Question

Atmospheric air at a pressure of 1 atm and dry bulb temperature of 28 C has a wet-bulb temperature of 20 C.

Answer 1

Answer:

a) $\phi = 48\%$, b) $\omega = 0.012\,\frac{kg\,H_{2}O}{kg\,Air}$, c) $h = 58\,\frac{kJ}{kg\,Air}$, d) $T = 17^{\textdegree}C$, e) $P_{v} = 1.831\,kPa$

Explanation:

a) The relative humidity is given by the intersection of the dry bulb and wet bulb temperatures:

$\phi = 48\%$

b) The humidity ratio is:

$\omega = 0.012\,\frac{kg\,H_{2}O}{kg\,Air}$

c) The enthalpy is:

$h = 58\,\frac{kJ}{kg\,Air}$

d) The dew-point temperature is:

$T = 17^{\textdegree}C$

e) The water vapor pressure is the product of the relative humidity and the saturation pressure evaluated at dry bulb temperature:

$P_{v} = \phi \cdot P_{sat}$

$P_{v} = 0.48\cdot (3.816\,kPa)$

$P_{v} = 1.831\,kPa$