Aircraft’s Cabin Door Commercial jet aircraft have pressurized cabins because they travel at high altitude where the atmosphere is thin.At the cruising altitude of 30,000 ft,the outside atmospheric pressure is only about30% of the sea-level value.The cabin is pressurized to the equivalent of

Question

Aircraft’s Cabin Door Commercial jet aircraft have pressurized cabins because they travel at high altitude where the atmosphere is thin. At the cruising altitude of 30,000 ft, the outside atmospheric pressure is only about 30% of the sea-level value. The cabin is pressurized to the equivalent of a mountaintop where the air pressure is about 70% that at sea level. Estimate the force that is applied to the door of the aircraft’s main cabin by this pressure imbalance. Treat the following information as “given” when making the order-of-magnitude estimate: (1) The air pressure at sea level is 14.7 psi, and (2) the force F on the door is the product of the door’s area A and the pressure difference Δp according to the expression [latex] F=A\Delta \rho [/latex]

Approach
We are tasked with approximating the amount of force exerted on the interior of an aircraft door during flight. The pressure information is given, but we have to make some assumptions about the door and cabin surroundings. We assume that:
• The size of the aircraft’s door is approximately 6 × 3 ft, or 18 [latex]ft^2[/latex]
• We can neglect the fact that the door is not precisely rectangular
• We can neglect the fact that the door is curved to blend with the shape of the aircraft’s fuselage
• We do not have to account for small changes in pressure due to the movement of passengers inside the cabin during flight

We will first calculate the pressure difference and then calculate the area of the door to find the total force.

Accepted Answer

The net pressure acting on the door is the difference between air pressures
inside and outside the aircraft.

[latex]\Delta ho = (0.7 – 0.3)(14.7 psi)= 5.88 psi[/latex]

Because [latex]Delta ho [/latex] has the units of pounds per square inch (Table 3.5),

Table 3.5 Certain Derived Units in the USCS

Quantity	Derived Unit	Abbreviation	Definition
Length	mil	mil	1 mil = 0.001 in.
	inch	in.	1 in. = 0.0833 ft
	mile	mi	1 mi = 5280 ft
Volume	gallon	gal	[latex] 1 gal = 0.1337 ft^3 [/latex]
Mass	slug	slug	[latex]1 slug = 1 (lb . S^2)/ft[/latex]
Mass	pound-mass	lbm	[latex] 1 lbm = 3.1081 imes 10^{-2}(lb s^{2})/ft [/latex]
Force	ounce	oz	1 oz = 0.0625 lb
Force	ton	ton	1 ton = 2000 lb
Torque, or moment of a force	foot-pound	ft · lb	—
Pressure or stress	[latex]pound/inch^2[/latex]	psi	[latex]1 psi = 1 lb/in^2[/latex]
Energy, work, or heat	foot-pound	ft · lb	—
Energy, work, or heat	British thermal unit	Btu	1 Btu = 778.2 ft . lb
Power	horsepower	hp	1 hp = 550 (ft . lb)/s
Temperature	degree Fahrenheit	[latex]°F[/latex]	[latex]°F = °R-459.67[/latex]
Although a change in temperature of [latex] 1°[/latex] Rankine also equals a change of 1 [latex]°F[/latex], numerical , numerical values are converted using the formula.

in order for the equation[latex]F=A\Delta ho [/latex] to be dimensionally consistent, the area must be converted to the units of square inches:

[latex]A=(18 ft^2)\left(12\frac{in.}{ft} ight)^2 [/latex]

[latex]= 2592(\cancel{ft^2})(\frac{in}{\cancel{ft^2}} ) =2592 in^2[/latex]

The total force acting on the door becomes

[latex]F = (2592 in^2)(5.88 psi) \longleftarrow \left[F=A\Delta ho ight] [/latex]

[latex] = 15,240(\cancel{in^2})(\frac{lb}{\cancel{in^2}} )=15,240 lb [/latex]

Discussion
First, we evaluate the order of magnitude of the solution. The forces created by pressure imbalances can be quite large when they act over large surfaces, even for seemingly small pressures. Therefore, our force seems quite reasonable. Second, our assumptions greatly simplified the problem. But since we need to estimate only the force, these assumptions are realistic. Third, recognizing the uncertainty in our estimate of the door’s area and the actual value of the pressure differential, we conclude that the pressure is in the range of 10,000 20,000 lb.

approximately 10,000 – 20,000 lb

Question 3.8: Aircraft’s Cabin Door Commercial jet aircraft have pressuriz...

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