Thermodynamics filled in class notes_Part_45

# Thermodynamics filled in class notes_Part_45 - 3.4...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3.4. GAS-VAPOR MIXTURES 97 Now for the absolute humidity (or specific humidity), one has ω = 0 . 622 P v P a , (3.280) = 0 . 622 2 . 3675 kPa 97 . 62 kPa , (3.281) = . 0152 kg H 2 O kg dry air . (3.282) Now for the masses of air and water, one can use the partial pressures: m a = P a V RT M a , (3.283) = P a V R a T , (3.284) = (97 . 62 kPa ) ( 75 m 3 ) parenleftBig 8 . 314 kJ kmole K 28 . 97 kg kmole parenrightBig (298 K ) , (3.285) = 85 . 61 kg. (3.286) m v = P v V RT M v , (3.287) = P v V R v T , (3.288) = (2 . 3675 kPa ) ( 75 m 3 ) parenleftBig 8 . 314 kJ kmole K 18 . 015 kg kmole parenrightBig (298 K ) , (3.289) = 1 . 3 kg. (3.290) Also one could get m v from m v = ωm a , (3.291) = (0 . 0152)(85 . 61 kg ) , (3.292) = 1 . 3 kg. (3.293) Now the dew point is the saturation temperature at the partial pressure of the water vapor. With P v = 2 . 3675 kPa , the saturation tables give T dew point = 20 . 08 ◦ C. (3.294) 3.4.1 First law The first law can be applied to air-water mixtures.The first law can be applied to air-water mixtures....
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

Thermodynamics filled in class notes_Part_45 - 3.4...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online