re: dehumidify basement?
5 jun 2002
>how do i decide that nick? it's 59f out today and 90% rh. if i bring
>that air in and warm it to 67, how much does the rh drop?
the outdoor vapor pressure pa = 0.9exp(17.863-9621/(460+59)) = 0.509 "hg,
approximately, and 67 f air at 100% rh has ps = 0.650 "hg, so warming
the outdoor air to 67 f would make the rh 100pa/ps = 75%, bordering on
you might only cool the house to 60% rh, where ps = pa/0.6 = 0.849
= exp(17.863-9621/(460+t)), and t = 9621/(17.863-ln(ps))-460 = 73.7 f.
which is the more comfortable condition?
does anyone have an accurate formula for the "heat index" that combines
dry bulb temp and rh into an equivalent temp, or a simplified form of
fanger's comfort equations? can we differentiate the heat index and set
the result to zero to find the most comfortable combination of temp and
rh that can be achieved by natural means?
in the situation above, you might consider cooling the house to 67 by
ventilation, then closing it up and running an ac as a dehumidifier,
with an extra counterflow heat exchanger inside the house that cools
house air going into the ac with cooler house air coming out of the
ac, while dumping the latent heat outdoors.
a 5k btu/h basement ac might move 150 cfm of cool air into a $28
50'x2' diameter poly film duct near the basement ceiling, inside
another poly film duct, with ntu = au/cmin = 314x0.75/150 = 1.6 (or
more, with condensation in the warm duct) and a 1.6/2.6 = 60% heat
exchanger effectiveness. with an efficient central dehumidifier,
a house with no ducts might use local humidifiers for zone cooling.