Evaporation issues caused by liquid ammonia leaks
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It is generally believed that when the flashing rate is >0.2, the unvaporized portion of the leaked liquid forms tiny droplets dispersed in the air, rather than forming a liquid pool that evaporates. If liquid ammonia is stored at high pressure and room temperature, its flash rate in the event of a leak is 0.183; only about 8% of the leaked liquid ammonia forms a pool, so this factor is generally not taken into consideration. However, if liquid ammonia is stored at low temperatures, a leak of liquid ammonia below its boiling point will not result in flashing. In the air or on a concrete surface, it absorbs heat and evaporates rapidly; due to this intense boiling, I believe that a large number of liquid droplets must also be dispersed in the air. How should the amount of the part that forms the liquid pool be considered? Dear experts, let’s discuss this. This post was last edited by sccj504 on 2008-2-26 06:57.]
No flashing occurs during storage at low temperatures; a liquid pool should form
Calculate using this formula: F = Cp * (T0 – Tb) / H. Where: Cp is the constant-pressure heat capacity of liquid ammonia, in KJ/KG·K； T0 – Temperature of liquid ammonia before leakage ； Tb-ambient temperature ； Vaporization heat of H-liquid ammonia, KJ/KG. When the flashing rate F is greater than 0.2, the unvaporized portion of the leaked liquid forms tiny droplets that are dispersed in the air, rather than forming a liquid pool for evaporation. At F=0.1, there is 50% flashing; use this to estimate the flashing rate for other values.
Question: In the above formula, should Tb represent the boiling point of the liquid at normal pressure, in K? , rather than the ambient temperature? Ask a question?