On the enthalpy value of unsaturated water vapor
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Experts, heat exchange using municipal steam at 0.8 Mpa and 156°C. I checked the steam table; it’s in a unsaturated state. Is there a mixture of vapor and liquid inside? What should the enthalpy value be?
To determine the enthalpy of unsaturated steam, its dryness ratio also needs to be known. For example: steam pressure is 0.8 Mpa, dryness degree is 0.95. The enthalpy of saturated steam at a pressure of 0.8 Mpa, as indicated by the enthalpy-entropy table, is h1. At this pressure and temperature, the enthalpy of saturated water h2 is given by h = 0.95h1 + 0.05h2
Thank you for your answer. But my current conditions are such that I only have this parameter: the steam inlet pressure is 0.8 Mpa, the temperature is 156°C; the steam enters the heat exchanger where it undergoes heat exchange, and exits as condensed water at 90°C. I want to calculate the change in enthalpy for this.
Based on a gauge pressure of 8 BAR, the enthalpy value is 17,000 W; I’m not sure if this is correct. If you know the answer, let me know
It should be! This is also what troubles me: the pressure is 8 kilograms, while the temperature is 156°C, which indicates that the quality of the steam is relatively low; the actual enthalpy value is estimated to be only 2000 kJ/kg. However, the specific steam condition is unclear. You're talking about the ideal situation; in reality, it shouldn't be possible to achieve that