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.

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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.

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