What force must the seals of a passenger door resist at 10000m altitude?

[richter1966@hotmail.]

a jet is kept at 70% of atmospheric pressure. at an altitude of 10000m pressure is 1/3 normal atmospheric pressure. what force must be resisted by seals of a passenger door dimensions 1m by 2m.
atmospheric pressure corresponds with 1.0 * 10^5Nm^-2

 

[Astronuc]

Force = Pressure x Area

or conversely Pressure = Force / area

 

[quicknote]

Based on the given information, we can calculate the pressure at an altitude of 10000m to be 1/3 * (1.0 * 10^5 Nm^-2) = 3.33 * 10^4 Nm^-2. This means that the pressure inside the jet at 70% of atmospheric pressure is 0.7 * 3.33 * 10^4 Nm^-2 = 2.33 * 10^4 Nm^-2.

To find the force that must be resisted by the seals of the passenger door, we can use the formula F = P * A, where F is the force, P is the pressure, and A is the area of the door. Plugging in the values, we get F = (2.33 * 10^4 Nm^-2) * (1m * 2m) = 4.66 * 10^4 N.

Therefore, the seals of the passenger door must be able to resist a force of 4.66 * 10^4 N to maintain the pressure inside the jet at 70% of atmospheric pressure. It is important for the seals to be strong and secure to prevent any loss of pressure and to ensure the safety and comfort of the passengers onboard.