growing cubes SOLAR.KnowledgePublications.com

growing cubes
31 oct 2001
people say small "envelope-dominated buildings" need heat in cool climates,
vs large "internally-load-dominated buildings" that need cooling most of
the year. where is the borderline? when does a cube change from "small" to
"large" as it grows?

an 8' cube with 64 ft^2 of floorspace and r25 (6" sip) walls has
384ft^2/r25 = 15 btu/h-f of thermal conductance. with 600 kwh/month
(2800 btu/h) of internal electrical energy use, it can keep itself
2800/15 = 187 f degrees warmer than the outdoors, ie at 65 f indoors,
its balance point temp is minus 122 f. 

a 16' cube with 512 ft^2 on 2 floors and r25 walls and 61 btu/h-f of
thermal conductance and 600 kwh/month (2800 btu/h) of internal gain
can keep itself 2800/61 = 46 f degrees warmer than the outdoors, ie
at 65 f indoors, its balance point temp is 17 f. 

a 24' cube with 138 btu/h-f and a 5600 btu/h internal gain has
a balance point temp of 65-2800/138 = 45 f.

so maybe people are wrong, or this only applies to commercial buildings 
with internal gains proportional to floorspace... 

if a typical family occupies about 2000 ft^2 and uses 600 kwh/month...

a 32' "2-family" cube has a balance point temp of 65-5600/246 = 42 f.

a 48' "8-family" cube has a 65-22400/553 = 25 f balance point.

a 64' "16-family" cube with 45k btu/h or 1.1 million btu/day has 19 f.
we might distribute this heat with hydronic floors, with 16 zones.

ok. that takes care of heating on an average 30 f january day in
philadelphia. what about daylight, you say? that's more helpful for
office buildings that are occupied during the day and more internal
gain, but if you like daylight and views, we could add a few windows
around the perimeter to light the rooms within, say 16', and add
a lightshaft down the middle with a reflective concentrating south
sunscoop on top and automatically-controlled diffuse reflective
"windows" with hinges beneath that open into the lightshaft for
daylighting the inner 32' core, and artificial lights that dim
to maintain constant illumination. 

direct sun is about 200x brighter than a well-lit room, so lighting
8x32x32' of floorspace takes about 41 ft^2 of solar aperture at 100%
efficiency. we might use a 4'x8' shaft under a 4x linear parabolic
concentrator with a 4' focal line at dawn. 

this would raise the balance point temp, because sunlight has about twice
as much light per watt (110 lumens) as conventional flourescents, and
windows lose more heat than walls. then again, there's solar gain.

for cooling, we might make the indoor and outdoor temps the same (76.7 f)
in july, less than the adaptive ashrae comfort standard temp, 54+0.31x76.7
= 77.8 f. removing 45k btu/h of heat with 14x64x64 = 57k ft^2 of cool floor
and ceiling surface with a slowly-moving airfilm conductance of 1.5 btu/h-f
requires a 0.5 f temperature difference. we might make the floors 75 f and
flood a shallow roofpond at night to cool the water.

nrel says the average humidity ratio w = 0.0133 pounds of water per pound
of dry air in july, and the average windspeed is 8 mph and the daily min
and max temps are 67.2 and 86.1 f. this corresponds to a dew point temp
tdp = 9621/(14.46+ln(1+0.62198/w))-460 = 64.9 f. if the night air temp to
= 72 f (halfway between the 24-hour average and daily min) and the outdoor 
water vapor pressure pa = 29.921/(1+0.62198/w) = 0.626 "hg, the wet bulb
temp twb (r) = 9621/(22.47-ln(460+to+100pa-twb), ie 67 f, after iteration.

phil niles says a pond loses 1.63x10^-9((t+460)^4-a(to+460)^4) btu/h-ft^2
by radiation, where a = 0.002056tdp+0.7378 = 0.87 in this case, so a 75 f
roof pond in 72 f air loses 19.8 btu/h-ft^2, and qc = (0.74+0.3v)(t-to)
= 2.9 btu/h-ft^2 more by convection, and qe = b(t-twb)-qc by evaporation,
where b = 3.01(0.74+0.3v)((t+twb)/65-1) = 3.01(0.87)1.14 = 2.98, so qe
= 2.98(75-67)-2.9 = 21 btu/h-ft^2. the total loss is 43.6 btu/h-ft^2, so
a 64'x64' roofpond might lose 64x64x43.6 = 178k btu/h, or 2.1 million btu
over a 12 hour night, almost twice the internal gain. 

a person respires about 1 lb/day of water vapor and perspires another.
other activities raise that to about 4 pounds per day, so we have to
remove at least 16x4x4 = 256 pounds of water per day, with 16 4-person
families. we might dehumidify by absorbing water vapor at night in
a licl pond on the roof, via the lightshaft. a square foot of an a ft^2
pond under the 4x concentrator might be 160 f for 8 hours per day (an
equilibrium temp for a 50% licl solution) and evaporate 256/a pounds
of water, losing about 8h(160f-80f)+256k/a btu/day while receiving about
3k btu/day of summer sun, which makes a = 108 ft^2.

we might use a 4'x32' pond under a 16'x32'x16' tall concentrator with
reflective endwalls and south glazing, leaving a 12'x32' 2-story party
space on the roof. with a transparent cover, the pond could also heat
water for showers.

nick