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A vessel is floating at 7.8m draft in DW of RD 1.010 . TPC is 18 and FWA is 250mm. the maximum permissible draft .
Solution:
Present draft = 7.8m,
RD of DW = 1.010 & TPC = 18 t/cm
FWA = 250mm = 25cm
Maximum permissible draft = 8.0m
As we know :
Change in draft =
(Change in RD )x(FWA)
0.025
= (1.010 – 1.020 ) x FWA /0 .025
=(0.015 x 25) /0 .025
= 15cm
When vessel enters SW, the draft will rise by 15 cm,
Hence ,actual draft in SW = (7.80 -0. 15)
=7.65m
Sinkage available = (8.00 – 7.65)
=0.35m
TPC = 18 ( given)
We know that :
TPC = (A/100) x ( density)
18 = (A/100) x 1.025
So, A = (18 x 1.025)/ 100
= 17.56 m2
Now , TPC at RD 1.010 = A / 100 x 1.010
= 17.56 x 1.010
= 17.73 t/cm.
DWT available =( sinkage x TPC)
= (35 x 17.73)
= 620.5t.
A ship loads in fresh water to her salt water marks and proceeds along a
river to a second port consuming 20 tonnes of bunkers. At the second port,
where the density is 1016 kg per cu. m, after 120 tonnes of cargo have been
loaded, the ship is again at the load salt water marks. Find the ship’s load
displacement in salt water.
Joe Bee. Can you please send me the solved question of the above question you posted. please
A mariner Enters port with 17,700 tons displacement,KG=28 feet, and GG=1.05 feet.A piece of Heavy machinery which weighs 180 tons is lifted by shore based cranes and placed on death so that its centre of gravity is 47 ft above the ship Khel and 19 feet two star board of centerline what will be the final mean draught gvm and angle of list when loading is complete?