Question

The humidity ratio of atmospheric air at 28°C dry bulb temperature and 760 mm of mercury is 0.016 kg / kg of dry air. Determine:

i. partial pressure of Water vapour; ii.relative humidity; iii. dew point temperature; iv. specific enthalpy; and v. wet bulb depression. ​

Answer 1

Answer:

Solution: Given: td= 28°C ; Pb= 760 mm of Hg ; W = 0.016 kg/ kg of dry air 1. Partial pressure of water vapour Let Pv= Partial pressure of water vapour. We know that humidity ratio (W), 12.16 - 0.016 Pv= 0.622 Pv or 0.638 Pv= 12.16 Pv= 12.16/0.638 = 19.06 mm of Hg = 19.06 x 133.3 = 2540.6 N/m2Ans

Explanation:

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Answer 2

Answer:

Example.3. The humidity ratio of atmospheric air at 28°C dry bulb temperature and760 mm of mercury is 0.016 kg / kg of dry air. Determine: 1. partial pressure of Watervapour; 2.relative humidity; 3. dew point temperature; 4. specific enthalpy; and 5.vapour density.Solution: Given:td= 28°C ;Pb= 760 mm of Hg ; W = 0.016 kg/ kg of dry air1.Partial pressure of water vapourLetPv= Partial pressure of water vapour.We know that humidity ratio (W),12.16 - 0.016Pv= 0.622Pvor 0.638Pv= 12.16Pv= 12.16/0.638 = 19.06 mm of Hg= 19.06 x 133.3 = 2540.6 N/m2Ans.2. Relative humidityFrom steam tables, we find that the saturation pressure of vapour corresponding to dry bulbtemperature of 28`C isPs= 0.03778 bar = 3778 N/m2∴Relative humidity,3. Dew point temperatureSince the dew point temperature is the saturation temperature corresponding to thepartial pressure of water vapour (Pv), therefore from steam tables, we find that correspondingto a pressure of 2540.6 N/m2(0.025406 bar), the dew point tempe