Question:
An oil tanker of breadth 20 m and having a displacement of 24000 tonnes in sea water (density of sea water = 1025 kg/m\(^3\)) is carrying oil of relative density 0.8 in 9 longitudinally distributed tanks which are all half-filled. Each longitudinal tank is 12 m long and 16 m wide. The apparent change in vertical center of gravity, due to the presence of oil in the tanks is .................... m (round off to one decimal place)
An oil tanker of breadth 20 m and having a displacement of 24000 tonnes in sea water (density of sea water = 1025 kg/m\(^3\)) is carrying oil of relative density 0.8 in 9 longitudinally distributed tanks which are all half-filled. Each longitudinal tank is 12 m long and 16 m wide. The apparent change in vertical center of gravity, due to the presence of oil in the tanks is .................... m (round off to one decimal place)
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The free surface effect is a critical stability calculation. Remember the formula for \(i_t\) for a rectangular tank is \(lw^3/12\) for transverse stability (it's \(wl^3/12\) for longitudinal). The width is cubed for transverse moments because the shift of liquid is across the width of the tank when the ship rolls.
Updated On: Aug 30, 2025
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Solution and Explanation
Step 1: Understanding the Concept:
This problem deals with the effect of "free surface" on the stability of a ship. When a tank is only partially filled with liquid, the liquid can shift as the vessel heels, creating a "free surface moment." This effect is accounted for by an apparent rise in the ship's vertical center of gravity (VCG), known as the free surface effect (FSE). The apparent change in VCG is the Free Surface Rise, \(GG_v\).
Step 2: Key Formula or Approach:
The free surface rise in the center of gravity (\(GG_v\)) for a single tank is given by: \[ GG_v = \frac{\rho_l . i_t}{\Delta} \] where
- \(\rho_l\) is the density of the liquid in the tank.
- \(i_t\) is the transverse second moment of area of the free surface of the liquid in the tank about the tank's centerline. For a rectangular tank, \( i_t = \frac{l . w^3}{12} \), where \(l\) is the length and \(w\) is the width of the tank. - \(\Delta\) is the displacement of the ship.
Since there are 9 identical tanks, the total free surface effect is the sum of the effects from each tank. \[ \text{Total } GG_v = \sum_{n=1}^{9} \frac{\rho_{oil} . i_t}{\Delta} = 9 \times \frac{\rho_{oil} . i_t}{\Delta} \] Step 3: Detailed Calculation:
Given values:
- Displacement, \(\Delta = 24000\) tonnes = \(24,000,000\) kg.
- Number of tanks, \(N = 9\).
- Tank length, \(l = 12\) m.
- Tank width, \(w = 16\) m.
- Relative density of oil = 0.8. Density of water \(\rho_w = 1000\) kg/m\(^3\). So, density of oil \(\rho_{oil} = 0.8 \times 1000 = 800\) kg/m\(^3\).
1. Calculate the second moment of area (\(i_t\)) for one tank:
\[ i_t = \frac{l . w^3}{12} = \frac{12 . (16)^3}{12} = 16^3 = 4096 \text{ m}^4 \] 2. Calculate the total free surface rise (\(GG_v\)):
\[ \text{Total } GG_v = N \times \frac{\rho_{oil} . i_t}{\Delta} \] \[ \text{Total } GG_v = 9 \times \frac{800 \text{ kg/m}^3 \times 4096 \text{ m}^4}{24,000,000 \text{ kg}} \] \[ \text{Total } GG_v = \frac{9 \times 800 \times 4096}{24,000,000} = \frac{7200 \times 4096}{24,000,000} \] \[ \text{Total } GG_v = \frac{29,491,200}{24,000,000} = 1.2288 \text{ m} \] 3. Round to one decimal place: \[ GG_v \approx 1.2 \text{ m} \] Step 4: Final Answer:
The apparent change in the vertical center of gravity is 1.2 m.
This problem deals with the effect of "free surface" on the stability of a ship. When a tank is only partially filled with liquid, the liquid can shift as the vessel heels, creating a "free surface moment." This effect is accounted for by an apparent rise in the ship's vertical center of gravity (VCG), known as the free surface effect (FSE). The apparent change in VCG is the Free Surface Rise, \(GG_v\).
Step 2: Key Formula or Approach:
The free surface rise in the center of gravity (\(GG_v\)) for a single tank is given by: \[ GG_v = \frac{\rho_l . i_t}{\Delta} \] where
- \(\rho_l\) is the density of the liquid in the tank.
- \(i_t\) is the transverse second moment of area of the free surface of the liquid in the tank about the tank's centerline. For a rectangular tank, \( i_t = \frac{l . w^3}{12} \), where \(l\) is the length and \(w\) is the width of the tank. - \(\Delta\) is the displacement of the ship.
Since there are 9 identical tanks, the total free surface effect is the sum of the effects from each tank. \[ \text{Total } GG_v = \sum_{n=1}^{9} \frac{\rho_{oil} . i_t}{\Delta} = 9 \times \frac{\rho_{oil} . i_t}{\Delta} \] Step 3: Detailed Calculation:
Given values:
- Displacement, \(\Delta = 24000\) tonnes = \(24,000,000\) kg.
- Number of tanks, \(N = 9\).
- Tank length, \(l = 12\) m.
- Tank width, \(w = 16\) m.
- Relative density of oil = 0.8. Density of water \(\rho_w = 1000\) kg/m\(^3\). So, density of oil \(\rho_{oil} = 0.8 \times 1000 = 800\) kg/m\(^3\).
1. Calculate the second moment of area (\(i_t\)) for one tank:
\[ i_t = \frac{l . w^3}{12} = \frac{12 . (16)^3}{12} = 16^3 = 4096 \text{ m}^4 \] 2. Calculate the total free surface rise (\(GG_v\)):
\[ \text{Total } GG_v = N \times \frac{\rho_{oil} . i_t}{\Delta} \] \[ \text{Total } GG_v = 9 \times \frac{800 \text{ kg/m}^3 \times 4096 \text{ m}^4}{24,000,000 \text{ kg}} \] \[ \text{Total } GG_v = \frac{9 \times 800 \times 4096}{24,000,000} = \frac{7200 \times 4096}{24,000,000} \] \[ \text{Total } GG_v = \frac{29,491,200}{24,000,000} = 1.2288 \text{ m} \] 3. Round to one decimal place: \[ GG_v \approx 1.2 \text{ m} \] Step 4: Final Answer:
The apparent change in the vertical center of gravity is 1.2 m.
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