Week 3 - Adiabatic Flame Temperature calculation

STUDY ON ADIABATIC FLAME TEMPERATURE

AIM:
⦁ To calculate the Adiabatic Flame Temperature(AFT) of Methane in a constant volume chamber for various Equivalence ratios.
⦁ To calculate the Adiabatic Flame Temperature(AFT) of Methane in a constant pressure chamber with H_loss = 0.35.
⦁ To calculate the Adiabatic Flame Temperature(AFT) for alkanes, alkenes and alkynes.
⦁ To calculate the Adiabatic Flame Temperature(AFT) with respect to number of carbon atoms in the alkane chain.

THEORY:
When we assume no heat loss during a combustion process, the heat generated is entirely used to raise the temperature of the products as a result of combustion and this temperature is known as Adiabatic Flame Temperature(AFT).
The equivalence ratio(`phi`) is defined as the ratio of the actual fuel/air ratio to the stoichiometric fuel/air ratio.
`phi = ((Air//Fuel)_(st))/(Air//Fuel)`
If the `phi=1`, then the air-fuel mixture is at stoichiometric quantity. Then the combustion equation is;
`C_xH_y+a(O_2+3.76N_2)=xCO_2+(y/2)H_2O+3.76aN_2`
If `phi<1`, it is said to be a 'lean mixture' where in the air-fuel proportion, air has a higher value compared to fuel ,i.e., excess air is supplied. Then the combustion equation is;
`C_xH_y+a(O_2+3.76N_2)=xCO_2+(y/2)H_2O+3.76aN_2+bO_2`
If `phi>1`, it is said to be a 'rich mixture' where in the air-fuel proportion, fuel has a higher value compared to fuel ,i.e., more products(toxic pollutants) are produced. Then the combustion equation is;
`C_xH_y+a(O_2+3.76N_2)=xCO_2+(y/2)H_2O+3.76aN_2+eCO`

PROCEDURE:
⦁ In a constant volume combustion process, the internal energy of the reactant will be equal to that of the product.
`U_(r)(T_iP_i)=U_(p)(T_(AFT)P_f)`
But, H = U + PV
and, PV = NRT
Therefore,
`H_r−H_p−R_u(N_rT_i−N_pT_(AFT))=0`; f(T) = 0
Based on these information, to determine the value of `T_(AFT)`, we can use Newton Raphson technique to find the root(T). This is given by,
`T_(AFT) = T_(AFT) - alpha( f(T_(AFT))/(f'(T_(AFT))))`
So, since we don't know the value of `T_(AFT)`, we give a guess value for T and iterate it based on the function for which a better or minimized value of T is obtained. This process is iterated until we reach a certain tolerance of error value for `T_(AFT)`.
This entire algorithm can be formulated using Python with the help of NASA thermodynamic data table.

The drive link for the python program is attached below:

https://drive.google.com/file/d/13OcpYCK9aRZ35-z2nSZuhVzXmPNmMP9G/view?usp=sharing

The drive link for the cantera program is attached below:

https://drive.google.com/file/d/10nvNNUfijopP3LelCsh-jy5E1I0mhjvo/view?usp=sharing

⦁ In case of constant pressure combustion process, the enthalpy of the reactant will be equal to that of the product.
`H_r = H_p` or `H_r-H_p=0`; f(T) = 0
The same process can be carried on for the constant pressure combustion also. In this, the f'(T) can be determined using Forward differencing technique;
`f'(T) = (f(T+dt) - f(T))/dt`

The drive link for the python program is attached below:

https://drive.google.com/file/d/1LB2ALGM4ypfjnTNi6FyfzQSYBKh7cMSg/view?usp=sharing

⦁ In case of Alk-ane, -ene & -yne combustion on AFT, the drive link for python program is below:

https://drive.google.com/file/d/12Hvnpgdq4mZf_dvKJAPT6r3FGjCxygH2/view?usp=sharing

⦁ In case of no of carbon atoms on AFT, the drive link for python program is below:

https://drive.google.com/file/d/1Dhy4AQ7XRTCj4S2K9bt6L0kdjoAsqdFO/view?usp=sharing

RESULTS:
⦁ The plots for AFT vs Equivalence ratio for Constant volume chamber is as follows:

According to the above plots, we can see that the Cantera results for AFT is lesser than that of Python results. This is because, in Python, we neglected most of the negligible products which forming the reaction as if we include them, the equation become more complex. But in Cantera, all the products are included in some definite proportions as a result of which some energy is utilized to heat those additional products as well. So, Cantera results for AFT will be lower than that of Python results.
⦁ The plot for AFT vs No. of iterations for constant pressure chamber with heat loss of 0.35 is as follows:

According to the results, we can see that the AFT is deopped significantly if we can compare it without healoss. This is because, with heatloss, there will be some amout of heat descipation to the surroundings as a result of which the Temperature inside the chamber is reduced, so the AFT is also reduced.
⦁ The plot for Alk-ane, -ene & -yne 's effect on AFT is as follows:

According to the above results, it can be ssen that the AFT is the highest for Alkynes, average for Alkene and the lowest for Alkane in comparison. This is because of the strong covalent bond or C-C bond which decipated a lot of head upon decomposition of these bonds during combustion. So, Alkyne, which has a triple bond structure has the highest AFT than Alkene, which has a double bond structure and Alkane, which has only single bonds in their structure.
⦁ The plot for effect of number of carbon atoms on AFT is as follows:

Similar to the above plot's reason, if we take some forms of Alkanes like Methane, Ethane and Propane, the higher the number of bonds, the higher is the value of AFT which is shown as a result of comparison.