Hybrid Car – More Fun with Less Gas

Re: New house - Should I consider gas heat?

register ::  Login Password  :: Lost Password?
Posted by nicksanspam on May 9, 2005, 1:51 pm
 


January is the worst-case month for solar house heating in Norfolk, with
the least "sun per degree day," ie 1080/(70-39.1) = 35.0 Btu/F in January
vs 1040/(70-43.8) = 39.7 in December. "Worst-case design" is common in
aerospace engineering, but rarely used for solar houses. If it were, more
would be close to 100% solar-heated. It isn't hard. If cloudy days are
coin-flips and a house can store enough solar heat for 1 cloudy day, it
can be at most 50% solar-heated... 2 days make 75% max, 3, 88%, 4, 94%,
and 5, 97%, but few store heat for more than 1 cloudy day.

First make sure the house can keep itself warm on an average day in the
worst-case month: a 48'x46'x8' tall house with 200 ft^2 of R4 windows and
R16 SIP walls and ceiling and 0.2 ACH of air leaks would have 200ft^2/R4
= 50 Btu/h-F of window thermal conductance + 1336ft^2/R16 = 84 for walls
+ 2304/16 = 144 for the ceiling + about 0.2x2304x8/60 = 61 for air leakage,
totaling 339 Btu/h-F. On an average January day in Norfolk, it would need
24h(65-39.1)339 = 210.7K Btu of heat.

A frugal 600 kWh/mo of indoor electrical use could provide 68.2K Btu/day.
NREL says 470, 480, and 210 Btu/ft^2 fall on east, west, and north walls, and
710 falls on the ground, so 100 ft^2 of south windows and 50 on the east
and west walls and 25 on the north with 50% solar transmission would gain
0.5x25(4x1080+2x470+480+210) = 74,275 Btu/day, leaving 210.7K-74,275 = 68.2K.
That might come from 68.2K/750 = 91 ft^2 of south sunspace glazing.

With no sun, the house needs 210.7K-68.2K = 142.5K Btu on a 39.1 F day,
or 712.5K for 5 cloudy days. At 70 F, with no electrical use, it needs
(70-39.1)339 = 10.5K Btu/h, eg 2304 ft^2 of R1 radiant floor with water
at 70+10.5K/2304 = 74.5 F from a 4'x8'x4' 7978 lb EPDM-lined plywood box
that cools from 74.5+712.5K/7978 = 164 F to 74.5 F over 5 cloudy days.

As an alternative to a radiant floor and tank, hot water might live in
poly film ducts on plywood shelves under the ceiling, with foil under the
shelves to avoid overheating the rooms by radiation and slow ceiling fans
and room temp thermostats to move warm air down as needed on cloudy days.

If 10.5K Btu/h flows from temp T water through 24'x48' of R0.27 shelf
surface into 70 F air with a 10.5K/(24x48/0.27) = 2.4 F temp drop, the
min usable water temp is 72.4 F. If it's initially 110 F, the ducts need
712.5K/(110-72.4) = 18.9K lb, ie 12x18.9K/(24x48x62.33) = 3.2 inches of
water (with 800' ($0) of 1/2" PE pipe inside to preheat pressurized water
for showers and a simple greywater-house air heat exchanger.) We already
figured the ceiling would lose 24h(65-39.1)2304/16 = 89.5K Btu on a cloudy
day. Over (110+72.4)/2 = 91.2 water for 5 cloudy days, it would lose
89.5K/day if 24h(91.2-39.1)2304/R = 89.5K, with an R = 32 ceiling SIP.

On an average day, it would lose about 24h(110-39.1)2304/32 = 122.5K Btu,
33K more than we figured before. A solar attic might collect 68.2K Btu
more than that, ie 101.2K Btu/day in 110 F water, to avoid the need for a
sunspace. The attic ridge might be 4' above the ceiling, with 8' of clear
corrugated polycarbonate Dyanglas roofing that slopes down to the south
wall at ceiling level and a 4'x24' draindown pond solar collector on the
6.92'x48' R32 floor. The north part of the attic could be a cathedral
ceiling with exposed rafters to form a truss that supports duct shelves.
A few clerestory windows in the 4'x48' reflective back wall of the attic
would add light and architectural drama.

On an average Jan day, the pond (a 4'x24' layer of poly film over 2" of
water over a 4'x24' piece of EPDM rubber) would collect approximately
0.9x24(4x710+0.9x2.27x1080) = 109K Btu of direct and reflected sun and
lose about 6h(110-T)4'x24'x1.5 = 864(110-T) Btu to T (F) attic air. The
non-pond attic would gain about 0.9x(236x710+234x1080) = 378.3K Btu plus
864(T-110) from the pond and lose about 6h(T-43)8x48/R1 = 2304(T-43),
which makes 378.3K + 864(110-T) = 2304(T-43), so T = 181 F, or less,
with some white attic floor and more radiation loss.


An interesting question. Most "solar" houses are only 30-60% solar-heated,
so the backup fuel cost is important. PE Norman Saunders estimates how
often his near-100% solar heated houses in cold, cloudy New England will
need "purchased energy" like other engineers estimate 100-year floods.
("Your house will need backup heat for 3 hours every 35 years.") And his
predictions come true. He's been designing solar houses since 1946, and
some have long track records with digital data loggers.

He suggests buying a 5 kW electric space heater for backup, figuring it's
inexpensive if rarely used, but some of his clients have never done that.
They prefer to wear sweaters indoors every 35 years :-)

Nick


This Thread
Bookmark this thread:
 
 
 
 
 
 
  •  
  • Subject
  • Author
  • Date
please rate this thread