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.

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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

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