Pressure relief from 15,000 psi to 0

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To relieve pressure from 15,000 psi to 0 psi in a 2.5" diameter tube using water at 75°F, the incompressibility of water complicates the calculation, as water's compressibility increases under high pressure. While typically considered incompressible, at 15,000 psi, water's volume may decrease by about 5%, equating to a reduction of approximately 3 inches in a 60-inch tube. The valve's design, particularly if it's a pressure relief valve, may also impact the ability to reach 0 psi without an atmospheric vent. The discussion highlights the need for additional tube length to accommodate this pressure drop, while also emphasizing the importance of understanding the medium's properties under extreme conditions. Accurate calculations are crucial for safe and effective pressure management in hydraulic systems.
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If, I have a 2.5" dia X 60 inches tube, at a internal pressure of 15,000 psi. If the valve is opened (orifice opening of 3/4" Dia), what will be the ( same dia) additonal length of tube that will be needed to relieve the pressure from 15,000 psi to 0, Using water as a medium at 75 deg F. The goal is to bring down the presure to 0 before it is existed out of the equipment.

if there are any other suggestions , I am open to it.

PP
 
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I'm missing something in your question.
Water is essentially incompressible, so any opening will relieve the pressure.
Tube size and valve throat diameter are not really needed afaik, but there may be more to the question than what you have given.
 
Thanks for replying so fast, example: take a hydraulic cylinder. when I compress the piston to a length of 60" ( ID of the cylinder is 2.5" dia ) I reach a pressure of 15,000psi. Now , when I let go the piston to retrieve back , assuming I have enough length for the piston to travel. what will be the length of the cylinder the piston has to travel so that the pressure in the cylinder will be 0 ?.
 
As etudiant said, water is essentially incompressible, this question therefore doesn't make much sense as it is stated. If you were talking about air (or any other compressible medium) the question would make sense.
 
pankajmala said:
If, I have a 2.5" dia X 60 inches tube, at a internal pressure of 15,000 psi. If the valve is opened (orifice opening of 3/4" Dia), what will be the ( same dia) additonal length of tube that will be needed to relieve the pressure from 15,000 psi to 0, Using water as a medium at 75 deg F. The goal is to bring down the presure to 0 before it is existed out of the equipment.

Others have pointed out that water is usually approximated as being incompressible. But at 15,000 psi this may no longer be an adequate approximation.

Per Wikipedia...

"The compressibility of water is a function of pressure and temperature. At 0 °C, at the limit of zero pressure, the compressibility is 5.1×10−10 Pa−1. At the zero-pressure limit, the compressibility reaches a minimum of 4.4×10−10 Pa−1 around 45 °C before increasing again with increasing temperature. As the pressure is increased, the compressibility decreases, being 3.9×10−10 Pa−1 at 0 °C and 100 MPa.

The bulk modulus of water is 2.2 GPa. The low compressibility of non-gases, and of water in particular, leads to their often being assumed as incompressible. The low compressibility of water means that even in the deep oceans at 4 km depth, where pressures are 40 MPa, there is only a 1.8% decrease in volume."

40 MPa is about 400 atmospheres. You are talking about 1000 atmospheres. So that's more like a five percent decrease in volume. Five percent of 60 inches is 3 inches.

That's more than I'd expected.
 
If the valve in question is a pressure relief valve they usually require a presssure differential to operate so the pressure will not reach 0kPa without a vent to atmosphere on the outlet.
 
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/
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