C L DUBEY – EXERCISE – 08 (DRY DOCKING)

= 24.14 tonnes m.

7. A vessel, with following particulars, enters a seawater drydock: length B.P = 140m, draft Fwd : 3.4m, Aft : 5.8 m, KG = 8.0m, KM = 9.02m,  displacement  =  8930 t, TPC = 21.9, LCF  =  2m from A.P, MCTC  =  162.5 tm. Find the virtual GM and the drafts F & A, when the level of water has fallen one metre after the stern has taken to the blocks, given the KM at this displacement = 9.18m.

Solution –

Clearly stern will take water level equal aft draft i.e 5.8m

Present level = 4.8m
Corn to aft draft = 1.238m, Hd( Hydrostatic draft ) = 4.562m
Since level ˃ Hd means stern has taken to block while forward end is shall afloat
Reduction in aft draft = 1.0m = 100cm
or, 100 =   P/21.9 + P ⨯ 72.2/162.5 ⨯ 72.2/140
Or, P = 363.90 t

GG₁ = P ⨯ KM/W
= 0.374m

Virtual GM  = ( old GM – Virtual loss of GM )
= (old GM – 0.374)
= (1.18 – 0.374)
= 0.806m

Trim caused = 1.617m,
Rise = 0.116m,
Ta = 0.834, T_f = 0.783