A aviation & planes forum. AviationBanter

If this is your first visit, be sure to check out the FAQ by clicking the link above. You may have to register before you can post: click the register link above to proceed. To start viewing messages, select the forum that you want to visit from the selection below.

Go Back   Home » AviationBanter forum » rec.aviation newsgroups » Piloting
Site Map Home Register Authors List Search Today's Posts Mark Forums Read Web Partners

Physics Quiz Question



 
 
Thread Tools Display Modes
  #11  
Old August 7th 07, 07:31 PM posted to rec.aviation.piloting
Jim Logajan
external usenet poster
 
Posts: 1,958
Default 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!
  #12  
Old August 7th 07, 08:32 PM posted to rec.aviation.piloting
Dallas
external usenet poster
 
Posts: 541
Default Physics Quiz Question

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
  #13  
Old August 7th 07, 09:19 PM posted to rec.aviation.piloting
Doug Semler
external usenet poster
 
Posts: 175
Default Physics Quiz Question

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.)


  #14  
Old August 8th 07, 01:43 AM posted to rec.aviation.piloting
Jim Logajan
external usenet poster
 
Posts: 1,958
Default Physics Quiz Question

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.
  #15  
Old August 8th 07, 04:48 AM posted to rec.aviation.piloting
Jim Logajan
external usenet poster
 
Posts: 1,958
Default Physics Quiz Question

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?
  #16  
Old August 8th 07, 02:28 PM posted to rec.aviation.piloting
Tony
external usenet poster
 
Posts: 312
Default Physics Quiz Question

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



  #17  
Old August 8th 07, 02:57 PM posted to rec.aviation.piloting
Doug Semler
external usenet poster
 
Posts: 175
Default Physics Quiz Question

On Aug 8, 12:25 am, Clark wrote:
Jim Logajan wrote :

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!"


Nit picking my nitpic? Hmmm, is anyone keeping score? :-) You are correct of
course but maybe we can take a closer look. Lets-see-here-now, we have
density in moles/L^3. Is that a nuclear density? Whatever it is it must be a
Chem E thing... I've seen flow rates in moles/hour, but this is new. Ya never
can trust those Chem E types anyways.


Nope, ya just can't. What the hell is a cubic liter anyway? I've
heard of cubic meters, but never cubic liters g. And just when you
think your all smart, they start throwing moles at you. Sheesh.

Seriously, though, I too have a tendency to skip steps/simplify things
when doing this sort of thing. The target audience here is not a
bunch of physicists, it's a bunch of pilots. While some of them might
care/want to know how it works in a general sense, it seems silly to
start bringing out all the math involved, and easier to make some wild
ass assumptions that are only valid in a theoretical model. The silly
theoretical model can help explain what's going on, and may help
someone understand that temperature, pressure, and density are all
related to each other. I try to indicate when I am hand waving a
"that difference is close enough to zero to not matter", but I
sometime forget. However, problems like this is like the high school
physics problems that "ignore friction". Understanding that friction
DOES play a part of an overall system is important; the principles
being taught are MORE important. (little delta/epsilon are implied).

I admit I am not a physicist, I *am* a mathemetician, and some upper
division calculus type classes are very closly linked with physics;
hell a whole class was devoted to the calculus of thermodynamics.
Most of the stuff I am writing is coming from memory 10 years old or
more, so it probably does have some holes in it, if I am wrong i am
sure to be corrected (as I was in r.a.s; i fell into the trap of not
fully explaining my meaning sigh). But that doesn't change the fact
that I don't think anyone in *this* group wants me to dust off the 'ol
applied calculus book and start quoting dry formulae about the
thermodynamics of an open system subject to uneven energy transfers,
since there are plenty of books out there that already do it.

  #18  
Old August 8th 07, 09:20 PM posted to rec.aviation.piloting
Dallas
external usenet poster
 
Posts: 541
Default Physics Quiz Question

On Wed, 08 Aug 2007 06:57:54 -0700, Doug Semler wrote:

But that doesn't change the fact
that I don't think anyone in *this* group wants me to dust off the 'ol
applied calculus book and start quoting dry formulae about the
thermodynamics of an open system subject to uneven energy transfers


Aw... come on... please?

:-)

Thanks to all the respondents

I've actually read everything here and at RAS, but I can see it will take
more than one read to wade through all this.

If we're taking a vote, it looks like the grossly oversimplified answer
winning answer is: b) False

--
Dallas
  #19  
Old August 8th 07, 09:44 PM posted to rec.aviation.piloting
d&tm[_2_]
external usenet poster
 
Posts: 4
Default Physics Quiz Question


"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


  #20  
Old August 9th 07, 04:46 AM posted to rec.aviation.piloting
Jim Logajan
external usenet poster
 
Posts: 1,958
Default Physics Quiz Question

Clark wrote:
Nit picking my nitpic? Hmmm, is anyone keeping score? :-) You are
correct of course but maybe we can take a closer look.
Lets-see-here-now, we have density in moles/L^3. Is that a nuclear
density?


The way I see it, if the units of the denominator is length cubed, it's a
density of some sort! That's all a man can ask for - and expect - some
days! ;-)

Whatever it is it must be a Chem E thing... I've seen flow
rates in moles/hour, but this is new. Ya never can trust those Chem E
types anyways.


Add physics grads to the types you can't trust. ;-) You do realize that
physicists consider 1, pi, and 4.9 to all equal 1, while 11, 33, and 49 are
all equal to 10? They love to round to the nearest power of 10, so they can
reduce all multiplcations and divisions to additions and subtractions of
powers of ten. We never learned math beyond adding and subtracting. :-)

I figured you chose to skip a few steps to get to the answer sooner
and avoid losing the audience. It's a tough call on the minor
technical details. I probably err in the other direction too often. I
figured this group has some folks that wouldn't mind seeing an
expanded explanation in a separate post.


I was sure people's eyes had glazed over before they finished reading.
What's a few lazy mistakes among friends, eh?

While I'd love to expand on my explanation, as I write this I believe a new
episode of Mythbusters comes on in ~15 minutes and I know I wont be able to
compose an accurate post in that amount of time. And my short attention
span means I'll never get round to it later. (Did I mention I did better in
QM than statistical and thermo physics and that I'm really rusty in both
now? Or as Scotty once said: "I canna change the laws of physics Captain! I
need 30 minutes!")
 




Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

vB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Forum Jump

Similar Threads
Thread Thread Starter Forum Replies Last Post
Sailplane quiz Mike[_8_] Soaring 5 April 3rd 07 03:45 PM
Aircraft guess quiz [email protected] Piloting 2 May 6th 06 10:03 AM
Pop Quiz Flyingmonk Piloting 19 April 26th 06 01:54 AM
Physics question Rich S. Home Built 62 September 14th 05 02:05 PM
Fun Quiz John Clonts Owning 2 August 14th 05 04:35 AM


All times are GMT +1. The time now is 06:13 PM.


Powered by vBulletin® Version 3.6.4
Copyright ©2000 - 2024, Jelsoft Enterprises Ltd.
Copyright ©2004-2024 AviationBanter.
The comments are property of their posters.