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#11
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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!") |
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