- #1
MexChemE
- 237
- 55
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:
[tex]y' = \frac{0.005 \ \textrm{lb water}}{1 \ \textrm{lb dry air}}[/tex]
Then I figured one can go from humidity ratio to specific humidity with the following conversion:
[tex]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}}[/tex]
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:
[tex]x' = \frac{0.5 \ \textrm{lb water}}{1 \ \textrm{lb dry pulp}}[/tex]
[tex]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}}[/tex]
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.
[tex]y' = \frac{0.005 \ \textrm{lb water}}{1 \ \textrm{lb dry air}}[/tex]
Then I figured one can go from humidity ratio to specific humidity with the following conversion:
[tex]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}}[/tex]
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:
[tex]x' = \frac{0.5 \ \textrm{lb water}}{1 \ \textrm{lb dry pulp}}[/tex]
[tex]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}}[/tex]
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.