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re: average electricity usage?
1 jun 2004
pete c   wrote:

>>consider two scenarios:
>>1. ac all day, with 24h(2000+(71.8-75)200) = 32.6k btu, about 6 hours
>>of window-rattling.
>>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...
>also some houses may allow some humidity to seep in overnight, so the
>indoor humidity might be nearly as high as the outdoor humidity by

there's no change in the absolute humidity above, and no dehum needed
for comfort, but even if so, that's not so bad, if it's only humidity,
vs incoming mist or condensation inside the house. a 2400 ft^2 1-story
house holds 19.2k ft^3 of air that weighs about 1440 pounds. at 75 f
and 50% rh, the humidity ratio w = 0.00942 pounds of water per pound
of dry air, so the air contains about 181 pounds of water vapor. if
the 61.8 f outdoor air had had 100% rh, it would have had w = 0.0119,
and we decided we could live with w = 0.0134:

>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. 

if we could somehow crank up the rh of the indoor air to 90% at 75 f
overnight, it would contain 325 pounds of water vapor. removing the extra
325-181 = 144 pounds takes takes 14.4k btu, which is still a lot less
than the 32.6k btu in scenario 1.

when is it better to keep the windows closed at night and ac vs vent,
based on humidity gain? it seems better to keep the windows closed if
the temperature of any part of the house is less than the dew point of
the outdoor air, in order to avoid condensation and house heat gain,
but what if the house is warmer than the dew point of the outdoor air?

if the house has volume v and thermal conductance g and continuous heat
gain p (v = 19.2k ft^3 and g = 200 btu/h-f and p = 2000 btu/h above), and
we want to maintain indoor temp ti and humidity ratio wi, and outdoor air
has a 24-hour average temp ta and wo > wi, and opening windows at night
eliminates the need for ac cooling, we need to dehum 0.075v(wo-wi)1000
btu/day, but we've saved 24h(p+(ta-ti)g) btu, so opening windows saves
24(p+(ta-ti)g)-0.075v(wo-wi)1000 btu/day, net. plugging in numbers gives 
24(2000+(71.8-75)200)-0.075(19.2k)(wo-0.0134)1000 = 51.9k - 1440kwo, which
is greater than 0 whenever wo < 0.0360, ie whenever the outdoor air temp
is less than 94 f at night. looks like opening windows always make sense
in this case, from a humidity point of view. 


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