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Given the following particulars, Find the GM fluid : W= 8800 t, tank of i = 1166 m4 is partly full of HFO of RD 0.95, KM 10.1m, KG 9.4m .
Solution:
W = 8800 t & i = 1166m4,
RD = 0.95,
KM = 10.1 m & KG = 9.0m.
GM (solid) = (KM – KG)
= (10.1 – 9.0)
= 1.1m
We know that:
FSC = (i di /W )
= (1166 x 0.95)/ 8800
= 0.126m
GM fluid = GM (solid) – FSC
= (1.1 – 0.126)
= 0.974m.
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On a vessel of W 16000 t, NO.4 port DB tank 20m long and 8m wide is partly full of DW ballast of RD 1.010. Find the FSC.
Solution:
W = 16000t,
L = 20m & B = 8m,RD = 1.010
We know that:
FSC = (i di / W)
FSC =( LB3 X di )/( 12 x W)
= (20 x 83 x 1.010)/(12 x 16000)
= 0.054m.
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..