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=
36 answers
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
Because integrating the first relation, I obtained:
P(final) = P(initial)*e^(-Mgz/RT)
and what is the value of M ???
Somebody with the steps, I can't understand this one!
Please some hint.
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!!!
for temp of 5 degrees celsius
b) 55465.76
c) 24 percent
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......
green tick
a)po=0.263
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 found a ) p=0.263
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))
I obtained this results for the problem above.
a) 0.26
b) 54715.5
c) I am not sure
0 answers