re: solar collector design for a solar closet?
9 dec 2002
>>we have had wind chills of -85f and absolute lows of -45f. i am afraid
>>of the solar space failing and turning the water wall into an iceberg :-)
>...how long will it take for an 8' cube of water with r20 walls to freeze
>solid at -45 f?
the freezing takes 8x8x8x64x144 = 4,718,592 btu. with a thermal conductance
of 6x8'x8'/r20 = 19.2 btu/h-f and a (32-(-45))19.2 = 1,728.4 btu/h heatflow,
it might freeze solid in 4,718,592/1,728.4 = 3,192 hours or 133 days in a
-45 f room. with a pound of salt or antifreeze, it might never freeze solid,
if it concentrates as pure water freezes out.
>...if we heated the house in the daytime from the solar space and used
>the hotter collector to store heat in a solar closet for the night,
>that would be a fairly simple system.
something like that, but a solar closet (tm) has a low charging rate and
a small mass surface-to-volume ratio, and it only provides heat on cloudy
days, vs average nights. a sunspace in a 3-temp system might:
a) heat a house during the day with air from the space between a first
sunspace glazing and a dark screen, and
b) heat the house at night with warmer air from a rock bin heated with
air from the north side of the screen, and
c) keep a solar closet with its own glazing hot.
>> we might have a sunspace on the south wall of a walk-out basement, with
>> a room full of 4" water tubes or cylindrical vertical rock gabions made
>> from 2"x4" welded-wire fencing for overnight heat storage on an average
>> day, and a waterwall solar closet with its own glazing for cloudy-day
>> heat storage. the floor above could have vents, and we could release warm
>> air from the rock room or hot air from the closet as needed to keep the
>> upstairs living space warm. this might work without fans.
>this is the kind of system i was kind of hoping would be workable. i
>am having a preference for 4" water tubes, as i could figure a way to
>dust them, where the rock gabions would be kind of difficult.
the basement ceiling might be a good place for the water tubes, exposed,
and tucked up between the rafters...
if a 32'x32'x16' house has r48 (12" sip) walls and ceiling and 88 ft^2
of r4 windows with 50% solar transmission, say 48 ft^2 on the south wall
and 16 ft^2 on the east and the west and 8 ft^2 on the north, and 0.2 air
changes per hour (vs a typical new american house with 1 ach, or a new
swedish house with 0.025 ach), the upstairs conductance is 1024ft^2/r48
= 21.3 btu/h-f for the ceiling plus 88/4 = 22 for the windows plus 936/48
= 19.5 for the walls plus about 27 for 27 cfm of air leaks, a total of
90 btu/h-f, so it needs about (65-21.2)90 = 3942 btu/h or 94.6k btu/day
to stay 65 f indoors on an average december day in helena...
>flat-plate collector facing south at fixed tilt=90, solar radiation,
s sun 810 btu/ft^2-day |
e/w 290 " |
n 130 " |
avg. min 11.2 f | nrel data for helena, mt
avg. 24h 21.2 f |
avg. max 31.3 f |
avg. day 26.3 f during a 6h solar collection time?
rec. min 2.0 f
if electrical energy use of 600 kwh/mo contributes 68.2k btu/day of that
and the windows collect 24.6k btu more, that leaves 94.6-68.2-24.6 = 1.8k.
with 616 kwh/mo or 0.25 people in residence or less air leakage or
another south window, it could keep itself warm on an average day.
let's make it cheaper, with r24 (6" sip) walls and a house conductance of
131, so it needs (65-21.2)131 = 5738 btu/h or 137.7k btu on an average day,
69.5k more than the electrical gain, and 50.1k more, counting gain from
the windows. it needs 103.3k of overnight heat. if 18h/24hx68.2k = 51k
comes from electrical use, that leaves 52.3k...
how many overhead basement 4"x10' pipes do we need if the air near
the basement ceiling is 80 f for 6 hours per day? c = 56 btu/f and
10 ft^2 at 1.5 btu/h-f-ft^2 make rc = 3.7 h for each pipe, so tmax
80+(tmin-80)exp(-6/3.7) = 64.2+0.198tmin.
at dawn, the house needs (60-21.2)131-2841(electrical) = 2242 btu/h
= (tmin-60)10x1.5n, with n pipes, so tmin = 60+149/n. if 52.3k btu
= 56n(tmax-tmin) = 56n(64.2-0.802tmin) = 56n(16.1-119/n), n = 66,
tmin = 62.2, and tmax = 76.5 f.
a rock gabion room with lots of airflow and a short time constant could
store overnight heat with tmin and tmax close to 60 and 80 f. solid
limestone weighs 156 lb/ft^3, with a 0.215 btu/lb-f specific heat, or
156x0.215 = 33.5 btu/ft^3-f by volume. a 3" rock with c = 33.5x0.00818ft^3
= 0.274 btu/f and 0.196 ft^2 of surface and 1.5 btu/h-f-ft^2 of slow-
moving airfilm conductance has g = 1.5x0.196 = 0.295 btu/h-f, so rc = c/g
= 0.93 hours. with a 40% void ratio, it can store 0.6x33.5 = 20.1 btu/ft^3-f,
and 52.3k = 20.1v(80-60) makes v = 130 ft^3, as in a 2'x8'x8' room.
...6 tons of rocks vs 2 tons of water. rocks are cheap, with lots of
surface and a shorter time constant, so they can store heat closer to
the endpoint temps, and they never need to be refilled, but as you say,
they may get dusty. we might keep them clean with an air filter, but
that might require a fan. gabions want to be large in diameter to stand up,
but small to allow free airflow. they might have some sort of channels
inside to help air flow. i'd say the pipes win, in this case.
a square foot of r1 vertical sunspace glazing with 90% solar transmission
could collect 0.9x810 = 729 btu/day. it might lose about 6h(80-26.3)/r1
= 332 of that, leaving 397. if the house needs 50.1k btu/day, net, it
might have 50.1k/397 = 126 ft^2 of sunspace glazing, eg 8'x16'.
the closet needs to store 5x69.5k = 347.5k btu for 5 21.2 f cloudy days
in a row. it might have 347.5k/(130-80) = 6950 pounds or 869 gallons or 16
55 gallon drums of water stacked 2-high in 2 4-drum rows in a 4'x8'x8' tall
box with a 16x25/(8'x8') = 6.25 mass/glass surface ratio, with most of
the box inside the house. on an average day, it might gain 0.9x0.9x810x8'x8'
= 42k btu of sun and lose about 6h(130-80)8'x8'/r1 = 19k from the glazed
south wall to the sunspace air during the day and 18h(130-21.2)64/r at night
and 24h(130-70)192/r from the rest of the box, so the closet walls need
the closet might be simpler with 6950/64 = 108 ft^3 of waterwalls vs drums,
eg 4 1'x4'x8 tall ns waterwalls with 1' between them, with a mass/glass
surface ratio of 4. the shelves for 55 gallon plastic film drum liners
could be made with 2x4s and bolts and 2"x4" welded wire fencing. the lower
shelves could hold some smaller containers of water to raise the ratio.