The Virtuosi (Posts about thermodynamics)https://thephysicsvirtuosi.com/enContents © 2019 <a href="mailto:thephysicsvirtuosi@gmail.com">The Virtuosi</a> Thu, 24 Jan 2019 15:05:03 GMTNikola (getnikola.com)http://blogs.law.harvard.edu/tech/rss- Letting Air Out of Tires IIhttps://thephysicsvirtuosi.com/posts/old/letting-air-out-of-tires-ii/Jesse<p>In a <a href="http://thevirtuosi.blogspot.com/2010/05/letting-air-out-of-tires.html">recent
post</a>
I calculated how cold air coming out of bike tires should feel. However,
at the end of the post, I did note that there are competing explanations
for why the air cools. There's the approach I took, which is adiabatic
cooling, but there's also something called the Joule-Thomson effect. The
Joule-Thomson effect has the interesting property that helium being let
out of a bike tire would actually be warmer, which suggests an immediate
way to test which effect is dominant. We pressurize a bike tire with
helium, and see if the valve gets cold or hot. This is exactly what I
did. With the help of Mark and Vince, our local equipment gurus, I was
able to pressurize a bike tire to 26 psi with helium. Using one of those
little thermal measurers you can buy at radio shack, we measured the
initial temperature of the valve as 80 F. We then released the helium,
and measured the temperature of the valve as 73 F. The adiabatic
approach is the winner! Our experiment confirms that the dominant effect
of the cooling is the adiabatic cooling I talked about yesterday. The
Joule-Thomson effect may be at play, but if so it takes a secondary role
to the adiabatic cooling. Now, some of you may be saying: wait a second,
you predicted the air would be -100 F! It doesn't feel that cold! Nor
did your valve cool down to -100 F! To which I reply: Yes, I did predict
very cold air. But you have to remember that it is mixing with a lot of
room temperature air, so it won't feel as cold as I predicted. Nor will
it transfer much heat to the valve (recall, we predicted this would be
an adiabatic process, with absolutely no heat transfer, something that
is obviously false). Also, I didn't have 60 psi of pressure. If we do
the calculation, 26 psi only gives a temperature of 250K = - 10 F.
Hopefully that answers your question. And now, dear reader, as I've
wanted to say for a while: Myth Busted!</p>bikebustedcoldexperimentgasthermodynamicstirehttps://thephysicsvirtuosi.com/posts/old/letting-air-out-of-tires-ii/Tue, 04 May 2010 14:52:00 GMT
- Letting Air Out of Tireshttps://thephysicsvirtuosi.com/posts/old/letting-air-out-of-tires/Jesse<p>Have you ever noticed how when you let air out of a bike tire (or, I
suppose, a car tire) it feels rather cold? Today we're going to explore
why that is, and just how cold it is. Many people consider the air
escaping from a tire as a classic example of an adiabatic process. What
is an adiabatic process? It is a process that happens so quickly there
is no time for heat flow to occur. For our air in the bike tire this
means we're letting it out of the tire so quickly that no energy can
move into it from the surrounding air. This may not be exactly true,
there may be a little energy flow, but there is little enough that we
can ignore it. Given that, how do we talk about temperature change?
Let's give a physical motivation first. Imagine a gas as a collection of
hard spheres, like baseballs. Envision this bunch of baseballs in a box.
Suppose you make the volume of the box smaller, you move the walls in.
The baseballs will start to bounce around faster. Having trouble
thinking of this? Think of a single baseball in a box. Imagine it hits a
wall moving towards it. What happens? That's just like what happens when
a baseball hits a baseball bat moving towards it, it goes flying. That
is, it starts moving faster. The speed with which these gas particles
are moving is what we measure as temperature. So shrinking our box
increases the temperature. Likewise, expanding our box will decrease the
temperature. The same principle holds here. Our gas is expanding from a
small volume (the bike tire) into a larger volume (the surrounding
world). Thus we expect the temperature to decrease.
Got all that? Good. Now for some math. It turns out that using the ideal
gas law, we can derive that for an (reversible) adiabatic process
$$PV^\gamma=constant$$
where P is the pressure of the gas, V is the volume of the gas, and
gamma is (for our purposes) just a number. The ideal gas law states that
$$PV=NkT$$
Where N is the number of gas molecules we have, T is the temperature of
the gas, and k is a constant. Since N doesn't change in the process
we're considering, we can use this to rewrite the above equation, by
substituting for V. This gives
$$P^{1-\gamma}T^{\gamma}=constant$$
Where the constant is not the same as above.
Because this is equal to a constant, we can say that our initial
pressure and temperature are related to our final pressure and
temperature by
$$P_i^{1-\gamma}T_i^{\gamma}=P_f^{1-\gamma}T_f^{\gamma}$$
We can solve this for the final temperature giving
$$T_f=T_i\left(\frac{P_i}{P_f}\right)^{\tfrac{1-\gamma}{\gamma}}$$
Finally, we can plug in some numbers. Gamma is 7/5 for diatomic gases
(which is most of air). If we assume the air is about room temperature,
20 C, and the tire is at 60 psi, this gives (1 atmosphere of pressure is
15 psi):
$$T_f=293K\left(\frac{60 psi}{15
psi}\right)^{\tfrac{1-7/5}{7/5}}=197K$$
Converting back from Kelvin to C, this is -76 C or -105 F. That's cold!
<strong>For the Expert:</strong>
There is actually a debate as to whether or not adiabatic cooling is
responsible for the chill of air upon being let out of a tire. The
argument for it is fairly straightforward. The released air does work on
the surrounding atmosphere as it leaves, lowering the energy of the gas.
If this is the primary effect, then the change in temperature is given
above. However, it is possible that we can consider this a free
adiabatic expansion. In a free adiabatic expansion (like a gas expanding
into a vacuum), there is no work done, because gas is not acting
'against' anything.
The other possibility is the <a href="http://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect">Joule-Thomson
effect</a>. I
don't claim to understand this effect very thoroughly, but it is another
mechanism for cooling when air is let out of a well insulated valve.
I've seen claims both ways as to which process is actually responsible
for cooling.
Fortunately, a simple experiment suggests itself. Helium heats up
through the Joule-Thomson effect (when it starts abot \~50K). It will
cool down through the above described adiabatic cooling. So, fill a bike
tire with Helium gas, and let it out. See if the valve/gas feels hot or
cold. This will determine the dominant effect. As an experimentalist,
this <a href="http://www.brightlywound.com/?comic=42">appeals greatly to me</a>.
But if any theorists out there have ideas, please speak up.</p>bikecoldexperimentgasthermodynamicstirehttps://thephysicsvirtuosi.com/posts/old/letting-air-out-of-tires/Mon, 03 May 2010 23:21:00 GMT
- The Beer Diethttps://thephysicsvirtuosi.com/posts/old/the-beer-diet/Jesse<p>I know it's been pretty quiet over here this week. The semester is
winding down (a week of classes left), and that means that things have
been kicked up into another gear. We've got four or five ideas bouncing
around at the moment, so hopefully we'll get some up soon. Today I'd
like to talk about the beer diet. A while back, there was a rumor going
around that if you drank ice cold beer your body would burn more
calories heating the beer than the beer contained. It turns out that
this false, and I think the claim relied on a lack of knowledge that the
American food Calorie is actually one thousand calories (note the
difference in capitalization). Let's prove this to ourselves. Let us
consider 12 fluid oz of beer. To any good approximation we can treat
this as 12 oz of water. Let us assume that the beer starts at 0C and our
body raises it to body temperature, 37C. This takes energy Q given by
$$Q=mc\Delta T$$ Where m is the mass of the beer, c the specific heat
of water, and the last term the change in temperature. We convert out of
the archaic units (thanks google!) to get 12 oz = .35 liters. The
density of water is 1000kg/m^3, which is 1kg/liter. This gives us the
mass of the beer as .35 kg. The specific heat of water is 1 kcal/kg<em>C
(note: kcal = kilocalorie = 1 Calorie, i.e. 1000 calories), so the
energy it takes to change the temperature of our beer is
$$Q=(.35kg)(1kcal/kg</em>C)(37C-0C)$$ $$Q=13kcal$$ If memory serves, some
beer company recently had ads toting how their light beer was 'under 100
Calories'. So, I imagine 100 kcal is a fairly good lower bound on the
calories in a 12 oz beer. So you can't lose weight just by drinking
beer. Sad, isn't it? Finally, we can see that if we didn't understand
the distinction between Calorie and calorie, we might think that, since
it took 1300 calories to heat up the beer, and beer only contains 100
Calories, this would be a great way to lose weight! I leave it as a
question for you, dear readers, as to how much ice water you'd have to
drink to have an effective ice water diet. If it does become the next
fad diet, you heard about it here first!</p>beerCaloriesdietthermodynamicshttps://thephysicsvirtuosi.com/posts/old/the-beer-diet/Fri, 30 Apr 2010 11:35:00 GMT