Chemical Engineering - Mass Transfer - Discussion
Discussion Forum : Mass Transfer - Section 12 (Q.No. 34)
34.
An air-water vapour mixture has a dry bulb temperature of 60°C and a dew point temperature of 40°C. The total pressure is 101.3 kPa and the vapour pressure of water at 40°C and 60°C are 7.30 kPa and 19.91 kPa respectively.The humidity of air sample expressed as kg of water vapour/kg of dry air is
Discussion:
4 comments Page 1 of 1.
Anish said:
5 years ago
Answers should be an option a because dew point in question is 40 deg.c.
So we use Pa=7.30 and Pt=101.3.
So, humidity= 7.3*18/(101.3)*29= 0.048 kg h20 vapor/kg of dry Air.
So we use Pa=7.30 and Pt=101.3.
So, humidity= 7.3*18/(101.3)*29= 0.048 kg h20 vapor/kg of dry Air.
(2)
Radhe said:
4 years ago
No, we should use dry bulb temperature of air, not the dew point temperature.
Sahar said:
3 years ago
The right answer should be 0.152.
H = 18(19.91)/29(101.3-19.91) = 0.152.
H = 18(19.91)/29(101.3-19.91) = 0.152.
Taisir said:
3 months ago
Humidity = (nH2O)/(n air) * (M H2O)/(Mair)
= 0.07766⋅(18.02/28.97),
= 0.0483 kg water/kg dry air.
Then, (nH2O)/(n air) = 7.3/101.3 - 7.3 = 0.07766.
= 0.07766⋅(18.02/28.97),
= 0.0483 kg water/kg dry air.
Then, (nH2O)/(n air) = 7.3/101.3 - 7.3 = 0.07766.
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