Flame Temperature calculation under various scenarios
In the project, the flame temperature of combustion in various scenarios are calculated using Python and compared.
Flame temperature is the temperature attained by products formed as a result of combustion reaction. When the combustion takes place in an adiabatic environment the calculated flame temperature is called the adiabatic flame temperature. For the project, flame temperature is calculated under constant volume and constant pressure. Under the constant volume setting the effect of the equivalence ratio on adiabatic flame temperature is investigated. The investigation is given in detail below:
Effect of equivalence ratio on adiabatic flame temperature under constant volume:
Under constant volume and adiabtic condition, from the first law of thermodynamics
`Q_v = DeltaU` and `Q_v = Q_p -RDeltanT`
as we know that,
`Q_p = DeltaH_p`
so,
`Q_v = DeltaH_p - RDeltanT`
as adiabatic condition is considered here,
`Q_v = 0`
`implies DeltaH - RDeltanT =0`
we also know that number of moles on the product side is equal to the number of moles of the reactant side so,
`implies H_r - H_p - nRDeltaT =0`
This is the equation for which the Newton-Raphson technique is used to find the adiabatic temperature in case of constant volume. For comparison, the same thing is done using Cantera also.
The setup is repeated for different equivalence ratios of methane. The balanced equation for different equivalence ratios are given below:
For `phi` = 1 the equation is,
`CH_4 + 2(O_2 + 3.76N_2) = (1)CO_2 + (2)H_2O + (7.52)N_2`
For phi<1 the equation is,
`(phi)CH_4 + 2(O_2 + 3.76N_2) = (phi)CO_2 + (2phi)H_2O + (7.52)N_2 + (2-2phi)O_2`
For phi>1 the equation is,
`(phi)CH_4 + 2(O_2 + 3.76N_2) = (4-3phi)CO_2 + (2phi)H_2O + (7.52)N_2 + (4phi-4)CO`
The code is uploaded to the Google drive and the link is given below:
https://drive.google.com/file/d/1bmcSSWzfY9F9RJOBtgG-LPI2Lxc4HTov/view?usp=sharing
The result obtained from the Newton-Raphson code and Cantera code as compared using plot and attached below:
Here the blue dots are found using Cantera and the red dots by Newton-Raphson method.
From the above, it is clear that maximum adiabatic flame temperature is reached when equivalence ratio is 1. The flame temperature found by Cantera is lesser than that of Newton-Raphson method, this is because more products are taken into account in case of Cantera which have the tendency to absorb heat and reduce the temperature. But as we increase the fuel above stoichiometric ratio, dissociation comes into picture which releases heat raising the adiabatic flame temperature.
Next, the setup the setup is changed so that flame temperature is calculated under constant pressure with heat loss.
Under constant pressure with heat loss:
From the first law of thermodynamics,
`DeltaU = Q_p - W_p = Q_p - pDeltaV`
we know that,
`H = U+pV`
so, `DeltaH=DeltaU+pDeltaV+VDeltap`
under constant pressure `Deltap = 0`
`implies DeltaH=DeltaU+pDeltaV`
under adiabatic conditions,
`Q_p = 0`,
so,
`DeltaH = 0`
`implies H_r = H_p`
This above equation is for adiabatic conditions, if heat loss is also to be included then the equation becomes,
`H_r =H_p + "heat loss"`
The maximum possible heat loss is the Lower Heating value of the fuel unless a condenser is present in which case Higher Heating Value, here since a condenser is not considered the Lower Heating value (LHV) is used.
so,
heat loss = heat loss fraction`xx`LHV
For the following program the heat loss fraction is taken as 0.35, the LHV is found by multiplying molecular mass in kg/mol units with lower heating value in MJ/Kg and number of moles. Finally the value obtained is converted into J. By this way, the LHV of CH4 is found as 802123 J.The balanced equation here is:
`CH_4 + 2(O_2 + 3.76N_2) = (1)CO_2 + (2)H_2O + (7.52)N_2`
The code for the program is linked below:
https://drive.google.com/file/d/105cArIAykxKMGAH4iwF1nGSQ8wSQjZLF/view?usp=sharing
From the program the flame temperature of CH4 at constant pressure with 35percent heat loss is found to be: 1678.7345891127416K.
Similarly the same setup is utilised for finding the flame temperature under constant pressure with 35 percent heat loss for C2H6, C2H4 and C2H2.
The LHV for one mole of C2H6 :1437346 J
The LHV for one mole of C2H4 :1323819.75 J
The LHV for one mole of C2H2 : 1299396 J
The balanced equation for C2H6 is,
`C_2H_6 + 3.5(O_2 + 3.76N_2) = (2)CO_2 + (3)H_2O + (13.16)N_2`
The balanced equation for C2H4 is,
`C_2H_4 + 3(O_2 + 3.76N_2) = (2)CO_2 + (2)H_2O + (11.28)N_2`
The balanced equation for C2H2 is,
`C_2H_2 + 2.5(O_2 + 3.76N_2) = (2)CO_2 + (1)H_2O + (9.4)N_2`
The link for the program is attached below:
https://drive.google.com/file/d/1aPZupQj72SEl4qDcYql-xLE0QuY3fwKR/view?usp=sharing
From the program the flame temperatures were obtained and the bar plot representing the flame temperature is attached below:
From the above plot we can find that the flame temperature of the C2H2 is the highest, next is C2H4 followed by C2H6, this is because number of moles of products formed increases as we move from alkynes to alkenes to alkanes, the products formed abosrb the heat released reducing the flame temperature especially water which absorbs heat to evaporate.
Now, under constant pressure with zero percent heat loss, the flame temperature(i.e., adiabatic flame temperature) is found out for CH4, C2H6 and C3H8.
The balanced equation for CH4 is:
`CH_4 + 2(O_2 + 3.76N_2) = (1)CO_2 + (2)H_2O + (7.52)N_2`
The balanced equation for C2H6 is:
`C_2H_6 + 3.5(O_2 + 3.76N_2) = (2)CO_2 + (3)H_2O + (13.16)N_2`
The balanced equation for C3H8 is:
`C_3H_8 + 5(O_2 + 3.76N_2) = (3)CO_2 + (4)H_2O + (18.8)N_2`
The program can be found in the link below:
https://drive.google.com/file/d/1jkA6yXHeeBi6tVKPw-yYO59o_QI476Mg/view?usp=sharing
The result obtained is attached below:
Here as we move from CH4 to C2H6 to C3H8 the mass increases resulting in more heat release. Although the number of products formed increases, amount of heat release increase is more which results in higher adiabatic flame temperature.