Question

A ship of 5000 tonnes displacement has a rectangular tank 6 m long and 10 m wide, half-filled with oil of relative density 0.8. The virtual reduction in the transverse metacentric height of the ship due to the free surface effect of the oil in the tank is ……….. cm.
 

Hint:

For problems involving the free surface effect: 1. Use the formula \( \Delta GM = \frac{\rho_{\text{fluid}} \cdot A \cdot h}{\Delta} \) carefully.
2. Ensure consistent units (e.g., displacement in kg, length in meters).
3. The free surface area (\( A \)) and the relative density (\( \rho_{\text{fluid}} \)) directly influence the reduction in metacentric height.

Answer 1

Step 1: Recall the formula for the free surface effect (FSE). 
The virtual reduction in the transverse metacentric height (\( \Delta GM \)) due to the free surface effect is given by: \[ \Delta GM = \frac{\rho_{\text{fluid}} \cdot A \cdot h}{\Delta}, \] where: - \( A \) is the free surface area of the tank, - \( h \) is the height of the fluid in the tank, - \( \Delta \) is the ship's displacement, - \( \rho_{\text{fluid}} \) is the density of the fluid in the tank relative to water. 

Step 2: Calculate the free surface area (\( A \)). 
The dimensions of the tank are: - Length = 6 m, - Width = 10 m. Thus, the free surface area is: \[ A = \text{Length} \times \text{Width} = 6 \cdot 10 = 60 \, \text{m}^2. \] 

Step 3: Substitute the given values into the formula. 
Given: - \( A = 60 \, \text{m}^2 \), - \( h = 0.5 \, \text{m} \) (since the tank is half-filled), - \( \Delta = 5000 \, \text{tonnes} = 5000 \times 1000 \, \text{kg} \), - \( \rho_{\text{fluid}} = 0.8 \). Substitute into the formula: \[ \Delta GM = \frac{\rho_{\text{fluid}} \cdot A \cdot h}{\Delta} = \frac{0.8 \cdot 60 \cdot 0.5}{5000}. \] Simplify: \[ \Delta GM = \frac{24}{5000} = 0.0048 \, \text{m}. \] Convert to centimeters: \[ \Delta GM = 0.0048 \cdot 100 = 8.0 \, \text{cm}. \] Conclusion: The virtual reduction in the transverse metacentric height is \( 8 \, \text{cm} \).