Physics, asked by tanvirana4534, 1 year ago

a sample of moist air at a total pressure of 85kpa has a dry bulb temperature of 30 deg.c (saturation vapour pressure of water =4.24kpa). if the air sample has relative humidity of 65% the absolute humidity (in grams) of watervapour per kg of dry air is

Answers

Answered by alinakincsem
2

Please find the answer below:


Keeping in view that; a sample of moist air at a total pressure of 85 kpa has a dry bulb temperature of 30 deg.c (saturation vapour pressure of water =4.24kpa).


If the air sample has relative humidity of 65% the absolute humidity (in grams) of water vapour per kg of dry air is 20.5-21 grams.


Answered by Anonymous
0

Explanation:

The relation between relative humidity, saturation vapour pressure and actual vapour pressure is

P_{w} = P_{w, s}\times \frac{RH}{100}P

w

=P

w,s

×

100

RH

Here, RH=65\%RH=65% ,P_{w, s} = 4.24 KPaP

w,s

=4.24KPa

Therefore,

P_{w} =4.24 KPa \times \frac{65}{100}P

w

P_{w}=2.76 KpaP

w

=2.76Kpa

Now the absolute humidity,

AH=622\times\frac{P_{w} }{P_{T} }

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