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Elasticity Pressure

  1. May 3, 2008 #1
    Elasticity!! Pressure!!

    1. The problem statement, all variables and given/known data
    A cylindrical steel pressure vessel with volume 1.3m^3 is to be tested. The vessel is filled with water, then a piston at one end pushes until the pressure inside increases by 2000kPa, then suddenly a safety plug on the top bursts.
    how many litres of water come out???
    B=.02x10^10Pa


    2. Relevant equations

    P1=P0+[tex]\rho[/tex]gh
    P1=-B([tex]\Delta[/tex]V/V)
    3. The attempt at a solution

    P1=P0+2000kPa
    [tex]\Delta[/tex]V=-V(P1/B)=-1.3(P0+2000/.2x10^10)m^3
    please i need help.... i got no idea where to go from now.... i dont have density, or a height, or youngs modulus of steel?? plz can someone help?
     
    Last edited: May 3, 2008
  2. jcsd
  3. May 3, 2008 #2
    am i missing any formulas???? coz im really stuck with this one
     
  4. May 3, 2008 #3

    alphysicist

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    Hi fredrick08,

    The container originally held 1.3 m^3 of water; once the pressure has increased (by the piston being pushed inwards) how much has that volume decreased? (This would be related to how far were they able to push in the piston, but you find it using the equation you have with [itex]\Delta V[/itex].) What number do you get?

    Once the safety plug burst, the water goes back to its original volume (since it's open to the atmosphere), but the container is still at its new volume. So how much water escapes?
     
  5. May 3, 2008 #4
    ok yes, but how can i find the distance?? coz i dont know the initial pressure??
     
  6. May 3, 2008 #5
    i mean how can i find the distance the piston goes in.... coz dont i need the initial pressure, coz the pressure increases by 2000kPa??
     
  7. May 3, 2008 #6

    alphysicist

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    I just mentioned the distance to show what was happening in the experiment; you cannot find the distance here.

    What you want to find first is the change in the volume, using the equation you had in your first post. What number do you get for the change in volume [itex]\Delta V[/itex]?
     
  8. May 3, 2008 #7
    i dont know, coz the [tex]\Delta[/tex]V=-1.3(P0+2000kPa/.2x10^10) i dont understand how i can find this because i dont know P0????
     
  9. May 3, 2008 #8

    alphysicist

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    P0 is the pressure before the piston began pushing, when the water was just poured into the container. So it would be atmospheric pressure.
     
  10. May 3, 2008 #9
    oh ya ok ty lol ok well [tex]\Delta[/tex]V=-1.3x10^-3m^3
     
  11. May 3, 2008 #10
    which is 1.3L??? is this right??
     
  12. May 3, 2008 #11

    alphysicist

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    It looks like your formula is a bit off. You just need the change in pressure, which is the 2000kPa. The formula is

    [tex]
    \Delta P = -B \frac{\Delta V}{V}
    [/tex]

    so

    [tex]
    \Delta V = -V (\Delta P)/B
    [/tex]
     
  13. May 4, 2008 #12
    ok thankyou, this book im using is hopeless..... btw that other tank question i got rite, ty for all help = )
     
  14. May 4, 2008 #13

    alphysicist

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    Sure, glad to help!
     
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