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Daily Question for Nov. 16, 2017

 

What is the reduction in metacentric height due to free

surface when a tank 60 feet long and 30 feet wide is

 partially filled with salt water, and is fitted with a

centerline bulkhead? (The vessel has a displacement

of 10,000 tons.)


A. 0.1 foot  B. 0.8 foot   C. 1.0 foot  D. 1.2 feet


Comments

  1. The best answer is A.

    Solution:

    GG' free surface = 2 x length x (1/2 width)^3 / 12 x 35 cubic feet per ton x 10,000 tons

    GG' free surface = 2 x 60 ft x (1/2 (30 ft))^3 / 12 x 35 cubic feet per ton x 10,000 tons

    GG' free surface = 0.09 ft, so the best choice is A.

    ReplyDelete
  2. I cannot follow your numbers. Sorry, I don't see it.

    ReplyDelete
  3. Let me try it again for you, this time I will expand the math and the explanation.

    First the free surface formula assumes a rectangular tank with a given width and length.

    Second, the moment of inertial of a rectangular shape is Length x Width x Width x Width divided by 12, which is a constant for the rectangular shape and the displaced volume. In this case the displaced volume is 10,000 tons x 35 cubic feet ton = 350,000 cubic feet.

    Third, in this problem the original tank has been subdivided down its centerline so there are now two smaller tanks with a width of 15 feet and a length of 60 feet. This is why I started my calculation with 2 x.

    So lets put it all together doing one of the two tanks first then simply doubling the answer:

    FS = LBBB/12Displaced Volume
    FS = 60ft(15ft)(15ft)(15ft) / 12 (350,000) cubic feet = 0.48 ft for one tank
    Now if we double it for two tanks we get 0.098 ft and Answer A. = 0.1 foot !

    I hope this makes the answer easier to see.

    ReplyDelete
    Replies
    1. Thanks...Much better. Now I see what you did.

      Delete

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