An airplane's emergency escape window is a rectangle measuring 95cm by 56cm.?

Question

If the cabin pressure is 0.78atm and the external pressure is 0.28atm , what force would be required to pull the window inward?

Answer 1

Pcab = 0.78 atm = 79,033.5 Pa = 79,033.5 N/m² = 79.0335 kPa

Pext = 0.28 atm = 28,371 Pa = 28,371 N/m² = 28.371 kPa



Awin = 95 cm x 56 cm = 5,320 cm² = 0.523 m²



Differential pressure = 79,033.5 Pa - 28,371 Pa = 50,662.5 Pa = 50,662.5 N/m² = 50.6625 kPa



F = force

P = pressure

A = area



P = F/A

50,662.5 N/m² = F/0.523 m²



F = (50,662.5 N/m²)(0.523 m²)

F = 26,496.5 N ≈ 26.5 kN ANSWER

Answer 2

Wouldn't you need to know the thickness of the window, the material it is made of, the type of gasket, the type of frame, etc



In other words, there are many other variables needed to answer this question.

Answer 3

0.78 - 0.28 = pressure of 0.5 atm. on window.

Atmospheric pressure @ STP = 101,325Pa.

Area of window = (0.95 x 0.56m) = 0.532m^2.

(101,325 x 0.5) = 50,662.5Pa.

(50,662.5 x 0.532) = 26,952.45N., or 26.95245kN., force required.