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To solve this problem, we need to determine the increase in both the diameter and the length of the cylindrical vessel due to internal pressure. This is a problem of cylindrical pressure vessels, which are subjected to internal pressure, leading to both radial and axial deformations. The formulas for these deformations are based on elasticity theory and use the given material properties and dimensions.

### Given Data:
- Pressure, \( p = 3.6 \, \text{MPa} = 3.6 \times 10^6 \, \text{Pa} \)
- Outer diameter, \( D = 2700 \, \text{mm} = 2.7 \, \text{m} \)
- Wall thickness, \( t = 18 \, \text{mm} = 0.018 \, \text{m} \)
- Length of cylinder, \( L = 5.0 \, \text{m} \)
- Young’s modulus, \( E = 193 \, \text{GPa} = 193 \times 10^9 \, \text{Pa} \)
- Poisson’s ratio, \( \nu = 0.27 \)

### 1. **Increase in Diameter:**
The increase in diameter due to internal pressure is given by the formula for the circumferential (or hoop) strain, which is:

\[
\Delta D = \frac{p D}{2 t E} \left( 1 - \nu \right)
\]

Substituting the given values:

\[
\Delta D = \frac{(3.6 \times 10^6) (2.7)}{2 (0.018) (193 \times 10^9)} (1 - 0.27)
\]

Now, calculating step-by-step:

\[
\Delta D = \frac{(9.72 \times 10^6)}{6.948 \times 10^9} (0.73)
\]

\[
\Delta D = (1.398 \times 10^{-3}) (0.73) = 1.02 \times 10^{-3} \, \text{m} = 1.02 \, \text{mm}
\]

So, the increase in the diameter is **1.02 mm**.

### 2. **Increase in Length:**
The increase in length due to internal pressure is given by the axial strain formula:

\[
\Delta L = \frac{p L}{2 t E} \left( \nu \right)
\]

Substituting the given values:

\[
\Delta L = \frac{(3.6 \times 10^6) (5.0)}{2 (0.018) (193 \times 10^9)} (0.27)
\]

Now, calculating step-by-step:

\[
\Delta L = \frac{18 \times 10^6}{6.948 \times 10^9} (0.27)
\]

\[
\Delta L = (2.59 \times 10^{-3}) (0.27) = 6.993 \times 10^{-4} \, \text{m} = 0.699 \, \text{mm}
\]

So, the increase in the length is **0.699 mm**.

### Summary:
- **Increase in diameter** = 1.02 mm
- **Increase in length** = 0.699 mm

Do you want any further details or clarifications?

Here are 5 related questions:
1. How would these values change if the wall thickness were doubled?
2. What effect would a different Poisson's ratio have on the cylinder's deformations?
3. How would the results differ for a cylinder made from a different material (e.g., aluminum)?
4. What would happen to the deformations if the internal pressure were increased by 50%?
5. How is the increase in diameter related to the increase in length?

**Tip:** The wall thickness plays a significant role in reducing deformation; thicker walls mean less strain under the same pressure.

A closed cylindrical vessel contains a fluid at a pressure of 3.6 MPa. The cylinder, which has an outside diameter of 2700 mm and a wall thickness of 18 mm, is fabricated from stainless steel [E = 193 GPa; ν = 0.27]. Assume L = 5.0 m. Determine the increase in both the diameter and the length of the cylinder.

Learn how to calculate the increase in diameter and length of a cylindrical vessel under internal pressure. Using elasticity theory, we solve for the deformations of a stainless steel vessel with 2700 mm diameter, 18 mm wall thickness, and 5 m length under 3.6 MPa of pressure. Material properties include E = 193 GPa and Poisson's ratio ν = 0.27.

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