Calculating the Adiabatic Flame Temperature for different hydrocarbon fuels at different conditions

Adiabatic flame temperature is the maximum temperature that can be achieved by a fuel-oxidizer mixture when there is no energy loss to the surrounding. The AFT can be calculated in two ways the constant volume AFT and the constant pressure AFT.

Calculate the constant pressure AFT value for Methane-Air mixture for different values of Equivalence ratios.

The combustion of methane with air is studied by varying the equivalence ratio. The AFT is calculated using a python program and it is compared using Cantera.

For a stoichiometric mixute of methane and air the equivalence ratio `phi` = 1 and the balanced chemical reaction can be written as :

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

For a lean mixture `phi` < 1 and the balanced chemical reaction can be written as:

`CH_4 + \\frac{2}{phi}(O_2 +3.76N_2)-> CO_2 + 2H_2O +(\\frac{2}{phi}-2)O_2 +\\frac{7.52}{phi}N_2`

For a lean mixture `phi` > 1 and the balanced chemical reaction can be written as:

`CH_4 + \\frac{2}{phi}(O_2 + 3.76N_2) -> (\\frac{4}{phi}-3)CO_2 + 2H_2O + (4 -\\frac{4}{phi})CO + \\frac{7.52}{phi}N_2`

The code for AFT calculation is written in python. The Independent program and Cantera are also used for the calculation. A comparison is also done between the two results. The plot obtained by comparing the two results is shown below.

The code is written is shared here:

https://drive.google.com/open?id=1v8Vm2Atz481-NRER614vPm16EY5oEUxm

The correct behavior of AFT is calculated using Cantera. The difference in calculated value is because when the program is written in python only the major products of combustion alone are considered but in Cantera, all the products of combustion are considered. From the above diagram, we can see that the maximum AFT is achieved at `phi`= 1, but in the real case some excess air is required for achieving complete combustion and therefore the maximum AFT is at a \'phi\' > 1. But when further excess air is added the AFT value drops since the excess oxygen and nitrogen which is not taking part in the reaction absorbs heat and thus the AFT falls.

For an equivalence ratio up to 0.6, we can see that the values calculated using python and Cantera almost match, but beyond that, till \'phi\' = 1 the actual value of AFT is less than the calculated value. Beyond \'phi\' > 1.25 the calculated value is less than the actual value.

It can be seen that when there was no sufficient air i.e \'phi\' < 1, the AFT was low since there was no complete combustion. But when there is excess air the AFT reduces sine the excess amount of matter absorbs heat energy and thereby reduces the final temperature. This is very important since in many practical applications there is a requirement of reducing the final temperature such that it doesn\'t damage the machine components.

Effect of heat loss to the Adiabatic flame temperature for Methane-Air mixture

The maximum possible heat loss is the lower heating value (LHV) of the fuel oxidiser mixture. The AFT value is calculated for the different values of heat loss which is a fraction of the LHV. The effect of heat loss on AFT is plotted below.

The trend seen is as expected where the AFT decreases as the heat loss increases.

https://drive.google.com/open?id=1oqxcJQ6_3iYe3fHg9IPv0e-tuEe1V8g_

The AFT of a general fuel of type `C_xH_y` for 35% of maximum possible Heat Loss

For a general fuel of the form \'C_xH_y\' the balanced chemical reaction can be written as

`C_xH_y + (x+\\frac{y}{4})(O_2 + 3.76N_2)-> xCO_2 +\\frac{y}{2}H_2O + (x+\\frac{y}{4})3.76N_2`

A program is written to evaluate the effect of the presence of double and triple bond in the AFT. For that the number of carbon atoma are kept constant as x = 2. But the no of hydrogen atoma are varied. The combustion process is evaluated for y = 6 which is ethane, y = 4 which ethene and y = 2 which is ethyne.

It is seen that, as the number of bonds increases the AFT increases. It is because the breaking of a double bond releases more energy than a single bond and the breaking of the triple bond releases more energy than a double bond. As the number of hydrogen atom in the molecule decreases the no of bonds between the carbon atom increases which result in a higher energy release during combustion and therefore the AFT increases as the number of hydrogen atom increases.

The code written is given below:

https://drive.google.com/open?id=1F5JrAEcUX_0Jx9UiijNTQqvGU1CgLSOf

The effect of AFT on the number of carbon atoms in a general alkane for no heat loss

The effect of number of atoms in the chain of the hydrocarbon is analysed. Methane, ethane and propane which contains one, two and three atoms of carbon respectively are studied.

The general balanced chemica reaction is written below.

\'C_xH_y + (x+\\frac{y}{4})(O_2 + 3.76N_2)-> xCO_2 +\\frac{y}{2}H_2O + (x+\\frac{y}{4})3.76N_2`

When x = 1 we get methane, x = 2 gives ethane and x = 3 gives propane and y = 2x + 2.

From the above diagram it is seen that as the number of carbon atoms increases the AFT increases. As number of carbon atom increases the hydrogen atom increases and therefore the number of bonds also increases. A large number of bond breaking causes a large release of energy which in turn increases the adiabatic flame temperature.

The code written for calculating the AFT for different alkanes is given below:

https://drive.google.com/open?id=12maJ0aHDnKG7RdqKp3TkR_d468-rSkNz

CONCLUSION

From running the code it can be seen that :

1. As the equivalence ratio is increased the ATF increases at first and then decreases.
2. If the heat is lost to the surrounding the AFT decreases.
3. If the number of bonds in the hydrocarbon increases then the AFT increases.
4. If the number of carbon atoms increases then the AFT increases.