Here you can get semester wise questions papers, Finally they made this available. All the best for your semester exams.
Tool Box meetings
- Tool Box meeting to be carried out prior commencement of every work.
- All Risk and control measures associated with particular task should be discussed in tool box meeting.
- Any Special Precautions to be taken, Procedure to be followed and who will do what, to be clarified.
- Any Doubts with regard to the Job to be clarified
- Permits required if any are should be identified and discussed.
What is DPA, CSO
DPA – Designated Person Ashore – Any Matters with regard to safety of the vessel, Individual or environment can be directly brought to the notice of the DPA.
When DPA is not reachable due to any reason, The Alternate DPA(ADPA) has to be contacted.
CSO – Company Security Officer – For any concerns in security of the vessel the CSO has to be contacted.
When CSO is not reachable, ACSO is to be contracted.
In a vessel of 8800 tonnes displacement and KG 6.2m, 200 tonnes of cargo was loaded in the lower hold 1.7m above the keel. Find the final KG.
From the following information , calculate the DWT available up to the tropical loadline SW :-Persent freeboards : port 1.6m, starboard 1.79m inRD 1.017. Tropical SW freeboard : 1.63m Tropical of SW draft : 9.6m FWA 150mm, TPC 20.4 .
Present freeboard port = 1.68cm
Stbd side = 1.79m
Water of RD = 1.017,
Tropical SW freeboard = 1.63c
Tropical SW draft = 9.6m
FWA = 150mm
TPC = 20.4
From the following details, Calculate the DWT available :- present free board: port 3.0m, starboard 2.9m in water of RD 1.020 FWA 200mm. TPC 30. Statutory summer freeboard 2.8m.
Solution :
For, Port Side = (3 -2.8) = 0.2m
= 20cm (above )
For , Starboard side =( 2.9 – 2.8) =0. 1m
= 10cm above
Hence ,Mean freeboard can be calculated as
= (20+ 10)/2
= 30 /2 cm
= 15cm available
As we know:
Change in draft =
(Change in RD )x(FWA)
0.025
= (1.025 – 1.020 ) x 20 /0 .025
= 4cm
Now, Total sinkage = (15 + 4)cm
= 19cm
TPC = 30( given)
TPC = (A/ 100) x density
30 = (A/100) x 1.025
A = 30/ 1.025
= 29.268m2
Again for calculating ,TPC at RD 1.020
= (A/100) x ( 1.020)
= 29.85 t/cm
Now , Cargo can be load =( TPC x sinkage)
= (29.85 x 19)
= 567.15 t
A vessel is loading in a SW dock and is lying with her starboard winter loadline 60mm above and her port winter loadline 20mm below the surface of water. if her summer draught in SW is 7.2m and TPC is 20,, find how many tonnes of cargo the vessel can load to bring to her down to her tropical loadline in SW.
Solution:
Starboard winter load line = 60mm above
Port winter load line = 20mm below
Mean draft = ( 60 – 20)/2
= (40/2)
= 20mm
= 2cm
Summer SW draft = 7.2m( given)
We know that:
Waterline (W) = (1/48 of summer draft)
=(1/48 x 7.2)
= 0.15m
= 15cm
Total sinkage available to bring the vessel to her tropical load line = ( 2+ 15 + 15)
= 32cm
TPC =20( given)
So, Cargo can be load = (TPC x sinkage)
=(20 x 32)
= 640t.
A vessel is floating in dockwater of RD 1.005with her starboard WNA mark 30mm below and her port WNA mark 60mm below the waterline. If her summer SW draught is 8.4m, TPC is 30 and FWA is 160 mm, calculate how much cargo can be loaded to bring the vessel to her vessel draught in SW.
Solution:
Dockwater RD = 1.005,
Stbd WNA mark = 30mm below
Port WNA mark = 60mm below
Mean draft = (30 + 60) / 2
= 45mm
=4.5cm,below the water line.
Distance from WNA to water(W) = 50mm
= 5cm
So, sinkage available = (50 – 45)
= 5mm
= 0.5cm
We know that :
Water (W) is 1/48 of summer draft
= (1/48 x 8.4)
= 0.175m
= 17.5cm
Sinkage = (17.5 + 0.5 )
= 18cm
Total sinkage = (18 + 12.8)
= 30.8cm
As we know:
Change in draft =
(Change in RD )x(FWA)
0.025
= (1.025 – 1.005 )x 16 / 0.025
= 12.8cm
Again, TPC = (A/100) x density
30 = (A/100) x 1.025
A = 30/ 1.025
A = 29.268 m2
Now , TPC for density of 1.005
TPC = (A/100) x 1.005
= (29.268 x 1.005)
= 29.41 t/cm.
Hence ,cargo can be loaded = (30.8 x 29.41)
= 905.82 t
A vessel is lying in a river berth of density 1.010 tonnes per m3 , with her summer loadline 20mm above the water on the starboard side and 50mm above the water on the port side. Find how much cargo she can load to bring her to her to her summer loadline in SW, if her summer displacement is 15000 tonnes and TPC is 25.
Solution:
RD of river = 1.010
Summer load line on starboard side = 20mm above
Port side = 50mm above
Mean draft = ( 20 + 50) / 2
= (70/2)
= 3.5cm above
W = 15000t,
FWA = (W/TPC)
= 15000/(40 x 25)
= 15cm
As we know:
Change in draft =
(Change in RD )x(FWA)
0.025
= (1.025 – 1.010 ) x 15 / 0.025
= 9cm
Total sinkage = (9 + 3.5 )
= 12.5cm
TPC = 25( Given)
We know that:
TPC = (A/100) x 1.025
= (25 / 1.025)
= 24.39 m2
Now, TPC for RD 1.010 = (A /100) x( 1.010)
= (24.39 x 1.010)
= 24.63t/cm
Hence ,Cargo can be load = (sinkage x TPC)
= (12.5 x 24.63)
= 307.875 t
A vessel’s statutory freeboard is 2.0m. She is loading in DW of RDD 1.015 and her freeboard is 2.1m . TPC =24. FWA = 200mm. find the DWT available .
Solution:
Statutory freeboard is = 2.0m
DW RD = 1.015 & Freeboard = 2.1m
TPC = 24, FWA = 200mm = 20cm
As we know :
Change in draft =
(Change in RD )x(FWA)
0.025
= (1.015 – 1.025 ) x 20 /0.025
= ( 0.010 x 20) /0 .025
= 8cm
Hence , actual freeboard available =( 2.1 + 0.08)m
= 2.18m
Sinkage available = ( 2.18 – 2.00)
= 0.18m
= 18cm
TPC = 24 ( Given)
24 = (A / 100) x(1.025)
A = (24 x 100)/ 1.025
A = (24 / 1.025)
= 23.414 m2
Now, TPC for RD 1.015 = (A/ 100) x(1.015)
= (23.414 x 1.015)
= 23.76 t/cm
DWT available = (sinkage x TPC)
= (18 x 23.76 )
= 427.68t.