Let's do the math.
Since November here we had about 2 hours total sun in 45 days, but we
will consider the example given by others.
My house, to be build next year, will take about 50,000 BTU/hr on cold
days.
For 14 days of no solar addition 50K x 24hr. x 14 days = 16,800K BTUs
If we need aprox 110F water to hydronically heat our house
We assume we can heat our storage tank to aprox 160F using solar.
We have our tank with a 160F-110F = 50 degress F. heat reserve
Now if raising 1 lb of water 1 deg. F takes 1 BTU then to store
16,800K BTU
we need 16,800K / 50F = 336K lb. water in tank
336 K lb water / 62.5 lb = 5376 cu feet of water in tank.
Now if my house is 1800 square feet we have a depth of
5376 cu ft. / 1800 ft^3 = 2.986 feet in depth.
This means I would need a sub-basement full of water the size of my
house and almost 3 feet deep. Now we haven't considered losses in the
ground, the mould, dampness, leaks, repairs, extra trusses for
supporting, the sheer weight to make the structure sink or maintenance
of the tank.
It was tried in many places back in the 80s and it never worked 100%.
You could heat your home, your children's and your grandchildren's
homes for cheaper with NG. We wouldn't have as much fun though.
"SJC" wrote in message
news
36dh.2019$ne3.567@trndny03...
Without evidence, quite frankly your opinion is worthless.
"Solar Flare" wrote in message
...
It has been tried and it never works.
"SJC" wrote in message
news:TJ2dh.2065$lb1.1849@trnddc05...
"Cosmopolite" wrote in message
news:9A2dh.29751$dX4.28149@clgrps13...
Nick Pine wrote:
People still doubt that houses can be close to 100% solar heated
outside of
the Southwest, inexpensively. Here's another 8' D-cube that
might be deployed
to regional Infestations of Doubt, with 2" double-foil
polyisocyanurate walls
and ceiling and an 8'x8' layer of Thermaglas Plus twinwall
polycarbonate over
the south wall and 0.44 inches of water under the ceiling and
6.86" of water
in a 6'x6' EPDM-rubber-lined cloudy day heat storage tank under
the floor. On an average day, thermosyphoning sunspace air would
heat the ceiling mass
to about 160 F, and it would cool to about 80 by dawn. A room
temp thermostat
with a 2-watt damper motor would rotate foil foamboard louvers
below the mass
to keep the room 70 F by radiation. On a cloudy day, a low-power
pump would
move hot water up from the floor tank through the ceiling. 20
TA=30'outdoor temp (F)
30 TI=70'indoor temp (F)
40 SS=1000'south sun (Btu/ft^2-day)
50 TAU=.8'south glazing solar transmission
60 UVG=.58'south glazing U-value
70 SUN=64*TAU*SS'sun in (Btu/day)
80 RVW=15.5'wall R-value
90 RVS=RVW+1/UVG'south wall R-value
100 HC=(TI-TA)*64*(1/RVS+3/RVW)'required ceiling heat (Btu/h)
110 GC=4*1.714E-09*(TI+5+460)^3'min ceiling rad conductance
(Btu/h-F-ft^2)
120 TMIN=TI+HC/GC/64'min ceiling temp (F)
130 SWNL=18*(TI-TA)*64/RVS'south wall night loss (Btu)
140 OWL=24*(TI-TA)*3*64/RVW'other wall loss (Btu/day)
150 RC=15.5'ceiling R-value
160 NETSUN=SUN-SWNL-OWL+6*TA*64*UVG-(TMIN/2-TA)*64/RC'net sun
(Btu/day)
170 TS=NETSUN/(6*64*UVG+12*64/RC)'sunspace day temp (F)
180 TC=(TS+TMIN)/2'est. average ceiling temp (F)
190 ONH=SWNL+.75*OWL+18*(TC-TA)/RC'overnight heat (Btu)
200 C=ONH/(TS-TMIN)'ceiling mass (Btu/F)
210 DW=12*C/62.33/64'ceiling water depth (inches)
220 CDH=(TI-TA)*64*(1/RVS+3/RVW)+(TMIN-TA)*64/RC'cloudy day heat
(Btu/h)
230 CD5=5*24*CDH'heat for 5 cloudy days (Btu)
240 CC=CD5/(TS-TMIN)'cloudy day capacitance (Btu/F)
250 CTD=12*CC/62.33/36'cloudy day floor tank depth (inches)
260 PRINT TS,TMIN
270 PRINT C,DW,CTD
sunspace min ceiling
temp (F) temp (F)
158.9396 79.58628
ceiling water depths (inches) cap (Btu/F) ceiling floor
147.4126 .443444 6.864856
Nick
What about 2 weeks of overcast sky and 10 below temp ?
In that case, use a large thermal store. I favor the earth under
the house.