a) A pipe of DN 250 mm nominal diameter, OD = 273.1 mm, yield strength 250 MPa, factor of safety 2.5, pressure 12 barg. Find the minimum thickness of the pipe.
Answer (1)
To find the minimum thickness of a pipe given the specified conditions, we can use the formula for the minimum wall thickness required for safety under internal pressure. The formula is derived from the hoop stress equation:
t = 2 × S × F + P P × D
Where:
t is the minimum wall thickness.
P is the internal pressure. Since the pressure provided is 12 barg, we need to convert it to Pascals (1 bar = 100,000 Pascals): P = 12 × 100 , 000 = 1 , 200 , 000 Pa
D is the outside diameter of the pipe: 273.1 mm = 0.2731 m.
S is the yield strength of the material: 250 MPa = 250,000,000 Pa.
F is the factor of safety: 2.5.
Substituting these values into the formula:
t = 2 × 250 , 000 , 000 × 2.5 + 1 , 200 , 000 1 , 200 , 000 × 0.2731
Calculating the values:
t = 1 , 250 , 600 , 000 + 1 , 200 , 000 327 , 720
t = 1 , 251 , 800 , 000 327 , 720
t ≈ 0.0002617 m
Converting meters to millimeters (1 m = 1000 mm):
t ≈ 0.2617 mm
Therefore, the minimum required thickness for the pipe is approximately 0.2617 mm.
This calculation ensures the pipe can withstand the internal pressure with the given safety factor, accounting for the material's yield strength.
