Effect of Adiabatic Flame Temperature on Equivalence ratio

Objectives:

• To calculate the effect of the equivalence ratio on adiabatic flame temperature for methane in a constant volume chamber.
• To calculate the AFT for methane with a heat loss of 0.35 for a constant pressure reactor.
• To calculate AFT for the -ane,-ene,-yne variants of 2 carbon atom chain and plot the results.
• For alkane plot the effect of number of carbon atoms on the final temperature with o heat loss.

Introduction:

Adiabatic Flame Temperature:

The Maximum temperature the combustion products can reach adiabatically (Without heat loss/transfer) theoritically is known as adiabatic flame temperature AFT.

Constant pressure process:

Isobaric process `Delta H=0`

In constant pressure process the enthalpy of the reactants and products are equal

`H_r-H_p=0`

Constant Volume process:

Isochoric process `Delta U=0`

The adiabatic flame temperature of constant volume process is higher than constant pressure process.

`H=U+PV`

`H_r-H_p-R(N_r*T_std-N_pT_(aft))`

Equivalence ratio(`phi`):

Equivalence ratio of a system is defined as the ratio of the fuel-to-oxidizer ratio to the stoichiometric fuel-to-oxidizer ratio .

`phi=((n_(fuel)/n_(o2)))/(n_(fuel)/n_(o2))_(st)` - in terms of moles

`phi=1`-Stoichiometric mixture results in the complete combustion reaction where no fuel is left and `CO_2 and H_2O` are the products with Nitrogen.

`phi<1`- Lean mixture is when there is less fuel quantity or there is more amount of air,it results in more number of products due to the excess air.

`phi<1`- Rich mixture is when there is more fuel quantity or there is less amount of air, it results in incomplete combustion and produces toxic emissions.

General stoichiometric reaction:

`C_xH_y + ar(O_2+3.76N_2) -> a CO_2 + bH_20+cN_2`

ar-Stoichiometric coefficient where a=x,b=y/2 &ar=(x+(y/4))

Combustion reaction for methane:

For `phi <=1`

`CH_4+2/phi(O_2+3.76*N_2)->CO_2+2/phiH_2O+7.52/phiN_2+(2/phi-(a+b/2))O_2`

For `phi=1`

`CH_4+2(O_2+3.76*N_2)->CO_2+2H_2O+7.52N_2`

For phi=2

`CH_4+(O_2+3.76*N_2)->0.5CO_2+0.5H_2O+3.76N_2+0.5CO+1.5H_2`

Python code:

1.AFT at constant volume:

https://drive.google.com/open?id=1sdh_JPgIJMVOZ7LxXtyG9SSN-gClZjpe

From the above plot the Adiabatic Flame temperature of python is quite higher Compared to that of cantera. AFT of cantera is low because cantera uses the Gibb\'s minimization energy for calculating the AFT and it calculates enthalpy for many products which are negliigble and results in reduction in heat loss.In python we use Newton Raphson method for root finding problem.

Newton\'s Raphson method:

`T_(aft)=T_(aft)-alpha f(T_(aft))/(fprime(T_(aft))`

AFT values for different equivalence ratio values.

Equivalence ratio	Python	Cantera
0.4	1567	1563
0.6	2040	2018
0.8	2452	2377
1.0	2817	2585
2.0	1977	1929

2.AFT at constant pressure:

Consider the stiochiometric combustion of methane in a constant pressure reactor with heat loss.

`CH_4+2(O_2+3.76*N_2)->CO_2+2H_2O+7.52N_2`

`H_r-H_p+0.35*Heat loss`

code:

https://drive.google.com/open?id=1uD80IMLeny78ydQ_J_b7l1V22Ldh42Aj

From the above plot the AFT varies linearly with heat loss fraction.As the heat loss increase the AFT of temperature decreases.

3.AFT for the -ane,-ene,-yne variants of 2 carbon atom chain:

Genralized reaction:

alkane:

`C_nH_(2n+2)+ (3n+1)/2(O_2+3.76N_2)->nCO_2+(n+1)/2H_2O+(3n+1)/2N_2`

Alkene:

`C_nH_(2n)+ (3n)/2(O_2+3.76N_2)->nCO_2+n/2H_2O+(3n)/2N_2`

Alkyne:

`C_nH_(2n-2)+ (3n-1)/2(O_2+3.76N_2)->nCO_2+(n-1)/2H_2O+(3n-1)/2N_2`

code:

https://drive.google.com/open?id=19jVrqRKgVHGx0N2jIVBqW-jxDZsq_WH8

Hydrocarbons	Flame Temperature
Ethane	1714
Ethene	1837
Ethyne	2065

From the plot Ethyne (Triple bond) has higher temperature than ethane because of decomposition of c bonds.Higher amount of energy is released during the breaking off carbon bonds by oxygen.

4.AFT for the alkane hydrocarbons:

Genralized reaction:

`C_nH_(2n+2)+ (3n+1)/2(O_2+3.76N_2)->nCO_2+(n+1)/2H_2O+(3n+1)/2N_2`

code:

https://drive.google.com/open?id=19jVrqRKgVHGx0N2jIVBqW-jxDZsq_WH8

Alkane	Flame Temperature
Methane	1826
Ethane	2325
Propane	2391

From the plot the Flame temperature of the propane is higher than methane due to more number of carbon atoms , the fuel content is higher so flame temperature is higher.

Conclusion:

Thus python program was coded to calculate the AFT of methane.