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Week 3 - Adiabatic Flame Temperature calculation

KKaustubh Awachat5 Jun 2021
CANTERACOMBUSTION

Objectives

To calculate adiabatic flame temperature,

  1. Adiabatic flame temperature variation with Equivalence ratio (In constant volume-chamber)
  2. Adiabatic flame temperature variation with Heat loss (In constant pressure-chamber)
  3. Adiabatic flame temperature variation with respect to Alkane, Alkene, Alkyne
  4. Adiabatic flame temperature variation with respect to a number of Carbon atom chains

1. Adiabatic flame temperature variation with Equivalence ratio (In constant volume-chamber) :

The adiabatic flame temperature is defined as the temperature attained when all of the chemical reaction heat released heats combustion products.

In the study of combustion, there are two types of adiabatic flame temperature depending on how the process is completed: the constant volume and constant pressure, both of which describe temperature that combustion products theoretically can reach if no energy is lost to the outside environment.

THE CONCEPT OF AIR-FUEL MIXTURE:

STOICHIOMETRIC COMBUSTION:

The air-fuel ratio for complete combustion is called the stoichiometric air-fuel ratio.

LEAN COMBUSTION:

If the amount of air is more than fuel for complete combustion. This is called lean combustion.

RICH MIXTURE:

If the amount of fuel is more than air for complete combustion. This is called rich combustion.

EQUIVALENCE RATIO:

The fuel-air equivalence ratio (phi) of a system is defined as the ratio of the fuel to oxidizer ratio to the stoichiometric fuel to oxidizer ratio.

`phi = ((m_(fuel)/(m_air))_(actual))/((m_(fuel)/(m_(air )))_(stio))`

Phi < 1, lean combustion

Phi = 1, stoichiometric combustion.

Phi > 1, rich combustion.

GENERAL EQUATION FOR ANY HYDROCARBON COMBUSTION:

CxHy + [x + (y/4)](O2 + 3.76N2) => xCO2 + (y/2)H2O + 3.76[x+(y/4)]N2

CHEMICAL REACTION FOR COMBUSTION OF METHANE:

STOICHIOMETRIC COMBUSTION:

CH4 + 2(O2 + 3.76N2) = CO2 + 2H2O + 7.52N2

LEAN COMBUSTION;

CH4 + (2/phi)(O2 + 3.76N2) = CO2 + 2H2O + (2/phi-2)O2 + (7.52/phi)N2

RICH COMBUSTION:

CH4 + (2/phi)(O2 + 3.76N2) = (4/phi-3)CO2 + 2H2O + (4-4/phi)CO + (7.52/phi)N2

CASE 1: EFFECT OF EQUIVALENCE RATIO ON AFT FOR COMBUSTION OF METHANE UNDER CONSTANT VOLUME COMBUSTION PROCESS:

For a constant volume process, the internal energy is equated for reactants and products to get AFT,

U_r – U_p = 0

So using the newton Raphson method the final temperature is obtained.

(Hproducts-n_products*R*T) – (Hreactants-n_reactants*R*Tstd)

CODING APPROACH:

USING NR METHOD:

  • Enthalpy of reactants and products were calculated from the NASA polynomial.

H = ((a1 + a2*T/2 + a3*pow(T,2)/3 + a4*pow(T,3)/4 + a5*pow(T,4)/5 + a6/T)*R*T)

The coefficients are obtained from the Berkeley website.

  • For the constant volume process, the internal energy is equated to get AFT.
  • Using the Newton Raphson method the final temperature is obtained.
  • The equivalence ratio was varied from 0.1 to 2.

CODE USING CANTERA:

  • The solution class of Cantera is used.
  • The gas object is created from the solution class.
  • The gas species were obtained from gri30.xml.
  • The equilibrate solver was used with constant volume and internal energy to obtain AFT.

Python code

# CH4 + 2(O2 + 3.76*N2) = CO2 + 2H2O + 7.5N2
import numpy as np
import matplotlib.pyplot as plt 
import math
def h(T,co_effs):
	#function to evaluate enthalpies
	R = 8.314
	a1 = co_effs[0]
	a2 = co_effs[1]
	a3 = co_effs[2]
	a4 = co_effs[3]
	a5 = co_effs[4]
	a6 = co_effs[5]
	return (a1 + a2*T/2 + a3*pow(T,2)/3 + a4*pow(T,3)/4 + a5*pow(T,4)/5 + a6/T)*R*T

CH4_coeffs_l = [5.14987613E+00, -1.36709788E-02, 4.91800599E-05, -4.84743026E-08, 1.66693956E-11, -1.02466476E+04, -4.64130376E+00]
O2_coeffs_l = [3.78245636E+00, -2.99673416E-03, 9.84730201E-06, -9.68129509E-09, 3.24372837E-12, -1.06394356E+03, 3.65767573E+00]
N2_coeffs_l = [0.03298677E+02, 0.14082404E-02, -0.03963222E-04, 0.05641515E-07, -0.02444854E-10, -0.10208999E+04, 0.03950372E+02]
N2_coeffs_h = [0.02926640E+02, 0.14879768E-02, -0.05684760E-05, 0.10097038E-09, -0.06753351E-13, -0.09227977E+04, 0.05980528E+02]
H2O_coeffs_h = [3.03399249E+00, 2.17691804E-03, -1.64072518E-07, -9.70419870E-11, 1.68200992E-14, -3.00042971E+04, 4.96677010E+00]
CO2_coeffs_h = [3.85746029E+00, 4.41437026E-03, -2.21481404E-06, 5.23490188E-10, -4.72084164E-14, -4.87591660E+04, 2.27163806E+00]
Heat_loss=0.35
def f(T):
	# function represents root finding problem
	Tstp =298.15 
	LHV = 800950 
	h_CO2_p = h(T,CO2_coeffs_h)
	h_N2_p = h(T,N2_coeffs_h)
	h_H2O_p = h(T,H2O_coeffs_h)
	H_products = h_CO2_p + 2*h_H2O_p + 7.52*h_N2_p
	print (H_products)
	#for reactants
	Tstd = 298.15
	h_CH4_r = h(Tstd,CH4_coeffs_l)
	h_O2_r = h(Tstd,O2_coeffs_l)
	h_N2_r = h(Tstd,N2_coeffs_l)
	H_reactants = h_CH4_r + 2*h_O2_r + 7.52*h_N2_r
	print (H_reactants) 
	Loss = Heat_loss* LHV
	return H_products - H_reactants+ Loss

def fprime(T):
	return (f(T + 1e-6) - f(T))/ 1e-6
T_guess = 1500
tol = 1e-3
ct = 0
alpha = 0.2
while (abs(f(T_guess)) > tol):
	T_guess = T_guess - alpha * ((f(T_guess))/(fprime(T_guess)))
	ct = ct+1
	plt.plot (ct,T_guess, '-*', color='yellow')
print(T_guess)
plt.xlabel('number of iterations')
plt.ylabel('Adiabatic flame temperature for 0.35 heatloss')
print(ct)
plt.show()

From the graph, we can see that at stoichiometric condition (phi =1) the AFT is high and as we move away from the stoichiometric condition the AFT starts decreasing because at stoichiometric condition all the fuel gets burned and produces more energy.

We can see that the Cantera result shows lower AFT than the newton Raphson method because Cantera uses all the products of the combustion so more energy is used to produce those products.

Hence lower AFT is obtained through Cantera which is the actual solution.

2. Adiabatic flame temperature variation with Heat loss (In constant pressure-chamber) :

For a constant pressure, process enthalpy is equated for reactants and products to get AFT,

H_r – H_p + Heat_loss = 0

the maximum possible heat loss is the lower heat value(LHV)

Heat_loss = Heat_loss factor * (LHV)

CODING APPROACH:

USING NR METHOD:

  • Enthalpy of reactants and products were calculated from the NASA polynomial.

H = ((a1 + a2*T/2 + a3*pow(T,2)/3 + a4*pow(T,3)/4 + a5*pow(T,4)/5 + a6/T)*R*T)

The coefficients are obtained from the Berkeley website.

  • For the constant pressure, process enthalpy is equated to get AFT.
  • Using the Newton Raphson method the final temperature is obtained.
  • The heat loss was varied from 0 to 1.

Python code

import numpy as np

import matplotlib.pyplot as plt


# Determining AFT using Newton Raphson Method

def h(T,Coeff):
	
	R= 8.314 
	a1=Coeff[0]
	a2=Coeff[1]
	a3=Coeff[2]
	a4=Coeff[3]
	a5=Coeff[4]
	a6=Coeff[5]
	a7=Coeff[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

# Defining the coefficients of the Elements 

ch4_coeffs_l = [5.14987613E+00,-1.36709788E-02, 4.91800599E-05,-4.84743026E-08, 1.66693956E-11,-1.02466476E+04,-4.64130376E+00]
	
o2_coeffs_l  = [3.78245636E+00,-2.99673416E-03, 9.84730201E-06,-9.68129509E-09, 3.24372837E-12,-1.06394356E+03, 3.65767573E+00]

o2_coeffs_h  = [3.28253784E+00,1.48308754E-03, -7.57966669E-07,2.09470555E-10, -2.16717794E-14,-1.08845772E+03, 5.45323129E+00]
	
n2_coeffs_l  = [0.03298677E+02, 0.14082404E-02,-0.03963222E-04, 0.05641515E-07,-0.02444854E-10,-0.10208999E+04, 0.03950372E+02]
	
n2_coeffs_h  = [0.02926640E+02, 0.14879768E-02,-0.05684760E-05, 0.10097038E-09,-0.06753351E-13,-0.09227977E+04, 0.05980528E+02]
	
co2_coeffs_h = [3.85746029E+00, 4.41437026E-03,-2.21481404E-06, 5.23490188E-10,-4.72084164E-14,-4.87591660E+04, 2.27163806E+00]    
	
h2o_coeffs_h = [3.03399249E+00, 2.17691804E-03,-1.64072518E-07,-9.70419870E-11, 1.68200992E-14,-3.00042971E+04, 4.96677010E+00]

co_coeffs_h =  [2.71518561E+00, 2.06252743E-03,-9.98825771E-07, 2.30053008E-10,-2.03647716E-14,-1.41518724E+04, 7.81868772E+00]


def f(T,psi):


	Tstp =298.15 #k
	R= 8.314 #Jol/ mol-k

	
	h_ch4_r = h(Tstp,ch4_coeffs_l)
	h_o2_r= h(Tstp,o2_coeffs_l)
	h_n2_r = h(Tstp,n2_coeffs_l)

	h_co2_p = h(T,co2_coeffs_h)
	h_h2o_p = h(T,h2o_coeffs_h)
	h_o2_p = h(T,o2_coeffs_h)
	h_n2_p = h(T,n2_coeffs_h)
	h_co_p = h(T,co_coeffs_h)

	if psi>1:  # Lean mixture
		h_reactants = h_ch4_r+ 2*psi*h_o2_r + 7.52*psi*h_n2_r
		h_products= h_co2_p+ 2*h_h2o_p +(2*psi-2)*h_o2_p + 7.52*psi*h_n2_p
		N_reactants= 1+ 2*psi+ 7.52*psi
		N_products=1+2 +(2*psi-2)+ 7.52*psi

	else:      # Rich Mixture 
		h_reactants = h_ch4_r+ 2*psi*h_o2_r + 7.52*psi*h_n2_r
		h_products= (4*psi -3)*h_co2_p+ 2*h_h2o_p +(4-4*psi)*h_co_p + 7.52*psi*h_n2_p
		N_reactants= 1+ 2*psi+ 7.52*psi
		N_products=(4*psi -3)+ 2+(4-4*psi)+ 7.52*psi

	return h_products - h_reactants - R*(N_products*T -N_reactants*Tstp)

def fprime(T, psi):
	return (f(T+ 1e-6, psi)- f(T,psi))/1e-6

T_guess=1500
tol = 1e-3
ct = 0
alpha = 0.2

psi=np.linspace(0.6,1.6,11)

T_aft=[]

for i in range (0,len(psi)):
	while (abs(f(T_guess,psi[i]))>tol):
		T_guess = T_guess - alpha*((f(T_guess,psi[i])/fprime(T_guess, psi[i])))
		ct=ct+1
	T_aft.append(T_guess)


# Determining AFT using Cantera
import cantera as ct 
gas = ct.Solution('gri30.xml')

T_aft_cantera=[]

for i in range (0, len(psi)):
	Total_mole = 1 + 2*psi[i] +7.52*psi[i]
	mole_fraction_ch4= 1/Total_mole
	mole_fraction_o2= (2*psi[i])/Total_mole
	mole_fraction_n2= (7.52*psi[i])/Total_mole

	gas.TPX= 298.15,101325,{'CH4':mole_fraction_ch4,'O2':mole_fraction_o2,'N2':mole_fraction_n2}
	gas.equilibrate('UV','auto')
	T_aft_cantera.append(gas.T) 

Equivalence_ratio = 1/psi
plt.plot(Equivalence_ratio,T_aft,color = 'blue')
plt.plot(Equivalence_ratio,T_aft_cantera,color = 'red')
plt.xlabel('Equivalence Ratio')
plt.ylabel('Adiabatic Flame Temperature(K)')
plt.grid()
plt.title('Effect of Equivalence ratio on the AFT')
plt.legend(['Python','Cantera'])
plt.show()

From the graph, we can infer that as the heat loss increases the AFT keeps decreasing because less energy is contained in the chamber as it decreases the temperature.

3. Adiabatic flame temperature variation with respect to Alkane, Alkene, Alkyne :

The same approach is carried out and tested for alkane, alkyne, and alkene for 2 carbon atom chains. The hydrocarbons are ethane, ethene and ethyne

Python Code

import matplotlib.pyplot as plt
import numpy as np
import cantera as ct


def h(T,Coeff):

	
	R= 8.314 #Jol/ mol-k
	a1=Coeff[0]
	a2=Coeff[1]
	a3=Coeff[2]
	a4=Coeff[3]
	a5=Coeff[4]
	a6=Coeff[5]
	a7=Coeff[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

# Defining the coefficients of the Elements taken from Nasa thermodynamic data

c2h6_coeffs_l =[4.29142492E+00,-5.50154270E-03  , 5.99438288E-05 ,-7.08466285E-08 ,2.68685771E-11,-1.15222055E+04 ,2.66682316E+00]

c2h4_coeffs_l =[3.95920148E+00,-7.57052247E-03 , 5.70990292E-05 ,-6.91588753E-08 ,2.69884373E-11 ,5.08977593E+03 ,4.09733096E+00]

c2h2_coeffs_l =[8.08681094E-01 ,2.33615629E-02,-3.55171815E-05 ,2.80152437E-08,-8.50072974E-12 ,2.64289807E+04 ,1.39397051E+01]

n2_coeffs_l  = [0.03298677E+02, 0.14082404E-02,-0.03963222E-04, 0.05641515E-07,-0.02444854E-10,-0.10208999E+04, 0.03950372E+02]
	
n2_coeffs_h  = [0.02926640E+02, 0.14879768E-02,-0.05684760E-05, 0.10097038E-09,-0.06753351E-13,-0.09227977E+04, 0.05980528E+02]
	
o2_coeffs_l  = [3.78245636E+00,-2.99673416E-03, 9.84730201E-06,-9.68129509E-09, 3.24372837E-12,-1.06394356E+03, 3.65767573E+00]

co2_coeffs_h = [3.85746029E+00, 4.41437026E-03,-2.21481404E-06, 5.23490188E-10,-4.72084164E-14,-4.87591660E+04, 2.27163806E+00]    
	
h2o_coeffs_h = [3.03399249E+00, 2.17691804E-03,-1.64072518E-07,-9.70419870E-11, 1.68200992E-14,-3.00042971E+04, 4.96677010E+00]



def f_Alkane(T):

	
	#C2H6 + 3.5O2 +3.5 *3.76 N2 = 2CO2 + 3H2O + 3.5 *3.76 N2 + 0.35* LHV



	Tstp =298.15 #k
	Heat_loss = 0.35
	LHV = 1428275 #J/mole
	#LHV=0
	
	h_c2h6_r= h(Tstp,c2h6_coeffs_l)
	h_o2_r= h(Tstp,o2_coeffs_l)
	h_n2_r = h(Tstp,n2_coeffs_l)
	h_n2_p = h(T,n2_coeffs_h)
	h_co2_p = h(T,co2_coeffs_h)
	h_h2o_p = h(T,h2o_coeffs_h)
	
	

	h_reactants = h_c2h6_r+ 3.5*h_o2_r +(3.5*3.76*h_n2_r)
	h_products= 2*h_co2_p+ 3*h_h2o_p +(3.5*3.76*h_n2_p)

	Loss = Heat_loss*LHV

	return h_products - h_reactants + Loss 

def fprime_Alkane(T):
	return (f_Alkane(T+ 1e-6)- f_Alkane(T))/1e-6

def f_Alkene(T):

	#C2H4 + 3O2 + 3 *3.76 N2 = 2CO2 + 2H2O + 3 *3.76 N2 + 0.35* LHV 

	Tstp =298.15 
	Heat_loss = 0.35
	LHV = 1323474 
	#LHV=0
	
	h_c2h4_r= h(Tstp,c2h4_coeffs_l)
	h_o2_r= h(Tstp,o2_coeffs_l)
	h_n2_r = h(Tstp,n2_coeffs_l)
	h_n2_p = h(T,n2_coeffs_h)
	h_co2_p = h(T,co2_coeffs_h)
	h_h2o_p = h(T,h2o_coeffs_h)
	

	h_reactants = h_c2h4_r+ 3*h_o2_r +(3*3.76*h_n2_r)
	h_products= 2*h_co2_p+ 2*h_h2o_p +(3*3.76*h_n2_p)

	Loss = Heat_loss*LHV

	return h_products - h_reactants + Loss 

def fprime_Alkene(T):
	return (f_Alkene(T+ 1e-6)- f_Alkene(T))/1e-6

def f_Alkyne(T):

	
	#C2H2 + 2.5O2 + 2.5 *3.76 N2  = 2CO2 + H2O + 3 *3.76 N2 + 0.35* LHV 

	Tstp =298.15 #k
	Heat_loss = 0.35
	LHV = 1258020 #J/mole
	#LHV=0
	
	h_c2h2_r= h(Tstp,c2h2_coeffs_l)
	h_o2_r= h(Tstp,o2_coeffs_l)
	h_n2_r = h(Tstp,n2_coeffs_l)
	h_n2_p = h(T,n2_coeffs_h)	
	h_co2_p = h(T,co2_coeffs_h)
	h_h2o_p = h(T,h2o_coeffs_h)
	
	h_reactants = h_c2h2_r+ 2.5*h_o2_r +(2.5*3.76*h_n2_r)
	h_products= 2*h_co2_p+ h_h2o_p +(2.5*3.76*h_n2_p)

	Loss = Heat_loss*LHV

	return h_products - h_reactants + Loss 

def fprime_Alkyne(T):
	return (f_Alkyne(T+ 1e-6)- f_Alkyne(T))/1e-6


T_Alkane=1500
T_Alkene=1500
T_Alkyne=1500
tol = 1e-3
alpha = 1


T_aft=[]

while (abs(f_Alkane(T_Alkane))>tol):
	T_Alkane = T_Alkane - alpha*(f_Alkane(T_Alkane)/fprime_Alkane(T_Alkane))
	

T_aft.append(T_Alkane)

while (abs(f_Alkene(T_Alkene))>tol):
	T_Alkene = T_Alkene - alpha*(f_Alkene(T_Alkene)/fprime_Alkene(T_Alkene))
	
T_aft.append(T_Alkene)

while (abs(f_Alkyne(T_Alkyne))>tol):
	T_Alkyne = T_Alkyne - alpha*(f_Alkyne(T_Alkyne)/fprime_Alkyne(T_Alkyne))

T_aft.append(T_Alkyne)

fuel=['Ethane','Ethene','Ethyne']
print(T_aft)
plt.bar(fuel,T_aft)
plt.xlabel('Type of Hydrocarbon')
plt.ylabel(' Flame Temperature(K)')
plt.title('Estimation of FT for a Heat Loss Factor of 0.35')
plt.show()

From the bar graph, we can obtain that the highest final temperature is obtained for ethyne and the lowest for ethane. A high amount of energy is obtained in the case of ethyne which is due to the triple C bond which requires more energy to break.

4. Adiabatic flame temperature variation with respect to a number of Carbon atom chains :

The carbon atoms are varied from 1 to 3 for the alkane group. The hydrocarbons are methane, ethane, and propane.

import matplotlib.pyplot as plt
import numpy as np
import cantera as ct

def h(T,Coeff):
	
	R= 8.314 
	a1=Coeff[0]
	a2=Coeff[1]
	a3=Coeff[2]
	a4=Coeff[3]
	a5=Coeff[4]
	a6=Coeff[5]
	a7=Coeff[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

ch4_coeffs_l = [5.14987613E+00,-1.36709788E-02, 4.91800599E-05,-4.84743026E-08, 1.66693956E-11,-1.02466476E+04,-4.64130376E+00]

c2h6_coeffs_l =[4.29142492E+00,-5.50154270E-03, 5.99438288E-05,-7.08466285E-08 ,2.68685771E-11,-1.15222055E+04 ,2.66682316E+00]

c3h8_coeffs_l =[0.93355381E+00, 0.26424579E-01,0.61059727E-05,-0.21977499E-07 ,0.95149253E-11 ,-0.13958520E+05 ,0.19201691E+02]

n2_coeffs_l  = [0.03298677E+02, 0.14082404E-02,-0.03963222E-04, 0.05641515E-07,-0.02444854E-10,-0.10208999E+04, 0.03950372E+02]
	
n2_coeffs_h  = [0.02926640E+02, 0.14879768E-02,-0.05684760E-05, 0.10097038E-09,-0.06753351E-13,-0.09227977E+04, 0.05980528E+02]
	
o2_coeffs_l  = [3.78245636E+00,-2.99673416E-03, 9.84730201E-06,-9.68129509E-09, 3.24372837E-12,-1.06394356E+03, 3.65767573E+00]

co2_coeffs_h = [3.85746029E+00, 4.41437026E-03,-2.21481404E-06, 5.23490188E-10,-4.72084164E-14,-4.87591660E+04, 2.27163806E+00]    
	
h2o_coeffs_h = [3.03399249E+00, 2.17691804E-03,-1.64072518E-07,-9.70419870E-11, 1.68200992E-14,-3.00042971E+04, 4.96677010E+00]

# Methane_reaction :CH4 + 2O2 + 2*3.76 N2 = CO2 + 2H2O + 2*3.76 N2 

def f_Methane(T):

	Tstp =298.15 
	
	h_ch4_r= h(Tstp,ch4_coeffs_l)
	h_o2_r= h(Tstp,o2_coeffs_l)
	h_n2_r = h(Tstp,n2_coeffs_l)
	h_n2_p = h(T,n2_coeffs_h)
	h_co2_p = h(T,co2_coeffs_h)
	h_h2o_p = h(T,h2o_coeffs_h)
	
	h_reactants = h_ch4_r+ 2*h_o2_r +(2*3.76*h_n2_r)
	h_products = h_co2_p+ 2*h_h2o_p +(2*3.76*h_n2_p)


	return h_products - h_reactants  

def fprime_Methane(T):
	return (f_Methane(T+ 1e-6)- f_Methane(T))/1e-6

#Ethane_reaction: C2H6 + 3.5O2 +3.5 *3.76 N2 = 2CO2 + 3H2O + 3.5 *3.76 N2 

def f_Ethane(T):

	Tstp =298.15 
	h_c2h6_r= h(Tstp,c2h6_coeffs_l)
	h_o2_r= h(Tstp,o2_coeffs_l)
	h_n2_r = h(Tstp,n2_coeffs_l)
	h_n2_p = h(T,n2_coeffs_h)
	h_co2_p = h(T,co2_coeffs_h)
	h_h2o_p = h(T,h2o_coeffs_h)
	h_reactants = h_c2h6_r+ 3.5*h_o2_r +(3.5*3.76*h_n2_r)
	h_products= 2*h_co2_p+ 3*h_h2o_p +(3.5*3.76*h_n2_p)
	return h_products - h_reactants 

def fprime_Ethane(T):
	return (f_Ethane(T+ 1e-6)- f_Ethane(T))/1e-6



#propane: C3H8 + 5O2 + 5*3.76 N2 = 3CO2 + 4H2O + 5 *3.76 N2 



def f_Propane(T):

	Tstp =298.15 
	
	h_c3h8_r= h(Tstp,c3h8_coeffs_l)
	h_o2_r= h(Tstp,o2_coeffs_l)
	h_n2_r = h(Tstp,n2_coeffs_l)
	h_n2_p = h(T,n2_coeffs_h)	
	h_co2_p = h(T,co2_coeffs_h)
	h_h2o_p = h(T,h2o_coeffs_h)
	
	h_reactants = h_c3h8_r+ 5*h_o2_r +(5*3.76*h_n2_r)
	h_products= 3*h_co2_p+ 4*h_h2o_p +(5*3.76*h_n2_p)

	return h_products - h_reactants 

def fprime_Propane(T):
	return (f_Propane(T+ 1e-6)- f_Propane(T))/1e-6

T_Methane=1500
T_Ethane=1500
T_Propane=1500
tol = 1e-3
alpha = 1


T_aft=[]

while (abs(f_Methane(T_Methane))>tol):
	T_Methane = T_Methane - alpha*(f_Methane(T_Methane)/fprime_Methane(T_Methane))
	

T_aft.append(T_Methane)

while (abs(f_Ethane(T_Ethane))>tol):
	T_Ethane = T_Ethane - alpha*(f_Ethane(T_Ethane)/fprime_Ethane(T_Ethane))
	
T_aft.append(T_Ethane)

while (abs(f_Propane(T_Propane))>tol):
	T_Propane = T_Propane - alpha*(f_Propane(T_Propane)/fprime_Propane(T_Propane))

T_aft.append(T_Propane)

fuel=['Methane','Ethane','Propane']
print(T_aft)
plt.bar(fuel,T_aft)
plt.xlabel('Type of Fuel')
plt.ylabel('Adiabatic Flame Temperature(K)')
plt.title('Effect of alkane carbon atoms on the AFT ')
plt.show()

From the graph, we can conclude that propane has the highest AFT whereas methane has a lower AFT.

The AFT increases with increases in no of the carbon atom and hydrogen atom for the alkane group of hydro carbons.

Conclusion

More nos. of carbon atoms means it will produce more heat, hence fuel with 3 carbon atoms gives us slightly higher AFT than that of fuel with lower carbon atoms of the alkane family. In this way, I have shown through this study that what is AFT, how this AFT will be impacted with changing equivalence ratio, type of fuel, nos. of carbon atoms of alkane, and also I have demonstrated the effect of heat loss coefficient on AFT of constant pressure chamber where combustion is being done by using METHANE.

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