Pressure swing adsorption time issue
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Everyone, in pressure swing adsorption, I’ve never been clear about what exactly is meant by the adsorption time
This post was last edited by leo_0088 on 2015-7-28 08:52: the total equalization time multiplied by the number of towers in use for adsorption
Take a look at how your PSA operates; for example, in an 8-2-4 process, the adsorption time is 2(t1+t2)
The key is that I don’t understand why the adsorption time can be related to the pressure equalization time
The adsorption time of PSA refers to the time during the \"adsorption\" step; in all PSA systems, this adsorption time is the sum of all times such as equalization time and reverse discharge time. The 824 mentioned on the 3rd floor refers to 8 adsorption towers, with 2 towers undergoing equalization processing 4 times each; the adsorption time in this case is the sum of the time required for those 4 equalization processes and the time needed for reverse discharge.
It mainly depends on how many towers you have for absorption! ! ! It should be written in this operating procedure! The 8-2-3 process is generally 2 (T1+T2), while 10-2-4 is: T1+T2
The adsorption time refers to the period from the end of the final rise to the start of pressure equalization in a single tower!
I see that everyone else is talking about 8-2-4. So I’ll explain how to determine the adsorption process for this method. First, the complete PSA process involves adsorption, followed by four stages of pressure reduction, then forward operation, reverse operation, flushing, and finally four stages of pressure increase, followed by three stages of pressure increase, two stages of pressure increase, one stage of pressure increase, with the final increase in pressure corresponding to the hydrogen product. For 8-2-4, that’s all the steps. The adsorption time is 2(T1+T2). If you want to calculate the entire cycle of a PSA, it is 4 times 2(T1+T2). I’m not good at speaking, but I hope what I’ve said can be helpful to you.