A futuristic Aircraft manufacturer has come up with the idea to build a supersonic passenger jet, which is supposed to perform cruise flight at 24 kilometres altitude. For passenger comfort, the cabin pressure should remain at least 72 percent of the sea level value. To give the aircraft an aerodynamic shape its fuselage is rather long and slender, meaning the fuselage diameter is 'just' 4.5 metres.
The engineers have asked you, now experts in the field of pressure cabins and fatigue, to determine the required skin thickness to deal with the circular stress (which may not exceed 75MPa) in the fuselage skin. To be on the safe side, they ask you to take into account a safety factor of 1.5.
What is the minimum required skin thickness for this aircraft (in milimeters)?
To calculate the minimum required skin thickness for the aircraft's fuselage, we can use the formula for stress in a thin-walled pressure vessel:
Stress = (PD) / (2t)
Where:
Stress = Maximum allowable stress (75 MPa)
P = Pressure difference (sea level pressure - cabin pressure)
D = Diameter of the fuselage
t = Skin thickness
First, we need to calculate the pressure difference. Since the cabin pressure should be at least 72% of the sea level value, we can determine the sea level pressure by dividing the desired cabin pressure by 0.72:
Sea level pressure = Cabin pressure / 0.72
Next, we can calculate the pressure difference:
Pressure difference = Sea level pressure - Cabin pressure
Finally, we rearrange the formula to solve for the skin thickness (t):
t = (PD) / (2 * Stress)
Let's plug in the given values to calculate the skin thickness.
Diameter of the fuselage (D) = 4.5 meters
Desired cabin pressure = 72% of sea level pressure
Sea level pressure = Desired cabin pressure / 0.72
Pressure difference (P) = Sea level pressure - Cabin pressure
Maximum allowable stress (Stress) = 75 MPa
Safety factor = 1.5
Skin thickness (t) = (PD) / (2 * Stress)
Calculating the above steps will give us the minimum required skin thickness for the aircraft.
First, we need to determine the maximum cabin pressure and the corresponding differential pressure that the fuselage skin will need to withstand.
From the given information, we know that the cabin pressure should remain at least 72% of the sea level value. At a cruise altitude of 24 km, the atmospheric pressure is approximately 22.4% of sea level pressure. Therefore, the cabin pressure should be at least 72% of 22.4% of sea level pressure, which is:
0.72 x 0.224 x 101325 Pa = 16424 Pa
To this cabin pressure, we need to add the external pressure at cruise altitude, which is:
101325 x 0.224 x exp(-0.00012 x 24000) = 11820 Pa
Therefore, the maximum differential pressure that the fuselage skin will need to withstand is:
16424 - 11820 = 4604 Pa
Next, we can use the formula for hoop stress in a thin-walled pressure vessel to calculate the required skin thickness:
hoop stress = P x D / (2 x t)
where P is the pressure, D is the diameter of the vessel, and t is the skin thickness.
Rearranging the formula, we get:
t = P x D / (2 x hoop stress)
Taking into account the safety factor of 1.5, we can use a maximum hoop stress of:
75 / 1.5 = 50 MPa
Substituting the values, we get:
t = (16424 + 11820) x 4.5 / (2 x 50 x 10^6) x 1.5
t = 0.0132 m = 13.2 mm
Therefore, the minimum required skin thickness for this aircraft is 13.2 mm.