Idaho houses

Posted by nicksanspam on March 21, 2008, 2:32 pm

NREL says 820 Btu/ft^2 of sun falls on a south wall on an average 29 F
January day in Boise. North, east, and west walls get 180, 340, and 350.

Working backwards, a \$00 cloudy-day heat storage tank and a \$5 1000 Btu/h-F
used car radiator with its 12V fans (36 watts, in series) can keep a house
with a G Btu/h-F thermal conductance 70 F on a 21.6 F morning with a min
water temp Tmin = 70+(70-21.6)G/1000. With 2 foil-polyiso foamboard dampers
with limit switches and \$0 windshield wiper motors, the radiator can also
collect and store heat from a sunspace.

If cloudy days are coin flips and we store heat for 5 days, the house can
be at most 100(1-2^-5) = 97% solar-heated. If a frugal 600 kWh/mo indoor
electrical use provides 68.2K Btu/day and a 4'x8'x3'-tall plywood tank with
a folded EPDM liner (and a 1/2% ACI-100 non-toxic corrosion inhibitor from
DW Davies) supplies 4x8x3x62.33(140-Tmin) = 5d(24h(65-29)G-68.2K) Btu to
the 65 F average house, G = 165 Btu/h-F max.

A 48'x48'x8' house with 180 ft^2 of R4 windows and no air leaks and a 30 cfm
80% air-air heat exchanger and an R40 ceiling and R30 walls would have
180/R4+48^2/R40+(1-0.8)30+(4x48x8-180)/R20 = 165 Btu/h-F, approximately.

If 20, 40, 40, and 80 ft^2 of north, east, west, and south windows with
50% solar transmission collect 0.5x20(180+2(340+350)+4x820) = 48.4K Btu
and electrical use adds 68.2K Btu and the house loses 8h(65-29)G = 47.5K
and the 140 F tank with R30 walls loses 8h(140-65)104/30 = 2.1K during
an average 8 hour solar collection day, the house mass needs to store
48.4+68.2-47.5+2.1 = 71.2K Btu of overnight heat, eg in C = 7120 Btu/F
(or less, with a waterwall near the south windows) with a 10 F temp swing
from 60 F at dawn to 70 at dusk.

RC = C/G = 43 hours lets the house cool to 60 F in -43ln(60-29)/(70-29))
= 12.1 hours, so it needs (16h-12.1h)(60-29)G = 20K Btu more night heat
from the tank and radiator, which can collect 22K Btu at 2.8K Btu/h from
a sunspace or a solar attic or a 140+2.8K/1000 = 143 F airspace inside
A ft^2 of R2 \$/ft^2 polycarbonate "solar siding" with 80% transmission
if 0.8x820A = 8h(143-29)A/2+22K, ie A = 110 ft^2. A \$0 1"x300' 13-gallon
PE plastic water pipe coil in the tank can provide hot water for showers,
with a few 4"x10' PVC pipes under the floor as a greywater heat exchanger.

Is this "passive"? Is it "green building"? Do we care? :-)

Nick

Posted by Morris Dovey on March 21, 2008, 1:41 pm

nicksanspam@ece.villanova.edu wrote:

<snip>

I'm curious about Marilyn's response to that. ;-)

--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/

Posted by nicksanspam on March 23, 2008, 11:15 am

Haven't heard back from her, which is typical developer behavior :-)

Meanwhilst, check out Riverdale, one of Canada's first 12 Net Zero

http://www.riverdalenetzero.ca/

http://www.cmhc.ca/en/inpr/su/eqho/index.cfm

Nick

Posted by Kitep on May 10, 2008, 9:54 pm

Nothing like responding to a post 1-1/2 months later :)

I noticed the above web site says Alberta pays \$.67/kWh.  I assume that's
Canadian, so it's only \$.66/kWh US.  I know it's a peak rate, but still
seems like a place where solar may be cost effective.

Posted by You on May 10, 2008, 11:10 pm

Hmmm, I would doubt that anyone on a BIG GRID is paying 67 Cents
by an avalanche last month, and had to go on purely Diesel Generated
Power and they went from \$.11/KwH, to a whopping \$.52/KwH.  The locals
are looking to let some Blood Flow, from the Politico's who handed the
local Power system over to a Private Company, a few years back, now that
their rates have gone up 4 fold. \$.67/kWh is way out of line for a
Grid Tied Power System... I would expect more in the range of \$.067/KwH.
Maybe you slipped a Decimal Point, or something.