Engineering guide
The VGAS module performs the complete combustion calculation for fuel gases. From the volumetric composition of the fuel gas — up to 10 components such as methane, ethane, hydrogen, carbon monoxide, carbon dioxide or nitrogen — the program sets up the combustion equations automatically and determines the minimum oxygen requirement, the minimum air requirement, and the minimum wet and dry flue gas quantities for stoichiometric combustion. Using the excess air ratio λ, the actual air quantity supplied and the actual flue gas quantity are then determined.
Anyone who needs to run a combustion calculation for natural gas or other gas mixtures — for example when sizing burners, furnaces, waste heat boilers or flue gas ducts — obtains all the essential process quantities: flue gas composition, water loading of the flue gas (kg of water per kg of dry flue gas), and the dew point temperature of the flue gas at the given absolute pressure. The water dew point is decisive for avoiding condensation and thus corrosion in flue gas paths, stacks and waste heat boilers — or for deliberately inducing it in condensing-boiler applications.
The module covers complete combustion with excess air ratios λ ≥ 1. Incomplete combustion with λ < 1 (partial combustion, gasification) is outside the scope of the calculation.
Scope
- Combustion calculation of gases
Typical workflow
Specify the fuel gas composition
The volume fractions of up to 10 gas components are entered; the program checks that the fractions sum to 100 %.
Set up the combustion equations
For each combustible component, the stoichiometric reaction equation with oxygen is set up (e.g. CH₄ + 2 O₂ → CO₂ + 2 H₂O). This yields the minimum oxygen requirement O₂,min per standard cubic meter of fuel gas.
Minimum air requirement and minimum flue gas quantity
From the oxygen requirement and the oxygen volume fraction of the combustion air (21 % for dry air), the minimum air requirement L_min follows. Balancing the reaction products and the nitrogen carried in with the air gives the minimum wet and minimum dry flue gas quantities.
Actual quantities via the excess air ratio
With the chosen excess air ratio λ, the air quantity supplied and the actual (wet and dry) flue gas quantities are calculated; the excess air appears as an additional air fraction in the flue gas.
Water loading and dew point
From the water vapor fraction of the flue gas, the loading in kg of water per kg of dry flue gas is determined; from the water vapor partial pressure at the specified absolute pressure, the dew point temperature of the flue gas follows.
Frequently asked questions
Why is the flue gas dew point temperature so important?
If the wall or flue gas temperature drops below the water vapor dew point, water condenses in the flue gas path. Together with CO₂ and possibly sulfur oxides, corrosive condensates form that attack boilers, ducts and stacks. In condensing boilers, on the other hand, condensation is exploited deliberately — the components must then be designed to resist the condensate. Note that for sulfur-bearing fuels, the acid dew point lies well above the water dew point calculated here.
What does the excess air ratio λ mean, and which values are common in practice?
The excess air ratio λ is the ratio of the air quantity actually supplied to the stoichiometric minimum air requirement. λ = 1 corresponds to exactly stoichiometric combustion. In practice, combustion is run with excess air to ensure complete burnout: gas burners typically operate at λ ≈ 1.05 to 1.2. Higher excess air ratios increase the stack loss, since more flue gas mass has to be heated; at the same time they lower the dew point, because the water vapor is diluted.
Can the module also calculate sub-stoichiometric combustion (λ < 1)?
No. At λ < 1, incompletely oxidized products such as CO and H₂ are formed, whose fractions can only be determined via equilibrium or kinetics calculations. The module covers complete combustion with λ ≥ 1 only; gasification and partial combustion processes require different calculation methods.
What do the calculated air and flue gas quantities refer to?
The quantities are given as specific volumes at standard conditions, i.e. per standard cubic meter of fuel gas. To convert to operating conditions (operating pressure, operating temperature) — for example when sizing fans and duct cross-sections — the ideal gas law must be applied.
Worked example
For a natural gas with 90 vol% CH₄, 5 vol% C₂H₆, 3 vol% CO₂ and 2 vol% N₂, determine the minimum air requirement and the flue gas quantities at an excess air ratio λ = 1.2 (dry combustion air with 21 vol% O₂, all quantities in standard m³ per standard m³ of fuel gas) — a worked example of a combustion calculation for gases.
Given
| Methane CH₄ | 90 vol% |
|---|---|
| Ethane C₂H₆ | 5 vol% |
| Carbon dioxide CO₂ | 3 vol% |
| Nitrogen N₂ | 2 vol% |
| Excess air ratio λ | 1.2 |
| O₂ content of the air | 21 vol% |
Solution
Minimum oxygen and air requirement
Combustion equations: CH₄ + 2 O₂ → CO₂ + 2 H₂O and C₂H₆ + 3.5 O₂ → 2 CO₂ + 3 H₂O.
O2,min = 0.90 · 2 + 0.05 · 3.5 = 1.975 m³/m³
Lmin = O2,min / 0.21 = 1.975 / 0.21 = 9.405 m³/m³
Minimum flue gas quantity (stoichiometric)
CO₂ in the flue gas: 0.90 · 1 + 0.05 · 2 + 0.03 = 1.030 m³/m³
H₂O in the flue gas: 0.90 · 2 + 0.05 · 3 = 1.950 m³/m³
N₂ in the flue gas: 0.02 + 0.79 · 9.405 = 7.450 m³/m³
VFG,min (wet) = 1.030 + 1.950 + 7.450 = 10.430 m³/m³
VFG,min,dry (dry) = 10.430 − 1.950 = 8.480 m³/m³
Actual quantities at λ = 1.2
Air quantity supplied: L = λ · Lmin = 1.2 · 9.405 = 11.286 m³/m³
Excess air: (λ − 1) · Lmin = 0.2 · 9.405 = 1.881 m³/m³
Actual flue gas quantity (wet): 10.430 + 1.881 = 12.31 m³/m³
Dry flue gas quantity: 8.480 + 1.881 = 10.36 m³/m³
Result
| Minimum oxygen requirement O₂,min | 1.975 m³/m³ |
|---|---|
| Minimum air requirement Lmin | 9.405 m³/m³ |
| Minimum flue gas quantity (wet) | 10.43 m³/m³ |
| Air quantity supplied (λ = 1.2) | 11.29 m³/m³ |
| Actual flue gas quantity (wet) | 12.31 m³/m³ |
| Dry flue gas quantity | 10.36 m³/m³ |
Technical content source: LV-Solved-Problems-Content 628d313247e7