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.
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)
If you are noticing some error in problems kindly comment below.Thanks
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
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
Tan#=mxd/^xgm
ReplyDelete=100x3.5/8000x.412
Kindly check this step because in Reed its given like this, let me know.
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DeleteThe question is angle of heel after 100t of ballast water pumped out. So that time displacement will be 8000 - 1000 = 7900 t
Delete