Mechanics - cylindrical steel pressure vessel

[designmind93]

mechanics -- cylindrical steel pressure vessel

Homework Statement

A thin-walled, closed-end, cylindrical steel pressure vessel, internal diameter 500 mm and wall-thickness 10 mm, has an internal volume 0.5 cubic metres. Find the additional volume of a compressible fluid that must be pumped in at a pressure of 20 bar and fill the vessel completely. Take for steel: E = 210 GPa, Nu = 0.25 and for the fluid K = 2500 MPa. If the flat end plates are replaced by two thin-walled hemi-spheres of similar internal diameter, what should be the correct ratio of thickness between the sphere and cylinder? How much additional volume of fluid (due to the strain expansion of the vessel) is required to fill the vessel completely at the pressure of 20 bar?

Homework Equations

3pd/4tE (1- Nu) + pv/K + 638.1 x 10^-6

The Attempt at a Solution

Additional volume = 638.1 x 10^-6 m^3

ratio = 3/7

thickness of sphere = 4.29 mm

How do i go about solving the bit in bold?

My lecturer gave me the formula, but I cannot make the answer equal to 731.4 x 10^-6 m^3 which he says it will...
if anyone can solve this and show how they did it i would very much appreciate it!

 

[KataKoniK]

I would approach this problem by first understanding the concept behind the formula given by your lecturer. This formula represents the change in volume of a pressure vessel due to external pressure and material properties. Let's break it down:

3pd/4tE (1- Nu) is the change in volume due to external pressure, where p is the pressure, d is the internal diameter of the vessel, t is the wall thickness, E is the Young's modulus of the material, and Nu is the Poisson's ratio.

pv/K represents the change in volume due to the compressibility of the fluid, where v is the volume and K is the bulk modulus of the fluid.

638.1 x 10^-6 is a constant that takes into account the units and conversion factors.

Now, let's plug in the given values:

p = 20 bar = 20 x 10^5 Pa
d = 500 mm = 0.5 m
t = 10 mm = 0.01 m
E = 210 GPa = 210 x 10^9 Pa
Nu = 0.25
v = 0.5 m^3
K = 2500 MPa = 2500 x 10^6 Pa

Substituting these values into the formula, we get:

3(20 x 10^5)(0.5)/(4(0.01)(210 x 10^9))(1-0.25) + (0.5)(20 x 10^5)/(2500 x 10^6) + 638.1 x 10^-6 = 731.4 x 10^-6 m^3

Therefore, the additional volume of fluid required to fill the vessel completely at a pressure of 20 bar is 731.4 x 10^-6 m^3.

For the second part of the problem, we need to find the ratio of thickness between the sphere and cylinder. This can be done by equating the change in volume due to external pressure for both the cylinder and the sphere. Let's call the ratio of thickness as x:

3(20 x 10^5)(0.5)/(4(0.01)(210 x 10^9))(1-0.25) = 3(20 x 10^5)(0.5)/(4(x)(210 x 10^9))(1-0.