Answer:
a) 48.154 millibar
b) 0.0475
Explanation:
Given:
dew point temperature = 90° F
we have relation °F = 9/5 °C + 32
thus,
in °C
= [tex]\frac{\textup{5}}{\textup{9}}(F - 32)[/tex]
= [tex]\frac{\textup{5}}{\textup{9}}(90 - 32)[/tex]
= 32.22° C
a) Now,
water vapor pressure is given as:
= [tex]6.11\times10^{\frac{7.5\times T_D}{237.3+T_D}}[/tex]
on substituting the respective values, we get
water vapor pressure = [tex]6.11\times10^{\frac{7.5\times32.22}{237.3+32.22}}[/tex]
or
water vapor pressure = 48.154 millibar
b) Mixing ratio is calculated as:
= [tex]\frac{\textup{Vapor pressure}}{\textup{Atmospheric pressure}}[/tex]
= [tex]\frac{\textup{48.154}}{\textup{1013}}[/tex]
= 0.0475
