Question

104. An air-water vapour mixture has a dry-bulb temperature of 333 K and a dew-point temperature of 313 K. The total pressure is 101.3 kPa and the vapour pressure of water at 313 K is 7.3 kPa and that at 333 K is 19.91 kPa. The humidity of air expressed as kg of water vapour per kg of dry air is (a) 0.048 (b) 0.079 (c) 0.122 (d) 0.152

Answer 1

Total pressure of air water vapor mixture Pt = 101.3 kpa

Air water vapor mixture has a dry bulb temperature T = 333K and vapor pressure Pv1 = 19.91 kpa

And dew point temperature Td = 313 K and vapor pressure Pv2 = 7.3 Kpa

And we know at dew point air-con water vapor mixture is saturated with water vapor and relative humidity is 1.

If we want humidity at 333k then vapor pressure of water vapor at 313 k is become partial pressure of water vapor at 333k.

Because we know if total pressure is constant and increase temperatue then its relative humidity decreases and have tendency to get absorb more water vapor.

Then at 333k, partial pressure of water vapor P1 = 7.3 kpa

Humidity of air water vapor mixture., H =( 18/29)*(7.3/101.3-7.3) = 0.0482 kg water vapor / kgdry air

Ans:

a) 0.048