Managed to find the humidity ratio

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    Humidity Ratio
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To determine the specific enthalpy of moist air with given conditions (760 mmHg, 80% relative humidity, and 30 degrees C), one can use a psychometric chart to find the corresponding specific enthalpy, which is 85.25 kJ/kg. The calculation involves using the equation h = h_air + ωh_water, where ω is the humidity ratio. Although the enthalpy value may not appear in standard steam tables for the specific temperature and pressure, the psychometric chart provides the necessary data. This method effectively accounts for the contributions of both dry air and water vapor in the moist air mixture. Understanding these principles is essential for accurate thermodynamic calculations in engineering applications.
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Moist air has these values; 760mmHg, 80%, 30 degrees C. I have managed to find the humidity ratio (21,65 g/kg), but how can I determine the specific enthalpy?
 
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TSN79 said:
Moist air has these values; 760mmHg, 80%, 30 degrees C. I have managed to find the humidity ratio (21,65 g/kg), but how can I determine the specific enthalpy?

Can you assume the water vapor is at saturation, then look up the enthalpy in a steam table for that temp/press combination?
 
No, the answer should be 85.25 kJ/kg, and this value is not in the table for this temp/pressure comb., so there must be some equation in use to figure it out...I guess.
 
Use a psychometric chart for water in air (One is presented in " Basic Principles and Calculations in Chemical Engineering" -David Himmelblau ,Fourth Edition- Prentice Hall, Inc. ) and intercept temperature ( 30 C) and relative humidity ( 80 %). Then you read the corresponding specific enthalpy 85. 25 kJ/kg.
 
What about if you prove with:

h=h_{air}+\omega h_{water}

where \omega is the humidity ratio?
 
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