Saab340 From United States of America, joined Sep 2001, 320 posts, RR: 2 Posted (12 years 9 months 3 weeks 6 days 2 hours ago) and read 2529 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!

Timz From United States of America, joined Sep 1999, 7231 posts, RR: 7
Reply 1, posted (12 years 9 months 3 weeks 6 days 1 hour ago) and read 2432 times:

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

MITaero From United States of America, joined Jul 2003, 497 posts, RR: 7
Reply 2, posted (12 years 9 months 3 weeks 6 days 1 hour ago) and read 2406 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.

FlightSimFreak From United States of America, joined Oct 2000, 720 posts, RR: 0
Reply 3, posted (12 years 9 months 3 weeks 6 days 1 hour ago) and read 2403 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

Bobrayner From United Kingdom, joined Apr 2003, 2227 posts, RR: 6
Reply 4, posted (12 years 9 months 3 weeks 6 days ago) and read 2402 times:

Not entirely irrelevant 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.

MITaero From United States of America, joined Jul 2003, 497 posts, RR: 7
Reply 5, posted (12 years 9 months 3 weeks 6 days ago) and read 2388 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.

Bobrayner From United Kingdom, joined Apr 2003, 2227 posts, RR: 6
Reply 7, posted (12 years 9 months 3 weeks 5 days 23 hours ago) and read 2362 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.

MITaero From United States of America, joined Jul 2003, 497 posts, RR: 7
Reply 8, posted (12 years 9 months 3 weeks 5 days 23 hours ago) and read 2351 times:

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

Saab340 From United States of America, joined Sep 2001, 320 posts, RR: 2
Reply 9, posted (12 years 9 months 3 weeks 5 days 21 hours ago) and read 2313 times:

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

PilotHighFlyer From Canada, joined Jul 2003, 220 posts, RR: 0
Reply 10, posted (12 years 9 months 3 weeks 5 days 21 hours ago) and read 2295 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.

Bobrayner From United Kingdom, joined Apr 2003, 2227 posts, RR: 6
Reply 11, posted (12 years 9 months 3 weeks 5 days 21 hours ago) and read 2283 times:

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

MITaero From United States of America, joined Jul 2003, 497 posts, RR: 7
Reply 12, posted (12 years 9 months 3 weeks 5 days 19 hours ago) and read 2265 times:

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