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Buoyancy of a capsized ship

  1. Jan 28, 2014 #1

    Maylis

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    1. The problem statement, all variables and given/known data

    The battleship Oklahoma was attached at Pearl Harbor by Japanese dive bombers. The ship was hit by multiple torpedoes and capsized (rolled 180 degrees so the bottom of the hull pointed straight up). The final orientation of the ship had 20 percent of its height above the sea level. Approximate the ship dimensions as a box 50 feet tall, 80 feet wide, and 500 feet long. Determine the ship’s weight on the morning of the attack. The specific gravity of sea water is 1.03.

    2. Relevant equations



    3. The attempt at a solution

    For this problem, I don't know if there is actually a buoyant force. I know the air underneath the water is compressed, thus its pressure increases. I also don't know where the water line underneath the tank should be relative to the sea level.
     

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  2. jcsd
  3. Jan 28, 2014 #2

    SteamKing

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    You don't really have enough information to calculate the size of any air bubble inside the hull.

    As a first cut, assume the hull is floating and calculate the displacement given the s.g. of the water and the dimensions given for the hull (including the height which remains above water).
     
  4. Jan 28, 2014 #3

    SteamKing

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    A quick note about your calculations: fresh water has a weight of 62.4 lbf / cu. ft., not lbm. Multiplication of the amount of water displaced by g = 32.2 ft/s^2 is not necessary.
     
  5. Jan 29, 2014 #4

    Maylis

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    Is there anyway to know if the water level inside the hull is the same as the water level of the ocean or if it is higher/lower?? You say there is not enough information to know the size of the air bubble in the hull, but shouldn't it just be the 10 ft times the length and width?

    Also, how come you suggest that I include the volume of the hull that is above the water line in finding the volume displaced of the sea water?
     
    Last edited: Jan 29, 2014
  6. Jan 29, 2014 #5
    I guess the problem statement intends for you to include the water contained inside the hull as part of the weight of the ship. Under these circumstances, the location of the air-water interface inside the ship is irrelevant to the solution.
     
  7. Jan 30, 2014 #6

    nvn

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    Maylis: The problem statement is slightly unclear. Therefore, we must try to interpret it using a logical process.

    (1) If the box top (which is down) is open, is there any way to know how much water the box took on as it capsized, without knowing the box weight? No, I currently do not think so, at least not easily.

    (2) Without knowing the initial water level in the upside-down box before the air became compressed, is there any way to know the final water level, inside the box, without knowing the box weight? No, I currently do not think so.

    (3) Is the water level inside the box the same as the ocean surface level? No, unless the box is weightless. If the box is not weightless, then I currently think the water level inside the box must be below the ocean surface. (I am stating this currently without a mathematical proof. But I think we might be able to prove it mathematically, if we tried. Here is a nonmathematical proof. We know the air pressure inside the box is higher than atmospheric pressure. Therefore, we know the water level inside the box must be below the ocean surface.)

    (4) Even if we blindly assume the initial water level in the upside-down box, before the air became compressed, is at the box top (as Maylis currently assumed in post 1), is there any way to know the final water level, inside the box, without knowing the box weight? No, I currently do not think so. We only know that the final water level would be below the ocean surface, but we do not know where.

    (5) Because the answer to questions 1, 2, and 4 is no, then it appears the only logical assumption is, the problem statement means the given box top (which is down) is closed. I.e., it means the given box is an airtight, closed box, and it contains only air, no water. If you use this interpretation, see if you can now solve the problem.

    Maylis: Even though I currently disagree with (and cannot match) your answer in post 1 because of my items 2, 3, and 4, above, I can at least tell you your exponent on the second to last number written on your attached file page appears to be a typographic mistake.
     
    Last edited: Jan 30, 2014
  8. Jan 30, 2014 #7

    SteamKing

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    I did not mean to imply this. I meant that the draft of the box was the number to be used in calculation displacement.
     
  9. Jan 30, 2014 #8

    Maylis

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    Nvn: To be clear, is your assumption that there is no water in the box, regardless of whether it is capsized or not, and it is just air? If that were the case, I imagine that it would be no different whether the ship was capsized or not, it'd be like flipping a box over in a pool, so nothing happened to it and the same dimensions would have been underwater even before it was capsized.

    I know when asking him for hints he drew the picture with the water line inside the hull, so when the water got into the hull it compressed the gas.
     
    Last edited: Jan 30, 2014
  10. Jan 30, 2014 #9

    SteamKing

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    Who he?
     
  11. Jan 30, 2014 #10

    Maylis

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    The professor
     
  12. Jan 30, 2014 #11

    nvn

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    Yes, it was my assumption in item 5 of post 6 ... at least until you posted post 8. :biggrin:

    Correct.

    OK, then this might set us back a little, and we will need to keep thinking, and try a little harder.

    OK, so you are claiming the box is open. If so, then should we assume the box contained only air before the air became compressed? I currently think this is a logical assumption.
     
    Last edited: Jan 30, 2014
  13. Jan 30, 2014 #12

    Maylis

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    Yes, that is correct. When the ship was right side up, it only contained air
     
  14. Jan 31, 2014 #13
    As I said before, the problem wants you to assume that the hull is partially full of water. The compressed air doesn't matter at all, since its density is negligible compared to the water, even when compressed. The air is assumed to be above the water in the hull, trapped in the hull. What you are trying to find is the weight of the ship, including the weight of the water inside. You just use Archimedes principle on the combination of hull plus water inside.
     
  15. Jan 31, 2014 #14

    nvn

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    Chestermiller: No, as Maylis implied in post 12, the problem statement appears to want you to assume the box contained only air (no water), before the air was compressed. After the air is compressed, then yes, the box partially contains water. But we do not know the final water level inside the box, until we solve for the final water level. True, the compressed air weight does not matter at all, but the compressed air matters a lot, because the compressed air pressure determines the final water level inside the box. True, the hull weight plus the water weight inside the box (after the air is compressed) gives you a total initially displaced water weight, but that, by itself, does not tell you how much of that total weight is water weight versus hull weight. You would still need to solve for the final water level inside the box, to be able to determine the hull weight. The problem statement in post 1 is asking for ship weight, before the attack, not ship weight plus water weight.

    Maylis: I currently agree with you that the problem statement intends for you to assume the box initially contained no water, before the air became compressed.

    I retract my above statement, quoted from post 6. The above statement was incorrect, and not true.

    Maylis: Let us use the following nomenclature.

    h2 = distance below the ocean surface of the final water level inside the box.
    p1 = initial absolute air pressure inside the box before the air is compressed = patm.
    p2 = final absolute air pressure inside the box after the air is compressed.
    p3 = atmospheric absolute air pressure above the upside-down box, and also atmospheric absolute air pressure on the ocean surface = patm.
    W = hull (ship) weight, not including water.
    V2 = final air volume inside box.
    L = box length.
    b = box width.
    t = box height = 15.240 m.
    A = box horizontal area = L*b.
    rhow = ocean water density = 1030 kg/m^3.
    go = earth gravitational acceleration = 9.81 m/s^2.​

    The final absolute air pressure at the water surface inside the box (p2) is transmitted directly upward, through the air, to the inside surface of the upside-down box. At the same time, outside atmospheric absolute air pressure (p3) presses downward on the outside surface of the box. Therefore, the buoyant force on the box is, Fb = p2*A - p3*A = (p2 - p3)*A.

    Summation of all vertical forces on the box is, summation(Fy) = 0 = Fb - W.

    How many unknown variables do you have? You have four unknown variables, which are p2, V2, h2, and W.

    How many equations do you have? You have four equations, as follows.

    (1) W = (p2 - p3)*A.
    (2) p1*V1 = p2*V2.
    (3) V2 = L*b*(0.20*t + h2).
    (4) __________.​

    We are not allowed to give you a complete solution, so you will need to fill in the blank for eq. 4.

    Hint 1: Eq. 4 is as follows. Do you know how to compute the difference in pressure between two elevations of water? I think you do. Therefore, write that equation for eq. 4. It is a very common equation. The two water surfaces here are separated by a vertical distance of h2. One water surface has pressure p3, and the other water surface has pressure p2.

    Now that you have four equations and four unknowns, see if you can solve the problem.
     
    Last edited: Jan 31, 2014
  16. Feb 1, 2014 #15
    Thanks nvn. I can see how you could interpret the problem statement this way. That's not how I interpreted it, but this makes for a much more interesting problem. With that said, I am in agreement with the way you set up the model in your previous post. I'm glad you left out Eqn. 4, since deriving that should pose an interesting challenge for the OP.

    Chet
     
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