Question

Example: Knock Nevis was the world's largest supertanker with a length of 485 m and a depth of 30m. It was made from steel (density 8050 kg m-3) with an average hull thickness of 3 cm Approximating it as a half cylinder what is the maximum possible mass it could carry before it sinks? Knock Nevis was rated to carry up to 565,000 tonnes (1 tonne - 1000 kg). How much of it would be underwater with this cargo?

Answer 1

Assume, the supertanker is a half-cylinder with length L = 485 m and radius R = 30 m.

The total surface area due to the curved surface and the top and bottom flat surface is $\pi RL + 2*(0.5*\pi R^2+0.5*\pi R^2) = \pi*485*30 + \pi*30^2 = 48537.606 \rm{~m^2}$ .

If the tackiness is 3 cm = 0.03 m, the total volume of the steel is 48537.606*0.03 = 1456.13 m^3.

As the density of steel is 8050 kg/m^3, the total mass of the supertanker is 1456.13*8050 kg = 11721846.5 kg.

Volume of the supertanker is $0.5*\pi R^2L$ = 685652.6 m^3.

The density of water is about 1000 kg/m^3. Therefore the mass of water which the volume of supertanker can occupy = 685652.6*1000 kg. Hence the tanker can carry a maximum of (685652600 - 11721846.5) kg = 673930753.5 kg before it sinks.

## Knot Nevis was rated to carry up to 565000000 kg. With this amount of load the total weight is (565000000 + 11721846.5) kg = 576721846.5 kg. = (576721846.5/673930753.5)*100% = 85.6 % of the volume would be underwater.