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Q2. A ship of length 160 m displaces 22862 tonne when floating at a draught of 8.526 m in sea water of density 1025 kg/m³.

The waterplane area is defined by half breadths as given in Table Q1.

The following tanks are partially full of liquid as indicated:

Rectangular tank 10.2 m long and 6 m wide, containing fresh water of density 1000 kg/m³.

Rectangular tank 7.4 m long and 5 m wide containing oil fuel of density 890 kg/m³.

When a mass of 20 tonne is moved a distance of 22 m across the deck, a deflection of 71 mm is recorded on a pendulum of 9.2 m length.

The height of the center of buoyancy above the keel (KB) may be determined using Morrish's formula as given below.

KB =

Calculate the KG of the ship in the above condition.

Q3. A vessel of 14000 tonne displacement floats at a draught of 8 m in sea water of density 1025 kg/m³.

Further hydrostatic data for the above condition are:

Center of buoyancy above the keel (KB) = 4.456 m

Transverse metacenter above the keel (KM) = 7.715 m

Tonne per centimeter immersion (TPC) = 20

The vessel in the above condition is unstable and heels to an angle of 6º.

To restore positive stability, ballast of 640 tonne is now loaded at a Kg of 0.6 m.

Calculate EACH of the following for the final condition:

(a) The transverse metacentric height;

(b) The righting moment when the vessel is heeled to an angle of 15º.

Note: The vessel may be considered wall-sided between the limits of draught, hence:

GZ = sin θ (GM + ½ BM tan2 θ)

Q4. The hydrostatic particulars given in Table Q3 are for a ship of length 150 m when floating in water of density 1025 kg/m³.

The ship floats in water of density 1015 kg/m³ with draughts of 7.6 m aft and 6.8 m forward.

Calculate EACH of the following:

(a) The displacement;

(b) The longitudinal position of the ship’s centre of gravity.

Q6. A spade-type rudder has an area of 6.33 m². At its maximum designed rudder angle of 35^o, the centre of effort is 0.12 m aft of the axis of rotation and 1.6 m below the lower edge of the rudder stock bearing.

The force on the rudder normal to the plane of the rudder is given by the expression:

F_n=18.32 A v²  α newtons

where:    A=rudder area (m²)

v=ship speed (m⁄s)

α=rudder angle ( degrees)

The equivalent twisting moment (T_E )  is given by :

T_E=M+ √(M²+ T² )

where :          M=bending moment

T=torque

The maximum stress in the rudder material is to be limited to 77 MN/m².

Calculate EACH of the following:

(a) the diameter of the rudder stock required for a ship speed of 16 knots (10)

(b) the speed to which the ship must be restricted, given that the effective diameter of the stock is reduced by wear and corrosion to 375 mm. (6)

Q7. A ship of length 140 m and breadth 18 m floats at a draught of 8 m in sea water of density 1025 kg/m³. In this condition the block coefficient (Cb) is 0.68.

At a speed of 15 knots the following data applies:

Delivered power  = 4720 kW

Quasi-propulsive coefficient (QPC) = 0.70

Ship correlation factor (SCF) = 1.18

Calculate the pull required to tow a similar model of length 5 m at the corresponding speed in fresh water density 1000 kg/m³.  (16)

Note:  The frictional coefficient to be used:

for the model in fresh water of density 1000 kg/m³ is 1.694

for the ship in sea water of density 1025 kg/m³ is 1.415

Speed in m/s with the speed index (n) for ship and model 1.825

Wetted surface area (S) = 2.57 √(DL )  (m²)

Q3. The ship data in Table Q6 have been derived from the results of model experiments.

Using the data in Table Q6, determine EACH of the following:

(a) The ship speed when the propeller is absorbing 5250 kW delivered power;

(b) The propeller speed (rev/sec) given that the propeller has a diameter of 5 m with a pitch ratio of 0.9 and is operating at a real slip of 32 %.