factor

3. AFT of Methane

A python program was written to calculate the AFT using the above equation (Please refer Appendix-1 for the code). The AFT for methane was found to be 2327.033K. The convergence tolerance was set to 0.001 and the effect of relaxation factor ( $α$ $alpha$ ) on convergence was studied. The number of iterations for convergence for varying $α$ $alpha$ is compared in Table1. Fig1 compares the variation of AFT as the solution converges. It can be concluded that it is best to use a relaxation factor close to one.

$α$ $alpha$	No of Iterations to converge
0.2	90
0.5	30
0.9	10
1.1	10
1	4
1.5	30

Table1: Effect of relaxation ( $α$ $alpha$ ) on the convergence of AFT.

Fig1: Comparison of convergence for different relaxation factors

4. AFT of Methane from Cantera

Cantera is a library which can be used to solve thermodynamic and chemical kinetics problems. This can be imported to Matlab, Python, etc. For the combustion of methane, the AFT was calculated using Cantera in Python, and it was found to be 2223.965K. This value is much lesser when compared to the value obtained using the python code above, which is 2327.033K. The reason for this difference is, in the python code, the products are assumed to be formed from stoichiometric combustion and enthalpies are equated. This assumption totally excludes the possibility of formation of other products (eg., reactions at high temperatures to produce NOX). But Cantera uses the principle of 'Minimization of Gibbs Free Energy' to find the products of the reaction and their respective mole fractions. This gives the value of AFT close to experimental values.

5. AFT of Methane in a constant pressure reactor with Heat loss.

Previously, methane was assumed to be combusting in an adiabatic container (i.e., no heat loss). To find the temperature of combustion of a mole of methane in a constant pressure reactor with a specified percentage of heat loss, another python code written. Please find the code in Appendix-2. For a heat loss of 35%, the AFT was found to be 1679.389K.

6. Comparison of AFT for Ethane, Ethene, and Ethyne

AFT for ethane, ethene, and ethyne are compared assuming stoichiometric combustion. AFT is calculated with the python code n Appendix-3. The reactions for all three fuels are given below.

$C_{2} H_{6} + 3.5 (O_{2} + 3.76 N_{2}) \to 2 C O_{2} + 3 H_{2} O + 13.16 N_{2}$ $C_2H_6 + 3.5(O_2 + 3.76N_2)-> 2CO_2 + 3H_2O + 13.16N_2$

$C_{2} H_{4} + 3 (O_{2} + 3.76 N_{2}) \to 2 C O_{2} + 2 H_{2} O + 11.28 N_{2}$ $C_2H_4 + 3(O_2 + 3.76N_2)-> 2CO_2 + 2H_2O + 11.28N_2$

$C_{2} H_{2} + 2.5 (O_{2} + 3.76 N_{2}) \to 2 C O_{2} + H_{2} O + 9.4 N_{2}$ $C_2H_2 + 2.5(O_2 + 3.76N_2)-> 2CO_2 + H_2O + 9.4N_2$

The values are tabulated in Table2 and represented in Fugure2. Ethyne has the highest and ethane has the lowest adiabatic flame temperature. This is because ethyne has the lowest number (1) of hydrogen atoms amongst the three, which produces one mole of water vapor. Ethene has two and ethane has three hydrogen atoms which produce two and three moles of water vapor respectively. Since water vapor has a relatively higher specific heat, it absorbs most of the heat and reduces the temperature (AFT).

Fuel	AFT (K)
Ethane	1981.136
Ethene	2104.841
Ethyne	2378.655

Table2: Comparison of AFT for Ethane, Ethene, and Ethyne

Fig2: Comparison of AFT for Ethane, Ethene, and Ethyne

7. Comparison of AFT for alkanes with different Carbon atoms

The effect of the number of carbon atoms on AFT is compared. In this project, only methane, ethane, and propane are studied. The general reaction for alkane which was used is given below. The code for calculating AFT is in Appendix-4

$C_{n} H_{2 n + 2} + (1.5 n + 0.5) (O_{2} + 3.76 N_{2}) \to n C O_{2} + (n + 1) H_{2} O + (1.5 n + 0.5) 3.76 N_{}$ $C_nH_(2n+2) + (1.5n+0.5)(O_2 + 3.76N_2)-> nCO_2 + (n+1)H_2O + (1.5n+0.5)3.76N_$

The values are tabulated in Table3 and compared in Fig3. It can be concluded that the AFT reduces with the increase in the number of Carbon atoms. This is because (n+1) moles of water vapor is produced for an alkane with n carbon atoms. Since water vapor has a higher heat carrying capacity, the temperature (AFT) is reduced.

Fuel	AFT (K)
Methane	2327.033
Ethane	1981.136
Propane	1807.621

Table3: Comparison of AFT for Methane, Ethane, and Propane

Fig3: Comparison of AFT for Methane, Ethane, and Propane

8.Conclusion

Python programs are written to solve for Adiabatic Flame Temperature (AFT) of different fuels using the Newton-Raphson method for the function $h_{r e a c \tan t} - h_{p r o d u c t} = 0$ $h_(reactant)-h_(pro duct) =0$ . The AFT value for methane was compared with that obtained by Cantera, and it was found that Cantera gave a lower value for AFT. This is because the code in this project only considers ideal combustion reaction. But Cantera uses the principle of minimization of Gibbs free energy and includes all possible reactions and products. It gives out the exact number of moles of each product. Cantera gives results close to experimental values and also it is efficient to use this library. Hence using Cantera library is better than scripting an entire program for each condition.

Appendix-1: Code to solve AFT of Methane without heat loss

"""
Program to calculate Adiabatic Flame Temperatre of Methane by Newton-Raphson Method.

CH4 + 2(O2 + 3.76N2) ---> CO2 + 2H20 + 7.52N2

"""

import matplotlib.pyplot as plt
import math

T_std = 298.15 #Temperature (K) at STP
AFT = 1500 #Guess value of AFT (K)
tol = 1e-3 #Convergence tolerance
i = 1 #Iteration count
alpha = 0.2 #Relaxation factor.

def h(T, C):
	"""
	This function returns the enthalpy in joule
	Inputs - T:Temperatre, 	C:NASA Polynomials
	"""
	R = 8.314 #J/Kmol
	a1 = C[0]
	a2 = C[1]
	a3 = C[2]
	a4 = C[3]
	a5 = C[4]
	a6 = C[5]
	a7 = C[6]
	return (a1 + a2*T/2 + a3*pow(T,2)/3 + a4*pow(T,3)/4 + a5*pow(T,4)/5 + a6/T)*R*T

#NASA Polynomials:
ch4_low = [5.14987613E+00,-1.36709788E-02,4.91800599E-05,-4.84743026E-08,1.66693956E-11,-1.02466476E+04,-4.64130376E+00]
o2_low = [3.78245636E+00,-2.99673416E-03,9.84730201E-06,9.68129509E-09,3.24372837E-12,-1.06394356E+03,3.65767573E+00]
n2_low = [0.03298677E+02,0.14082404E-02,-0.03963222E-04,0.05641515E-07,-0.02444854E-10,-0.10208999E+04,0.03950372E+02]
co2_high = [3.85746029E+00,4.41437026E-03,-2.21481404E-06,5.23490188E-10,-4.72084164E-14,-4.87591660E+04,2.27163806E+00]
h20_high = [ 3.03399249E+00,2.17691804E-03,-1.64072518E-07,-9.70419870E-11,1.68200992E-14,-3.00042971E+04,4.96677010E+00]
n2_high = [0.02926640E+02,0.14879768E-02,-0.05684760E-05,0.10097038E-09,-0.06753351E-13,-0.09227977E+04,0.05980528E+02]
co2_low = [2.35677352E+00,8.98459677E-03,-7.12356269E-06,2.45919022E-09,-1.43699548E-13,-4.83719697E+04,9.90105222E+00] 
h20_low = [4.19864056E+00,-2.03643410E-03,6.52040211E-06,-5.48797062E-09,1.77197817E-12,-3.02937267E+04,-8.49032208E-01]

def f(T):
	#This function represents root finding problem 
	H_reactant = h(T_std,ch4_low) + 2*h(T_std,o2_low) + 7.52*h(T_std,n2_low)
	H_product = h(T,co2_high) + 2*h(T,h20_high) + 7.52*h(T,n2_high)
	return (H_product - H_reactant)

def fprime(T): #Numerical derivative of f(T) using FDM
	return(f(T+1e-6)-f(T))/1e-6

while (abs(f(AFT))>tol):
	AFT = AFT - alpha*(f(AFT)/(fprime(AFT)))
	plt.plot(i,AFT,'*',color = 'red')
	i = i+1

print('Solution converged in ' + str(i) + ' iterations')
print('Adiabatic Flame Temperature without heat loss is ' + str(AFT) + 'K')
plt.xlabel('Iteration Count')
plt.ylabel('Adiabatic Flame Temperatre (K)')
plt.title('Relaxation Factor =' + str(alpha))
plt.show()

Appendix-2: Code to solve AFT of Methane with heat loss

"""
Program to calculate AFT of Methane with Heat Loss.

CH4 + 2(O2 + 3.76N2) ---> CO2 + 2H20 + 7.52N2

"""

import matplotlib.pyplot as plt
import math

T_std = 298.15 #Temperature (K) at STP
AFT = 1500 #Guess value of AFT (K)
tol = 1e-3 #Convergence tolerance
i = 1 #Iteration count
j = 1 #Iteration count
alpha = 1.1 #Relaxation factor.
H_loss = 0.35 #Ratio of heat loss

def h(T, C):
	"""
	This function returns the enthalpy in joule
	Inputs - T:Temperatre, 	C:NASA Polynomials
	"""
	R = 8.314 #J/Kmol
	a1 = C[0]
	a2 = C[1]
	a3 = C[2]
	a4 = C[3]
	a5 = C[4]
	a6 = C[5]
	a7 = C[6]
	return (a1 + a2*T/2 + a3*pow(T,2)/3 + a4*pow(T,3)/4 + a5*pow(T,4)/5 + a6/T)*R*T

#NASA Polynomials:
ch4_low = [5.14987613E+00,-1.36709788E-02,4.91800599E-05,-4.84743026E-08,1.66693956E-11,-1.02466476E+04,-4.64130376E+00]
o2_low = [3.78245636E+00,-2.99673416E-03,9.84730201E-06,9.68129509E-09,3.24372837E-12,-1.06394356E+03,3.65767573E+00]
n2_low = [0.03298677E+02,0.14082404E-02,-0.03963222E-04,0.05641515E-07,-0.02444854E-10,-0.10208999E+04,0.03950372E+02]
co2_high = [3.85746029E+00,4.41437026E-03,-2.21481404E-06,5.23490188E-10,-4.72084164E-14,-4.87591660E+04,2.27163806E+00]
h20_high = [ 3.03399249E+00,2.17691804E-03,-1.64072518E-07,-9.70419870E-11,1.68200992E-14,-3.00042971E+04,4.96677010E+00]
n2_high = [0.02926640E+02,0.14879768E-02,-0.05684760E-05,0.10097038E-09,-0.06753351E-13,-0.09227977E+04,0.05980528E+02]
co2_low = [2.35677352E+00,8.98459677E-03,-7.12356269E-06,2.45919022E-09,-1.43699548E-13,-4.83719697E+04,9.90105222E+00] 
h20_low = [4.19864056E+00,-2.03643410E-03,6.52040211E-06,-5.48797062E-09,1.77197817E-12,-3.02937267E+04,-8.49032208E-01]

def f(T):
	#This function represents root finding problem 
	H_reactant = h(T_std,ch4_low) + 2*h(T_std,o2_low) + 7.52*h(T_std,n2_low)
	H_product = h(T,co2_high) + 2*h(T,h20_high) + 7.52*h(T,n2_high)
	return (H_product - H_reactant)

def fprime(T): #Numerical derivative of f(T) using FDM
	return(f(T+1e-6)-f(T))/1e-6

while (abs(f(AFT))>tol):
	AFT = AFT - alpha*(f(AFT)/(fprime(AFT)))
	plt.plot(i,AFT,'*',color = 'red')
	i = i+1

print('Solution for no heat loss converged in ' + str(i) + ' iterations')
print('Adiabatic Flame Temperature without heat loss is ' + str(AFT) + 'K')

H_high = h(AFT,co2_high) + 2*h(AFT,h20_high) + 7.52*h(AFT,n2_high)
H_low = h(T_std,co2_low) + 2*h(T_std,h20_low) + 7.52*h(T_std,n2_low)
Max_H = H_high - H_low
Lost_H = (H_loss*Max_H)


def g(T):
	#This function represents root finding problem with heat loss
	H_reactant = h(T_std,ch4_low) + 2*h(T_std,o2_low) + 7.52*h(T_std,n2_low)
	H_product = h(T,co2_high) + 2*h(T,h20_high) + 7.52*h(T,n2_high)
	return (H_product + Lost_H - H_reactant)

def gprime(T): #Numerical derivative of g(T) using FDM
	return(g(T+1e-6)-g(T))/1e-6

while (abs(g(AFT))>tol):
	AFT = AFT - alpha*(g(AFT)/(gprime(AFT)))
	plt.plot(j,AFT,'*',color = 'blue')
	j = j+1

print('Solution with heat loss converged in ' + str(i) + ' iterations')
print('Adiabatic Flame Temperature with heat loss of ' + str(H_loss) + ' is ' + str(AFT) + 'K')

plt.xlabel('Iteration Count')
plt.ylabel('Adiabatic Flame Temperatre (K)')
plt.title('Relaxation Factor =' + str(alpha))
plt.show()

Appendix-3: Code to solve AFT of Ethane, Ethene, and Ethyne

"""
Program to calculate AFT of Ethane, Ethene and Ethyne.

C2H6 + 3.5(O2 + 3.76N2) ---> 2CO2 + 3H20 + 13.16N2
C2H4 + 3(O2 + 3.76N2) ---> 2CO2 + 2H20 + 11.28N2
C2H2 + 2.5(O2 + 3.76N2) ---> 2CO2 + H20 + 9.4N2
"""

import matplotlib.pyplot as plt
import math

T_std = 298.15 #Temperature (K) at STP
AFT_a = 1500 #Guess value of AFT for Ethane
AFT_e = 1500 #Guess value of AFT for Ethene 
AFT_y = 1500 #Guess value of AFT for Ethyne
tol = 1e-3 #Convergence tolerance
i = 1 #Iteration count
j = 1 #Iteration count
k = 1 #Iteration count
alpha = 1.1 #Relaxation factor.

def h(T, C):
	"""
	This function returns the enthalpy in joule
	Inputs - T:Temperatre, 	C:NASA Polynomials
	"""
	R = 8.314 #J/Kmol
	a1 = C[0]
	a2 = C[1]
	a3 = C[2]
	a4 = C[3]
	a5 = C[4]
	a6 = C[5]
	a7 = C[6]
	return (a1 + a2*T/2 + a3*pow(T,2)/3 + a4*pow(T,3)/4 + a5*pow(T,4)/5 + a6/T)*R*T

#NASA Polynomials:
c2h6_low = [4.29142492E+00,-5.50154270E-03,5.99438288E-05,-7.08466285E-08,2.68685771E-11,-1.15222055E+04,2.66682316E+00]
c2h4_low = [3.95920148E+00,-7.57052247E-03,5.70990292E-05,-6.91588753E-08,2.69884373E-11,5.08977593E+03,4.09733096E+00]
c2h2_low = [8.08681094E-01,2.33615629E-02,-3.55171815E-05,2.80152437E-08,-8.50072974E-12,2.64289807E+04,1.39397051E+01]
o2_low = [3.78245636E+00,-2.99673416E-03,9.84730201E-06,9.68129509E-09,3.24372837E-12,-1.06394356E+03,3.65767573E+00]
n2_low = [0.03298677E+02,0.14082404E-02,-0.03963222E-04,0.05641515E-07,-0.02444854E-10,-0.10208999E+04,0.03950372E+02]
co2_high = [3.85746029E+00,4.41437026E-03,-2.21481404E-06,5.23490188E-10,-4.72084164E-14,-4.87591660E+04,2.27163806E+00]
h20_high = [ 3.03399249E+00,2.17691804E-03,-1.64072518E-07,-9.70419870E-11,1.68200992E-14,-3.00042971E+04,4.96677010E+00]
n2_high = [0.02926640E+02,0.14879768E-02,-0.05684760E-05,0.10097038E-09,-0.06753351E-13,-0.09227977E+04,0.05980528E+02]


def a(T):
	#This function represents root finding problem for Ethane
	c2h6_reactant = h(T_std,c2h6_low) + 3.5*h(T_std,o2_low) + 13.16*h(T_std,n2_low)
	c2h6_product = h(T,co2_high) + 3*h(T,h20_high) + 13.16*h(T,n2_high)
	return (c2h6_reactant-c2h6_product)

def e(T):
	#This function represents root finding problem for Ethene
	c2h4_reactant = h(T_std,c2h4_low) + 3*h(T_std,o2_low) + 11.28*h(T_std,n2_low)
	c2h4_product = h(T,co2_high) + 2*h(T,h20_high) + 11.28*h(T,n2_high)
	return (c2h4_reactant-c2h4_product)

def y(T):
	#This function represents root finding problem for Ethyne
	c2h2_reactant = h(T_std,c2h2_low) + 2.5*h(T_std,o2_low) + 9.4*h(T_std,n2_low)
	c2h2_product = h(T,co2_high) + h(T,h20_high) + 9.4*h(T,n2_high)
	return (c2h2_reactant-c2h2_product)

def aprime(T): #Numerical derivative of a(T) using FDM
	return(a(T+1e-6)-a(T))/1e-6
def eprime(T): #Numerical derivative of e(T) using FDM
	return(e(T+1e-6)-e(T))/1e-6
def yprime(T): #Numerical derivative of y(T) using FDM
	return(y(T+1e-6)-y(T))/1e-6

while (abs(a(AFT_a))>tol):
		AFT_a = AFT_a - alpha*(a(AFT_a)/(aprime(AFT_a)))
		i= i+1

while (abs(e(AFT_e))>tol):
		AFT_e = AFT_e - alpha*(e(AFT_e)/(eprime(AFT_e)))
		j= j+1

while (abs(y(AFT_y))>tol):
	AFT_y = AFT_y - alpha*(y(AFT_y)/(yprime(AFT_y)))
	k= k+1


print('Adiabatic Flame Temperature for Ethane is ' + str(AFT_a) + 'K. Solution converged in ' + str(i) + ' iterations')
print('Adiabatic Flame Temperature for Ethene is ' + str(AFT_e) + 'K. Solution converged in ' + str(j) + ' iterations')
print('Adiabatic Flame Temperature for Ethyne is ' + str(AFT_y) + 'K. Solution converged in ' + str(k) + ' iterations')

AFT =[AFT_a,AFT_e,AFT_y]
Fuel = ['Ethane','Ethene','Ethyne']
plt.bar(Fuel,AFT)
plt.xlabel('Fuel Type')
plt.ylabel('Adiabatic Flame Temperatre (K)')
plt.title('Comparison of AFT for fuels with two Carbon atoms')
plt.show()

Appendix-4: Code to solve AFT of Methane, Ethane, and Propane

"""
Program to calculate Adiabatic Flame Temperatre of an Alkane

CnH2n+2 + 1.5n+0.5(O2 + 3.76N2) ---> n CO2 + (n+1) H20 + (1.5n+0.5)*3.76 N2

"""

import matplotlib.pyplot as plt
import math

T_std = 298.15 #Temperature (K) at STP
AFT1 = 1500 #Guess value of AFT for Methane
AFT2 = 1500 #Guess value of AFT for Ethane
AFT3 = 1500 #Guess value of AFT for Propane
tol = 1e-3 #Convergence tolerance
i = 1 #Iteration count
j = 1 #Iteration count
k = 1 #Iteration count
alpha = 1.1 #Relaxation factor.

def h(T, C):
	"""
	This function returns the enthalpy in joule
	Inputs - T:Temperatre, 	C:NASA Polynomials
	"""
	R = 8.314 #J/Kmol
	a1 = C[0]
	a2 = C[1]
	a3 = C[2]
	a4 = C[3]
	a5 = C[4]
	a6 = C[5]
	a7 = C[6]
	return (a1 + a2*T/2 + a3*pow(T,2)/3 + a4*pow(T,3)/4 + a5*pow(T,4)/5 + a6/T)*R*T

#NASA Polynomials:
ch4_low = [5.14987613E+00,-1.36709788E-02,4.91800599E-05,-4.84743026E-08,1.66693956E-11,-1.02466476E+04,-4.64130376E+00]
c2h6_low = [4.29142492E+00,-5.50154270E-03,5.99438288E-05,-7.08466285E-08,2.68685771E-11,-1.15222055E+04,2.66682316E+00]
c3h8_low = [0.93355381E+00,0.26424579E-01,0.61059727E-05,-0.21977499E-07,0.95149253E-11,-0.13958520E+05,0.19201691E+02]
o2_low = [3.78245636E+00,-2.99673416E-03,9.84730201E-06,9.68129509E-09,3.24372837E-12,-1.06394356E+03,3.65767573E+00]
n2_low = [0.03298677E+02,0.14082404E-02,-0.03963222E-04,0.05641515E-07,-0.02444854E-10,-0.10208999E+04,0.03950372E+02]
co2_high = [3.85746029E+00,4.41437026E-03,-2.21481404E-06,5.23490188E-10,-4.72084164E-14,-4.87591660E+04,2.27163806E+00]
h20_high = [ 3.03399249E+00,2.17691804E-03,-1.64072518E-07,-9.70419870E-11,1.68200992E-14,-3.00042971E+04,4.96677010E+00]
n2_high = [0.02926640E+02,0.14879768E-02,-0.05684760E-05,0.10097038E-09,-0.06753351E-13,-0.09227977E+04,0.05980528E+02]

def f1(T):
	#This function represents root finding problem for Methane
	ch4_reactant = h(T_std,ch4_low) + 2*h(T_std,o2_low) + 7.52*h(T_std,n2_low)
	ch4_product = h(T,co2_high) + 2*h(T,h20_high) + 7.52*h(T,n2_high)
	return (ch4_product - ch4_reactant)

def f2(T):
	#This function represents root finding problem for Ethane
	c2h6_reactant = h(T_std,c2h6_low) + 3.5*h(T_std,o2_low) + 13.16*h(T_std,n2_low)
	c2h6_product = h(T,co2_high) + 3*h(T,h20_high) + 13.16*h(T,n2_high)
	return (c2h6_reactant-c2h6_product)

def f3(T):
	#This function represents root finding problem for Propane
	c3h8_reactant = h(T_std,c3h8_low) + 5*h(T_std,o2_low) + 18.8*h(T_std,n2_low)
	c3h8_product = h(T,co2_high) + 4*h(T,h20_high) + 18.8*h(T,n2_high)
	return (c3h8_product - c3h8_reactant)

def f1prime(T): #Numerical derivative of f1(T) using FDM
	return(f1(T+1e-6)-f1(T))/1e-6
def f2prime(T): #Numerical derivative of f2(T) using FDM
	return(f2(T+1e-6)-f2(T))/1e-6
def f3prime(T): #Numerical derivative of f3(T) using FDM
	return(f3(T+1e-6)-f3(T))/1e-6

while (abs(f1(AFT1))>tol):
	AFT1 = AFT1 - alpha*(f1(AFT1)/(f1prime(AFT1)))
	i = i+1
while (abs(f2(AFT2))>tol):
	AFT2 = AFT2 - alpha*(f2(AFT2)/(f2prime(AFT2)))
	j = j+1
while (abs(f3(AFT3))>tol):
	AFT3 = AFT3 - alpha*(f3(AFT3)/(f3prime(AFT3)))
	k = k+1

print('Adiabatic Flame Temperature for Methane is ' + str(AFT1) + 'K. Solution converged in ' + str(i) + ' iterations')
print('Adiabatic Flame Temperature for Ethane is ' + str(AFT2) + 'K. Solution converged in ' + str(j) + ' iterations')
print('Adiabatic Flame Temperature for Propane is ' + str(AFT3) + 'K. Solution converged in ' + str(k) + ' iterations')

AFT =[AFT1,AFT2,AFT3]
Fuel = ['Methane','Ethane','Propane']
plt.bar(Fuel,AFT)
plt.xlabel('Fuel Type')
plt.ylabel('Adiabatic Flame Temperatre (K)')
plt.title('Comparison of AFT for fuels with two Carbon atoms')
plt.show()