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re: sizing a heating system
11 jun 2000
news  wrote:

>> some people define a "solar house" as
>> "one with no other form of heat."

>a bit of a misnomer as most need some form no matter how little.

most, but not all. some houses can store solar heat for a cloudy week,
or even a whole cloudy winter or two, as in this village system...

  the lyckebo [sweden] system is a cavern of 100,000 m^3 capacity,
  cut out of bedrock using standard mining methods, of cylindrical
  shape, with a central column of rock left to support overhead rock.
  the cavern is about 30 m high and its top is about 30 m below ground
  level. it is water filled, and inlet and outlet pipes can be moved
  up and down to inject and remove water from controlled levels. the
  water is highly stratified with top to bottom temperatures of about
  80 to 30 c... it takes several years of cycling through the annual
  weather variations for a storage system of this size to reach a
  "steady periodic" operation. in the second year of its operation,
  while it was still in a "warm-up" stage, 74% of the energy added
  to the store was recovered. 

     from p 404, "seasonal storage," of _solar engineering of
     thermal processes_, by john a duffie and william a beckman,
     2nd edition, 1991, wiley-interscience isbn 0-471-51056-4

on a smaller scale, john hait's passive annual heat storage houses
have modified soil microclimates which take about 3 years to warm up.
not exactly tract housing :-) www.montana.com/rmrc (?)

hait says cover the dirt around and over an underground house with an
insulating "umbrella/watershed" made with several layers of foamboard
and plastic film to keep rain from washing stored heat down and out
of the dirt. the umbrella slopes downward to shed water and extends
some 20' outwards from the house, about 2' underground. it contains
about 4" of foamboard at the thickest point. it is warmed and cooled
by natural convection using earth tubes. 

[use courier font]    earth     earth     earth     earth
                    uuuuuuuuuuuuuuuuuuuuuumbrella          earth
          solar gain  --------------------- <-> uuuuuuuuuuuuuuuuuuu
  earth   uu      \  |                     |   \ upper earth tube uuu
        uu           |        house        |                      \ uu
 lower earth tube<-->|                     |                        \
                      ---------------------

the lower tube slopes up to the house (making an igloo-like heat trap)
and enters at the floor level. the upper tube enters at ceiling level,
and slopes down to exit at the same outdoor level as the lower. when
the house is slightly warmer than the surrounding earth, outside air
naturally enters the lower tube and warm air flows out the upper one.
when the house is cooler than the earth, flow reverses. in either case,
heat exchanges between earth tube air and surrounding earth, storing
excess house heat in the earth or removing it to heat the house.

>even superinsulated houses need some form of heat
>to bring the place up to temperature after no use.

a house with lots of thermal mass may not need additional heat.
some of the solar houses designed by norman saunders, pe, have
long honest digital track records with no backup heating systems,
even in cold cloudy new england. they store heat for about a week.
some use the heat store to make domestic hot water, which can be 
the second-largest energy need for a house, after space heating.

thermal mass also enables natural cooling--ventilate the house
at night, and button it up during the day.
 
>> conservation of energy is a basic requirement...
>> energy in = energy out.

>> for instance, an airtight 2 meter cube with metric
>> r4 walls has a total u-value of 6x2mx2m/r4 = 6 w/c.
>> keeping it 20 c inside on a 5 c day requires
>> (20c-5c)6w/c = 90 watts, ie 2160 wh/day. this
>> might come from 2 100 watt people living inside
>> for 10.8 hours per day.
 
>...what i am not certain about is the how the insolation is calculated
>and brought into the equation.

dat cube doan need no steenkeeng eensolation :-) nor thermal mass.
windows would make it colder inside on a cloudy day, if they have
less thermal resistance than the walls (metric r4 or us r23.) 

>suppose a south facing wall is 6 meters high and 8 metres long
>with a total of 8 metres of double glazing low "e".

with say, 8m^2/r0.8 = 10 w/c for the glazing and 40m^2/r4 = 10 w/c
for the rest of the wall... ?

>there will be heat gain.

low-e "hardcoat" might have 50% solar transmission...

>i am inclined to take the worst case scenario in jan/feb and calculate
>the heat gain in those months and size a heating system to suit. 

i'm so inclined too, but i'd say "size a solar heat storage system." 

>what tables would i use to assess the heat gain of the walls
>and glazing?

you might use nrel weather data if you lived in the us, eg 30-year
monthly average temperatures and amounts of sun for 239 us locations.
nrel cds have hourly weather data for a typical year and actual hourly
weather data for 30 years, if you'd like to do solar house heating
simulations. you might use the european solar atlas in the uk, the
older paper edition or the newer multiple-cd version.

i'd ignore the solar heat gain of the walls, since it is usually small
compared to the solar heat gain of the glazing, ie the insolation times
the solar transmittance as specified by the glazing manufacturer.

>...what calcs are used.

here's an example: nrel says the average temp is -0.9 c in january
where i live near philadelphia, pa, usa (on the east coast, about
100 miles south of new york city), and on an average january day,
3.3 kwh/m^2 of sun falls on a south wall. (the uk is mostly warmer
and cloudier.) that 2 meter windowless cube with 2 100 watt people
inside for 10.8 hours per day has 2160 wh/day of internal heat gain.
if it were t degrees c inside for 10.8 hours a day, it would lose 
10.8h(t-(-0.9c))6w/c = 2160 wh, so t = -0.9+2160/(10.8x6) = 32.4 c,
(90.4 f), rather warm. with lots of thermal mass inside, t = 14.1 c
for 24 hours per day, rather cool. keeping it 20 c (68 f) inside 
for 24 hours per day requires 24(20-(-0.9))6 = 3010 wh, eg 2 100 w
people inside for 15 hours per day, or 2 10.8 hour 100 w people and
850 wh/day (26 kwh/month) of internal electrical usage.

replacing the cube's south wall by a 2mx2m r0.8 window raises the
cube's thermal conductance to 2mx2m/r0.8 = u5 for the window plus
5x2mx2m/r4 = u5 for the other 5 surfaces, ie 10 w/c, which lowers
its 24-hour "balance point temperature" to t = -0.9 + 2160/(24x10)
= 8.1 c, with lots of thermal mass and 2 100 w people inside for
10.8 hours on a -0.9 c cloudy day.

on an average day, 0.5x2mx2mx3.3kwh/m^2 = 6.6 kwh of solar heat
flows in through the window. if the solar energy that flows into
the cube over an average day plus the internal heat gain equals
the heat energy that flows out, with cube temp t, 6660 + 2160 =
24h(t-(0.9))10, so we could make the cube 36 c (100 f) indoors,
over a long string of average days. 

now suppose the cube contains 500 liters of water in smallish
containers, and we keep it 20 c on average days. each liter releases
about 1.16 wh of heat (4.19kj/kgc/3600ws/wh) as it cools 1 c, so
500 liters cooling 1 c releases about 582 wh. the cube needs
24(20-(0.9))10 = 5 kwh per day to stay 20 c for 24 hours. if the
occupants provide 2.16 kwh of that, the water loses about 2.84 kwh 
on the first cloudy day, cooling 2.84kwh/(582wh/c) = 4.9 c to 15.1.
over the second day, the cube loses about 24h(15.1-(0.9))10 = 3840 wh,
cooling the water (3840-2160)/582 = 2.9 c to 12.2 c. coolish. after
a long string of cloudy days, it cools to -0.9+2160/24/10 = 8.1 c. 

there's a quicker and more accurate way to calculate this. the cube
has a natural cooling time constant rc = 582 wh/c/(10w/c) = 58.2 hours.
after n cloudy days, it cools to 8.1+(20-8.1)exp(-24n/58.2), eg 16 c
after 1 day or 9.6 c (very coolish) after 5 cloudy days. exp() is the
"e-to-the-x" key on a $9.99 casio fx-260 scientific calculator.

making the window into a low-thermal mass sunspace or air heater over
the original r4 wall helps. that way, it stays cool overnight and on
cloudy days, with little heat loss from the house. this raises the
cube's cloudy day balance point to -0.9+2160/(24x6) = 10.8 c and makes
its cloudy day time constant rc = 582/6 = 97 hours, so it cools to
10.8+exp(20-10.8)exp(-5x24/97) = 13.5 c (56 f.) keeping the cube 22 c
on average days makes this 14.1. still coolish. 

if we separate the heat store from the living space, we can raise the
average-day heat store temperature more and store more heat for cloudy
days without worrying about occupant comfort, and eliminate the cloudy
day living-space temperature droop. the heat store can keep the living
space at a constant temperature for a few cloudy days, until the heat
store gets close to the living space temperature.

so let's move the water containers into the sunspace and insulate them
with r4 walls all round and assume circulate sunspace air through this
box during the day so the water is about the same temp as the sunspace
over say, 6 sunny hours on an average winter day. (i wouldn't call this
heat store a "solar closet" unless it were warmer than the sunspace and
had its own separate glazing.) suppose the sunspace provides heat for
the living space during those 6 hours, and the heat store heats the
living space over the next 18 hours on an average day, and on cloudy days.
the living space needs about 24(20-(-0.9))6 = 3010 wh on an average day,
of which 2160 comes from people, leaving 850 for the sunspace and heat
store. if the sunspace has a temperature t during the day, and the amount
of solar energy that flows into the sunspace equals the amount of heat
energy that flows out, 6600 = 850+6(t-(-0.9))2mx2m/r0.8, so t = 191 c :-)
well no, since water boils at 100 c, but now we can reduce the sunspace
area or the amount of insulation or thermal mass. 

let's keep the sunspace the same and reduce the insulation to r2, which
means the cube needs 6020 wh on an average day, of which 2160 comes from
people, leaving 3860 for the sunspace and heat store, which lowers the
average day water temp to 90 c (so it can still supply water for showers,
with more plumbing.) keeping the cube 20 c for five -0.9 cloudy days in
a row requires 5x(6020-2160) = 19.3 kwh from the heat store. if the heat
store can keep the cube 20 c until it reaches 25 c (this depends on the
minimum heat transfer rate between store and living space on a cold day),
we need 19.3k/(1.16x(90-25)) = 256 liters (66 gallons) of water, using
these linear calculations.

that's one basic approach. other configurations are possible, as well as
refinements in these calculations, which are mostly based on "ohm's law
for heatflow" aka "newton's law of cooling." 

>also is there a book that is worth buying
>that explains with examples of how to do it?

duffie and beckman is good, but it contains lots of calculus, and
more calculations than strategies. i don't know of a simpler good
book on this subject. i've thought about writing one.

nick




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