Question

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 = 4.24 kpa). if the air sample has a relative humidity of 65%, the absolute humidity (in gram) of water vapour per kg of dry air is

Answer 1

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

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

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

Therefore,

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

$P_{w}=2.76 Kpa$

Now the absolute humidity,  

$AH=622\times\frac{P_{w} }{P_{T} }$ gm (of water)/mg .d.a

Here, $P_{T} = 85 KPa$

Therefore,

$AH=622\times \frac{2.76}{85}$

$AH=20.17$gm (of water)/mg .d.a

Answer 2

Answer:

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

=4.24KPa×

100

65

P_{w}=2.76 KpaP

w

=2.76Kpa

Now the absolute humidity,

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

P

T

P

w

gm (of