On my planet, it takes about 1000 Btu to evaporate a pound of water.
At -10 F, wo = 0.0004608 max. At 68 F and 42% RH, wi = 0.006198. Raising
154 cfm of outdoor air to 42% RH for the previous 2304x8 ft^3 house with
0.5 ACH and 192 Btu/h-F requires evaporating 60x154x.075(wi-wo) = 3.976
pounds of water per hour, which requires about 3976 Btu/h of heat.
The ASHRAE 55 standard says people are equally comfy at 68.0 and 42% RH
and 69.7 F and 10% RH. Turning the thermostat up from 68 to 69.7 would
only add (69.7-68)192 = 326 Btu/h to the fuel bill.
Then again, you might make 42% indoor air by caulking until 60C0.075(wi-wo)
= 2x8.33/24h, ie C = 27 cfm, saving 3282 Btu/h in humidification energy plus
about (154-27)(68-(-10)) = 9906 Btu/h in heating energy, for a net savings of
13.2K Btu/h or 316.5K Btu ($5 in oil or $9 in electricity) on a -10 F day.
firstname.lastname@example.org wrote in message
No quarrel with the rest of your analysis, but if ASHRAE 55 says that,
then it is horses**t. Hardly anybody is comfortable at 10% RH under
any conditions, and prolonged exposure to humidity that low,
especially in the absence of adequate air exchange (as can be created
by overzealous attention to insulation), is well known to encourage
upper respiratory infections.
> Then again, you might make 42% indoor air by caulking until 60C0.075(wi-wo)
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