Potential Energy Stored in Hydrostatic Test Unit

In summary, a pressure vessel of volume 1 m^3 with atmospheric pressure of 101 kPa and temperature of 22 C is slowly pumped with liquid water until the internal pressure reaches 15 MPa. The process is isothermal and the potential energy stored in the tank of water is calculated as the difference between the total internal energy at the maximum pressure and atmospheric conditions. It is important to evacuate all air from the vessel to ensure that the majority of the energy is stored in the compressed gas rather than the liquid or vessel walls. However, some strain energy is still present in the case and can be significant depending on the strength and modulus of the material. To prevent potential bursting, a sufficient gas bubble should be present in the tank to counter
  • #1
free
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0
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.
 
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  • #2
Good question - how is potential energy normally calculated, in general?
 
  • #3
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.

http://en.wikipedia.org/wiki/Compressed_air_energy_storage#Isothermal_storage
 
  • #4
How is potential energy normally calculated

Simon Bridge said:
Good question - how is potential energy normally calculated, in general?

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.
 
  • #5
Baluncore said:
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.

http://en.wikipedia.org/wiki/Compressed_air_energy_storage#Isothermal_storage

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.
 
  • #6
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?
 
  • #7
free said:
Assuming the energy stored in the case is negligible.

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%.
 
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  • #8
Correction

free said:
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.

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.
 
  • #9
Baluncore said:
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?

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.

AlephZero said:
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%.

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.
 
  • #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.
 
  • #11
free said:
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.

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!
 
  • #12
ceej said:
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!
free has been gone from PF for almost 18 months. I wouldn't get my hopes up for a timely response.
 
  • #13
SteamKing said:
free has been gone from PF for almost 18 months. I wouldn't get my hopes up for a timely response.
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...
 

1. What is the concept of potential energy stored in a hydrostatic test unit?

The potential energy stored in a hydrostatic test unit is the energy that is stored in the unit due to the difference in height between the water level inside the unit and the level of the water outside the unit. This energy can be released and used to perform work, such as pressurizing the system for testing purposes.

2. What factors affect the potential energy stored in a hydrostatic test unit?

The potential energy stored in a hydrostatic test unit is affected by the height of the water column, the density of the water, and the gravitational acceleration at the location of the unit. Additionally, the volume and shape of the unit can also impact the potential energy stored.

3. How is potential energy stored in a hydrostatic test unit calculated?

The potential energy stored in a hydrostatic test unit can be calculated using the formula PE = mgh, where PE is the potential energy, m is the mass of the water, g is the gravitational acceleration, and h is the height of the water column.

4. What is the difference between potential energy and kinetic energy in a hydrostatic test unit?

Potential energy in a hydrostatic test unit refers to the energy stored in the unit due to the water column's height, while kinetic energy refers to the energy of the moving water as it is released from the unit. Kinetic energy is a form of energy in motion, while potential energy is a form of stored energy.

5. How is the potential energy stored in a hydrostatic test unit used in testing procedures?

The potential energy stored in a hydrostatic test unit is used in testing procedures to pressurize the system being tested. As the water is released from the unit, the potential energy is converted into kinetic energy, which creates pressure within the system. This pressure allows for testing of the system's strength and integrity.

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