The Deaerator Vent Rate Calculator helps determine the amount of non-condensable gases, primarily oxygen and carbon dioxide, 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.
By calculating the vent rate, operators can:
- Ensure the deaerator is functioning correctly.
- Optimize oxygen removal efficiency.
- Reduce corrosion-related maintenance costs.
- Improve the longevity and efficiency of boilers and steam systems.
Formula of Deaerator Vent Rate Calculator
The 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
Where:
- Feedwater Flow (kg/hr or lb/hr) = The mass flow rate 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
Below is a reference table 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 |
This table provides estimated vent rates to help engineers and boiler operators quickly assess deaerator performance.
Example of Deaerator Vent Rate Calculator
Let’s say a power plant is deaerating 50,000 kg/hr of feedwater, and the initial oxygen content in the water is 6 ppm (parts per million). The target oxygen removal efficiency is 99.5%.
- Convert ppm to kg/kg:
- Since 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/hr
This means the deaerator must vent 0.302 kg/hr of gases to achieve the desired oxygen removal efficiency.
Most Common FAQs
Venting is essential to remove dissolved oxygen and carbon dioxide from feedwater, which helps prevent corrosion in boilers and steam turbines.
If the vent rate is too low, oxygen may not be fully remove, leading to 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.