Pressure drop due holes in pipeline ? 1

[Rizkyffq]

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.

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 and after screw broke ?

 

[RVAmeche]

I don't really understand what you're asking. If there's atmospheric pressure outside the pipe, the pressure drop from inside to outside is 4000 psi. If there's no way to close the holes, your system will vent through those holes until there's no more pressure.

 

[MJCronin]

What are you trying to accomplish that cannot be accomplished by a power-operated relief valve and a pressure transmitter ?

The relief valve is a repeatable, proven precision device.

Your two set screws ...????//

MJCronin
Sr. Process Engineer

 

[LittleInch]

OK, this post is missing some crucial details.

1) What is to the left of this drawing? If your 4,000 psi is created by liquid pressurising a fixed volume then yes your pressure will fall quite dramatically due to the extra vvolume created by movement of the piston. but this depends on the volume of fluid in total versus the volume of the piston.
2) Is there a pump suppying the pressure?
3)( can you creat a steady state flow out of the holes?
4) Forget Bernoulli
5 4000 psi on a lot fo sq inches is a big force.

From a practical level there is nothing shown at to how to stop the piston from simply flying out of the end.

If this is a break system what's wrong with a bursting disc? Well known proven technology.

you are going to get some transient effects (shock wave etc) that will need a transient analysis.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Compositepro]

The pressure will depend on the the resistance to flow upstream of the holes compared to the flow resistance of the holes. That is, two resistances in series. The only way that I can see this concept be useful is if your pipe is fed through an orifice significantly smaller than your holes.

 

[Rizkyffq]

thanks for your response guys, ok then i will add some detail,

1.there is additional tank with no fluid inside (empty tank) that will full fill by fluid when the holes opened (detail at the additional picture below)
2. assume that i need to know volume of the tank to make pressure drop holds until particular time, example i need 1 minute to hold pressure drop until it back normal to 4000 psi because tank already filled by water
3. the obstacle or relief valve is only once use system, after the screw broke the valve will no back to cover the hole, it will continously broke and let the hole open
3. fluid will be generated by particular pump and the pressure produced by the pump toward fluid is 4000 psi cause we assume that in 4000 psi pressure it will gives screw broke and relief valve will open the holes and let the pressure inside pipeline reduces, this why we need to know how to calculate pressure drop due size hole ?

so basically our goal is to know what size of the hole to drop pressure of fluid from 4000 psi to 450 psi (what equation ?) and hold the pressure drop until 1 minute so what volume of tank we use to hold it until time we already determined ?

 

[RVAmeche]

Instead of this overly complicated circumferential hole/tank arrangement with a calibrated set screw cap you're showing, why not install a tee/branch/whatever with a valve (and hose or something to a tank if needed).

Knowing the initial volume of your system @ 4000 psi, you can calculate how much volume can be bled to get to 450 psi. Size the collection tank appropriately. If you need that pressure drop to occur within a given time frame, iteratively calculate the flow through the valve/hose (or use software) to determine the size needed.

 

[LittleInch]

This gets more and more complex as it goes on.

I assume these extra tanks are sealed and hence the pressure in them starts at 0 psig, but eventually goes to 4000 psig?? as the fluid enters over a period of time

~In which case you have a transient analysis on your hands and no equation will solve this issue for you.

Your pump had bette rbe a positive displacement pump of some sort otherwise it won't like operating at between 450 to 4000 psig

The size of your hole will need to know even in the first seconds what the flowrate from your pump is.

This flow through an orifice will be critical flow, but only until the pressure drop from one side to the other reaches a certain point, then it changes.
The volume of your tanks will impact the pressure rise.

I think you have too many variables so you will need to fix one thing or another and then try to calculate the others.

But a bursting disc and a pressure regulating valve of some sort sounds a much better plan....

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Rizkyffq]

Thanks for you response sir, so probably to make it easier actually mechanism we want to create is actually inspaired by this mechanism,

https://www.youtube.com/watch?v=vu65sdhKzu0

The pipeline is relatively short with fluid is water / incompressible fluid, wall of pipeline is thick.
pressure is 4000 psi that 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 fluid that pass through holes, 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 that in isolated state finally may flow because there is leakage because of holes ? is it right ?

and the variable we have :
1. Diamater of holes : 20 mm
2. Diameter of pipe : 100 mm
3. Length of pipe until relief valve : 5000 m
4. Diameter of relief valve : 80 mm
5. Length of relief valve : 100 m
6. Flow rate produced by pump : 1.4 Litre/s
7. Pressure 1 (P1) : 4000 PSI


and what we want to seek is
1. by what calculation and what variable that pressure from 4000 PSI may drop to 450 PSI ? is it right because there is leakage from holes that open by relief valve ? if it is true than what equation to calculate it ?
2. What volume of tank if i want to hold my pressure drop 450 PSI in 1 minute ? in our imagination after relief valve opened the holes and let the fluid which isloated in 4000 PSI finally may flow out through holes and pressure will drop to 450 psi (need your validation about it) and to hold the pressure drop and stay at 450 PSI because flow rate that from isloated state finally may release...accordingly with flow rate ..
3. And also we have to know how much exactly flow rate that goes out from the holes, so we can calculate what volume of tank appropirate with flow rat, let say i know that flow rate is 1 litre/minute and i want to holds drop pressure in 5 minutes then i need volume of tank is 5 litre

 

[LittleInch]

This will be a transient event and only really possible to calculate anything using a transient simulator.

At 20mm your holes are relatively big compared to the pipeline and will send a shock wave back down the pipe at sonic velocity.

Your initial flow will be choked flow but will be fairly short lived as the pressure decays in your pipe. My guess is that it will fall within 15 to 20 seconds, but might be more.

Then your pressure drop will stabilise based on the flow rate from your pump.

However it will still vary as your tanks fill up with liquid a start to pressurize the gas inside. Now to be fair, it won't be until you're about 90% full that the pressure will rise to anything worth worrying about, but even so....

If what you want to achieve is 450 psi after one minute then your variable is the size of the hole., but only by iterative runs in a transient simulator will you be able to determine what this.

Also consider that the velocity through your holes will be sonic velocity. This will erode the hole very quickly unless you make it very strong and hard material.

I still really don't understand what you're trying to do but I can't see how you're going to calculate this as opposed to studying it using dynamic simulation software.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[GD2]

Rizkiffq,

If I understand correctly, your objective is the following:
1. Reduce the pressure from 4000 psi to 450 psi.
2. Hold 450 psi at the tank for 1 min.

Correct?

I have an alternate suggestion open for discussion.


i) Ignore the pins, obstacle etc and instead install a Pressure Reducing Valve (PRV) set at 450 psi between the pipe and the tank.
ii) Install a solenoid at the tank to cut-out the pump after 1 min. of holding. Is it something you want to attain or something else?



GDD
Canada