Pressure drop due to holes in a pipeline?

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Pressure drop due holes in pipeline ?
We have a project that relating with pipeline system and pressure control, the project is actually having goal to knowing how much pressure drop will occurs in pipe line system that in initial condition the pressure inside the pipeline staying at steady or constant in particular psi and at the time in sudden, pressure will drop because of there are holes opened by a system that we have created before.

ASK.png



as you see above, that is probably the system or mechanism we created, so there is an obstacle (brown coloured) to blocking flow rate from upstream. The obstacle fixed by screws (red coloured) that will broken if there is a particular pressure of fluid in constant continously impacting the obstacle, we assume in 4000 PSI screws will be broken and in automatically the obstacle will goes forward and let the holes open and slowly flow rate of fluid pass through the holes and pressure of fluid will drop accordingly with area of those holes.

If
1. Diameter of hole : 20 mm
2. Diameter of pipe : 100 mm
3. Diameter of obstacle : 80 mm
4. P1 : 4000 PSI

how can we calculate the pressure drop ? which equation we have to use ? maybe someone may give us some advice, cause we have tried to calculate it with equation of bernoulli where (P1/rho + V1^2/2 = P2/rho + V2^2/2) but also we have a concern on about the volume changes due the obstacle, cause from the obstacle in first position where still fixed by screw and the obstacle after screw broken, is it also affecting 4000 PSI ? Cause there is volume changes from before screw broke and after ?
 

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  • #2
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This is a relief valve. The pressure conditions in the pipe (at the entrance to the relief port) depend on the characteristics of whatever is supplying the line, the pipe length, the fluid, the size of the 'relief holes,' and a number of other (minor) things. You need more information (and it's not a single 'canned' equation).
 
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  • #3
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This is a relief valve. The pressure conditions in the pipe (at the entrance to the relief port) depend on the characteristics of whatever is supplying the line, the pipe length, the fluid, the size of the 'relief holes,' and a number of other (minor) things. You need more information (and it's not a single 'canned' equation).
we assume that;
pipe length : 510 mm
fluid : water
size of the hole : 1256,63 mm2

actually the goal is predict how much pressure drop relating with size of the hole ?
in this case we want to reduce pressure from 4000 PSI in sudden to 450 PSI, does it can ?

so if we cannot use only by one equation, what calculation we have to use ?
 
  • #4
Baluncore
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If the pipe is short with a thick wall, and the water incompressible, then the pressure may be reduced extremely rapidly.

If the pipeline is long with an elastic wall, with some dissolved gas, or maybe air bubbles trapped in the water, then it could take quite some time for the pressure to be reduced, and a large volume of water to be released. That problem is concerned with evaluating possible modes of energy storage.

The release of a high pressure jet of liquid will cause local destruction to the surrounding building, and traumatic injury or death to anyone impacted by the jet.

A 2000 psi pressure wave, like water hammer when a valve is closed, propagating through the water in the pipeline, would arrive at the obstruction where it will be reflected. That reflection will produce a momentary double pressure pulse, to 4000 psi, that will break the bolts and trigger release. You may need baffles in the pipe to prevent that pressure wave reaching the relief valve.

You might consider spring loading the piston, and having a number of small holes progressively uncovered by movement of the piston. A spring loaded piston will self recover from an impulse.
 
  • #5
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If the pipe is short with a thick wall, and the water incompressible, then the pressure may be reduced extremely rapidly.

If the pipeline is long with an elastic wall, with some dissolved gas, or maybe air bubbles trapped in the water, then it could take quite some time for the pressure to be reduced, and a large volume of water to be released. That problem is concerned with evaluating possible modes of energy storage.

The release of a high pressure jet of liquid will cause local destruction to the surrounding building, and traumatic injury or death to anyone impacted by the jet.

A 2000 psi pressure wave, like water hammer when a valve is closed, propagating through the water in the pipeline, would arrive at the obstruction where it will be reflected. That reflection will produce a momentary double pressure pulse, to 4000 psi, that will break the bolts and trigger release. You may need baffles in the pipe to prevent that pressure wave reaching the relief valve.

You might consider spring loading the piston, and having a number of small holes progressively uncovered by movement of the piston. A spring loaded piston will self recover from an impulse.

The pipeline is relatively short with fluid is water / incompressible fluid, wall of pipeline is thick.
pressure is 4000 psi is generated by particular pump at upstream and fluid with that pressure will broke the screw and open the relief valve permanently (relief valve system is only once use system) and let holes open because of relief valve changes position. After holes open, pressure that from 4000 psi will reduced by holes because fluid that compressed inside pipeline with 4000 psi pressure that generated by pump finally may flow out through holes and 4000 psi will drops until point of pressure we already determined "450 psi".

so my question again sir, how can i make the pressure from 4000 psi drop into 450 psi ? is it relating with area of holes right ? because in logic, fluid that compressed in isolated volume with particular pressure, in quick periods occurs volume changes because there is holes open up and the pressure will reduced because that volume changes, or fluid isolated state finally may flow because there is leakage because of holes ? is it right ?

then what variable that affecting pressure from 4000 psi may reduce into 450 psi in quickly ? is it volume changes because there is holes then these area of holes determine how much pressure will drop ?

maybe you can watch this video,
actually we inspired by that mechanism, our project is to know what variable that determining the pressure drop, and we have plan to make pressure drop from 4000 psi tp 450 psi stay in 5 minutes, maybe some advice from you..
 
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  • #6
Baluncore
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... pressure is 4000 psi is generated by particular pump at upstream and fluid with that pressure will broke the screw ...
1. What is the flow rate from the upstream pump when pipeline pressure is 450 psi ?
2. What is the volume of pipeline between pump and this relief valve ?
 
  • #7
Tom.G
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pressure is 4000 psi is generated by particular pump at upstream
For the pressure to increase, either the pump is still running or something else is trying to compress the water.
  • If the pump is still running, you will have to use the pump operating curves to find the flow rate that will keep the pressure where you want it.
  • If something else is creating the pressure, you have to determine what it does when the pressure decreases.
  • As @Baluncore stated above, if there is gas (air) dissolved in the water, you need to account for the gas expansion as the pressure is released.
  • Also as noted, even a thick walled container will expand at 4000psi. Please take that into account for needed volume to expel.
Then calculate the hole sizes to allow the needed flow.

There are others here that can help you with the needed calculations.

Cheers,
Tom
 
  • #8
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1. What is the flow rate from the upstream pump when pipeline pressure is 450 psi ?
2. What is the volume of pipeline between pump and this relief valve ?

1. we assume that continuity will occurs during pressure drop, so probably Q1 = Q2 then our pump may produced 1.4 Litre/s at 4000 psi also at 450 psi...but does it works in reality ? i dont know if there is pressure drop from 4000 psi to 450 psi in those two conditions will be having same flow rate ? maybe some consideration from you..

2. for the volume from upstream until relief valve, the diameter of pipe is 100 mm with length 5000 mm and diameter of valve is 80 mm with length 100 mm, then the volume total is 399772562.99 mm3 or simplified 0.039 m3 (assume that pipeline is plain pipe and it is not like pipeline in the vide above)
 
  • #9
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For the pressure to increase, either the pump is still running or something else is trying to compress the water.
  • If the pump is still running, you will have to use the pump operating curves to find the flow rate that will keep the pressure where you want it.
  • If something else is creating the pressure, you have to determine what it does when the pressure decreases.
  • As @Baluncore stated above, if there is gas (air) dissolved in the water, you need to account for the gas expansion as the pressure is released.
  • Also as noted, even a thick walled container will expand at 4000psi. Please take that into account for needed volume to expel.
Then calculate the hole sizes to allow the needed flow.

There are others here that can help you with the needed calculations.

Cheers,
Tom

The soure of pressure we assume is only from pump, so pump continuosly produces flow rate 1,4 l/s until whole of pipeline that blocked by relief valve is full filled by water and pump still running, until the pressure reach 4000 psi and makes screw that fixed relief valve broke because of this pressure.

that is how 4000 psi produced, because fluid that isolated in pipeline continously pressed by pump that still running while fluid already full filled inside pipeline, and what we want to know is, drop pressure after relief valve open the holes, in our expectation when screw broke and relief valve in quick period open the hole, 4000 psi will suddenly drop and what we want is, it will drop until 450, did our prediction right if the variable that determining how much pressure drop is from area of holes, "if area of holes more larger then pressure drop also more higher ?" is it right or not ?

and then after we know what variable that determining pressure drop, we trying to to hold that pressure drop 450 psi will stay for 1-2 minutes so we add some tank, for detail you can see below:

34343433434.PNG

So we want to control the flow rate that through holes, if we want to hold pressure drop in 1 minutes what volume of tank we have to use?


and for there is gas in the water, we assume that there is no gas/air inside water because for simplify the case
 
  • #10
Baluncore
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So we want to control the flow rate that through holes, if we want to hold pressure drop in 1 minutes what volume of tank we have to use?
and for there is gas in the water, we assume that there is no gas/air inside water because for simplify the case
Water is not compressible, and there are no air bubbles, so when a hole is uncovered, the internal pressure will drop instantly to match the external pressure. No fluid will need to flow. The size of the hole is NOT important.

You have simplified the situation beyond reason. We do not know enough about the external conditions or environment to predict system performance.

If the device is in a bore hole it will be surrounded by hydrostatic pressure. So long as you keep pumping, water will continue to flow into the bore until it fills and overflows the hole.
Will you use this to fracture rock, or for something else?
 
  • #11
JBA
Science Advisor
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As a note, "frac" pumps are positive displacement pumps and so their delivery rate is not determined by the orifices' count or or size nor by the back pressure in the tanks. If the tanks are "empty" then they must contain air or some other vapor so the tank volume and pump delivery rate will determine the point at which 425 psig is reached and quickly exceeded. On the other hand, if you assume the tanks are completely evacuated then the fluid will fill the tanks at the pump's delivery rate with no increase in tanks' pressure until it becomes completely fluid filled; and, at that point, the tank and system materials' modulus and stress will determine the point at which your 425 psig pressure point is reached and then quickly exceeded, whereupon, depending upon the set pressure of the relief valves at the pump, either the pump's relief valve will open or a material failure of the lowest strength element in the piping and tank system will occur; and, during this whole process the pump will continue to deliver at its initial set flow rate.
 

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