re: average electricity usage?
31 may 2004
>> >...in some areas, the humidity is a major part of the a/c load, so
>> >opening the windows in the evening when it's cooler (thinking this
>> >helps reduce a/c can actually let in humid air and can actually
>> >raise your usage.
>> the ac energy needed to remove a houseful of humid air seems negligible
>> compared to the potential coolth gain, as long as condensation does not
>> occur inside the house.
>if the house is 'cool' during the day, and then you open the windows in the
>evening, you *don't* gain any 'coolth'.
by opening windows in the evening, you can keep the house and its thermal
mass closer to the average daily min than the 24-hour average temp (or
higher, given some internal heat gain and sun in the windows.) this seems
>but you do raise the humidity level from a nice 50% back up to 90%.
possibly. 'pends on where you live. of course that's relative humidity,
with no condensation.
>and so the a/c has to run to remove that as
>well as maintain the house cool tomorrow.
maybe, but the energy required to dehumidify a houseful of humid air (once)
is small compared to the potential coolth gain. it could be large if water
vapor condensed inside a house, but that would be bad thermal management.
>leave the windows closed at night and the a/c doesn't have to remove the
>moisture, just the heat conducted/radiated into the house tomorrow. a much
>lower number amount of energy.
not necessarily. with no night venting, the ac has to remove more heat...
say you live near phila, like me, and it's an nrel average 71.8 june day
with an 81.7 average daily max and a 61.8 average daily min and a constant
(day and night) absolute humidity ratio w = 0.0080 pounds of water per
pound of dry air. suppose the house has a 200 btu/h-f thermal conductance
and 2000 btu/h of internal heat gain (adding 10 f to the outdoor temp) and
you want to keep it at 75 f max. suppose (arguendo) the outdoor temp is
61.8 all night and 81.7 all day.
the 1992 ashrae-55 standard comfort zone has an upper 68 f wet bulb temp.
at 68 f and 100% rh has pw = e^(17.863-9621/(460+68)) = 0.699 "hg, so
(using bowen's equation) we are in the zone if 100(0.699-pa)/(68-75) = -1,
ie pa <= 0.229 "hg, ie w = 0.0080 <= 0.62198x0.629/(29.921-0.629) = 0.0134,
so no dehumidification is needed.
consider two scenarios:
1. ac all day, with 24h(2000-(71.8-75)200) = 32.6k btu, about 6 hours
2. open the windows at night, cool a thermal mass c to 65 f, and let it
warm to 75 f over 12 hours, so 75 = 91.7+(65-91.7)e^(-12/rc), which makes
rc = -12/(ln(75-91.7)/(65-91.7)) = 25.6 hours = c/200, so c = 5114 btu/f,
with 0.00 btu of air conditioning.