DieselShip's Online Content Library

Q10. With reference to 3-phase transformers;

(a) Discuss the essential and desire conditions to be fulfil for operating two three-phase transformers in parallel.

(b) What are the advantages of transformer bank of three 1-phase transformers over a unit three phase transformer of the same kVA rating?

(c) In a 25 kW, 3300/233V, 1- phase transformer the iron and full load copper losses are 350 watts and 400 watts respectively. Calculate the efficiency  at half load, 0.8 p.f.

Q10. A. List the factors that determine the starting torque of the three-phase induction motor. How does this torque generally compare with the value of the rated torque? B. A twelve-pole, three-phase, delta-connected alternator runs at 600 rev/min and supplies a balanced star-connected load. Each phase of the load is a coil of resistance 35 ohm and inductive reactance 25 ohm. The line terminal voltage of the alternator is 440V. Determine (a) frequency of supply, (b) current in each coil, (c) current in each phase of the alternator, (d) total power supplied to the load.

Q10. A. The capacitor-start induction run motor has a much higher starting torque than the resistance split-phase motor, Explain.

B. An eight-pole armature is wound with 480 conductors. The magnetic flux and each conductor is capable of carrying a full load current of 100A. Calculate the terminal voltage on no load. The output current on full load and the total power generated on full load when the armature is 1. Lap connected 2. Wave- Connected.

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.

Q10: A. Explain the term volumetric heeling moments. B. A box barge 33m long and 5.5m wide has a light displacement and KG of 216 tonne and 2.75 m respectively. 80 tonne of machinery are put on board and in order to maintain stability 54 tonne of Ballast are at Kg 0.15m.  The final GM is 0.13 m. Calculate the Kg of the added machinery.

Q7. A) Why is an inclining experiment carried out? Write short account of the method adopted.

b) An inclining experiment was carried out on a ship of 8000 tonne displacement. The inclining ballast was moved transversely through 12 m and the deflections of a pendulum 5.5m long, measured from the centreline, were as follows

3 tonne port to starboard 64 mm S

3 Tonne 116 mm S

Ballast restored 3 mm S

3 Tonne starboard to port 54 mm P

3 tonne starboard to port 113 mm P

Calculate the metacentric height of the vessel. (10)

Q8. A. Explain the purpose of non-watertight longitudinal subdivision of tanks.

B. A ship 160m long and 8700-tonne displacement floats at a waterline with ½ ordinates of 0, 2.4, 5.0, 7.3, 7.9, 8.0, 8.0, 7.7, 5.5, 2.8, 0m respectively. While floating at this waterline, the ship develops a list of 10° due to instability. Calculate the negative metacentric height when the vessel is upright in this condition.