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A vessel has a deep tank on the starboard side 12m long 9m wide which is partly full of coconut oil of RD 0.72 . If W = 1200 t , KM = 9m and KG = 8.5m , find the GM fluid .
Solution:
L = 12m, B = 9m, RD = 0.72
W = 12000 t, KM = 9m & KG = 8.5m
GM (solid) = (KM – KG)
= (9 – 8.5)
= 0.5m
We know that:
FSC = (i di /W )
= (LB3 x di ) / (12 x W )
= (12 x 93 x 0.72) / (12 x 1200)
= 0.044m
Fluid GM = GM (solid) -FSC
= (0.5 – 0.044)
= 0.456m.
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A vessel displacing 8000t , has a rectangular deep tank 10m long 8m wide and 9m deep full of SW . The KM is 7m and KG 6.2m. Find the GM when 1 / 3 of this tank as pumped out.
Solution:
W = 8000t.
L = 10m, B = 8m & D = 9m
RD = 1.025, KM = 7m, KG = 6.2m
Mass of the tank = (Volume x Density)
= (L x B x D) x 1.025
= (10 x 8 x 9) x (1.025)
= 738t
According to question, 1/3 of water pumped out
= (1/3 x738)
= 246t
After pumping out the water the KG of the tank = (6 + 1.5)
= 7.5m
In question #2 doesn’t it need to convert into 0.050??
Where is the solution for que 18&19 ?
yes, question number 20, solid gm is 0.9, subtract fsc and get fluid gm and then apply RM=WGMSinx
the breath of the load water plan of a ship 90m long measured at equal intervals from forward are as follow 0,3.96,8.53,11.58,12.19,12.5,11.58,5.18,3.44, and 0.3m respectively.if the load draft is 5m and the block coefficient is 0.6. find the freshwater allowance and position of the centre of floatation
Very helpful to solve questions
Sir . There is some mistake in the solution of 20th question of FSC..
Thanks for the easy calculations..