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help for Canadian engineers
Canadian
First Class ME
Naval Architecture
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:
 |
a)
the area of the waterplane
|
|
b)
the position of the longitudinal center of flotation
|
|
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:
|
a)
the mass of water of density 1025 kg/m3 added as ballast
|
|
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:
|
a)
the area waterplane
|
|
b)
distance of LCF from midships
|
|
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:
|
a)
the area of the waterplane
|
|
b)
the position of the longitudinal center of flotation
|
|
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:
|
a)
the mass of water of density 1025 kg/m3 added as ballast
|
|
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:
|
a)
the area of the waterplane
|
|
b)
the position of the longitudinal center of flotation
|
|
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:
|
15
tonnes 38 m aft of midships
|
|
55
tonnes 30 m aft of midships
|
|
30
tonnes 3 m aft of midships
|
|
60
tonnes 10 m forward of midships
|
|
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:
|
cargo 10000 tonnes LCG 3.0
m forward of midships
|
|
fuel 1600 tonnes LCG 2.0
m aft of
midships
|
|
water 400 tonnes LCG 8.0 m
aft of
midships
|
|
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:
|
a)
length
|
|
b)
breadth
|
|
c)
draught
|
|
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:
|
a) M.C.T. 1cm
|
|
b)
the longitudinal metacentric height
|
|
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:
|
a)
the distance of the longitudinal center of flotation (LCF) from midships
|
|
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:
|
a)
the distance of the longitudinal center of flotation (LCF) from midships
|
|
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:
|
a)
the distance of the longitudinal center of flotation (LCF) from midships
|
|
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:
|
a)
displacement
|
|
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:
|
a)
the distance of the longitudinal center of flotation (LCF) from midships
|
|
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:
|
a)
the distance of the longitudinal center of flotation (LCF) from midships
|
|
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:
|
a)
the distance of the longitudinal center of flotation from midships
|
|
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.
|
a)
if the admiralty co-efficient is 620, calculate the shaft power required
at 19 knots
|
|
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:
|
a)
the corresponding speed and wetted surface of a similar ship of 2500
tonnes displacement.
|
|
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:
|
a)
when the tank is pressed up
|
|
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:
|
a)
the mass of oil used
|
|
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:
|
a)
filled with sea water
|
|
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:
|
a)
the load is just clear of the tank top
|
|
b)
the load is lifted to the deck head
|
|
c)
the load is landed on a deck 12 m above the tank top
|
|
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:
|
a)
tonnes per centimeter immersion
|
|
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:
|
a)
tonnes per centimeter immersion
|
|
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.
|
a)
does the GM alter during the operation?
|
|
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:
|
a)
the mass of oil used
|
|
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:
|
a)
fuel co-efficient for the ship
|
|
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:
|
a)
no longitudinal bulkheads
|
|
b)
a longitudinal centerline bulkhead
|
|
c)
twin longitudinal bulkheads forming three equal tanks
|
|
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:
|
a)
the propeller pitch
|
|
b)
real slip
|
|
c)
quasi propulsive co-efficient if the Taylor wake fraction and thrust
deduction factor are 0.31 and 0.2 respectively.
|
|
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:
|
a)
the quasi propulsive co-efficient (Q.P.C.)
|
|
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:
|
a)
the effective power
|
|
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:
|
a)
the displacement
|
|
b)
the area of immersed midships section
|
|
c)
the prismatic co-efficient
|
|
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:
|
a)
the displacement
|
|
b)
the area of immersed midships section
|
|
c)
the prismatic co-efficient
|
|
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