Subject: A roofpond house

From: [email-masked] (Nick Pine)
Date: 5 Aug 1996 09:41:46 -0400

Medford, OR, doesn't have a lot of sun in the winter.
Here are some NREL climatic average data... 

          24 hour   daytime  nightime  average sun on  
          avg temp  high     low       s wall  horiz surface

December      38 F  44 F     31 F      550     390 Btu/ft^2/day
July          73    91       55        1020    2450

In July, the average windspeed is 6 mph, and the average humidity ratio
of the air is 0.0076 #water/#dry air. The elevation is 1299', and the
air pressure is 14.1 psia, vs 14.7 at sea level.

A 2-story 2000 ft^2 house in Medford with 3000 ft^2 of walls and ceiling with
an average US R-value of 10 might need 3000 ft^2/R10 = 300 Btu/hr-F, ie
(91 F - 75 F) x 300 Btu/hr-F = 4800 Btu/hour of cooling when the house is 75 F
inside and it's 91 F outside in July. Over the 4600 heating degree day winter,
it might need 24 hours x 4600 DD x 300 = 33 million Btu/year of heat, say 300
gallons of oil per year, at a cost of $300, or $900 in propane? Could we use
a sunspace and solar closet for space heat and domestic hot water? 

Some R1 sunspace glazing, eg a single layer of polyester film, with something
white on the ground in front of it, will receive about 550x4/3 = 733 Btu/ft^2
of solar heat, on an average December day, and lose about 6 hours x (80F - 44F)
x 1 ft^2/R1 = 216 Btu, for a net gain of 517 Btu/day. The house needs about 
24 hours (68 F - 38 F) 300 Btu/hr-F = 216K Btu to stay warm on that average
day, ie 216K/517 = 418 ft^2 of shallow sunspace glazing, eg a lean-to sunspace
16' tall and 32' long. (Moving the house to El Paso, Texas, would reduce that
sunspace glazing area to 16' tall x 8' long.)

Storing space heat for 5 cloudy days would require 5x216K/25K = 43 55 gallon
plastic drums full of 130 F water, each 2' diameter x 3' tall. We might use
50 of them in 5 layers, each about 10' wide x 4' deep in a solar closet, or
stack fewer drums on top of a concrete septic tank that is used for high
temperature sewage treatment. And partition the store into two sections, 
a lower temp one for space heating and water preheating, and a higher temp
one for second-stage water heating.  

Spraying water on the roof would keep it close to the summer wet bulb temp
of 50 F, reducing the heat load for the house by an amount that depends on
the roof area and the amount of insulation between the house and the roof.
It takes about 1000 Btu to evaporate a pound of water, so 5 pounds of water
per hour would cool it by 5,000 Btu/hour. Spraying more water and collecting
it in a gutter and recycling it through a water-air heat exchanger in the 
house would cool the house more. Magic Aire (a division of United Electric
in Galveston, TX) makes a nice all-copper 2' x 2' SHW 2347 duct heat exchanger
that costs about $150 and transfers 45K Btu/hour between 125 F water and 68 F
air at 1400 cfm, with a 0.1" H20 pressure drop. This would make about 16K
Btu/hour of cooling with a 20 F air/water delta T, equivalent to about 3
window ACs, but with less dehumidification ability. Could the 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 the inside at dawn? We probably don't need this in Medford.

It would help if the house had lots of thermal mass inside so we could
ventilate it at night and button it up during the day, for good cooling.

A flat roof with a 2" deep roof pond might need less materials and electrical 
cooling power, with winter heating and fire protection advantages, and make
a natural gravity-pressurized rainwater supply for the house. A 1000 ft^2 roof
would collect an average of 67 gallons of water per day with a 40" annual
rainfall and a 2" pond would store 1250 gallons of water. A 50' long x 12'
wide x 4' deep pond made from a single piece of EPDM rubber under a north
overhang of the house would increase rainwater collection to over 100 gallons
per day and make a total cooling and water supply storage of over 20,000
gallons. Richard Komp's house in Maine has a 1000 gallon cistern and a
roof/rainwater supply. One does have to conserve water, this way. Could we
evaporate and recondense greywater with a shallow pond inside the sunspace?
It would be nice to do the condensing inside the house somehow in winter.

With an average 38 F December temperature and an average 390 Btu/ft^2/day of
sun that falls on a horizontal surface, we might float an R1 pool cover on the
roof in winter, so the pond water at temperature T might lose 24 hours x (T-38)
x 1 ft^2/R1 on an average day, so T = 38 + 390/34 = 54 F, not counting the
heat that leaks up from the house through the roof. This would reduce the
heating load through the roof. (Perhaps this should be a one story house?)
Say we designed the roof to hold up 20 pounds per square foot (with 12' 2x6
or 20' 2x10 rafters on 2' centers?) 10 pounds for the 2" deep pond, and 10
more pounds to allow 2' of snow on the roof. (Which might be helpful extra
insulation, during a very cold spell. It would take about 1660 Btu/ft^2 to
freeze the roof pond, starting at T = 54 F, eg about 1660/38 = 44 sunless
hours at Medford's record low temp of -6 F.)

If the snow on the roof got too deep (we might use a microswitch activated by
roof deflection to sense this) we could pump some groundwater up through the
roof pond to melt it. How much 55 F groundwater would we need to keep up with
a very heavy snowfall, eg 3" per hour? I think that's equivalent to about
0.3" of ice per hour, ie 1.6 pounds of ice, or 230 Btu/hr-ft^2, since it takes
144 Btu to melt a pound of ice, at 32 F. For a 1000 ft^2 roof, that's 230K 
Btu/hour. Each pound of water supplies 55-32 = 23 Btu, so 230K/23 = 10K pounds
of water per hour could keep up with that snowstorm, ie 21 gallons per minute,
a big pump, so... we might want to set the switch to turn on with 1' of snow,
in anticipation. And keep a big post handy to prop up the roof inside, once
in a great while :-), especially if the roof starts to sag in the middle,
making the pond deeper and increasing the load in the middle... But it seems
to me all this would rarely be required, especially since the house would be
providing some heat through the roof too.

I've heard there's a building in a park in Mt. Ranier, WA that works this way,
with a flat roof, even tho they get 20'/year of snow there. I doubt they use
groundwater for snow melting. If we have to do this seldom enough we might use
stored solar heat or even electric resistance heat... 

In summer, without the roof pond cover, we might pump some 70 F water through
the roof pond at night, when the average low is 55 F, in July. With a rough
surface thermal conductivity of 2 + 6 mph/2 = 5 Btu/hr-ft^2-F from the wind,
a 1000 ft^2 pond might lose 4 hours (70 F - 55 F) 1000 ft^2 x 5 = 300K Btu
per day, vs the 38K Btu/day needed for cooling the house for 8 hours a day.
That's good, but it gets better...

At sea level, the ratio of evaporative to convective pond cooling is about
1000(Pp-Pa)/(Tp-Ta)), independent of windspeed, where Pp and Pa are vapor
pressures of water in the pond and air in inches of mercury, and Tp and Ta
are Farenheit pond and air temperatures. The vapor pressure of water in 70 F
air is 0.74 "Hg (Clausius-Clapeyron: ln(Pw) = 17.863-9621/Tabs), and air with
a 0.0076 humidity ratio, W, has Pa = 29.92/(1+0.62198/W) = 0.36 "Hg. Medford's
elevation increases this ratio by another factor of 14.7/14.1, so the pond
might lose 25 times more heat by evaporation than by convection on an average
July night. A total of almost 8 million Btu/day, over 200 times more cooling
than needed...

Nick