105.NA November 2020 Q.9(b)

A ship 85 m long displaces 8100 tonne when floating in seawater at draughts of 5.25 m forward & 5.55 m aft. TPC 9.0, GML 96 m, LCF 2 m aft of midships. It is decided to introduce water ballast to completely submerge the propeller & a draught aft of 5.85 m is required. A ballast tank 33 m aft of midships is available. Find the least amount of water required & the final draught forward.

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 Given

L = 85 m

Δ = 8100 t

df = 5.25 m

da = 5.55 m

TPC = 9

GML = 96m

LCF = 2 m

 

To find

i) Amount of water required to make draught aft 5.85 m

ii) Final Forward draught

 

Solution

 We know that moment to change trim one cm, MCT1 cm = (Δ x GML) / (100 x L)

                                                                                             = (8100 x 96) / (100 x 85)

                                                                                             = 91.48 t m

 

We know that trimming moment = mass moved x distance moved

In question it is given that ballast tank is 33 m aft of midship and LCF is 2 m aft of midship.

i) So distance at which water has to be taken = 33 - 2 = 31 m

So trimming moment = m x 31 

Change in trim = trimming moment / MCT1 cm

                         = (31 x m) / 91.48

                      t  =  0.339 x m -----------------------------------------------(1)

We know that bodily sinkage = mass added / TPC

                                                = m / 9 

                                                = 0.111 x m -------------------------------(2)

Distance from LCF to aft = (85/2) - 2 (As 85/2 will be mid ship. then 2 m from mid to LCF)

                                  WF = 40.5 m

Change in draught aft = (t / L) x WF

                                    = {(0.339 x m) / 85} x 40.5

                                    = 0.161 x m ------------------------------------------(3)

So new draught aft = old draught + bodily sinkage + change in draught

5.85 = 5.55 + 0.111 m + 0.161 m

5.85 = 5.55 + 0.272 m

0.272 m = 0.3

m = 1.10 = 110 t (Bodily sinkage and change in trim are in cm)

So amount of water required = 110 t

ii) So Distance from LCF to fwd = (85/2) + 2 (As 85/2 will be mid ship. then 2 m from mid to LCF)

                                                FL = 44.5 m

Change in draught forward = - (t / L) x FL

                                            = -{(0.339 x m) / 85} x 44.5

                                            = - {(0.339 x 110) / 85} x 44.5

                                            = -  19.52 cm

So new draught fwd = old draught + bodily sinkage + change in draught

                                 = 5.25 + (0.111 x 110) - 0.195

                                 = 5.25 + 0.122-0.195

                                 = 5.177 m