5.4 Combustion Stoichiometry, Flue Gas Analysis & Thermochemistry

Combustion Stoichiometry, Flue Gas Analysis & Thermochemistry

Combustion is a rapid chemical reaction in which fuel oxidizes, releasing stored chemical bond energy as high-temperature thermal energy. On the NCEES PE Mechanical exam, combustion questions test stoichiometric balancing, theoretical and excess air requirements, mass-based Air-Fuel ($AF$) ratios, Orsat flue gas interpretations, heating values ($HHV$ vs $LHV$), and adiabatic flame temperatures.

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1. Fuel Classes & Atmospheric Air Composition

Power generation fuels encompass diverse physical forms and chemical compositions:

+-----------------------------------------------------------------------------------------+
|                           COMMON ENGINEERING FUEL TYPES                                 |
|                                                                                         |
|   GASEOUS FUELS:     Methane (CH4, natural gas ~90-95%), Propane (C3H8), Butane (C4H10) |
|   LIQUID FUELS:      Gasoline (model: Octane C8H18), Diesel/Fuel Oil (model: C12H23)   |
|   SOLID FUELS:       Coal (Anthracite, Bituminous, Lignite), Biomass, Refuse-Derived    |
+-----------------------------------------------------------------------------------------+

Atmospheric Combustion Air

Dry atmospheric air is composed of approximately $21\% \text{ O}_2$ and $79\% \text{ N}_2$ by volume (molar basis). For every mole of oxygen involved in combustion, there are:

$$\frac{0.79 \text{ kmol } \text{N}_2}{0.21 \text{ kmol } \text{O}_2} = 3.76 \frac{\text{kmol } \text{N}_2}{\text{kmol } \text{O}_2}$$ $$M_{air} = 0.21(32.00) + 0.79(28.013) = 28.97 \text{ kg/kmol} \approx 28.97 \text{ lbm/lbmol}$$

Nitrogen is assumed to act as an inert diatomic spectator gas in standard stoichiometric calculations, absorbing heat without reacting directly.

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2. Stoichiometric Balancing & Excess Air Calculations

Stoichiometric (Theoretical) Air is the exact minimum quantity of oxygen required to completely oxidize all carbon to carbon dioxide ($\text{CO}_2$), all hydrogen to water ($\text{H}_2\text{O}$), and all sulfur to sulfur dioxide ($\text{SO}_2$), leaving zero unburned fuel and zero free oxygen in the exhaust products.

+-----------------------------------------------------------------------------------------+
|                    GENERAL STOICHIOMETRIC HYDROCARBON BALANCING                         |
|                                                                                         |
|   For a generic hydrocarbon C_n H_m:                                                    |
|                                                                                         |
|   C_n H_m + a_th (O_2 + 3.76 N_2) ---> n CO_2 + (m/2) H_2O + 3.76 a_th N_2              |
|                                                                                         |
|   Balancing Oxygen:   a_th = n + m/4                                                    |
|                                                                                         |
|   For an oxygenated fuel C_n H_m O_p:                                                   |
|   a_th = n + m/4 - p/2                                                                  |
+-----------------------------------------------------------------------------------------+

Excess Air & Equivalence Ratio

Industrial boilers and gas turbines always operate with Excess Air to ensure complete combustion, avoid deadly carbon monoxide ($\text{CO}$) and soot formation, and moderate peak flame temperatures:

$$\text{Percent Excess Air} = \frac{\dot{m}_{air,actual} - \dot{m}_{air,stoich}}{\dot{m}_{air,stoich}} \times 100 = (\text{Percent Theoretical Air}) - 100$$ $$\text{Actual Moles of Air } a_{act} = \left(1 + \frac{\text{Percent Excess Air}}{100}\right) a_{th}$$

The Equivalence Ratio ($\phi$) relates the actual fuel-to-air ratio to the stoichiometric ratio:

$$\phi = \frac{(F/A)_{actual}}{(F/A)_{stoich}} = \frac{(A/F)_{stoich}}{(A/F)_{actual}}$$

• $\phi = 1.0$: Stoichiometric mixture
• $\phi < 1.0$: Fuel-lean (excess air, common in boilers $\sim 10-25\%$ and gas turbines $\sim 100-300\%$)
• $\phi > 1.0$: Fuel-rich (excess fuel, incomplete combustion, generates $\text{CO}$ and $\text{H}_2$)

Air-Fuel ($AF$) Ratio Formulations

$$(A/F)_{molar} = \frac{N_{air}}{N_{fuel}} = \frac{a_{act} (1 + 3.76)}{1.0} = 4.76 \cdot a_{act}$$ $$(A/F)_{mass} = \frac{m_{air}}{m_{fuel}} = \frac{N_{air} \cdot M_{air}}{N_{fuel} \cdot M_{fuel}} = \frac{4.76 \cdot a_{act} \cdot 28.97}{M_{fuel}}$$

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3. Flue Gas Analysis (Orsat) & Flue Acid Dew Point

An Orsat Gas Analyzer measures the volumetric (molar) concentrations of combustion products on a dry basis, because water vapor is intentionally condensed and drained from the sample line before chemical absorption.

+-----------------------------------------------------------------------------------------+
|                         DRY TO WET FLUE GAS CONVERSIONS                                 |
|                                                                                         |
|   Let the dry gas mole numbers be:  N_dry = N_CO2 + N_CO + N_O2 + N_N2                  |
|   Total combustion products:        N_total = N_dry + N_H2O                             |
|                                                                                         |
|   Wet Molar Fraction:   y_i,wet = N_i / N_total = y_i,dry * (1 - y_H2O,wet)             |
|   Water Vapor Fraction: y_H2O,wet = N_H2O / (N_dry + N_H2O)                             |
+-----------------------------------------------------------------------------------------+

Flue Gas Acid Dew Point Prevention

The partial pressure of water vapor in wet flue gas is:

$$P_v = y_{\text{H}_2\text{O},wet} \cdot P_{total}$$

The water dew point is the saturation temperature $T_{sat}(P_v)$. If exhaust gas cools below this temperature, liquid water condenses in the stack.

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4. Thermochemistry: Enthalpy of Formation & Heating Values

For a steady-flow reactive system with zero work and negligible kinetic/potential energy changes, the First Law is expressed as:

$$\dot{Q}_{out} = \sum_{\text{reactants}} \dot{N}_r \left(\bar{h}_f^\circ + \bar{h} - \bar{h}^\circ\right)_r - \sum_{\text{products}} \dot{N}_p \left(\bar{h}_f^\circ + \bar{h} - \bar{h}^\circ\right)_p$$

Where:

• $\bar{h}_f^\circ$ is the Standard Enthalpy of Formation at reference state ($25^\circ\text{C} / 298.15\text{ K}, 1\text{ atm}$). For stable elements ($\text{O}_2, \text{N}_2, \text{H}_2, \text{C}_{graphite}$), $\bar{h}_f^\circ = 0$.
• $(\bar{h} - \bar{h}^\circ)$ accounts for sensible enthalpy differences between state temperature $T$ and the $298.15\text{ K}$ reference.

+-----------------------------------------------------------------------------------------+
|                       STANDARD ENTHALPIES OF FORMATION (298 K)                          |
|                                                                                         |
|   Substance             Formula      Enthalpy of Formation (kJ/kmol)                    |
|   --------------------  ----------   ------------------------------                     |
|   Carbon Dioxide (g)    CO2          -393,520                                           |
|   Carbon Monoxide (g)   CO           -110,530                                           |
|   Water Vapor (g)       H2O(g)       -241,820                                           |
|   Water Liquid (l)      H2O(l)       -285,830                                           |
|   Methane (g)           CH4           -74,850                                           |
|   Propane (g)           C3H8         -103,850                                           |
|   n-Octane (l)          C8H18(l)     -249,950                                           |
+-----------------------------------------------------------------------------------------+

Higher Heating Value (HHV) vs Lower Heating Value (LHV)

The heat of combustion represents the enthalpy released when a fuel is completely burned at standard conditions:

       Higher Heating Value (HHV)                     Lower Heating Value (LHV)
   +--------------------------------+             +--------------------------------+
   | H2O in combustion products is  |             | H2O in combustion products is  |
   | fully condensed to LIQUID      |             | fully in VAPOR phase           |
   | (Releases latent heat h_fg)    |             | (Carries away latent heat)     |
   +--------------------------------+             +--------------------------------+

$$HHV = LHV + \left(\frac{m_{\text{H}_2\text{O}}}{m_{fuel}}\right) h_{fg,\text{H}_2\text{O}}$$

Where $h_{fg,\text{H}_2\text{O}}$ is the enthalpy of vaporization of water evaluated at the standard reference temperature of $25^\circ\text{C}$ ($2442.3 \text{ kJ/kg} = 1050 \text{ BTU/lbm} = 44,010 \text{ kJ/kmol } \text{H}_2\text{O}$).

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5. Adiabatic Flame Temperature ($T_{ad}$)

The Adiabatic Flame Temperature is the maximum theoretical temperature achieved during combustion when zero heat is lost to the environment ($\dot{Q} = 0$), zero shaft work is produced ($\dot{W} = 0$), and changes in kinetic and potential energy are negligible:

$$H_{products}(T_{ad}) = H_{reactants}(T_i)$$ $$\sum_{\text{products}} N_p \left[\bar{h}_f^\circ + \bar{h}(T_{ad}) - \bar{h}^\circ\right]_p = \sum_{\text{reactants}} N_r \left[\bar{h}_f^\circ + \bar{h}(T_i) - \bar{h}^\circ\right]_r$$

+-----------------------------------------------------------------------------------------+
|                    FACTORS INFLUENCING ADIABATIC FLAME TEMPERATURE                      |
|                                                                                         |
|   1. EXCESS AIR: Peak T_ad occurs at stoichiometric conditions (phi = 1.0). Excess air  |
|      acts as thermal ballast (inert mass absorbing heat), drastically reducing T_ad.    |
|                                                                                         |
|   2. AIR PREHEAT: Preheating combustion air in an air preheater (AH) directly increases |
|      reactants enthalpy H_reactants, driving T_ad significantly higher.                 |
|                                                                                         |
|   3. HIGH-TEMP DISSOCIATION: Above ~1800 K, endothermic chemical dissociation           |
|      (CO2 <--> CO + 0.5 O2, H2O <--> H2 + 0.5 O2) absorbs heat, clamping maximum T_ad.  |
+-----------------------------------------------------------------------------------------+

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6. Step-by-Step Worked Problem: Methane Combustion with 20% Excess Air

Problem: Industrial natural gas modeled as pure methane ($\text{CH}_4$) enters a boiler burner at $25^\circ\text{C}$ ($77^\circ\text{F}$) and $1\text{ atm}$ and is combusted with $20\%$ excess dry air entering at $25^\circ\text{C}$. Calculate:

1. The actual balanced chemical reaction equation
2. The actual mass-based Air-Fuel ratio ($(A/F)_{mass}$)
3. The percentage of carbon dioxide ($\text{CO}_2$) in the dry flue gas (Orsat basis)
4. The Higher and Lower Heating Values ($HHV$ and $LHV$ in $\text{kJ/kg}$)

+-----------------------------------------------------------------------------------------+
|                        METHANE COMBUSTION CALCULATION STEPS                             |
|                                                                                         |
|   STEP 1: Stoichiometric Reaction Balancing (CH4 + a_th*(O2 + 3.76 N2))                 |
|           Carbon Balance:   1 CO2  ==> n = 1                                            |
|           Hydrogen Balance: 4 H ==> 2 H2O ==> m = 4                                     |
|           Oxygen Balance:   a_th = n + m/4 = 1 + 4/4 = 2.0 kmol O2                      |
|           Theoretical:      CH4 + 2.0 (O2 + 3.76 N2) ---> CO2 + 2 H2O + 7.52 N2         |
|                                                                                         |
|   STEP 2: Actual Reaction with 20% Excess Air (a_act = 1.20 * 2.0 = 2.40 kmol O2)       |
|           CH4 + 2.40 (O2 + 3.76 N2) ---> 1 CO2 + 2 H2O + 0.40 O2 + 9.024 N2             |
|                                                                                         |
|   STEP 3: Actual Mass Air-Fuel Ratio (AF_mass)                                          |
|           M_fuel = 12.011 + 4*(1.008) = 16.043 kg/kmol                                  |
|           m_air = 2.40 * 4.76 kmol air * 28.97 kg/kmol = 330.95 kg air                  |
|           (A/F)_mass = 330.95 kg air / 16.043 kg fuel = 20.63 kg air / kg fuel          |
|           (Note: Stoichiometric (A/F)_stoich = 2.0 * 4.76 * 28.97 / 16.043 = 17.19)     |
|                                                                                         |
|   STEP 4: Dry Flue Gas Orsat Analysis (Exclude 2 moles of H2O)                          |
|           N_dry = N_CO2 + N_O2 + N_N2 = 1.0 + 0.40 + 9.024 = 10.424 kmol                |
|           % CO2 (dry) = (1.0 / 10.424) * 100% = 9.59%                                   |
|           % O2 (dry)  = (0.40 / 10.424) * 100% = 3.84%                                   |
|           % N2 (dry)  = (9.024 / 10.424) * 100% = 86.57%                                 |
|                                                                                         |
|   STEP 5: Heating Value Calculations (HHV & LHV)                                        |
|           LHV_molar = -Delta H_comb (H2O vapor)                                         |
|           LHV_molar = -[1*(-393520) + 2*(-241820) - 1*(-74850)]                         |
|           LHV_molar = -[-393520 - 483640 + 74850] = 802,310 kJ/kmol CH4                 |
|           LHV_mass  = 802,310 / 16.043 = 50,010 kJ/kg CH4                               |
|                                                                                         |
|           HHV_molar = -Delta H_comb (H2O liquid)                                        |
|           HHV_molar = -[1*(-393520) + 2*(-285830) - 1*(-74850)]                         |
|           HHV_molar = -[-393520 - 571660 + 74850] = 890,330 kJ/kmol CH4                 |
|           HHV_mass  = 890,330 / 16.043 = 55,496 kJ/kg CH4                               |
|                                                                                         |
|           Difference: HHV - LHV = 55,496 - 50,010 = 5,486 kJ/kg CH4                     |
|           (Matches: 2 kmol H2O * 44,010 kJ/kmol / 16.043 kg = 5,486 kJ/kg)              |
+-----------------------------------------------------------------------------------------+

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7. Common Exam Traps & PE Pro-Tips

• Trap 1 — Dry vs. Wet Flue Gas in Orsat Problems: Orsat gas analyzers do not measure water vapor because moisture is intentionally condensed before testing. When converting dry Orsat volume percentages to true wet stack compositions, always multiply by $(1 - y_{\text{H}_2\text{O},wet})$.
• Trap 2 — Nitrogen Multiplier in Combustion Air: Air contains $3.76\text{ kmol } \text{N}_2$ per $\text{kmol } \text{O}_2$, which corresponds to $4.76\text{ kmol air} / \text{kmol } \text{O}_2$. When computing air moles from oxygen moles, multiply $a_{act}$ by $4.76$, not $3.76$.
• Trap 3 — HHV vs LHV Water State: Remember that Higher Heating Value assumes water in the combustion products is fully condensed into liquid at $25^\circ\text{C}$ (recovering latent heat $h_{fg}$), while Lower Heating Value assumes water remains in the vapor phase.