Question

The partial pressure of water vapor in a moist air sample of relative humidity 70% is 1.6 kPa, the total pressure being 101.325 kPa. Moist air may be treated as an ideal gas mixture of water vapor and dry air. The relation between saturation temperature (T_s in K) and saturation pressure (p_s in kPa) for water is given by In(p_s/p_o) = 14.317 – 5304/T_s, where p_o = 101.325 kPa. The dry bulb temperature of the moist air sample (in °C) is _______

Answer 1

ϕ  = 70%, P_v = 1.6 kPa, P₀ = 101.325 kPa

\(\ln\;\left( {\frac{{{P_s}}}{{{P_0}}}} \right) = 14.317 - \frac{{5304}}{{{T_s}}}\)................(Given)

\(ϕ = \frac{{{P_v}}}{{{P_s}}} \Rightarrow 0.7 = \frac{{1.6}}{{{P_s}}} \Rightarrow {P_s} = 2.286\;kPa\)

\(\Rightarrow \ln\;\left( {\frac{{2.286}}{{101.325}}} \right) = 14.317 - \frac{{5304}}{{{T_s}}}\)

⇒ T_s = 292.9 K

T_s = (292.9 - 273)^oC = 19.89^oC