Challenge

Week 3 - Adiabatic Flame Temperature calculation

RRattanvir Singh6 Jul 2020

Adiabatic Flame Temperature is the maximum temperature reached during the combustion under adiabatic conditions. For a constant pressure process, It is calculated by equating the enthalpies of formation for products and reactants and thus solving the root equation . For constant volume process the internal energies are equated and root equation is solved .

Effect of equivalence ratio on AFT

Standard equation for methane combustion is as follows

For phi<1

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

For phi >1

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

For constant volume reactor the root equation will be Internal Energy (U = H – nRT), which will be solved using Newton Raphson method to calculate the AFT. Python code for calculation is given below.

import matplotlib.pyplot as plt

import math

import numpy as np

import cantera as ct

R = 8.314 #J/mol-K

def h(T,coeffs):

a1 = coeffs[0]

a2 = coeffs[1]

a3 = coeffs[2]

a4 = coeffs[3]

a5 = coeffs[4]

a6 = coeffs[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]

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,phi):

"""

Function to calculate the difference in the enthalpy between the Reactants and the Products

"""

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 phi<=1: # Lean mixture

H_reactants = h_ch4_r+ (2/phi)*(h_o2_r + 3.76*h_n2_r)

n_reactant = 1 + (2/phi) + 3.76*(2/phi)

H_products = h_co2_p + (7.52/phi)*h_n2_p + 2*h_h2o_p + (2/phi-2)*h_o2_p

n_product = 1 + 2 + (2/phi-2) + (7.52/phi)

else: # Rich Mixture (phi > 1)

H_reactants = h_ch4_r+ (2/phi)*(h_o2_r + 3.76*h_n2_r)

n_reactant = 1 + (2/phi) + 3.76*(2/phi)

H_products = (4/phi-3)*h_co2_p + (7.52/phi)*h_n2_p + 2*h_h2o_p + (4-4/phi)*h_co_p

n_product = (4/phi-3) + 2 + (4-4/phi) + (7.52/phi)

return H_products - H_reactants - R*(n_product*T -n_reactant*Tstp)

def fprime(T, phi):

return (f(T+ 1e-6, phi)- f(T,phi))/1e-6

T_guess=1500

tol = 1e-4

alpha = 0.2

phi= np.linspace(0.5,1.5,101)

T_AFT=[]

for i in range (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])))

T_AFT.append(T_guess)

Equivalence_ratio = phi

T_AFT_cantera= []

gas = ct.Solution('gri30.xml')

for i in range (len(phi)):

Total_mole = 1 + 2/phi[i] +7.52/phi[i]

moles_CH4 = 1

moles_O2 = 2/phi[i]

moles_N2= 7.52/phi[i]

gas.TPX= 298.15,101325,{'CH4':1 ,'O2':moles_O2,'N2':moles_N2}

gas.equilibrate('UV','auto')

T_AFT_cantera.append(gas.T)

print (T_AFT_cantera)

print (T_AFT)

plt.plot(Equivalence_ratio,T_AFT,color = 'green')

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()

Results

Adiabatic Flame Temperature = 2817.83 K using Python

Adiabatic Flame Temperature = 2603.31 K using Cantera

Observation:

Higher AFT using Python as in cantera more species are involved in the combustion process, thus consuming more energy which led to lower AFT. Cantera considers this loss of energy and also works on Minimization of Gibbs energy principle.

Constant Pressure Reactor

In the constant pressure reactor, enthalpies are equated for root equation . In this case , heat loss factor is also considered as some percentage of LHV of the fuel which is the maximum possible heat of the fuel , is lost to the surroundings.

The effect of the percentage of this heat loss is given as follows:

import matplotlib.pyplot as plt

import math

import numpy as np

import cantera as ct

R = 8.314 #J/mol-K

LHV = 50 #MJ/kg

m_fuel = 16.04 #gram

Q_heat = LHV*m_fuel*1000

H_loss = np.linspace(0,1,21)