Combustions

This session discusses the thermodynamics of combustion.

In combustion chambers operating in an open system, it is generally assumed that the outlet pressure equals the inlet pressure. If you want to refine the analysis, you can consider that there is a small pressure drop, but it may be neglected in first approximation.

It follows that the reference process is the isobar.

Boilers and combustion chambers in which combustion takes place baing fixed, the useful work put in is zero.

Moreover, at the entrance and exit of these components, the velocities of fluids are always relatively low, so that the kinetic energies are negligible.

Under these conditions the first principle writes simply for combustion chambers: Δh = Q

Calculate the heat brought into play in these components is to determine the enthalpy change of the fluid flowing through them.

Course reference:

• “Technological Components / Combustion”

To follow the presentation, go to next step

• textbook: combustion (pdf, 47 Kb)
• textbook: boilers (pdf, 86 Kb)

(Session realized on 06/16/11 by Renaud Gicquel)

COMBUSTION THERMODYNAMICS

Component	molar fraction %	molar mass kg/kmole	mass fraction %
N2	0.781	28	0.756
Ar + CO2	0.009	40	0.012
O2	0.210	32	0.232
N2 atmosph.	0.790	28.15	0.768

COMBUSTION THERMODYNAMICS

• Stoechiometry

$C{H}_a+\left(1+\frac{a}{4}\right)\left(O_2+\mathrm{3,76}{N}_2\right)\to C{O}_2+\frac{a}{2}{H}_{2}O+\mathrm{3,76}\left(1+\frac{a}{4}\right)N_2$

• highest combustion temperature

COMBUSTION THERMODYNAMICS

• Non stoichiometric
• excess air: e
• richness: R = 1 / λ
• air factor: λ = 1 + e

• λ > 1

  $C{H}_a+\lambda \left(1+\frac{a}{4}\right)\left(O_2+\mathrm{3,76}{N}_2\right)\to C{O}_2+\frac{a}{2}{H}_{2}O+\left(\lambda -1\right)\left(1+\frac{a}{4}\right)O_2+\mathrm{3,76}\lambda \left(1+\frac{a}{4}\right)N_2$

COMBUSTION THERMODYNAMICS

• Incomplete combustion

$C{H}_a+\lambda \left(1+\frac{a}{4}\right)\left(O_2+\mathrm{3,76}{N}_2\right)\to \left(1-k_1\right)C{O}_2+k_{1}CO+\left(1-k_2\right)\frac{a}{2}{H}_{2}O+k_2\frac{a}{2}{H}_2+\left(\left(\lambda -1\right)\left(1+\frac{a}{4}\right)+\frac{k_1}{2}+k_2\frac{a}{4}\right)O_2+\mathrm{3,76}\lambda \left(1+\frac{a}{4}\right)N_2$

COMBUSTION THERMODYNAMICS

• CO₂ dissociation

CO₂ + H₂ ↔ CO + H₂O

$K_p=\frac{\left[CO\right]\left[H_{2}O\right]}{\left[C{O}_2\right]\left[H_2\right]}=\frac{k_1\left(1-k_2\right)\frac{a}{2}}{\left(1-k_1\right)k_2\frac{a}{2}}=\frac{k_1\left(1-k_2\right)}{\left(1-k_1\right)k_2}$

COMBUSTION THERMODYNAMICS

• CO₂ dissociation

$K_p=exp\left(\mathrm{3,387}-\frac{3753}{T}\right)$

• quenching temperature:
• temperature to be used in the law of mass action to get the actual exhaust gas composition

Combustion calculation steps

• analyze the compositions of the fuel and oxidizer
• Set λ, T_quench and the CO₂dissociation rate
• determine the composition of products
• calculate the heat of reaction involved
• calculate the end of combustion temperature

COMBUSTION THERMODYNAMICS

• Higher and Lower Heating Values
• Higher Heating Value (HHV) water contained in the fume is liquefied
• Lower Heating Value (LHV) water produced remains in the vapor state

COMBUSTION THERMODYNAMICS

• Heat of reaction
• ν₁ A₁ + ν₂ A₂ ↔ ν₃ A₃ + ν₄ A₄

• $-\Delta H_r={\nu }_1{h}_{\mathrm{f1}}+{\nu }_2{h}_{\mathrm{f2}}-{\nu }_3{h}_{\mathrm{f3}}-{\nu }_4{h}_{\mathrm{f4}}$(4.6.30)

  • h_f enthalpies of formation (1 bar, 25 °C)

    • (0 for pure substances)
• ΔH_r function of the ν_i and thus of the x_i

Variation of the heat of reaction in case of lack of air

Let us calculate for example the heat of reaction of octane for a stoichiometric reaction:

C₈H₁₈ + 12.5 O₂ + 3.76 x 12.5 N₂ →8 CO₂ + 9 H₂O + 3.76 x 12.5 N₂

• ΔH_r = h_fC8H18 + 12.5 h_fO2 - 8 h_fCO2 - 9 h_fH2O
• ΔH_r = - 208450 + 0 - 8 (-393520) - 9 (-285830) = 5 512 180 kJ/kmole

The molar mass of octane being 114.22 kg / kmol, its heat of reaction mass is equal to:

• Q_p = 48 259 kJ/kg

If the reaction is not stoichiometric, but occurs in the absence of air, the formation of carbon monoxide decreases the heat of reaction.

For richness equal to 1.25, for example, the reaction becomes:

C₈H₁₈ + 10 O₂ + 37.6 N₂ → 3 CO₂ + 5 CO + 9 H₂O + 37.6 N₂

• ΔH_r = h_fC8H18 + 10 h_fO2 - 3 h_fCO2- 5 h_fCO - 9 h_fH2O
• ΔH_r = - 208450 + 0 - 3 (-393520) - 5 (- 110530)- 9 (-285830) = 4 097 230 kJ/kmole

i.e. a decrease of 25.7% over the stoichiometric reaction.

COMBUSTION THERMODYNAMICS

• End of combustion temperature
• adiabatic flame temperature

• Basic equation:

  • the heat released by the combustion is used to heat the reaction products
  • maximal at stoichiometry
  • function of the air factor λ
  • quite complex calculation

COMBUSTION THERMODYNAMICS

• Adiabatic flame temperature

$h_r=h_{r_0}+\underset{T_0}{\overset{T_r}{\int }}{\mathrm{Cp}}_r\left(t\right)dt$

$-\Delta H_r=h_{r_0}-h_{p_0}$

$h_p=h_{p_0}+\underset{T_0}{\overset{T_{\mathrm{ad}}}{\int }}\mathrm{Cp}\left(t\right)dt=h_r$

COMBUSTION THERMODYNAMICS

• Calculation of combustion in Thermoptim
• either in the form CHa (enter the values of a and hf0 relative to the formulation Cha)
• or as a compound gas (chemical formulas are decoded)

COMBUSTION THERMODYNAMICS

Combustions

This session discusses the thermodynamics of combustion.

We have shown that:

the reference process is the isobar

the stoichiometry corresponds to complete combustion without excess air

the air factor lambda characterizes the deviation from stoichiometry

taking into account unburnt hydrocarbons can be done by considering the CO₂dissociation

Combustion process setting

Setting this process being a bit more complex than others, we suggest you refer to the Thermoptim documentation for further explanations.

The downloadable document below shows the information provided orally with the previous screen.

• Combustion calculation options (pdf, 32 Kb)