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: