Determining the Adiabatic Flame Temperature for Complete Combustion of Liquid Octane
Liquid octane at 25°C, 1 atm enters a well-insulated reactor and reacts with air entering at the same temperature and pressure. For steady-state operation and negligible effects of kinetic and potential energy, determine the temperature of the combustion products for complete combustion with (a) the theoretical amount of air, (b) 400% theoretical air.
Known Liquid octane and air, each at 25°C and 1 atm, burn completely within a well-insulated reactor operating at steady state.
Find Determine the temperature of the combustion products for (a) the theoretical amount of air and (b) 400% theoretical air.
Schematic and Given Data:
Engineering Model
1. The control volume indicated on the accompanying figure by a dashed line operates at steady state.
2. For the control volume, \dot{Q}_{ cv }=0, \dot{W}_{ cv }=0, and kinetic and potential effects are negligible.
3. The combustion air and the products of combustion each form ideal gas mixtures.
4. Combustion is complete.
5. Each mole of oxygen in the combustion air is accompanied by 3.76 moles of nitrogen, which is inert.
Analysis At steady state, the control volume energy rate balance Eq. 13.12b reduces with assumptions 2 and 3 to give Eq. 13.21a:
\frac{\dot{Q}_{ cv }}{\dot{n}_{ F }}-\frac{\dot{W}_{ cv }}{\dot{n}_{ F }}=\bar{h}_{ P }-\bar{h}_{ R } (13.12b)
\sum_{ P } n_{e} \bar{h}_{e}=\sum_{ R } n_{i} \bar{h}_{i} (13.21a)
\sum_{ P } n_{e} \bar{h}_{e}=\sum_{ R } n_{i} \bar{h}_{i} (1)
When Eq. 13.9 and table data are used to evaluate the enthalpy terms, Eq. (1) is written as
\bar{h}(T, p)=\bar{h}_{ f }^{\circ}+\left[\bar{h}(T, p)-\bar{h}\left(T_{ ref }, p_{ ref }\right)\right]=\bar{h}_{ f }^{\circ}+\Delta \bar{h} (13.9)
\sum_{ P } n_{e}\left(\bar{h}_{ f }^{\circ}+\Delta \bar{h}\right)_{e}=\sum_{ R } n_{i}\left(\bar{h}_{ f }^{\circ}+\Delta \bar{h}\right)_{i}
On rearrangement, this becomes
\sum_{ P } n_{e}(\Delta \bar{h})_{e}=\sum_{ R } n_{i}(\Delta \bar{h})_{i}+\sum_{ R } n_{i} \bar{h}_{ f i}^{\circ}-\sum_{ P } n_{e} \bar{h}_{ f e}^{\circ}
which corresponds to Eq. 13.21b. Since the reactants enter at 25°C, the (\Delta \bar{h})_{i} terms on the right side vanish, and the energy rate equation becomes
\sum_{ P } n_{e}(\Delta \bar{h})_{e}=\sum_{ R } n_{i} \bar{h}_{ f i}^{\circ}-\sum_{ P } n_{e} \bar{h}_{ f e}^{\circ} (2)
a. For combustion of liquid octane with the theoretical amount of air, the chemical equation is
C _{8} H _{18}( l )+12.5 O _{2}+47 N _{2} \rightarrow 8 CO _{2}+9 H _{2} O ( g )+47 N _{2}
Introducing the coefficients of this equation, Eq. (2) takes the form
8(\Delta \bar{h})_{ CO _{2}}+9(\Delta \bar{h})_{ H _{2} O ( g )}+47(\Delta \bar{h})_{ N _{2}}
=\left[\left(\bar{h}_{ f }^{\circ}\right)_{ C _{8} H _{18}(l)}+12.5\left(\bar{h}_{ f }^{\circ \nearrow0}\right)_{ O _{2}}+47\left(\bar{h}_{ f }^{\circ\nearrow0}\right)_{ N _{2}}\right]
-\left[8\left(\bar{h}_{ f }^{\circ}\right)_{ CO _{2}}+9\left(\bar{h}_{ f }^{\circ}\right)_{ H _{2} O ( g )}+47\left(\bar{h}_{ f }^{\circ \nearrow0}\right)_{ N _{2}}\right]
The right side of the above equation can be evaluated with enthalpy of formation data from Table A-25, giving
8(\Delta \bar{h})_{ CO _{2}}+9(\Delta \bar{h})_{ H _{2} O ( g )}+47(\Delta \bar{h})_{ N _{2}}=5,074,630 kJ/kmol(fuel)
Each \Delta \bar{h} term on the left side of this equation depends on the temperature of the products, T_{ P}. This temperature can be determined by an iterative procedure.
The following table gives a summary of the iterative procedure for three trial values of T_{ P}. Since the summation of the enthalpies of the products equals 5,074,630 kJ/kmol, the actual value of T_{ P} is in the interval from 2350 to 2400 K. Interpolation between these temperatures gives T_{ P }=2395 K.
| T _{ P } | 2500 K | 2400 K | 2350 K |
| 8(\Delta \bar{h})_{ CO _{2}} | 975,408 | 926,304 | 901,816 |
| 9(\Delta \bar{h})_{H_{2}O (g)} | 890,676 | 842,436 | 818,478 |
| 47(\Delta \bar{h})_{N_{2}} | 3,492,664 | 3,320,597 | 3,234,869 |
| \sum_{ P } n_{ e }(\Delta \bar{h})_{e} | 5,358,748 | 5,089,337 | 4,955,163 |
Alternative Solution
The following IT code can be used as an alternative to iteration with table data, where hN2_R and hN2_P denote the enthalpy of N _{2} in the reactants and products, respectively, and so on. In the Units menu, select temperature in K and amount of substance in moles.
Using the Solve button, the result is TP = 2394 K, which agrees closely with the result obtained above.
b. For complete combustion of liquid octane with 400% theoretical air, the chemical equation is
\begin{aligned}& C _{8} H _{18}( l )+50 O _{2}+188 N _{2} \rightarrow \\&\quad 8 CO _{2}+9 H _{2} O ( g )+37.5 O _{2}+188 N _{2}\end{aligned}
The energy rate balance, Eq. (2), reduces for this case to
\begin{aligned}&8(\Delta \bar{h})_{ CO _{2}}+9(\Delta \bar{h})_{ H _{2} O ( g )}+37.5(\Delta \bar{h})_{ O _{2}}+188(\Delta \bar{h})_{ N _{2}} \\&\quad=5,074,630 kJ / kmol \text { (fuel) }\end{aligned}
1 Observe that the right side has the same value as in part (a). Proceeding iteratively as above, the temperature of the products is T_{ P }=962 K. The use of IT to solve part (b) is left as an exercise.
1 The temperature determined in part (b) is considerably lower than the value found in part (a). This shows that once enough oxygen has been provided for complete combustion, bringing in more air dilutes the combustion products, lowering their temperature.
Skills Developed
Ability to…
• apply the control volume energy balance to calculate the adiabatic flame temperature.
• evaluate enthalpy values appropriately.
Quick Quiz
If octane gas entered instead of liquid octane, would the adiabatic flame temperature increase, decrease, or stay the same? Ans. Increase.
TR = 25 + 273.15 // K
// Evaluate reactant and product enthalpies,
hR and hP, respectively
hR = hC8H18 + 12.5 * hO2_R + 47 * hN2_R
hP = 8 * hCO2_P + 9 * hH2O_P + 47 * hN2_P
hC8H18 = −249910 // kj/kmol (value from Table
A-25)
hO2_R = h_T(“O2”,TR)
hN2_R = h_T(“N2”,TR)
hCO2_P = h_T(“CO2”,TP)
hH2O_P = h_T(“H2O”,TP)
hN2_P = h_T(“N2”,TP)
// Energy balance
hP = hR