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Potential Energy Stored in Hydrostatic Test Unit

  1. Jul 2, 2014 #1
    I have a pressure vessel of volume V= 1 m^3 . The atmospheric pressure Patm = 101 kPa and Tatm = 22 C. I slowly pump liquid water into the pressure vessel until the internal pressure reaches P1 = 15 MPa. Also, the process is isothermal because the water was pumped in slowly and heat is exchange with the atmosphere. What is the potential energy stored in the tank of water? For example, how much energy could possible be released with the case burst? Assuming the energy stored in the case is negligible.
    Last edited: Jul 2, 2014
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  3. Jul 3, 2014 #2

    Simon Bridge

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    Good question - how is potential energy normally calculated, in general?
  4. Jul 4, 2014 #3


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    It is normal to test pressure vessels such as steam boilers with cold water. Healthy cold boilers are not as well behaved as when they are hot. It is critically important that there be very little air present in the vessel being tested since the energy is stored in the gas, with very little stored in the liquid or the vessel walls.

  5. Jul 4, 2014 #4
    How is potential energy normally calculated

    I am defining the potential energy as the difference between the total internal energy U2 at (Pmax, Tam) and U1 at (Tatm, Patm). Pmax is the maximum pressure, Tatm is the atmospheric temperature, and Patm is the atmospheric pressure. My first thought was to just look up the internal energy from the compressed liquid water tables since the state is fixed. However, the specific internal energy actually decreases as the pressure is isothermally increased. I must be reading the tables incorrectly because the total internal energy certainly must increase as the pressure increases at a fixed temperature.
  6. Jul 4, 2014 #5
    Your response does not answer the question and is misleading. Of course, as much air as possible should be evacuated. However, to say that little energy is stored in the liquid or vessel walls is wrong and dangerous. It all depends on the situation. Just look at the strain energy for a long cylinder under pressure [itex]U=\frac{V}{2E} \left(\sigma_\theta^2+\sigma_z^2\right)≈\frac{1.5rtL}{E}\frac{pr}{t}[/itex]. If the case has a high strength and low modulus the energy stored in the case can be much larger than that stored in a small pocket of air. My question is about the energy stored in the compressed liquid. I would like to keep the discussion focused on the topic because there are many worked example of the energy store in compressed gas.
  7. Jul 4, 2014 #6


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    I am sorry to have wrongly interpreted your question. I assumed by isothermal you started with 1m3 of air at 101kPa and compressed it to 15MPa by using water as a liquid piston. From my experience, more than 99% of the energy would then have been stored in the compressed gas. The worked answer to that question was in the wikipedia article I linked. It does not really matter if you extract that energy pneumatically or hydraulically, it is still mostly stored in the gas.

    Without a sufficient gas bubble present, the pressures in a closed tank can burst the tank due to daily temperature changes. How much air do you have in your tank when you start to pump in the water? Is it at 101kPa or do you pull a vacuum first?
  8. Jul 4, 2014 #7


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    If you assume that, you are being inconsistent. If there is no energy stored in the case, it won't break. But since there are no perfectly rigid materials, there always WILL be strain energy stored in the case, and that is what eventually breaks it.

    If the casing is strong enough, you can store a lot of energy in the compressed liquid. For example at the deepest point in the ocean (about 10 km) the pressure is about 1000 bar (15,000 psi), and the water volume is compressed by about 5%.
    Last edited: Jul 4, 2014
  9. Jul 5, 2014 #8

    I forgot to square the stress so the expression for the total strain energy was incorrect. It should be
    [itex]U=\frac{V}{2E} \left(\sigma_\theta^2+\sigma_z^2\right)≈\frac{1.25rtL}{E}\frac{p^2r^2}{t^2}[/itex],

    which then has the correct units of force times distance.
  10. Jul 5, 2014 #9
    There is no air in the tank. I first pull a perfect vacuum and pump water into the tank. This is a thought experiment so I can do things like pull a perfect vacuum. Also, there are no temperature changes. As I stated in the problem the process is isothermal.

    I am not being inconsistent. I stated as an assumption that the energy stored in the case is negligible so please take it to be true. The reason I made this assumption is not because it is a practical assumption in the "real world", but because it is trivial to calculate the strain energy stored in the case. I only care about the energy stored in the liquid. Further, this is a thought experiment so I can make the young's modulus of the case arbitrarily large and thus make the strain energy in the case approach zero in the limit.
  11. Jul 5, 2014 #10
    Found the Solution

    I would like to thank everyone, especially @Baluncore for their responses. It is great to see people interested in the same topic. I was able to get the correct solution even though I still do not understand the compressed liquid water tables. The solution is found by determining the ideal work it would take to compress the liquid water. The trick of course is knowing the equation of state for the liquid water. I took some searching, but the equation of state and a nice example was given in Fluid Mechanics by Granger.
  12. Jan 14, 2016 #11
    Do you still happen to have the equation for this? Or could you post the two sources you reference for ideal work and the equation of state? Thanks!
  13. Jan 14, 2016 #12


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    free has been gone from PF for almost 18 months. I wouldn't get my hopes up for a timely response.
  14. Jan 16, 2016 #13
    ah thanks. Does anyone else know a quick source or formula for ideal work to compress water? and the equation of state for water? I am certainly an amateur so if these are complex I may be out of my league....
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