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water vapor, rocks, air leakiness and relative humidity
6 feb 1997
david r. smith wrote:
>i agree with van ardoy about the common sense of using stone for building.
what are the dessicant properties of various kinds of stones, including the
man-made kind?
house air at 70 f and 50% relative humidity has a dew point of about 50 f
(9121/(9121/(70+460)-ln(0.5))-460.) perhaps a pool of water in a 70 f room
gains the same amount of heat from the room as it loses by evaporation.
a shallow uninsulated pool with about 1 ft^2 of 50 f surface and 1 ft^2 of
bottom exposed to still air might gain about (70f-50f)2ft^2x1.5 = 60 btu/hr
or 1440 btu/day, ie evaporate about 1.4 pounds of water per day, vs a green
plant that evaporates 1 pound of water per day or a person who evaporates
3 pounds of water per day. that seems high for the pool of water...
insulating the pool or making the sides higher (as in a toilet) so cool humid
air can collect inside, or covering the pool with a lid to raise the relative
humidity near the water surface would decrease this loss and evaporation.
a house with 3 toilets, 4 people and 10 green plants might generate 1 pound
of water vapor per hour, or more from showers, cooking, cleaning, etc. how
would this increase the relative humidity inside the house in the winter?
suppose the house makes an average of 3 lb of water vapor per hour and leaks
10,000 ft^3 of air per hour to the outside (about 0.5 ach in a 32x32' 2-story
house) and the average humidity ratio of the outdoor air is w = 0.0025 lb of
water vapor per lb of dry air (an outdoor relative humidity of 0.0025/0.0035
= 71%, since the 30.5 f outdoor air can hold a maximum of 0.0035 lb water/lb
of dry air.) since 70 f air can hold at most 0.0158 lb water/lb of dry air,
heating that outdoor air to 70 f lowers its relative humidity to about
0.0025/0.0158 = 16%. but adding 3 lb of water to 10,000 ft^3 of air that
weighs 13.35 ft^3 per lb, ie 750 pounds, raises its humidity ratio by 3/750
to 0.0065 lb water per pound of dry air and raises the relative humidity to
about 0.0065/0.0158 = 41%.
evaporating 3 lb of water vapor per hour takes about 72k btu/day. if the vapor
leaves the house as moist air, that heat is lost. it seems to me that most of
it can be recovered in a condensing air-air heat exchanger. should we try to
humidify incoming air with outgoing air using dessicant wheels, etc? i suspect
not, if that most of the heat of evaporation is recovered, as well as the
sensible heat in the outgoing air. it seems simpler to just let the outgoing
water drain away, and let indoor plants and people continue to humidify winter
house air. the ashrae hof says
the total evaporative heat loss (latent heat) from the human body due to
both respirative losses and skin losses can be measured directly from
the body's rate of mass loss as observed by a sensitive scale...
one way to control moisture and store heat in a solar house might be to use
building materials with dessicant properties, like concrete. suppose the walls
of a solar house sourced water vapor to house air as the house warmed up,
storing heat of evaporation and keeping house air cooler, and condensed water
vapor as the house tried to cool off at night, and kept the house warmer...
liquid polyethylene glycol can absorb 11 times its weight in water. expanded
shale used in lightweight "insulating" concrete can absorb about 13% of its
weight in water, ie about 10 lb water/ft^3 of material. an 8" x 32' x 8' house
wall of this material might absorb 170 lb of water on a cloudy day, ie about
170k btu, possibly keeping the house warm with little change in temperature.
this is about 5 times more heat than we might store in the mass of a plain
concrete wall with a 10 f temperature change, ie 10f(8"/12"x32'x8'x22btu/ft^3
= 37k btu. perhaps we should think of solar houses as storing heat in humidity
swings as well as temperature swings, with phase-change cycles, like clouds.
nick
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