Physics Quiz Question

Dallas wrote:
The origin of the question comes from the current edition of the
Jeppesen Private Pilot manual.
[...]

http://img.photobucket.com/albums/v1...allas/Jep2.jpg

Context helps. The kicker, and misleading, aspect is this:

"...assuming that all other variables remain constant...."

First, what the Jeppesen manual states is self-evident from the ideal gas
law:

P*V = n*R*T

So their text is technically correct, but really incorrect as far as
describing what *really* happens when some part of the atmosphere heats
up more than the surrounding atmosphere.

In the real atmosphere the volume V0 that is occupied by some n moles of
gas does not "remain constant". But neither does a volume V0 of gas
undergo "free expansion" as the temperature rises from T0 to T1, as I've
seen some people suggest. To be clear: "free expansion" means the total
energy of the n molecules in V0 remains the same as the volume goes to V1
- that is, as the gas expands it does no work, and no work is done on it.
(Or put mathematically, "free expansion" says that P0*V0 = P1*V1; that is
because the product of pressure and volume yields units of energy!) Of
course it is not the case that the energy remains constant because as the
gas expands it has to do work against gravity (i.e. it has to push
against the surrounding atmosphere). So in general what actually happens
is something _in between_ what happens when V0 = V1 (volume held
constant) and P0*V0 = P1*V1 (energy held constant). (And to determine P1
and V1 is why thermodynamics texts are filled with imposing looking
differential equations! :-))

BUT THE BEST FORMULA (and most relevant to aviators) DOESN'T DEAL WITH
VOLUME CHANGES! IT DEALS WITH DENSITY CHANGES! Here is how it is derived:

First divide both sides of the ideal gas law by V and rearrange
variables:

P = R*T*(n/V)

But n/V is just a density! So density = n/V and we get:

P = R*T*density

But R is "just" a conversion constant to make sure all the units work
out. If we only want to understand proportionalities we can discard the
R. Then dividing both sides by T yields:

density = P/T

*** SO IMHO THE BEST FORMULATION OF THE IDEAL GAS LAW FOR PILOTS IS: ***
**** ****
****** DENSITY = PRESSURE/TEMPERATURE ******

So when Jeppesen said "assuming all other variables remain constant" it
was basically saying "assuming the density remains constant." But air
density is the ultimate variable of interest to pilots! So Jeppesen was
effectively posing an example that said "assuming pressure and
temperature vary such that it doesn't affect the lift produced by your
wings!" Now *that* is a misleading (and useless) example IMHO!

On Tue, 07 Aug 2007 18:31:41 -0000, Jim Logajan wrote:

So when Jeppesen said "assuming all other variables remain constant" it
was basically saying "assuming the density remains constant."

Are you pretty comfortable with that statement? I can't imagine why
Jeppesen would bother to publish the paragraph if the assumption was that
the density would remain constant, which is basically impossible outside
the laboratory, (at least, as far as I know) and makes whole statement of
no value to a pilot in the real world.

"Assuming all other variables remain constant". - I picture two barometers
a few miles apart on a consistent, flat surface. There is no wind and the
sky is overcast. A hole in the overcast opens up and heats the area around
the first barometer. If I correctly interpret what Jeppesen appears to be
saying, the pressure in the area of the heated barometer will rise above
the barometer in the shade.


--
Dallas

On Aug 7, 3:32 pm, Dallas wrote:
On Tue, 07 Aug 2007 18:31:41 -0000, Jim Logajan wrote:
So when Jeppesen said "assuming all other variables remain constant" it
was basically saying "assuming the density remains constant."

Are you pretty comfortable with that statement? I can't imagine why
Jeppesen would bother to publish the paragraph if the assumption was that
the density would remain constant, which is basically impossible outside
the laboratory, (at least, as far as I know) and makes whole statement of
no value to a pilot in the real world.

"Assuming all other variables remain constant". - I picture two barometers
a few miles apart on a consistent, flat surface. There is no wind and the
sky is overcast. A hole in the overcast opens up and heats the area around
the first barometer. If I correctly interpret what Jeppesen appears to be
saying, the pressure in the area of the heated barometer will rise above
the barometer in the shade.

OK, I think I understand what's going on. You are interpreting
something that, while technically correct, is aimed at non-physists.
All it is trying to say is that if you have the same mass of air
contained in the same volume, but with the air at different
temperatures, the pressures will be different. One of the main issues
of contention that I have is that we are talking about energy
transfers that affect multiple different component variables (e.g.
mass, volume, and temperature) in an open system, and trying to close
the system AND hold all but one of them constant (the old "ignoring
friction" routine).

I still don't like their use of "exerting pressure" on the surrounding
atmosphere, because I really don't think it is. However, I guess you
could demonstrate their statement by using an infinitely thin, capped,
and flexible column surrounding a mass of air (think condom shape).
if you heated the air inside of it, you would see the column walls bow
to the pressure changes. But the walls bowing is a demonstration of
the pressure differential and attempt to establish equalibrium more
than anything else, and handwaves the fact that pressure isn't defined
this way.

I think Jim is right; a more important relation is how a particular
temperature reading at a particular pressure reading relates to the
density of the air at altitude. This is probably one of those cases
where cursory hand wave...it's close enough about how it works is
less important than the effects that it has on an aircraft's
performance at a given altitude for different conditions.

(see also further example in r.a.s.)

Dallas wrote:
On Tue, 07 Aug 2007 18:31:41 -0000, Jim Logajan wrote:

So when Jeppesen said "assuming all other variables remain constant"
it was basically saying "assuming the density remains constant."

Are you pretty comfortable with that statement?

Pretty comfortable. Although I admit I did better in quantum mechanics in
college than thermo and statistical phsyics. :-)

I can't imagine why
Jeppesen would bother to publish the paragraph if the assumption was
that the density would remain constant, which is basically impossible
outside the laboratory, (at least, as far as I know) and makes whole
statement of no value to a pilot in the real world.

Well, it looks like they were trying to make a point and be technically
accurate. I believe that typically requires making text-book simplifying
assumptions.

"Assuming all other variables remain constant". - I picture two
barometers a few miles apart on a consistent, flat surface. There is
no wind and the sky is overcast. A hole in the overcast opens up and
heats the area around the first barometer. If I correctly interpret
what Jeppesen appears to be saying, the pressure in the area of the
heated barometer will rise above the barometer in the shade.

Your interpretation matches exactly what they say: "On the other hand, a
warmer temperature increases atmospheric pressure, all else being equal."

But the problem is that "all else" _doesn't_ remain equal in your
scenario. That is why I think the Jeppesen paragraph is misleading, since
it will lead students to believe certain values remain constant when they
really don't.

In your scenario the air over the sunny area may try to increase in
pressure to, for example, 14.8 lb/in^2 with surrounding air at, say, 14.7
lb/in^2. The pressure difference will cause the heated air to balloon
outward till the pressures at the imaginary boundary equalize. What you
get is an outflowing "wind" as the hotter air balloons out. The hot area
should also cool a little. The details get ugly, but suffice to say that
almost immediately after the heating begins, the volume starts expanding
so the density starts dropping and the barametric pressure will appear,
in this example, to be between 14.7 and 14.8.

So in general if it is going to be a hot day, the air density will _tend_
to be either the same or less than on a cooler day. So wing lift and
oxygen content/intake is slightly reduced on the hotter day.

But dang it, nothing in physics seems to rule out weather events
conspiring so that one gets a hot day and a high pressure system such
that the air density is much higher than average. Such is the nature of a
dynamic atmosphere that experiences unequal heating.

Clark wrote:
A minor nitpick on a previous post from Jim:

It was stated that n/V was equal to density.

A reasonable nit, but I was careful enough to say it was "a" density, not
"the" density. My words we "But n/V is just a density!"

While the stated equality
isn't, the jist of what was said was ok. The equality would be

(Mass/Molecular Weight)/V = density

Since the molecular weight of air is a constant it can be combined
with the other constant in the ideal gas law when used for atmospheric
calcs.

Thanks for the nit pick and clarification.

Nowadays I sometimes qualify things that don't need qualification, and fail
to qualify things I should. For example, you might find me stating "It is
my understanding that 1 + 1 = 2 for most values of 1 and 2," rather than
the unequivical "1 + 1 = 2". Otherwise somebody will go:

"Bzzzzt! Wrong! Try again. 1 + 1 = 3. And here's the proof...."

And by gosh the proof would look valid. Or they'd direct me to a .gov web
page where it states that 1 + 1 = 3. And since I like to use as
"authoritative" web sites as I can find to support my own assertions, and I
consider .gov sites reasonably authoritative, I'd be forced to admit my
ignorance and stupidity for making such an unqualified assertion. It
becomes tiring admitting to ignorance and stupidity too often (at least it
is for me!), and I suspect that after a while the only people left reading
my posts are those who are equally ignorant and stupid. At that point I
think my posting becomes futile/redundant since my ignorance and stupidity
has reached equilibrium with the equally ignorant and stupid readers of my
posts.

What, me ramble?

"Clark" wrote in message
...
A minor nitpick on a previous post from Jim:

snip

It was stated that n/V was equal to density. While the stated equality
isn't,
the jist of what was said was ok. The equality would be

(Mass/Molecular Weight)/V = density

Since the molecular weight of air is a constant it can be combined with
the
other constant in the ideal gas law when used for atmospheric calcs.


a minor nitpick on your minor nitpick is that the molecular weight of air is
not constant. It is affected by humidity.
density is better expressed as d = PM/RT
where d will be denity in units of mass/volume
(M is molecular weight), R the gas constant ,P and T press and Temp . For
us SI people R=8.31, P in Pa and T in Kelvin gives density in kg /m3 (
using M in kg/mole)Terry

Think of it this way. If the entire atmosphere's temperature was
increased by say 10 degrees, the average pressure at the surface would
be as it had been, each square inch supporting about 15 pounds of air.
The 15 pounds doesn't change bcause it's hotter.



On Aug 6, 10:05 am, Dallas wrote:
Brought over from RAS:

Assuming that all other variables remain constant:

An increase in temperature will result in a higher atmospheric pressure - a
higher temperature speeds up the movement of the air molecules, thereby
raising the pressure they exert on the surrounding atmosphere.

A) True
B) False

--
Dallas

I seem to have come late to this scrum...

Anyway, the last poster missed a thing or three... Like Euler, like
Boyles law, like the equation for lift, like Bernouille, etc..

The biggest thing he missed is if you heat a gas and keep its pressure
the same then it has to expand (That old Boyle guy again)
Expanded gas means the molecules are farther apart... Molecules being
farther apart means less density, less density means less lift...

Ya know in the ancient times of BI (before internet) you had to truck
down to the library and dig through piles of books to find this
information... Today, in 0.002 seconds your web browser will pull up
the articles and equations on this subject... It is highly recommended
by me...

denny

The statement (taking into account the later added context, i.e.
Jeppesen's text) is technically true, although not very useful in the
aviation environment.

As Jim L. pointed out, the presure is proportional to gas density and
temperature. That's all there is to it (the pressure is NOT the weight
of the the column of air). So when Jeppesen says "all else being
equal" while analyzing temperature and pressure, that "else" has to be
density. If density stays constant, higher temperature will result in
higher pressure.

But airplanes don't fly in an enclosed container where one can keep
the density constant. For example: higher temperature (all else being
equal :-) ) will tend to deacrease the density. And then there are
winds, ...

So the statement (although technically correct) is at best pointless,
and at worst (dangerously ?) misleading.

- Tom

On Mon, 6 Aug 2007 12:05:17 -0500, Dallas
wrote:

Brought over from RAS:


Assuming that all other variables remain constant:

An increase in temperature will result in a higher atmospheric pressure - a
higher temperature speeds up the movement of the air molecules, thereby
raising the pressure they exert on the surrounding atmosphere.

A) True
B) False

On Mon, 6 Aug 2007 12:05:17 -0500, Dallas wrote:

An increase in temperature will result in a higher atmospheric pressure - a
higher temperature speeds up the movement of the air molecules, thereby
raising the pressure they exert on the surrounding atmosphere.

Most respondents discounted this statement as incorrect or a flawed
problem. And yet it appears very simply stated in this form as a test
question on the private pilot written:

How do variations in temperature affect the altimeter?

A. Higher temperatures expand the pressure levels and the indicated
altitude is higher than true altitude.
B. Lower temperatures lower the pressure levels and the indicated altitude
is lower than true altitude.
C. Pressure levels are raised on warm days and the indicated altitude is
lower than true altitude.


Answer (C) is Correct - On warm days, the atmospheric pressure levels are
higher than on cold days. Your altimeter will indicate a lower than true
altitude. Remember, "Low to high, clear the sky."

:-/

--
Dallas