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Daily Question for Jan 31, 2018

A tank 36 ft. by 36 ft. by 6 ft. is filled with water to a depth of 5 ft. If a bulkhead is placed in the center of the tank running fore-and-aft along the 36-foot axis, how will the value of the moment of inertia of the free surface be affected?

A. The moment of inertia would remain unchanged.

B. The moment of inertia would be 1/4 its original value.

C. The moment of inertia would be 1/2 the original value.

D. None of the above
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  1. UnknownJanuary 31, 2018 at 5:14 PM

    C

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  2. Friends of STABILITY AND TRIM FOR THE SHIP'S OFFICERJanuary 31, 2018 at 7:43 PM

    Let's do the math and see how a centerline bulkhead will affect Free Surface Correction.

    The normal Free Surface Correction = r x{(l x b^3) /12 x (Displaced Volume)}
    Where:

    r = the ratio of the denisty of the liquid in the tank to that of the water the vessel is floating in

    l = the length of the tank in feet or meters

    b = the width of the tank in feet or meters

    12 is a constant used for a rectangular tank

    Displaced Volume the the Displacement in Long Tons x 35 cubic feet / LT
    or
    Displacement in Metric tons / ( 1.025 MT/ cubic meter)

    All things being equal we can calculate the difference in FS Correction by evaluating the with of the tanks alone because everything else will be equal.

    The width of the tank without the Centerline Bulkhead = b

    the width of the two small tanks formed by the Centerline Bulkhead = b/2 each!

    Keep in mind the width of the tank is cubed, and the Centerline Bulkhead has now produced two smaller tanks each with width b/2 !

    So we can compare (b^3) without Centerline Bulkhead
    to
    (2 x (b/2)^3) = 2x(b^3/8) = 1/4 b^3 With the Centerline Bulkhead.

    So the Centerline Bulkhead will reduce FS correct to 1/4th its original value.

    The Best and only answer for this question is choice B.
    Do you Agree or Disagree?

    ReplyDelete
  3. UnknownFebruary 7, 2018 at 2:48 AM

    Bravo; 1/4th

    Looking at the surface area (L x B) the breadth is cubed here.
    The breadth of a tank has three times the effect on virtual rise of the fluids CG.
    A slosh plate is commonly used to mitigate this effect. This is a good question
    to prove this point.

    (36 x 36 x 36 x 36) = 1679616

    now divide the same tank into two parts

    2(36 x 18 x 18 x 18) = 419904

    with this (419904) / (1679616) = .25 or 1/4

    Viola, there you have it !

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