Could someone help me calculate how much water can be pumped per hour?
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Today I asked the manufacturer about a piece of equipment, a vacuum dryer. I asked them how much water this vacuum dryer can remove per hour for our company’s use. The manufacturer said they were not sure; they only knew that at a residual pressure of 40 mbar, it can remove 1.3 million kcal of heat per hour. They didn’t know how to calculate how much water can be removed per hour, or what additional data would be needed to obtain an approximate figure
It takes 100 calories of energy for 1 gram of water to rise from 0°C to 100°C, while it takes 539 calories to turn 1 gram of water at 100°C into steam at the same temperature. The amount of water is 1300000/10000 = 130 grams (as I understand it). The amount of energy required varies depending on the temperature.
In this simple estimation, the general calculation approach is outlined on the second floor, namely sensible heat and latent heat. What I need to add here is that the operating conditions must not be overlooked; when the residual pressure is 40 mbar, there are corresponding values for the vaporization temperature of water and its latent heat. By taking this into account, such an estimation becomes simpler. :lol
Agree with the person above! :handshake
It seems that 1 calorie equals 1000 calories; is that not a difference of 1000 times?
It is known that the temperature of water corresponding to a vapor pressure of 40 mbar is 28.7 degrees Celsius, and the latent heat of vaporization is 2426.6 kJ/kg. Assuming the temperature of the incoming material is also 28.7, the amount of water evaporated = 1300000×4.18/2426.6 = 2239 kg/h. (Of course, a difference of 10 or 20 degrees makes little difference.) For an accurate calculation: (1300000×4.18 – C×m0×(t0–28.7)) = m×2426.6, where mo is the total amount of material supplied per hour in kg/h, C is the average specific heat of the material, t0 is the temperature of the material, and m is the amount of water that evaporates.
I’d like to ask about the 6th floor: the calculation process is flawless, but 1.3 million kcal of heat seems not to should be equated directly with the latent heat of vaporization. Based on the results, a vacuum device used for drying removes 2,239 kg of water per hour, which is an extremely high value!
Yes, I also think this value seems quite high (I believe that’s because the temperature of dry materials is usually higher than the temperature during drying, and moreover, the drying process is far from reaching equilibrium).
It is known that the temperature of water corresponding to a vapor pressure of 40 mbar is 28.7 degrees Celsius, and the latent heat of vaporization is 2426.6 kJ/kg. Assuming the temperature of the incoming material is also 28.7, the amount of water evaporated = 1300000×4.18/2426.6 = 2239 kg/h. (Of course, a difference of 10 or 20 degrees makes little difference.) For an accurate calculation: (1300000×4.18 – C×m0×(t0–28.7)) = m×2426.6, where mo is the total amount of material supplied per hour in kg/h, C is the average specific heat of the material, t0 is the temperature of the material, and m is the amount of water that evaporates. =========================================================== If it’s 2426.6 kJ/kg, then wouldn’t it all turn into steam? Well, it should be saturated water or a mixture of water and steam, I guess. . I’m not sure exactly how much it should be. Moreover, if the difference is 10 or 20 degrees, the difference should be quite significant. Although the vaporization latent heat is the same, the steam humidity must differ significantly. . I’m not really sure either, haha
Analyze each specific case individually; as long as the heat balance is correct, the results won’t differ much. The example above is only a very rough estimate; to carry out a precise calculation, one must use the formula provided below. That “2426.6 kJ/kg” is the latent heat of vaporization at 27.8 degrees Celsius. This vacuum drying process is one in which water evaporates under low pressure; generally, the heat consumed is primarily used for the evaporation of moisture.