Le Calculateur du taux de ventilation du dégazeur helps determine the amount of non-condensable gases, primarily oxygène et dioxyde de carbone, that must be vented from feedwater in a deaerator system. Deaerators are used in power plants, boiler systems, and industrial steam applications to remove dissolved gases from feedwater to prevent corrosion and improve efficiency.
En calculant le vent rate, les opérateurs peuvent :
- Ensure the deaerator is functioning correctly.
- Optimize oxygen removal efficiency.
- Reduce corrosion-related maintenance costs.
- Améliorer le longévité and efficiency of boilers and steam systems.
Formula of Deaerator Vent Rate Calculator
Le vent rate is determined by the amount of non-condensable gases removed from the feedwater using the following formula:
Vent Rate (kg/hr) = (Feedwater Flow × Oxygen Content) / Desired Oxygen Removal Efficiency
Où :
- Feedwater Flow (kg/hr or lb/hr) = Le débit massique of the incoming water.
- Oxygen Content (kg/kg or ppm) = The dissolved oxygen content in the feedwater before deaeration.
- Desired Oxygen Removal Efficiency (%) = The fraction of oxygen to be remove, typically expressed as a percentage.
This formula helps determine how much gas needs to be vented to achieve the desired deaeration level in a given system.
Deaerator Vent Rate Reference Table
Voici une table de référence for common deaeration scenarios:
| Feedwater Flow (kg/hr) | Initial Oxygen Content (ppm) | Desired Removal Efficiency (%) | Required Vent Rate (kg/hr) |
|---|---|---|---|
| 10,000 | 8 | 99.5% | 0.04 |
| 20,000 | 6 | 99.0% | 0.12 |
| 30,000 | 7 | 99.5% | 0.11 |
| 50,000 | 5 | 98.0% | 0.25 |
| 100,000 | 4 | 99.8% | 0.08 |
Ce tableau fournit estimated vent rates to help engineers and boiler operators évaluer rapidement deaerator performance.
Example of Deaerator Vent Rate Calculator
disons un centrale is deaerating 50,000 kg/hr of feedwaterainsi que, initial oxygen content dans l'eau est 6 ppm (parties par million). La cible oxygen removal efficiency is 99.5%.
- Convertir ppm to kg/kg:
- Depuis que 1 ppm = 1 mg/L,
- 6 ppm = 6 mg/kg = 0.000006 kg/kg.
- Apply the formula:Vent Rate = (50,000 kg/hr × 0.000006 kg/kg) / 0.995
Vent Rate = (0.3 kg/hr) / 0.995
Vent Rate ≈ 0.302 kg / h
Cela signifie que le deaerator must vent 0.302 kg/hr of gases pour atteindre le desired oxygen removal efficiency.
FAQ les plus courantes
Venting is essential to remove dissolved oxygen and carbon dioxide from feedwater, which helps prévenir la corrosion in boilers and steam turbines.
If the vent rate is too low, oxygen may not be fully remove, menant à corrosion in the steam system and reducing boiler lifespan.
Yes, excessive venting can waste energy and steam, reducing the efficiency of the deaerator system and increasing operating costs.
