Final answer:
To achieve a safety factor of 5, the wall thickness of the cylindrical pressure vessel must be increased by a factor of approximately 7.77.
Step-by-step explanation:
To determine by what factor the wall thickness needs to be multiplied to achieve a safety factor of 5 for the cylindrical pressure vessel, we can use the formula for the hoop stress induced by an internal pressure in a thin-walled cylinder:
σ = P * r / t where σ is the stress, P is the internal pressure, r is the radius of the cylinder, and t is the wall thickness.
Given that the cylinder's yield strength is 200 MPa, we want the working stress to be 1/5th of this value (200 MPa / 5 = 40 MPa) for our safety factor of 5.
First, convert the internal pressure from atm to Pa:
P = 35 atm * 101,325 Pa/atm
= 3,546,375 Pa
The current stress σ can be calculated:
σ = (3,546,375 Pa * 0.14 m) / 0.0016 m
= 310,559,375 Pa (or 310.56 MPa)
To achieve a stress of 40 MPa, the thickness must be:
t_needed = P * r / σ_desired = (3,546,375 Pa * 0.14 m) / 40,000,000 Pa
≈ 0.01243 m or 12.43 mm
Therefore, the factor by which we need to multiply the original thickness (1.6 mm) to get to 12.43 mm is:
Factor = t_needed / t_original = 12.43 mm / 1.6 mm
= 7.77
We must increase the thickness by a factor of approximately 7.77 to achieve a safety factor of 5.