27.NA January 2019 Q.4(B)

A ship of 8000 tonne displacement floats upright in seawater. KG = 7.6m and GM = 0.5m. A tank, KG is 0.6m above the keel and 3.5m from the centreline, contains 100 tonne of water ballast. Neglecting the free surface effect, calculate the angle which the ship will heel, when the ballast water is pumped out.

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Given

Δ = 8000 t
KG =  7.6m
GM = 0.5m

For tank
KGtank = 0.6m
m = 100 t

To find

Angle of heel

Solution

KM = KG + GM 
      = 7.6 + 0.5
      = 8.1 m

New KG = (Δ x KG - m x KGtank) / (Δ - m)
              = ((8000 x 7.6) - (100 x 0.6) / (8000 - 100)

              = 7.688 m

New GM = KM - New KG
               = 8.1 - 7.688
               = 0.412 m

Heeling moment of water = m x distance
                                         = 100 x 3.5
                                         = 350

Heeling moment of ship   = New displacement x New GM
                                         = 7900 x 0.412
                                         = 3254.8

                               tan θ  = Heeling moment of water / Heeling moment of ship
                                         = 350 / 3254.8
                                         = 0.1075

               Angle of heel θ  = 60131