Calculation of Adiabatic Flame Temperature
Problem Definition:
1) Effect of equivalence ratio on Adiabatic Flame temperature (AFT) and comparing with Cantera for methane
2) Constant pressure chamber with Heat loss
- Calculation of AFT for Alkane, Alkene, Alkyne with the Heat loss of 35 percent
- Comparison of AFT with no of carbon atoms in Alkane assuming no heat loss
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General Stoichiometry equation:
`C_xH_y + a(O_2 + 3.76N_2) -> xCO_2 + (y/2)H_2O + 3.76aN_2`
`a = x + y/4`.
Calculation of Stoichiometric Air-Fuel Ratio
`(A/F)_"stoic"= (m_"air"/m_"fuel")_"stoic" = 4.76a (MW_"air")/(MW_"fuel")`
where,
MW is the molecular weight of the air and the fuel
As air contains 79 percent of Nitrogen and 21 percent of oxygen (`79/21 = 3.76`) which represents that for one mole of oxygen there are 3.76 moles of nitrogen that gives the total of 4.76.
Equivalence ratio(`phi`)
`phi = (A/F)/(A/F)_"stoic"`
`phi <1` is a lean mixture
`phi > 1` is a rich mixture which means that the fuel is more than the air
Constant Volume Adiabatic Flame Temperature:
`U_"reat"(T_"init",P_"Init") = U_"prod"(T_"ad",P_"f")`
`H_"react"-H_"prod" - R_u (N_"react"T_"init"-N_"prod"T_"ad") = 0`
where,
H - standardized enthalpy of reactants and products, which is the sum of enthalpy of formation at a standard reference temperature (298.15k) and the sensible enthalpy change form reference temperature to the adiabatic temperature
R - gas constant
N - Total no of moles in the Products and the reactants
To calculate the Adiabatic Flame Temperature with varying the Equivalence ratio, the generalized form of the equation for rich and lean mixture can be written as:
`C_xH_y +a(O_2+3.76N_2) -> bCO_2 +dH_2O+eH_2+fO_2+3.76N_2`
`a = (x+y/4)/phi`
`b = x`
Calculation of coefficients for Lean or stoichiometric mixture(`phi<=1`)
`c = 0`
`d = y/2`
`e =0`
`f = ((1-phi)/phi)(x+y/4)`
calculation of coefficients for rich mixture(`phi>1`)
`c =x-b`
`d=2a-b-x`
`e=-2a+b+x+y/2`
`f=0`
Calculation of Enthalpy:
NASA polynominals:
`H/RT = a1 + a2 T /2 + a3 T^2 /3 + a4 T^3 /4 + a5 T^4 /5 + a6/T`
where, a1,a2,a3,a4,a5,a6 are coefficient which is obtained through GRI mech and AFT is calculated through Newton-Raphson method
Results:
Comparison of the effect of equivalence ratio of methane
In this, the AFT(Adiabatic Flame Temperature) is calculated for the equivalence ratio from 0.25 to 2.0. The maximum temperature gained at the `phi = 1` is about 2800k approx. It is clearly seen that below and above`phi = 1` there is a change in temperature. For the case, `phi<1`, which means a fuel lean mixture where the all the fuel burnt successfully and remaining air is used for minimizing the temperature or it will reduce the maximum temperature. On Contrary to this when `phi > 1`which is a rich mixture and more fuel is present when compared to the stoichiometric ratio of air. So, as a result, there is no complete combustion of the fuel leads to less Adiabatic Flame temperature. In the graph, it is seen that the temperature decreases as the equivalence ratio of the fuel increase above `phi = 1`. Meanwhile, when `phi <1`, the AFT decreases with respect to decreasing of the equivalence ratio.
This graph is the calculation of Adiabatic Flame Temperature using Cantera library. The subtle difference between the two graphs is, in the former one Enthalpy is calculated only for the products such as `H_2O`,`CO_2`,`N_2`,`CO`,`H_2`,`O_2` but this is not the real case. As the atmospheric air consists of various elements which could cause more reaction, as a result, there is a decrease in the temperature. Cantera takes account of all the possible reaction that could take place. which is why there is a difference in the temperature between the two graphs.
Constant Pressure Adiabatic Flame Temperature
Calculation of Lower Heating Value
the lower heating value can be calculated by, `H_"react"-H_"prod"=0` at the standard temperature (T = 298.15k`
Calculation of AFT using heat loss = `H_"react"-H_"prod" = LHV*heat loss` `
where `H_" react"`at standard temperature (T = 298.15 k) and the `H_"prod"`is at adiabatic temperature by which using Newton Raphson method, we can find the adiabatic Flame Temperature.
Calculation of AFT for various alkanes, alkenes, alkynes:
Alkane -`C_"2n"H_"2n+2"`
Alkene - `C_"2n"H_"2n"`
Alkyne - `C_"2n"H_"2n-2"`
which are `C_2H_2,C_2H_4,C_2H_6`
The graph elucidates the relation between the alkyne, alkene, alkane. Coming to the type of bonds that the three species have are Alkane - it consists of a single bond or covalent bond between the two carbon atoms and between the carbon and hydrogen. Alkene consists of a double bond between the carbon atom and covalent bond between the carbon and hydrogen. On the other hand, Alkyne consists of a triple bond between the two carbon atoms. Comparing the strength of the bond or bond energy which is nothing but the average value of gas-phase dissociation energy. The triple bond which is seen in the alkyne is much stronger than the double bond and covalent bond, which tells that the triple bond has much bond strength or energy than the double bond and single bond. This substantiates the graph as the alkyne has a greater Adiabatic flame temperature. Meanwhile, the double bond is stronger than the covalent bond. This gives that the alkene has higher AFT compared to the alkane.
Calculation of AFT with respect to the no of carbon atoms in Alkane
The general form of Alkane:
`C_nH_"2n+2"`
so, the calculated alkanes are `CH_4`,`C_2H_6`,`C_3H_8`
The graph clearly shows that an increase in the carbon atom in alkanes will result in higher adiabatic flame temperature. As the alkanes have a covalent bond between the carbon and carbon, carbon and hydrogen. As number of carbon atoms increases the no of bond increases, as a result, this has a high bonding strength which increases the Adiabatic Flame Temperature.
python code :
https://drive.google.com/open?id=1o4cpeOH3ntyN6MHwYX-WqE9jca8t-xt_
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Refrences:
1) "An Introduction to Combustion -Concepts and Application", by Stephen R Turns.
2) "Fundamentals of Combustion", by D.P. Mishra.