n 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 15 ∘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=
9 answers
T = 288 - 6.5 z for z <11 km
when z = 11 km, T = 288-6.5(11) = 216.5 K
T remains at 216.5 K until z = 20 km
At z = 10 km
T = 288 - 65 = 223 K
The column of air above ground 1 meter in cross section weighs 1.01235 *10^5 N at sea level (given)
To get the weight of the column above z = 10 km we must either find the mass between z = 0 and z = 10 or from z = 10 to z = infinity.
Let's try from 0 to 10 km
that means we first need n/V in PV = nRT
n/V = P/RT
the weight of a cubic meter is then
W/V = n*.029 kg/mol*10 m/s^2 = .29n Newtons
Now at sea level how much does a cubic meter weigh ?
P = 1.01315*10^5
V = 1 m^3
T = 288 K
R = 8.314
so
n = PV/RT = 1.01315*10^5 /(8.314*288)
= 4.23 * 101 = 42.3 moles/m^3
W = .29 n = 12.27 Newtons
so my pressure decreases by 12.27 N/m^2 when I go up one meter
Of course I could repeat that process for ever meter up from ground to 10,000 meters but that would be tiresome.
That should get you started.
( We better start doing this in meters, not km of course )
dP/P = [.29/(RT)] dh
dP/dh = -.29 n = -.29 P /(RT)
( We better start doing this in meters, not km of course )
dP/P = -[.29/(RT)] dh
Unfortunately T is a function of h
if h is meters h = 10^3 z with z in km
T = 288 - 6.5 * 10^-3 h
so
dP/P = -[.29/(R(288-.0065h)] dh
integrate that
or
ln p + c = ln -(.0065 h -288)^(.29/.0065R)
e^p = const e^ -(.0065 h -288)^(.29/.0065R)
p = constant(288-.0065h)^.29/.0065R
put in initial conditions of p and T
p = 1.01325 *10^5 (1 -.0065h/288)^(.29/.0065R)
36 answers