Alternative Power sources - Another Deployable Doubt Dispeller

Author

Another Deployable Doubt Dispeller

Nick Pine

2006-12-04, 9:25 am

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

2006-12-04, 8:25 pm

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 ?

SJC

2006-12-04, 8:25 pm

"Cosmopolite" <anywhere@anywhen.net> wrote in message news:9A2dh.29751$dX4.28149@clgrps13...
> Nick Pine wrote:
>
>
> 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.

Chris Friesen

2006-12-04, 8:25 pm

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

Solar Flare

2006-12-04, 8:25 pm

It has been tried and it never works.

"SJC" <sjc_paul_1@yahoo.com> wrote in message
news:TJ2dh.2065$lb1.1849@trnddc05...
>
> "Cosmopolite" <anywhere@anywhen.net> wrote in message
> news:9A2dh.29751$dX4.28149@clgrps13...
>
> In that case, use a large thermal store. I favor the earth under
> the house.

Anthony Matonak

2006-12-04, 8:25 pm

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

SJC

2006-12-05, 3:25 am

Without evidence, quite frankly your opinion is worthless.

"Solar Flare" <solaerfart@hootmail.invalidated> wrote in message news:zKidnf-rHr9QTenYnZ2dnUVZ_sOdnZ2d@golden.net...
> It has been tried and it never works.
>
> "SJC" <sjc_paul_1@yahoo.com> wrote in message
> news:TJ2dh.2065$lb1.1849@trnddc05...
>
>

Morris Dovey

2006-12-05, 3:25 am

Anthony Matonak (in 6i4dh.63269$si3.52067@tornado.socal.rr.com) 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

Solar Flare

2006-12-05, 3:25 am

Ditto

"SJC" <sjc_paul_1@yahoo.com> wrote in message
news:D36dh.2019$ne3.567@trndny03...[color=darkred]
> Without evidence, quite frankly your opinion is worthless.
>
> "Solar Flare" <solaerfart@hootmail.invalidated> wrote in message
> news:zKidnf-rHr9QTenYnZ2dnUVZ_sOdnZ2d@golden.net...

Solar Flare

2006-12-05, 3:25 am

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" <sjc_paul_1@yahoo.com> wrote in message
news:D36dh.2019$ne3.567@trndny03...[color=darkred]
> Without evidence, quite frankly your opinion is worthless.
>
> "Solar Flare" <solaerfart@hootmail.invalidated> wrote in message
> news:zKidnf-rHr9QTenYnZ2dnUVZ_sOdnZ2d@golden.net...

nicksanspam@ece.villanova.edu

2006-12-05, 3:25 am

Cosmopolite <anywhere@anywhen.net> wrote:

>Nick Pine wrote:

[color=darkred]
[color=darkred]
> 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

nicksanspam@ece.villanova.edu

2006-12-05, 9:25 am

Anthony Matonak <anthonym40@nothing.like.socal.rr.com> wrote:

>Nick Pine 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. 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

nicksanspam@ece.villanova.edu

2006-12-05, 9:25 am

Solar Flare <solaerfart@hootmail.invalidated> 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

Anthony Matonak

2006-12-05, 9:25 am

nicksanspam@ece.villanova.edu wrote:
> Anthony Matonak <anthonym40@nothing.like.socal.rr.com> wrote:
>
> 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.

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

Since always, I would imagine.

http://www.jm.com/insulation/buildi..._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.

>
> 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

nicksanspam@ece.villanova.edu

2006-12-05, 9:25 am

Anthony Matonak <anthonym40@nothing.like.socal.rr.com> wrote:

>
>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.

>
>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

nicksanspam@ece.villanova.edu

2006-12-05, 9:25 am

Chris Friesen <cbf123@mail.usask.ca> wrote:

>
>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

nicksanspam@ece.villanova.edu

2006-12-05, 1:25 pm

Chris Friesen <cbf123@mail.usask.ca> wrote:

>
>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

daestrom

2006-12-05, 8:25 pm

"Nick Pine" <nick@acadia.ece.villanova.edu> wrote in message
news:el1aqo$31q@acadia.ece.villanova.edu...
> 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

Solar Flare

2006-12-05, 8:25 pm

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.

<nicksanspam@ece.villanova.edu> wrote in message
news:el3fd8$3k6@acadia.ece.villanova.edu...
> Solar Flare <solaerfart@hootmail.invalidated> wrote:
>
>
> Goody.
>
>
> You live above the arctic circle? :-)
>
>
> Wow. What will it take on an average December day? Why not add more
> insulation and airseal and zone and make it smaller?
>
>
> Sounds unrealistic. Where will this house be?
>
>
> 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
>

Anthony Matonak

2006-12-06, 3:25 am

nicksanspam@ece.villanova.edu wrote:
> Anthony Matonak <anthonym40@nothing.like.socal.rr.com> wrote:
>
>
> 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.

>
> 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.

>
> 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

nicksanspam@ece.villanova.edu

2006-12-06, 9:25 am

daestrom <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote:

[color=darkred]
>Nick, I believe the last term in line 160 above should be,
>"...-24*(TMIN/2-TA)*64/RC"...

I think you are correct. Thanks :-) I'll fix that. I was wondering why raising
RC didn't raise the water temp much. With a warmer-than-room temp, the ceiling
seems like a good place for some extra insulation.

I also added lines 110 and 120 to the Liu and Jordon isotropic sky calc below
(which didn't change the answer, in this case.) This is described on pages
102-105 of the third (2006) edition of Duffie and Beckman's Solar Engineering
of Thermal Processes. It's a way to estimate how much sun falls on a south
wall (Rb = 5 times more beam sun, below), if we know how much falls on
the ground, eg from an Energy Plus weather stat file for Saskatoon.

10 SCREEN 9:KEY OFF:PI=4*ATN(1)
20 LD=52+10/60'Saskatoon 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 (radians)
80 BETAD=90'vertical surface tilt (degrees)
90 BETA=BETAD*PI/180'radians
100 X=-TAN(L-BETA)*TAN(DEC)
110 HSRP=-ATN(X/SQR(-X*X+1))+PI/2'sunrise hour angle (radians)
120 IF HSRP>HSR THEN HSRP=HSR'Hsrp is the min
130 NUM=COS(L-BETA)*COS(DEC)*SIN(HSRP)+HSRP*SIN(L-BETA)*SIN(DEC)
140 DEN=COS(L)*COS(DEC)*SIN(HSR)+HSR*SIN(L)*SIN(DEC)
150 RB=NUM/DEN'tilted surface to horizontal beam rad rat
160 RD=(1+COS(BETA))/2'tilted to horizontal diff rad rat
170 RHOG=.6'ground reflectivity
180 RR=RHOG*SIN(BETA/2)*SIN(BETA/2)'ground reflectance factor
190 IGLOH=1.007*317.1'global horizontal radiation (Btu/ft^2)
200 IDIFH=.472*317.1'diffuse horizontal radiation (Btu/ft^2)
210 IBEAMH=IGLOH-IDIFH'beam horizontal radiation (Btu/ft^2)
220 SS=RB*IBEAMH+RD*IDIFH+RR*IGLOH'south sun (Btu/ft^2-day)...

>You don't seem to account for any heat to 'charge' the under-floor tank.

I figure that's already accounted for, since most of the heat that leaks
from the tank moves up through the floor. Some leakage is desirable, since
that lowers the amount of heat the ceiling needs to provide, which lowers
the min usable ceiling water temp, so the cloudy-day store lasts longer.

It would also make sense to open a 2-watt motorized damper with a room
temp thermostat to let warm air flow up from the space between the top
of the tank and the bottom of the floor, instead of pumping water up
through the ceiling with more electrical power. A night/unoccupied setback
would also make sense, and a PV/battery-powered microcontroller.

>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.

On an average day, the static ceiling mass would provide 100% of the heat
for the cube, so the pump would only run long enough to make up for the heat
leakage from the tank. The tank could also be heated with Big Fins or
fin-tube pipe in the sunspace, if the ceiling mass were not already there.

>If it takes five days to 'charge' it for five cloudy days, wouldn't that
>about double the heat load on the sun space?

Much less, IMO, because long strings of cloudy days are unlikely.
Less warmup time would be desirable for a portable cube.

>Is the ratio of sunny/cloudy days in Philadelphia about 50/50?

Something like that. So cloudy day strings are like coin flips, 2 in a row
with probability 1/4, 3 with 1/8, 4 with 1/16, and 5 with 1/32th. For more
precision, we can do a simple simulation with TMY2 hourly weather data.

>Granted, the ground is warmer, it isn't 70F.

Deep ground is 54.3 F in Phila, and some ASHRAE people figure it's R10
for downwards heatflow. We might put 2" of foamboard under the 6'x6'x6"
tank and surround it with 1 foot of peat moss, with some deadmen under
the tank for wind overturning resistance. Stratification would help.

>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??

With more insulation, they'd need no sun :-)

>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.

Sounds good to me. The new PA ICC building code finally allows building
a house with no windows (as they have always allowed for commercial
buildings), if someone wants to do that, with flat screen TVs and outdoor
cameras for views, CFs for light, insulated doors for fire escapes, and
an exhaust fan with a humidistat for ventilation. Windows and their framing
are expensive, and they can leak water and heat and bugs and burglars and
baseballs... 2ACHx1200ft^2x8'/24h/60m/h = 13 cfm sounds good for natural
air leakage.

>Would such a 'practical' house still perform in the 'regional infestations
>of doubt'??

Sure, but it might never get built, if nobody believes it can heat itself,
inexpensively. Almost all architects and newspaper reporters and housing
developers around Phila seem firmly entrenched and espouse disbelief, with
little knowledge of basic physics.

>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)

A door would be nice for serious doubters. And a small window, so crowds
can peer in to see that it's still 70 F on a big thermometer on a wall,
after a few cloudy 30 F days. I'd like to see "solar shrines" in public
places, eg the Franklin Institute, who are building an outdoor science
park with funding from McDonalds. I'd make them stark, pure science plays,
so people don't miss the point and start arguing about aesthetics.
Lots of people know how to make buildings pretty.

Nick

nicksanspam@ece.villanova.edu

2006-12-06, 9:25 am

Solar Flare <solaerfart@hootmail.invalidated> wrote:

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

You might be able to spell "retort"? :-)
[color=darkred]
>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.
>
><nicksanspam@ece.villanova.edu> wrote in message
>news:el3fd8$3k6@acadia.ece.villanova.edu...

daestrom

2006-12-06, 5:25 pm

<nicksanspam@ece.villanova.edu> wrote in message
news:el6bvl$46e@acadia.ece.villanova.edu...
> daestrom <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote:
<snip>
>
>
> I figure that's already accounted for, since most of the heat that leaks
> from the tank moves up through the floor. Some leakage is desirable, since
> that lowers the amount of heat the ceiling needs to provide, which lowers
> the min usable ceiling water temp, so the cloudy-day store lasts longer.
>
> It would also make sense to open a 2-watt motorized damper with a room
> temp thermostat to let warm air flow up from the space between the top
> of the tank and the bottom of the floor, instead of pumping water up
> through the ceiling with more electrical power. A night/unoccupied setback
> would also make sense, and a PV/battery-powered microcontroller.
>

What you seem to be describing are ways to get the heat *out* of the tank
and into the cube (sorry, I just can't bring myself to call what you're
heating (8' cube with no windows/doors) a 'house').

What I was referring to is that how does the tank get heated in the first
place? If it's from circulating water from the overhead storage, then your
calcs for the TS (temperature of the sun space) don't consider this.

If the house uses 'X' BTU/day, and the floor tank can supply heat for five
days, then it stands to reason that it holds '5X' BTU. So at the end of a
cloudy period, to 'recharge' the tank over the next five 'sunny' days, you
would need to put at *least* 'X' BTU into the tank each day (more,
considering any losses it has).

So to 'recharge' in five days, your system has to develop '2X' BTU/day. 1X
for heating the house, and 1X for dumping into storage. You have to get
those '5X' BTU from *somewhere* over some number of days, it isn't going to
'recharge' itself.

>
> On an average day, the static ceiling mass would provide 100% of the heat
> for the cube, so the pump would only run long enough to make up for the
> heat
> leakage from the tank. The tank could also be heated with Big Fins or
> fin-tube pipe in the sunspace, if the ceiling mass were not already there.
>

Now you're speculating about a different design, putting the storage in the
sun space. If the storage is 'discharged', do you think it would 'recharge'
fully in one day? If not, what's heating the cube while the storage is
'recharging'?

>
> Much less, IMO, because long strings of cloudy days are unlikely.
> Less warmup time would be desirable for a portable cube.
>
>
> Something like that. So cloudy day strings are like coin flips, 2 in a row
> with probability 1/4, 3 with 1/8, 4 with 1/16, and 5 with 1/32th. For more
> precision, we can do a simple simulation with TMY2 hourly weather data.
>

On sunny days the energy collected in the sunspace would just 'break-even'
to maintain cube temperature, but nothing left to go into storage. On
cloudy days, energy would come from storage. But with no excess energy on
sunny days, the storage would soon be depleted and not replenished.

Your sun-space has to collect *more* than what is just needed to maintain
the cube temperature. The excess is what goes into the storage. If the
amount of energy collected is only *slightly* more than the cube's heating
requirements, it may take many sunny days to 'recharge' the storage. And
each cloudy day in between sunny days just drains the storage again.

>
> Deep ground is 54.3 F in Phila, and some ASHRAE people figure it's R10
> for downwards heatflow. We might put 2" of foamboard under the 6'x6'x6"
> tank and surround it with 1 foot of peat moss, with some deadmen under
> the tank for wind overturning resistance. Stratification would help.
>
>
> With more insulation, they'd need no sun :-)
>
>
> Sounds good to me. The new PA ICC building code finally allows building
> a house with no windows (as they have always allowed for commercial
> buildings), if someone wants to do that, with flat screen TVs and outdoor
> cameras for views,

Nah, not a chance. Besides, that's a constant electrical drain. Although
windows do lose heat in the winter, they are useful in many ways. For other
seasons, ventilation. Additional solar gain. Mental health. An above
ground, windowless house would never catch on.

>
> Sure, but it might never get built, if nobody believes it can heat itself,
> inexpensively.

What do you mean by 'sure'? That a 1200 ft^2, four window, low-leakage
house could be built? I have no doubt.

But would a south-facing wall/sunspace, with overhead and underfloor storage
keep such a house 70F in a 30F winter? That I doubt. The windows and
air-changes would add a significant heating and storage requirement.
*Thats* the question I was asking, "Would such a house still perform [with
your sunspace/storage idea]?"

daestrom

nicksanspam@ece.villanova.edu

2006-12-06, 5:25 pm

daestrom <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote:

>
>What I was referring to is that how does the tank get heated in the first
>place? If it's from circulating water from the overhead storage, then your
>calcs for the TS (temperature of the sun space) don't consider this.

I disagree. The tank loss heats the cube. Try mentally drawing a box around
the cube and balancing the solar energy that enters with the heat energy
that leaves.

>If the house uses 'X' BTU/day, and the floor tank can supply heat for five
>days, then it stands to reason that it holds '5X' BTU.

Sure.

>So at the end of a cloudy period, to 'recharge' the tank over the next five
>'sunny' days, you would need to put at *least* 'X' BTU into the tank each day

Sure, if you wanted to recharge the tank completely in 5 days,
after 5 cloudy days in a row, but is that a requirement?

>So to 'recharge' in five days, your system has to develop '2X' BTU/day. 1X
>for heating the house, and 1X for dumping into storage.

Sure, in the above scenario, which is unlikely to be required.

>
>Now you're speculating about a different design, putting the storage in the
>sun space...

No. Just moving the small solar collector from the ceiling to the sunspace,
with the cloudy-day tank still under the floor. In that case, we forget
the ceiling mass and its louvers and add enough higher temp isolated mass
to store overnight heat (eg 200 lb of water cooling from 150 to 80 F) in
a small closet near the south wall.

>If the storage is 'discharged', do you think it would 'recharge' fully
>in one day?

Probably not. Then again, it doesn't have to.

>On sunny days the energy collected in the sunspace would just 'break-even'
>to maintain cube temperature, but nothing left to go into storage.

Nonono. That's what happens on an average vs sunny day. We can model solar
weather as binary coin flips with cloudy (no sun) and sunny (2X sun) days
or as ternary coin flips with cloudy and average (1X) and sunny (2X) days.

>Your sun-space has to collect *more* than what is just needed to maintain
>the cube temperature.

No... It costs nothing to maintain part of the cube at a higher temp, if
the heat that leaks from that part heats the cube. If sun were to shine
in through a window and heat some water inside several nested aquaria to
100 F, that wouldn't change the amount of solar heat the cube needs to
stay 70 F on a 30 F day.

>What do you mean by 'sure'? That a 1200 ft^2, four window, low-leakage
>house could be built? I have no doubt.

I agree. A larger solar house has more available heat storage volume, with
a lower surface to volume ratio. It's harder to build small solar houses.
A 2' D-cube would be almost impossible, even with R40 per inch evacuated
aerogel insulation. We might have a contest with a prize for the smallest
D-cube.

>But would a south-facing wall/sunspace, with overhead and underfloor storage
>keep such a house 70F in a 30F winter?

No problem, if it is well-designed.

Nick

Jon Elson

2006-12-06, 5:25 pm

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>
>
><nicksanspam@ece.villanova.edu> wrote in message
>news:el3fd8$3k6@acadia.ece.villanova.edu.
>
>
>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

>
>

darrylvan

2006-12-06, 8:25 pm

Chris Friesen wrote:
> Nick Pine wrote:
>
> 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

jgraber@ti.com

2006-12-06, 8:25 pm

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"Solar Flare" <solaerfart@hootmail.invalidated> 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?

[color=darkred]
> <nicksanspam@ece.villanova.edu> wrote in message
> news:el3fd8$3k6@acadia.ece.villanova.edu...
[color=darkred]

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'
[color=darkred]

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?
[color=darkred]
[color=darkred]

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 R>6.8 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?

--

Solar Flare

2006-12-06, 8:25 pm

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

<nicksanspam@ece.villanova.edu> wrote in message
news:el6c49$475@acadia.ece.villanova.edu...
> Solar Flare <solaerfart@hootmail.invalidated> wrote:
>
>
> You might be able to spell "retort"? :-)
>
>

Solar Flare

2006-12-06, 8:25 pm

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 bullshit 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" <jmelson@artsci.wustl.edu> wrote in message
news:457750E3.6080904@artsci.wustl.edu...
> 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
>
>

.p.jm@see_my_sig_for_address.com

2006-12-06, 8:25 pm

On Wed, 6 Dec 2006 19:55:54 -0500, "Solar Flare"
<solaerfart@hootmail.invalidated> 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 bullshit 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" <jmelson@artsci.wustl.edu> wrote in message
>news:457750E3.6080904@artsci.wustl.edu...
>

--
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/

Solar Flare

2006-12-06, 8:25 pm

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" <jmelson@artsci.wustl.edu> wrote in message
news:457750E3.6080904@artsci.wustl.edu...
> 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
>
>

jgraber@ti.com

2006-12-07, 3:25 am

nicksanspam@ece.villanova.edu writes:

> daestrom <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote:
>
>
> I disagree. The tank loss heats the cube. Try mentally drawing a box around
> the cube and balancing the solar energy that enters with the heat energy
> that leaves.

Nick, you keep talking about tank heat loss.
Daestrom keeps asking about tank heat gain,
but you dont answer much.

>
> Sure.
>
>
> Sure, if you wanted to recharge the tank completely in 5 days,
> after 5 cloudy days in a row, but is that a requirement?

Yes.
There has to be some recharge requirement,
otherwise, we can assume the heat tank is charged at the beginning
of the winter, and once we have 5 cloudy days, the heat tank
is discharged and useless for the rest of the year.

If the recharge rate is less than the depletion rate,
then this design does not fulfill the design goal
in any location where there are more cloudy days than sunny days.
So a complete recharge in 5 days sounds like the minimum requirement.

What is the ratio of cloudy days to sunny days in Philly?

What is the rate at which the heat tank can be recharged on a sunny day?

>
> Sure, in the above scenario, which is unlikely to be required.

How do you calculate it is an unlikely requirement?

I'll parse that as agreement.
[color=darkred]
>
> Probably not. Then again, it doesn't have to.

If you dont permit 1 sunny day to recharge for 5 cloudy days,
then what do you permit?

>
> Nonono. That's what happens on an average vs sunny day. We can model solar
> weather as binary coin flips with cloudy (no sun) and sunny (2X sun) days
> or as ternary coin flips with cloudy and average (1X) and sunny (2X) days.

You are the model guy Nick. Can you do better than coin flip for
modeling the frequency and run-length of cloudy days in Philly,
by using some weather/insolation database?

>
> No... It costs nothing to maintain part of the cube at a higher temp, if
> the heat that leaks from that part heats the cube. If sun were to shine
> in through a window and heat some water inside several nested aquaria to
> 100 F, that wouldn't change the amount of solar heat the cube needs to
> stay 70 F on a 30 F day.

You missed the point again Nick.
By this point, I'm beginning to think it is deliberate.

On a cloudy day, there is no solar gain, so all the heat comes
from the tank, right?

On an Average day, the solar gain equals the loss,
so there is no net change in the heat tank, right?

On a sunny day (only), the heat tank charge can be increased, right?

How much more energy is available on a sunny day,
than on an average day, that can be used to recharge,
so we can tell how many days it takes to recharge after 5 cloudy days.

>
> I agree. A larger solar house has more available heat storage volume,
> with a lower surface to volume ratio.

Lets check the surface to volume ratio. Ok, thats right.
Dcube has 6x8x8 surface to 8x8x8 volume = 3:4 = 0.75 ratio
A 42x42x8 house has 42x42x2+42x8x4= 3108 surface to 42x42x8 volume = 0.22 ratio

What about the ratio of collector surface vs the entire surface?
Dcube 8x8x8 = 1/6 = 0.167
42x42x8 house = 42x8/3108 = 0.11, a smaller ratio, so even average days
will draw down the heat store,
and more Sunny days will be required to recharge it,
unless the design parameters like R value are change to compensate.

>
> No problem, if it is well-designed. Nick

Circular definition alert.

--

nicksanspam@ece.villanova.edu

2006-12-07, 9:25 am

<jgraber@ti.com> wrote:

>

Where is here? How much is about? How long is November?
[color=darkred]
>
>
>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

nicksanspam@ece.villanova.edu

2006-12-07, 9:25 am

<jgraber@ti.com> wrote:

>nicksanspam@ece.villanova.edu writes:
>
>
>Nick, you keep talking about tank heat loss.
>Daestrom keeps asking about tank heat gain...

The tank won't need much heat, since it only needs to provide heat on
cloudy days, ie it is rarely used. We might pump tank water up through
the ceiling to keep the tank hot. Or heat it with Big Fins or fin-tube
in a sunspace, or use a closet with some overnight heat storage mass.
If the natural tank heat loss heats the cube on an average day, that
loss is automatically included in the average-day solar heating budget.

>
>Yes.

Why?

>There has to be some recharge requirement,

Sure.

>otherwise, we can assume the heat tank is charged at the beginning
>of the winter, and once we have 5 cloudy days, the heat tank
>is discharged and useless for the rest of the year.

If it takes 2 average weeks to recharge, how much does that lower
the maximum solar heating fraction?

>If the recharge rate is less than the depletion rate,
>then this design does not fulfill the design goal
>in any location where there are more cloudy days than sunny days.
>So a complete recharge in 5 days sounds like the minimum requirement.

You might think harder about this. What is "the depletion rate"? How often
do strings of 5 cloudy days occur? This is like "the gambler's ruin," with
a little gain, vs even odds.

>What is the ratio of cloudy days to sunny days in Philly?

About 1:1.

>What is the rate at which the heat tank can be recharged on a sunny day?

About twice the average-day rate.

>
>How do you calculate it is an unlikely requirement?

Cloudy days are like coin flips. Long strings are unlikely. You might try
this: Flip a coin 365 times. Add 2 to the heat store if it comes up heads.
Subtract 1 every day (or 0.8, if you want to be more conservative) to heat
the house. Limit the store to 5 days max. If it contains no heat, add 1
to the yearly backup fuel bill...

>
>If you dont permit 1 sunny day to recharge for 5 cloudy days,
>then what do you permit?

.... 5 or 6 average vs sunny days for a complete recharge seems to work well.

>
>You are the model guy Nick. Can you do better than coin flip for
>modeling the frequency and run-length of cloudy days in Philly,
>by using some weather/insolation database?

Sure. A simulation using NREL's Phila TMY2 hourly measured weather data file.

>
>You missed the point again Nick.
>By this point, I'm beginning to think it is deliberate.

You might think harder about this :-) If sun shines in through a window
and heats a black spot on the floor to 100 F, that does not increase
the amount of heat the cube needs to stay 70 F on a 30 F day...

> On a cloudy day, there is no solar gain, so all the heat comes
>from the tank, right?

Right (in this simple model, with no internal heat gains.)

> On an Average day, the solar gain equals the loss,
>so there is no net change in the heat tank, right?

I think we need a little tank gain on an average day.

> On a sunny day (only), the heat tank charge can be increased, right?

On a sunnier or warmer than average day, when there is excess solar heat.

>How much more energy is available on a sunny day,
>than on an average day, that can be used to recharge,

About 2:1, on a clear day.

>so we can tell how many days it takes to recharge after 5 cloudy days.

What's the requirement, 1/(1-0.82) = 5.6 days, as in the calc below?
Perhaps that depends on economics.

>
>Lets check the surface to volume ratio. Ok, thats right.
>Dcube has 6x8x8 surface to 8x8x8 volume = 3:4 = 0.75 ratio
>A 42x42x8 house has 42x42x2+42x8x4= 3108 surface to 42x42x8 vol = 0.22 ratio

Counting the floors, which I ignore. How about a 24' cube?

>What about the ratio of collector surface vs the entire surface?
> Dcube 8x8x8 = 1/6 = 0.167
> 42x42x8 house = 42x8/3108 = 0.11, a smaller ratio, so even average days
>will draw down the heat store,
>and more Sunny days will be required to recharge it,
>unless the design parameters like R value are change to compensate.

Different sizes and shapes need different R-values. Is that surprising?

Nick

20 N=100000!'simulate days
30 FRAC=.82'house heat/average day gain fraction
40 FOR SMAX=2 TO 8'store size (days)
50 HEAT=0:STORE=SMAX
60 FOR FLIP=1 TO N
70 IF RND>.5 THEN STORE=STORE+2'sunny day
80 STORE=STORE-FRAC'heat house
90 IF STORE>SMAX THEN STORE=SMAX'store smax days
100 IF STORE<0 THEN STORE=0:HEAT=HEAT+FRAC'purchase heat
110 NEXT FLIP
120 HEATFRAC=HEAT/N'non-solar heat fraction
130 COINFRAC=2^-SMAX'coin flip estimate
140 PRINT SMAX,HEATFRAC,COINFRAC,HEATFRAC/COINFRAC
150 NEXT SMAX

store size non-solar coin-flip fraction
(days) fraction fraction ratio

2 .1333603 .25 .5334413
3 7.934428E-02 .125 .6347543
4 5.306072E-02 .0625 .8489715
5 3.220179E-02 .03125 1.030457
6 2.200014E-02 .015625 1.408009
7 1.580112E-02 .0078125 2.022543
8 1.066005E-02 3.90625E-03 2.728973

nicksanspam@ece.villanova.edu

2006-12-07, 9:25 am

<jgraber@ti.com> wrote:

>
>Sure. A simulation using NREL's Phila TMY2 hourly measured weather data file.

Here's a direct gain version:

10 SCREEN 9:KEY OFF:CLS:PI=4*ATN(1)
20 LINE (0,0)-(639,349),,B:XDF=.073:YDF=3.88
30 FOR TR= 60 TO 80 STEP 10'temp ref lines
40 LINE (0,349-YDF*(TR-10))-(639,349-YDF*(TR-10)):NEXT
50 OPEN "13739.tm2" FOR INPUT AS #1'NREL TMY2 file name (Phila)
60 LINE INPUT#1,S$'read header
70 CITY$=MID$(S$,8,25)
80 LAT=VAL(MID$(S$,40,2))+VAL(MID$(S$,43,2))/60
90 L=PI*LAT/180'Phila latitude (radians)
100 RHOG=.6'ground reflectance
110 CCUBE=42700!'cube thermal capacitance (Btu/F)
120 GCW=4*64/15.5'cube wall and ceiling conductance (Btu/h-F)
130 TC=70'initialize cube temp (F)
140 TMIN=100'initialize min cube temp (F)
150 FOR H=1 TO 8760'hour of year
160 LINE INPUT#1,S$
170 MONTH=VAL(MID$(S$,4,2))'month of year (1-12)
180 DAY=VAL(MID$(S$,6,2))'day of month
190 HOUR=VAL(MID$(S$,8,2))-.5'hour of day
200 N=1+H/24'day of year (1 to 365)
210 TDB=VAL(MID$(S$,68,4))*.18+32'dry bulb temp (F)
220 'PSET(XDF*H,349-YDF*(TDB-10))'plot dry bulb temp
230 IF DAY=1 AND HOUR=.5 THEN LINE (XDF*H,349)-(XDF*H,345)'tick months
240 IGLOH=VAL(MID$(S$,18,4))*.317'global horizontal radiation (Btu/ft^2)
250 IDIF=VAL(MID$(S$,30,4))*.317'diffuse horizontal radiation (Btu/ft^2)
260 IDIR=VAL(MID$(S$,24,4))*.317'direct normal radiation (Btu/ft^2)
270 T=HOUR'solar time (EST)
280 D=PI*23.45/180*SIN(2*PI*(284+N)/365)'declination (radians)
290 W=2*PI*(T-12)/24'hour angle (radians)
300 X=COS(D)*SIN(L)*COS(W)-SIN(D)*COS(L)
310 THETAI=-ATN(X/SQR(-X*X+1))+PI/2'incidence angle to s wall (radians)
320 IF THETAI>=PI/2 THEN THETAI=PI/2'beam sun behind wall
330 SS=IDIR*COS(THETAI)+IDIF/2+IGLOH*RHOG/2'sun on south wall (Btu/ft^2)
340 SGAIN=.8*64*SS-(TC-TDB)*.58*64'solar gain/loss (Btu/h)
350 WLOSS=(TC-TDB)*GCW'cube wall loss (Btu)
360 TC=TC+(SGAIN-WLOSS)/CCUBE'new cube temp (F)
370 IF TC>80 THEN TC=80'limit max cube temp (F)
380 PSET(XDF*H,349-YDF*(TC-10))'plot cube temp
390 IF MONTH=12 AND TC<TMIN THEN TMIN=TC
400 NEXT H
410 PRINT TMIN

70.01348 (F)

It needs 42,700 pounds of water (685 ft^3, vs 512 for an 8' cube :-)
to stay at least 70 F in December. How much would it need with
indirect gain?

Nick

nicksanspam@ece.villanova.edu

2006-12-07, 1:25 pm

<jgraber@ti.com> wrote:

>
>Here's a direct gain version:
>
>110 CCUBE=42700!'cube thermal capacitance (Btu/F)...
>
>It needs 42,700 pounds of water (685 ft^3, vs 512 for an 8' cube :-)
>to stay at least 70 F in December. How much would it need with
>indirect gain?

10 SCREEN 9:KEY OFF:CLS:PI=4*ATN(1)
20 LINE (0,0)-(639,349),,B:XDF=.073:YDF=3.88
30 FOR TR= 60 TO 80 STEP 10'temp ref lines
40 LINE (0,349-YDF*(TR-10))-(639,349-YDF*(TR-10)):NEXT
50 OPEN "13739.tm2" FOR INPUT AS #1'NREL TMY2 file name (Phila)
60 LINE INPUT#1,S$'read header
70 CITY$=MID$(S$,8,25)
80 LAT=VAL(MID$(S$,40,2))+VAL(MID$(S$,43,2))/60
90 L=PI*LAT/180'Phila latitude (radians)
100 RHOG=.6'ground reflectance
110 CCUBE=5500!'cube thermal capacitance (Btu/F)
120 OWG=4*64/15.5'non-south wall and ceiling conductance (Btu/h-F)
130 SWG=64/(15.5+1/.58)'non-south wall and ceiling conductance (Btu/h-F)
140 TC=70'initialize cube temp (F)
150 TMIN=100'initialize min cube temp (F)
160 FOR H=1 TO 8760'hour of year
170 LINE INPUT#1,S$
180 MONTH=VAL(MID$(S$,4,2))'month of year (1-12)
190 DAY=VAL(MID$(S$,6,2))'day of month
200 HOUR=VAL(MID$(S$,8,2))-.5'hour of day
210 N=1+H/24'day of year (1 to 365)
220 TDB=VAL(MID$(S$,68,4))*.18+32'dry bulb temp (F)
230 'PSET(XDF*H,349-YDF*(TDB-10))'plot dry bulb temp
240 IF DAY=1 AND HOUR=.5 THEN LINE (XDF*H,349)-(XDF*H,345)'tick months
250 IGLOH=VAL(MID$(S$,18,4))*.317'global horizontal radiation (Btu/ft^2)
260 IDIF=VAL(MID$(S$,30,4))*.317'diffuse horizontal radiation (Btu/ft^2)
270 IDIR=VAL(MID$(S$,24,4))*.317'direct normal radiation (Btu/ft^2)
280 T=HOUR'solar time (EST)
290 D=PI*23.45/180*SIN(2*PI*(284+N)/365)'declination (radians)
300 W=2*PI*(T-12)/24'hour angle (radians)
310 X=COS(D)*SIN(L)*COS(W)-SIN(D)*COS(L)
320 THETAI=-ATN(X/SQR(-X*X+1))+PI/2'incidence angle to s wall (radians)
330 IF THETAI>=PI/2 THEN THETAI=PI/2'beam sun behind wall
340 SS=IDIR*COS(THETAI)+IDIF/2+IGLOH*RHOG/2'sun on south wall (Btu/ft^2)
350 SGAIN=.8*64*SS-(TC-TDB)*.58*64'solar gain/loss (Btu/h)
360 IF SGAIN<0 THEN SGAIN=0:GCUBE=OWG+SWG ELSE GCUBE=OWG
370 WLOSS=(TC-TDB)*GCUBE'cube loss (Btu)
380 TC=TC+(SGAIN-WLOSS)/CCUBE'new cube temp (F)
390 IF TC>80 THEN TC=80'limit max cube temp (F)
400 PSET(XDF*H,349-YDF*(TC-10))'plot cube temp
410 IF MONTH=12 AND TC<TMIN THEN TMIN=TC
420 NEXT H
430 PRINT TMIN

70.01386

About 8X less, which would actually fit into the 8' cube :-)

How much for the ceiling mass version with a higher temp swing?

Nick

Jeff

2006-12-08, 3:25 am

<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?
>

.p.jm@see_my_sig_for_address.com

2006-12-08, 3:25 am

On Fri, 08 Dec 2006 07:28:20 GMT, Jeff <dont_bug_me@all.uk> wrote:

><snip>
>
>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.

[color=darkred]
>
> 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
>

--
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/

nicksanspam@ece.villanova.edu

2006-12-08, 9:25 am

<jgraber@ti.com> wrote:
[color=darkred]

Here's a ceiling mass version:

10 SCREEN 9:KEY OFF:CLS:PI=4*ATN(1)
20 LINE (0,0)-(639,349),,B:XDF=.073:YDF=3.88
30 FOR TR= 60 TO 80 STEP 10'temp ref lines
40 LINE (0,349-YDF*(TR-10))-(639,349-YDF*(TR-10)):NEXT
50 OPEN "13739.tm2" FOR INPUT AS #1'NREL TMY2 file name (Phila)
60 LINE INPUT#1,S$'read header
70 CITY$=MID$(S$,8,25)
80 LAT=VAL(MID$(S$,40,2))+VAL(MID$(S$,43,2))/60
90 L=PI*LAT/180'Phila latitude (radians)
100 RHOG=.6'ground reflectance
110 CCEIL=1217!'ceiling capacitance (Btu/F)
120 CEG=64/15.5'ceiling conductance (Btu/h-F)
130 OWG=3*64/15.5'non-south wall conductance (Btu/h-F)
140 SWG=64/(15.5+1/.58)'non-south wall and ceiling conductance (Btu/h-F)
150 TR=70'constant room temp (F)
160 TC=70'initial ceiling temp (F)
170 TMIN=100'initial min ceiling temp (F)
180 FOR H=1 TO 8760'hour of year
190 LINE INPUT#1,S$
200 MONTH=VAL(MID$(S$,4,2))'month of year (1-12)
210 DAY=VAL(MID$(S$,6,2))'day of month
220 HOUR=VAL(MID$(S$,8,2))-.5'hour of day
230 N=1+H/24'day of year (1 to 365)
240 TDB=VAL(MID$(S$,68,4))*.18+32'dry bulb temp (F)
250 'PSET(XDF*H,349-YDF*(TDB-10))'plot dry bulb temp
260 IF DAY=1 AND HOUR=.5 THEN LINE (XDF*H,349)-(XDF*H,345)'tick months
270 IGLOH=VAL(MID$(S$,18,4))*.317'global horizontal radiation (Btu/ft^2)
280 IDIF=VAL(MID$(S$,30,4))*.317'diffuse horizontal radiation (Btu/ft^2)
290 IDIR=VAL(MID$(S$,24,4))*.317'direct normal radiation (Btu/ft^2)
300 T=HOUR'solar time (EST)
310 D=PI*23.45/180*SIN(2*PI*(284+N)/365)'declination (radians)
320 W=2*PI*(T-12)/24'hour angle (radians)
330 X=COS(D)*SIN(L)*COS(W)-SIN(D)*COS(L)
340 THETAI=-ATN(X/SQR(-X*X+1))+PI/2'incidence angle to s wall (radians)
350 IF THETAI>=PI/2 THEN THETAI=PI/2'beam sun behind wall
360 SS=IDIR*COS(THETAI)+IDIF/2+IGLOH*RHOG/2'sun on south wall (Btu/ft^2)
370 SGAIN=.8*64*SS-(TC-TDB)*.58*64'solar gain/loss (Btu/h)
380 IF SGAIN<0 THEN SGAIN=0:GCUBE=OWG+SWG ELSE GCUBE=OWG
390 CLOSS=(TR-TDB)*GCUBE+(TC-TDB)*CEG'cube loss (Btu)
400 TC=TC+(SGAIN-CLOSS)/CCEIL'new ceiling mass temp (F)
410 IF TC>130 THEN TC=130'limit max ceiling temp (F)
420 PSET(XDF*H,349-YDF*(TC-10))'plot ceiling temp
430 IF MONTH=12 AND TC<TMIN THEN TMIN=TC
440 NEXT H
450 PRINT TMIN

70.02401

Only 1200 pounds, here assumed all under the ceiling, although putting most
of it under the floor would be more practical and efficient. With about
18h(70-30)20 = 14.4K Btu of overnight heat, the ceiling only needs about
14.4K/(130-70) = 240 pounds of water, or less all around, if the cube is
70 F for 12 hours per day and 50 for the other 12 hours.

Nick

nicksanspam@ece.villanova.edu

2006-12-08, 9:25 am

><jgraber@ti.com> wrote:
>
>
>Here's a ceiling mass version:
>
>10 SCREEN 9:KEY OFF:CLS:PI=4*ATN(1)
>20 LINE (0,0)-(639,349),,B:XDF=.073:YDF=3.88
>30 FOR TR= 60 TO 80 STEP 10'temp ref lines
>40 LINE (0,349-YDF*(TR-10))-(639,349-YDF*(TR-10)):NEXT
>50 OPEN "13739.tm2" FOR INPUT AS #1'NREL TMY2 file name (Phila)
>60 LINE INPUT#1,S$'read header
>70 CITY$=MID$(S$,8,25)
>80 LAT=VAL(MID$(S$,40,2))+VAL(MID$(S$,43,2))/60
>90 L=PI*LAT/180'Phila latitude (radians)
>100 RHOG=.6'ground reflectance
>110 CCEIL=1217!'ceiling capacitance (Btu/F)...

>Only 1200 pounds, here assumed all under the ceiling, although putting most
>of it under the floor would be more practical and efficient. With about
>18h(70-30)20 = 14.4K Btu of overnight heat, the ceiling only needs about
>14.4K/(130-70) = 240 pounds of water, or less all around, if the cube is
>70 F for 12 hours per day and 50 for the other 12 hours.

Only 879 pounds altogether, with the night setback.

110 CCEIL=879!'ceiling capacitance (Btu/F)...
390 IF T>6 AND T<18 THEN TR=70 ELSE TR=50'night setback temp
400 CLOSS=(TR-TDB)*GCUBE+(TC-TDB)*CEG'cube loss (Btu)...
460 PRINT TMIN

70.00746

Nick

daestrom

2006-12-08, 9:25 am

<nicksanspam@ece.villanova.edu> wrote in message
news:el8o02$4pg@acadia.ece.villanova.edu...
> <jgraber@ti.com> wrote:
>
>
> Where is here? How much is about? How long is November?
>
>
> I didn't count the floor area.
>
>
> M. Flare might use 8" SIPs, and put most of the windows on a sunspace.
>
>
> Another 15 Btu/h-F for 15 cfm, or less, with a heat exchanger.
>
>
> 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 bullshit".

daestrom

daestrom

2006-12-08, 1:25 pm

<nicksanspam@ece.villanova.edu> wrote in message
news:el7iqu$4m9@acadia.ece.villanova.edu...
> daestrom <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote:
>
>
> I disagree. The tank loss heats the cube. Try mentally drawing a box
> around
> the cube and balancing the solar energy that enters with the heat energy
> that leaves.
>
>
> Sure.
>
>
> Sure, if you wanted to recharge the tank completely in 5 days,
> after 5 cloudy days in a row, but is that a requirement?
>
>
> Sure, in the above scenario, which is unlikely to be required.
>

You specified a storage capable of supplying up to five days of heat for
'cloudy' days. Yet you have *not* given any clue as to how that storage
will be 'charged', nor how long it would take to 'charge' it.

Are we to assume that you take 360 days to 'charge' it, and then 'discharge'
it in 5 ? Without including the storage in the energy balance, you have
nothing. No basis for 'charging' the tank means there is no viable way to
use the storage.

<snip>
>
> Probably not. Then again, it doesn't have to.
>

Now that's just admitting you don't know. You're just *guessing* how well
the storage will be recharged.

>
> Nonono. That's what happens on an average vs sunny day. We can model
> solar
> weather as binary coin flips with cloudy (no sun) and sunny (2X sun) days
> or as ternary coin flips with cloudy and average (1X) and sunny (2X) days.
>

To do that, you must get away from the 'average' energy influx to specific
daily levels. You're suggestion about using coin-flips between 0x and 2x
isn't really very good. Look at weather data and you'll see the probability
of a cloudy day *increases* if the previous day was cloudy. "Cloudy days"
are not completely independent events (such as a coin flip). So you are
more likely to have two cloudy days in a row (and some other time have two
sunny days in a row).

>
> No... It costs nothing to maintain part of the cube at a higher temp,

Except increased losses.
> if
> the heat that leaks from that part heats the cube. If sun were to shine
> in through a window and heat some water inside several nested aquaria to
> 100 F, that wouldn't change the amount of solar heat the cube needs to
> stay 70 F on a 30 F day.
>

To get anything meaningful, you *have* to model the heat storage. That
means accounting for the transfer of heat from the sunspace to the storage,
the constant loses from the storage (either to the house, the ambient, or
both depending how you arrange things), and the supply from storage to the
house. That would include some 'if-checks' to turn on/off the various
modes.

daestrom

nicksanspam@ece.villanova.edu

2006-12-08, 1:25 pm

daestrom <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote:

>
>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 bullshit".

I disagree.

Nick

nicksanspam@ece.villanova.edu

2006-12-08, 1:25 pm

daestrom <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote:

>You specified a storage capable of supplying up to five days of heat for
>'cloudy' days. Yet you have *not* given any clue as to how that storage
>will be 'charged', nor how long it would take to 'charge' it.

Incorrect. Your reading skills and logic seem to have deteriorated lately.

Nick

daestrom

2006-12-08, 1:25 pm

<nicksanspam@ece.villanova.edu> wrote in message
news:el8tu1$4r3@acadia.ece.villanova.edu...
> <jgraber@ti.com> wrote:
>
>
> The tank won't need much heat, since it only needs to provide heat on
> cloudy days, ie it is rarely used.

*THAT* makes no sense. You've been saying the ratio of cloudy/sunny days is
1:1, yet you say the storage tank '..is rarely used'. Care to explain that
a bit?

If the *average* solar gain is 1000 BTU/ft^2-day, then if you assume 0
BTU/ft^2-day for cloudy days, you must also be *assuming* 2000 BTU/ft^2-day
on 'sunny' days. How hot does the sunspace get on such days, and how much
are the losses increased by this higher temperature?

If you're not storing some of that heat for the 'cloudy' day, then how cold
does the d-cube get on a cloudy day? Using your original code, with zero
solar gain for just 24 hours it gets pretty chilly in your d-cube. Your
obvious reply is to use the heat stored in the under-floor tank, but you
haven't put any heat *into* the tank. Besides, now we hear that it's only
'rarely used'. Meaning not on a cloudy day that happened to occur after a
sunny day?

> We might pump tank water up through
> the ceiling to keep the tank hot. Or heat it with Big Fins or fin-tube
> in a sunspace, or use a closet with some overnight heat storage mass.

Yes, you *might* do that. But your model doesn't take that into account, so
as far as dispelling any doubt, it hasn't done much.

<snip>
>
> Why?
>

Well, you haven't provided *ANY* recharging of the tank, so what are we to
expect? That Maxwell's daemons do it for you?

If cloudy/sunny days are 50/50, it seems quite reasonable that a 5-day
cloudy storage should be recoverable in 5 sunny days. But you haven't shown
this a possibility in your model. You haven't shown *any* recovery, so how
long would it take in your design?

>
> Sure.
>
>
> If it takes 2 average weeks to recharge,

Would you have evidence for this article of faith? Will the tank recharge
at all?

> how much does that lower
> the maximum solar heating fraction?
>

You started this thread with 100% solar. Are you now suggesting some
fraction less is acceptable?

>
> You might think harder about this. What is "the depletion rate"? How often
> do strings of 5 cloudy days occur? This is like "the gambler's ruin," with
> a little gain, vs even odds.
>

The 'gambler' is working with perfectly independent events. Each toss of
the coin; roll of the die; or other 'game of chance' is independent of all
previous runs. The idea that a 'streak' of good luck or bad luck happens to
gamblers is false in games of chance. I think we can agree on that.

But weather patterns aren't completely independent events. If a storm front
heads towards Phila from the west and you have your first cloudy day, the
chances of tomorrow *also* being cloudy are much higher than 50/50.

You're trying to apply statistics for indpendent events to weather
prediction. And any fool knows that doesn't work. After all, if it snows
on average 90 days in a year, that doesn't mean the chances of snow on July
4 is 90/365.

Consider this. You say that cloudy days are about 50/50 in your area. So
if a winter storm heads into the area Monday morning, you're methodology
*assumes* that the chances of clouds on Tuesday are completely independent
of Monday's storm. That's a good one, go on, pull the other leg...

>
> About 1:1.
>
>
> About twice the average-day rate.

Have you any evidence of this article of faith? We haven't seen you present
any. Recharging at 'twice the average-day rate'?? And the average-day rate
of charging is zero?

>
>
> Cloudy days are like coin flips.

No, they are not. I've explained this to you in previous threads, yet you
persist in this fantasy. Just because you can't be bothered with the higher
statistics to model cloudy/sunny days more accurately, doesn't make your
gross assumption correct.

<snip>
>
> You might think harder about this :-) If sun shines in through a window
> and heats a black spot on the floor to 100 F, that does not increase
> the amount of heat the cube needs to stay 70 F on a 30 F day...
>
>
> Right (in this simple model, with no internal heat gains.)
>
>
> I think we need a little tank gain on an average day.
>

FINALLY! Now, just how much tank gain is needed? Above you suggested 5 to
6 days to recharge the tank 'seems to work well'. So, to recharge the tank
in 5 to 6 days, how much 'little tank gain' is that? More than a 'little'
isn't it?

>
> On a sunnier or warmer than average day, when there is excess solar heat.
>

Changing the conditions again. What a joke! Heck Nick, you can gain a lot
in July!! But for every 'sunnier' or 'warmer than average day', there is a
'cloudier' or 'colder than average day'. Consider the definition of
'average'.

Are you now conceding that your idea won't work in 'below average' weather?

>
> Counting the floors, which I ignore. How about a 24' cube?

It's still nothing more than a 'doghouse' with no practical living
potential. Two and a half stories tall? Still no
windows/doors/ventilation?

What about the fact that if you start out with a depleted storage tank and
you have several sunny days in a row, the first day will increase the
storage temperature by X degrees. But the second day the sunspace will be
operating at higher temperatures so it will only raise the storage
temperature by (1-<deltaloss> )*X. And the third day by (1-<deltaloss>^2)*X
and so on. The hotter the tank gets the fewer BTU's that will be collected
and stored.

daestrom

daestrom

2006-12-08, 5:25 pm

<nicksanspam@ece.villanova.edu> wrote in message
news:elc9d6$5pi@acadia.ece.villanova.edu...
> daestrom <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote:
>
>
> I've done that.
>

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

>
> 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

daestrom

2006-12-08, 5:25 pm

<nicksanspam@ece.villanova.edu> wrote in message
news:elc9k5$5q1@acadia.ece.villanova.edu...
> daestrom <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote:
>
>
> Incorrect. Your reading skills and logic seem to have deteriorated lately.
>

All you've done is assume that days that are sunnier than average will put
enough energy into the storage tank to meet the needs of days that are less
sunny than average. And you know what happens when you assume.

Do you have any evidence to support this article of faith?

daestrom

nicksanspam@ece.villanova.edu

2006-12-09, 8:25 pm

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 SGAIN<0 THEN SGAIN=0:GCUBE=OWG+SWG ELSE GCUBE=OWG
440 IF T>6 AND T<18 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 TC>TR+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 TC<TT+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 TT>170 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 MONTH<12 GOTO 670'capture final month stats
610 PUPT=PUPT+PU'up pump total (pounds per month)
620 IF PU>PUMAX THEN PUMAX=PU'max up pump rate (lb/h)
630 PDNT=PDNT+PD'down pump total (pounds per month)
640 IF PD>PDMAX THEN PDMAX=PD'max down pump rate (lb/h)
650 IF TC<TCMIN THEN TCMIN=TC'min closet temp (F)
660 IF TT<TTMIN 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

daestrom

2006-12-10, 8:25 pm

<nicksanspam@ece.villanova.edu> wrote in message
news:elfl9h$6kp@acadia.ece.villanova.edu...
> 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

Solar Flare

2006-12-10, 8:25 pm

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" <dont_bug_me@all.uk> wrote in message
news:ou8eh.7987$sf5.950@newsread4.news.pas.earthlink.net...[color=darkred]
> <snip>