The Virtuosi (Posts about radiation)https://thephysicsvirtuosi.com/enContents © 2019 <a href="mailto:thephysicsvirtuosi@gmail.com">The Virtuosi</a> Thu, 24 Jan 2019 15:05:02 GMTNikola (getnikola.com)http://blogs.law.harvard.edu/tech/rss- Freezing in Space II - Turn On The Sun!https://thephysicsvirtuosi.com/posts/old/freezing-in-space-ii-turn-on-the-sun-/Jesse<p>Yesterday I considered how long it would take a human to <a href="http://thevirtuosi.blogspot.com/2010/05/freezing-in-space-i-blackest-night.html">freeze in
space</a>.
However, I considered only what would happen if you were not absorbing
any radiation from nearby sources. Today we consider what happens if you
do have hot objects nearby. Namely, the sun. The sun provides a lot of
energy, even as far away from it as we are. It keeps the earth at a
comfortable \~20 C, good for us humans, and provides the energy for life
on earth, also good for us humans. That's a lot of energy. So maybe the
sun can keep you alive when you're adrift in space. Or at least keep you
warm. I still think you'll asphyxiate. From here on out we're going to
assume that we are adrift in space near earth. You were out for a
joyride in that new spaceship of yours and something went horribly
wrong. We could go through a whole song and dance of calculating how
much power the sun delivers to the earth, but we won't (if you're
interested, let me know, an I can do that later). Instead, we'll quote
the known result, that the sun delivers \~1370W/m^2 in the vicinity of
the earth. To find out what temperature we would cool to we set the
power we absorb from the sun equal to the power we radiate
$$P_{sun}=P_{rad}$$
$$1370W/m^2<em>A_{ab}</em>e_{ab}=e_{rad}<em>A_{rad}</em>\sigma<em>T^4$$ Where
A_ab is the surface area absorbing the suns power, e_ab is a factor
between 0 and 1 that indicates how much of the incident power we
actually absorb, and e_rad is the emissivity of us, while A_rad is our
radiating area. Note that the emitting and absorbing areas are not the
same! Take a simple example. If you put a sheet in space, and face the
flat side towards the sun, it will only absorb energy from the sun on
one side, but it will radiate energy from both sides. Likewise e_ab and
e_rad are not necessarily equal because we are radiating and absorbing
at different wavelengths. We can solve the above equation for T, giving
$$T=\left(\frac{A_{ab}}{A_{rad}}\right)^{1/4}\left(\frac{e_{ab}}{e_{rad}}\right)^{1/4}\left(\frac{1370W/m^2}{\sigma}\right)^{1/4}$$
For a first pass, we'll make the simplifying assumption that
e_ab=e_rad. Given this,
$$T=394K</em>\left(\frac{A_{ab}}{A_{rad}}\right)^{1/4}$$ Now, the
absorption area of an object is just the shape of the object flattened
into the plane the incident radiation is perpendicular to. That is, the
absorption area of a sphere is a circle (a sphere projected to 2D is
always a circle), while the absorption area of a cylinder could be a
sheet or a circle, or something stranger. The best area ratio we can
ever have is that of a flat sheet, which gives 1/2. For a sphere, like
the earth, the ratio is 1/4. As an aside, this gives an equilibrium
temperature of the earth as \~5C, which is too cold. It turns out that
we shouldn't neglect either the emissivity ratio or the natural
greenhouse effect in the case of the earth. Now, we need to figure out
the area ratio for us. In a <a href="http://thevirtuosi.blogspot.com/2010/05/human-radiation.html">previous
post</a> I
modeled myself as a cylinder with height 1.8 m and radius .14 m. Let us
assume we are facing the sun dead on, beating down on our chests. This
gives the cross sectional area of a sheet with width 2<em>.14 m =.28 m and
height 1.8 m. This is an area of .5 m^2, while my total surface area is
1.7 m^2. This gives an area ratio of \~.3, or an equilibrium
temperature of $$T=394K</em>(.3)^{1/4}\approx292K$$ That is an
equilibrium temperature of 19 C. Not too cold, but certainly not body
temperature! So the sun will not save us. We also have to factor in the
fact that we reflect better in the visual that we do in the infrared, so
the emissivity ratio we set to 1 probably is less than that, reducing
our equilibrium temperature even more. It is interesting to note,
though, that if we model a human as a two sided sheet instead of a
cylinder, we can bring our equilibrium temperature up to 331 K. That's
\~58 C! So in our model our geometrical assumptions change whether or
not we freeze or die of heat stroke. Finally, since it looks like the
sun may not save us, lets see how much it might slow down our
temperature loss. Instead of a net loss of 860W at body temperature, as
we calculated yesterday, sticking with our cylindrical human we'll have
a net loss of $$860W-1370W/m^2(.5m^2)=175W$$ Similarly at our lowered
body temperature of \~30 C, we'll be losing a net of \~105 W. Once again
taking a geometric average gives an average power loss of \~135 W. Using
the energy to cool we found yesterday it would take 16300 s, or 4.5
hours to freeze! Also note that if you're getting too hot or cold, given
how much the geometry plays into things, by changing your orientation to
the sun you'll be able to have a certain amount of control over how much
you heat up or cool down. Also, make sure you rotate yourself so that
you end up evenly heated, and not roasted on one side and frozen on the
other!</p>blackbodyfreezehumanpowerradiationspacehttps://thephysicsvirtuosi.com/posts/old/freezing-in-space-ii-turn-on-the-sun-/Thu, 13 May 2010 23:38:00 GMT
- Freezing in Space I - Blackest Nighthttps://thephysicsvirtuosi.com/posts/old/freezing-in-space-i-blackest-night/Jesse<p>In the last post I made, I discussed the fact that <a href="http://thevirtuosi.blogspot.com/2010/05/human-radiation.html">humans radiate
energy</a>.
In that post I calculated that we actually radiate quite a lot of power.
This immediately raises a few questions, the most obvious one being: How
long would it take you to freeze in space? This question is
multifaceted, and I'm going to split it between two parts. This first
part, 'Blackest Night' is how quickly we'd freeze if we were completely
lost in space, nothing anywhere near. The second part, 'Turn On The
Sun!' will address what would happen in near earth orbit. We need to
clarify what we mean when we say freezing in space. The fatal
temperature change for a human is (according to the all knowing
internet), roughly a drop of 7C. If you remember my previous post, we
calculated that we would radiate about 860 W. Now, we have to ask how
much energy it takes to change our temperature by 7 C. Well, as I've
discussed a
<a href="http://thevirtuosi.blogspot.com/2010/04/beer-diet.html#more">few</a>
<a href="http://thevirtuosi.blogspot.com/2010/04/falling-water-hot-or-cold.html">times</a>,
the energy it takes to change the temperature is given by $$Q=mc\Delta
T$$ The mass of a human is \~75 kg. We're mostly water, so let's just
assume that we are all water. This gives us a specific heat of 4.2
kJ/kg<em>K. The energy needed for our 7 C temperature change is then
$$Q=(75kg)(4.2kJ/kg</em>K)(7K)=2.2 MJ$$ By the time we have dropped to 30
C, we are only radiating a power of $$P=eA\sigma
T^4\=(.97)(1.7m^2)(5.67\cdot10^{-8}W/m^2K^4)(303K)^4=790W$$ Let
us assume that the average power radiated is the geometric average of
these two powers, $$P_{avg}=\sqrt{(860W)(790W)}\approx825W$$ This
gives us a time to freeze of $$t=\frac{2.2MJ}{825W}\approx2700s$$ This
is \~45 minutes. So you've got 45 minutes until a deadly freeze in deep
space. Seems a rather long time, does it not? I'm fairly certain that
you'd freeze much faster in antarctica than deep space. Why? Because in
Antarctica you have more cooling mechanisms that just radiation, you
have conduction in the air around you and convection of that warmer air
away from your skin. If you're adrift in space, for all that it is
rather cold, the good news is that you'll asphyxiate before you freeze!
All of this was done assuming that there's no energy gain from anywhere.
That is, that we're stuck somewhere in the deepest space, the blackest
night. Tomorrow we'll consider what would happen if you were in near
earth orbit, with all of these lovely energy sources around,
particularly the sun.</p>blackbodyfreezehumanpowerradiationspacehttps://thephysicsvirtuosi.com/posts/old/freezing-in-space-i-blackest-night/Wed, 12 May 2010 22:26:00 GMT
- Human Radiationhttps://thephysicsvirtuosi.com/posts/old/human-radiation/Jesse<p>Things are still busy here at the Virutosi. Hopefully in a week or so
we'll be back to normal, and much more active than we've been recently
Anyways, today I'd like to consider human radiation. It is well known
that any object will radiate energy based on its temperature. Even more
interesting, we radiate at all wavelengths, though at the human body
temperature our radiation is sharply peaked in the infrared. Even so, we
still put out some x-ray radiation. As a professor of mine once said,
consider that next time you sleep with someone! Given all this, the
question on my mind today is: how does the energy we radiate daily
compare to the energy we consume? That is, why don't I lose weight
sitting here typing on the computer? We physicists call perfect
radiators black bodies (something that radiates perfectly also must
absorb perfectly). For perfect radiators, the power radiated is given by
$$P=\sigma A T^4$$ where sigma is the stephan-boltzman constant, A is
the surface area of the radiator, and T is the temperature of the
radiator. For objects that are not perfect emitters or perfect
absorbers, we through in a fudge factor, e the emissivity, which is
between 0 and 1. This makes the power emitted $$P=e \sigma A T^4$$ To
figure out the power radiated by a human, we need to know three things.
The first is the emissivity of human skin. It turns out this is .97. The
second is the temperature of a human body, \~37 C. The third is the
surface area of a human. This requires a little estimation. I'm about
180 cm tall, and I wear 35" waist pants, so my radius is \~14 cm.
Modeling myself as a cylinder, I have a surface area of \~1.7 m^2. Now
we can estimate my power output: $$P=.97<em>5.67\cdot 10^{-8}W/m^2K^4
</em> 1.7 m^2 * (310K)^4$$ $$P \approx 860W$$ I'm powerful! That's
about 14 (60W) lightbulbs! We'd like to compare that to our daily energy
intake, so we need to turn this power into an energy. Well, there are
86400 s in a day. So we radiate 74<em>10^6 J per day. If you read my
<a href="http://thevirtuosi.blogspot.com/2010/04/beer-diet.html">beer diet</a>
post, you'll know the conversion between J and Calories (note the
capitol C). If not, suffice it to say that 1 Calorie = 4.2</em>10^3 J. If
I consume about 2000 Calories a day (typical, right?), then I take in
about 8.4<em>10^6 J per day. So, dear reader, why haven't I lost weight
while typing this post? There are a few answers. I'm wearing insulating
layers, clothing which keeps in some of my radiated heat. Also, our skin
temperature is lower than our internal temperature of 37 C. But more
than that, I'm absorbing energy from the surroundings. The earth is a
fairly good blackbody radiator, with an average temperature of \~20 C.
This means that my net power loss, with no clothing, would be about
$$\Delta P \approx (5.67\cdot 10^{-8}
W/m^2K^4)(1.7m^2)((310K)^4-(293K)^4)$$ $$\Delta P = 180W$$ This is
\~15</em>10^6 J per day. While still more than my intake, this is much
closer, and you can imagine that the rest of the difference can be made
up by our lower skin temperature, and clothing and such instruments of
men. Of course, these numbers are rough, so I don't recommend the 'naked
diet', where you try to lose weight by walking around naked. Or if you
do try it, don't say I told you to when you're taken in for indecent
exposure!</p>blackbodyhumanpowerradiationhttps://thephysicsvirtuosi.com/posts/old/human-radiation/Sun, 09 May 2010 00:24:00 GMT