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roofponds in austin?
25 aug 1996
> mike rice wrote:
>>i'm designing a tight, well insulated house with an open floor plan.
>>i need 18k btuh to cool the house...
would that be the peak cooling load? i wonder about the average...
i guess we are more concerned with cooling than heating in austin, tx.
here are some nrel climatic average data...
24 hour daytime nightime average sun on
avg temp high low s wall horiz surface
jan 49 f 59 f 39 f 1200 940 btu/ft^2/day
aug 85 96 74 820 2010
in august, the average windspeed is 7.5 mph, and the average humidity ratio
of the air is w = 0.0153 #water/#dry air, corresponding to saturated air at
69 f, according to the ashrae hof, which also says the 5%-ile design wet bulb
temp in austin is 77f. does nightime radiant cooling improve this? where are
the ashrae gurus? the elevation is 620', and the air pressure is 14.4 psia,
vs 14.7 at sea level.
a 1-story 2000 ft^2 house in austin with 1400 ft^2 of walls with an average
us r-value of 10, and 2000 ft^2 of r1 ceiling, under a shallow 69 f roofpond
with some sort of shading above the south side, might have a simple electrical
model that looks like this, when it's 95 f outside:
/-- resistors --\
95 ---www-------------------------www--- 69
0.0071 = 10/1400 | 1/2000 = 0.0005
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thouse
the heat that flows from the walls through the roof (with some help from a
ceiling fan) might be (95-69)/(0.0071+0.0005) = 3401 btu/hour, so the
temperature difference between the inside of the house and the ceiling might
be 3401 x 1/2000 = 1.7 f, ie the house might be 70.7 f inside in august, so
it might not need any other form of summer cooling. it takes about 1000 btu
to evaporate a pound of water, so 3.4 lb water/hr, ie 0.007 gpm or 10 gallons
per day of water might cool the roof in august.
for a long, hot, humid spell, we might store some cool water in the house, and
design the ceiling with an airgap over a radiant barrier, with a large hole
in it, to allow warm air to flow up. or we might not change the ceiling. the
ashrae hof says the 5%-ile design dry bulb temp is 97 f in austin, with a mean
coincident wet bulb temp of 74 f, and note c says this only applies to the
month of july. perhaps this means that a worse heating load combination of
outdoor temperature and humidity happens less than 5% of the time, in july,
ie less than 1.5 days. let's call it 2 days. now suppose the radiant barrier
has an r-value of 14 for downward heatflow. how much 69 f water would we have
to store in the house to keep it at, say, 75 f, for those 2 days?
if the house is warmer than 74 f, this model might apply:
97 ---www-------------------------www--- 74
0.0071 = 10/1400 | 1/2000 = 0.0005
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thouse
in this case, with no thermal mass in the house, 3009 btu/hr flow up through
the ceiling, and the house temp would be 75.5 f. oh my. we may have exceeded
our comfort limit. we might have to take off our 3-piece suits. how can we
prevent this catastrophe? we need a house capacitor:
97 ---www-------------------------www--- 74
0.0071 = 10/1400 | 14/2000 = 0.007
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----- c
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how big does c have to be, to keep the house at 72 f for two days, starting
with 69 f water? the temperature rise is 3 f, and q = c dt, so q = 3c, and
over two days, the house will gain 48(97-72)/0.007 = 171k btu of heat from
the walls and 48(74-72)/0.0071 = 13.5k btu of heat from the ceiling, a total
of 185k btu, so c = 62k pounds or 7700 gallons or 961 ft^3 or a 10' cube of
water. or we might just live with the higher temperature, or go for a swim in
the roofpond, or make the walls of adobe, or strawbales with a masonry floor.
could this house be dehumidified with a dessicant sunspace floor (a thin
layer of reinforced concrete, with some foam underneath?), vented to the
outside during the afternoon and to the house just before dawn?
a flat roof with a 2" deep roof pond might need less materials than a
conventional roof, and add fire protection, reducing insurance costs. with
an average 49 f january temperature and an 940 btu/ft^2/day of sun falling
on a horizontal surface, we might float an r1 pool cover on the roof, so
the pond water at temperature t might lose 24 hours x (t-49) x 1 ft^2/r1
on an average day, so t = 49 + 940/24 = 88 f. so this house might not need
any other form of winter heating, in austin.
nick
ps: our solar closet paper is at http://leia.ursinus.edu/~physics/solar.html
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