Challenge

Week 3 - Adiabatic Flame Temperature calculation

SSiddhesh Sakpal1 Jan 2022

Assignment no. 1

Assignment name- Adiabatic Flame Temperature (python and cantera)

Objective of this assignment-

Case 1- Determine effect of equivalence ratio on AFT in a constant volume chamber for methane.

Write the program in python and compare with cantera

Case 2-To calculate the AFT for methane combustion with heat loss in a constant pressure chamber

Case 3- Adiabatic flame temperature variation with respect to Alkane, Alkene, Alkyne

Case 4- Adiabatic flame temperature variation with respect to number of Carbon atom chains

Theory and introductions

Adiabatic Flame temperature-

Final temperature obtained after the fuel is completely burned called as adiabatic flame temperature
When the combustion is takes place energy is released to the combustion product. If no heat loss in this process, the temperature of combustion product is known as adiabatic flame temperature
There are two types of Adiabatic flame temperature depending upon how is process completed i.e. constant volume/constant pressure

Significance-

It is characteristics the quality of fuel
It is used for determine which fuel is suitable for particular application
The adiabatic flame temperature is very important in design and scale up procedures where heat transfer is significant consideration

Some important concept

Stoichiometric combustion-

It is a theoretical position in which the optimal amount of oxygen and fuel mix generates the most heat possible, achieving maximum combustion efficiency. There is no unburnt combustibles and no excess air. It is something to strive for, though in reality it does not exits
In stoichiometry is founded on the law of conservation of mass where the total mass of the reactants equals to total mass of products.

Lean combustion-

If the amount of air is more than required for complete combustion, there will be excess air after combustion

Rich combustion

If the amount of air is less than required for complete combustion, part of the fuel would not be fully oxidized such that either will be excess fuel after 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

General equations

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

Chemical reaction pf methane

Stoichiometric combustion

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

Lean combustion

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

Rich combustion

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

Constant volume process

For constant volume process internal energy is equated for reactants and product to get the AFT

Ur-Up=0

After we will used newton Rapson method and we get final temperature

(H_product-n_prod*R*T)-(Hreactant-N-reactants-n_react*R*Tstd)=0

Coding approach-

1.enthalpy of product and reactants calculated from formula that means nasa polynomial

2.coefficient a1-a7 is directly obtained from Berkeley website

3.For the constant volume method process internal energy is equated for reactants and product to get the AFT

Ur-Up=0

4.by using newton Raphson technique and this is iterative technique below term equated to 0

(H_product-n_prod*R*T)-(Hreactant-N-reactants-n_react*R*Tstd)=0

5.put appropriate value of R=8.314 J/mol.k , Tstd=298.15k

6.equivalence ratio is varied according to situation

7.compare the result with cantera

8.coding approach using cantera

For getting result from cantera solution class containing methane and air
The gas species obtained from gri30.cti
Equilibrate solver was used with constant volume and internal energy obtain AFT

Case 1

Assignment number 1

Adbatic temperture

using python and cantera

"""

#ch4 + 2(O2+3.76N2)=CO2+2H2O+7.52N2 stichiometric combustion

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

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

#date-29 dec 2021

import matplotlib.pyplot as plt

import math

import numpy as np

def h(T,co_effs):

R=8.314 #J/mol.k

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]

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]

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_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_CO2_p=h(T,CO2_coeffs_h)

h_H2O_p=h(T,H2O_coeffs_h)

h_N2_p=h(T,N2_coeffs_h)

h_O2_p=h(T,O2_coeffs_h)

h_CO_p=h(T,CO_coeffs_h)

R=8.314

if phi<1:

hproduct=h_CO2_p +2*h_H2O_p+(7.52/phi)*h_N2_p+(2/phi-2)*h_O2_p

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

if phi==1:

hproduct=h_CO2_p+2*h_H2O_p+7.52*h_N2_p

n_product= 1 + 2 + 7.52

if phi>1:

hproduct=(4/phi -3)*h_CO2_p +2*h_H2O_p +(7.52/phi)*h_N2_p+(4-4/phi)*h_CO_p

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

Tstd=298.15

h_CH4_r=h(Tstd,CH4_coeffs_l)

h_N2_r=h(Tstd,N2_coeffs_l)

h_O2_r=h(Tstd,O2_coeffs_l)

hreactant=h_CH4_r + (2/phi)*h_O2_r +(7.52/phi)*h_N2_r

n_reactant=1+ (2/phi) +(7.52/phi)

return (hproduct-n_product*R*T)-(hreactant-n_reactant*R*Tstd)

#by using forward differencing

def fprime(T):

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

T_guess=1500

tol = 1e-3

ct=0

alpha=0.2

phii= np.arange(0.1,2,0.05)

T_adibatic_temp=[]

for phi in phii:

while (abs(f(T_guess))>tol):

T_guess=T_guess -alpha*(f(T_guess)/fprime(T_guess))

ct=ct+1

T_adibatic_temp.append(T_guess)

plt.plot(phii,T_adibatic_temp,color='red',label="newton raphson itrative technique")

plt.grid ()

# import cantera as ct

import cantera as ct

T_adibatic_temp_cantera=[]

for phi in phii:

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

gas.TPX= 298.15,101325,{'CH4':1,'O2':2/phi,'N2':7.52/phi}

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

T_adibatic_temp_cantera.append(gas.T)

plt.plot(phii,T_adibatic_temp_cantera,color='blue',label="cantera")

plt.grid ()

plt.xlabel('equivalence ratio')

plt.ylabel('adiabatic flame tempertaure')

plt.title('effect of equivalence ratio on AFT')

plt.legend()

plt.show()

Result

From graph we easily conclude that at equivalence ration equal to 1 stoichiometric condition The AFT is highest. After Stoichiometric AFT is decreased. This is because of in this condition all fuel is burned and then producing more energy

Result got from cantera shows lower AFT than the code using newton Rapson method. The Reason for this cantera considered all the product of combustion which would occure in the actual case.so more energy used to produce products.

Hence AFT is obtained through cantera which is actual solution

Case 2

Coding approach-

The combustion is considered at constant pressure chamber.

For constant constant enthalpy process equation obtaines is

hproduct-hreactant-heatloss=0

heat loss=heat loss factor*LHV

vary heat loss factor varies between 0 to 1

"""

Adaibatic temperature

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

"""

import matplotlib.pyplot as plt

import math

import numpy as np

def h(T,co_effs):

R=8.314 #J/mol.k

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]

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]

N2_coeffs_h=[0.02926640E+02 ,0.14879768E-02,-0.05684760E-05 ,0.10097038E-09,-0.06753351E-13,-0.09227977E+04, 0.05980528E+02]

def f(T):

h_CO2_p=h(T,CO2_coeffs_h)

h_H2O_p=h(T,H2O_coeffs_h)

h_N2_p=h(T,N2_coeffs_h)

hproduct=h_CO2_p +2*h_H2O_p+7.52*h_N2_p

Tstd=298.15

LHV=800000

h_CH4_r=h(Tstd,CH4_coeffs_l)

h_N2_r=h(Tstd,N2_coeffs_l)

h_O2_r=h(Tstd,O2_coeffs_l)

hreactant=h_CH4_r + 2*h_O2_r +7.52*h_N2_r

return hproduct - hreactant +L*LHV

def fprime(T):

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

T_guess=1500

tol = 1e-3

ct=0

alpha=0.2

for L in np.arange(0,1,0.1):

while (abs(f(T_guess))>tol):

T_guess=T_guess -alpha*(f(T_guess)/fprime(T_guess))

ct=ct+1

plt.plot(L,T_guess,'o',color='blue')

plt.grid ()

plt.xlabel('heat loss factor')

plt.ylabel('adiabatic flame tempertaure')

plt.title('effect of heat loss factor on AFT')

plt.show()

Results

From the graph AFT decreases as heat loss factor increases and this is because of less energy is contained in the chamber thereby reducing the temperture

Case 3

Coding approach

The same approach is consider was formed and tested for -ane,ene,yne variant of 2 carbon atom chain. The hydrocarbon are ethane, ethane, ethyne

"""

Adaibatic temperature

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

"""

import matplotlib.pyplot as plt

import math

import numpy as np

def h(T,co_effs):

R=8.314 #J/mol.k

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]

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]

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_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]

N2_coeffs_h=[0.02926640E+02 ,0.14879768E-02,-0.05684760E-05 ,0.10097038E-09,-0.06753351E-13,-0.09227977E+04, 0.05980528E+02]

def f(T):

h_CO2_p=h(T,CO2_coeffs_h)

h_H2O_p=h(T,H2O_coeffs_h)

h_N2_p=h(T,N2_coeffs_h)

hproduct=x*h_CO2_p +(y/2)*h_H2O_p+(x+y/4)*3.76*h_N2_p

Tstd=298.15

h_C2H6_r=h(Tstd,C2H6_coeffs_l)

h_N2_r=h(Tstd,N2_coeffs_l)

h_O2_r=h(Tstd,O2_coeffs_l)

h_C2H4_r=h(Tstd,C2H4_coeffs_l)

h_C2H2_r=h(Tstd,C2H2_coeffs_l)

if y==y_ane:

LHV=1428600

hreactant=h_C2H6_r + (x+y/4)*h_O2_r +(x+y/4)*3.76*h_N2_r

if y==y_ene:

LHV=1227070

hreactant=h_C2H4_r + (x+y/4)*h_O2_r +(x+y/4)*3.76*h_N2_r

if y==y_yne:

LHV=1157784

hreactant=h_C2H2_r + (x+y/4)*h_O2_r +(x+y/4)*3.76*h_N2_r

return hproduct - hreactant +0.35*LHV

def fprime(T):

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

T_guess=1500

tol = 1e-3

ct=0

alpha=0.5

x=2

y_ane=2*x+2

y_ene=2*x

y_yne=2*x-2

y1=[y_ane,y_ene,y_yne]

fuel=['ethane','ethene','ethyne']

T_adibatic=[]

for y in y1:

while (abs(f(T_guess))>tol):

T_guess=T_guess -alpha*(f(T_guess)/fprime(T_guess))

ct=ct+1

T_adibatic.append(T_guess)

print (ct)

print (T_guess)

plt.bar (fuel,T_adibatic)

plt.xlabel('hydrocarbon')

plt.ylabel('adiabatic flame tempertaure')

plt.title('effect of types hydrocarbon on AFT')

plt.show()

The AFT for ethane=1714 k, ethane=1890k, ethyne=2133k was obtained

From the bar graph AFT is higher in case ethyne and lowest in case ethane. Because ethyne has triple bond higher energy is required to produce product

Case 4

The carbon atoms are varied from 1 to 3 for the alkane group and the AFT is compared between them. The hydrocarbon is Methane, ethane and propane. Heat loss Factor taken is 0