I have a question for everyone: How much heat is required to vaporize 100 kg of ammonia at 20°C completely to a temperature above 100°C?

What is the ammonia concentration? Why not use liquid ammonia?

To ask this question, there needs to be pressure and concentration, you know

I’m sorry; it’s ammonia solution with a concentration of 20%, at atmospheric pressure

For denitration; it is necessary to vaporize ammonia water

When 1 kg of ammonia at 100°C turns into steam at the same temperature, the heat absorbed for vaporization is 539 calories, and this value depends on the temperature.

When 1 kg of ammonia at 100°C turns into steam at the same temperature, the heat absorbed for vaporization is 539 calories, and this value depends on the temperature.

Here are my calculations; please take them as a reference! Calculation of the heating energy required for the ammonia solution in the evaporator: Ammonia requirement = 100.00 kg/h. The weight percentage of ammonia is 20%, so the amount of ammonia = 20.00 kg/h, while the amount of water = 80.00 kg/h. The heating temperature ranges from 20 ℃ to 100 ℃. The boiling point of ammonia is 38 ℃, and that of water is 100 ℃. The specific heat capacity of liquid ammonia is 4.325 kJ/kg·K, and its vaporization heat is 1731 kJ/kg. The specific heat capacity of ammonia vapor is 2.5852 kJ/kg·K. The energy required to heat the ammonia solution to its boiling point is 7785 kJ/h. The energy needed to convert the ammonia solution into ammonia vapor is 34620 kJ/h. The energy required to heat the ammonia vapor to the desired temperature is 3206 kJ/h. The specific heat capacity of water is 4.184 kJ/kg·K, and the latent heat of vaporization of water is 2675 kJ/kg. The specific heat capacity of water vapor at 1.1 BarG is 2.05 kJ/kg·K. The energy required to heat water to its boiling point is 26778 kJ/h, and the energy needed for water to evaporate is 214000 kJ/h. The energy required to heat water vapor to the desired temperature is 0 kJ/h. The total heating energy required is 286388 kJ/h, which is equivalent to 68416 kcal/h, or 80 kW

The calculations upstairs are detailed but incorrect; they fail to take into account that the reaction of ammonia in water generates heat amounting to 4757 kJ/kg. Additionally, when denitration takes place, the ammonia is injected in gaseous form, so all the water should evaporate. Simplified algorithm: The heat absorbed during the dissociation of 20 kg of ammonia is 20*4757=95140 KJ/h. The heat required to vaporize 80 kg/h of water is 80*2250=180000 KJ/h. To raise the temperature of 20 kg of ammonia from 20 degrees to 100 degrees, 20*4.325*80=6920 KJ/h is needed. The heat required to raise the temperature of water is 80*4.18*80=26752 KJ/h. The total amount of heat is 308812 KJ/h, which corresponds to 85.8 KW. This is a rough algorithm; it ignores factors such as specific heat and enthalpy related to temperature changes. It serves as a reasonable baseline for heat calculation during design. The area of the heater calculated in this way should be increased by 25–40%, as room needs to be left for adjusting the amount of ammonia used.