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A vessel is lying in a river berth of density 1.010 tonnes per m3 , with her summer loadline 20mm above the water on the starboard side and 50mm above the water on the port side . Find how much cargo she can load to bring her to her to her summer loadline in SW, if her summer displacement is 15000 tonnes and TPC is 25.
Solution:
RD of river = 1.010
Summer load line on starboard side = 20mm above
Port side = 50mm above
Mean draft = ( 20 + 50) / 2
= (70/2)
= 3.5cm above
W = 15000t,
FWA = (W/TPC)
= 15000/(40 x 25)
= 15cm
As we know:
Change in draft =
(Change in RD )x(FWA)
0.025
= (1.025 – 1.010 ) x 15 / 0.025
= 9cm
Total sinkage = (9 + 3.5 )
= 12.5cm
TPC = 25( Given)
We know that:
TPC = (A/100) x 1.025
= (25 / 1.025)
= 24.39 m2
Now, TPC for RD 1.010 = (A /100) x( 1.010)
= (24.39 x 1.010)
= 24.63t/cm
Hence ,Cargo can be load = (sinkage x TPC)
= (12.5 x 24.63)
= 307.875 t
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?