On a vessel of 9,000 tons displacement there are two slack deep tanks of palm oil (SG .86). Each tank is 40 feet long and 30 feet wide.
What is the reduction in metacentric height due to free surface with the vessel in sea water (SG 1.025)? .
A. 0.27 ft
B. 0.48 ft
C. 0.57 ft
D. 0.74 ft
I get "B".
ReplyDeleteThis question is different from the one posted on Nov. 16th because the SG or specific gravity or density of the liquid in the tanks and that of the water the vessel is floating in has been given.
ReplyDeleteFirst 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 9,000 tons x 35 cubic feet ton = 315,000 cubic feet. Now because the SG's were given we need to calculate "r" which is the ratio of the SG of the liquid in the tank to that of the water the vessel is floating.
r = 0.86/1.025 = 0.8390
Third, in this problem there are two tanks with a width of 30 feet and a length of 40 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 = rLBBB/12Displaced Volume
FS = (0,8390)(40ft)(30ft)(30ft)(30ft) / 12 (315,000) cubic feet = 0.24 ft for one tank
Now if we double it for two tanks we get 0.48 ft and Answer B. = 0.48 foot !