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

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 ?

"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.

Nick Pine wrote:
People still doubt that houses can be close to 100% solar heated outside of
the Southwest, inexpensively.

I'd like to see the numbers for Saskatoon, SK, Canada.

Winter temps down to sub -40, but we do get a fair bit of sun.

Chris

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.

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.

It seems like an odd design. Why not make the water tank a cube or
cylinder shape and use existing hydronic heating technology to warm
the house. That is to say, circulating hot water through plastic
pipe in the floors. This eliminates .44 inches of water under the
ceiling and 6.86" of water under the floor and the motorized foamboard
louvers. I will note that normal foamboard is a fire hazard. If you
need hot water in the walls and ceiling then all it requires is
more plastic pipe.

Then all you need is a method to heat the water using sunlight. I
believe there are plenty of existing technologys to do this. For
backup use a natural gas, electric or propane water heater.

Ok, if we want to go with unusual ideas then how about this one?
Part of how people perceive temperature is based on the temperature
of the walls, ceiling and floor. What if the walls, ceiling and
floor were covered in an infrared reflecting coating? Specifically,
something that would reflect IR roughly back where it came from,
much like how road reflectors reflect visible light.

Anthony

Anthony Matonak (in ) said:

| Ok, if we want to go with unusual ideas then how about this one?
| Part of how people perceive temperature is based on the temperature
| of the walls, ceiling and floor. What if the walls, ceiling and
| floor were covered in an infrared reflecting coating? Specifically,
| something that would reflect IR roughly back where it came from,
| much like how road reflectors reflect visible light.

Fantastic! Construct the room as a rectangular parallelepipedon and
bond aluminum foil (shiny side in) to all six surfaces...

It sounds a lot like a turkey roaster :-)

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

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.

Ditto

"SJC" wrote in message
news36dh.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.

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
news36dh.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.

Nick-sans has a new account? Gawd, you're tiresome. Tom

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

Cosmopolite wrote:

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...

230 CD5=5*24*CDH'heat for 5 cloudy days (Btu)...

What about 2 weeks of overcast sky and 10 below temp ?

That's very unlikely in Phila, where it's above 14 F 97.5% in wintertime
and above 10 99% of the time, ie for all but 22 hours per year. NREL's
30-year record min is 7 below. A TMY2 simulation would be interesting.

Nick

Anthony Matonak wrote:

Nick Pine wrote:

People still doubt that houses can be close to 100% solar heated outside of
the Southwest, inexpensively...

Why not make the water tank a cube or cylinder shape and use existing
hydronic heating technology to warm the house.

Sounds more expensive and less portable. We could get rid of the ceiling mass
and louvers and heat the cloudy day tank with some fin-tube pipe near the top
of the air heater, and let the top of the tank warm air that warms the cube
on a cloudy day, with a thermal chimney and a motorized damper to control
the tank airflow.

... I will note that normal foamboard is a fire hazard.

Since 1995? :-) Foil-faced polyiso seems fine to me.

Ok, if we want to go with unusual ideas then how about this one?
Part of how people perceive temperature is based on the temperature
of the walls, ceiling and floor. What if the walls, ceiling and
floor were covered in an infrared reflecting coating? Specifically,
something that would reflect IR roughly back where it came from,
much like how road reflectors reflect visible light.

Foil-faced foamboard could be a start. You seem to be describing something
more mirrorlike, in a corner reflector. Where can we buy that? :-)

Nick

Solar Flare wrote:

Let's do the math.

Goody.

Since November here we had about 2 hours total sun in 45 days...

You live above the arctic circle? :-)

My house, to be build next year, will take about 50,000 BTU/hr on cold days.

Wow. What will it take on an average December day? Why not add more
insulation and airseal and zone and make it smaller?

For 14 days of no solar addition 50K x 24hr. x 14 days = 16,800K BTUs

Sounds unrealistic. Where will this house be?

Now if my house is 1800 square feet...

And it needs 50K Btu/h, with 3200 ft^2 of R32 walls and ceiling, and
it's T (F) outdoors, with 50K = (70-T)3200ft^2/R32, T = minus 430 F.
That's a very cold day :-)

Nick

wrote:
Anthony Matonak wrote:
Nick Pine wrote:
People still doubt that houses can be close to 100% solar heated outside of
the Southwest, inexpensively...

Why not make the water tank a cube or cylinder shape and use existing
hydronic heating technology to warm the house.

Sounds more expensive and less portable.

Portability isn't much of an issue as a water tank can be made
out out plastic or fiberglass and doesn't weigh much when empty.
For the floors, you are discussing an 8x8 foot floor which could
be made out of two 4x8 foot sections. Each section could have
a length of plastic tubing with some form of quick-connect.
Compressed air could be blown through the tubes to empty them of
water prior to shipping.

... I will note that normal foamboard is a fire hazard.

Since 1995? :-) Foil-faced polyiso seems fine to me.

Since always, I would imagine.

http://www.jm.com/insulation/buildin...foil-faced.pdf
: As with all foam plastics, this product will burn. ... AP sheathing
: requires an interior finish of a minimum 1/2" (13 mm) gypsum board
: or equivalent 15-minute fire barrier.

What if the walls, ceiling and
floor were covered in an infrared reflecting coating?

Foil-faced foamboard could be a start. You seem to be describing something
more mirrorlike, in a corner reflector. Where can we buy that? :-)

I would imagine that some material could be found that operates in
this fashion. Road signs and/or road paint use very small glass beads.
Some visibility tape and garments use plastic embossed on a microscopic
scale. Perhaps foil embossed with microscopic corner reflectors or paint
mixed with some mineral that has IR reflecting properties.

This outfit sells IR reflecting paint, for instance.
http://www.ntt-at.com/products_e/atshield/index.html

Anthony

Anthony Matonak wrote:

Why not make the water tank a cube or cylinder shape and use existing
hydronic heating technology to warm the house.

Sounds more expensive and less portable.

Portability isn't much of an issue as a water tank can be made
out out plastic or fiberglass and doesn't weigh much when empty.

One might say the same for EPDM rubber.

For the floors, you are discussing an 8x8 foot floor which could
be made out of two 4x8 foot sections. Each section could have
a length of plastic tubing with some form of quick-connect.

On a cloudy day, this cube would need about (70-30)20 = 800 Btu/h of heat,
which might come from 72 ft^2 of tank and floor surface with thermosyphoning
air with a 1.5 Btu/h-F-ft^2 film conductance at 70 + 800/72//1.5 = 77.4 F.
On an average day, this could work with a 18x800/20 = 720 Btu/F mass with
a 20 F day/night temp swing, vs a 0 F swing with the 154 Btu/F ceiling mass.

How would we build the floor with tubing and what would it cost and
what would the water temp be for the same heat transfer? Sounds like
it would need a pump.

... I will note that normal foamboard is a fire hazard.

Since 1995? :-) Foil-faced polyiso seems fine to me.

Since always, I would imagine.

IIRC, uncovered foamboard became a fire hazard fairly recently.
Aluminum louvers would also work, with radiative ceiling heat,
as in Zomeworks Architectural Cool Cells.

Nick

Chris Friesen wrote:

People still doubt that houses can be close to 100% solar heated outside of
the Southwest, inexpensively.

I'd like to see the numbers for Saskatoon, SK, Canada.

Energy Plus stats say 1007 kWh/m^2 of sun (472 diffuse) falls on the ground
in December in Saskatoon, when the average outdoor temp is -13.7 C. The NRC
Solarium Workbook says 2903 Wh/m^2 (921 Btu/ft^2) of sun falls on a south
wall on an average -11 C (12 F) December day in Swift Current, SK, which
leads to a D-cube with another inch of foamboard and more mass:

20 TA=12'outdoor temp (F)
30 TI=70'indoor temp (F)
40 SS=921'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=22.7'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=RVW'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'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 tank depth (inches)
260 PRINT TS,TMIN
270 PRINT C,DW,CTD

sunspace min ceiling
temp (F) temp (F)

137.3602 79.56306

ceiling water depths (inches)
cap (Btu/F) ceiling floor

201.4366 .6059579 9.249586

Nick

Chris Friesen wrote:

People still doubt that houses can be close to 100% solar heated outside of
the Southwest, inexpensively.

I'd like to see the numbers for Saskatoon, SK, Canada.

Energy Plus stats say 1007 Wh/m^2 of sun (472 diffuse) falls on the ground
on an average -13.7 C day in December in (52 N lat) Saskatoon... which leads
to a D-cube with another inch of foamboard and a little more mass:

10 SCREEN 9:KEY OFF:PI=4*ATN(1)
20 LD=52+10/60'north latitude (degrees)
30 L=LD*PI/180'radians
40 DECD=-23'December declination (degrees)
50 DEC=DECD*PI/180'radians
60 X=-TAN(L)*TAN(DEC)
70 HSR=ATN(X/SQR(-X*X+1))-PI/2'sunrise hour angle -arccos(x) (radians)
80 BETAD=90'surface tilt (degrees)
90 BETA=BETAD*PI/180'radians
100 NUM=COS(L-BETA)*COS(DEC)*SIN(HSR)+HSR*SIN(L-BETA)*SIN(DEC)
110 DEN=COS(L)*COS(DEC)*SIN(HSR)+HSR*SIN(L)*SIN(DEC)
120 RB=NUM/DEN'tilted surface to horizontal beam rad rat
130 RD=(1+COS(BETA))/2'tilted to horizontal diff rad rat
140 RHOG=.6'ground reflectivity
150 RR=RHOG*SIN(BETA/2)*SIN(BETA/2)'ground reflectance factor
160 IGLOH=1.007*317.1'global horizontal radiation (Btu/ft^2)
170 IDIFH=.472*317.1'diffuse horizontal radiation (Btu/ft^2)
180 IBEAMH=IGLOH-IDIFH'beam horizontal radiation (Btu/ft^2)
190 SS=RB*IBEAMH+RD*IDIFH+RR*IGLOH'south sun (Btu/ft^2-day)
200 TAU=.8'south glazing solar transmission
210 UVG=.58'south glazing U-value
220 SUN=64*TAU*SS'sun in (Btu/day)
230 RVW=22.7'wall R-value
240 RVS=RVW+1/UVG'south wall R-value
250 TA=1.8*(-13.7)+32'outdoor temp (F)
260 TI=70'indoor temp (F)
270 HC=(TI-TA)*64*(1/RVS+3/RVW)'required ceiling heat (Btu/h)
280 GC=4*1.714E-09*(TI+5+460)^3'min ceiling rad conductance (Btu/h-F-ft^2)
290 TMIN=TI+HC/GC/64'min ceiling temp (F)
300 SWNL=18*(TI-TA)*64/RVS'south wall night loss (Btu)
310 OWL=24*(TI-TA)*3*64/RVW'other wall loss (Btu/day)
320 RC=RVW'ceiling R-value
330 NETSUN=SUN-SWNL-OWL+6*TA*64*UVG-(TMIN/2-TA)*64/RC'net sun (Btu/day)
340 TS=NETSUN/(6*64*UVG+12*64/RC)'sunspace day temp (F)
350 TC=(TS+TMIN)/2'est. average ceiling temp (F)
360 ONH=SWNL+.75*OWL+18*(TC-TA)/RC'overnight heat (Btu)
370 C=ONH/(TS-TMIN)'ceiling mass (Btu/F)
380 DW=12*C/62.33/64'water depth (inches)
390 CDH=(TI-TA)*64*(1/RVS+3/RVW)+(TMIN-TA)*64/RC'cloudy day heat (Btu/h)
400 CD5=5*24*CDH'heat for 5 cloudy days (Btu)
410 CC=CD5/(TS-TMIN)'cloudy day capacitance (Btu/F)
420 CTD=12*CC/62.33/36'cloudy day tank depth (inches)
430 PRINT TS,TMIN
440 PRINT C,DW,CTD

sunspace min ceiling
temp (F) temp (F)

151.1164 80.33141

ceiling water depths (inches)
cap (Btu/F) ceiling floor

177.7378 .5346678 8.159242

The next step might be a TMY2 simulation with hourly Energy Plus data.

Nick

"Nick Pine" wrote in message
...
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)

Nick, I believe the last term in line 160 above should be,
"...-24*(TMIN/2-TA)*64/RC". If the RC is in the usual units of 'R-value'
(it seems so, considering its usage in line 220), then what you have in the
term above is only BTU/hr versus BTU/day in all the other terms. But this
only lowers performance slightly (TS comes out to 155.5F instead of 158.9F)

You don't seem to account for any heat to 'charge' the under-floor tank.
With it in/under the floor, you would have to run the pump on sunny days to
'charge' it, as well as on cloudy days to 'discharge' it. If it takes five
days to 'charge' it for five cloudy days, wouldn't that about double the
heat load on the sun space? Is the ratio of sunny/cloudy days in
Philadelphia about 50/50?

Heat losses through the floor? Granted, the ground is warmer, it isn't 70F.

Anyway, all well and good if you want to live in an 8' cube with no windows
or doors. A pair of St. Bernards' doghouse?? What about a 'practical'
house, say a very modest 1200 sq ft single story with only a few windows,
say 48 ft^2 of double glazed (that's about one double-hung window in each
wall). And a paltry 2 air-changes per day.

Would such a 'practical' house still perform in the 'regional infestations
of doubt'?? What you've posted here has very little to do with heating what
most people would call "a house". (even a dog house has a door)


daestrom

If I thought you were actually serious with any part of your retirt I
would address some of the bad math and knowledge you displayed. I am
sure if you think about it you will see what I have siad has been
based on actual figures.

wrote in message
...
Solar Flare wrote:

Let's do the math.

Goody.

Since November here we had about 2 hours total sun in 45 days...

You live above the arctic circle? :-)

My house, to be build next year, will take about 50,000 BTU/hr on
cold days.

Wow. What will it take on an average December day? Why not add more
insulation and airseal and zone and make it smaller?

For 14 days of no solar addition 50K x 24hr. x 14 days = 16,800K
BTUs

Sounds unrealistic. Where will this house be?

Now if my house is 1800 square feet...

And it needs 50K Btu/h, with 3200 ft^2 of R32 walls and ceiling, and
it's T (F) outdoors, with 50K = (70-T)3200ft^2/R32, T = minus 430 F.
That's a very cold day :-)

Nick

wrote:
Anthony Matonak wrote:

Why not make the water tank a cube or cylinder shape and use existing
hydronic heating technology to warm the house.

Sounds more expensive and less portable.

Portability isn't much of an issue as a water tank can be made
out out plastic or fiberglass and doesn't weigh much when empty.

One might say the same for EPDM rubber.

Sure, you can make water pools or bladders out of EPDM rubber
but you have to make them while water tanks are off the shelf
items. It just seems easier to me to use existing technology.

For the floors, you are discussing an 8x8 foot floor which could
be made out of two 4x8 foot sections. Each section could have
a length of plastic tubing with some form of quick-connect.

How would we build the floor with tubing and what would it cost and
what would the water temp be for the same heat transfer? Sounds like
it would need a pump.

I don't have blueprints if that is what you are asking. I know
all kinds of companies sell stuff to put water heating in floors
so I can't believe it's terribly difficult or expensive.

I don't see a major problem with having a pump and your original
plan suggested the use of a pump on some days anyhow.

... I will note that normal foamboard is a fire hazard.

Since 1995? :-) Foil-faced polyiso seems fine to me.

Since always, I would imagine.

IIRC, uncovered foamboard became a fire hazard fairly recently.

I can't believe a product suddenly "becomes" a fire hazard.
I can believe that it wasn't always widely recognized as being
the fire hazard that it is.

Anthony

Solar Flare wrote:

If I thought you were actually serious with any part of your retirt...

You might be able to spell "retort"? :-)

I would address some of the bad math and knowledge you displayed. I am
sure if you think about it you will see what I have siad has been
based on actual figures.

wrote in message
...
Solar Flare wrote:

Let's do the math.

Goody.

Since November here we had about 2 hours total sun in 45 days...

You live above the arctic circle? :-)

My house, to be build next year, will take about 50,000 BTU/hr on
cold days.

Wow. What will it take on an average December day? Why not add more
insulation and airseal and zone and make it smaller?

For 14 days of no solar addition 50K x 24hr. x 14 days = 16,800K
BTUs

Sounds unrealistic. Where will this house be?

Now if my house is 1800 square feet...

And it needs 50K Btu/h, with 3200 ft^2 of R32 walls and ceiling, and
it's T (F) outdoors, with 50K = (70-T)3200ft^2/R32, T = minus 430 F.
That's a very cold day :-)

Nick

"Solar Flare" writes:

If I thought you were actually serious with any part of your retirt I
would address some of the bad math and knowledge you displayed. I am
sure if you think about it you will see what I have siad has been
based on actual figures.

While Nick has been less than useful in his other recent replies,
it might be useful to the rest of us for you to take the
high road by providing the actual figures or math
that is the source of your 50K BTU/hr.

Nick's post has math. Do you have math to rebut it?


wrote in message
...
Solar Flare wrote:

Since November here we had about 2 hours total sun in 45 days...

Subjectively, it sounds like my memory of Seattle.
Objectively, on what days and times did "here" have sun?
If you dont have the datalog,
then perhaps this figure of 'about 2 hours' is not 'actual'

My house, to be build next year, will take about 50,000 BTU/hr on
cold days.

If your house is not yet built, how can this figure be actual?

Is this 50K BTU/hr the assembly cost during cold days?
Or the replacement heat needed to keep a constant internal
temperature delta above a constant external temperature?

For 14 days of no solar addition 50K x 24hr. x 14 days = 16,800K
BTUs
Now if my house is 1800 square feet...

And it needs 50K Btu/h, with 3200 ft^2 of R32 walls and ceiling, and
it's T (F) outdoors, with 50K = (70-T)3200ft^2/R32, T = minus 430 F.
That's a very cold day :-)

Nick, where does 3200 ft^2 come from?
A square house 42' on a side has 1764 ft^2 = ~1800ft^2 floorspace.
With 8' tall walls, there are 4*42*8=1344ft^2 walls.
A 42*42 floor plus 42*42 ceiling is 3528 ft^2
for a total of 4872 ft^2 surface area.

If we assume T=0 outdoors, and use your same eqn,
50,000 BTU/hr = (70-0)*4872ft^2/R = 341040/R and solve for R,
we see that 50K BTU/hr fits this eqn model for such a house
insulated to R=6.8, (or R6.8 with some allowance for air exchange.)

Conversely, If the 1800 ft^2 house is insulated to
Nick's proposed R=32 (with no allowance for air exchange),
we see that it will take only ~11K BTU/hr (vs original 50K)
to maintain 70F while its 0F outside.

Regarding Nick's DDD,
How much insulation is between the interior floor,
and the heat store, the same R=32?

Would moveable louvers and thermosiphon air (instead of a pump)
provide heating from the heatstore during cloudy days?

Assume that to attract passersby, we need a window to view the
circle chart recorder sitting on a table inside,
(next to the cage with the asphyxiated gerbil.)
The chart recorder shows inside and outside temperatures.
What type of window glazing should be used?

To keep the gerbil alive, suppose we require a 0.0004 ft^2 hole
(about 1/4 in ^2) in the wall for air exchange.
How does that ECN affect the design?

--

wrote in message
.


Now if my house is 1800 square feet...


And it needs 50K Btu/h, with 3200 ft^2 of R32 walls and ceiling, and
it's T (F) outdoors, with 50K = (70-T)3200ft^2/R32, T = minus 430 F.
That's a very cold day :-)



Obviously, his walls are NOT R32, or even close. With windows,

even good thermopane ones, it is hard to come anywhere near the insulation
level of the wall. Unless you can live with tiny view-ports in your walls,
it is a bit tough to get anywhere near an R32 average for the entire
wall area.
(There are special construction techniques like double walls, etc. that can
get there.)

Also, maybe the 50 K BTU/Hr is the capacity of the furnace, and hopefully
it will rarely have to run flat out.

As for 2 hrs sunshine/Mo. I would not want to live there!

Jon

Chris Friesen wrote:
Nick Pine wrote:
People still doubt that houses can be close to 100% solar heated
outside of
the Southwest, inexpensively.

I'd like to see the numbers for Saskatoon, SK, Canada.

Winter temps down to sub -40, but we do get a fair bit of sun.

Chris


Chris,

Surely even in Saskatoon sub -40 is EXTREMELY rare (unless your
including wind chill). In winnipeg I'd be shocked if it happened once or
twice a winter.

The heating degree days in winnipeg is around 10,000; i'm guessing
similar to Saskatoon.

-darryl

I figured it was just a troll. I was correct on that one.

wrote in message
...
Solar Flare wrote:

If I thought you were actually serious with any part of your
retirt...

You might be able to spell "retort"? :-)

I would address some of the bad math and knowledge you displayed. I
am
sure if you think about it you will see what I have siad has been
based on actual figures.

wrote in message
...
Solar Flare wrote:

Let's do the math.

Goody.

Since November here we had about 2 hours total sun in 45 days...

You live above the arctic circle? :-)

My house, to be build next year, will take about 50,000 BTU/hr on
cold days.

Wow. What will it take on an average December day? Why not add
more
insulation and airseal and zone and make it smaller?

For 14 days of no solar addition 50K x 24hr. x 14 days = 16,800K
BTUs

Sounds unrealistic. Where will this house be?

Now if my house is 1800 square feet...

And it needs 50K Btu/h, with 3200 ft^2 of R32 walls and ceiling,
and
it's T (F) outdoors, with 50K = (70-T)3200ft^2/R32, T = minus 430
F.
That's a very cold day :-)

Nick

Sounds like a few people get it.

Yes, I am not going to live in a dark box, I will have windows. If
nick has ever done a complete heat study for a building he would know
his figures are uneducated bull**** and his whole retort was just a
troll, not worth discussing. I am not going to dispute the heat study
done. The 50K was a little high and actually was in the order of 45K
at 95F dT.

3200ft^2 in a 1800ft^2 building. I cannot live without a floor or
windows, come to think of it.

The whole point was this long term heat storage is an old topic and
tried by many design engineers in the 70 and 80s and never worked 100%
even in some moderate climates. The storage costs too much and most
don't have the property to support the size.

"Jon Elson" wrote in message
...


wrote in message
.

Now if my house is 1800 square feet...

And it needs 50K Btu/h, with 3200 ft^2 of R32 walls and ceiling,
and
it's T (F) outdoors, with 50K = (70-T)3200ft^2/R32, T = minus 430
F.
That's a very cold day :-)


Obviously, his walls are NOT R32, or even close. With windows,

even good thermopane ones, it is hard to come anywhere near the
insulation
level of the wall. Unless you can live with tiny view-ports in your
walls,
it is a bit tough to get anywhere near an R32 average for the entire
wall area.
(There are special construction techniques like double walls, etc.
that can
get there.)

Also, maybe the 50 K BTU/Hr is the capacity of the furnace, and
hopefully
it will rarely have to run flat out.

As for 2 hrs sunshine/Mo. I would not want to live there!

Jon

On Wed, 6 Dec 2006 19:55:54 -0500, "Solar Flare"
wrote:

Sounds like a few people get it.

Yes, I am not going to live in a dark box, I will have windows. If
nick has ever done a complete heat study for a building he would know
his figures are uneducated bull**** and his whole retort was just a
troll, not worth discussing. I am not going to dispute the heat study
done. The 50K was a little high and actually was in the order of 45K
at 95F dT.

3200ft^2 in a 1800ft^2 building. I cannot live without a floor or
windows, come to think of it.

You merely need to gain the insite to understand Nick's
concept of 'modern living', which include the ability to levitate,
exceptional night vision, and most often a herd of ASHRAE Standard
bunnies in the basement. Then his ideas ....... ummmmm..... STILL
don't make any sense :-)



The whole point was this long term heat storage is an old topic and
tried by many design engineers in the 70 and 80s and never worked 100%
even in some moderate climates. The storage costs too much and most
don't have the property to support the size.

"Jon Elson" wrote in message
...


wrote in message
.

Now if my house is 1800 square feet...

And it needs 50K Btu/h, with 3200 ft^2 of R32 walls and ceiling,
and
it's T (F) outdoors, with 50K = (70-T)3200ft^2/R32, T = minus 430
F.
That's a very cold day :-)


Obviously, his walls are NOT R32, or even close. With windows,

even good thermopane ones, it is hard to come anywhere near the
insulation
level of the wall. Unless you can live with tiny view-ports in your
walls,
it is a bit tough to get anywhere near an R32 average for the entire
wall area.
(There are special construction techniques like double walls, etc.
that can
get there.)

Also, maybe the 50 K BTU/Hr is the capacity of the furnace, and
hopefully
it will rarely have to run flat out.

As for 2 hrs sunshine/Mo. I would not want to live there!

Jon





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http://www.theanimalrescuesite.com/

Paul ( pjm @ pobox . com ) - remove spaces to email me
'Some days, it's just not worth chewing through the restraints.'
'With sufficient thrust, pigs fly just fine.'
HVAC/R program for Palm PDA's
Free demo now available online http://pmilligan.net/palm/

On a quick glance of the calculations the doors and wondows will
consume about 12K BTU/h of the heat losses.

No, the dark weather is rare here. It has been an exeptional year for
bleak. Michigan needs to shut Ohio down or move further away.

"Jon Elson" wrote in message
...


wrote in message
.

Now if my house is 1800 square feet...

And it needs 50K Btu/h, with 3200 ft^2 of R32 walls and ceiling,
and
it's T (F) outdoors, with 50K = (70-T)3200ft^2/R32, T = minus 430
F.
That's a very cold day :-)


Obviously, his walls are NOT R32, or even close. With windows,

even good thermopane ones, it is hard to come anywhere near the
insulation
level of the wall. Unless you can live with tiny view-ports in your
walls,
it is a bit tough to get anywhere near an R32 average for the entire
wall area.
(There are special construction techniques like double walls, etc.
that can
get there.)

Also, maybe the 50 K BTU/Hr is the capacity of the furnace, and
hopefully
it will rarely have to run flat out.

As for 2 hrs sunshine/Mo. I would not want to live there!

Jon

wrote:

Solar Flare wrote:

Since November here we had about 2 hours total sun in 45 days...

Where is here? How much is about? How long is November?

Now if my house is 1800 square feet...

And it needs 50K Btu/h, with 3200 ft^2 of R32 walls and ceiling, and
it's T (F) outdoors, with 50K = (70-T)3200ft^2/R32, T = minus 430 F.
That's a very cold day :-)

Nick, where does 3200 ft^2 come from?
A square house 42' on a side has 1764 ft^2 = ~1800ft^2 floorspace.
With 8' tall walls, there are 4*42*8=1344ft^2 walls.
A 42*42 floor plus 42*42 ceiling is 3528 ft^2
for a total of 4872 ft^2 surface area.

I didn't count the floor area.

...If the 1800 ft^2 house is insulated to Nick's proposed R=32...

M. Flare might use 8" SIPs, and put most of the windows on a sunspace.

(with no allowance for air exchange),

Another 15 Btu/h-F for 15 cfm, or less, with a heat exchanger.

Regarding Nick's DDD,
How much insulation is between the interior floor, and the heat store,
the same R=32?

The 8' Phila D-cube had R15.5 walls and ceiling. R22.7 worked for Saskatoon.
In Phila, we might let floor leakage supply half the heat on a 30 F day, ie
about (70-30)10 = 400 Btu/h. If the 6'x6' store is 150 F on an average day,
it can supply 400 Btu/h through a (150-70)6'x6'/400 = R7.2 floor.

Would moveable louvers and thermosiphon air (instead of a pump)
provide heating from the heatstore during cloudy days?

Sure. We might store overnight heat in a 1'x1'x8'-tall closet, with no
ceiling mass or ceiling louvers, and allow warm air from the space between
the tank and the floor to convect up through that closet on cloudy days,
using a thermostat with a damper that only uses 2 watts when it is moving.
The closet might contain 200 pounds of water in a 3" PVC pipe loop that
is also used to collect heat for the tank.

Assume that to attract passersby, we need a window to view the
circle chart recorder sitting on a table inside,

How quaint. I'd hang a large dial thermometer on the wall.
Hobos (tm) might be nice, eg individual homeless shelters.

(next to the cage with the asphyxiated gerbil.)

The Franklin Institute might have a smiling Ronald McDonald. He's plastic,
so he cannot die (although children might find meltdowns disturbing.)
The Institute just hosted that German exhibit of plasticized human
corpse art, perhaps inspired by Dr. Mengele and his friends.

The chart recorder shows inside and outside temperatures.
What type of window glazing should be used?

Small, eg 2 1'x2' layers of flat polycarbonate, facing east, and a narrow
clerestory slot with a transparent motorized damper and a reflective
lightshelf just inside the twinwall polycarbonate south wall.

Nick

snip

Assume that to attract passersby, we need a window to view the
circle chart recorder sitting on a table inside,

Nick has never been big on windows, I recall a thread suggesting
replacing windows with video display panels.

Might I suggest instead, a periscope, with perhaps a snorkel for the
fresh air supply. Such things can no doubt be bought cheaply as military
surplus. Perhaps just buy the whole submarine and either wrap it in
polyurethane foam or submerge it to the desired climate.

Jeff

(next to the cage with the asphyxiated gerbil.)
The chart recorder shows inside and outside temperatures.
What type of window glazing should be used?

To keep the gerbil alive, suppose we require a 0.0004 ft^2 hole
(about 1/4 in ^2) in the wall for air exchange.
How does that ECN affect the design?

On Fri, 08 Dec 2006 07:28:20 GMT, Jeff wrote:

snip

Assume that to attract passersby, we need a window to view the
circle chart recorder sitting on a table inside,

Nick has never been big on windows, I recall a thread suggesting
replacing windows with video display panels.

I recall a thread on replacing NICK with a video display
panel.



Might I suggest instead, a periscope, with perhaps a snorkel for the
fresh air supply. Such things can no doubt be bought cheaply as military
surplus. Perhaps just buy the whole submarine and either wrap it in
polyurethane foam or submerge it to the desired climate.

Jeff

(next to the cage with the asphyxiated gerbil.)
The chart recorder shows inside and outside temperatures.
What type of window glazing should be used?

To keep the gerbil alive, suppose we require a 0.0004 ft^2 hole
(about 1/4 in ^2) in the wall for air exchange.
How does that ECN affect the design?


--
Click here every day to feed an animal that needs you today !!!
http://www.theanimalrescuesite.com/

Paul ( pjm @ pobox . com ) - remove spaces to email me
'Some days, it's just not worth chewing through the restraints.'
'With sufficient thrust, pigs fly just fine.'
HVAC/R program for Palm PDA's
Free demo now available online http://pmilligan.net/palm/

wrote in message
...
wrote:

Solar Flare wrote:

Since November here we had about 2 hours total sun in 45 days...

Where is here? How much is about? How long is November?

Now if my house is 1800 square feet...

And it needs 50K Btu/h, with 3200 ft^2 of R32 walls and ceiling, and
it's T (F) outdoors, with 50K = (70-T)3200ft^2/R32, T = minus 430 F.
That's a very cold day :-)

Nick, where does 3200 ft^2 come from?
A square house 42' on a side has 1764 ft^2 = ~1800ft^2 floorspace.
With 8' tall walls, there are 4*42*8=1344ft^2 walls.
A 42*42 floor plus 42*42 ceiling is 3528 ft^2
for a total of 4872 ft^2 surface area.

I didn't count the floor area.

...If the 1800 ft^2 house is insulated to Nick's proposed R=32...

M. Flare might use 8" SIPs, and put most of the windows on a sunspace.

(with no allowance for air exchange),

Another 15 Btu/h-F for 15 cfm, or less, with a heat exchanger.

Regarding Nick's DDD,
How much insulation is between the interior floor, and the heat store,
the same R=32?

The 8' Phila D-cube had R15.5 walls and ceiling. R22.7 worked for
Saskatoon.
In Phila, we might let floor leakage supply half the heat on a 30 F day,
ie
about (70-30)10 = 400 Btu/h. If the 6'x6' store is 150 F on an average
day,
it can supply 400 Btu/h through a (150-70)6'x6'/400 = R7.2 floor.


Not until you tell us how you're replenishing the heat in the under-floor
storage. It's easy for someone to just say that heat leakage up from the
under-floor storage will *help* heat the building. But a *true* engineer
would have some detailed figures on that heat storage. Like how it gets
heated, how much it looses to the ground, how long it takes to warm it up on
sunny days.

Until you do a *real* study of the under-floor storage and its energy
'cycle', claiming any benefit from it is just 'smoke and mirrors' trying to
"baffle 'em with bull****".

daestrom

daestrom wrote:

In Phila, we might let floor leakage supply half the heat on a 30 F day,
ie about (70-30)10 = 400 Btu/h. If the 6'x6' store is 150 F on an average
day, it can supply 400 Btu/h through a (150-70)6'x6'/400 = R7.2 floor.

Not until you tell us how you're replenishing the heat in the under-floor
storage.

I've done that.

It's easy for someone to just say that heat leakage up from the under-floor
storage will *help* heat the building. But a *true* engineer would have
some detailed figures on that heat storage.

You got 'em :-)

Until you do a *real* study of the under-floor storage and its energy
'cycle', claiming any benefit from it is just 'smoke and mirrors' trying to
"baffle 'em with bull****".

I disagree.

Nick

wrote in message
...
daestrom wrote:

In Phila, we might let floor leakage supply half the heat on a 30 F day,
ie about (70-30)10 = 400 Btu/h. If the 6'x6' store is 150 F on an
average
day, it can supply 400 Btu/h through a (150-70)6'x6'/400 = R7.2 floor.

Not until you tell us how you're replenishing the heat in the under-floor
storage.

I've done that.


Maybe in your head, but you haven't posted them here.

It's easy for someone to just say that heat leakage up from the
under-floor
storage will *help* heat the building. But a *true* engineer would have
some detailed figures on that heat storage.

You got 'em :-)

Not in your listings that you've posted so far. Or are you confusing
another thread again?

Your original listing in this thread calculated the mass of water that would
be needed to provide 5 days of heat with no solar input, but it did not
calculate how long to charge such a tank with the sunspace outlined in the
design.

When circulating water from the tank to the sunspace, or overhead storage to
'charge' the underfloor storage, what is the temperature of the sunspace?
Obviously cooler than your listing calculated. How much cooler? Will the
overhead storage be heated enough for maintaining temperature during the
night if the bulk of the solar heat collected is pumped into the storage?
If the overhead isn't heated to around 155F, it won't have enough heat to
carry through till morning. Haven't seen any figures on that at all.

daestrom

Here's an 8' D-cube simulation with explicit pumping between an overnight
heat storage closet containing 220 pounds of water and an underfloor tank
containing 185 pounds of water. The tank would be very hot for all but 2
months per year, so with a $60 300' coil of 1" PE pipe as a heat exchanger,
it could likely provide hot water for showers for most of the year... 5 gpm
up and down pumps would run a few hours per month, heating the tank from
the closet when the closet is warmer and heating the closet from the tank
when it needs heat. This lets the closet run cooler and more efficiently,
with less total mass than a single ceiling mass.

10 SCREEN 9:KEY OFF:CLS:PI=4*ATN(1)
20 TDMAX=180'max display temp (F)
30 LINE (0,0)-(639,349),,B:XDF=.073:YDF=350/(TDMAX-10)
40 FOR TR= 60 TO 80 STEP 10'temp ref lines
50 LINE (0,349-YDF*(TR-10))-(639,349-YDF*(TR-10)):NEXT
60 OPEN "13739.tm2" FOR INPUT AS #1'NREL TMY2 file name (Phila)
70 LINE INPUT#1,S$'read header
80 CITY$=MID$(S$,8,25)
90 LAT=VAL(MID$(S$,40,2))+VAL(MID$(S$,43,2))/60
100 L=PI*LAT/180'Phila latitude (radians)
110 RHOG=.6'ground reflectance
120 CCLOS=220!'closet capacitance (Btu/F)
130 ACLOS=60'closet heat transfer area (ft^2)
140 CTANK=185'floor tank capacitance (Btu/F)
150 OWG=4*64/15.5'non-south wall conductance (Btu/h-F)
160 SWG=64/(15.5+1/.58)'south wall conductance (Btu/h-F)
170 TSR=1/(ACLOS*1.5)+1/(.58*64)'Thevenin sunspace resistance (F-h/Btu)
180 TC=70'initialize closet temp (F)
190 TCMIN=500'initialize min closet temp (F)
200 TT=70'initialize tank temp (F)
210 TTMIN=500'initialize min tank temp (F)
220 FOR H=1 TO 8760'hour of TMY2 year
230 LINE INPUT#1,S$
240 MONTH=VAL(MID$(S$,4,2))'month of year (1-12)
250 DAY=VAL(MID$(S$,6,2))'day of month
260 HOUR=VAL(MID$(S$,8,2))-.5'hour of day
270 N=1+H/24'day of year (1 to 365)
280 TDB=VAL(MID$(S$,68,4))*.18+32'dry bulb temp (F)
290 'PSET(XDF*H,349-YDF*(TDB-10))'plot dry bulb temp
300 IF DAY=1 AND HOUR=.5 THEN LINE (XDF*H,349)-(XDF*H,345)'tick months
310 IGLOH=VAL(MID$(S$,18,4))*.317'global horizontal radiation (Btu/ft^2)
320 IDIF=VAL(MID$(S$,30,4))*.317'diffuse horizontal radiation (Btu/ft^2)
330 IDIR=VAL(MID$(S$,24,4))*.317'direct normal radiation (Btu/ft^2)
340 T=HOUR'solar time (EST)
350 D=PI*23.45/180*SIN(2*PI*(284+N)/365)'declination (radians)
360 W=2*PI*(T-12)/24'hour angle (radians)
370 X=COS(D)*SIN(L)*COS(W)-SIN(D)*COS(L)
380 THETAI=-ATN(X/SQR(-X*X+1))+PI/2'incidence angle to south wall (rad)
390 IF THETAI=PI/2 THEN THETAI=PI/2'beam sun behind wall
400 SS=IDIR*COS(THETAI)+IDIF/2+IGLOH*RHOG/2'sun on south wall (Btu/ft^2)
410 TST=TDB+.8*SS/.58'Thevenin sunspace temp (F)
420 SGAIN=(TST-TC)/TSR'solar gain (Btu/h)
430 IF SGAIN0 THEN SGAIN=0:GCUBE=OWG+SWG ELSE GCUBE=OWG
440 IF T6 AND T18 THEN TR=70 ELSE TR=50'night setback temp
450 CLOSS=(TR-TDB)*GCUBE'cube loss (Btu)
460 TC=TC+(SGAIN-CLOSS)/CCLOS'new closet temp (F)
470 IF TCTR+5 GOTO 520'no pumping from floor tank
480 PU=CCLOS*(TR+5-TC)/(TT-TC)'pump up PU lb H2O
490 PSET(XDF*H,349-YDF*(30-10))'plot pump up
500 TT=(PU*TC+(CTANK-PU)*TT)/CTANK'new tank temp (F)
510 TC=TR+5'new closet temp (F)
520 IF TCTT+5 GOTO 570'no pumping to floor tank
530 PD=CTANK*CCLOS*(TT+5-TC)/(CCLOS+CTANK)/(TT-TC)'pump down PD lb H2O
540 PSET(XDF*H,349-YDF*(20-10))'plot pump down
550 TC=(PD*TT+(CCLOS-PD)*TC)/CCLOS'new closet temp (F)
560 TT=TC-5'new tank temp (F)
570 IF TT170 THEN TT=170'limit max tank temp (F)
580 'PSET(XDF*H,349-YDF*(TC-10))'plot closet temp
590 PSET(XDF*H,349-YDF*(TT-10))'plot tank temp
600 IF MONTH12 GOTO 670'capture final month stats
610 PUPT=PUPT+PU'up pump total (pounds per month)
620 IF PUPUMAX THEN PUMAX=PU'max up pump rate (lb/h)
630 PDNT=PDNT+PD'down pump total (pounds per month)
640 IF PDPDMAX THEN PDMAX=PD'max down pump rate (lb/h)
650 IF TCTCMIN THEN TCMIN=TC'min closet temp (F)
660 IF TTTTMIN THEN TTMIN=TT'min tank temp (F)
670 NEXT H
680 PRINT TCMIN,PDMAX/8.33/60,PDNT/8.33/5/60
690 PRINT TTMIN,PUMAX/8.33/60,PUPT/8.33/5/60

min closet max down down pump
temp (F) pump (gpm) time (h/mo)

55.07849 .1821537 11.89652

min closet max up up pump
temp (F) pump (gpm) time (h/mo)

75.3698 .1863623 6.075693

Nick

wrote in message
...
Here's an 8' D-cube simulation with explicit pumping between an overnight
heat storage closet containing 220 pounds of water and an underfloor tank
containing 185 pounds of water. The tank would be very hot for all but 2
months per year, so with a $60 300' coil of 1" PE pipe as a heat
exchanger,
it could likely provide hot water for showers for most of the year... 5
gpm
up and down pumps would run a few hours per month, heating the tank from
the closet when the closet is warmer and heating the closet from the tank
when it needs heat. This lets the closet run cooler and more efficiently,
with less total mass than a single ceiling mass.

10 SCREEN 9:KEY OFF:CLS:PI=4*ATN(1)
20 TDMAX=180'max display temp (F)
30 LINE (0,0)-(639,349),,B:XDF=.073:YDF=350/(TDMAX-10)
40 FOR TR= 60 TO 80 STEP 10'temp ref lines
50 LINE (0,349-YDF*(TR-10))-(639,349-YDF*(TR-10)):NEXT
60 OPEN "13739.tm2" FOR INPUT AS #1'NREL TMY2 file name (Phila)
70 LINE INPUT#1,S$'read header
80 CITY$=MID$(S$,8,25)
90 LAT=VAL(MID$(S$,40,2))+VAL(MID$(S$,43,2))/60
100 L=PI*LAT/180'Phila latitude (radians)
110 RHOG=.6'ground reflectance
120 CCLOS=220!'closet capacitance (Btu/F)
130 ACLOS=60'closet heat transfer area (ft^2)
140 CTANK=185'floor tank capacitance (Btu/F)
150 OWG=4*64/15.5'non-south wall conductance (Btu/h-F)
160 SWG=64/(15.5+1/.58)'south wall conductance (Btu/h-F)
170 TSR=1/(ACLOS*1.5)+1/(.58*64)'Thevenin sunspace resistance (F-h/Btu)
180 TC=70'initialize closet temp (F)
190 TCMIN=500'initialize min closet temp (F)
200 TT=70'initialize tank temp (F)
210 TTMIN=500'initialize min tank temp (F)
220 FOR H=1 TO 8760'hour of TMY2 year
230 LINE INPUT#1,S$
240 MONTH=VAL(MID$(S$,4,2))'month of year (1-12)
250 DAY=VAL(MID$(S$,6,2))'day of month
260 HOUR=VAL(MID$(S$,8,2))-.5'hour of day
270 N=1+H/24'day of year (1 to 365)
280 TDB=VAL(MID$(S$,68,4))*.18+32'dry bulb temp (F)
290 'PSET(XDF*H,349-YDF*(TDB-10))'plot dry bulb temp
300 IF DAY=1 AND HOUR=.5 THEN LINE (XDF*H,349)-(XDF*H,345)'tick months
310 IGLOH=VAL(MID$(S$,18,4))*.317'global horizontal radiation (Btu/ft^2)
320 IDIF=VAL(MID$(S$,30,4))*.317'diffuse horizontal radiation (Btu/ft^2)
330 IDIR=VAL(MID$(S$,24,4))*.317'direct normal radiation (Btu/ft^2)
340 T=HOUR'solar time (EST)
350 D=PI*23.45/180*SIN(2*PI*(284+N)/365)'declination (radians)
360 W=2*PI*(T-12)/24'hour angle (radians)
370 X=COS(D)*SIN(L)*COS(W)-SIN(D)*COS(L)
380 THETAI=-ATN(X/SQR(-X*X+1))+PI/2'incidence angle to south wall (rad)
390 IF THETAI=PI/2 THEN THETAI=PI/2'beam sun behind wall
400 SS=IDIR*COS(THETAI)+IDIF/2+IGLOH*RHOG/2'sun on south wall (Btu/ft^2)
410 TST=TDB+.8*SS/.58'Thevenin sunspace temp (F)
420 SGAIN=(TST-TC)/TSR'solar gain (Btu/h)

Seems like you're assuming in line 410 that the sunspace will quickly reach
the temperature where losses out the glazing equals the solar input. Just
taking dry-bulb and adding the necessary temperature difference needed to
lose 80% of the solar input. (the 80% accounting for the transmissivity of
the glazing).

But then in 420, you use that same temperature to transfer heat to the
'closet'. If there is heat flow from the sunspace to the closet as well as
the outdoors, then the calculation in 410 is wrong. With heat flow from the
sunpace to the 'closet' and outdoors, the equilibrium temperature, where
heat outflows (to both closet and ambient) equal energy inflow from solar,
would have to be lower than what 410 calculates. And that equilibrium
temperature would rise/fall as the closet temperature rose/fell (but not
degree for degree).

Granted, if TST (the way you've calculated it here) is at least = TC then
there will be some positive SGAIN. But you're overestimating SGAIN with
this since it's assuming that heat transfered to the closet doesn't lower
the temperature of the sunspace.

daestrom

The trouble is you can't get the submarine to submerge when you warp
it in foam.

Ohhhh Yeah...more thermal mass with weight will get you down...

"Jeff" wrote in message
news
snip

Assume that to attract passersby, we need a window to view the
circle chart recorder sitting on a table inside,

Nick has never been big on windows, I recall a thread suggesting
replacing windows with video display panels.

Might I suggest instead, a periscope, with perhaps a snorkel for
the fresh air supply. Such things can no doubt be bought cheaply as
military surplus. Perhaps just buy the whole submarine and either
wrap it in polyurethane foam or submerge it to the desired climate.

Jeff

(next to the cage with the asphyxiated gerbil.)
The chart recorder shows inside and outside temperatures.
What type of window glazing should be used?

To keep the gerbil alive, suppose we require a 0.0004 ft^2 hole
(about 1/4 in ^2) in the wall for air exchange.
How does that ECN affect the design?

daestrom wrote:

Seems like you're assuming in line 410 that the sunspace will quickly reach
the temperature where losses out the glazing equals the solar input.

Tst is not the actual sunspace temperature. Here's a clarification:

120 CCLOS=220!'closet capacitance (Btu/F)
130 ACLOS=60'closet heat transfer area (ft^2)
140 CTANK=185'floor tank capacitance (Btu/F)
150 OWG=4*64/15.5'non-south wall conductance (Btu/h-F) [16.5]
160 SWG=64/(15.5+1/.58)'south wall conductance (Btu/h-F)
170 TSR=1/(ACLOS*1.5)+1/(.58*64)'Thevenin sunspace resistance (F-h/Btu)

400 SS=IDIR*COS(THETAI)+IDIF/2+IGLOH*RHOG/2'sun on south wall (Btu/h-ft^2)
410 TST=TDB+.8*SS/.58'Thevenin sunspace temp (F)
420 SGAIN=(TST-TC)/TSR'solar gain (Btu/h)
430 IF SGAIN0 THEN SGAIN=0:GCUBE=OWG+SWG ELSE GCUBE=OWG
440 IF T6 AND T18 THEN TR=70 ELSE TR=50'night setback temp
450 CLOSS=(TR-TDB)*GCUBE'cube loss (Btu)
460 TC=TC+(SGAIN-CLOSS)/CCLOS'new closet temp (F)

"Rushd45" RushD45@iron... wrote:

Do you have a schematic of your 8' cube?

Here's one, when the sun is shining:

Ts (F) Tc ~60 F Tr = 50 F
--- | | --- |
|---|--|-----*--------www-------*----|--|-------www--- Tdb ~30 F
--- | 1/(1.5x60) | --- 1/16.5
0.8x64xss | =1/90 | Iheat
Btu/h | | ---
| |----|--|--------- Tt ~100 F
Tdb---www----- | --- |
1/(0.58x64) = 1/37 | pump down |
| --- |
|----|--|------|
| --- |
| pump up --- 185 Btu/F
| ---
--- 220 Btu/F |
--- -
|
-

If 100 Btu/h-ft^2 of sun is shining (ss = 100), the sun is a 0.8x64ft^2x100
= 5120 Btu/h current source, so we can model it like this:

Ts (F) Is Tc ~60 F Tr = 50 F night setpoint
| ---- | --- |
---www----*--------www-------*----|--|-------www--- Tdb ~30 F
| 1/37 1/90 | --- 1/16.5
| | Iheat
| Tst=30+5120/37 | ---
--- = 168 F |----|--|--------- Tt ~100 F
- | --- |
| | pump down |
- | --- |
|----|--|------|
| --- |
| pump up --- 185 Btu/F
| --- floor tank
--- 220 Btu/F |
--- closet -
|
-

Is = (168-60)/(1/37+1/90) = 2832 Btu/h and Iheat = (50-30)x16.5 = 330 Btu/h,
so 2832-330 = 2502 Btu/h flows into the closet mass, raising its temperature
to 60+2502/220 = 71 F (a big jump--maybe this little cube needs smaller-
than-one-hour time steps.) If the closet had been cooler than 55 F, we'd have
pumped some warmer tank water up through it. If it had been warmer than 105,
we'd have pumped some cooler tank water up through the closet. I sized the
closet mass to avoid pumping on an average winter day, large enough so it
never quite touches a 55 F daily min temp. I made the floor tank mass large
enough to clear a yearly min 75 F temp.

The next refinement might be to model thermosyphoning airflow between
the sunspace and the closet, and the closet and the room. If the closet
air temp is Ta (F), we have something like this:

Ts (F) Ta (F) Tc ~60 F water temp
| --- | |
---www----*-------|--|-------www---*
| 1/37 --- 1/90 |
| I |
| Tst = 168 F ---
--- ---
- |
| |
- -

One empirical formula says Q = 16.6Avsqrt(HdT) cfm flows in a chimney with
2 Av ft^2 vents with an H' height diff and a dT (F) temp diff, and a Q cfm
airflow with a dT (F) temp diff moves about QdT Btu/h, so Av = 1 ft^2 and
H = 8' would make I = 47(Ts-Ta)^1.5 Btu/h, approximately. If Ts = 168-I/37
and Ta = 60+I/90, I = 47(168-I/37-60-I/90)^1.5 = 47(108-0.0381I)^1.5
= (47^(2/3)(108-0.0381I)^1.5 = (1406-0.497I)^1.5, so I = 2832-2.01I^(2/3).
Plugging in I = 2832 on the right makes I = 2411 on the left. Repeating
makes I = 2470, 2464, then 2465...

A Rochester NY TMY2 simulation with more insulation (R22.7 vs 15.5 walls)
and twice the total mass required no backup heat. Smaller cubes are harder.
Phila cubes can heat themselves down to 2' (with 21 pounds of water :-)

Nick

JUST DO IT. Post the results.

You are all talk, all theory.

Just for once prove you concept.


wrote:
daestrom wrote:

Seems like you're assuming in line 410 that the sunspace will quickly reach
the temperature where losses out the glazing equals the solar input.

Tst is not the actual sunspace temperature. Here's a clarification:

120 CCLOS=220!'closet capacitance (Btu/F)
130 ACLOS=60'closet heat transfer area (ft^2)
140 CTANK=185'floor tank capacitance (Btu/F)
150 OWG=4*64/15.5'non-south wall conductance (Btu/h-F) [16.5]
160 SWG=64/(15.5+1/.58)'south wall conductance (Btu/h-F)
170 TSR=1/(ACLOS*1.5)+1/(.58*64)'Thevenin sunspace resistance (F-h/Btu)

400 SS=IDIR*COS(THETAI)+IDIF/2+IGLOH*RHOG/2'sun on south wall (Btu/h-ft^2)
410 TST=TDB+.8*SS/.58'Thevenin sunspace temp (F)
420 SGAIN=(TST-TC)/TSR'solar gain (Btu/h)
430 IF SGAIN0 THEN SGAIN=0:GCUBE=OWG+SWG ELSE GCUBE=OWG
440 IF T6 AND T18 THEN TR=70 ELSE TR=50'night setback temp
450 CLOSS=(TR-TDB)*GCUBE'cube loss (Btu)
460 TC=TC+(SGAIN-CLOSS)/CCLOS'new closet temp (F)

"Rushd45" RushD45@iron... wrote:

Do you have a schematic of your 8' cube?

Here's one, when the sun is shining:

Ts (F) Tc ~60 F Tr = 50 F
--- | | --- |
|---|--|-----*--------www-------*----|--|-------www--- Tdb ~30 F
--- | 1/(1.5x60) | --- 1/16.5
0.8x64xss | =1/90 | Iheat
Btu/h | | ---
| |----|--|--------- Tt ~100 F
Tdb---www----- | --- |
1/(0.58x64) = 1/37 | pump down |
| --- |
|----|--|------|
| --- |
| pump up --- 185 Btu/F
| ---
--- 220 Btu/F |
--- -
|
-

If 100 Btu/h-ft^2 of sun is shining (ss = 100), the sun is a 0.8x64ft^2x100
= 5120 Btu/h current source, so we can model it like this:

Ts (F) Is Tc ~60 F Tr = 50 F night setpoint
| ---- | --- |
---www----*--------www-------*----|--|-------www--- Tdb ~30 F
| 1/37 1/90 | --- 1/16.5
| | Iheat
| Tst=30+5120/37 | ---
--- = 168 F |----|--|--------- Tt ~100 F
- | --- |
| | pump down |
- | --- |
|----|--|------|
| --- |
| pump up --- 185 Btu/F
| --- floor tank
--- 220 Btu/F |
--- closet -
|
-

Is = (168-60)/(1/37+1/90) = 2832 Btu/h and Iheat = (50-30)x16.5 = 330 Btu/h,
so 2832-330 = 2502 Btu/h flows into the closet mass, raising its temperature
to 60+2502/220 = 71 F (a big jump--maybe this little cube needs smaller-
than-one-hour time steps.) If the closet had been cooler than 55 F, we'd have
pumped some warmer tank water up through it. If it had been warmer than 105,
we'd have pumped some cooler tank water up through the closet. I sized the
closet mass to avoid pumping on an average winter day, large enough so it
never quite touches a 55 F daily min temp. I made the floor tank mass large
enough to clear a yearly min 75 F temp.

The next refinement might be to model thermosyphoning airflow between
the sunspace and the closet, and the closet and the room. If the closet
air temp is Ta (F), we have something like this:

Ts (F) Ta (F) Tc ~60 F water temp
| --- | |
---www----*-------|--|-------www---*
| 1/37 --- 1/90 |
| I |
| Tst = 168 F ---
--- ---
- |
| |
- -

One empirical formula says Q = 16.6Avsqrt(HdT) cfm flows in a chimney with
2 Av ft^2 vents with an H' height diff and a dT (F) temp diff, and a Q cfm
airflow with a dT (F) temp diff moves about QdT Btu/h, so Av = 1 ft^2 and
H = 8' would make I = 47(Ts-Ta)^1.5 Btu/h, approximately. If Ts = 168-I/37
and Ta = 60+I/90, I = 47(168-I/37-60-I/90)^1.5 = 47(108-0.0381I)^1.5
= (47^(2/3)(108-0.0381I)^1.5 = (1406-0.497I)^1.5, so I = 2832-2.01I^(2/3).
Plugging in I = 2832 on the right makes I = 2411 on the left. Repeating
makes I = 2470, 2464, then 2465...

A Rochester NY TMY2 simulation with more insulation (R22.7 vs 15.5 walls)
and twice the total mass required no backup heat. Smaller cubes are harder.
Phila cubes can heat themselves down to 2' (with 21 pounds of water :-)

Nick


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