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Fluid Mechanics - Water, Ridiculous Answer

  1. Sep 10, 2012 #1
    Water at standard atmospheric pressure and temperature fills a sealed, rigid container. If the temperature of the water is increased by 50°C, what is the pressure?

    V = volume, P = pressure, T = temperature

    Coefficient of compressibility, kappa = (-1/V)(dV/dP)
    Coefficient of thermal expansion, beta = (1/V)(dV/dT)

    I solved for dV/V from the coefficient of thermal expansion equation:
    dV/V = beta*dT.

    Plugging into coefficient of compressibility equation:
    kappa = -(beta*dT)/dP.

    Solved for dP = -(beta*dT)/kappa, changed the d's into deltas.

    Plugging kappa = 5.18 * 10^-10 and beta = 1.48 * 10^-4 into delta P equation gave me delta P = -92.3 MPa. Is this reasonable at all? I mean, the pressure difference seems a bit large.
     
  2. jcsd
  3. Sep 11, 2012 #2
    Anyone?
     
  4. Nov 22, 2012 #3
    Hi eurekameh,

    Now without even looking through your equations I can tell you that they are wrong, or you have made a mistake somewhere.

    An increase in temperature will cause an increase in pressure, vice versa... Look up how refrigerators and heat pumps work, also think of your deodorant can - it loses pressure and gets colder.

    I'm afraid I can't help you much more as my brain is currently fried from work.
     
  5. Nov 22, 2012 #4
    The magnitude of your answer is correct, but the sign is wrong. Write the equation of the change in V in terms of the changes in P and T. Then set the change in V equal to zero in the equation.
     
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