Posted by Nick Pine on September 27, 2003, 10:43 pm
It's easy. Ask some questions, s'il te plait. A low-performance mass and
glass passive solar house has an optimal glass/heat load ratio: too few
south windows means too little solar heat. Too many means too much backup
heat on cloudy days.
Bris-soleil is French for "sun breaker," an array of slats that let low
winter sun in and block high summer sun. You can see them over the back
windows of sports cars.
Posted by Doug on September 28, 2003, 9:13 pm
email@example.com (Nick Pine) wrote in message
Merci beaucoup, Nick.
Mais, je pense que ce n'est pas tres facile.
The house has been designed with approx 360 sq ft of south facing
windows. The floor area is approx 1800 sq ft. There is also a loft
of about 400 sq ft. The floor of the loft is 3" thick Douglas fir
(acts as the floor of the loft as well as the ceiling of the
livingroom/kitchen area). On the south east corner is a sunroom about
3 ft lower than the livingroom. The windows of the sunroom are
included in the 360 sq ft. All the floors are cement with no covering
that would impede heat flow. The floors will be acid etched a medium
dark colour. Under the 6" of cement is approx 3" of polystyrene
insulation. I estimate that about 700 sq ft of the foor area will be
in direct sun. The railing of the loft is a kneewall made of drywall
- it will be in direct sun. There will be a rock/adobe fireplace
approx 9' high and 3' by 5' that will be in the sunny area although
the 3' wide side is facing south. The house is a relatively open
concept allowing the warm air to flow throughout the house as well as
reradiated heat to warm the area in shade. I also want to be able to
pump water through the pex radiant floor tubing when the sun shines on
the livingroom floor so that this heat can be ditributed throughout
The windows we selected are not high performance - they have a SHGC of
.51 and a U factor of .32. We had a solar pathfinder test done on the
site that showed that we get approx 90% of the available sunshine on
the south of the house in Jan. Hmmmm, what else... The house is just
outside of Ottawa.
If you need any further information, let me know.
Posted by Nick Pine on September 29, 2003, 3:08 pm
Exact ways involve charts and tables or a TMY2 or SBIC simulations, but
why bother? Direct gain (aka "direct loss") houses are such poor performers,
compared to other passive solar heating configurations. For one thing, the
thermal mass has a small temp swing, so it doesn't store much heat, compared
to the same mass with a larger temp swing. For another, a lot of heat leaks
out of the glazing at night and on cloudy days. Isolated sunspaces and air
heaters can gain more heat during the day and lose a lot less at night or
on cloudy days.
You don't want to collect more heat than you can store, because extra
glazing adds to the heat loss but doesn't help much on an average day.
You might remove its thermal mass and add an insulated wall to close it off
from the rest of the living space at night and on cloudy days.
How about some thermal mass in ceilings, eg a layer of sand above the
ceiling drywall or plywood and below the attic floor insulation, where
it can be warmer than the living space?
The 1979 NRC Solarium Design Guide for Canadian homes implies December is
the worst-case month for solar heating in Ottowa, when 2.257 kWh/m^2 (715
Btu/ft^2) of sun falls on a south wall on an average -8 C (17.6 F) day, so
your 360 ft^2 of windows would admit 360x0.51x715 = 131.3K Btu of solar heat
on an average day. A frugal 600 kWh/mo of indoor electrical usage would add
2842 Btu/h to that, even on cloudy days. With a G Btu/h-F house conductance
to outdoors, electrical usage has the effect of raising the outdoor temp to
17.6+2842/G, even on cloudy days. For instance, 200 Btu/h-F effectively
raises the outdoor temp to 17.6F+2842Btu/h/(200Btu/h-F) = 31.8 F.
You might imagine the house is 70 F at dusk on an average day and 60 F at
dawn, and figure out how much thermal mass with a 10 F day-night swing is
needed to store 18 hours of heat. The house might be 80 F at dusk on a
clear day. You might weight the mass based on its position and depth. Dan
Chiras says each square foot of glazing in excess of 7% of the floorspace
requires 5.5 ft^2 of uncovered sunlit floor mass or 40 ft^2 of unlit floor
mass or 8.3 ft^2 of wall mass... Then again, that's just a rule of thumb.
It really depends on the amount of insulation and the climate. You might put
5 vertical 8'x4" sealed PVC water pipes in each 2'x2x6 inside wall cavity,
with holes at the top and bottom for ventilation...
You might figure cloudy days are like coin flips, with 2x715 Btu/ft^2 on
sunny days and 0 on cloudy days. A house that can store 1 day's worth of heat
can have a 50% max solar heating fraction, with 75% for 2 days, 88 for 3, and
100(1-2^(-N))% for N. But real weather changes more slowly, so you might use
a die instead of a coin and toss for cloudy days (1 or 2), average days (3
or 4), and clear days (5 or 6), remembering that more glass loses more heat
on cloudy days.
Posted by Eric Jacobsen on September 29, 2003, 5:04 pm
interesting. but watch out when you go to put a nail in the wall to hang a
Posted by Doug on September 30, 2003, 1:29 am
That was a bit glib (and lazy) - sorry Nick. I really would like to
know how to do this, though.
I'm guessing that Ottawa gets ~700 Btu/ft^2 on a south wall in January
with an average temp ~ 25F. With a SHGC of .5, that'd be ~122K5
Btu's. The 6" cement slab with ~700 ^ft in direct sun would gain
700x6/12x25 = 8750 Btu/F. Let's assume a factor of .5 for the rest of
the cement slab out of the direct sunshine... 0.5x1100x6/12x25 = 6875
Btu/F degree. A swing of ~10 F degrees would yield 156250 Btu's. So
it looks like the mass of the cement slab can store the available heat
But now I'm a bit stuck. If you take sq ft divided by the R factor,
you get Btu/h-F degrees. The house is being built with ICF which has
3" of polysturene on the outside and on the inside with 6" of cement
sandwiched in between. Sadly, the total is only R24. However, this
is the effective R value and is "supposed to be equivalent to R??"
with normal construction. Guess R30 or R35? Dunno.
However, just taking the south windows of 360 ft^2 with an R of ~3, we
lose 120 Btu's /hour x F degree. If the ambient inside temp is 68
degrees, the differential would be 43 degrees, giving a loss of 123K
Looks like a write-off.
Comments would be appreciated, s'il te plait ;-).