How to calculate the volume ratio after the evaporation of 20% ammonia solution?

This post was last edited by Qingcheng Jun on 2016-6-7 at 14:56. The concept presented by the original poster is too vague; there’s a lot to learn. For 20% ammonia solution, the saturated pressure of liquid ammonia at 132°C is 11.1 Mpa, so the saturated pressure for 20% ammonia solution is 2.2 Mpa. If the container pressure does not exceed 2.2 Mpa, ammonia is completely vaporized at 150°C. Assuming the volume of the original ammonia solution is 1 L, and its density at that time is 0.92 g/ml, the mass of ammonia is 184 g. The amount of substance is 184/17 = 10.8 mol. The standard volume after vaporization is 10.8 x 22.4 = 242 L, representing a volume expansion of 242 times. Since gases are compressible, the actual volume depends on temperature and pressure, as calculated by the formula PV=nRT

This post was last edited by yggswl on 2016-6-7 at 15:33. 20% ammonia solution is at atmospheric pressure, not liquid ammonia? I mainly want to calculate what the individual volumes and their volume ratio will be when 20% ammonia solution completely vaporizes at high temperature

This post was last edited by yggswl on 2016-6-7 15:45. How is the calculation done for water vaporization? For example, still at 1L, the mass of water = 1000*0.92*80% = 736g; the amount of substance is 736/18 = 40.89 mol. The standard volume after vaporization is 40.89*22.4 = 915.9L, so the volume increases by a factor of 915.9? The volume of the mixed gas after complete vaporization under standard conditions is 242 + 915.9 = 1157.9 times the volume under standard conditions (at normal pressure, with no consideration for temperature changes). Is that correct? Is the volume ratio after complete vaporization 915.9:242? In the fully gasified mixed gas, what is the volume percentage of ammonia? Approximately 20%?

The calculation is correct, but the assumption does not hold. In practice, the degree of vaporization of ammonia and water in ammonia solution is different, and there are no conditions to maintain a proportional vaporization of these two substances. It’s pointless to get stuck over the question as posed; it won’t solve the real problem you have. You should just state what it is you actually want to do. Your foundational knowledge is too weak; the questions you raise are far from those that truly need to be addressed. If others spend too much time struggling with the same issues as you, they will inevitably get led astray by you

This post was last edited by yggswl on 2016-6-7 at 21:55. The question is: When 20% ammonia solution is atomized using compressed air and then evaporated with hot air, since the temperature of the hot air is above 200 degrees, both the water and ammonia gas in the ammonia solution are carried away by the hot air. There are two scenarios: 1. As proven through experiments, all of the ammonia vaporizes and is carried away by the hot air; for example, if the amount of ammonia is 500 liters, and it is required that the volume proportion of ammonia gas in the mixture not exceed 5%, then how many cubic meters of hot air are needed? In this case, all the water is carried away as water vapor along with ammonia. 2. If the temperature of the hot air is reduced so that it carries away only ammonia, with the volume ratio after mixing still not exceeding 5%, what is the required volume in cubic meters of hot air? In this case, the water in ammonia does not get carried away along with the ammonia gas; only the ammonia gas evaporates and is carried away by the air. @Qingcheng Jun

Is hot air introduced into water to evaporate the liquid? Is it the mixing of hot air with the atomized mixture that causes all the liquid droplets to vaporize? Introducing it into water is equivalent to combining evaporation and stripping, which enables better separation of ammonia from water. This model is quite demanding; modeling it is difficult, and hiring professionals for the modeling calculations comes at a high cost. It’s better to conduct experiments, as the experimental data are effective

The hot air mixes with the atomized mixture, causing all the liquid droplets to vaporize; with an adequate amount of hot air, all the droplets are vaporized

Also seeking a solution to this problem. . . . . I’ve noticed that some people in the company handle algorithms in a rather casual manner. . . Of course, there are also those who make calculation mistakes and end up failing. I’m currently thinking about preparing a calculation sheet for this ammonia evaporator as well

Dude, have you done this calculation? If you have, let’s exchange ideas

In fact, it’s quite simple to calculate using formulas: the safety factor for ammonia gas and hot air should be kept at less than 5%, and then the required amount of hot air can be determined. Generally, for ammonia evaporation in denitration processes, secondary hot air from boilers is used; electric heating can also be employed, and the appropriate option can be chosen by performing thermal calculations