Posted by nicksanspam on December 6, 2006, 12:10 pm
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 LDR+10/60'Saskatoon north latitude (degrees)
40 DECD=-23'December declination (degrees)
70 HSR=-ATN(X/SQR(-X*X+1))+PI/2'sunrise hour angle (radians)
80 BETAD'vertical surface tilt (degrees)
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
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)...
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.
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.
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.
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, 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
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.
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.
Posted by daestrom on December 6, 2006, 9:35 pm
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
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
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.
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.
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]?"
Posted by nicksanspam on December 6, 2006, 11:13 pm
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
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.
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.
Probably not. Then again, it doesn't have to.
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.
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.
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
No problem, if it is well-designed.
Posted by jgraber on December 7, 2006, 1:05 am
Nick, you keep talking about tank heat loss.
Daestrom keeps asking about tank heat gain,
but you dont answer much.
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?
How do you calculate it is an unlikely requirement?
I'll parse that as agreement.
If you dont permit 1 sunny day to recharge for 5 cloudy days,
then what do you permit?
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?
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.
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.
Circular definition alert.
Posted by nicksanspam on December 7, 2006, 11:29 am
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.
If it takes 2 average weeks to recharge, how much does that lower
the maximum solar heating fraction?
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.
About twice the average-day rate.
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...
... 5 or 6 average vs sunny days for a complete recharge seems to work well.
Sure. A simulation using NREL's Phila TMY2 hourly measured weather data file.
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.
On a sunnier or warmer than average day, when there is excess solar heat.
About 2:1, on a clear day.
What's the requirement, 1/(1-0.82) = 5.6 days, as in the calc below?
Perhaps that depends on economics.
Counting the floors, which I ignore. How about a 24' cube?
Different sizes and shapes need different R-values. Is that surprising?
20 N0000!'simulate days
30 FRAC=.82'house heat/average day gain fraction
40 FOR SMAX=2 TO 8'store size (days)
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