Calculation of AFT in constant pressure and constant volume condition

1. Effect of Equivalence Ratio on Adiabatic Flame Temperature

The python program was written to calculate AFT using Newton-Raphson Optimisation of the following function :

f = (Enthalpy of products) - (Enthalpy of reactants) - R*((Number of moles of reactants)*T_stp - (Number of moles of products*T_ad)  = 0

(Refer to appendix for resources explaining the above formula)

Plot for normal code : 

Plot for cantera code:

It can be clearly seen that for lean mixture, as phi increases AFT increases and vice-versa is true for rich mixture. However, Cantera code gives out lesser magnitude values in general compared to the traditional code written.

Normal Code : https://drive.google.com/open?id=1awECA2xZraBaXf-zHeypBwFaWr6Moa2W

Cantera Code : https://drive.google.com/open?id=1xAIC_SfEPfRV-aYgVQfKN1pVaowYv4z1

2.1. Constant pressure reactor with heat loss in Ethane, Ethene and Ethyne

The python program was written to calculate AFT using Newton-Raphson Optimisation of the following function :

f = (Enthalpy of products) - (Enthalpy of reactants) - (Heat Loss Factor * Maximum possible heat loss) = 0

The hydrocarbons considered are as follows : 

1. C2H6 (Ethane) - AFT : 2304.7 K

2. C2H4 (Ethene) - AFT : 2834.77 K

3. C2H2 (Ethyne) - AFT : 2747.1 K

(Considering Heat Loss Factor = 0)

The AFT for Ethene is the highest amongst the three.

Graph : 

Code : https://drive.google.com/open?id=1c38DhkOmqA3fARBLWlwSTyqVOO-hx8U8

2.2 Constant Pressure AFT in Alkanes with varying number of C atoms

The following Hydrocarbons(Alkanes) have been considered : 

1. Methane (C = 1) - AFT = 2325.59 K

2. Ethane (C = 2) - AFT = 2379.84 K

3. Propane (C = 3) - AFT = 2392.09 K

As number of carbon atoms increases, the value of AFT increases.

Code : https://drive.google.com/open?id=174XJxwxZDIT0hWKYAMmCX9PRo2phVZ7Z

Appendix:

1. https://www.researchgate.net/file.PostFileLoader.html?id=549d631ad3df3e311a8b4603&assetKey=AS%3A273658277236765%401442256626135