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
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
This is similar to a question already posted.
ReplyDeleteLet'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)}
The Free Surface Moment of Inertia is simply everything above except the Displaced Volume!
The Free Surface Moment of Inertia = r x{(l x b^3) /12}
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
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?