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re: btu - formula
3 jan 1998
chris wilkenshoff wrote:
>does someone have the formula to figure out the btus needed to heat a house.
sure. divide each external surface area ai by its r-value ri, and add up
each ai/ri number to find a sum of areas/r-values, sar = sum(ai/ri).
then find the volume v of the house in cubic feet, multiply v by the number
of air changes per hour (ach = 0.2 for a good superinsulated house, 1 for
a good new house, 2 for an older house in good condition, and 4 for an old
air-leaky house), divide vxach by 60 to find the number of cubic feet per
minute of air leaks, and add that to the sar to find a thermal conductance
u = sar+vxach/60 btu/h-f.
then multiply u by the indoor-outdoor temperature difference, to find the
power p = (tindoors-toutdoors)u in btu per hour needed to keep the house
warm. some of that heat power may come from the occupants and their pets
and electrical energy consumption, and the sun...
for instance, a 10' r20 cubical "house" with 6" of fiberglass insulation
and 1 ach has 6x10x10ft^/r20 = 30 btu/h-f of thermal conductance and
10x10x10x1/60 = 16.7 cfm of air leaks, with an effective air infiltration
conductance of about 16.7 btu/h-f, so the total cube conductance u is about
46.7 btu/h-f, and keeping it 70 f inside when it's 30 f outside takes about
(70-30)46.7 = 1,534 btu/hour, ie 1,534/3.41 = 450 watts, so this cube
can be kept warm with 3 50 watt light bulbs and 3 100 watt people.
replacing one side of the cube with an 10'x10' double-glazed r2 sliding
glass door with a thermal conductance of 10'x10'/r2 = 50 btu/h-f leaves
a conductance of 5x10x10/r20 = 25 for the other 5 walls, and the same
16.7 btu/h-f for air leaks, making the total 91.7 btu/h-f, so that cube
needs (70-30)91.7 = 3,668 btu/h to stay warm, just over 1.076 kw, which
might come (at night) from 4 100 watt light bulbs, 4 100 watt people, a
50 watt dog, 4 25 watt cats, and 5 10 watt rabbits. if the glass wall
faces south, it passes about 750 btu/ft^2 of sun on an average january
day where i live in pennsylvania, ie it collects 10x10x750 = 75k btu/day
or a 24-hour average of 3,125 btu/h, or a continuous 916 watts, so with
enough thermal mass inside, eg masonry walls with the insulation outside
the walls, the glazing can keep the cube warm on an average jan day with
only a 50 watt bulb and a 100 watt person who make an extra 512 btu/hr.
after a few cloudy days, the indoor temperature of that cube would drop
to 30f+ 512btu/h/(91.7btu/h-f) = 35.6 f. this might be improved by making
another masonry wall inside the sliding glass doors, with r20 insulation
between the masonry and the glass, and an air gap between the insulation
and the glass, with a black window screen in the air gap to help absorb
the sun. the glass wall still gathers 75k btu/day of sun, with warm air
circulating between the air gap and the cube during the day, but the
air gap behind the glass can quickly get cold, so little heat is lost
to the outdoors, if the air circulation is stopped at night...
with an indoor temperature of t degrees f and average 6 hour winter solar
collection day, this structure loses about 6h(t-30)100ft^2/r2= 300(t-30)
btu from the south wall during the day, 18h(t-30)100ft^2/r22 = 82(t-30)
from the south wall at night, and 24(t-30)(500ft^2/r20+16.7) = 1001(t-30)
btu from the other 5 walls and air leaks over a 24 hour day, ie the cube
loses a total of 1383(t-30) btu. if all this heat comes from the sun,
75k = 1383(t-30), so t = 30+75k/1383 = 84 f. a small exhaust fan with
a 75 f thermostat could keep the indoor temperature comfortable.
suppose the 4 walls were made from 30 lb 8x8x16" cement blocks with a
specific heat of 0.16 btu/f, ie 0.16x30x144/(8x16)= 5.4 btu/f-ft^2, ie
the cube has a heat capacitance c = 4x10'x10'x5.4 = 2,160 btu/f, and the
occupants keep it 75 f inside on an average day. over one cloudy 30 f day,
with no air circulating through the south wall, the cube will lose about
24hx1,534btu/h = 36.8k btu, so the interior temperature will drop by about
36.8k/2,160 = 17 f. more insulation or thermal mass can reduce the drop:
if each 3-hole block has an interior volume of 3x4"x6"x8"/12^3 = 0.33 ft^3,
ie 0.33x144/128 = 0.375 ft^3 per ft^2 of wall, we might fill most of the
blocks with sand (about 18 btu/f-ft^3) to raise the cube capacitance by
400ft^2x0.375x18 = 2700 btu/f to a total of 4,860, and lower the drop to
36.8k/4,860 = 8 f. (some of the blocks need to remain empty, to provide
sufficient surface for the warm air from the "sunspace" to efficiently heat
the masonry as it flows upwards through some vertically lined-up holes in
the blocks, with a few holes into the blocks at the top and bottom of the
wall, and the aid of 4 small exhaust fans near the top of the wall that
blow air from the wall cavities into the room.)
the south masonry wall might be thicker, and insulated on the inside as
well as the outside, so it can store heat at a higher temperature than
the living space, to allow better temperature control of the living space
temperature on cloudy days, and make the stored heat last longer...
suppose the south wall is about 16" thick, using the same kind of blocks,
with say, 2" of foam insulation on the inside, and 6" of fiberglass on
the outside, and a heat capacity c = 100ft^2x13 btu/f-ft^2 = 1,300 btu/f.
suppose the heat that isn't required to keep the cube warm on an average
january day ends up in the south wall (using something like a small fan
with a 70 f cooling thermostat) : 75k btu/day arrive from the sun, and the
cube needs about 1383(70-30) = 55k to stay 70 f on an average day, so 20k
btu/day go into the south wall, at an average temperature tc, where
18h(tc-30)100ft^2/r20 = 90(tc-30) = 20k btu, roughly, so tc = 30+20k/90
= 252 f. i don't think it will get much warmer than about 130 f without a
selective surface, so let's assume tc = 130 f. then the south wall stores
about (130-80)c = 65k btu of useful space heat, enough to keep the indoor
temperature exactly 70 f for more than 24 hours.
a selective surface would help here, maybe some sort of rusty galvanized
iron. or we might make the walls of the cube out of 55 gallon drums full
of water, or use more insulation...
if the cube walls are made with 48 55 gallon drums, we have c = 48x55x8
= 21,120 btu/f, and r = 1/46.7, so rc = 452 hours or 19 days. if that cube
starts out a 30 f cloudy week at 75 f, at the end of the week it will be
about 30+(75-30)exp(-7/19) = 61 f. at that point, we might burn some wood,
or add more insulation.
we might get equivalent performance by just replacing the south wall
masonry by higher-temperature drums, and heat water for showers too.
a houses loses heat through its surface, and living space increases with
volume, so bigger houses are easier to solar heat, with a smaller ratio
of thermal storage to living space volume.
i hope this answers your question.
nick
nicholson l. pine system design and consulting
pine associates, ltd. (610) 489-0545
821 collegeville road fax: (610) 489-7057
collegeville, pa 19426 email: nick@ece.vill.eedeeyou
computer simulation and modeling. high performance, low cost, solar heating and
cogeneration system design. bsee, msee. senior member, ieee. registered us
patent agent. solar closet paper: http://leia.ursinus.edu/~physics/solar.html
web site: http://www.ece.vill.edu/~nick
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