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Week 3 - Adiabatic Flame Temperature calculation
AIM: To calculate the Adiabatic Flame Temperature. OBJECTIVES: 1) Plot the effect of equivalence ratio on the final adiabatic flame temperature and further compare the results with Cantera. 2) Find out the Adiabatic Flame Temperature of Methane Air Combustion in constant pressure with heat loss. 3) To examine the Adiabatic…
Vishavjeet Singh Yadav
updated on 07 Apr 2021
AIM: To calculate the Adiabatic Flame Temperature.
OBJECTIVES:
1) Plot the effect of equivalence ratio on the final adiabatic flame temperature and further compare the results with Cantera.
2) Find out the Adiabatic Flame Temperature of Methane Air Combustion in constant pressure with heat loss.
3) To examine the Adiabatic Flame Temperature of Alk-ane, Alk-ene, Alk-yne with no heat loss.
4) Find out the Adiabatic flame temperature variation with respect to a number of carbon atom chains.
Adiabatic Flame Temperature:
The chemical process in which a substance reacts rapidly with oxygen and gives off heat called Combustion. The burning substance is called fuel and the source of oxygen is called an oxidizer and fuel can be solid, liquid, or gas. Also, the oxidiser can be solid, liquid, or gas.
''Adiabatic Flame Temperature is referred to as the maximum amount of heat that is generated when an air-fuel mixture is burnt completely without loss of heat''.
Theoretically, it can be calculated by taking out the differences between the enthalpies of the products and the reactant's side. The enthalpies of a substance can be calculated with the help of the polynomial function provided by NASA for a particular temperature.
where h = enthalpy
a(i) = coefficient of each species (http://combustion.berkeley.edu/gri-mech/version30/files30/thermo30.dat)
T = Temperature
R = Gas Constant
Coefficients are classified into high-temperature coefficients and low-temperature coefficients here. The Newton Raphson scheme is used to solve the enthalpy because it is a polynomial function. Using this method we will find out the roots of the function using the initial guess and iterate till we reach the given tolerance value. It is defined as
where T_new: new T value after iteration,
T_guess: guess T value,
alpha is the relaxation factor
CASE1. Adiabatic flame temperature variation with Equivalence ratio (In constant volume-chamber)
The equivalence ratio can be defined as the actual air-fuel ratio to the stoichiometric air-fuel ratio.
If the equivalence ratio is 1 then the combustion is called the stoichiometric combustion else if it is < 1 then the combustion is called the lean combustion where the air is in excess else if the ratio is > 1 the combustion is called the rich combustion where fuel is in excess.
The stoichiometric combustion of the methane can be seen below:
For the case of equivalence ratio the reaction changes to:
Lean Mixture (phi<1)
Rich Mixture (phi>1)
We know that the reactions will be taking place under constant pressure.
Enthalpy balance equation will be as follows:
The relaxation factor is taken as 0.2 and it can be increased up to 0.9 and the behavior change of the solution can be verified. Further, if we increase the relaxation factor above 1 the solution may overshoot and diverge. The program for python can be seen below:
"""
Adiabatic flame temperature at different equivalence ratiaos
ch4+2(o2+3.76n2)=co2+2h2o+7.52n2
"""
import matplotlib.pyplot as plt
import math
import numpy as np
R=8.314 #(J/mol-k)
def h(T, co_effs):
#enthalpy function
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_h = [0.02926640E+02, 0.14879768E-02, -0.05684760E-05, 0.10097038E-09, -0.06753351E-13, -0.09227977E+04, 0.05980528E+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]
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]
o2_coeffs_h = [3.28253784E+00, 1.48308754E-03, -7.57966669E-07, 2.09470555E-10, -2.16717794E-14, -1.08845772E+03, 5.45323129E+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,phi):
# calculating enthalpies of products and reactants
if(phi<=1):
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_o2_p = h(T,o2_coeffs_h)
h_products = h_co2_p + 2*h_h2o_p + (7.52/phi)*h_n2_p + (2/phi-2)*h_o2_p
n_products = 1 + 2 + 3.76*(2/phi) + (2/phi-2)
elif (phi>1):
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_co_p = h(T,co_coeffs_h)
h_products = (4/phi - 3)*h_co2_p + 2*h_h2o_p + (7.52/phi)*h_n2_p + (4 - 4/phi)*h_co_p
n_products = (4/phi - 3) + 2 + (4-4/phi) + (7.52/phi)
print(h_products)
T_std = 298.15
h_ch4_r = h(T_std,ch4_coeffs_l)
h_o2_r = h(T_std,o2_coeffs_l)
h_n2_r = h(T_std,n2_coeffs_l)
h_reactants = h_ch4_r + (2/phi)*h_o2_r + (7.52/phi)*h_n2_r
n_reactants = 1 + (2/phi) + 3.76*(2/phi)
print(h_reactants)
return(h_products-h_reactants)-R*(n_products*T-n_reactants*T_std)
def fprime(T,phi):
return (f(T+1e-6,phi) - f(T,phi))/1e-6
T_guess = 1500
tol = 1e-3
ct=0
alpha=0.2
phi = np.linspace(0.1,1.5,20)
for i in range (0,len(phi)):
while(abs(f(T_guess,phi[i])) > tol):
T_guess = T_guess - alpha*(f(T_guess,phi[i])/fprime(T_guess,phi[i]))
plt.plot(phi[i],T_guess,'*',color='blue')
ct=ct+1
print(T_guess,phi)
print(T_guess)
plt.xlabel('Equivalence Ratio')
plt.ylabel('Adiabatic Flame Temperature')
plt.title('Effect of Equivalnce Ratio on AFT')
plt.grid('on')
plt.show()
We run the code and get this graph successfully. It can be noticed that the temperature increases with an increase in equivalence ratio until 1 and after that, it starts to decrease. So it can be concluded that AFT is maximum at the stoichiometric mixture and gets low as we increase the fuel or air quantity .i.e. make the mixture lean or rich. The maximum observed temperature for the stoichiometric mixture is 2800K.
Cantera:
Cantera is a collection of object-oriented tools for problems involving chemical kinetics, thermodynamics, and transport processes. gas.TPX is an inbuilt function in Cantera and used to initialize the temperature, pressure, and mole fraction for the species. gs.set_equivalence_ratio is used to give the equivalence ratio for the species and gas. equilibrate is used to initialize the flame temperature. We run the same problem in Cantera below:
import matplotlib.pyplot as plt
import matplotlib.pyplot as plt
import cantera as ct
# ch4+2(o2+3.76N2)=co2+2h2o+7.52n2
gas= ct.Solution('gri30.xml')
phi = 0.1
for i in range (0,15):
gas.TPX = 298.15,101325,{'CH4':1, 'O2':2/phi, 'N2':7.52/phi}
gas.set_equivalence_ratio(phi, 'CH4', 'O2:1, N2:3.76')
gas.equilibrate('UV','auto')
print(phi,gas.T)
plt.plot(phi,gas.T,'*',color='blue')
phi = phi+0.1
plt.xlabel('Equivalence Ratio')
plt.ylabel('Adiabatic Flame Temperature')
plt.title('Effect of Equivalence ratio on AFT using Cantera')
plt.grid('on')
plt.show()
It can be showed in the graph that the maximum temperature attained in Cantera is 2600K. The decrease in the temperature can be observed as compared to python, This is due to the process through which the Cantera solves, it uses the Gibbs free energy to reach the equilibrium state.
Other than that many other species are formed in Cantera that are not considered in the Python program for different mixtures.
CASE:2
Heat loss considered for this case in the constant pressure chamber, thus the reaction is not going to be an adiabatic reaction.
The stoichiometric reaction for methane-air combustion can be written as:
The heat loss in the program can be calculated from the Lower heating value LHV. LHV can be obtained from the difference of products and reactants at STP condition
The heat loss factor is varied from no heat loss to entire heat loss as a multiplier of LHV. Further after the heat loss, the enthalpy equation changes to:
(where Loss: heat_loss*LHV)
The program in python can be seen below:
"""
adiabatic flame temperature with heat loss
ch4+2(o2+3.76n2)=co2+2h2o+7.52n2
"""
import matplotlib.pyplot as plt
import math
import numpy as np
R=8.314 #J/mol-k
def h(T, co_effs):
#enthalpy function
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_h = [0.02926640E+02, 0.14879768E-02, -0.05684760E-05, 0.10097038E-09, -0.06753351E-13, -0.09227977E+04, 0.05980528E+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]
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]
#o2_coeffs_h = [3.28253784E+00, 1.48308754E-03, -7.57966669E-07, 2.09470555E-10, -2.16717794E-14, -1.08845772E+03, 5.45323129E+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,h_loss):
T_std = 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_ch4_r = h(T_std,ch4_coeffs_l)
h_o2_r = h(T_std,o2_coeffs_l)
h_n2_r = h(T_std,n2_coeffs_l)
h_products = h_co2_p + 2*h_h2o_p + 7.52*h_n2_p
h_reactants = h_ch4_r + 2*h_o2_r + 7.52*h_n2_r
loss = h_loss*LHV
return h_products - h_reactants + loss
def fprime(T,h_loss):
return (f(T+1e-6,h_loss) - f(T,h_loss))/1e-6
T_guess = 1500
tol = 1e-3
ct=0
alpha=0.2
h_loss = np.linspace(0,1,20)
T_aft = [] #required temp
for i in range (0,len(h_loss)):
while(abs(f(T_guess,h_loss[i])) > tol):
T_guess = T_guess - alpha*(f(T_guess,h_loss[i])/fprime(T_guess,h_loss[i]))
ct=ct+1
T_aft.append(T_guess)
print(T_aft)
#print(T_guess)
plt.plot(h_loss,T_aft,'*',color='blue')
plt.xlabel('Heat Loss Factor')
plt.ylabel('Flame Temperature (k)')
plt.title('Effect of Heat Loss on Flame Temperature')
plt.grid('on')
plt.show() 
It can be noticed from the above figure that with an increase in heat loss the temperature of the flame decreases because heat is getting lost from the system.
When heat loss is zero the flame temperature is 2325.59K and when the heat loss is 1 the flame temperature drops to 299.45K.
Case III
In this case, we have to consider an alkane, alkene, and an alkyne for the reaction including the heat loss of 35%.
The reactions can be written as:
Alkane
Alkene
Alkyne
The program in python can be seen below:
"""
adiabatic flame temperature using alkane alkene alkynewith heat loss of 35%
ethane
c2h6+3.5(o2+3.76n2)=2co2+3h2o+3.5*3.76n2
ethene
c2h4+3(o2+3.76n2)=2co2+2h2o+3*3.76n2
ethynec
2h2+2.5(o2+3.76n2)=2co2+h2o+2.5*3.76n2
"""
import matplotlib.pyplot as plt
import math
import numpy as np
R=8.314 #J/mol-K
def h(T, co_effs):
#enthalpy Function
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
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]
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_h = [0.02926640E+02, 0.14879768E-02, -0.05684760E-05, 0.10097038E-09, -0.06753351E-13, -0.09227977E+04, 0.05980528E+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]
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]
#o2_coeffs_h = [3.28253784E+00, 1.48308754E-03, -7.57966669E-07, 2.09470555E-10, -2.16717794E-14, -1.08845772E+03, 5.45323129E+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]
"""
Ethane
"""
def f_ane(T):
# calculating enthalpies of products and reactants
T_std = 298.15
LHV = 1428275
#lower heating value
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_c2h6_r = h(T_std,c2h6_coeffs_l)
h_o2_r = h(T_std,o2_coeffs_l)
h_n2_r = h(T_std,n2_coeffs_l)
h_products = 2*h_co2_p + 3*h_h2o_p + (3.76*3.5*h_n2_p)
h_reactants = h_c2h6_r + 3.5*h_o2_r + (3.5*3.76*h_n2_r)
# print(h_reactants)
return h_products - h_reactants + (0.35*LHV)
def fprime_ane(T):
return (f_ane(T+1e-6) - f_ane(T))/1e-6
"""
Ethene
"""
def f_ene(T):
# calculating enthalpies of products and reactants
T_std = 298.15
LHV = 1323474 #lower heating value
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_c2h4_r = h(T_std,c2h4_coeffs_l)
h_o2_r = h(T_std,o2_coeffs_l)
h_n2_r = h(T_std,n2_coeffs_l)
h_products = 2*h_co2_p + 2*h_h2o_p + (3.76*3*h_n2_p)
h_reactants = h_c2h4_r + 3*h_o2_r + (3*3.76*h_n2_r)
# print(h_reactants)
return h_products - h_reactants + (0.35*LHV)
def fprime_ene(T):
return (f_ene(T+1e-6) - f_ene(T))/1e-6
"""
Ethyne
"""
def f_yne(T):
# calculating enthalpies of products and reactants
T_std = 298.15
LHV = 1258020 #lower heating value
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_c2h2_r = h(T_std,c2h2_coeffs_l)
h_o2_r = h(T_std,o2_coeffs_l)
h_n2_r = h(T_std,n2_coeffs_l)
h_products = 2*h_co2_p + h_h2o_p + (3.76*2.5*h_n2_p)
h_reactants = h_c2h2_r + 2.5*h_o2_r + (2.5*3.76*h_n2_r)
# print(h_reactants)
return h_products - h_reactants + (0.35*LHV)
def fprime_yne(T):
return (f_yne(T+1e-6) - f_yne(T))/1e-6
T_ane_guess = 1500
T_ene_guess = 1500
T_yne_guess = 1500
tol = 1e-3
#ct=0
alpha=0.2
T_aft = []
#phi = np.linspace(0.1,1.5,20)
#for i in range (0,length(phi)):
#ane
while(abs(f_ane(T_ane_guess)) > tol):
T_ane_guess = T_ane_guess - alpha*(f_ane(T_ane_guess)/fprime_ane(T_ane_guess))
T_aft.append(T_ane_guess)
#ene
while(abs(f_ene(T_ene_guess)) > tol):
T_ene_guess = T_ene_guess - alpha*(f_ene(T_ene_guess)/fprime_ene(T_ene_guess))
T_aft.append(T_ene_guess)
#yne
while(abs(f_yne(T_yne_guess)) > tol):
T_yne_guess = T_yne_guess - alpha*(f_yne(T_yne_guess)/fprime_yne(T_yne_guess))
T_aft.append(T_yne_guess)
fuel = ['Ethane', 'Ethene', 'Ethyne']
print(T_aft)
plt.bar(fuel,T_aft)
plt.xlabel('Type of hydrocarbon')
plt.ylabel('Adiabatic Flame Temperature')
plt.title('Changes in AFT with Respect to Hydrocarbon Type')
#plt.grid('on')
plt.show()
It can be noticed that the flame temperature is minimum for the alkane and maximum for an alkyne. This can be due to the higher bonds C-C bonds thus leading to the higher energy generation. The flame temperature for ethane is 1714.17 K, ethene is 1836.93 and ethyne is 2064.78.
Case IV
In this case, we increased the carbon atoms in alkane and observe the change in the flame temperature.
Thus the fuel used will be methane, ethane, and propane.
The reactions can be written as:
Methane
Ethane
Propane
The program in python can be seen below:
"""
adiabatic flame temperature wrt c= 1 to 3
with heat loss of 35%
methane
ch4+2(o2+3.76n2)=co2+2h2o+2*3.76n2
ethane
c2h6+3.5(o2+3.76n2)=2co2+3h2o+3.5*3.76n2
propane
c3h8+5(o2+3.76n2)=3co2+4h2o+5*3.76n2
"""
import matplotlib.pyplot as plt
import math
import numpy as np
R=8.314 #J/mol-K
def h(T,co_effs):
#enthalpy Function
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]
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]
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_h = [0.02926640E+02, 0.14879768E-02, -0.05684760E-05, 0.10097038E-09, -0.06753351E-13, -0.09227977E+04, 0.05980528E+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]
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]
#o2_coeffs_h = [3.28253784E+00, 1.48308754E-03, -7.57966669E-07, 2.09470555E-10, -2.16717794E-14, -1.08845772E+03, 5.45323129E+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]
"""
Methane
"""
def f_methane(T):
# calculating enthalpies of products and reactants
T_std = 298.15
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_ch4_r = h(T_std,ch4_coeffs_l)
h_o2_r = h(T_std,o2_coeffs_l)
h_n2_r = h(T_std,n2_coeffs_l)
h_products = h_co2_p + 2*h_h2o_p + (3.76*2*h_n2_p)
h_reactants = h_ch4_r + 2*h_o2_r + (2*3.76*h_n2_r)
# print(h_reactants)
return h_products - h_reactants
def fprime_methane(T):
return (f_methane(T+1e-6) - f_methane(T))/1e-6
"""
ethane
"""
def f_ethane(T):
# calculating enthalpies of products and reactants
T_std = 298.15
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_c2h6_r = h(T_std,c2h6_coeffs_l)
h_o2_r = h(T_std,o2_coeffs_l)
h_n2_r = h(T_std,n2_coeffs_l)
h_products = 2*h_co2_p + 3*h_h2o_p + (3.76*3.5*h_n2_p)
h_reactants = h_c2h6_r + 3.5*h_o2_r + (3.5*3.76*h_n2_r)
# print(h_reactants)
return h_products - h_reactants
def fprime_ethane(T):
return (f_ethane(T+1e-6) - f_ethane(T))/1e-6
"""
Propane
"""
def f_propane(T):
# calculating enthalpies of products and reactants
T_std = 298.15
# LHV = 1258020 #lower heating value
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_c3h8_r = h(T_std,c3h8_coeffs_l)
h_o2_r = h(T_std,o2_coeffs_l)
h_n2_r = h(T_std,n2_coeffs_l)
h_products = 3*h_co2_p + 4*h_h2o_p + (3.76*5*h_n2_p)
h_reactants = h_c3h8_r + 5*h_o2_r + (5*3.76*h_n2_r)
# print(h_reactants)
return h_products - h_reactants
def fprime_propane(T):
return (f_propane(T+1e-6) - f_propane(T))/1e-6
T_methane_guess = 1500
T_ethane_guess = 1500
T_propane_guess = 1500
tol = 1e-3
#ct=0
alpha=0.2
T_aft = []
#phi = np.linspace(0.1,1.5,20)
#for i in range (0,length(phi)):
#methane
while(abs(f_methane(T_methane_guess)) > tol):
T_methane_guess = T_methane_guess - alpha*(f_methane(T_methane_guess)/fprime_methane(T_methane_guess))
T_aft.append(T_methane_guess)
#ethane
while(abs(f_ethane(T_ethane_guess)) > tol):
T_ethane_guess = T_ethane_guess - alpha*(f_ethane(T_ethane_guess)/fprime_ethane(T_ethane_guess))
T_aft.append(T_ethane_guess)
#propane
while(abs(f_propane(T_propane_guess)) > tol):
T_propane_guess = T_propane_guess - alpha*(f_propane(T_propane_guess)/fprime_propane(T_propane_guess))
T_aft.append(T_propane_guess)
fuel = ['methane', 'Ethane', 'Propane']
print(T_aft)
plt.bar(fuel,T_aft)
plt.xlabel('Type of hydrocarbon')
plt.ylabel('Adiabatic Flame Temperature')
plt.title('Changes in AFT with Respect to number of carbon')
#plt.grid('on')
plt.show()
It can be observed here, that the temperature increases in ascending order from methane to propane, this can be due to the increase in carbon atoms and hydrogen atoms.
The temperature of Hydrocarbons are:
methane: 2325.598K ethane; 2379.84K propane: 2392.09K
Conclusion:
@ The maximum flame temperature for a mixture achieved in the case of a stoichiometric mixture. For rich and lean mixture the flame temperatures are always less than that of the stoichiometric mixture.
@ We use two approaches here (Python and Cantera) and there is a difference in results of Cantera and python, This is because Cantera solves by reducing the Gibbs free energy of the equation. Also, it produces more intermediate byproducts for the reaction mechanism than we produce manually.
@ If there is a heat loss then we also get a decrease in the flame temperature.
@ The flame temperature is more for an alkyne than that for an alkene and alkane (same number of carbon atoms) respectively. This is due to more carbon bonds in an alkyne as compared to other structures.
@ In case of an increase in the number of carbon atoms in an alkane, we got the high flame temperature. This is due to an increase in hydrogen and carbon atoms.
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