Q1. A ship 126 m floats at a draught of 7.5 m and in this condition the immersed cross sectional areas and waterplane areas are as given in Tables Q1(A) and Q1(B). The equivalent base area (Ab) is required because of the fineness of the bottom shell.
(a) the equivalent base area value Ab; (8)
(b) the longitudinal position of the centre of buoyancy from midships; (4)
(c) the vertical position of the centre Of buoyancy above the base. (4)
Q2. The ‘ wall sided formula ‘ gives an expression for righting lever ( GZ) as follows:
GZ=sinθ (GM+ 1/2 BM tan2 θ)
(a) Using the wall sided formula, derive an expression for the angle of loll of a ship which is initially unstable in still water.(5)
(b) A box shaped vessel is 40 m long, 10 m wide and floats at a draught of 4 m in sea water of density 1025 kg/m3 with a KG of 3.883 m.
A beam wind acts on the exposed area of the vessel causing it to heel to an angle of 12o.
The heeling moment caused by the wind is given by the expression:
Heeling moment = 850 v2 cos2 θ Nm
where: v = wind speed in m/s
θ= angle of heel in degrees
Calculate the wind speed using the wall sided formula for GZ. (11)
Q3. The following particulars apply to a ship of length 140 m when floating in sea water of density 1025 kg/m3 at an even keel draught of 7.265 m. displacement = 15800 tonne centre of gravity above the keel (KG) = 7.8 m centre of buoyancy above the keel (KB) = 4.05 m waterplane area = 2146 m2 centre of flotation from midships (LCF) = 3.0 m aft second moment of area of the waterplane about a transverse axis through midships = 2.305 × 106 m4 (a) Calculate the value of the moment to change trim by one centimetre (MCT1cm) in the above condition. (6) (b) The ship in the above condition now undergoes the following changes in loading: 352 tonne added at an Lcg of 10.5 m forward of midships 110 tonne removed from an Lcg of 2.0 m aft of midships 150 tonne restowed at a new position 52.7 m aft of its original position. Calculate the new end draughts of the ship. (10)
Q11. Fig Q6 shows the results of progressive speed trials on a ship at a load displacement of 22350 tonne in sea water of density 1025 kg/m3 with a wetted surface area of 4860 m.
Using the data given below, calculate the shaft power required to achieve a service speed of 17 knots with a geometrically similar ship having a load displacement of 29245 tonne in sea water.
Propulsive coefficient based upon shaft power for both trial and service conditions = 0.68
Allowance for appendages and weather in trial condition = 8%
Allowance for appendages and weather in service condition = 20%
Note: Frictional coefficient for the 22350 tonne ship in sea water is 1.410
Frictional coefficient for the 29245 tonne ship in sea water is 1.406
Speed is in m/s with index (n) = 1.825
Q8. A vessel of 10500 tonne displacement is fitted with a propeller Of 5.5 m diameter
and pitch ratio 0.9.
During a fuel consumption trial of 6 hours duration, a steady shaft speed of 1.8 revs/sec was maintained and 7.54 tonne of fuel was consumed.
The following results were also recorded:
real Slip ratio = 0.34
Taylor wake fraction = 0.32
shaft power = 6050 kW
transmission losses = 3%
quasi-propulsive coefficient (QPC) = 0.71
propeller thrust = 680 kN
Calculate EACH of the following:
(a) the speed of the ship; (4)
(b) the apparent Slip ratio; (1)
(c) the propeller efficiency; (3)
(d) the thrust deduction fraction; (3)
(e) the fuel coefficient; (3)
(f) the specific fuel consumption. (2)
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