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RT assumes the use of 120-degree steam for interwall heat exchange of the material. Under negative pressure conditions: In Case 1, the pressure is 20 kp and the boiling point of the material is 58 degrees; in Case 4, the pressure is 55 kp and the boiling point is 70 degrees. In the case with a higher pressure, more 120-degree steam is consumed per unit of time, while this condition results in a better effect of evaporation and concentration of the material. Determine this and provide the reasoning
Is this meant to test how the sensible heat and latent heat of your material change with pressure?
Theoretically, the greater the temperature difference, the greater the thermodynamic drive, and thus the faster the heat transfer~
But the temperature is high, and the air temperature is also high – doesn’t that mean not much heat is carried away?
Reduced pressure facilitates distillation. The purpose of installing a vacuum pump in the final stage of multi-effect evaporation is to lower the pressure of the material in order to enhance evaporation. An operating pressure of 20 kPa results in less steam consumption and better concentration effects. Meanwhile, if the material being processed is volatile, the evaporation loss of this material will be greater at 20 kPa compared to 45 kPa
Big brother! Are you also working on MVR projects?
Sure, the greater the temperature difference, the easier it is to exchange heat
This post was last edited by Organometal on 2015-6-11 09:34. However, in a boiling state, the enthalpy of phase change should be taken into consideration. The greater the u+pv pressure, the more heat is required
It is indeed a two-effect evaporator. I know that a large temperature difference leads to better heat transfer, but in all cases it’s under boiling conditions with high pressure. Since the atmospheric temperature is also high, shouldn’t the amount of heat removed through evaporation be greater? Wouldn’t that result in more steam being consumed?