Answer:
a) [tex]\phi = 48\%[/tex], b) [tex]\omega = 0.012\,\frac{kg\,H_{2}O}{kg\,Air}[/tex], c) [tex]h = 58\,\frac{kJ}{kg\,Air}[/tex], d) [tex]T = 17^{\textdegree}C[/tex], e) [tex]P_{v} = 1.831\,kPa[/tex]
Explanation:
a) The relative humidity is given by the intersection of the dry bulb and wet bulb temperatures:
[tex]\phi = 48\%[/tex]
b) The humidity ratio is:
[tex]\omega = 0.012\,\frac{kg\,H_{2}O}{kg\,Air}[/tex]
c) The enthalpy is:
[tex]h = 58\,\frac{kJ}{kg\,Air}[/tex]
d) The dew-point temperature is:
[tex]T = 17^{\textdegree}C[/tex]
e) The water vapor pressure is the product of the relative humidity and the saturation pressure evaluated at dry bulb temperature:
[tex]P_{v} = \phi \cdot P_{sat}[/tex]
[tex]P_{v} = 0.48\cdot (3.816\,kPa)[/tex]
[tex]P_{v} = 1.831\,kPa[/tex]