In the lecture, we discussed the case of an isothermal atmosphere where the temperature is assumed to be constant. In reality, however, the temperature in the Earth's atmosphere is not uniform and can vary strongly and in a non-linear way, especially at high altitude. To a good approximation, the temperature T drops almost linearly with altitude up to 11 km above sea level, at a constant rate:
dTdz=−α forz≤11 km
where α=6.5 K/km (Kelvin per km) and z is the height above the sea level. The temperature stays then approximately constant between 11 km and 20 km above sea level.
Assume a temperature of 20 ∘C and a pressure of 1 atm at sea level (1 atm = 1.01325 ×105 N/m^2). Furthermore, take the molecular weight of the air to be (approximately) 29 g/mol. The universal gas constant is R=8.314 JK−1mol−1 and the acceleration due to gravity is g=10 m/s2 (independent of altitude). Assume that air can be treated as an ideal gas.
(a) Under the assumptions above, calculate the atmospheric pressure p (in atm) at z= 10 km above sea level for the case of a linear temperature drop.
p=
(b) The cruising altitude of a commercial aircraft is about 33'000 ft (or 10 km). Assume that the cabin is pressurized to 0.8 atm at cruising altitude. What is the minimal force Fmin (in Newton) per square meter that the walls have to sustain for the cabin not to burst? Use the atmospheric pressure found in (a).
Fmin=
(c) We close a plastic bottle full of air inside the cabin when the aircraft is at cruising altitude of z= 10 km. The volume of the bottle is V1, the pressure and temperature inside the cabin are 0.8 atm and T1=27 ∘C, respectively. Assume that at sea level the atmospheric pressure is 1 atm, and the temperature is decreased by 20 Kelvin with respect to the cabin's temperature.
What is the magnitude of the percentage change in volume of the air inside the bottle when it is brought to sea level? (Enter the magnitude of the percentage change in volume in
∣∣∣ΔVV1∣∣∣×100=
This is an easy MITX 8.01 final exam problem, it is a shame that you have no honor and found it necessary to cheat
First Solve for the function T(z)
T = (-alpha)*z + T(initial)
T = -6.5z+293
Now plug that in the hydrostatic pressure equation
dP/P = (-M*g/RT)dz
integrate that
and you will get
P(final) = P(initial)*(1-(alpha*z/293))^(Ma*g/(R*alpha))
Part B is easy, get the answer from part a then subtract, don't forget to convert to Newtons per m^2 or pascal cause the given is in atm.
part C. use the following relationship:
P1V1/T1 = P2V2/T2 and you are good
have you solve the problem?
In the equation of P(final), what does ''a'' mean???
Because integrating the first relation, I obtained:
P(final) = P(initial)*e^(-Mgz/RT)
and what is the value of M ???
Please some help with this exercise....
the value of m isn't the molecular weight of the gas? in this case 29g/mol?
Somebody with the steps, I can't understand this one!
Anonymous, I can't solve the problem.....
Please some hint.
some help....anybody????
Please help with this exercise, is the only one that I need......please.
I obtained:
a) 0.86 atm
b) 5503.4 Pa
c) 7%
But all are wrong and I don't know why!!!
And I have only one submission more.
Please help!!!
c) 24 percent
fred
for temp of 5 degrees celsius
b) 55465.76
c) 24 percent
Thanks a lot abc, and what about a) and b)
I did it again and now I obtained
a)0.83 atm
b)3061.5 Pa
But it's my last chance and I am not sure......
P(final) =P(initial)*(1-(alpha*z/293))^(M*g/(R*alpha))
green tick
but what is the answer for b and in this problem
a)po=0.263
b and c
some help....anybody????
c) 25.33 percent
If you have a)
you only need to rest from b)
..... but I am continue suffering with a)
did you replace all the values without converting to other unid system???
i need b and c for the problem above (20oC etc) as it is i have only one left
i found a ) p=0.263
give your values and i ll tell you the answer
fred to the same system but be careful with the units, give me temperatutr and i ll tell you a)
it is midnight here in Greece and i'm feeling realy tired ....last call for help for b and c as i have only one try left..
for a temperature of 20c, a)0.260269. b)54688.24358 and c) 25.33%
and vivipop rescues the day (or better night!)
thx a lot vivipop!!!
fred
h=10km=10000m and temp.T= 273+20=293, M=0.029 alpha=0.0065 and M=0.029
and the formula for a) is
P(final) = P(initial)*(1-(alpha*z/293))^(Ma*g/(R*alpha))
glad i could help :)
P(final) = P(initial)*(1-(alpha*z/293))^(M*g/(R*alpha))
Greco, thanks a lot!!
I obtained this results for the problem above.
a) 0.26
b) 54715.5
c) I am not sure