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A double bottom tank 20m x 10.5m x 1.0m has a block a coefficient of 0.82. Calculate how much fuel oil of RD 0.9, it can hold .
Solution :
Given : – (L x B x H) of double bottom tank =( 20 x 10 .5 x 1)
Cb = 0.82 , RD = 0.95
We know that:
Cb = (volume of tank) /( L x B x depth)
0.82 = (volume of tank) /(20 x 10.5×1)
volume of tank = 0.82 x ( 20 x 10.5x 1)
= 172.2 m3
Weight of oil can be calculated as :- (volume of tank) x(density)
= (172.2 x 0.95)
= 163.59 t
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A ship floating in SW at a draft of 8m is 110m long and 14m wide at the water line . If her block coefficient is 0,72, find her displacement. If her load displacement is 12000 t, find the DWT available .
Solution:
Given: – (L x B) = 110m x 14m,
Present draft = 8m
Cb = 0.72
Load displacement = 12000 t ,
We know that :-
Cb = (u/w volume ) / (Lx B x D)
0.72 = (u/w volume) / (110 x 14 x8)
u/w volume = 0.72 x ( 110 x 14 x 8 )
= 8870.4 t
DWT available = (Load displacement – present displacement)
= (12000 – 8870.4)
= 3129.6t.
many questions have wrong answers in various exercises,,at least book ke end me answers ko chk kar lete…anyways thanks for this website
many questions have wrong answers in various exercises,,at least book ke end me answers ko chk kar lete…
what is the draft of a box shape steel (L=4 meters , B= 4 meters , Depth = 0.60 meters)
floating in a salt water density 1.028 and the total weight of the steel box with steel on top of it is 1500 kg
question 5 is wrong. you forget to take the u/w vol and multiply by the sw density.
sir exercise no 4 problem no 7 draft you had taken is wrong it is not 4.1 it is 6.1 mtr