Formula
Simple oxygen method (complete combustion assumed):
\(EA_{O_2} = \frac{y_{O_2}}{0.2095 - y_{O_2}}\)
ASME PTC 4 nitrogen balance with CO correction:
\(y_{N_2} = 1 - y_{O_2} - y_{CO_2} - y_{CO} \qquad EA_{PTC4} = \frac{y_{O_2} - \tfrac{1}{2} y_{CO}}{\tfrac{0.2095}{0.7905}\, y_{N_2} - \left(y_{O_2} - \tfrac{1}{2} y_{CO}\right)} \qquad \lambda = 1 + EA_{PTC4}\)
Variables
| Symbol | Meaning | SI unit |
|---|---|---|
| \(y_{O_2}\) | oxygen in dry flue gas, volume fraction | – |
| \(y_{CO_2}\) | carbon dioxide in dry flue gas, volume fraction | – |
| \(y_{CO}\) | carbon monoxide in dry flue gas, volume fraction (1 ppm = 10⁻⁶) | – |
| \(y_{N_2}\) | nitrogen (with argon) in dry flue gas, by difference | – |
| \(EA_{O_2}\) | excess air fraction by the simple method | – |
| \(EA_{PTC4}\) | excess air fraction by the PTC 4 nitrogen balance | – |
| \(\lambda\) | air ratio, actual air over stoichiometric air | – |
Method and standards
Air is taken as 20.95 % O₂ and 79.05 % atmospheric nitrogen (N₂ + Ar) by volume, as in ASME PTC 4-2013. The nitrogen in the flue gas all came with the air (fuel nitrogen is neglected), so the air supplied is y_N₂ / 0.7905 and the oxygen that came with it is 0.2095/0.7905 · y_N₂. The oxygen left over, after crediting half a mole of O₂ per mole of CO that did not burn, divided by the oxygen consumed gives the excess air.
Assumptions and limits
- Dry-basis readings; convert wet readings first.
- Fuel nitrogen is neglected in the balance (it adds to the "air" nitrogen; the error is below 0.5 % EA for coal, nil for gas and oil).
- Undefined when O₂ reaches 20.95 % or when the readings leave no nitrogen.
References
- ASME PTC 4-2013 §5.11.4
- U.S. Standard Atmosphere (20.95 % O₂)
Questions and answers
Why does 3 % O₂ mean about 17 % excess air?
Air is 20.95 % oxygen, so 3 % O₂ left in the dry flue gas means 3/(20.95 − 3) ≈ 0.167 of the stoichiometric air was surplus. Nitrogen dominates the gas, so a small O₂ change is a large excess-air change.
When do the two methods differ?
When CO is significant or the fuel carries much nitrogen or CO₂. The PTC 4 balance credits the oxygen that would have burned the CO and uses the measured nitrogen; the simple formula assumes complete combustion.
Is the reading wet or dry basis?
Most analysers condition the sample and read dry. If yours reads wet, convert first: y_dry = y_wet / (1 − y_H₂O).