Naval Architecture Numerical Solutions
Numerical 1: What is Prismatic Coefficient (Cp). (a). Derive the formula Cp=Cb/Cm; where Cb = Coefficient of fineness and Cm= midship section area coefficient.(b). The length of a ship Is 18 times the draught while the breadth is 2.1 times the draught. At the load water plane, the water plane area coefficient is 0.83 and the difference between the TPC in sea water and the TPC in fresh water is 0.7. Determine the length of the ship and the TPC in freshwater.
Numerical 2: With respect to Ship Propulsion: (a). Explain the various efficiencies associated with propeller and shafting arrangement. (b). When a propeller of 4.8 m pitch turns at 110 rpm, the apparent slip is found to be —S and the real slip is 1.5 S. If the wake speed is 25% of the ship speed, calculate the ship speed, apparent slip and the real slip.
Numerical 3: A ship 150 m In length, 24 m breadth, displaces 25000 tonne when floating at a draught of 9 m in sea water of density 1025 kg/m3. The ship's propeller has a diameter of 5.8 m, a pitch ratio of 0.9 and a blade area ratio of 0.45. With the propeller operating at 2 rev/sec, the following results were recorded: Apparent slip ratio = 0.06 Thrust power = 3800 kW Propeller efficiency 64% The Taylor wake fraction is given by: Wt= 0.5Cb -0.05.Calculate EACH of the following for the above condition:(a) The ship's speed;(b) The real slip ratio;(c) The thrust per unit area of blade surface;(d) The torque delivered to the propeller.
Numerical 4: The speed of a ship is increased to 18% above normal for 7.5 hours, and then reduced to 9% below normal for 10 hours. The speed is then reduced for the remainder of the day so that the consumption for the day is the normal amount. Find the percentage difference between the distance traveled in that day and the normal distance traveled per day.
Numerical 5: With reference to fixed pitch propellers: (a). Explain Propeller Slip and Propeller Thrust (b). The shaft power of a ship is 3000 KW, the ship speed V is 13.2 knot Propeller rps is 1.27. Propeller pitch is 5.5m and the speed of advance is 11 Knots. Find: (i). Real Slip (ii). Wake fraction (iii). Propeller thrust, when its efficiency, η = 70 %.
Numerical 6: The following data applies to a ship operating on a particular voyage with a propeller of 6 m diameter having a pitch ratio of 0.95: propeller speed =1.8 revs/s. Real slip = 34 %, Apparent slip = 7%, Shaft power =10000 kW, Specific fuel consumption = 0.22 kg/kW-hr. Calculate EACH of the following: (a) The ship speed in knots; (b) The Taylor wake fraction; (c) The reduced speed at which the ship should travel in order to reduce the voyage consumption; (d) The voyage distance if the voyage takes 3 days longer at the reduced speed; (e) The amount of fuel required for the voyage at the reduced speed.
Numerical 7:A ship of length 140m , Breadth of 18.5m, draught of 8.1m and a displacement of 17,025 tonnes in sea water, has a face pitch ratio of 0.673. Diameter of the Propeller is 4.8m. The results of the speed trial show that true slip may be regarded as constant over a range of 9 to 13 knots and is 30%, w = 0.5Cb-0.05. If fuel used is 20t/day at 13 knots and fuel consumption/day varies as cube of speed of ship, Determine the fuel consumption, when the propeller runs at 110 rpm.
Numerical 8: A box shaped vessel, 50 metres long × 10 metres wide, floats in salt water on an even keel at a draft of 4 metres. A centre line longitudinal watertight bulkhead extends from end to end and for the full depth of the vessel. A compartment amidships on the starboard side is 15 metres long and contains cargo with permeability 30%. Calculate the list if this compartment is bilged. KG = 3 metres.
Numerical 9: A ship 120m long floats has draughts of 5.50m forward and 5.80m aft; MCT1 cm 80 tonne m, TPC 13, LCF 2.5m forward of midships. Calculate the new draughts which a mass of 110 tonne is added 24m aft of midships.
Numerical 10: An oil tanker 160m long and 22m beam floats at a draught of 9m in seawater. Cw is 0.865. The midships section is in the form of a rectangle with 1.2m radius at the bilges. A midships tank 10.5m longhas twin longitudinal bulkheads and contains oil of 1.4 m3/t to a depth of 11.5m. The tank is holed to the sea for the whole of its transverse section. Find the new draught.
Numerical 11: A ship of 14900 tonne displacement has a shaft power of 4460 Kw at 14.55 knots. The shaft power is reduced to 4120 kW and the fuel consumption at the same displacement is 541 kg/h. Calculate the fuel coefficient for the ship.
Numerical 12: A ship of 15000 tonne displacement has an Admiralty Coefficient, based on shaft power, of 420. The 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 2550kW. Calculate: (i) Indicated power; (ii) Effective power; (iii) Ship speed.
Numerical 13: 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.
Numerical 14: The immersed cross-sectional area of a ship 120m long, commencing from aft are 2,40,79,100,103,104,104,103,97, 58 and 0 m2 calculate – (i) Displacement; (ii) Longitudinal position of the centre of buoyancy.
Numerical 15:A forward deep tank 12 m long extends from a longitudinal bulkhead to the ship’s side. The widths of the tank surface measured from the longitudinal bulkhead at regular intervals are 10, 9, 7, 4 and 1 m. Calculate the second moment of area of the tank surface about a longitudinal axis passing through its centroid.
Numerical 16. A ship of length 140m , Breadth of 18.5m, draught of 8.1m and a displacement of 17,025 tonnes in sea water, has a face pitch ratio of 0.673. Diameter of the Propeller is 4.8m. The results of the speed trial show that true slip may be regarded as constant over a range of 9 to 13 knots and is 30%, w = 0.5CB- 0.05. If fuel used is 20t/day at 13 knots and fuel consumption/day varies as cube of speed of ship, Determine the fuel consumption, when propeller runs at 110 rpm.
Numerical 17: A. Describe briefly the significance of the factor of subdivision.B. 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.
Numerical 18: A ship of displacement 10,010 tones has a container of 10t at KG = 7.5m. The container is shifted transversely. A pendulum of length 7.5m defects through 13.5m. GM of ship = 0.76m, KM = 6.7m. Find the distance through which the container shifted. Also find the new KG if the container is removed.
Numerical 19: A ship of 15000 tonne displacement has righting levers of 0, 0.38, 1.0, 1.41 and 1.2 m at angles of hell of 0°, 15°, 30°, 45° and 60° respectively and an assumed KG of 7.0 m. The vessel is loaded to this displacement but the KG is found to be 6.80m and GM 1.5m – (i) Draw the amended stability curve; (ii) Estimate the dynamic stability at 60°.
Numerical 20: A ship of 11200 tonne displacement has a double bottom tank containing oil. Whose centre of gravity is 16.5m forward and 6.6m below the centre of gravity of the ship? When the oil is used the ship’s centre of gravity moves 380mm calculate – (a) The mass of oil used; (b) The angle which the centre of gravity moves relative to the horizontal.
Numerical 21. With the aid of sketches: a. Explain various lines plan. b. The half -breadths of water plane of a ship of 120m length ad 15m breadth are given below:b. The half -breadths of water plane of a ship of 120m length ad 15m breadth are given below:Station 0 1 2 3 4 5 6 7 8. Half breadths 1.6 2.8 5.5 6.4 7.3 6.2 4.2 2.0 0 Calculate i) Water plane area ii) TPC in salt water iii) Cw iv) LCF from Mid-ship.
Numerical 22: A ship 100 m long floats at a draught of 6 m and in this condition the immersed cross-sectional areas and water plane areas are as given below.The equivalent base area (Ab) is required because of the fineness of the bottom shell.
Calculate EACH of the following:
(a) The equivalent base area value Ab.
(b) The longitudinal position of the center of buoyancy from midships.
(c) The vertical position of the centre of buoyancy above the base.
Numerical 23: A ship travelling at 15.5 knots has a propeller of 5.5 m pitch turning at 95 rev/min. The thrust of the propeller is 380 KN and the delivered power 3540 KW. If the real slip is 20% and the thrust deduction factor O.198, calculate the Quasi Propulsive Coefficient (QPC) and the wake fraction.
Numerical 24: A ship of 8,000 tonnes displacement takes the ground on a sand bank on a falling tide at an even keel draft of 5.2 metres. KG 4.0 metres. The predicted depth of water over the sand bank at the following low water is 3.2 metres. Calculate the GM at this time assuming that the KM will then be 5.0 metres and that mean TPC is 15 tonne.
Numerical 25: A ship of length 120 m displaces 11750 tonne when floating in sea water of density 1025 kg/m3-cube. The center of gravity is 2m above the center of buoyancy and the water plane is defined by the following equidistant half-ordinates given in Table station AP 1, 2, 3, 4, 5, 6, 7 .FP½ breadth (m) 3.3, 6.8, 7.6, 8.1, 8.1, 8.0, 6.6, 2.8, 0. 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).The center of gravity is 2m above the center of buoyancy.
Numerical 26: A box shaped vessel is 80 m long, 12 m wide and floats at a draught of 4m. A full width midships compartment 15 m long is bilged and this results in the draught increasing to 4.5 m. Calculate EACH of the following,
(a) The permeability of the compartment;
(b) The change in metacentric height due to bilging.
Numerical 27: A ship 100 m long floats at a draught of 6 m and in this condition the immersed cross-sectional areas and water plane 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.The equivalent base area (Ab) is required because of the fineness of the bottom shell.Section AP 1 2 3 4 5 FPImmersed cross section area(m2) 12 29 64 78 70 48 0Draught (m) 0 0.6 1.2 2.4 3.6 4.8 6.0Waterplane area (m2) Ab 560 720 876 942 996 1028Calculate EACH of the following:(a) The equivalent base area value Ab; (b) The longitudinal position of the center of buoyancy from midships; (c) The vertical position of the center of buoyancy above the base.
Numerical 28: A box shaped vessel 65m X 12m X 8m has KG 4m and is floating in sea water upright on an even keel at 4m draft F and A. Calculate the moments of statically stability at i. 5 degrees and ii. 25 degrees heel.
Numerical 29: A propeller 4.6m diameter has a pitch of 4.3m and boss diameter of 0.75. The real slip is 28% at 95 rev/min. Calculate the speed of advance, thrust and thrust power.
Numerical 30: A ship 130m long displaces 14000 tonne when floating at draughts of 7.5m forward and 8.10m aft. GML –125m, TPC – 18,LCF-3m aft of midships. Calculate the final draughts when a mass of 180 tonne lying 40m aft of midships is removed from the ship.
Numerical 31: A ship of 9.900 tonnes displacement has KM 7.3m, and KG 6.4m. She has yet to load two 50 tonne lifts her own gear and the first lift is to be placed on deck on the inshore side (KG = 9m and centre of gravity = 6m, out from centre line). When the derrick plumbs the quay, its head is 15m above the keel and 12m out from centre line. Calculate the maximum list during the operation.
A box barge 45 m long and 15 m wide floats at a level keel draught of 2 m in sea water, the load being uniformly distributed over the full length. Two masses, each of 30 tonne, are loaded at 10 m from each end and 50 tonne is evenly distributed between them. Sketch the shear force diagram and give the maximum shear force.
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