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re: in floor heat
7 mar 2000
mario wrote:
>...in the same way you design a heat delivery system which is coupled
>with a thermal solar system, you design with the same components but
>operate in a controlled but reverse direction... an insulated liquid
>storage tank) would absorb the energy from the floor mass (via pumps)
>and then... transfer to the air (liquid to air ht-ex)...
sounds like the 10-day time constant house that physicist tim ellison
described at the '99 portland, me ases conference (a direct gain house
with a hydronic floorslab used for heat collection and distribution, and
"three 3,000 gallon water tanks in an outdoor strawbale sculpture.")
tim said "one can only add so much insulation to a house, and only
so much thermal mass in a floorslab, because of its internal thermal
resistance," and decided to use the tanks to increase the slab's
capacitance while barely increasing its series resistance.
consider a 32'x32'x8' tall house in philadelphia with r32 walls and
ceiling and a ft^2 of r4 south windows with 50% solar transmission
and no air leaks or internal heat gain and a thermal conductance
g = 32+a/r4+(1024-a)/r32 = 64+7a/32 btu/h-f, so the house needs
24h(65f-30f)g = 500a to stay 65f indoors on an average 30 f day
in january with 1,000 btu/ft^2 of sun that falls on a south wall.
a = 170 ft^2, min. like other direct gain houses, this one has an
optimal window area. too little means the solar heat gain is small
and it becomes a superinsulated house. too much window area increases
the cloudy day heat loss without increasing the floorslab/water temp
much (which is limited by the max room temp) and increases the house
temperature droop over say, 5 cloudy days. let's use 170 ft^2, which
makes g = 101 btu/f, so the house needs about 5x24(65-30)101 = 424k
btu for 5 cloudy days in a row.
four 2,100 gallon (8'dx8'h) tanks in the house make c = 67,200 btu/f.
this water would cool about 424kbtu/67.2kbtu/f = 6.3 f over 5 days.
let's say the room air to slab conductance is 1.5 btu/h-ft^2, in series
with an equal slab-to-water conductance, and 85k btu of sun arrives over
6 hours in january at a rate of 14.2k btu/h. here's one equivalent
thermal circuit for the house (in fixed courier font):
tr
| 1/101
|-------www-------30f
| b
w
w 1/1500 (0 f outdoors and no sun makes
w 1/101+1/1500+1/1500 = 0.01123.
--- | disconnecting the slab makes
|---|-->|---| 30+14.2k(1/1500+1/101) = 180 f.)
--- |
14.2k w which simplifies with a (thevenin)
btu/h w 1/1500 equivalent circuit above point a:
(day) w
| --- rt = 0.01123
| a --- ---www---
| --- | |
|--tw day/night | |--tw
| 180f/30f | |
----- 67.2k btu/f ----- ----- 67.2k btu/f
----- = c --- -----
| | |
| | |
--- --- ---
- - -
on an average day, q = 6h(180-tw)/rt btu flows into the water, roughly,
warming it dt = q/c above the dawn temp twd. then it falls by the same
amount dt = 18h(twd+dt-30)/(rtc), so dt = 90-4twd/3 = 6h(180-twd)/(rtc),
and twd = 66.8 f. dt = 90-4(66.8)/3 = 0.9 f, which makes the dusk water
temp 67.7. the room air is 30f+(66.8-30)/rt/g = 62.4 f at dawn. another
equivalent circuit below point b helps find the dusk air temp, 71.6 f.
tr = 30+(77.2-30)/rt/101 = 71.6 f.
|
| 1/101
|-------www-------30f
| b
w
w 1/1500+1/1500 = 1/750
w
|
|-- 67.7+14.2k/1500 = 77.2 f
|
-----
---
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---
-
the average day room air temp swing is 71.6-62.4 = 9.2 f, not bad...
but rtc = 755 hours (about 31 days), so after 5 30 f cloudy days in a row,
the water cools to 30f+(66.8f-30f)exp(-5x24/755) = 61.4, and the room air
becomes 30+(61.4-30)/rt/101 = 57.7 f, which is fairly cool, because this
giant house capacitor has such a small long-term temperature swing...
with a higher temp (eg 120 f) and a larger temp swing, the house capacitor
can be a lot smaller, and the room air temp can be constant until the cap
runs out of useful heat, at say, 75 f.
nick
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