Converting humidity ratio to specific humidity

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SUMMARY

The discussion centers on the conversion between humidity ratio and specific humidity in psychrometry. Humidity ratio is defined as the mass of water vapor per unit mass of dry air, while specific humidity is the mass of water vapor per unit mass of moist air. The user presents a calculation to convert humidity ratio to specific humidity, yielding a specific humidity value of 0.00498 lb water per lb moist air. Despite the user's confidence in their calculation, their professor disputes its validity, claiming that the conversion does not apply to the air-vapor system.

PREREQUISITES
  • Understanding of psychrometry and its terminology
  • Familiarity with humidity ratio and specific humidity definitions
  • Basic knowledge of mass balance calculations
  • Ability to interpret psychrometric charts
NEXT STEPS
  • Research the differences between humidity ratio and specific humidity in detail
  • Learn about psychrometric charts and their applications in HVAC
  • Explore the principles of material and energy balance in thermodynamics
  • Investigate authoritative sources on psychrometry for accurate definitions and calculations
USEFUL FOR

Students in thermodynamics, HVAC engineers, and professionals involved in environmental control systems who seek to deepen their understanding of humidity measurements and conversions.

MexChemE
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Hello PF! I have a very simple question on psychrometry. I had an argument with one of my professors regarding these two quantities, humidity ratio and specfic humidity. Humidity ratio is defined as mass of water vapor per unit mass of dry air. It's the one you can read from a psychrometric chart. Specific humidity is mass of water vapor per unit mass of moist air. Suppose you read a value from the chart of:
y' = \frac{0.005 \ \textrm{lb water}}{1 \ \textrm{lb dry air}}
Then I figured one can go from humidity ratio to specific humidity with the following conversion:
y = \frac{0.005 \ \textrm{lb water}}{1 \ \textrm{lb dry air} + 0.005 \ \textrm{lb water}} = \frac{0.00498 \ \textrm{lb water}}{1 \ \textrm{lb moist air}}
However, my professor said this was incorrect and gave me a bad grade in an exam, even though he once performed the following calculation in class:
x' = \frac{0.5 \ \textrm{lb water}}{1 \ \textrm{lb dry pulp}}
x = \frac{0.5 \ \textrm{lb water}}{1 \ \textrm{lb dry pulp} + 0.5 \ \textrm{lb water}} = \frac{0.33 \ \textrm{lb water}}{1 \ \textrm{lb moist pulp}}
However, he said this type of calculation did not apply with the air-vapor system, therefore my calculation is incorrect. I don't find that logical at all, what do you think?

Thanks in advance for any input!
P.S. I did not post this in order to keep arguing with my professor, it's just for the sake of my knowledge.
 
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From this link, it looks like you are right: http://www.daytonashrae.org/psychrometrics_hum.html. However, I have seen other links where specific humidity is defined as the same thing as humidity ratio.

Chet
 
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I knew there was nothing wrong with my calculation. Anyway, the grade doesn't reflect how much I've learnt. The class was Material and Energy Balance, btw. Thanks, Chet!
 

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