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Naval Architect

  1. Dec 20, 2016 #1
    hi Im a newbie here,
    from Philippines,

    I need you help guys i want to know how to calculate the lightweight of the barge since i only have the lenght =28 m, breadth=10m, depth=3m, light draft=0.8m,debsity of sea water=1.025t/m3, cb=0.9 and grt=237 tons. im so confused how to compute since i dont know also the deadweight and displacement of the barge.

    Thank you for immediate response.

    [Mentor's note: Thread moved from engineering forum to homework forum]
     
    Last edited by a moderator: Dec 22, 2016
  2. jcsd
  3. Dec 20, 2016 #2
    help me also to get the deadweight and the displacement especially light weight of this barge,the barge is box type only. please help me guys.

    thank you.
     
  4. Dec 21, 2016 #3

    Baluncore

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    Science Advisor

    Displacement of unladen vessel = lightweight ?
    A rectangular box displaces volume = length * beam * draft = 28 * 10 * 0.8 = 224 cubic metres.
    Then convert volume to weight. volume * density = 224 * 1.025 = 229.6 tonne.

    What is cb=0.9 ?
     
  5. Dec 22, 2016 #4
    Cb=block coefficient of the barge,
    yes im looking for a lightweight, of the barge.
    but, my problem is i already used that formula but still my answer is still big for the light weight of that barge, so im confused for the correct computation of lightweight of the barge.
     
  6. Dec 22, 2016 #5

    Baluncore

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    Science Advisor

    What is the correct lightweight of the barge ?
     
  7. Dec 22, 2016 #6
    I don't know. only I know my Boss told me the lightweight i give to him is to big for that barge, and he didn't give me any estimated value of lightweight, that's why im so confused. the exact value of lightweight for that barge. ; (
     
    Last edited by a moderator: Dec 23, 2016
  8. Dec 23, 2016 #7

    gneill

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    Staff: Mentor

  9. Dec 23, 2016 #8

    Baluncore

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    Science Advisor

    The empty hull has a lightweight draft of 0.8 m, the total displacement will then be ;
    28 * 10 * 0.8 * 0.9 * 1.025 = 206.64 tonne lightweight.
    If your Boss thinks that is wrong then ask him to explain why.

    The GRT is the internal volume measured in multiples of 100 cubic feet = 2.832 m3.
    GRT = 237 means 237 * 2.832 = 671.184 m3 internal volume.
    The external dimensions give an external volume of 28 * 10 * 3 * 0.9 = 756 m3
    The 85 m3 difference is the volume of the hull structure and the unusable space.
    Those figures are consistent, which suggests the 3 m depth is the height of the barge wall, not the fully laden draft.

    Remember that GRT is a volume, not a weight.
    But if the GRT volume available was loaded with a deadweight of 237 tonne,
    then the total mass would be 206.64 + 237 = 443.64 tonne displacement.
    The draft would then be 443.64 / (28 * 10 * 0.9 * 1.025) = 1.717 m. There would be 1.28 m freeboard.

    Without better defined information we cannot tell anything more about this problem.
     
  10. Dec 23, 2016 #9
    Good Morning,

    Thank you so much Baluncore, I really appreciate your help for me to solve this problem.

    Have a Merry Christmas :-)
    God Bless you
     
  11. Dec 23, 2016 #10
    Good morning,

    Thank you also Gneill. Have a Merry Christmas ☺
     
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