The relationship between burner fuel consumption and temperature
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If the inlet air temperature of a 4-ton burner is increased (10 degrees), how much can 1 ton of fuel be saved? Is there any formula to calculate it?
I calculated it based on light oil, and the calorific value was calculated based on 42900KJ. Ten degrees can save about 0.0026%, which means one ton can save 2.6KG. The calculation is as follows. Energy is conserved, and the oil saved is the amount required to heat the air ten degrees. Check the enthalpy thermometer. For a temperature difference of 10°C, the enthalpy difference is approximately 13.5. The amount of air required per KG of oil is about 11.15Nm3/KG, so the heat reduction per KG of oil is 13.5 * 11.15=155.25KJ, divided by the calorific value of oil 42900KJ/KG, the fuel saving percentage is 0.036%. Calculated based on oil consumption 1T. Can save 1000 * 0.0036=3.6KG. The enthalpy difference in the temperature fluctuation range and the selection of excess air coefficient will affect the calculation. If you want to save energy, you need to consider 1: The cost of the burner split machine is very high. 2: Generally, the heat transfer oil of industrial boilers will be equipped with an air preheater (the exhaust gas temperature is high). The steam is not needed. The condenser is enough to save energy, and there is no need to change the split machine burner. 3: See the picture below
Because the copied pictures cannot be uploaded, haha
Can you explain in detail? I don't understand something. If the calorific value is 9300, how to put it into the formula? How did you calculate these values?