How to calculate the leak rate based on pressure drop

AI Thread Summary
The discussion revolves around calculating the leak rate and fluid loss from pressurized vessels filled with water over a 30-minute period. The intern is using Boyle's Law but is encountering difficulties and is exploring alternative methods, including hydrostatic pressure calculations based on water height changes. Key questions include the presence of headspace above the water, the nature of any cracks, and their locations on the vessel. The intern has established the vessel's volume and hydrostatic pressure but seeks confirmation on the viability of their approach. Overall, the conversation emphasizes the need for accurate pressure drop and flow rate relationships in leak rate calculations.
blinkz303
Messages
2
Reaction score
0
TL;DR Summary
I am currently working on a project trying to calculate leak rates from pressurised vessels filled with water.
I am currently working on a project trying to calculate leak rate and fluid loss from pressurised vessels filled with water over a time period of 30 minutes. So far I have used the Boyles Law which assumes that PV/T = constant which gets us P1 * V1 = P2 * V2 and PdV+VdP = 0 I formed this over to PdV=VdP. and divided by the change in time to get dV/dt = - VdP/Pdt

Could anyone help me and tell me if I am on the right track here or have I overlooked something.
 
Engineering news on Phys.org
Is it liquid water or water vapor or are both present in the vessel. You need to establish the pressure drop vs flow rate relationship for the valve.
 
To clarify further: I am an intern at a company and I have been tasked with finding a method to calculate the leak rate and amount of fluid lost through possible cracks in a vessel. It can be assumed that the vessel is completely filled with water. I know the volume of the vessel and can figure out the hydrostatic pressure at the bottom of the vessel based on the volume. I have tried calculating the volume loss by calculating the change in the height of the water WH=P/rho*g
P being Pressure, rho being the density of water, g being the acceleration due to gravity.
I know the volume of water per meter of height and can therefore calculate the amount of water lost.
Would this be a viable method? I realized I am running into way to many problems using Boyle's law.
 
blinkz303 said:
To clarify further: I am an intern at a company and I have been tasked with finding a method to calculate the leak rate and amount of fluid lost through possible cracks in a vessel. It can be assumed that the vessel is completely filled with water. I know the volume of the vessel and can figure out the hydrostatic pressure at the bottom of the vessel based on the volume. I have tried calculating the volume loss by calculating the change in the height of the water WH=P/rho*g
P being Pressure, rho being the density of water, g being the acceleration due to gravity.
I know the volume of water per meter of height and can therefore calculate the amount of water lost.
Would this be a viable method? I realized I am running into way to many problems using Boyle's law.
Is there head space above the liquid water in the vessel? If so, what is in the head space, pure water vapor? Are the cracks exclusively on the bottom of the vessel, or are there cracks along the sides as well? how thick is the wall, and what are the approximate lengths of the cracks. Do you have any idea what the size of the crack openings are?
 
Hi all, I have a question. So from the derivation of the Isentropic process relationship PV^gamma = constant, there is a step dW = PdV, which can only be said for quasi-equilibrium (or reversible) processes. As such I believe PV^gamma = constant (and the family of equations) should not be applicable to just adiabatic processes? Ie, it should be applicable only for adiabatic + reversible = isentropic processes? However, I've seen couple of online notes/books, and...
I have an engine that uses a dry sump oiling system. The oil collection pan has three AN fittings to use for scavenging. Two of the fittings are approximately on the same level, the third is about 1/2 to 3/4 inch higher than the other two. The system ran for years with no problem using a three stage pump (one pressure and two scavenge stages). The two scavenge stages were connected at times to any two of the three AN fittings on the tank. Recently I tried an upgrade to a four stage pump...
Back
Top