Question

Two 30 m long bilge keels of mass 40 tonnes each, are fitted at the turn of the bilge on port and starboard sides of a ship. The cross section of the bilge keel is shown in the following figure. Assume density of water = 1000 kg/m\(^3\). If the TPC (tonnes per centimeter) immersion of the ship is 50, then the change in the mean draft is .................... cm 

 

• A. 1
• B. 0.8
• C. 0.6
• D. 1.6

Hint:

When adding or removing items that are fully or partially submerged, always remember to work with the *net* change in weight. The net change is the weight of the object minus the buoyant force it generates. This is a common trick in naval architecture problems.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem deals with the effect of adding a weight to a ship on its mean draft. The change in draft is related to the added weight and the ship's TPC (Tonnes Per Centimetre immersion). However, when the added item is submerged, its own buoyancy must be accounted for. The net weight added to the ship is the weight of the item in air minus the weight of the water it displaces (its buoyancy).
Step 2: Key Formula or Approach:
1. Calculate the total mass added: Two bilge keels, each 40 tonnes.
2. Calculate the volume of the bilge keels: The cross-section is a triangle. Volume = (Cross-sectional Area) \(\times\) Length.
3. Calculate the buoyant force of the bilge keels: Buoyancy = (Volume of bilge keels) \(\times\) (density of water).
4. Calculate the net weight added: Net Weight = Total Mass - Buoyancy.
5. Calculate the change in draft: Change in draft (\(\delta T\)) in cm = Net Weight Added (in tonnes) / TPC.
Step 3: Detailed Calculation:
1. Total mass added: \[ \text{Mass}_{added} = 2 \times 40 \text{ tonnes} = 80 \text{ tonnes} \] 2. Volume of one bilge keel:
The cross-section is a triangle with base = 0.5 m and height = 2.0 m.
\[ \text{Area}_{cs} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 0.5 \text{ m} \times 2.0 \text{ m} = 0.5 \text{ m}^2 \] The length of each keel is \( L = 30 \) m. \[ \text{Volume}_{onekeel} = \text{Area}_{cs} \times L = 0.5 \text{ m}^2 \times 30 \text{ m} = 15 \text{ m}^3 \] Total volume of two keels: \[ \text{Volume}_{total} = 2 \times 15 \text{ m}^3 = 30 \text{ m}^3 \] 3. Buoyancy of the bilge keels: Density of water \(\rho_w = 1000\) kg/m\(^3\) = 1 tonne/m\(^3\). \[ \text{Buoyancy} = \text{Volume}_{total} \times \rho_w = 30 \text{ m}^3 \times 1 \text{ tonne/m}^3 = 30 \text{ tonnes} \] 4. Net weight added: \[ \text{Net Weight} = \text{Mass}_{added} - \text{Buoyancy} = 80 \text{ tonnes} - 30 \text{ tonnes} = 50 \text{ tonnes} \] 5. Change in draft: Given TPC = 50 tonnes/cm. \[ \delta T \text{ (cm)} = \frac{\text{Net Weight}}{\text{TPC}} = \frac{50 \text{ tonnes}}{50 \text{ tonnes/cm}} = 1 \text{ cm} \] Step 4: Final Answer:
The change in the mean draft is 1 cm.