MeoscsIV Emr

95SCS-4 Sr. No. EMR EXAMINATION OF MARINE ENGINEER OFFICER Function: Controlling Operation of ship and care for persons on board Operational Level SHIP CONSTRUCTION & STABILITY CLASS IV (Time allowed - 3 hours) Afternoon Paper

INDIA (2002) N.B. -

at

Total Marks 100

(1) All Questions are compulsory (2) All Questions carry Equal Marks. (3) Neatness in handwriting and clarity in expression carries weightage

1. A vessel has the following dimensions : length 120.0m; block coefficient, 0.78; KB 3.2m. ½ Ordinates of the waterplane at the 6.0m draft are: Station 0 1 2 3 4 5 6 ½ Ordinate (m) 0.1 14.0 15.0 15.0 15.0 9.5 1.0 If the vessel is floating on an even keel at a draft of 6.0m, calculate the vessel’s KM. 2. A vessel about to complete loading in a summer zone is expected to enter a winter zone after steaming from the loading port for 10 days. On passage, fuel consumption is expected to be 30 tonnes per day and water consumption 15 tonnes per day. The ship is at present floating in water of density 1.013 tonnes/m3 at a draft of 9.0m. Summer load draft 9.475m FWA 203mm TPC 30 tonnes/cm Find A. The maximum permissible draft on completion of loading. B. The amount of cargo which the vessel can load. (Note. A vessel passing from a summer zone to a winter zone, or from a tropical zone to summer zone, must arrive at the zone boundary at the draft appropriate to that zone.) 3. A vessel has displacement 6200 tonnes KG, 8.0m. Distribute 9108 tonnes of cargo between spaces Kg, 0.59m and 11.45m, so that the vessel completes loading with a KG of 7.57m. 4. A floating body has square cross section of side 1 m and KG always equal to 0.5 m. Find the range of relative densities over which the body will be (a) stable and (b) unstable, when floating in fresh water. What will be the minimum GM and at which draft does it occur.

5. A box vessel; length, 140m; breadth, 20m; depth, 14m; is floating at draft 5.8m, KG, 8.00m. A midship compartment length, 24m; breadth, 7m, has a watertight flat at a height 6m above the keel.

Find the bilged draft and the GM in the intact and bilged condition. 6. A ship of 5000 tonnes displacement has GM = 0.5 meters. The still water period of roll is 20 seconds. Find the new period of roll when a mass of 100 tonnes is discharged from a position 14 metres above the centre of gravity. 7. Reserve buoyancy is the _____________. A. unoccupied space below the waterline B. volume of intact space above the waterline C. excess of the buoyant force over gravitational force D. difference between buoyancy in salt and fresh waters Briefly Justify Your Answer 8. Adding the transverse free surface correction to the uncorrected height of the center of gravity of a vessel yields ____________. A. "FSCT" B. "KG" C. "KGT" D. "GMT" Briefly Justify Your Answer 9. The difference between the average of the forward and aft drafts is _____________. A. list B. heel C. trim D. flotation Briefly Justify Your Answer 10. In the absence of external forces, adding weight to one side of a floating vessel, will cause the vessel to ____________. A. heel until the angle of loll is reached B. list until the center of buoyancy is aligned vertically with the center of gravity C. trim to the side opposite TCG until all moments are equal D. decrease draft at the center of flotation Briefly Justify Your Answer -----------------------------X-----------------------------

95SCS-4 Sr. No. EMR EXAMINATION OF MARINE ENGINEER OFFICER Function: Controlling Operation of ship and care for persons on board Operational Level SHIP CONSTRUCTION & STABILITY CLASS IV (Time allowed - 3 hours) Afternoon Paper

INDIA (2002) N.B. -

at

Total Marks 100

(1) All Questions are compulsory (2) All Questions carry Equal Marks. (3) Neatness in handwriting and clarity in expression carries weightage

Answers Answer For Question No. 1 Under water volume = 16850 m Function Iɠ = 41405 BM = 10.95m KM = 14.14m

3

Iɠ

= 184.022m4

Answer For Question No. 2 (a) Allowance between summer and winter load lines = summer draft x 1000 48 = 9.475 x 1000 = 197mm 48 As DWA = FWA (1025 - ) 25  DWA = 245 (1025 – 1013) 25 = 97mm To calculate the change of draft while on passage to the boundary of the winter zone, the total consumption of the fuel and water must be calculated and divided by TPC, as follows: Change of draft on passage = 30 x 10 + 15 x 10 = 15cm 30 Knowing the DWA and the allowance to make for fuel and water consumed on passage to the zone boundary, the maximum permitted draft may now be calculated, as follows: Summer draft = 9.475m S-W allowance = 0.197m Winter draft = 9.278m DWA draft = 0.097m ‘In dock’ draft = 9.375m Passage Allowance = 0.150m Max. draft = 9.525m Present draft = 9.000m Permitted sinkage

= 0.525m

The calculation so far has established that the vessel may load to a draft of 9.525m in dock in order to be at the appropriate winter draft on arrival at the winter zone boundary. There is another limitation which must be considered, i.e. will the vessel be deeper than the summer draft plus the DWA when at this draft in the dock? If so, then the figure of 9.525m will have to be reduced accordingly and the permitted sinkage will similarly decrease. Summer draft = 9.475m DWA = 0.097m Max. draft 9.572m As this figure is greater than 9.525m, then 9.525 is clearly the limiting draft to which the vessel can be loaded in the dock. The amount of cargo to load must now be determined using the permitted sinkage of 0.525m or 52.5cm, again correcting the TPC for the density of the dock water as follows: (b) TPCD = TPCS D S = 30 x 1013 = 29.65 tonnes/cm 1025  Cargo to load = sinkage x TPCD = 52.5 x 29.65 = 1556.6 tonnes at maximum draft of 9.525m Weight (tonnes) 6200.0 9108.0 – w W 15308.0 KG = moment of weight

KG 8.0 0.59 11.45 12.0

Answer to Question No. 3 Moment 49600.0 5373.7 – 0.59w + 11.45w 54973.7 – 10.86w

Displacement 7.57 = 54973.7 – 10.86w 15308.0 115881.6 = 54973.7 – 10.86w w = 5612 tonnes Load 5612 tonnes at Kg 11.45m Load 3496 tonnes at Kg 0.59m Answer for Question No. 4 KG = 0.5m KM = d + B2 2 12d KM = d + 1 2 12d For the vessel to be stable KM > KG. Putting KG = KM 0.5 = d + 1 2 12d 6d = 6d2 + 1 0= 6d2 – 6d + 1 d = 0.789m or 0.211m By inspection of box shape KM curve, vessel will be stable for RD < 0. 211 and RD > 0.789 Vessel will be unstable for 0.211  RD  0.789

KM minimum occurs at d = B = 1 = 0.408m 6 6 at this draft KM = d = 0.408 KG = 0.500 GM = - 0.092 Answer for Question No. 5 Intact volume before bilging = intact volume after bilging LBdi = LBdb – lBdh 140 x 20 x 5.8 = 140 x 20 x db – 24 x 7 x 6 16240 = 2800db – 1008 db = 17248 / 2800 = 6.16m KM intact KM = di + B2 = 5.8 + 202 = 2.9 + 5.747m = 8.647m 2 12 di 2 12 x 5.8 GM = KM – KG = 8.647 – 8.000 = 0.647m KM bilged In this case the KB must take in to account because of the fact that the underwater shape is not symmetrical. KB must be found by taking moment about the keel. Volume Kb Moment L x B x db 17248 3.08 53123.8 lxbxh - 1008 3.00 - 3024.0 16240 50099.8 KB = moment

= 50099.8 = 3.085m

Volume 16240 In this case the waterplane is intact BM = I = LB3 = B2 = 202 = 5.747m V 12LBdi 12di 12 x 5.8 GM1 = KB + BM – KG = 3.085 + 5.747 – 8.000 = 0.832m Answer for Question No. 6

15.6 secs

Answer for Question No. 7

Correct Answer : B

Answer for Question No. 8

Correct Answer : C

Answer for Question No. 9

Correct Answer : C

Answer for Question No. 10

Correct Answer : B