On Sun, Jul 20, 2003 at 10:12:11PM +0000, Robert J. Chassell wrote:
> Thank you. Am I right in thinking that for air, this is the heat
> conductivity for still air, and not the heat transfer capabilities of
> moving air?
It is the thermal conductivity of air, which is almost the same whether
the air is still or moving (it results from excited gas molecules
bumping into adjacent molecules and exciting them in turn). However, if
the air is moving, heat can be transferred by the moving gas molecules
-- the molecules themselves have energy, and when the molecules move,
they carry that energy which can excite other molecules at a different
location, thus transferring heat. That could be called a convection
current. Since the thermal conductivity of air is so low, in most cases
more heat can be transferred by moving air than by conduction through
the air.
> I can tell you by personal experience that a cold wind can chill a
> human body fast.
Cold water will conduct heat out of a body much more efficiently than
cold air can conduct heat away. Of course, if the air is moving, it will
be more efficient at taking the heat away from the body as compared to
still air. A windbreaker shell covering a fluffy layer is thus most
effective (when dry) at keeping you warm -- the outer layer stops the
wind from carrying heat away, and the fluffy layer traps air, which has
a very low thermal conductivity, to slow the transfer of body heat by
conduction.
Heat transfer by natural and forced air convection are complicated
processes. I'm not aware of any first principle calculations that are
useful for modeling heat transfer between surfaces and air. There are
some phenomenological formulas, however:
natural convection
------------------
W/m^2 = h (delta_T)^(5/4)
h ~ 1.3-2.5 W/m^2 K^(5/4)
(value depends on shape and orientation of surfaces)
forced air convection
---------------------
W/m^2 = eta delta_T
eta ~ 5-60 W/m^2 K
eta depends on airspeed and slightly on delta_T
Here's an example comparing heat transfer from a body to either (windy)
air, or to cold water.
The normal human skin temperature around the torso is about 33C, which
is 306 Kelvin. We'll look at heat transfer from the skin to either a
moderate wind at 7C, or to still water at 7C (280K). Assume the surface
area of the skin is 2 m^2 (rough estimate assuming all the skin is
exposed) and that half of the skin is in a wind shadow (opposite side of
body from the wind).
Then for forced air convection, assuming eta=50 for this windspeed and
delta T, the heat transferred per second is:
50*1*(306 - 280) = 1300 W
Compared to radiation cooling, which would be about 150W, this is
quite large. If we model the body as being made up of water, with
a mass of 80kg, then the heat capacity of the body is 4216*80 J/K
= 337000J/K. Dividing the heat transfer rate 1300 J/s, by the heat
capacity 337000 J/K, gives a temperature drop of about 0.004 K/s, or
about 0.23 degree Celsius per minute. This assumes zero cooling by
evaporation of sweat. This is just a rough estimate, since the core of
the body is somewhat insulated from the skin, the real problem is quite
a bit more complex. But this should be roughly correct. I found a paper
http://www.arbetslivsinstitutet.se/publikationer/pdf_ah/2003-04.pdf that
gives 1319 W/m^2 for wind chill in 4 m/s wind.
For a body in 7C water, to make the math simple I again have to make
some approximations. In addition to the ones already mentioned, I am
going to assume that the distance the heat must travel through the body
and water as it moves away from the body is 4cm (I chose 4cm because all
the water within 4cm of the body weighs about 80kg, same mass as assumed
for the body). Since the thermal conductivity of water is 0.6 W/m K,
then the heat transfer rate is
0.6*2/.04 * (306-280) = 780 W
which translates to a temperature drop of 0.14 deg C per minute. I
found a link http://www.heat.uk.net/cooling2.htm that gives body cooling
by ice water bath immersion of 0.2 C per min, so this estimate isn't too
bad. If the water were moving (or the person were moving in the water),
I wouldn't be surprised if the actual rate could be more than twice that
calculated above.
So, standing naked in a moderate wind at 7C and being submersed in cold
water at 7C will result in roughly the same body chilling by convection
and conduction. Of course, few people stand naked on a windy cold
day, and clothing makes a big difference in keeping a layer of warm
air trapped near the body, drastically reducing the wind cooling. In
contrast, clothing won't help much in the cold water (unless you are
wearing something like a neoprene wetsuit, which keeps the water away
from the skin).
> Hence the practical effect in any environment in which air can move,
> such as the interior of a spinning space habitat, is that air has a
> much higher possible heat transfer capacity than might be indicated by
> its still air heat conductivity value.
Yes.
> Are you saying that there will be no air movements in the spinning
> space habitat?
No, just the opposite. If there weren't air currents, then no
equilibrium would be possible.
> Certainly, air at the top of a mountain I went to Friday did not gain
> temperature from heat conducted through the rock that day from the
> warm valley floor.
How do you know that?
> That air at the top of the mountain was cool.
Did you measure the temperature of the rock at the surface and, say, 1
meter below the surface? Did you measure the air temperature near the
rock and far away (miles) from the rock?
When in sunlight, the rock will be heated by sunlight much more than the
air (the atmosphere only absorbs about 25% of sunlight incident on the
top of the atmosphere) and will thus be warmer than the air. Therefore
the rock will transfer heat to the air as long as it remains warmer
than the air, which will be for quite a while since the rock has a much
larger mass and higher heat capacity than the air. Haven't you ever felt
the outside of a brick or stone building after sundown, when it has
been in the sun during the day? It is quite warm compared to the air
temperature.
> The rocks at the top of the mountain were cool, too, they could not
> transfer much heat to the air.
If the rock was even 1 degree warmer than the air, then heat will
transfer from rock to air, and since the mass of the rock is so much
greater than the air, it can make a significant difference in the air
temperature. You even noticed this yourself -- you said that you see
thermals above fields more than forests. The same goes for rock.
> > I would expect ... that with sufficiently good insulation, the
> > surface of the end caps would come to the same temperature as
> > the air fairly quickly and would neither contribute nor take
> > much heat from the air.
>
> Same temperature as which air? The air on the axis is colder than the
> air at the rim.
>
> I meant the rock and the air at the same distances from the axis.
If the lapse rate is 8 C / km (very rough approx), then at 5km height
the air would be 40 C colder. If we assume the habitat has an inside
length of 6km, then let's look at 3km of air being heated by one endcap.
The volume is about 235 km^3. Very roughly, the air mass is 200e9kg. Air
has a specific heat of about 1000 J/kg K, so the heat capacity of the
entire air mass is about 2e14 J/K. If we put 1m of fiberglass insulation
on the endcaps, thermal conductivity 0.03 W/m K, and assume the asteroid
itself is all at the same temperature, then very roughly the heat flow
to the air will be .03/1*pi*5^2*1e6*40 = 9e7 W, which results in a
temperature change in the air of 0.04 C per day due to heat transfer
from the endcaps. Quite low.
However, the thermal conductivity of the air is comparable to the
thermal conductivity of the fiberglass insulation, so it is unrealistic
to assume that the entire 3km length of air is heated by thermal
conduction from the endcap. If that 3km length of air is all at the same
temperature, there would need to be quite a bit of air movement along
the axis. If we looked at air 3m from the endcap instead of 3km, it
could be heated by 40 C per day if it was not moving. Quite high.
So perhaps 1m of insulation is insufficient. Make it 100m thick, and
the air 3m from the endcap (if still) can only be heated by 0.4C per
day. Probably not enough to create strong air currents.
> Yes, but what is the rate of heat flow from the interior of the rock
> through its surface into the air? How would it compare to heat
> transfered by moving masses of air? What wind speed would it produce?
How could this be answered without a simulation? Do you know of a simple
model for calculating conduction, convection, and wind speeds?
> Would the heat from the sides flow to the end caps and then be
> radiated into space? Would the air by the end caps get chilled by the
> heat being radiated into space? Or is the insulating quality of rock
> and dirt good enough to make that fairly irrelevant?
If the walls are 1km thick and have an average thermal conductivity of
80 W/m K, and if the asteroid has the same temperature/depth profile at
all points on the surface, then radiation and conduction must be equal
and we get (assuming emissivity of 1)
(300K - T0) * 80 / 1000 = 5.67e-8 * (T0^4 - 3^4)
assuming the outside temperature is 3K. The solution is T0=125K, which
results in heat radiation of 14 W/m^2. The sunlight reaching the top
of the Earth's atmosphere is about 1400 W/m^2. Even if only about one
quarter of this is absorbed by the habitat (due to shadowed areas and
imperfect absorption), it still seems the habitat would have more of a
problem staying cool than staying warm, if it orbited at 1 A.U. from
the sun. The habitat would need either to be in a further orbit (about
5 A.U.'s), or to have high reflectivity on its outer surface (99%), or
to have lots of radiative thermal panels that stick out only on the
shadowed side, enough to increase the effective surface area by about
100 (but that might be difficult since it rotates they would have to go
in and out).
--
"Erik Reuter" <[EMAIL PROTECTED]> http://www.erikreuter.net/
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