i need to calculate roughly the hot water evaporation rate from open top vessel (mm/day).
known data:
water temp is 70 degree cel.
aversge wind speed is 2 meter/sec.
aversge air humidity is 50 %.
Irving Langmuir developed a way to measure vapor pressure
by measuring the evaporation rate, which we will now flip around backwards. His reasoning is that the rate at which molecues are lost due to evaporation to a gas with no partial pressure of the evaporating substance is the same as the rate at which molecules of the substance would hit the surface if it were in equilibrium with the vapor (because in equilibrium the evaporation rate and re-condensation rate cancel each other out). His expression is:
Solar radiation, geometry of the open top vessel, and the progressive cooling of the hot water batch also are factors in the estimation of time of evaporation.
you forumla for the chaps case is
mass loss rate)/(unit area) = (vapor pressure - ambient partial pressure)*sqrt( (molecular weight)/(2*pi*R*T) )
at 70C P vap for water is 0.3 bara. so working through
(0.3 - 1) * ( 18 / 2* 3.142 * R * 293)
.. gives a negative number. So unless I've miss understood your method here you cannot use this formula.
To james1030bruce, as I see it, owg's formula (right or wrong) speaks of ambient "partial" pressure (meaning the vol% of moisture in atmospheric air). Thus, there is no negative result.