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Absolute Pressure

  1. Apr 29, 2017 #1
    1. The problem statement, all variables and given/known data
    A 410kg bluefin tuna rests motionless at a depth of 15m below the ocean's surface. The absolute pressure at this depth is 251,700 Pa. What is the volume of the fish?

    2. Relevant equations
    absolute P=gauge Pressure+atmospheric pressure
    1atm=1.03e5Pa
    P=density x g x height
    density=mass/volume
    3. The attempt at a solution
    I first converted to atmospheres. So 251,700 Pa is 2.444 atm.
    2.44atm=(410/V)(9.8)(15)+1atm
    1.44atm=410/V x 9.8 x 15
    V=41854 m3

    This cannot be right...
     
  2. jcsd
  3. Apr 29, 2017 #2

    haruspex

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    You need to stop and think what that expression means. You are taking the tuna's density and finding the pressure at a 15m depth in a liquid of that density. I.e. you are taking the tuna as reaching all the way to the surface.

    This is a somewhat tricky problem. In general, if body floats neutrally in a fluid, what does that say about the body's density in relation to the fluid?
     
  4. Apr 29, 2017 #3
    Ok, I think that would mean the body would have the same density as the fluid. But why does the above mean the tuna is reaching all the way to the surface? Are we calling where the tuna is as h=0?
     
  5. Apr 29, 2017 #4

    haruspex

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    Well, you explain to me the basis for multiplying the tuna's density by g and its depth.
    You applied the equation P=density x g x height, but you cannot just apply equations because you happen to have variables of the right kind. An equation only applies in a specified context, where the variables in it are related in a specific way. In F=ma, F is the net force on an object an m is the mass of the object, not the mass of something else that's lying around. In P=ρgh, what do ρ, h and P represent? ρ is the density of what? h is what distance?
    Right, so consider this: if the water has a uniform density all the way down, what has the depth to do with anything? If it can float neutrally at one depth it can float neutrally at any depth.
     
  6. Apr 29, 2017 #5
    I guess I do not understand the equation well. In P=pgh, density would be referring to that of the object, Pressure would be referring to that of the incompressible fluid, and h the height of the column of fluid above location. This is what I have in my notes. If we multiply the tuna's density x g x depth, would we be finding the tuna's pressure and not the ocean?

    So because the tuna is not floating, its density must be different from the water. Do we not need to include the depth?
     
  7. Apr 29, 2017 #6

    haruspex

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    No. That equation is independent of any immersed object. It gives you the pressure at depth h in a fluid of density ρ.
    It is floating, not at the surface but fully immersed. It still counts as floating.
     
  8. Apr 29, 2017 #7
    Since we are given the pressure at the depth, is it redundant to include it in the equation?

    ok, I see what you mean about water not changing density with depth. But how do we solve for the fish volume without dealing with the tuna's density?
     
  9. Apr 30, 2017 #8

    haruspex

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    Since the fish is floating immersed it has neutral buoyancy. What is the relationship between the tuna's density and that of the surrounding water?

    How can you find the density of water at that depth?
     
  10. Apr 30, 2017 #9
    The tuna's density must be the same as the surrounding water. Would we do this:

    absolute pressure of water=density of water x gravity x depth and solve for density of water?

    we could then equate density=m/v and we have the mass of the fish.

    2.44atm=density x 9.8 x 15m
    density=.01698

    .01698=410kg/V
    V=24701 m^3

    this still can't be right.
     
  11. Apr 30, 2017 #10

    haruspex

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    Watch the units! You have calculated density in units of atm s2m-2. Doesn't seem useful.
    Try not converting to atm in the first place.
     
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