Resultant Force and CP of Window @ -40C

[Dmuldrock]

Problem States
If the air pressure inside the plane is
atmospheric, what is the resultant force on the
window (assume a rectangle) and where is the
CP if the outside air temperature is -40 C.
Ideal gas law, P = ρRT
Window is 350mm * 450mm

My work
P inside = Atmospheric = 101325Pa
Pc inside = 101325Pa*0.45mm*0.35mm = 15958.69N
Ixc =(b*h^3)/12 = (0.35*0.45^3)/12 = 2.658*10^-3
Cp = Ixc/(A*yc) = 2.658*10^-3/((0.35*0.45)*(0.45/2))

P outside = ρRT
what ρ should I use? R = 287 and T = -40+273 = 233
I'm a bit lost as to how to tackle pressure for outside as there is no height.

 

[LightHero]

AnswerThe pressure outside the plane is calculated using the ideal gas law, P = ρRT. Since the temperature outside the plane is -40 °C, the absolute temperature (T) is 233 K. The density of air at this temperature can be looked up from a density chart, and the pressure outside the plane is calculated as P = ρRT = (1.225 kg/m³)(287 J/kgK)(233 K) = 84435 Pa.The resultant force on the window is the difference between the pressure inside and outside the plane, i.e. F = Pinside - Poutside = 101325 - 84435 = 16890 N. The centroid of the rectangle window is located at (0.35/2, 0.45/2) = (0.175, 0.225). Therefore, the CP of the window is located at (0.175, 0.225).