Pressure Vessel Calculation

This solution determines the permissible pressure for a steel cylindrical pressure vessel using the given allowable strain.

Given Parameters

• Diameter, $D = 3$ m
• Wall thickness, $t = 15$ mm $= 0.015$ m
• Young's modulus, $E = 210$ GPa $= 210 \times 10^9$ Pa
• Poisson's ratio, $\nu = 0.3$
• Allowable normal strain (hoop strain), $\epsilon_h = 0.00034$

Formulas for Cylindrical Pressure Vessels

For a thin-walled cylindrical pressure vessel, the principal stresses are hoop stress ($\sigma_h$) and longitudinal stress ($\sigma_l$).

• Hoop stress: $\sigma_h = \frac{pD}{2t}$
• Longitudinal stress: $\sigma_l = \frac{pD}{4t}$

The relationship between stress and strain (generalized Hooke's law) for hoop strain ($\epsilon_h$) is:

$\epsilon_h = \frac{1}{E}(\sigma_h - \nu \sigma_l)$

Deriving Permissible Pressure

Substitute the stress formulas into the hoop strain equation:

$\epsilon_h = \frac{1}{E} \left( \frac{pD}{2t} - \nu \frac{pD}{4t} \right)$

Simplify the expression:

$\epsilon_h = \frac{pD}{4Et} (2 - \nu)$

Rearrange the formula to solve for pressure ($p$):

$p = \frac{4Et\epsilon_h}{D(2 - \nu)}$

Calculation

Substitute the known values into the derived formula:

$p = \frac{4 \times (210 \times 10^9 \, \text{Pa}) \times (0.015 \, \text{m}) \times (0.00034)}{(3 \, \text{m}) \times (2 - 0.3)}$

$p = \frac{4 \times 210 \times 10^9 \times 0.015 \times 0.00034}{3 \times 1.7}$

$p = \frac{4,284,000}{5.1}$ Pa

$p \approx 840,000$ Pa

Final Result

Convert the pressure from Pascals (Pa) to kilopascals (kPa):

$p = \frac{840,000}{1000} \, \text{kPa}$

$p = 840.0$ kPa

The permissible pressure is 840.0 kPa.