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Q8. A. Explain the purpose of interlopes and state their magnetic polarity relative to the main poles of both generators and motors.

A 200V, long-shunt compound-wound generator has a full-load output of 20kW. The various resistances are as follows; armature (including brush contact) 0.15 ohm, series field 0.025 ohm, interpole field 0.028 ohm, shunt field (including the field-regulator resistance) 115 ohm. The iron losses at full load are 780W, and the friction and wind age losses 590W. Calculate the efficiency at full load.

Q10. (a) What is meant by Quasi Propulsive Coefficient?

A ship of 15000 tonne displacement has mechanical efficiency of the machinery is 83%, shaft losses 6%, propeller efficiency 65% and QPC 0.71. At a particular speed the thrust power is 2550 KW. Calculate: (i) Indicated Power; (ii) Effective power.

Q8. A. Why is it important in a tender ship to keep the double bottom tanks pressed up?

B. A ship of 6000 tonne displacement has a wetted surface area of 2500 m2 and a speed of 15 knots.

(i) Calculate the corresponding speed and wetted surface area of as similar ship of 2000 tonne displacement.

(ii) If the ship resistance is of the form R=0.45 S V1.83 N; find the resistance of the 6000 tonne Ship.

Q7. A ship of length 120 m displaces 11750 tonne when floating in sea water of density 1025 kg/m3. The center of gravity is 2m above the center of buoyancy and the waterplane is defined by the following equidistant half-ordinates given in Table Q1:

Calculate EACH of the following:

(a) The area of the waterplane;

(b) The position of the centroid of the waterplane from midships;

(c) The second moment of area of the waterplane about a transverse axis through the centroid;

(d) The moment to change trim one centimeter (MCT1cm).

Q2. The ‘wall sided formula’ gives an expression for righting lever (GZ) as follows:  GZ = sin θ (GM + ½ BM tan2 θ) (a) Derive an expression for the `angle of loll' of a ship which is initially unstable in still water, using the wall sided formula. (b) A box shaped vessel designed to carry timber, is 80 m long, 12 m wide and floats at a draught of 5 m in sea water of density 1025 kg/m3 with a KG of 4.815 m. A beam wind acts on the exposed area of the vessel causing it to heel to an angle of 15°. The heeling moment caused by the wind is given by the expression: Heeling moment = 0.85 A b v2 cos2 θ Nm

where:  A = exposed area = 627 m2 b = lever = 6.5 m      v = wind speed in m/s                                        θ = angle of heel in degrees.

Calculate the wind speed in knots, using the wall sided formula for GZ.

Q8. A. Describe the procedure for speed, Power and fuel consumption trials. B. An oil tanker has LBP 142m, beam 18.8m and draught 8m. It displaces 17000 tonne in seawater of 1.025 1/m3. The face pitch ratio of the propeller is 0.673 and the diameter 4.8m. The results of the speed final show that the true slip may be regarded as constant over a range of speeds of 9 to 12 knots and is 35% the wake fraction may be calculated from the equation: W=0.5Cb = 0.05. If the vessel uses 20 tonne of fuel per day at 12 knots and the consumption varies as speed. Find the consumption per day at 100 rev/min.

Q8. A box barge of 60m long and 10m wide floats at a level keel draught of 3m. Its centre of gravity is 2.5m above the keel. Determine the end draughts if an empty, fore end compartment 9m long is laid open to the sea.

Q8. A. Explain the use of KN curves.

B. The half breadths of the load waterplane of a ship 150 m long commencing from aft, are 0.3. 3.8. 6.0, 7.7, 8.3.9.0,8.4.7.8, 6.9.4.7 and 0 m respectively. Calculate: (a) area of waterplane (b) distance of centroid from amidships (c) second moment of area about a transverse axis through centroid.

Q9. A. Explain the term volumetric heeling moments.

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

Q9. A ship of 6000 tonne displacement has a wetted surface area of 2500 m2 and a speed of 15 knots.  Calculate the corresponding speed and wetted surface area of as similar ship of 2000 tonne displacement. (i) If the ship resistance is of the form R=0.45 S V1.83 N; find the resistance of the 6000 tonne Ship.

Q9: A. Explain how trim occurs, and explain the effect of trim on tank soundings. B. A ship 150 metres long arrives at the mouth of a river with drafts 5.5m.  F and 6.3m A MCT 1 cm. 200 tonnes-m.  TPC 15 tonnes.  Centre of flotation is 1.5m. aft of amidships.  The ship has then to proceed up the river where the maximum draft permissible is 6.2m.  It is decided that SW ballast will be run into the forepeak tank to reduce the draft aft to 6.2m.  If the canter of gravity of the forepeak tank is 60 metres forward of the center of flotation, find the minimum amount of water which must be run in and also find the final draft forward.

Q8. (a) Explain what is meant by left and right handed propellers, and also explain the rotation of propellers in a twin-screw ship.

b) A ship 120m long displaces 8000 tonne, GML is 102m, TPC 17.5 and LCF 2m aft of midships. It arrives in port with draughts of 6.3m forward and 6.6m aft.

During the voyage the following changes in loading have taken place:

Fuel used          200 tonne         18m forward of midships

Water used        100 tonne         3m aft of midships

Stores used        10 tonne           9m aft of midships

Ballast added     300 tonne         24m forward of midships

Calculate the original draughts. (10)