I have a question regarding the cooling of wet air
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To take on a project, it is necessary to condense saturated humid air at 80 degrees into saturated humid air at 40 degrees, using circulating water for cooling. There is a question now: as moist air cools down, certain amounts of water vapor condense. When performing heat balance calculations, in addition to accounting for the heat loss resulting from the enthalpy of the dry air, is it necessary to also consider the heat loss that occurs due to the condensation of water vapor? I wonder if any expert knows.
Since the temperature drop process involves saturated states at two temperature points, the load associated with the temperature drop of moist air must include the load resulting from the condensation of water components at these two temperature points.
Of course. . . You can directly calculate h=1.01t+0.001d (2501+1.85t) using the formula for the enthalpy of moist air ； t is temperature, d is humidity content
Then I have a question: just as in calculating the heat required for incoming and outgoing gases during drying, it is necessary to divide this amount into two parts – one part is the heat resulting from the change in enthalpy of the gas at different humidity levels, and the other part is the heat required for the evaporation of water vapor. This feeling is the reverse process of what I mentioned
It feels a bit strange... It’s like when water is heated to 100 degrees, there’s latent heat involved; once it exceeds 100 degrees, the temperature difference is used for calculation – basically they are separate concepts. Let’s not talk about cooling for now... just focus on the saturated water vapor content beyond that temperature... condensation will naturally occur then... Intentional cooling must be due to the fact that the saturated water vapor content varies at different temperatures, so condensation occurs when a certain temperature is reached... That’s my opinion... but... I’m not sure... Just offering an idea