A vessel’s statutory freeboard is 2.0m. She is loading in DW of RDD 1.015 and her freeboard is 2.1m . TPC =24. FWA = 200mm. find the DWT available .

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Solution:

Statutory freeboard is = 2.0m
DW RD = 1.015 & Freeboard = 2.1m
TPC = 24, FWA = 200mm = 20cm

As we know :

Change in draft = 

(Change in RD )x(FWA)
             0.025

= (1.015 – 1.025 ) x 20 /0.025
= ( 0.010 x 20) /0 .025
= 8cm

Hence , actual freeboard available =( 2.1 + 0.08)m
= 2.18m

Sinkage available = ( 2.18 – 2.00)
= 0.18m
= 18cm

TPC = 24 ( Given)

24  =  (A / 100) x(1.025)
A = (24 x 100)/ 1.025
A   = (24 / 1.025)
= 23.414 m2

Now, TPC for RD 1.015 = (A/ 100) x(1.015)
= (23.414 x 1.015)
= 23.76 t/cm

DWT available = (sinkage x TPC)
= (18 x 23.76 )
= 427.68t.

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