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Question Fom Readers # 1: How Much More Cargo Can Loaded

How much more cargo can she load in order to be at her summer load-line on reaching salt water?

One of the readers of this blog named MarCo, has submitted a question for you and me to answer his question. This is probably a response from the author's invitation, which is yours truly, who encourage readers to submit questions so that you and me may help solve their problem, so to speak. And this particular question is regarding on cargo loading.

Before anything else, I have to say that I am no expert, so I am open to be corrected. However, I will try to answer your question to the best of my knowledge. Here is the question:

A vessel has a summer draft of 11.184m, TPC 46.99tons and a FWA of 254mm. She is loaded in a berth where the water RD is 1.011 and her present mean draft is 10.98m. How much more cargo can she load in order to be at her summer load-line on reaching salt water?

Analysis of the problem:


If a ship is coming from the sea going to fresh or brackish water, what will happen to the draft is that it will increase. And this will result to decrease to ship's freeboard.

While a ship from fresh or brackish water going to the sea will decrease her draft, thus increasing ship's freeboard as in the case of this problem.

So now, we have to get the dock water allowance for our ship's summer draft. Here's the formula for getting dock water allowance:

         Dock Water Allowance = 254 (1.025 - 1.011)
                                                             25
                                              = 254 x 0.014   
                                                        25
                                              = 3.556
                                                    25
         Dock Water Allowance = 0.14224 meters

So with our dock water allowance above, it means that with regards to the density of water where our ship is berthed and loaded, we can add 0.14224 m to the summer draft of 11.184 m.

            So, 11.184 m
               +   0.14224 m
                  11.32624 m     Now our mean draft is 10.98 m. So how much more tons shall we load?
               - 10.98 m           We have to subtract to get the difference.
                   0.34624 m 
               x         100         To convert it to centimeter so that we can apply the TPC.
                   34.624 cm      Now since tons per centimeter immersion is 46.99 tons, so multiply it by
               x   46.99          
                   1626.98 tons is the additional cargo to be loaded.

I hope this answer to the question. But again, I am open to correction. So if my answer is wrong, correct me.

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10 comments:

  1. MarCoMay 14, 2014 at 9:30 PM

    The answer on the reviewer says 1603.7 tons and im still trying to figure out the solution, i guess i should move on in my review and later find the solution. thank you anyway for answering, i got the same answer when i solved it at first. thanks again!

    ReplyDelete
    Replies
    1. NautaMay 14, 2014 at 10:46 PM

      I can relate because that was also my predicament during my review that there were answers that is not exact.

      So I asked my instructor on how to deal if I could encounter those kind. And his advice to me is to select the nearest answer to your solution. For the instruction in board examination is not to select the exact, but rather to select the best answer.

      Delete
    2. MarCoMay 14, 2014 at 11:09 PM

      noted mate! thanks!

      Delete
    3. UnknownJune 25, 2014 at 2:48 PM

      thanks for the solution :) great help mate :)

      Delete
    4. UnknownJanuary 5, 2021 at 12:55 PM

      You still need to convert the TPC into the brackish water density. Which will be = TPC× (r.d/1.025)

      Delete
    5. NautaJanuary 6, 2021 at 9:11 AM

      Thank you

      Delete
  2. Lifes of the WomansFebruary 5, 2021 at 5:20 PM

    This comment has been removed by a blog administrator.

    ReplyDelete
  3. CaptainAugust 20, 2021 at 3:56 AM

    the solution to the problem solved above is wrong.
    because:
    summer draft 11.184 ships with a TPC value of 46.99 tons. Since the TPC is 1.025 tons/cub.m, the TPC in DW should be multiplied by 34.624 cm by correcting RD 1.011.
    solution
    TPC (sw 1.025) = 46.99 tons TPC (dw 1.011) = 46.35 tons
    The cargo to be loaded will be 1,604.76 tons

    ReplyDelete
  4. AnonymousApril 19, 2024 at 7:39 PM

    You are in a barge transiting a river with brackish water of 1.015 density, your draft is 5.70 m and the under-keel clearance required for you to pass the next phase of your destination is 2.56 m with almost the same draft in a river with what water density?
    Just want to ask the solution of this.thank you

    ReplyDelete