Note: The second question on the afternoon of the second day
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It’s the second question in the afternoon’s case study on directly-cooled atmospheric condensers; even after finishing the exam, I’m still wondering what that device looks like and how it works I searched on Baidu but got no results. Does anyone who knows know anything? Thank you.
It must be a cooling tower, the kind of hyperbolic tower used in power plants
In fact, it has a structure similar to that of an absorption tower: steam enters at the bottom, while cooled water enters at the top. Some of the steam at the bottom inlet cools down as it rises; to maintain an operating pressure of 22 kPa, a small amount of air must be introduced. At the gas-phase outlet, the partial pressure of this introduced air is equal to the vapor pressure of the water discharged from the bottom, and this can be determined using the law of partial pressures.
Oh, thank you. It’s just hot air coming into direct contact with cold water
Thank you so much. It’s indeed a very rare type of equipment; I’ve never seen such a device in any refining plant before
Actually, I don’t know either; I just looked it up on Baidu. I thought it was a air cooler; I really have no idea what to think.
The chemical industry also uses cooling towers, but they require power. The hyperbolic cooling towers used in power plants rely on natural convection of air; no power is needed. However, in this context it seems to refer to another method of cooling the condensate water after the steam turbine in power plants – indirect cooling using air instead of circulating water. Then the circulating water is cooled by air as well. This is what I understand by direct cooling
Well, thanks. I still don’t quite understand it; it seems like this question is rather obscure. I haven’t grasped the principles behind the equipment, so I’m not sure how to do the calculations
Well, thanks. I still don’t quite understand it; it seems like this question is rather obscure. I haven’t grasped the principles behind the equipment, so I’m not sure how to do the calculations
Well, I gave up on this question too; I wondered why such an obscure topic was included in the exam. . .