Force on an airplane window, given altitude and area of the window.

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SUMMARY

The force exerted on a Boeing 747 window at cruising altitude can be calculated using the pressure difference between cabin pressure and external atmospheric pressure. The cabin pressure is maintained at the equivalent of 2,000 m, while the cruising altitude is 11,500 m. The relevant formula is ΔF = ΔP * A, where A is the window area of approximately 0.085 m². The pressure at altitude can be derived using the formula P = P_o * exp(-Mgh/RT), accounting for the non-constant density of air with altitude.

PREREQUISITES
  • Understanding of atmospheric pressure and its variation with altitude
  • Familiarity with the ideal gas law and isothermal atmosphere concepts
  • Basic knowledge of pressure difference calculations
  • Ability to manipulate exponential equations in physics
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  • Learn the derivation of the isothermal atmosphere formula for pressure calculations
  • Study the effects of altitude on air density and pressure
  • Explore the implications of cabin pressure maintenance in commercial aviation
  • Investigate the physics of forces acting on aircraft windows during flight
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Aerospace engineers, physics students, and anyone interested in the mechanics of aircraft design and cabin pressure management will benefit from this discussion.

monnapomona
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Homework Statement


The modern Boeing 747 passenger jet airplane cruises at a speed up to 920 km/hr, at an altitude of 11,500 m. The cabin pressure is maintained equivalent to the atmospheric pressure at an altitude of 2,000 m. This is called the "cabin altitude".

I couldn't find information on the area of the passenger windows in the 747, so let's assume they're about 0.085 m2.
If the airplane were stationary at its cruising altitude, what would be the force on the window?

Air density = 1.29 kg/m^3

Homework Equations


ΔF = ΔP*A

The Attempt at a Solution


Needed to find the pressure difference so I tried using this formula to find pressures at the given altitudes:
P2 = P1 + ρgh
But it gave me an unrealistic value at height 11000 m (since pressure decreases with increasing height). I'm not really sure which formula to use to find the pressure at the given altitudes. :s
 
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monnapomona said:

Homework Statement


The modern Boeing 747 passenger jet airplane cruises at a speed up to 920 km/hr, at an altitude of 11,500 m. The cabin pressure is maintained equivalent to the atmospheric pressure at an altitude of 2,000 m. This is called the "cabin altitude".

I couldn't find information on the area of the passenger windows in the 747, so let's assume they're about 0.085 m2.
If the airplane were stationary at its cruising altitude, what would be the force on the window?

Air density = 1.29 kg/m^3

Homework Equations


ΔF = ΔP*A

The Attempt at a Solution


Needed to find the pressure difference so I tried using this formula to find pressures at the given altitudes:
P2 = P1 + ρgh
But it gave me an unrealistic value at height 11000 m (since pressure decreases with increasing height). I'm not really sure which formula to use to find the pressure at the given altitudes. :s

Density of air won't be constant. Air thins as you go up.
 
For clarity, I would like to point out that, in the formula P2 = P1 + ρgh, h in this equation it is not really the height of the point whose pressure you want to know - e.g., at sea level it is not zero, h here is the height of the column of air above that point, which is smaller for higher altitudes (since there is less air above the point). I am under the impression that you got that mixed up, sorry if I am wrong.

Besides that, the formula you mention, P2 = P1 + ρgh, is only valid when the density of the fluid is constant. As stated above, density of air is not a constant (i.e. it is not the same for the entire atmosphere, it changes with height. Air thins as you go up, as stated above), it is also a function of height

Neglecting temperature change in the atmosphere and assuming air as an ideal gas, the formula for the pressure is

P = P_o\exp{(-\dfrac{Mgh}{RT})},

where Po is atmospheric pressure at sea level (101325 Pa), M is the molar mass of dry air (0.0289 kg/mol), R the universal gas constant (8.13 J/molK), T sea level standard temperature (around 290K) and h the height. (in this formula, h is really the height above the ground)

If you are interested in the derivation of that formula, just google up "isothermal atmosphere", it is pretty straightforward :)
 
Last edited:

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