Managed to find the humidity ratio

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    Humidity Ratio
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SUMMARY

The discussion focuses on calculating specific enthalpy for moist air with given conditions: 760 mmHg, 80% relative humidity, and 30 degrees Celsius. The humidity ratio is established at 21.65 g/kg. To determine specific enthalpy, users are advised to utilize a psychometric chart and apply the equation h = hair + ωhwater, where ω represents the humidity ratio. The specific enthalpy value derived from these calculations is 85.25 kJ/kg.

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  • Understanding of psychometric charts and their applications
  • Familiarity with specific enthalpy calculations
  • Knowledge of humidity ratio and its significance
  • Basic principles of thermodynamics related to moist air
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  • Study the use of psychometric charts in HVAC applications
  • Learn about specific enthalpy calculations for various mixtures
  • Explore steam tables and their limitations in specific conditions
  • Investigate advanced thermodynamic equations for moist air
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Engineers, HVAC professionals, and students in thermodynamics or chemical engineering who are involved in calculating properties of moist air and optimizing thermal systems.

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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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