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Vessel in same condition as in question 10 consume’s the following on a passage :
- All HFO (i.e 150t) from NO2 port .
- Part HFO (i.e 50t ) from NO.3c .
- All FW (i.e. 100 t ) from NO4 S.
Find the GM fluid on arrival port .
Solution:
Ship’s wt |
KG |
VM |
FSM |
| 10000 | 8.954 | 89540tm | 2087.4 |
| (-) 150t | 0.65 | (-) 97.5tm | (-) 684.0 |
| (-) 50t | 0.60 | (-) 30tm | (+) 1140.0 |
| (-) 100t | 0.70 | (-)70 tm | (-)300.0 |
Final W =9700 t FVM = 89342.5 tm FSM= 2243.5
We know that :
Final KG = (Final VM/ Final W)
= (89342.5/9700)
Final KG = 9.21m
Again, FSC = (FSM / Final W)
= (2243.5 /9700)
= 0.231m
GM (Solid) = KM – KG
= (9.9 – 9.21)
= 0.69m
GM fluid = GM (solid) – FSC
= (0.69 – 0.231)
= 0.459m.
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..