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The humidity ratio of atmospheric air at 27 °C dry bulb temperature and 760 mm of Hg is 0.015 kg/kg of dry air. The vapour density (in kg/m3 of dry air) is?
  • a)
    0.0172
  • b)
    0.172
  • c)
    0.0192
  • d)
    0.154
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The humidity ratio of atmospheric air at 27 °C dry bulb temperature a...
0.015 =
⇒ Pv = 173896 mm of Hg
Vapour of density,
=
= 0.0172kg/m2 of d.a.
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Most Upvoted Answer
The humidity ratio of atmospheric air at 27 °C dry bulb temperature a...
Given data:
Dry bulb temperature (Tdb) = 27°C
Pressure (P) = 760 mmHg
Humidity ratio (W) = 0.015 kg/kg of dry air

We need to find the vapour density (ρv) in kg/m3 of dry air.

Formula:
The vapour density (ρv) is given by the following formula:

ρv = (W×P)/(R×Tdb)

where R is the gas constant of dry air = 0.287 kJ/kg-K

Calculation:
Substituting the given values in the above formula, we get:

ρv = (0.015×760)/(0.287×(27+273)) = 0.0172 kg/m3 of dry air

Therefore, the vapour density of atmospheric air at 27°C and 760 mmHg is 0.0172 kg/m3 of dry air.

Answer: (a) 0.0172
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The humidity ratio of atmospheric air at 27 °C dry bulb temperature and 760 mm of Hg is 0.015 kg/kg of dry air. The vapour density (in kg/m3 of dry air) is?a)0.0172b)0.172c)0.0192d)0.154Correct answer is option 'A'. Can you explain this answer?
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