Certification Assistance for Marine Engineers

Canadian First Class ME 
Naval Architecture

In Canada, Transport Canada administers the Marine Engineering examination process; visit the Training Page for details on the process. The actual exam consist of nine (9) questions randomly drawn from a question bank of the various subject. Six (only) must be answered in a 3.5hrs time frame. The exam questions are similar to these, presented below, and are drawn heavily from similar question in the Reed's Marine Engineering series of books. 
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Disclaimer
Transport Canada has ask us to advise users of this webpage to keep in mind that these questions are not the exact questions found in their exams. Martin's Marine Engineering Page - www.dieselduck.net is not affiliated with Transport Canada and these questions have been gathered from various sources.

1 What are the three principal sources of shipboard vibration. Explain how you would trace the forces causing these vibrations and the measures that could be taken to reduce the severity of the vibrations.

2 Describe with the aid of sketches a keyless propeller showing how it is fitted to the tail shaft. Discuss the advantages of this design. Describe the method of driving the propeller on to the shaft and how it is locked in position.

3 Derive the Admiralty Coefficient formula and show how this may be modified to suit a fast ship. B) A ship of 14500 tone displacement requires 24000KW to drive it at 24 knots. Using the modified Admiralty Coefficient formula, calculate the shaft power required for a similar ship of 12000 tone displacement at 20 knots.

4 Describe the procedure for calculating the righting moments for a vessel for a) small angles of heel b) large angles of heel. A homogeneous block of wood (relative density 0.64) 3.0m long, 0.9 m wide and 0.5m deep floats in water of density 1025kg/m3. Calculate the righting moment when it is heeled to an angle of 8.5 degrees.

5 What is meant by 'stability'? a) what factors does a Master need to know before commencing a voyage b) what factors affect stability c) a vessel has take a sudden list in calm seas, what could be the reasons?

the 5 above submitted Aug 2005

C01. For a ship 55 m in length, floating in water of 1025 kg/m3 the equally spaced half-breadths of the waterplane commencing from the after perpendicular are: 0.0  6.9  7.2  6.0  0.0  m respectively.  Calculate the position of the longitudinal center of flotation from midships and the moment to change trim one centimeter

note: assume KB = KG

 

C02. For a ship 61 m in length floating in water of 1025 kg/m3 the equally spaced half-breadths of the waterplane commencing from the after perpendicular are: 0.0  7.0  7.4  6.1  0.0 respectively.  Calculate the position of the longitudinal center of flotation from midships and the moment to change trim one centimeter. 

note: assume KB = KG.

 

C03. For a ship 65 m long the half ordinates starting from aft are:  0  2  5  3  0 .  Find the LCF.  Given KB = KG find the moment required to change the trim 1 cm.

 

C04. For a ship 65 m in length floating in water of 1025 kg/m3 the equally spaced half-breadths of the waterplane commencing from the after perpendicular are: 0.0  7.6  8.0  6.7  0.0 m respectively.  Calculate the position of the longitudinal center of flotation from midships and the moment to change trim one centimeter.

Assume KB = KG

REF:JAN91

 

C05. For a ship 85 m long the load waterplane is defined by the following equidistant half breadths commencing from the after perpendiculars:

Stn  AP   1    2    3    4     5     6     7    8    9   FP
Half  0   5  10.5  15  16.3  16.5  15.2  12.9  7.7  3.4  0

breadths (m)

Determine:

bullet

a) the area of the waterplane bullet

b) the position of the longitudinal center of flotation bullet

c) the second moment of area of the waterplane about the center of flotation.

REF:FEB82

 

C06. A ship of 90.0 m length between perpendiculars contains ballast water in a forward compartment and has the following equidistant half areas of immersed sections commencing at the

after perpendicular. 0.3  7.5  21.3  33.4  40.7  45.4  48.3  51.9  51.0  34.3  0.0

respectively. If prior to ballasting the ship's displacement was 5600 tonnes and the position of the longitudinal center of buoyancy (LCB) was 4.6 m forward of midships, calculate:

bullet

a) the mass of water of density 1025 kg/m3 added as ballast bullet

b) the distance of the center of gravity of the ballast water contained in the forward compartment from midships.

 

C07. A ship 90 m long with half breadths of the load waterplane

commencing from aft are: 0.3  3.8  6.0  7.7  8.3  9.0  8.4  7.8  6.9  4.7  0.0

respectively.  Calculate:

bullet

a) the area waterplane bullet

b) distance of LCF from midships bullet

c) 2nd moment of area about a transverse axis through the LCF

REF:83 REF:FEB90

 

C08. A ship 90 m long displaces 8300 tonnes when floating in seawater at draught of 5.2 m forward and 5.6 m aft.  GML = 97, TPC = 10.0, LCF = 3.0 m aft of midships. It is decided to

ballast the ship to a draught aft of 5.95 m to submerge the propellor.  If ballast tank 34 m aft of midships is available.  Calculate the least amount of water required and final draught forward.

REF:FEB82 REF:APR91

 

C09. For a ship 92 m long the load waterplane is defined by the following equidistant half breadths commencing from the after perpendiculars:

AP    1     2     3     4     5     6     7    8    9    
FP 0.0  5.5  11.0  15.5  16.8  17.0  15.7  13.4  8.2  3.9  0.0

Determine:

bullet

a) the area of the waterplane bullet

b) the position of the longitudinal center of flotation bullet

c) the second moment of area of the waterplane about the center of flotation

 

C10. A ship of 93 m length between perpendiculars contains ballast water in a forward compartment and has the following equidistant half areas of immersed sections commencing at the

after perpendicular 0.6  7.8  21.6  33.7  50.0  45.7  48.6  52.2  51.3  34.6  0.0  meters2.  If prior to ballasting the ship's displacement was 6000 tonnes and the positions of the longitudinal center of buoyancy  (LCB) was 4.6 m forward of midships, calculate:

bullet

a) the mass of water of density 1025 kg/m3 added as ballast bullet

b) the distance of the center of gravity of the ballast water contained in the forward compartment from midships.

 

C11. A ship 100 m long and 15 m beam floats at a mean draft of 3.5 m.  The half ordinates of waterplane at equal intervals are: 0.0  3.0  5.5  7.3  7.5  7.5  7.5  7.05  6.10  3.25  0.0 respectively.  The section amidships is constant and parallel for 20 m and the submerged cross sectional area is 50 m2 at this section.  Calculate the new mean draft when a midships

compartment 15 m long is opened to the sea.  Assume the vessel to be wall sided in the region of waterplane.

REF:83 REF:JAN87

REF:Reeds P300 Q33

 

C12. For a ship 100 m long the load waterplane is defined by the following equidistant half breadths commencing from the after perpendiculars:

AP    1     2     3     4     5     6     7    8    9    FP
0.0  6.0  11.5  16.0  17.3  17.5  16.2  13.9  8.7  4.4  0.0

Determine:

bullet

a) the area of the waterplane bullet

b) the position of the longitudinal center of flotation bullet

c) the second moment of area of the waterplane about the center of flotation

REF:OCT92

 

C13. A ship 100 m long floats at draughts of 4.15 m forward and 4.5 m aft.  MTC 1 cm = 60 t/m, TPC = 10, LCF = 2.0 m forward of midships.  Calculate the new draughts after the following masses have been placed onboard:

bullet

15 tonnes 38 m aft of midships bullet

55 tonnes 30 m aft of midships bullet

30 tonnes 3 m aft of midships bullet

60 tonnes 10 m forward of midships bullet

20 tonnes 35 m forward of midships

 

C14. A ship 110 m long displaces 10500 tonnes and has a wetted surface area of 2900 m2 At 15 knots the shaft power is 4000 kW, propulsive coefficient = 0.6 and 60% of the thrust is available to overcome the frictional resistance. Calculate the shaft power required for a similar ship 140 m long at the corresponding speed.  Given that f = 0.42 and n = 1.825

REF:MAR89

 

C15. A ship 120 m long and 18 m beam floats at a mean draught of 3.8 m  The semi-ordinates of the waterplane at equal intervals are: 0.0  3.2  5.6  7.5  7.8  7.8  7.2  6.0  3.1  0.0m

respectively.  The midships section is constant and parallel for 24 m, the immersed cross-sectional area at this section is 58 m2.  Assuming the ship to be wall-sided in the region of the waterplane, Calculate the new mean draught when a  midships compartment 20 m long is opened to the sea.

 

C16. A ship 120 m long has a light displacement of 4000 tonnes and LCG in this condition 2.5 m aft of midships.  The following items are then added:

bullet

cargo  10000 tonnes LCG  3.0 m forward of midships bullet

fuel    1600 tonnes LCG  2.0 m aft     of midships bullet

water    400 tonnes LCG  8.0 m aft     of midships bullet

stores   100 tonnes LCG 10.0 m forward of midships

Using the following hydrostatic data, calculate the final draughts.

DRAUGHT DISPLACEMENT MCTI  LCB from midships  LCF from midships

8.50    16650        183   1.94 forward      1.20 aft
8.00    15350        175   2.10 forward      0.06 forward

REF:Reeds P301 Q41

 

C17. A ship of length, 120 m floats in water of relative density 1.025, the forward draft is 6.8 m and the after draft is 7.4 m.  For the conditions stated, the hydrostatic data is: 

TPC  = 18.1

MTCI = 125 tm

LCF  = 2 m aft of midships

Calculate the distance from amidships at which a mass may be added to the ship without altering the draft aft.  Assume the position of the LCF remains unaltered

REF:Specimen 3

 

C18. A ship 120 m long floats at draughts of 8.0 m forward and 8.8 m aft.  MCTI cm = 150 tm, TPC = 28 and LCF = 1.5 m aft of midships.  It is desired to ballast to an even keel and there is a tank available 55 m forward of midships.  Find the mass of water required to ballast to the even keel and the new draughts.

REF:FEB89

 

C19. A vessel 125 m long floats at a draught of 8.5 m in sea water of density 1025 kg/m3 and has the following hydrostatic data:

Draught     Displacement (tonne)

8.5                 14500
8.0                 13430

At a draught of 8.5 m the KB is 4.46 m and the KM is 7.35 m. Calculate the KB and KM at the 8.0 m draught, assuming the vessel is wall-sided between the above two waterlines.

 

C20. A ship 125 m in length, 18.5 m breadth and 7.5 m draught in water of density 1025 kg/m3 has a block co-efficient of 0.5.  During trials the following results were obtained:

Ship speed (knots)    16    17    18    19
Effective power (kW) 2420  3000  3750  4600(naked)

Determine for a ship of similar form having a displacement of 15500 tonnes and operating at a

corresponding speed of 19.5 knots:

bullet

a) length bullet

b) breadth bullet

c) draught bullet

d) the effective power

REF:FEB82 REF:JAN87

REF:Munro Smith P3.159 Q30

 

C21. A ship 130 m long displaces 9600 tonnes.  It loads in fresh water of 1000 kg/m3 to a level keel draught of 7 m.  It then moves into sea water of 1025 kg/m3.  T.P.C. in sea water is 17.  MCT 1 cm is 124 t/m.  LCF is 0.5 m aft of midships and LCB is 2.4 m forward of midships.  Calculate the forward and aft draughts in sea water.

 

C22. A ship 130 m long displaces 12200 tonnes.  When a mass of 105 tonnes is moved 75 m from forward to aft, there is a change in trim (by the stern) of 60 cm.  Calculate:

bullet

a) M.C.T. 1cm bullet

b) the longitudinal metacentric height bullet

c) the distance moved by the C. of G. of the ship

REF:FEB89

 

C23. A ship 130 m long displaces 14500 tonnes when floating at draughts of 7.7 m forward and 8.2 m aft.  GML = 127 m TPC = 19, LCF = 2.5 m aft of midships. Calculate the final draughts when a mass of 190 tonnes lying at 35 m aft of midships is removed from the ship.

 

C24. For a ship 135 m in length, 16.0 m breadth the values of tonnes per centimeter immersion (TPC) in water of density 1025 kg/m3 are as follows:

Draught  1.2   1.8    2.4   3.0    3.6    4.2   4.8
TPC     14.6  14.83  15.1  15.36  15.54  15.7  15.82

The displacement of the ship below the 1.2 m draught is 1200 tonnes.  If at a draught of 4.8 m  the position of the longitudinal center of buoyancy below the metacenter (BML) is 140 m and the second moment of area of the waterplane about midships is 935000 m4, calculate:

bullet

a) the distance of the longitudinal center of flotation (LCF) from midships bullet

b) the block coefficient

 

C25. For a ship 135 m in length, 15 m breadth the values of tonnes per centimeter immersion (TPC) in water of density 1025 kg/m3 are as follows:

Draught  1.3   2.1    2.9    3.7
TPC     15.2  15.46  15.63  15.72

The displacement of the ship below the 1.3 m draught is 1400 tonnes.  If at a draught of 3.7 m the position of the longitudinal center of buoyancy below the metacenter (BML) is 138 m and the second moment of area of the waterplane about midships is 710000 m4, Calculate:

bullet

a) the distance of the longitudinal center of flotation (LCF) from midships bullet

b) the block coefficient

 

C26. For a ship 137 m long the equally spaced half-breadths of the waterplane commencing from the after perpendicular are: 0.0  6.71  8.9  9.45  9.6  9.6  9.6  9.56  8.85  5.18  0.0 m respectively.  Calculate the change in end draughts of the ship if a mass of 300 tonnes is loaded on the center line at a position 30 m aft of the mid-ships.  The moment to change

trim 1 cm is 147 tonnes-m.

 

C27. For a ship 137 m in length, 15.3 m breadth the values of tonne per centimeter immersion (TPC) in water of density 1025 kg/m3 are as follows:

Draught 1.5    2.3    3.1    3.9
TPC    15.32  15.58  15.75  15.84

The displacement of the ship below the 1.5 m draught is 1500 tonnes.  If at a draught of 3.9 m the position of the longitudinal center of buoyancy below the metacenter (BML) is 140 m and the second moment of area of the waterplane about midships is 730000 m4, calculate:

bullet

a) the distance of the longitudinal center of flotation (LCF) from midships bullet

b) the block coefficient

 

C28. The immersed cross-sectional areas of a ship 140 m long, commencing from aft are:

3  42  81  103  107  109  109  107  94  52  0  m2

Calculate:

bullet

a) displacement bullet

b) longitudinal position of the center of buoyancy from midships

note: sea water density is 1025 kg/m3

 

C29. For a ship 140 m in length, 15.5 m breadth the values of tonne per centimeter immersion (TPC) in water of density 1025 kg/m3 are as follows:

Draught (m)  1.7   2.5    3.3    4.1
TPC         15.5  15.76  16.95  16.1

The displacement of the ship below the 1.7 m draught is 1650 tonnes.  If at a draught of 4.1 m the position of the longitudinal center of buoyancy below the metacenter (BML) is 145 m and the second moment of area of the waterplane about midships is 780000 m4, calculate:

bullet

a) the distance of the longitudinal center of flotation (LCF) from midships bullet

b) the block co-efficient.

 

C30. A ship 140 m long, 16 m beam floats at a draught of 7.6 m in sea water of 1025 kg/m3.  It's block co-efficient is 0.74.  Calculate the power required to overcome frictional resistance at 18 knots if f = 0.422, n = 1.825 and wetted surface area = 2.55 times sqrt of delta L.

 

C31. For a ship 140 m in length, 18 m breadth the values of tonnes per centimeter immersion (TPC) in water of density 1025 kg/m3 are as follows:

Draught 1.2   1.8   2.4    3.0   3.6    4.2   4.8
TPC    15.0  15.4  15.82  16.1  16.36  16.6  16.78

The displacement of the ship below the 1.2 m draught is 1600 tonnes.  If at a draught of 4.8 m the position of the longitudinal center of buoyancy below the metacenter (BML) is 145 m and the second moment of area of the waterplane about midships is 1050000 m4, calculate:

bullet

a) the distance of the longitudinal center of flotation (LCF) from midships bullet

b) the block coefficient

 

C32. For a ship 150 m in length, 20 m breadth the values of tonnes per centimeter immersion (TPC) in water of density 1025 kg/m3 are as follows:

Draught (m) 1.2   2.1   3.0   3.9   4.8   5.7   6.6
TPC        16.5  18.7  19.4  20.0  20.5  21.1  21.7

The displacement of the ship below the 1.2 m draught is 2100 tonnes.  If at a draught of 6.6 m the position of the longitudinal center of buoyancy below the metacenter (BML) is  140 m and the second moment of area of the waterplane about midships is 17750000 m4,  calculate:

bullet

a) the distance of the longitudinal center of flotation from midships bullet

b) the block co-efficient

 

C33. A ship 150 m long has 1/2 ordinates of waterplane of: 1.8  6.0  9.2  10.4  10.8  10.8  10.8  9.8  7.6  4.5  0.0 respectively.  Calculate the second moment of area of the waterplane about the centerline.

 

C34. A ship 150 m long displaces 8200 tonnes when floating in sea water with a density of 1025 kg/m3.  The half ordinates of the waterline are: 0.0  2.4  4.8  7.1  7.5  7.7  7.7  7.6  5.3  2.6  0.0 respectively. While floating at this waterline the ship develops a list of 8o due to instability.  Calculate the negative metacentric height when the vessel is upright in

this condition.

 

C35. A ship 160 m long and 9 m draught has a rudder whose area is one sixtieth of the middle-line plane, the rudder stock is 340 mm diameter.  The distance from the center of the stock to the center of effort of the rudder is 1 m and the maximum rudder angle is 35o.  If the stress in the stock is not to exceed 75 MN/m2.  Calculate the maximum ship's speed.

note: Fn = 580 Av2 (N)

 

C36. A ship 165 m long and 21 m beam floats at a draught of 8.5 m in sea water of 1024 kg/m3 the block co-efficient is 0.7.

bullet

a) if the admiralty co-efficient is 620, calculate the shaft power required at 19 knots bullet

b) if the speed is now increased to 22 knots, and within this speed range resistance varies as the speed cubed, find the new shaft power.

 

C37. The draughts of a ship 180 m long are 7.0 m forward and 7.85 m aft.  MCT 1 cm 310 t/m, TPC 30, LCF 2.5 m forward of midships.  Calculate the new draughts after the following changes in loading.  165 tonnes added 62 m aft of midships, 195 tonnes added 30 m forward of midships, 120 tonnes removed 78 m aft of midships, 75 tonnes removed 15 m aft of midships.

 

C38. A ship of 4800 tonne displacement in salt water of density 1025 kg/m3 has a double bottom tank 15 m long.  The half-breadths of the top of the tank are: 7.5  7.0  5.0  and 3 m respectively.  The tank has a watertight centerline division.  Calculate the free surface effect if the tank is partially full of fresh water on one side only.

 

C39. A ship displacing 4800 tonnes has a rudder area of 11m2. The distance between the center of lateral resistance and the center of the rudder is 1.7 m and the ship's metacentric height is 0.26 m.  If the ship is travelling at 15 knots, calculate the initial angle of heel if the rudder is put over to 35o.  note: Fn = 580 Av2 sin alpha N.

 

C40. For a vessel 5080 tonnes displacement the KM is 6.4 m. The vessel is inclined in sea water of density 1025 kg/m3 by moving a mass of 5.0 tonnes transversely a distance of 14.6 m, causing a pendulum 6.1 m long to deflect 10.2 cm. During the inclining experiment a double-bottom tank 7.3 m long, 9.1 m broad and 1.22 m deep, contains sea water to a depth of 0.61 m.  Determine the KG of the light vessel, if the only dead-weight on the vessel during the experiment is the water in the double bottom tank.

REF:MAR8?

 

C41. For a ship of 5100 tonnes displacement, 120 m in length, the even keel draught in water of density 1025 kg/m3 is 7 m and the center of gravity is 2.5 m  above the center of buoyancy.  Calculate the moment to change trim one centimeter (MCT 1 cm) if the ship's load waterplane is defined by the following half breadths commencing from the after perpendicular:

Station AP    1    2    3    4    5    6    7    FP
Half    3.3  6.8  7.6  8.1  8.1  8.0  6.6  2.8  0.0 breadths

 

C42. A ship of 5500 tonnes displacement has a KM of 6.5 m. When 6 tonnes are moved 16 m across the ship a pendulum 5 m long has a deflection of 10 cm.  A double bottom tank 8.0 m long, 9 m wide and 1.5 m deep is half full of sea water.  For a sea water density of 1025 kg/m3  calculate the KG of the light ship.

 

C43. A ship of 5600 tonnes displacement in sea water has three rectangular double bottom tanks, A is 11 m long and 15 m wide,  B is 13 m long and 14 m wide,  C is 14 m long and 15 m wide.  Calculate the free surface effect for any one tank and state in which order the tanks should be filled with sea water when making use of them for stability correction.

note: sea water density is 1025 kg/m3

 

C44. For a ship of 5600 tonnes displacement, 128 m in length, the keel draught in water of density 1025 kg/m3 is 7 m and the center of gravity is 3.0 m above the center of buoyancy.

Calculate the moment to change trim one centimeter (MCT 1 cm) if the ship's load waterplane is defined by the following half breadths commencing from the after perpendicular:

Station       AP    1    2    3    4    5    6    7    FP
Half breadths 3.7  7.2  8.0  8.5  8.5  8.4  7.0  3.2  0.0

 

C45. A ship of 5800 tonnes displacement in seawater has a rectangular double bottom tank 9.5 m wide and 13 m long, half full of sea water.  Calculate the reduction in metacentric height due to free surfaces.

note: sea water density is 1025 kg/m3

 

C46. A ship of 6800 tonnes displacement in sea water has it's center of gravity 6.1 m above the keel and its transverse metacentre 6.7 m above the keel.  A rectangular double bottom tank 11 m long, 13 m wide and 1.5 m deep is now half filled with sea water.  Calculate the metacentric height.

note: density of seawater = 1025 kg/m3

 

C47. A ship of 6900 tonnes displacement has a K.G. of 3.9 m and a K.M. of 4.7 m.  A mass of 55 tonnes is now lifted from the quay by one of the ship's derricks whose head is 18 m above the keel.  The ship heels a maximum of 9.750 when the mass is lifted.  Calculate the outreach of the derrick from the ship's center line.

 

C48. A ship of 7000 tonnes displacement has a wetted surface area of 2800 m2 and a speed of 16 knots.  Calculate:

bullet

a) the corresponding speed and wetted surface of a similar ship of 2500 tonnes displacement. bullet

b) if the skin resistance is of the form R = 0.45 SV1.83 N, find the resistance of the 7000 tonne ship.

 

C49. A ship of 7200 tonne displacement in sea water has a double bottom tank 15 m long, 11 m wide and 1.5 m deep full of sea water.  The center of gravity is 6.6 m above the keel and the metacentric height is 0.5 m.  Assuming that the K.M. remains constant, calculate the new G.M. if half of the water is pumped out of the tank.

note: density of sea water is 1025 kg/m3.

 

C50. A ship of 7500 tonnes displacement, 105 m long, floats in sea water of 1025 kg/m3 at draughts of 5.6 m forward and 6 m aft.  The TPC is 16, LCB is 0.5 m aft of midships, LCF is 3

m aft of midships and MCT 1 cm is 63 t/m.  The ship now moves into river water of 1008 kg/m3.  Calculate the distance a mass of 60 tonnes must be moved to bring the ship to an even keel and determine the final draught.

 

C51. A ship of 8100 tonnes displacement floats upright in sea water.  KG = 7.5 m and GM = 0.45 m.  A tank, whose center of gravity is 0.5 m above the keel and 4 m from the centerline contains 100 tonnes of water ballast.  Neglecting free surface effect, calculate the angle of heel when the ballast is pumped out.

REF:APR89

 

C52. A ship of 8400 tonnes displacement floats upright in sea water of density 1025 kg/m3.  KG = 7.8 and GM = 0.6 m.  A tank, whose center of gravity is 0.75 m above the keel and 5 m from the centerline, contains 120 tonnes of water ballast.  Neglecting free surface effect, calculate the angle of heel when the ballast is pumped out.

REF:OCT92

 

C53. For a ship of 8500 tonnes displacement in water of density 1025 kg/m3 the metacentric height is 0.6 m. Calculate the final effective G.M. of the ship is 100 tonnes of oil fuel of density 900 kg/m3 is pumped from an initially full rectangular double bottom tank 24.3 m long by 12.2 m wide by 1.2 m deep into an initially empty tank 9.15 long, 9.15 m wide by 7.6 m deep, situated on top of the double bottom tank.

REF:JAN91

 

C54. A ship of 9000 tonne displacement in sea water has it's center of gravity 4.8 m above the keel and transverse metacentre 5.4 m above the keel when a rectangular tank 8 m long and 15 m wide contains sea water.  A mass of 12 tonne is moved 13 m across the deck.  Calculate the angle of heel:

bullet

a) when the tank is pressed up bullet

b) if the water does not fill the tank

note: sea water density 1025 kg/m3

 

C55. A vessel of 10600 tonnes displacement floats upright at a draught of 6.7 m in water of density 1025 kg/m3. A mass of 140 tonnes is loaded by the ship's derrick which has a maximum outreach of 9.14 m.  The point of suspension is at 18.29 m above the keel.  Estimate using the hydrostatic data below, the angle of heel after the derrick at maximum outreach, has just lifted the mass from the quay.  The ship's hydrostatic data before lifting the mass is: Center of buoyancy above the keel  (KB) = 3.4  m Center of gravity above the keel   (KG) = 3.66 m Tonnes per centimeter immersion   (TPC) = 20.0

note: second moment of area of the waterplane about the centerline is 22788 m4

 

C56. A ship whose displacement is 10800 tonne when floating in sea water of density 1025 kg/m3 has a double-bottom tank containing oil, whose center of gravity is 15.6 m forward and 6.2 m below the center of gravity of the ship.  When the oil is use the ship's center of gravity moves 360 mm.  Calculate:

bullet

a) the mass of oil used bullet

b) the angle which the center of gravity moves relative to the horizontal

REF:FEB82

 

C57. A ship of 11000 tonnes displacement in sea water has KM = 8.1 m and GM = 0.5 m.  A rectangular double bottom tank is 1.5 m deep, 19 m long and 14 m wide.  Assuming that the KM remains constant, calculate the new GM when the tank is:

bullet

a) filled with sea water bullet

b) half filled with sea water

note: sea water density 1025 kg/m3

 

C58. A ship of 12200 tonnes displacement and length 110 m floats in seawater density 1025 kg/m3.  The half ordinates of the load water plane at equal intervals commencing from aft are: 0.0  5.27  7.2  7.72  7.45  5.5  0.0   respectively.  Given that KB = 5.7 and KG = 6.1 m.  Calculate:

a) the T.P.C.

b) the GM

 

C59. A ship of 12000 tonnes has a locomotive of 180 tonnes loaded in the center on the tank top below a derrick whose head is 18 m above the center of gravity of the load.  Find the shift in center of gravity when:

bullet

a) the load is just clear of the tank top bullet

b) the load is lifted to the deck head bullet

c) the load is landed on a deck 12 m above the tank top bullet

d) the load is swung 14 m outboard

REF:MAY81

 

C60. A ship of 12500 tonnes displacement has a rudder 16 m2 in area, whose center is 5.2 m below the waterline.  The ship's GM is 0.4 m and the center of buoyancy is 3.4 m below the waterline.  When travelling at 22 knots the rudder is turned through 300 allowing 20% for the race effect,

calculate the initial angle of heel given that Fn = 577 AV2 sin alpha (N).               REF:MAR89

 

C61. A ship of 12500 tonnes displacement and length 110 m floats in sea water of density 1025 kg/m3.  The half ordinates of the load waterplane at equal intervals commencing from aft are: 0.0  5.4  7.35  7.9  7.6  5.85  0.0 respectively.  Given that KB = 5.8 m and KG = 6.0 m.  Calculate:

bullet

a) tonnes per centimeter immersion bullet

b) metacentric height.

 

C62. A ship of 12190 tonnes displacement in seawater of density 1025 kg/m3 has the following hydrostatic data:

KB = 3.66 m KM = 7.31 m

KG = 6.1 m

Four hundred tonnes of liquid cargo of density 890 kg/m3 is pumped into a midships rectangular center deep tank 21.34 m long by 15.2 m wide by 2.44 m high.  If the bottom of the deep tank is 6.7 m above the keel, calculate the new metacentric height of the ship.  Assume the KB is

proportional to the displacement and the BM is inversely proportional to the displacement.

 

C63. A ship of 13000 tonnes displacement and length 120 m floats in sea water of density 1025 kg/m3.  The half ordinates of the load waterplane at equal intervals commencing from aft are:

0.0  5.5  7.45  8.0  7.7  5.8  0.0  m respectively.  Given that KB = 5.9 and KG = 6.2 m.  Calculate:

bullet

a) tonnes per centimeter immersion bullet

b) metacentric height

REF:FEB90

 

C64. A ship of 13000 tonnes displacement and 16 m beam has a GM of 1.4 m.  A mass of 90 tonnes is lifted from its position in the center of the lower hold by one of the ship's derricks, and placed on the quay 3 m from the ship's side. The ship heels to a maximum angle of 3.5o when the mass is being moved.

bullet

a) does the GM alter during the operation? bullet

b) calculate the height of the derrick head above the original center of gravity of the mass.

 

C65. The following ordinates define the GZ curve of a ship of 13000 tonnes displacement:

Angle of inclination    0    15   30    45    60    75    90
Righting lever GZ (mm)  0   140  315   430   250  -200  -800

Calculate the angle of list and the range of stability if, due to adverse weather conditions, 500 tonnes of cargo shifts horizontally, a distance of 5 m and vertically, a distance of 3 m downwards.

REF:Specimen 1

 

C66. A ship of 15000 tonnes displacement has a G.M. of 0.4 m, it's center of lateral resistance is 4.5 m above the keel. The ship's rudder has an area of 21 m2 and it's centroid is 2.4 m above the keel.  Calculate the angle of heel of the ship due to the force on the rudder if the rudder is put hard over to port to a maximum angle of 35 degrees when the ship is travelling at 23 knots.

note: Fn = 580 Av2

REF:APR91

 

C67. A ship of 15000 tonnes displacement is 130 m long and floats at draughts of 8.1 m forward and 8.6 m aft.  The TPC is 20, GML is 120 m and LCF is 4 m forwards of midships.  It is required to bring the ship to an even keel draught of 8.6 m.  Calculate the mass which must be added and the distance of the center of the mass from mid-ships.

 

C68. A ship of displacement 20000 tonnes and length 150 m floats at draughts forward and aft of 6.92 m and 7.69 m respectively, in water of density 1025 kg/m3.  Given that the LCF forward of midships = 3.46 m, LCB forward of midships = 0.52 m, TPC = 16.2, MTC 1 cm = 183 tonnes/m.  Calculate the draughts forward and aft when the ship passes into the water of density 1000 kg/m3.

REF:MAY91

 

C69. A vessel of 22000 tonnes displacement floats in sea water of density 1025 kg/m3.  A deep wing tank 18 m long and 9.25 m deep has a constant cross sectional area defined by the following equidistant breadths: 4.4  4.25  4.0  3.35  2.6 m. One side of the tank forms a longitudinal bulkhead parallel to and 5.0 m from the ship's centerline, while the other side forms part of the ship's shell.  The bottom of the tank is 2.0 m above the keel.  The ship has a K.M. of 5.0 m and a G.M. of 0.61 m when the tank is full of oil of density 900 kg/m3.  Calculate the ship's K.G. and the angle of heel when half of the oil in the wing tank has been used. The free surface effect can be neglected and the K.M. assumed to remain constant.

 

C70.  A ship of 80000 tonnes displacement has a double bottom tank which holds fuel oil. The center of gravity is 15.8 m forward and 7.5 m below the center of gravity of the ship. When the oil is used the ships center of gravity moves 365 mm.  Calculate:

bullet

a) the mass of oil used bullet

b) the angle which the center of gravity moves relative to the horizontal

REF:FEB90

 

C71. A tanker of 42660 tonnes displacement gives the following results when on trial:

Speed (knots)   13.5  14.0  15.0   16.0   17.0   17.5

Propeller RPM   86.1  89.5  96.7  104.8  113.6  118.5

The ship is taken on a six hour continuous run during which time 15.25 tonnes of fuel are used and the engine makes 38520 revolutions. Calculate:

bullet

a) fuel co-efficient for the ship bullet

b) the fuel required per day for a similar ship 35000 tonnes displacement which is fitted with a similar type engine and which runs at the corresponding speed.

REF:MAY91

 

C72. An oil tanker 27 m wide displaces 28500 tonne in sea water of density 1025 kg/m3 when loaded in nine equal tanks each 10 m long, with oil R.D. 0.85.  Calculate the total free

surface effect with:

bullet

a) no longitudinal bulkheads bullet

b) a longitudinal centerline bulkhead bullet

c) twin longitudinal bulkheads forming three equal tanks bullet

d) twin longitudinal bulkheads, the center tank having a width of 13 m

 

C73. The displacement of a ship at draughts of :

0    1    2    3    and  4  m  are:
0  201  456  710    and 998 tonnes.

Calculate the distance of the center of buoyancy above the keel when floating at a draught of 4 m, given:

VCB below waterline = Area between Displacement curve and Draught axis Displacement

 

C74. A ships speed is increased 12% above normal for 11 hours then the speed is reduced to 10% below normal for 9 hours and for the remainder of the day at normal speed.  If the fuel consumption is normal for the day, find the percentage difference for the day's run.

REF:FEB89

 

C75. The following data applies to a ship operating at a speed of 15 knots:

Shaft power = 3050 kW        Propeller speed = 1.58 rev/sec

Propeller thrust = 360kN     Apparent slip   = 0.05

Calculate:

bullet

a) the propeller pitch bullet

b) real slip bullet

c) quasi propulsive co-efficient if the Taylor wake fraction and thrust deduction factor are 0.31 and 0.2 respectively. bullet

d) briefly comment on the significance of negative apparent slip.

 

C76. A ship travelling at 15 knots has a metacentric height of 0.4 m.  The distance between the center of gravity and the center of lateral resistance is 2.5 m.  If the vessel turns in a circle of 1300 m diameter, calculate the angle to which it will heel.

 

C77. A ship is driven through the water at a rate of 15.5 knots by a propeller of 5.5 m pitch rotating at 95 R.P.M..  The power delivered by the propeller amounts to 3540 kW when the thrust is 380 kN.  The thrust deduction factor is 0.198 and the actual slip is 20%.  Calculate:

bullet

a) the quasi propulsive co-efficient (Q.P.C.) bullet

b) the wake fraction

 

C78. A ship travelling at 18 knots turns with a radius of 470 m when the rudder is put hard over. The center of gravity is 7.2 m above the keel, the transverse metacentre is 7.6 m above the keel and the center of buoyancy is 3.9 m above the keel.  If the centripetal force is assumed to act at the center of buoyancy, neglecting the rudder force calculate the angle of heel when turning.

 

C79. A ship with a maximum speed of 17 knots has a rudder area of 24 m2.  The center of effort is 1.2 m from the stock centerline when the rudder is turned to 35o.  Allowing 18% for race effect calculate the diameter of the stock if the maximum allowable stress is 72 MN/m2.  If the effective diameter of the stock is reduced to 370 mm, calculate the maximum speed that the ship may travel so that the above stress is not exceeded.

note: Fn = 577 AV2 sin a (N)

 

C80. A ship, whose maximum speed is 20 knots, has a rudder area of 28 m2.  The distance from the center of the stock to the center of the rudder is 1.5 m and the maximum rudder angle is 350.  If the maximum allowable stress in the rudder stock is 88 MN/m2 calculate the diameter of the stock. The rudder force parallel to the centerline of the ship (Fn) = 580 Av2 (N)

REF:MAY91

 

C81. A ship floats with an even keel draught of 8 m in sea water of density 1025 kg/m3.  The longitudinal center of flotation is at midships MCT 1 cm is 185 t/m and TPC = 16. Calculate the magnitude and position of a mass to be added in order to bring the ship to draughts of 8.3 m aft and 7.9 m forward.

 

C82. A ship model is towed at the rate of 3.6 knots through fresh water in a towing tank installation.  The model is 6 m long and the total resistance during this operation is 40 N. A ship designed along the same lines as the model is to be 180 m long with a displacement of 29400 tonnes.  Calculate the effective power (PE) for the ship when operating at the corresponding speed in sea water.

f(model) F.W. = 0.492        n = 1.825

f(ship)  S.W. = 0.421        S = 257 times sqrt delta L

Where L = length in meters, S = wetted area (m2) = displacement (t)

 

C83. A ship model 5 m long has a wetted surface area of 4.98 m2.  The measured tow rope pull of the model, when towed in water of density 1000 kg/m3 at the corresponding speed of a similar ship, is 25.57 N.  Determine using the given data the effective power (naked) for a ship 180 m in length when sailing at 18.25 knots in water of density 1025 kg/m3. The following data is given:

Frictional co-efficient of the model in water of density 1000 kg/m3 is 1.72

Frictional co-efficient of the ship in water of density 1025 kg/m3 is 1.42

Speed in m/sec with index (n) for ship and model 1.83

 

C84. A ship model 7 m long has a total resistance of 44 N when towed at 3.5 knots in fresh water.  The ship itself is 185 m long and displaces 22000 tonnes. The wetted surface area may be calculated from the formula S= 2.57 sqrt delta L Calculate the effective power (naked) for the ship at its corresponding speed in sea water given:

f(model) FW = 0.492     f(ship) SW = 0.421

n = 1.825               density of sea water = 1025 kg/m3

 

C85. A 1/25 scale model of a ship which is to be 140 m long has a wetted surface area of 5.67 m2 and when towed through water of relative  density, 1.0 at a speed of 1.39 m/s, the total resistance is 22.9 N and the frictional resistance is estimated at 17.3 N.  The following data apply to the ship:

Propulsive coefficient = 0.64

Transmission loss = 3.1%

Appendage and weather allowance = 16%

Frictional coefficient = 1.44 (when ship floats in water of relative density 1.025

Speed index, n = 1.825 (when speed is in m/s)

When the actual ship moves through water of relative density 1.025 at a speed related to the corresponding model speed, calculate:

bullet

a) the effective power bullet

b) the quasi-propulsive coefficient

REF:Specimen 4

 

C86. A vessel of 120 m length and 13 m breadth floats at a draught of 6.2 m in water of density 1025 kg/m3 The waterplane area is defined by the following equidistant half-breadths, taken from the after perpendicular: 0.0  2.84  4.92  6.3  6.5  6.28  4.8  3.0  0.0m  respectively Given that the midships area co-efficient = 0.9 and the block co-efficient = 0.62, calculate:

bullet

a) the displacement bullet

b) the area of immersed midships section bullet

c) the prismatic co-efficient bullet

d) the position of the transverse metacenter above the center of buoyancy

REF:FEB82

 

C87. A vessel of 126 m length and 14 m breadth floats at a draught of 6.5 m in water of density 1025 kg/m3. The waterplane area is defined by the following equidistant half-breadths taken from the after perpendicular: 0.0  3.16  5.12  6.46  7.0  6.34  4.98  3.04  0.0 m respectively.  Given that the midships area co-efficient = 0.94 and the block co-efficient = 0.65 calculate:

bullet

a) the displacement bullet

b) the area of immersed midships section bullet

c) the prismatic co-efficient bullet

d) the position of the transverse metacenter above the center of buoyancy

 

C88. A vessel of 8750 tonnes displacement has 80 tonnes of cargo on the deck.  It is lifted by a derrick whose head is 11 m above the center of gravity of the cargo and placed in the hold 8 m below the deck and 15 m forward of its original position.  Calculate the horizontal and vertical shift in the vessel's center of gravity from its original position when the cargo is:

bullet

a) just clear of the deck bullet

b) suspended 5 m below the derrick head bullet

c) in it's final position bullet

d) the angle which the center of gravity moves relative to the horizontal

 

C89. A vessel of 10500 tonnes displacement is 120 m long and has a wetted surface area of 3000 m2.  When delivering 4100 kW at the shaft the speed is 15 knots, the propulsive co-efficient = 0.6 and 55% of the thrust is available to overcome frictional resistance. Calculate the shaft power for a similar ship 140 m in length operating at a corresponding speed.

note: F = 0.42    n = 1.825

 

C90. A vessel of 14000 tonnes displacement has a load waterline length of 145 m.  The waterline lengths at 1 m intervals of draught below this are: 144  143  141.5  139.5  137  134m respectively.  If the center of lateral resistance is at the centroid of this immersed area, calculate the angle to which the ship will heel due to centrifugal force when the ship turns in a circle of 500 m diameter at a speed of 15 knots with K.G.= 5.5 m and G.M. = 0.5 m.

 

C91. The following data reflect the results of a tanker of 54300 tonnes displacement when undergoing a 6 hour continuous trial run:

Speed (knots)    13.5  14    15     16     17     17.5

Propeller speed  86.0  89.6  96.8  105.0  114.0  119.0 (revs/min)

Fuel consumption = 15.89 tonnes

Engine revolutions = 39530

Calculate:

  1. the fuel coefficient

  2. the fuel consumption (t/day) for a similar ship of 36000 tonnes displacement fitted with a similar type of engine and operating at a corresponding speed.

REF:Specimen 5

 

C92. A vessel of constant triangular cross-section floats apex down with the keel just touching mud on the bottom of the sea, the width of the waterplane is 8 m.  If the water level now drops 2 m, calculate how far the vessel will sink into the mud given that the R.D. of the mud is twice that of the water.

note: the original draught of the vessel = 4 m.

 

C93. The master of a vessel wishes to introduce water ballast into a tank located 33 m aft of midships in order to completely submerge the propeller.  The following data is applicable to the ship Displacement = 8100 t. Initial draft fwd = 5.25 m Length = 85 m. Initial draft aft = 5.55 m

TPC = 9. Final draft aft = 5.85 m. Longitudinal center of flotation (LCF) 2 m aft midships

Longitudinal metacentric height GML = 96 m Calculate:

  1. the least amount of ballast required

  2. the final draft forward

 

C94. For a barge 9 m in length, 4.5 m breadth floating at a draught of 2.3 m in water of density 1025 kg/m3 the KG is 2.15 m.  The barge is divided longitudinally into three compartments each 1.5 m wide.  The following cargoes are loaded:

-Oil of specific gravity 0.9 is loaded into the center compartment to a depth of 1.5 m

-Gasoline of specific gravity 0.72 is loaded into two outer compartments to a depth of 1.2 m.

Assuming the KM remains constant, calculate the metacentric height taking into consideration the free surface of the cargo.  Describe briefly what effect transverse subdivision and longitudinal subdivision has on free surface.

REF:MAR89 REF:FEB89

 

C95. A box barge 34 m long and 6 m wide has a light displacement of 220 tonne and a K.G. of 2.8 m when floating in sea water of density 1025 kg/m3.  85 tonne of cargo is put on board and in order to maintain stability 50 tonne of ballast is added at K.G. 0.16 m.  The final G.M. is 0.15 m. Calculate the K.G. of the added cargo.

 

C96. A box shaped barge is 46 m long, 7.6 m wide and has a central full -width compartment 9.14 m long which is partially filled with oil of density 920 kg/m3. In this condition the barge floats upright at a draught of 1.83 m in water of density 1025 kg/m3 and has a K.G. of 2.44 m.

Calculate the angle of heel which results from lifting a mass of 5 tonnes already onboard and moving it through a horizontal distance of 4.9 m by means of a derrick fixed on the centerline of the barge.  Assume the mass is originally located on the centerline of the barge and remains suspended from the derrick in its final positions. The derrick head is 9.1 m above the original position of the vertical center of gravity of the mass.

  

C97. A box barge 60 m long and 7.5 m wide has a vertical KG of 2.15 m.  The barge floats at an even keel draught of 2.5 m in sea water of density 1025 kg/m3. A forward compartment is formed by a transverse bulkhead 6 m from the forward end. Determine the new draughts forward and aft when 100 tonnes of sea water is pumped into this forward compartment.

 

C98. A box barge 85 m long, 18 m beam and 6 m draught floats in sea water of 1.025 t/m3.  A midships compartment 18 m long contains cargo stowing at 1.8 m3/t and having a density of

1.600 t/m3  There is a watertight flat 6 m above the keel. Calculate the new draught if this compartment is bilged below the flat.

REF:APR89

 

C99. A box barge is 120 m long and floats at a draught of 6 m.  It has a mid-length compartment 15 m long extending right across the barge, but sub-divided by a horizontal watertight flat 4 m above the keel.  The G.M. is 0.95 m. Calculate the new draught and G.M. if the compartment is

bilged below the flat.

REF:FEB90

 

C100. A box barge 120 m long, 14 m beam and 5 m draught has a compartment at the extreme aft end 9 m long, sub-divided by a horizontal watertight flat 2 m above the keel.  KG is 3 m. Calculate the end draughts if the compartment is holed above the flat.

 

C101. A box barge 120 m long, 14 m beam and 5 m draught has a compartment at the extreme after end 9 m long, sub-divided by a horizontal watertight flat 2 m above the keel.  KG is 3 m.

Calculate the end draughts if the compartment is holed above the flat with water entering both compartments.

 

C102. A box shaped barge 137 m long, 18.3 m wide has a K.G. of 4.57 m when floating at an even keel draught of 7.32 m in water of density 1025 kg/m3.  Calculate the forward and aft draughts if a full breadth aft end compartment 13.7 m long is bilged.

 

C103. For a box shaped barge 140 m long, 20 m wide the KG is 5 m and the even keel draught 8 m when floating in sea water of density of 1025 kg/m3.  The barge is fitted with a transverse watertight bulkhead 14 m from the forward end. Calculate the draughts forward and aft if the space forward of the bulkhead is bilged.  Assume the permeability of the space to be 85% and the second moment of area of the intact waterplane about the center of flotation to be 3.4 x 106 m4.

 

C104. A box barge of 580 tonne displacement, 36.6 m long, 6.4 m breadth and 3 m deep floats in water of density 1025 kg/m3.  Calculate the permeability of a full depth midships compartment 9.1 m long by 6.4 m wide if the compartment is bilged and the draught becomes 2.8 m.

 

C105. A barge with semi-circular ends is 22 m long overall, 6 m breadth and 2.5 m in depth.  The barge is flat bottomed with vertical sides and ends.  The rectangular section of the barge is separated from the semi-circular end sections by two full depth transverse watertight bulkheads.  It floats at an even keel draught of 1.4 m in water of density 1025 kg/m3 the KG of the barge and cargo being 1.5 m.  Neglecting the effect of free surface calculate the transverse metacentric height if water of density 1000 kg/m3 is pumped into both end compartments to a depth of 2 m.

 

C106. A raft is made from two cylinders each 2 m in diameter and 8 m long.  The distance between the centers of the cylinders is 4 m.  If the draught is 1 m, calculate the transverse B.M.

 

C107. A propeller of 4.4 m pitch has an efficiency of 66% when turning at 130 R.P.M., the real slip is 38% and the delivered power is 2500 kW.  Calculate the thrust of the propeller.

 

C108. A propeller of 4.6 m diameter has a pitch of 4.3 m and boss diameter of 0.75 m.  The real slip is 28% at 95 rev/min. Calculate:

  1. the speed of advance

  2. the thrust

  3. the thrust power

REF:APR89

 

C109. The delivered power to a propeller is 2980 kW.  The propeller of pitch 5.5 m, rotating at 1.33 rev/sec with a speed of advance of 11 knots, has a propeller efficiency of 70 per cent. If the thrust deduction factor and Taylor wake fraction are 0.2 and 0.29 respectively, calculate:

  1. the true slip

  2. the propeller thrust

  3. the effective power

REF:APR91

 

C110. The power delivered to a propeller is 3540 kW at a ship speed of 15.5 knots.  The propeller rotates at 1.58 rev/sec, develops a thrust of 378 kN and has a pitch of 4.87 m.  If

the thrust deduction fraction is 0.24, real slip 30 per cent and transmission losses are 3 per cent, calculate:

  1. the effective power

  2. the Taylor wake fraction

  3. the propulsive co-efficient

  4. the quasi propulsive co-efficient, assuming the appendage and weather allowance is 15 per cent.

 

C111. A vessel is propelled at a speed of 16 knots by a propeller of diameter, 5 m: pitch ratio, 0.95; rotating at 1.85 rev/sec while absorbing a torque of 290 kNm at an efficiency of 64.2%.  Calculate:

  1. the real slip, when the Taylor wake fraction is 0.3

  2. the ratio of effective power (naked) to shaft power when the thrust deduction factor and the transmission losses are 0.2 and 2% respecitvely.

note: Thrust deduction factor =     Total resistance without 1 propeller / Thrust of the propeller.

REF:Specimen 2

 

C112. A double bottom tank is 25 m long.  The half breadths of the top of the tank are: 5.7  4.8  4.4  3.8  3.0  respectively. When floating in sea water of density 1025 kg/m3 the ship displaces 5400 tonnes, the loss of metacentric height due to the free surface is 0.24 m.  Calculate the density of the liquid in the tank.

 

C113. The half-ordinates of the waterplane of a ship 110 m long are as follows:

Station AP    .5   1    2    3    4    5    6    7   7.5   FP
1/2     0.7  2.9  4.8  6.4  6.9  6.9  6.9  5.9  4.6  2.0  0.0 Ordinate.

Calculate the waterplane area and the distance of the center of flotation from midships.

 

C114. The half ordinates of a waterplane 120 m long are: 0.7  3.3  5.5  7.2  7.5  7.5  6.8  4.6  2.2  0.0 respectively. The ship displaces 11000 tonnes.  The relative density of the sea water is 1025 kg/m3 Calculate the BM.

REF:83 REF:JAN87

 

C115. Half ordinates of the waterplane of a ship 140 m long are: 0.8  3.5  5.8  7.6  8.0  8.0  8.0  7.2  4.9  2.5  0.0  m respectively.  The ship displaces 13000 tonnes when floating in sea water of density 1025 kg/m3. Calculate the transverse B.M.

 

C116. The half ordinates 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  0.0 respectively.  Calculate:

  1. the area of waterplane

  2. distance of centroid from midships

  3. the second moment of area about the transverse axis through the centroid.

REF:APR89

 

C117. A rectangular bulkhead 8 m wide has water of density 1000 kg/m3 to a depth of 7 m on one side and on the other side oil of density 850 kg/m3 to a depth of 4 m.  Calculate:

  1. the resultant force on the bulkhead

  2. the position of the resultant center of pressure

REF:FEB90 REF:MAR8?

 

C118. A rectangular watertight bulkhead 9 m high and 14.5 m wide has sea waterof density 1025 kg/m3  on both sides, the height of the water on one side being four times that on the other side.  The resultant center of pressure is 7 m from the top of the bulkhead.  Calculate:

  1. the depths of water

  2. the resultant load on the bulkhead

REF:FEB89

 

C119. A bulkhead is 8 m high and had vertical stiffeners 0.7 m apart.  It is filled on one side only with sea water of density 1025 kg/m3 . The stiffeners are fastened at the bottom with 8 rivets of 25 mm diameter.  Calculate:

  1. the sheer force top

  2. the sheer force bottom

  3. the position of zero sheer

  4. sheer stress in rivets

  5. draw load and sheer force diagrams

REF:FEB90

 

C120. A bulkhead is in the form of a trapezoid 13 m wide at the deck 10 m wide at the tank top and 7.5 m deep. Calculate the load on the bulkhead and the position of the center of pressure if it is flooded to a depth of 5 m with sea water on one side only.

REF:APR89

 

C121. A collision bulkhead is defined by the following equally spaced widths, 1.53 m apart, commencing at the top: 6.1  5.49  4.8  3.36  2.44 m.  The bulkhead is loaded to the top, on one side only, with water of density 1025 kg/m3. Determine the load on the bulkhead and the position of the center of pressure from the top of the bulkhead.

 

C122. A rectangular block of wood of square cross-section of sides S and length L greater than S is to float in fresh water with it's horizontal axis parallel to the waterline, it's sides vertical and in a condition of neutral equilibrium.  Calculate the relative density of the wood.

 

C123. An 8 m model of a ship has a wetted surface area 10 m2, and when towed in fresh water at 4 knots, has a total resistance of 70 N.  Calculate the effective power of the ship, 140 m long, at it's corresponding speed given:

n = 1.825,              ship correlation factor = 1.15

f = 0.417 + 0.773 _     density sea water = 1.025 t/m3

          L + 2.862

 

C124. A bulkhead 8 m wide and 6 m deep has seawater on one side to a depth of 5 m and fresh water on the other side to a depth of 4 m.  Calculate the resultant load and position of

the center of pressure.

note: the saltwater density is 1025 kg/m3.

 

C125. The after bulkhead of an oil fuel bunker is 9.5 m wide and 13 m high.  Find the total load and the position of the center of pressure relative to the top of the bulkhead if the

tank is filled with oil of relative density 0.85:

  1. to the top edge

  2. with a 4 m head to the top edge

 

C126. A fuel tank bulkhead is made in the shape of a trapezoid 13 m wide at the top, 10 m wide at the bottom and 7.5 m deep.  When the tank is filled with fuel to a depth of

5 m, calculate:

  1. the load on the bulkhead

  2. the position of the center of pressure relative to the surface of the fuel

note: R.D. of fuel = 0.8

INA  (rectangle) = BD3 12

INA  (triangle)  = BD3 36

REF:OCT92

 

C127. A bulkhead is in the form of a trapezoid 14 m wide at the top, 10 m wide at the bottom and 8 m deep.  Calculate the load on the bulkhead and the position of the center of pressure if it is flooded to a depth of 6 m width sea water of density 1025 kg/m3 on one side only.

 

C128. A triangular bulkhead is 6 m wide at the top and 8 m deep.  When flooded with sea water of density 1025 kg/m3 on one side only the load on the bulkhead is 700 kN.  Calculate the height of the water level relative to the top of the bulkhead.

 

C129.  A triangular bulkhead is 8 m wide at the top and has a vertical depth of 9 m and forms part of a tank.  Calculate the position of the center of pressure and the load on the bulkhead if the tank is filled with sea water of density 1025 kg/m3 on one side of the bulkhead only:

  1. to the top edge

  2. with 3 m head to the top edge

REF:MAR89 REF:OCT90

REF:OCT92

 

C130. A large tanker is 400 m long with a beam of 50 m at the water line.  The immersed cross-section areas, equally spaced from fore and aft are: (in m2) 130.0  226.4  587.0  825.0  825.0  825.0  587.2  262.1  139.8 Calculate:

  1. the displacement

  2. the prismatic co-efficient

  3. the draught

note: Relative density of saltwater = 1.03 midships section area co-efficient (Cm) = 0.9649

 

C131. The frictional resistance of a ship in fresh water at 4 m/s is 12 Nm2 The ship's wetted surface area is 2800 m2 and the frictional resistance is 74% of the total resistance and varies as speed1.92 If the effective power is 1200 kW, calculate the speed of the ship in sea water of 1.025 t/m3.

 

C132. The speed of a ship is increased to 12% above normal for 11 hours then reduced to 10% below normal for 9 hours. For the remainder of the day the speed is such that the day's consumption will remain normal.  Calculate the percentage difference in the days run.

 

C133. The T.P.C. values of a ship at 1.75 m intervals of draught, commencing at the keel are:

4.5  6.7  8.0  9.4  10.8  12.0  1.8 respectively.  At a draught of 10.5 m calculate:

  1. displacement

  2. K.B.

 

C134. The tonnes per centimeter (TPC) of a ship is 27.5 when floating in sea water of relative density 1.026.  The maximum allowance draught is 8.25 m in this sea water and 8.5 m in fresh water.  The vessel now loads to a draft of 8.44 m at a river port where the relative density of the water is 1.012 and then moves out to sea where the freeboard is checked by the Port Warden and found to be insufficient.

a) How much mass must be removed to make the vessel seaworthy

b) describe the different methods that may be employed to lighten the ship stating the limitations imposed by each approach.

REF:MAY81

 

C135. The fuel consumption of a ship at 18 knots is 52 tonnes/day.  The speed is reduced and the consumption is reduced to 24 tonnes/day, which is 20% more than the theoretical consumption.  Find the reduced speed and the percentage saving on a voyage of 4000 nautical miles.

 

C136. The immersed cross sectional areas of a ship 90 m long, commencing from aft are: 0.0  11.8  27.4  39.0  44.5  45.6  45.0  39.4  26.7  14.6 0.0 m2 respectively.  Calculate for a sea water density of 1025 kg/m3:

  1. displacement

  2. distance of center of buoyancy from midships

  3. prismatic co-efficient

 

New Questions posted April 2002

The underwater portion of a vessel is divided by transverse sections 12 ft., apart of the following areas commencing from forward;- 0.4, 25, 53.4, 78.6, 90.2, 88.3, 65.9, 26.8, 4 sq. ft. respectively. Find the position of the centre of buoyancy relative to the middle section, the displacement of the vessel in sea water, and the prismatic co-efficient.

A vessel whose length is 330 ft., has measurements of half girth from water line to keel as given by the following;- 7.13; 18; 29; 42; 44.1; 44.1; 43.8; 39; 24; 16 and 0 ft., respectively, all measured at regular intervals from ford. Given that the approximate wetted surface can be obtained from W.S. = 15.5/of •.L. sq., ft. Where • = displacement in tons and L = length of vessel in ft. Estimate the displacement of the vessel.


The following table gives the draught from the keel and the corresponding T.P.1" values of a vessel floating in sea water of 64 lbs., per cu. ft.
Draught (ft.) 3. 6. 9. 12. 15.
T.P.1" 33 39 43.5 46.5 49
The appendage below the 3 ft., draught displaces 600 tons and its center of buoyancy is 1 ft., below the 3 ft., draught line. Determine the displacement of the vessel and the vertical position of the centre of buoyancy from the keel when the vessel is floating at a draught of 15 ft.


A vessel has half ordinates of water plane at equal intervals of 40 ft., as follows;- Fwd.;- 0.4; 13.8; 25.7; 32.5; 34.7; 35; 34.9; 34.2; 32.1; 23.9; and 6.9 ft., respectively. Calculate the water plane area and the second moment of the water plane about the longitudinal center line.

The following table gives the half ordinates of the water plane of a vessel whose length between perpendiculars is 360 ft. Calculate the water plane area and the centre of flotation from the midship section. 
Station;- Ford.P. 1-1/2 2 3 4 5 6 7
Half ordinates (ft.) 0 7.5 13.4 18.0 23.0 25 25 25

8 8-1/2 Aft P.
16 12 0

A watertight door is 4 ft., high by 2 ft. 6 ins., wide and has a still of 2 ft., depth measured from the tank top. Sea water rises to a height of 10 ft., on one side and to a height of 5 ft., on the other side, both heights being measured from the tank top. Draw a load diagram and from it determine (a) the resultant force in tons acting on the door and (b) the position from the top of the door at which this force may be assumed to act.

A vertical rectangular bulkhead 25 ft., high is supported by vertical channel form stiffeners spaced 30 ins., apart. The stiffeners are connected to the tank top by 10 rivets each 0.875 ins., diameter. Sea water rises to the top of the bulkhead. Calculate (a) the shear force at the top of each stiffener, (b) the position of zero shear force measured from the tank top and (c) the shear stress in the tank top rivets. Sketch the load distribution and shear force diagrams for one stiffener. Assume each stiffener is simply supported at each end.

A triangular section pontoon floats at a draught of 16 ft., and has a length of 210 ft., and at any draught the breadth is twice the draught. Assuming that in all conditions the pontoon floats apex down, draw the draught/displacement curve and using it show that the vertical distance of the center of buoyancy from the water line is given by the expression;- Area enclosed by curve. Displacement

The semi-ordinates of a ships’ water plane at equal intervals of 15 ft., starting from the bow are;- 0.5; 4.0; 8.2; 10.4; 18.0; 17.8; 10.4; 7.0; and 1.0 ft., respectively. Calculate (a) the T.P.1" of this water plane, (b) the position of the center of flotation from amidships and (c) the second moment of area of the water plane about the transverse axis through the center of flotation. Note;- the second moment of area about any axis YY, which is parallel to an axis through the centroid and at a distance of X from it is found as follows;- Iyy = Igg + AX² Where Igg = second moment about centroid and A = area.

A rectangular block of wood of square cross section of side S and length L is to float in fresh water with its horizontal axis parallel to the water line, its sides vertical and in a condition of neutral equilibrium. What is the specific gravity of the wood to male this possible? The length L is greater than the side S.

A completely decked in scow of box form is 90 ft., long, 18 ft., beam and 8 ft., deep. She is so loaded that she draws 6 ft., in fresh water, and the center of gravity is 6 ft., above the bottom of the vessel. If the vessel is upright and on an even keel, what will be the angle of heel if a weight of 10 tons on the fore and aft center line of the vessel is shifted 9 ft., athwart ships? Define the terms “center of gravity” and “center of buayancy”.

A triangular form fore peak bulkhead has a breadth at the top of B ft., and a vertical depth of D ft. Find, using Simpsons First Rule, the total load on the bulkhead, in terms of B and D, if the bulkhead is flooded to the top with sea water on one side. Also determine the position of the center of pressure, if the depth of the center of pressure from the free surface of the water is given by ;- C of P = 2nd moment of area of immersed surface 1st moment of area of immersed surface Both moments are taken about the free surface of the water.

A rectangular deep tank whose height os 9 times the vertical stiffener spacing, is filled with sea water. The maximum allowable shear load on one stiffener is 22 tons. Assuming that the stiffeners are simply supported at the ends, calculate (a) the depth of the tank, (b) the total fluid load on the end of the tank, (c) the shear force at the top and bottom of each stiffener, and (d) the position of zero shear force on each stiffener.

An inclining experiment was carried out on a vessel of 2,750 tons displacement and 13 ft. 8 ins., draught, having a BM of 4 ft. 2 ins., the distance of B from the water line being 5 ft. 8 ins. A weight of 3.5 tons was moved 27 ft., transversely to give a deflection of 11 ins. The vessel was then ballasted 6 ft., below the original CG and the same experiment carried out, this time the deflection being 5.75 ins. Due to ballasting the GM increased by 6 ins. Find;- (a) the position of the original centre of gravity of the vessel before ballasting, measured from the keel, (b) the final GM, (c) weight of the ballast in tons.

A vessel of 10,000 tons displacement commences to fill a double bottom tank which is 60 ft., long 50 ft., wide and 5 ft., deep. Initially the centre of gravity of the vessel was 12 ft., above the keel and the transverse metacentric height was 2 ft. Assuming that the position of the transverse metacentric relative to the keel remains unaltered find the metacentric height (a) when the tank is half full and (b) when the tank is full. The vessel is in sea water and the tank is filled with sea water.

A box shaped vessel 200 ft., long, and 31 ft., beam floats in sea water. Calculate the distance of the centre of buoyance (B) and the transverse metacentre (M) from the keel for draughts up to 15 ft., in intervals of 3 ft. Plot these values in the form of a graph against a base of draught. With the aid of the graph find the transverse metacentric height of the vessel when the draught is 14 ft., given that the centre of gravity is then 9 ft., above the keel.

What do the following terms mean? “Stable equilibrium”, “neutral equilibrium” and “unstable equilibrium”? A ship of displacement 1,722 tons is inclined by shifting 6 tons of ballast across the deck through 22.25 ft. A mean deviation of 10.5" is obtained with pendulums 15 ft., long. The transverse metacentre is 15.28 ft., above the keel. Find (a) the position of the centre of gravity of the ship with reference to the keel, and (b) the moment of statical stability if the ship is steadily inclined at an angle of 8 degrees.

State briefly what is meant by ;- (a) the stability of a vessel and (b) the transverse metacentric height. A ship of 10,000 tons displacement with a beam of 50 ft., lies close to a quay. A weight of 75 tons is in the lower hold and on the fore and aft centre line of the vessel. The weight is to be lifted to a point on the quay 5 ft., from the ship’s side by the ship’s derrick. The head of the derrick is 65 ft., above the centre of gravity of the weight in the hold. Does any alteration of the metacentric height of the ship take place when the derrick takes the load and if so, calculate the alteration. What is the maximum angle of the heel during the operation given that the original metacentric height is 2 ft. 9 ins.

Deduce a formula for calculating the loss of GM due to free liquid surface in a ships tanks. A vessel of 14,000 tons displacement has a double bottom tank 68 ft., broad and 60 ft., long partly filled with water. Find the alteration in the transverse metacentric height of the vessel due to this free surface.

A ship with a displacement of 10,000 tons has a double bottom tank which contains 150 tons of oil. The centre of gravity of the tank is 100 ft., forward and 20 ft., below the centre of gravity of the ship. What distance in ins., does the centre of gravity of the ship move through due to this quantity of oil being consumed?

A ship of 8,500 tons displacement has a draught of 7 ft., 6 ins., a water plane area of 48,000 sq. ft., which may be considered constant and its centre of gravity is 22 ft., above the keel. The following changes in weight distribution occur;- 125 tons are lowered a distance of 16 ft.; 1,500 tons placed 15 ft., above the keel are discharged; 270 tons and 750 tons are loaded at pints 11 ft., and 9 ft., respectively above the keel. Calculate the new position of the centre of gravity of the vessel above the keel and the new draught.

A box form barge has a length of 300 ft., a breadth of 42 ft., and is floating at a draught of 16 ft., in water of density 63 lbs., per cu. ft. A weight of 80 tons, already on board, is moved from ford to aft, a distance of 100 ft. Find (a) the angle of trim and (b) the draught ford and aft. Assume the centre of flotation is amidships.

A ship whose displacement is 8,000 tons has draughts of 17 ft. 9-1/2 ins., ford and 18 ft. 9-1/2 ins., aft. It is required to trim the ship to obtain further immersion of the propeller by ballasting a tank, whose centre of gravity is 110 ft., aft of amidships. If the final maximum draught is not to exceed 19 ft. 6 ins., calculate the minimum amount of ballast water required. The following particulars refer to the ship at its present draught;- Length between perpendiculars 285 ft.; G.M.(longitudinally) 350 ft.; T.P.1" 21; Centre of Flotation 8 ft., aft of amidships.

Explain clearly how the formula “Moment to change trim 1 inch = W.GM tons ft., is arrived at. 12L A vessel 450 ft., long and 7,000 tons displacement has a longitudinal metacentric height of 630 ft. Find the change of trim caused by moving a weight of 15 tons already on board through a distance of 300 ft., from forward to aft.

A box shaped vessel is 200 ft., long, 30 ft., beam and floats at a draught of 10 ft., in sea water. The centre of gravity of the vessel is 8 ft., above the keel and amidships. An end compartment 30 ft., long and extending the full width of the vessel is now opened to the sea and flooded. Find the ford and after draughts of the vessel.

The draught of a ship at the completion of a voyage is 21 ft., ford and 22 ft. 6 ins., aft. During the voyage the following weight changes took place;- 400 tons of fuel consumed from 100 ft., ford of amidships 10 tons of stores consumed from 250 ft., ford of amidships. 230 tons of water consumed from 150 ft., aft of amidships. 300 tons of ballast added at 50 ft., ford of amidships. If the centre of flotation is 6 ft., aft of amidships, the T.P.1" is 52 and the Moment to change trim 1 inch is 1,100 tons ft., find the draught at the commencement of the voyage. The length of the ship is 600 ft.

A floating crane has a box shaped pontoon with the following dimensions;- length 180 ft., beam 80 ft.; even draught 16 ft., when floating in water of density 35 cu. ft. per ton. The centre of gravity is 20 ft., above keel. Calculate the longitudinal metacentric height. If a weight of 80 tons, already on aboard, is lifted from the deck, moved forward 80 ft., and remains suspended from the crane, calculate the fore and aft draughts. The jib head of the crane is 100 ft., above the deck.

A ship whose length is 505 ft., has a longitudinal metacentric height of 360 ft., and a displacement of 12,000 tons at draughts of 25 ft. 4 ins. ford and 27 ft. 8 ins., aft. Estimate the value of the weight and its position from midships which must be placed aboard the ship in order to bring it to a draught of 27 ft. 8 ins., ford and aft. The T.P.1" of the ship over the draughts concerned is 59 and the centre of flotation is amidships.

A vessel of 13,700 tons displacement has a Moment to change trim 1 inch of 1,280 tons ft., and a T.P.1" of 45. Its centre of floatation is 6 ft., aft of a amidships and its centre of buoyance is 1 ft., ford of amidships. When in water of 1,025 ozs. per cu.ft., the vessel is on an even keel at a draught of 26 ft. 3 ins. Calculate the fore and after draughts when the vessel is moved into water of 1008 ozs. per cu.ft.

A box barge of 100 ft., length, 30 ft., beam floats at a draught of 2 ft. 6 ins., in water of 35 cu.ft., per ton when empty. In a partly loaded condition the draught alters by 1.25 ins., when the barge passes from water of 1,000 ozs., per cu.ft., to water of 1,025 ozs., per cu.ft. Calculate the weight of the cargo in the barge.

The following table gives the draughts and corresponding T.P.1" values for a vessel when floating in sea water of 35 cu.ft., per ton.
Draught (ft.) 0, 4, 8, 12, 16, 20, 24.
T.P.1" 0, 30, 31.4, 32, 32.5, 32.8, 33.
Find the displacement of the vessel when floating in water of density 1,012 ozs. per cu.ft., at a draught of 20 ft., and also calculate the change in draught when the vessel moves into water of 1,024 ozs. per cu .ft.

An oil tanker is 525 ft., long, 72 ft., beam and floats at a draught of 32 ft., in sea water of 64 lbs., per cu. ft. The water plane area co-efficient at this draught is 0.88. The midship section is rectangular except for bilges of 4 ft., radius. The ship is divided transversely by bulkheads 35 ft., apart and longitudinally by two fore and aft bulkheads to form oil compartments. The compartments are filled with oil of 50 cu.ft. per ton to a depth of 38 ft. If the midship compartments, full width, were open to the sea what would be the draught?

A vessel of constant triangular section floats in fresh water at a draught of 12 ft., with the keel just touching mud at the bottom. The keel is the apex of the section and the axis is vertical. In the condition above the breadth at the water line is 24 ft. If the water level now drops 6 ft., calculate how far the vessel will sink into the mud given that the specific gravity of the mud is 2.

A vessel having a water plane area of 20,760 sq.ft., arrives in a harbour where the density of the water is 1,026 ozs., per cu.ft. On arrival she pumps out 288 tons of ballast and then proceeds to a river where the density is 1,006 ozs. Per cu.ft. Her draught was taken on arrival in harbour before discharging ballast and again on entering the river. The change is found to be 3 ins. Find the original displacement of the ship in tons and state whether the 3 inch change was an increase or decrease in draught.

A ship, while traveling at 12 knots, turns in a circle of a radius of 2,000 ft. Metacentric height of the vessel is 10 ins., and the vertical distance between the entre of gravity and the centre of buoyancy is 14 ft. Assuming that centripetal force acts at the centre of buoyancy what is the angle of heel produced?

A ship at a certain draught passes from sea water density 1,027 ozs., per cu.ft. To water of density 1,007 ozs., per cu.ft. If 200 tons of cargo have to be discharged to maintain the original draught, what was the initial displacement of the ship?

A ship passes from sea water into fresh water and the draught increases ‘a’ ins., due to change in density. The ship now discharges cargo while in fresh water and the draught decreases ‘b’ ins. The ship now returns to sea and the draught decreases ‘c’ ins. Assuming the water plane is constant, prove that ;- density of sea water = 1,000 x b ozs., per cu.ft. b + c - a

The fluid force in lbs., exerted on the rudder when the helm angle is i degrees and the ship is moving at V knots is given by;- F (lb.) = 1.12AV²Sini where A is the rudder area is sq.ft. A ship when travelling at 16 knots puts on a helm angle of 35 degrees. The area of the rudder is 250 sq.ft., and the centre of fluid pressure from the centre line of the rudder stock is 5 ft. Find the diameter of the rudder stock if the maximum shear stress is not to exceed 10,000 lbs./sq.in.

Assuming that the resistance to motion of a vessel varies directly as the speed squared, show that the consumption of fuel per day varies directly as the speed cubed and that the consumption per voyage varies directly as the speed squared. For a certain vessel it was found that by reducing the consumption of fuel by 41 tons per day the speed was reduced by 2 knots. If there was a saving of 24% in fuel on a voyage of 5,000 nautical miles, what was the original consumption per day and the original speed?

A vessel has a rudder area of 220 sq.ft., and the distance from the centre line of the rudder stock to the centre of fluid pressure on the rudder is 3 ft. 6 ins. Calculate the twisting moment in lbs.ins., in the rudder stock when the helm angle is 35 degrees and the speed of the vessel is 15 knots given that;- fluid pressure on rudder (lbs.) = 1.12 AV²Sini Where A = rudder area (sq.ft.), V = ships speed (knots), i degree = helm angle. If the shear stress in the rudder stock is not to exceed 900 lbs., per sq.in., calculate the rudder stock diameter.

A vessel of 12,000 tons displacement, having a metacentric height of 1 ft., and a centre of buoyancy 11 ft., below the water line, is steaming at 20 knots when 30 degrees of helm is applied. Find the angle of heel produced. The centre of pressure of the rudder, which is 150 sq.ft., in area, is 16 ft., below the water line. The fluid force produced on the rudder is;- where A = area rudder (sq.ft.) , V = speed of ship (knots) , i degrees = angle of helm.

A ship’s speed is increased by 21% above the normal for 7.5 hours and then decreased by 9% below the normal for 10 hours. The ship’s speed is then adjusted for the remainder of the day so that its daily consumption is normal. Determine the percentage variation from normal in its daily run.

Explain, with the aid of sketches, how the ship’s structure is made stronger in spaces used for machinery such as main engines and boilers.

A vessel has a wetted surface area of 14,670 sq. ft. When travelling at 23.5 knots in water of density 1,025 ozs., per cu. ft., the effective horse power was 19,200. If the resistance of a plate towed edgewise through water of density 1,008 ozs., per cu. ft., at 10 ft. sec., was 0.28 lbs., per sq. ft., and given that frictional resistance varies as the speed to the 1.96 power, find the percentage power used to over come frictional resistance.

The total resistance of a ship model 16 ft., long is found to be 4.1 lbs., when tested in a tank. The wetted surface of the model was 52 sq. ft. Estimate the effective horse power necessary to drive the ship designed on this model if the length of the ship is 400 ft. The speed of the model in the tank test was 2.5 knots and the model was towed in fresh water. The ship will operate in sea water of density 1.025. The laws of Dynamic similarity apply between the speed, residual resistance and, linear dimensions of the ship and model and frictional resistance can be found from;- F(lb.) = fsV6 Where f = 0.01 for model and 0.0088 for ship. S = wetted surface in sq. ft. V = speed in knots. n = 1.825 Frictional resistance % fluid density.

Explain how stresses are set up in a vessel’s hull due to “hogging” and “sagging” in a heavy sea-way. How is the machinery affected under these conditions?

The resistance of a flat plate when drawn through water of density 1,013 ozs., per cu. ft., at a speed 10 ft. sec., was found to be 0.28 lbs. per sq. ft. Estimate the speed of a ship whose effective horse power is 13,500 when the wetted surface area is 52,000 sq. ft., and it is in water of density 1,026 ozs., per cu., ft. It may be assumed that skin friction constitutes 72% of the total resistance of the ship to motion and that resistance due to friction % speed to the exponent 1.9 x fluid density.

State the Laws of Dynamic similarity which are applicable to the residual or wave resistance on the hull of a ship. A model 12 ft., long is towed at 300 ft., min., in a test tank and has a computed wave resistance of 0.49 lbs. What will be the corresponding speed and wave horse power required for a similar ship of 400 ft., length?

With regard to ballast tanks, what effect has free surface on the stability of a ship?

I reference to stability of ships explain what is meant by centre of gravity, centre of buoyancy, meta centre and metacentric height. Under practical conditions can the metacentric height be increased?

What is meant by the terms “Metacentric Height”? What effect would pumping out a double bottom ballast tank have on the metacentric height? What effect would it have on the metacentric height to allow excess bilge water to accumlate?

If the Port and Starboard deep tanks of a sip were only partially filled with oil, explain, with the aid of sketches, how the stability of the ship would be affected when rolling in a seaway.

What forces act on a ship’s transverse framing when, (a) the ship is at sea, and (b) when the ship is in dry dock. What parts of the ship’s structure absorb these forces?

Describe, with the aid of sketches, how the applicable forces act to set up the moment of statical stability when a vessel heels. What governs the position of the transverse meta centre and what factors govern the metacentric height?

For what purpose is an inclining experiment carried out? Describe briefly how the experiment is carried out.

Explain clearly what the following scantling are, what their function is , where they are found in a modern cargo ship;- shear strake; deck girder; beam knee; frame; floor; margin plate; deck stringer; reverse frame;

Define the following terms;- (a) garboard strake, (b) sheer strake, (c) keelson, (d) stringer, (e) floor, (f) intercostal.

What stresses are produced in a ship’s side plating during heavy weather conditions? What members of the ship’s structure normally resist these stresses and loads?

What is meant by the following terms;- (a) block co-efficient, (b) midship co-efficient, (c) prismatic co-efficient, (d) gross tonnage, (e) displacement?

Explain with the aid of a sketch, how the deck of a ship is strengthened to compensate for the weakness caused by the hatchway openings.

Describe briefly the principal structure features of longitudinal framing commonly adopted in tankers. Why is it specially adopted for tankers?

What do you understand by the following terms used in ship construction;- (a) tumble home, (b) rise of floor, (c) freeboard, (d) sheer, (e) camber, (f) scantlings.

Explain what is meant by the following;- (a) bilge keel, (b) sheer strake, (c) tumble home, (d) deck camber, (e) breast hook.

Explain what “half beams” are. Show by a sketch where they are used and how they are attached to the ships structure.

Describe with the aid of sketches;- (a) Duct keel, (b) Bilge keel. What are the functions of these?

Sketch a section through a Hatch Coaming. Show how the half beams are connected to the coaming.

Describe the various types of beam sections used in ship’s construction and how are they secured to the other parts of the ship’s structure. You may illustrate your answer with sketches.

Describe the construction of the double bottom tanks as fitted to a dry cargo vessel with engines amidships. Explain how the double bottom is strengthened in way of the machinery space and the ford, part of the vessel.

Sketch and describe a double bottom tank that is used for the storage of oil fuel. What fittings are provided for sounding, filling, emptying ect.; How do the arrangements differ from a tank which is used for water ballast only?

Sketch a transverse section through a ship showing clearly the double bottom construction and indicate on your sketch the position of ; - keel, garboard strake, tank margin plate, and frame and reverse bar.

For what purpose are “ summer” tanks fitted in bulk oil carriers? Make a sketch of the transverse section of a tanker showing how these tanks are arranged.

A vessel has a wetted surface area of 14,670 sq.ft. When travelling at 23.5 knots in water of density 1,025 ozs., per cu.ft., the effective horse power was 12,200. If the resistance of a plate towed edgewise through water of density 1,008 ozs., per cu.ft.,at 10 ft.sec., was 0.28 lbs., per sq.ft., and given that frictional resistance varies as the speed to the 1.96 power, find the percentage power used to overcome frictional resistance.

A vessel has a draught of 25 ft.., and a wetted surface area of 48,000 sq.ft. The effective horse power is 9,250 and the skin friction is 72% of the total resistance of the ship to motion. Calculate the speed of the vessel in knots when in sea water of 64 lbs./cu.ft. Resistance of flat plate is 0.28 lbs., per sq.ft., when towed through fresh water at 600 ft. min. It may be assumed that the frictional resistance is proportional to the speed raised to the power of 1.98 and is proportional to the density of the fluid.

A vessel consumes 47 tons of fuel per day at 17 knots. By reducing speed the consumption drops to 22 tons per day. The consumption per I.h.p., hour at the reduced speed is increased by 13.5%. Find the reduced speed and the percentage saving in fuel on a voyage of 3,000 miles.

A vessel consumes 20 tons of fuel per day for the main engines at a speed of 11 knots, and the consumption for auxiliary purposes is 5 tons per day, which is constant. When 1,100 nautical miles from port it is found that 60 tons of fuel remain in the bunkers. Estimate the reduced speed in order to arrive in port with 5 tons of fuel remaining in the bunkers and find the percentage reduction in total fuel consumption per day.

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