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Volume Stress and Bulk Modulus

  1. Jun 19, 2009 #1
    1. The problem statement, all variables and given/known data

    A cylindrical steel pressure vessel with volume 1.31 m^3 is to be tested. The vessel is entirely filled with water, then a piston at one end of the cylinder is pushed in until the pressure inside the vessel has increased by 1000 kPa. Suddenly, a safety plug on the top bursts. How many liters of water come out?

    2. Relevant equations

    B=0.2x10^10N/m^2
    P1=P0+(rho)gh
    P1=-B((Delta V)/V)

    3. The attempt at a solution

    I am assuming that I have to look for delta V as that would be the water that comes out causing the change in volume.

    Delta V=-V(Delta P)/B=-1.31(1000)/(0.2x10^10)
    Delta V= 6.55*10^-7

    But this is not the right answer. I am confused as to where I'm going wrong.
    ------------------------------------------------------------------------------------------------------------------
    A second problem

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

    A long, thin rod of mass M and length L is standing straight up on a table. Its lower end rotates on a frictionless pivot. A very slight push causes the rod to fall over.

    a) As it hits the table, what is the angular velocity of the tip of the rod?
    b) What is the speed of the tip of the rod?

    2. Relevant equations
    v = wr
    a = v^2/r=(w^2)r
    v= (2(pi)(r))/T

    3. The attempt at a solution

    I am having trouble picturing what the question is asking.

    a) w = v/r = (2(pi)r)/T/r = 2pi/T...don't know where to go from here

    b) I think after I find w in a i just use v = wr
     
    Last edited: Jun 19, 2009
  2. jcsd
  3. Jun 19, 2009 #2

    LowlyPion

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    For 1) don't you have 1000 kPa and not 1000 Pa?
     
  4. Jun 19, 2009 #3

    LowlyPion

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    For 2) consider the moment of inertia of the bar. Consider too the energy in the system as it starts to fall. The total potential energy of the bar should be rotational kinetic energy when it's at the horizontal shouldn't it?
     
  5. Jun 20, 2009 #4
    Yeah its kpa, so in the formula I should use 1 000 000pa and I get 6.55*10^-4 m^3 which will be 0.655L...is that right?
     
  6. Jun 20, 2009 #5

    LowlyPion

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    If your bulk modulus for water is good.

    I might use 4.58 * 10-10 for compressibility k, where k = 1/B, which yields a slightly smaller result.

    http://hyperphysics.phy-astr.gsu.edu/hbase/tables/compress.html#c1
     
  7. Jun 20, 2009 #6
    Ok I made potential energy = rotational energy

    so i get:

    Ug = Krot
    MgL = 1/2 (1/3ML^2)w^2
    GL = 1/6 (L^2)(w^2)
    sqrt(6g/L) = w

    so I put the answer as sqrt (6g/L) on the online system and its wrong, but I get this message:
    "Your answer either contains an incorrect numerical multiplier or is missing one."

    Where did I go wrong?
     
  8. Jun 20, 2009 #7
    We are supposed to use the bulk modulus from our textbook, and that one is 0.2*10^10. Anything else would give a wrong answer in the system. So with this bulk modulus, is 0.655L right?
     
  9. Jun 20, 2009 #8

    LowlyPion

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    For 2) your potential energy will be the height of the center of mass won't it?
     
  10. Jun 20, 2009 #9

    LowlyPion

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    That what it looks like to me then.
     
  11. Jun 20, 2009 #10
    oh so it should be sqrt (3g/L) right?
     
  12. Jun 20, 2009 #11

    LowlyPion

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    Yes.
     
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