Effect of air cooler design temperature on heat exchange area
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Friends, in areas with high climate temperatures, the design temperature of air coolers should be increased. Why does a higher design temperature require a larger heat exchange area for the air cooler? As the design temperature increases, which parameters change.....
As the temperature rises, the temperature difference for heat transfer certainly decreases, resulting in a reduced driving force; under the same amount of heat, the heat exchange area must increase.
The higher the design temperature, the larger the heat exchange area of the air cooler must be. Naturally, the temperature difference for heat transfer decreases, and the driving force diminishes; under the same amount of heat, the heat exchange area must be larger.
I am calculating an air cooler for condensing water vapor. When water vapor at 60 degrees condenses and the design temperature changes from 33°C to 35°C, the heat transfer temperature difference does not increase significantly. . Why, in practical applications, a rise of two degrees in temperature results in an area increase of more than double.
Hello, senior: The hot fluid is 60°C steam undergoing condensation. The design temperature T1 is 33°C, with an exit temperature of 50°C, giving a logarithmic temperature difference of 17.11°C. The design temperature T2 is 35°C, also with an exit temperature of 50°C, resulting in a logarithmic temperature difference of 16.37°C. Thus, an increase of two degrees in temperature leads to only a slight increase in the temperature difference. But in reality, the heat transfer area changes by a lot. WHY?
Hehe, you should look into the theory of heat transfer calculations; I can’t remember the details exactly, as I’m used to using software for such things. In simple terms, the closer the temperature between the hot fluid and the cold fluid is, the greater the impact of temperature changes in the cold fluid on heat transfer.
:L:L:L I seek your in-depth guidance, master!
Study it properly* – the calculation formulas
:'(:'(:'(Could it be that there’s an error in my formula... Could you please explain this to me, master? How about I send you a private message?