Blog, 4th Dec 2019, MUMBAI

I am thinking of writing my day to day experience so that people know how it feels in merchant navy line. Yes making videos will be more convenient for you guys to watch but that is a bit time consuming but i will definitely start making videos.

It is very fascinating as it offers 6 month of job and 6 month of rest or free time. But the actual story is quiet different initially as a deck cadet i will get a contract of 9 Months (+- 1 month). And this is really a long time almost a year. So it is going to be difficult as i never stayed away from home for this long. Have stayed Maximum of 6 months and that felt long at the end.

There are so many Questions poping in my mind right now.

  1. Is Merchant Navy a Right choice ?
  2. Is it a good choice for long term like 10-12 years?
  3. How it feels on board?
  4. Work culture?
  5. Away from family & friends for 9 months?

Answering all these questions require Experience, So what i will basically do is wait & experience things.And put everything on this blog column of the website and even try putting videos for everyone.

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.

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

qes 18

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.

qes 17

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