A liquid mud tank measures 30 ft. long, 15 ft. wide, 6 ft. deep. The vessel displaces 968 tons. The specific gravity of the mud is 1.8. What is the reduction in GM if two of this tank are slack?
Given:
Lenght = 30 ft
Width/beam = 15 ft
S.G. of mud = 1.8
displacement = 968 tons
What is asked:
Reduction in GM
Solution:
Free surface correction formula:
FSC = r L B3
(420 x displacement)
1. First we have to solve for relative density. "r" can be solve by dividing the specific gravity of the fluid inside tank by the density of water where the vessel floats. Since the problem does not specify, let us assume that it is sea water.
r = 1.8 = 1.76
1.025
2. We can now solve for the reduction in GM.
FSC = r L B3
(420 x displacement)
= 1.76 x 30 x (15)3
(420 x 968)
= 1.76 x 30 x 3375
(420 x 968)
= 178200
406560
= 0.43831
x 2 (since there two slack tanks)
0.87 is the final answer
Given:
Lenght = 30 ft
Width/beam = 15 ft
S.G. of mud = 1.8
displacement = 968 tons
What is asked:
Reduction in GM
Solution:
Free surface correction formula:
FSC = r L B3
(420 x displacement)
1. First we have to solve for relative density. "r" can be solve by dividing the specific gravity of the fluid inside tank by the density of water where the vessel floats. Since the problem does not specify, let us assume that it is sea water.
r = 1.8 = 1.76
1.025
2. We can now solve for the reduction in GM.
FSC = r L B3
(420 x displacement)
= 1.76 x 30 x (15)3
(420 x 968)
= 1.76 x 30 x 3375
(420 x 968)
= 178200
406560
= 0.43831
x 2 (since there two slack tanks)
0.87 is the final answer
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Where did you get the 420?
ReplyDeleteI think the 420 part of the formula. That's what I see in the formula given by the author
Deletewhat is the formula if given are in meters? Is it the same?
ReplyDelete((L*(B^3)*rel. density of cargo)) / (12 * Displacement)
Delete