A vessel whose TPC is 12.3 is drawing 4 meters. A rectangular midship compartment 12 meters long, 10 meters bread, and 6 meters depth has a permeability of 20%. What would be the mean draught if this compartment was bilged?
Given:
TPC = 12.3
Draft = 4 meters
Lenght of the compartment = 12 meters
Breadth of the compartment = 10 meters
Depth of the compartment = 6 meters
Permeability = 20%
What is asked:
mean draught if this compartment was bilge?
Solution:
1. We have to calculate the displacement. But first we have to convert 4 meters to centimeters.
4 m x 100 cm/1m = 400 cm
Now this is the formula to getting displacement:
Displacement = TPC X draft in centimeters
= 12.3 x 400 cm
= 4,920 tons
2. Then we have to solve for the additional weight added due to flooding.
Weight = lenght x bread x draft x density
= 12 x 10 x 4 x 1.025
= 492 Tons
But with 20% of permeability, so 492 tons x 20% = 98.4 tons
3. So now we are going to add the original displacement and the additional weight to get the total displacement.
Original displacement = 4 ,920 tons
Additional weight = 98.4 tons
Total displacement = 5, 018.4 tons
4. Now the final step which is solving for the ship's draft.
Draft = displacement / TPC
= 5,018.4 tons / 12.3
= 408 cms
= 4.08 meters is the final answer
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