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newton's law of cooling
23 jun 1996
david williams quotes:
>-> there seems to be some serious disagreement over how long it takes an
>-> unisolated heat store to cool down
ignorant people have serious disagreements over surprisingly simple things.
one might resolve this with a teacup and a watch: start with boiling water,
put a lid over the cup to slow evaporation, and recall that things at 98 f
feel neither warm nor cool to the touch.
>->...let me propose a very hypothetical situation...
>-> i have this infinitely large room with no windows at a constant
>-> temperature of 70 f. into it, moses-like, i place a wall of water,
>-> 10' high, 10' wide and 1' thick.
sounds more like an anti-moses :-) doesn't he play antipodal trombone
with bill parker of the red sea pedestrians? let's ignore evaporation...
>-> the top, bottom and short sides of
>-> this wall are insulated to an r value of infinity. i heat the water
>-> in this wall to a uniform temperature of 100 f.
>-> how long before the temperature of the water in the wall is 70 f?
>answer: never. the temperature of the water will decline exponentially
>towards the temperature of its surroundings - the 70 f room. it will
>never *quite* get there.
correct but stupid. a more interesting question is how long the water wall
will stay 15 f warmer than the room. the thermal resistance of a still air
film is about 2/3. the wall has two sides exposed to room air, and a cubic
foot of water weighs about 63 lb, so the rc time constant of the wall is
about 2/3/2ft^2x63 = 21 hours, over which time the initial wall-room temp
difference will decrease by a factor of 1/e, about 1/3. the wall temp at
time t will be t(t) = 70 + (100-70)exp(-t/21), so after time t in hours,
85=70+30exp(-t/21) ==> ln(15/30) = -t/21 or t = 14.6 hours, if the room were
kept warm by other means. but why should that be important? we want to keep
the room warm with the wall, not vice-versa.
if the water wall were half as thick, with about the same surface area and
volume as will's water walls, it would have rc = 10.5 hours, so it would
cool to 85 f in -10.5 ln(1/2) = 7.3 hours, if the room were kept at 70 f
by other means ("keep the home fires burning" :-)
if the room containing the wall needed 150 btu/f-hr to stay warm, like will's
house, in the dark, 24 hours a day, with the perfect ritual thermal curtains
so carefully drawn (who makes those, and how much do they cost?) and if it
were 40 f outside, what would the steady-state temp of the room be, if the
wall were 100 f? this has a simple electrical analog:
troom
| + 40 f -
100 f -------www-------|---------www----------40 f
1/3/100 1/150
the total series resistance is 1/300 + 1/150 = 0.01, so the heat that flows
from left to right is (100-40)/0.01 = 6000 btu/hr. this heat flowing through
1/150 "ohm" makes a temperature difference of 6000/150 = 40 f, so the room
temperature would be 40 + 40 = 80 f. warmish.
the rc time constant of the whole room, including the wall with 3200 lb of
water, might be about 0.01x3200 = 32 hours, so if there were no other source
of heat, the room temp t(t) after t hours might be
t(t) = 40 + (80-40)exp(-t/32)
80 0
59 24
49 48
...
how long would it take the room to reach 70 f? 70 = 40 + (80-40)exp(-t/32)
==> 30 = 40exp(-t/32) or t = -32ln(3/4) = 9.2 hours. if the room were 80 f at
3 pm on a mild winter day, what would the temp be at 9 am the next morning?
t(18) = 40 + 40exp(-18/32) = 63 f.
how long could the wall keep the room at 70 f, if its heat loss could be
controlled, vs uncontrolled natural cooling? the room needs (70-40)x150
= 4500 btu/hr to stay 70 f, so the wall could keep the room at 70 f until
the wall temp reached tmin = 70 + 4500/300 = 85 f. if the wall were 100 f to
start, and the room were 70 f, ie if the wall were somehow better insulated
to begin with, the wall could store (100-85)3200 = 48k btu of usable heat,
and the wall could keep the room warm for about 48k/4500 = 10.7 hours.
a little better, but still not long.
if the same wall were inside an attic warmstore or solar closet, and started
out at 130 f, it might keep the house warm for (130-85)3200/4500 = 32 hours.
that's a higher-performance solar house. a solar closet might also dry food
or clothes, serve as a sauna, or make domestic hot water, which seems nice.
one might even build a solar oven into one's solar closet, and cook for a week
with no sun.
adding a 36 watt 560 cfm fan might lower the minimum usable space heating temp
to 75 f, extending the cloudy day heat storage time to (130-75)3200/4500 = 39
hours. replacing the $2000 (?) window-blocking water walls with 220 recycled
5 gallon plastic paint pails, with lids, or 20 55 gallon drums in a 4' x 10'
x 6' tall solar closet would store (130-75)10,000 = 550k btu of heat, for a
solar storage time of 550k/4500 = 122 hours or about 5 days, which would make
it a much nicer solar house, in my opinion.
using a setback thermostat to make the house temp 60 f at night would lower
the average daily heating requirement to (8(60-40)+16(70-40))150 = 96k/day,
extending the solar storage time to 5.8 cloudy days. that can't be done with
exposed water walls. if the room were used as an office or workshop or school
or church, and it had little thermal mass, it might be 45 f at night and 70 f
for 8 hours a day, requiring only (16(45-40)+8(70-40))150 = 48k btu/day, so a
well-insulated solar heat store would last for 11.5 days without sun. better
still, but impossible using exposed water walls...
nick
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