Combustion Efficiency and Excess Air Optimization

First, determine the current excess air: At 8% O2 dry, excess air ≈ 8 / (21 - 8) * 100 = 61.5%. At 3% O2, excess air ≈ 3 / (21 - 3) * 100 = 16.7%. The efficiency improvement from reducing excess air can be estimated using the rule of thumb that each 1% reduction in O2 (or roughly 5% excess air) improves efficiency by about 0.5–1%. More accurately, use the flue gas loss method: The dry flue gas loss is approximately proportional to the excess air and the temperature difference. The current flue gas loss can be calculated from the stack temperature and excess air. Using typical natural gas combustion data: At 8% O2, the flue gas loss is about 18% (for 240°C stack, 20°C ambient). At 3% O2, the loss would be about 12%. So reducing excess air could improve efficiency by about 6 percentage points, from 78% to 84% (HHV). However, this is an estimate; a more precise calculation would use the ASME PTC 4 method. For the economizer: Reducing stack temperature from 240°C to 150°C recovers sensible heat. The flue gas mass flow can be estimated from fuel input. First, calculate fuel input: Boiler output = 5 MW = 5,000 kW. At 78% efficiency (HHV), fuel input = 5,000 / 0.78 = 6,410 kW (HHV). Annual fuel energy = 6,410 kW * 6,000 h = 38,460,000 kWh. Annual fuel cost = 38,460,000 * $0.05 = $1,923,000. The economizer would recover heat from the flue gas. The flue gas loss reduction from cooling 240°C to 150°C: The specific heat of flue gas is about 1.05 kJ/kg·K. The flue gas mass flow: For natural gas, approximately 1.2 kg flue gas per kWh fuel (HHV) (or more accurately, use stoichiometry). Alternatively, use the rule of thumb: Each 20°C reduction in stack temperature improves efficiency by about 1%. So reducing by 90°C gives about 4.5% improvement. So efficiency becomes 78% + 4.5% = 82.5% (if excess air unchanged). But if both measures are implemented, the combined effect is not simply additive; reducing excess air also reduces flue gas mass, so the economizer benefit may be slightly less. However, for simplicity, we can estimate separately. Fuel savings from excess air reduction: Efficiency from 78% to 84% (6 points) means fuel input reduces to 5,000 / 0.84 = 5,952 kW, saving 458 kW. Annual savings = 458 kW * 6,000 h = 2,748,000 kWh, at $0.05 = $137,400. Fuel savings from economizer alone: Efficiency from 78% to 82.5% (4.5 points) gives fuel input = 5,000 / 0.825 = 6,060 kW, saving 350 kW, annual savings = 350 * 6,000 = 2,100,000 kWh, $105,000. If both are implemented, the combined efficiency might be around 88% (since reducing excess air also reduces the flue gas mass, making the economizer slightly less effective, but the overall savings are substantial). The simple payback for the economizer alone: Cost $150,000 / $105,000 per year = 1.43 years. With both measures, the payback is even better. However, the economizer cost may include installation; still, payback is attractive. Note: The economizer must be designed to avoid condensation of water vapor (acid dew point) by keeping the metal temperature above the dew point, typically by bypassing or using a corrosion-resistant material. With natural gas, the dew point is about 60°C, so 150°C stack is safe.

Additional measures: 1) Reduce scale: The fireside scale of 0.5 mm can cause significant efficiency loss (approximately 2–5% depending on thickness). Perform fireside cleaning and implement a regular inspection and cleaning schedule. Also, investigate the cause of scale (likely incomplete combustion or dust) and adjust the burner to minimize soot. 2) Optimize blowdown: The continuous blowdown of 5% may be excessive if water quality is good. Conduct a water analysis to determine the minimum blowdown required to maintain dissolved solids within limits. Reducing blowdown from 5% to 3% saves energy because blowdown water is at saturation temperature and its heat is lost. The energy savings can be estimated: Blowdown heat loss = blowdown rate * (enthalpy of saturated water at 8 barg - enthalpy of feedwater). At 8 barg, saturated water enthalpy is about 721 kJ/kg; feedwater at 105°C is about 440 kJ/kg. So each kg of blowdown loses 281 kJ. At 5% blowdown, blowdown flow = 0.05 * steam flow. Steam flow = 5,000 kW / (2,700 kJ/kg) ≈ 1.85 kg/s = 6,667 kg/h. Blowdown = 333 kg/h. Heat loss = 333 * 281 = 93,600 kJ/h = 26 kW. Reducing to 3% saves about 10 kW, or 60,000 kWh/year, $3,000/year. Also, consider installing a blowdown heat recovery system to preheat feedwater. 3) Improve feedwater temperature: The feedwater is at 105°C; if a deaerator is used, it typically operates at 105°C, which is already good. However, if there is no economizer, the feedwater could be heated by flue gas to higher temperature, but that is the economizer. 4) Reduce radiation and convection losses: Ensure the boiler insulation is in good condition; repair any damaged insulation. 5) Install an oxygen trim system to continuously optimize excess air, rather than a fixed damper. This can maintain low excess air despite load changes. 6) Implement a burner tuning and maintenance program: Tune the burner at least annually, and after any fuel or load change. The current burner was last tuned 2 years ago; a tune-up could improve combustion efficiency. 7) Consider adding a stack economizer (condensing) if the return water temperature is low enough, but for a steam boiler, a conventional economizer is more appropriate. 8) Use a blowdown controller to automate blowdown based on conductivity, ensuring optimal blowdown. 9) Train operators on efficiency best practices and monitor key parameters (stack temperature, O2, CO, etc.) regularly. 10) Conduct an energy audit to identify other opportunities, such as steam trap maintenance, condensate return, and insulation of steam lines. All these measures together can improve overall boiler efficiency by 10–15%.