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Physics Problem  
User currently offlineSaab340 From United States of America, joined Sep 2001, 320 posts, RR: 2
Posted (11 years 1 month 1 week 5 days 10 hours ago) and read 1979 times:

Came across this problem which I cannot figure out in Jeppesen Sanderson PHYSICS FOR AVIATION.

Here it goes:
"The air pressure and density at a point on the wing of a B 747 flying at an altitude of 2900 m are 71.0 kPa, and 0.919 kg/m^3 respectively. What is the temperature at this point on the wing in degrees Centigrade?"

I would very much appreciate the process taken to figure this problem out shown for understanding. Thank you all very much for your help!

Paul

12 replies: All unread, jump to last
 
User currently offlineTimz From United States of America, joined Sep 1999, 6835 posts, RR: 6
Reply 1, posted (11 years 1 month 1 week 5 days 10 hours ago) and read 1882 times:

Presumably the 747 and the altitude are irrelevant? At a specified pressure and specified density the temperature is determined by them alone?

User currently offlineMITaero From United States of America, joined Jul 2003, 497 posts, RR: 8
Reply 2, posted (11 years 1 month 1 week 5 days 9 hours ago) and read 1856 times:

You just need the pressure and the density. If you assume that the air is ideal, and find the "molar mass" of air (about 29 g/mol, look it up online), you can just use the ideal gas law in terms of density to find temperature.

PV=nRT (where P=pressure, V=vol, n=# of moles, R=const, T=temp)
PV=(g/MM)*RT (where g is mass of gas, MM is molar mass)
P=(g/V)(RT/MM)

since g/V is density (rho),

P=rho*RT/MM

so

P*(MM) = rho*R*T

be careful with units - if you need more help, just ask again.


User currently offlineFlightSimFreak From United States of America, joined Oct 2000, 720 posts, RR: 0
Reply 3, posted (11 years 1 month 1 week 5 days 9 hours ago) and read 1853 times:

It's been all summer since I've been in school, so don't crucify me if I get something wrong, just correct me.

OK, for reference, P is pressure in Pascals, T is temperature in Kelvin, D is density in Kg/m^3 and R is the ideal gas constant, or 287 K Kg/ Pa M^3. The equation I'm going to use is the Ideal gas law solved for T.

T=P/(D*R)

We know P, D, and R... So it's a simple matter now of "plug and chug" as my math teacher used to say... Plug it in to the calculator, and chug out the answer...

71.0 kPa is 71000 Pascals...

T=71000/(.919*287)

or...

269.19 K or -3.96 C



Phew, I hope I didn't mess that one up, that was a doozie


User currently offlineBobrayner From United Kingdom, joined Apr 2003, 2227 posts, RR: 6
Reply 4, posted (11 years 1 month 1 week 5 days 9 hours ago) and read 1852 times:

Not entirely irrelevant  Big grin as altitude increases, molar mass goes down a little, which means the ideal gas law will produce different numbers. At 2900m, air composition is pretty normal.

Anyway. Let's set up some variables and constants.
P = Pressure (Pascals, not atmospheres or mmHg or whatever);
D = Density;
V = Volume;
T = Temperature (Kelvin);
n = Number of moles of gas in a given volume;
R = universal gas constant (8.3144 J per mole degree)
M = Mass of a given volume of gas

Ideal gas law: PV=nRT
Also, V=M/D

So, (P / D).(M / n) = RT. The (M / n) is the value that would slowly change with altitude, or indeed if you were flying over Venus instead of Earth, but for this case we can use the typical molar mass of air, hence 28.92g/mole . R is always 8.3144 in these units.

Therefore, T = (P / D) . (3.478)

Be careful with units, significant figures, &c &c  Smile



Cunning linguist
User currently offlineMITaero From United States of America, joined Jul 2003, 497 posts, RR: 8
Reply 5, posted (11 years 1 month 1 week 5 days 9 hours ago) and read 1838 times:

I know the ideal gas law is an approx. for air, but..

>as altitude increases, molar mass goes down a little

the mass of one mole, or 6.02e23 molecules or whatever it is should not change, no matter the state of the gas.

any two thermodynamic properties of a gas define the thermodynamic state of the gas (see http://web.mit.edu/16.unified/www/FALL/thermodynamics/thermo_2.htm for verification). In this case, density and pressure are enough.


User currently offlineTimz From United States of America, joined Sep 1999, 6835 posts, RR: 6
Reply 6, posted (11 years 1 month 1 week 5 days 8 hours ago) and read 1834 times:

I guess he means the air has a higher percentage of nitrogen at higher altitudes. Which would of course lower the molar mass.

User currently offlineBobrayner From United Kingdom, joined Apr 2003, 2227 posts, RR: 6
Reply 7, posted (11 years 1 month 1 week 5 days 8 hours ago) and read 1812 times:

the mass of one mole, or 6.02e23 molecules or whatever it is should not change, no matter the state of the gas.

Avogadro's number does not change. The molar mass varies with the composition of the gas. This is a bit of an irrelevance, since the atmosphere is well-mixed below 80km (IE anywhere you're likely to find an airliner) so composition changes with altitude are quite small, although not always negligible.

any two thermodynamic properties of a gas define the thermodynamic state of the gas

Presuming you're dealing with the same gas at all times.



Cunning linguist
User currently offlineMITaero From United States of America, joined Jul 2003, 497 posts, RR: 8
Reply 8, posted (11 years 1 month 1 week 5 days 7 hours ago) and read 1801 times:

Hey Bobrayner, you're right about those two points. Hopefully he doesn't have to discuss that in his answer.  Smile

User currently offlineSaab340 From United States of America, joined Sep 2001, 320 posts, RR: 2
Reply 9, posted (11 years 1 month 1 week 5 days 6 hours ago) and read 1763 times:

Thank you all for the input. Can we call the answer -4 c????

Paul


User currently offlinePilotHighFlyer From United States of America, joined Jul 2003, 220 posts, RR: 0
Reply 10, posted (11 years 1 month 1 week 5 days 5 hours ago) and read 1745 times:

This reminds me of chem...wow
No don't round to 4, keep the answer out to hundredth or thousandths for accuracy, especially if you have to use the answer in another problem.

~Robert


User currently offlineBobrayner From United Kingdom, joined Apr 2003, 2227 posts, RR: 6
Reply 11, posted (11 years 1 month 1 week 5 days 5 hours ago) and read 1733 times:

Your answer is as accurate as the least accurate input - in this case, the pressure. Round accordingly?  Smile


Cunning linguist
User currently offlineMITaero From United States of America, joined Jul 2003, 497 posts, RR: 8
Reply 12, posted (11 years 1 month 1 week 5 days 4 hours ago) and read 1715 times:

Yeah, use 3 significant digits. (71.0 kPa and .919 kg/m^3 both have 3 sig figs)

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