A vessel is floating in dockwater of RD 1.005with her starboard WNA mark 30mm below and her port WNA mark 60mm below the waterline. If her summer SW draught is 8.4m, TPC is 30 and FWA is 160 mm, calculate how much cargo can be loaded to bring the vessel to her vessel draught in SW.

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Solution:

Dockwater  RD = 1.005,
Stbd WNA mark = 30mm below
Port WNA mark = 60mm below
Mean draft = (30 + 60) / 2
= 45mm
=4.5cm,below the water line.

qes 17

Distance from WNA to water(W) = 50mm
= 5cm

So, sinkage available = (50 – 45)
= 5mm
= 0.5cm

We know that :
Water (W) is 1/48 of summer draft
= (1/48 x 8.4)
= 0.175m
= 17.5cm

Sinkage = (17.5 + 0.5 )
= 18cm

Total sinkage = (18 + 12.8)
= 30.8cm

As we know:

Change in draft = 

(Change in RD )x(FWA)
             0.025

= (1.025 – 1.005 )x 16 / 0.025
= 12.8cm

Again, TPC = (A/100) x density
30 = (A/100) x 1.025
A  =  30/ 1.025
A  = 29.268 m2

Now , TPC for density of 1.005
TPC = (A/100) x 1.005
= (29.268 x 1.005)
= 29.41  t/cm.

Hence ,cargo can be loaded = (30.8 x 29.41)
= 905.82 t

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