Horizontal balance calculation for the ammonia-based flue gas desulfurization system in boilers
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The components of the flue gas are 77.3% nitrogen, 14.2% carbon dioxide, 3.42% water vapor, 5% oxygen, and 0.08% sulfur dioxide. The inlet flue gas temperature is 110 degrees, while the outlet temperature is 55 degrees. How can the amount of water that needs to be added to the system be calculated? It’s best to include the formulas, thank you.
If only a simple calculation is performed, ignoring factors such as thermal equilibrium and the ammonia addition reaction, that is, the dry basis amount in the flue gas remains unchanged, and the emitted flue gas is at saturated humidity. At 55°C, the vapor pressure of water is 121 mmHg, which accounts for about 15.92% of the gas; thus, the dry basis constitutes 84.08%. It should not be difficult to proceed with further calculations from there.
Please write out the calculation process; I don’t understand it. Thank you
What is the volumetric flow rate of the smoke gas? Unit: Nm3/h
This post was last edited by jlwuli on 2015-7-4 23:00. Assuming the flue gas flow rate is 100,000 Nm3/h and the moisture content is 3.42%, the water vapor volume is 3,420 Nm3/h; thus, the dry basis flow rate is 96,580 Nm3/h. At 55°C, the vapor pressure of water is 121 mmHg, accounting for approximately 15.92% of the gases; thus, the dry basis constitutes 84.08%, and the volume of flue gas emitted is 114,870 Nm3/h. The water vapor content in the flue gas is 15.92%, which corresponds to 18290 Nm3/h. The amount of water that needs to be added is 18290 – 3420 = 14870 Nm3/h. Dividing this by 22.4 gives 663.7 kmol/h. Since the molecular weight of water is 18, the amount of water required is 663.7 * 18 = 11936 kg/h, or 11.94 tons/h
Can it be calculated using Avogadro’s law, P1/P2=N1/N2? By using the enthalpy values of the flue gas components, the result is 6.98 tons.
Wasn’t the amount of flue gas calculated earlier based on Avogadro’s law? And the molar volume under standard conditions was used later
Why is the vapor pressure of water at 55°C 121 mmHg, accounting for about 15.92% of the gas? How did it get here?
The vapor pressure of water at 55°C can be found in data sheets. According to the textbook \"Principles of Chemical Engineering\" that I studied before, it is 121 mmHg. The standard atmospheric pressure is 760 mmHg; therefore, at saturation at 55°C, the moisture content in the air is = 121/760*100% = 15.92%