Determine the Rate of Change of Pressure Across a Valve

[Purple_Dan]

Except for the change in volume and to a much smaller extent the compressibility of the water there would be no flow as you open the valve. The change in volume is providing the flow.

Damn, when you're right, you're right. No matter how small the change in volume is, it's still the reason for the flow in the first place. Where did all my Physics knowledge go, and why am I so stubborn?

I have a plan, I'll find out how the volume in the system changes by measuring the water that comes out at different pressures. Then I can determine what the volume will be based on what the flow coefficient tells me will come out! Aha! I have the solution and all it took was a load of people smarter than me to tell me why I was thinking about it the wrong way!

Thank you for your patience.

Now that I am seeing sense. Am I right in thinking that, as I have a temperature sensor, I can work out the pressure from the volume that way?
So it's at 23000psi, from my measurements, I know that the volume is X gallons, from the flow coefficient, I know that Y gallons will come out, from the temperature, I know that the new pressure will be Zpsi. Or is that relationship only for gases?

Thanks again for all your help!

 

[jim hardy]

if it's an irregular shaped volume look out for trapped gas. It'll be quite compressible compared to the water.
Ever bleed brakes?

Good Luck !

 

[montoyas7940]

Damn, when you're right, you're right. No matter how small the change in volume is, it's still the reason for the flow in the first place. Where did all my Physics knowledge go, and why am I so stubborn?

I'll bet if I had listened carefully I would have heard the forehead smack from across the Atlantic.

I have a plan, I'll find out how the volume in the system changes by measuring the water that comes out at different pressures. Then I can determine what the volume will be based on what the flow coefficient tells me will come out! Aha! I have the solution and all it took was a load of people smarter than me to tell me why I was thinking about it the wrong way!

That should work but I think the data is probably already sitting on someones desk somewhere.

Thank you for your patience.

I'm glad to have participated!

Now that I am seeing sense. Am I right in thinking that, as I have a temperature sensor, I can work out the pressure from the volume that way?
So it's at 23000psi, from my measurements, I know that the volume is X gallons, from the flow coefficient, I know that Y gallons will come out, from the temperature, I know that the new pressure will be Zpsi. Or is that relationship only for gases?

I don't think that will work. At least not measurably by practical means.

Thanks again for all your help!

Hey, just stick around and help someone else.

 

[HuskyNamedNala]

I looked into this some more. Are you sure 23ksi is correct? The only application I could find with these pressures is used for metal cutting water jets. It seems odd from a design standpoint to require a valve that can continuously operate as low as 50psi all the way up to 23ksi. Usually there is some type of interface, given that whatever piping this will have to go through will have to withstand 23ksi too, which will require one hell of a thickness.

 

[Purple_Dan]

I looked into this some more. Are you sure 23ksi is correct? The only application I could find with these pressures is used for metal cutting water jets. It seems odd from a design standpoint to require a valve that can continuously operate as low as 50psi all the way up to 23ksi. Usually there is some type of interface, given that whatever piping this will have to go through will have to withstand 23ksi too, which will require one hell of a thickness.

The test pressure they decide to put in depends on what valve they are testing, the maximum they can input is actually 22500psi but API criteria states that the pressure can go as high as 5% above the target pressure.
The client decided that there should be a cap of 500psi on this upper limit, so any target pressure above 10000psi just has 500 added to it to find the upper limit. This means they are tighter that API criteria and can brag about that to their customers.

 

[Tatersoup]

"Also, am I right in thinking that if the Cv is 0.004 at 100%, then the Cv at 50% will be 0.002?"

No.

"Pressure drop across valve can be made up as this is variable within the system, let's assume 500psi for this example.
Assuming it's pure water at 25°C with a specific gravity of 1 (because I don't want to get into all that mess)."

I don't know how you can make up a pressure drop across the valve without the missing component of velocity. If there is no movement of the water, then the pressure loss is zero. As example, assume the water is moving at 10 feet per second, or V = 10. Then the pressure drop across the valve would be Cv times velocity squared divided by (2 times the gravitational constant). So, assuming 10'/second, the loss (in feet of head) would be 0.006 feet; minimal, which suggests a good free flowing valve design.

Let's assume now that the valve is half closed, or more clearly, that the open area is 1/2 of the open area of an open valve. In this case, the valve is now materially functioning as an in-line orifice in the system. In this case, an orifice equation would be the best approach, solving for H. From memory, the equation would generally take the form of Q (flow rate, often in cubic feet per second) = C (orifice coefficient) x A (area of orifice) x the square root of [2 x g (gravitational constant) x H (the head loss you are solving for)].

So, depending on why you need to know this information, using the area of the opening and a C of, say, 0.6, you should be able to get close enough. If you can find out the area of the valve opening at different settings, I think you could determine head loss (pressure loss) pretty readily.

"I need to control the valve to maintain a constant psi/sec drop. I was going to do a simple, "if drop is too big, close valve a bit, else open valve a bit" logic. But I thought it would be nicer to try and do it properly."

If the pressure drop is too big, making the orifice bigger will reduce it; likewise, if pressure drop is too small, you would want to close the valve a bit, making the opening smaller.

If I understand you correctly, you are regulating flow to achieve a specific pressure drop across the valve.

"Yes, I know the change across the valve. I want to know, if I open the valve x%, what will the pressure be after X seconds. That way I can just open the valve to the required %, rather than guessing and inching it open. I'd rather not guess when there could be 23000psi in the system!"

If you can determine the area of the opening of the valve at different % opening, you can use the above orifice equation very easily, if you know the flow rate. The pressure change should be fairly rapid.

 

[Purple_Dan]

If the pressure drop is too big, making the orifice bigger will reduce it; likewise, if pressure drop is too small, you would want to close the valve a bit, making the opening smaller.

If I understand you correctly, you are regulating flow to achieve a specific pressure drop across the valve.

I am trying to maintain a specific rate of pressure drop, opening the valve would increase the rate the pressure drops, and vice versa.

Also, the pressure drop across the valve is variable at a maximum of 23000 psi, I already know what this is, I'm just trying to figure out what it will be.

 

[Purple_Dan]

I don't think that will work. At least not measurably by practical means.

Aww, why not? It can never just be relatively simple can it?
I have to think about the possibility of trapped gas as well now!

 

[montoyas7940]

Aww, why not? It can never just be relatively simple can it?
I have to think about the possibility of trapped gas as well now!

I know you don't get to design the test but it would be simple enough to pull a vacuum on the "Christmas tree" before the liquid fill. Then TaDa! No trapped gas!

How important is the rate of depressurization? Is it part of the test or just a controlled rate for the sake of maintaining good control?
Maybe it just doesn't matter that much.

 

[Purple_Dan]

I know you don't get to design the test but it would be simple enough to pull a vacuum on the "Christmas tree" before the liquid fill. Then TaDa! No trapped gas!

How important is the rate of depressurization? Is it part of the test or just a controlled rate for the sake of maintaining good control?
Maybe it just doesn't matter that much.

I actually hadn't thought to ask how important it is. It's not part of the test.

What is surprising is the way they depressurise the pumps is to just open a valve and let it burst out. Even at 23000 psi. And all the valves are failsafe open. So if someone hits the E-Stop, it all just opens.

 

[Purple_Dan]

Hello all!

On closer inspection of the valve specifications, the orifice size is only 1.57mm (0.062") at max, that's tiny!

Anyway, we've decided to go with proportional control. If we're way off the target, open/close the valve quickly. If we're closeish to the target, open/close the valve slowly.
That is until the client decides this is not effective enough and decides to do it the right way!

At the moment, they're dropping the pressure at about 2000 psi/sec after a test! That's mental! I think any improvement is an improvement and the control valve should definitely improve the way we control this!

In any case, when the customer decides they don't want to do it this way, I'll use what I've learned from all of you as my guide to improving the system.

But there's still one small question that hasn't been answered:

I don't think that will work. At least not measurably by practical means.

Why wouldn't Boyle's Law work? If I knew the temperature and the pressure, could I not work out the volume? Is it a special case at 23000psi? Or am I just being silly?

Also, thank you all for contributing to my first post and making it last week's most popular Mechanical Engineering post!

 

[montoyas7940]

Boyle's ideal gas law won't work for you in this application. It is for gases only.

Someone else here may be able to help devise a practical way to implement your idea. I don't know...

 

[Purple_Dan]

Boyle's ideal gas law won't work for you in this application. It is for gases only.

Someone else here may be able to help devise a practical way to implement your idea. I don't know...

Okay, so that's interesting. No one has come up with an equation for the relationship between Volume, Pressure and Temperature in a liquid. Someone should do that now, get to work scientists!

I spoke with the client today and there's no trapped gas because they pump water through until they see it coming out the other end and close the end valve whilst it's still pumping. So that's not a factor anymore, which is nice.

But I won't know any more about the system to shed some light on it, and hopefully figure this thing out, until I go on site in a few months. Which is annoying...

 

[Tatersoup]

I am trying to maintain a specific rate of pressure drop, opening the valve would increase the rate the pressure drops, and vice versa.

Also, the pressure drop across the valve is variable at a maximum of 23000 psi, I already know what this is, I'm just trying to figure out what it will be.

I understand your point, it's your nomenclature that is confusing. Instead of saying, "opening the valve would increase the rate the pressure drops", I would say, "opening the valve would reduce the pressure drop"

The system pressure has nothing to do with it. The rate of flow through the valve does. If you go look at my original post, I tried to suggest a method to estimate the flow for different valve openings using a basic orifice equation.

 

[Purple_Dan]

I understand your point, it's your nomenclature that is confusing. Instead of saying, "opening the valve would increase the rate the pressure drops", I would say, "opening the valve would reduce the pressure drop"

The system pressure has nothing to do with it. The rate of flow through the valve does. If you go look at my original post, I tried to suggest a method to estimate the flow for different valve openings using a basic orifice equation.

Ah, I see where the confusion could arise. I have the flow coefficient at different valve openings, from which I can calculate the flow rate without using orifice equations.

It's where I go from there that I'm having trouble with.

 

[russ_watters]

You should also look into the issue of cavitation:
http://www.valmatic.com/pdfs/Cavitation_in_Valves_7-22-08.pdf

 

[jim hardy]

I still sense a disconnect here.

Clearly from previous posts they're depressurizing a test rig that's been filled with a test fluid(water?) nearly purged of air.
It's depressurized through an orifice that's tiny compared to the size of the test chamber.
In petro world, water is considered "slightly compressible", and when compared to pure water steel itself is compliant.

So to exaggerate for purpose of seeing what's going on, refer way back to PurpleDan's water bottle analogy. He's depressurizing a water bottle through a hypodermic needle.

With so much not known about the system , the only approach i can envision would be to record pressure vs time for a known valve opening and catch the fluid released in a beaker.
One could then come up with a term for the total compliance ΔV_olume/ΔP_ressure of the combined water-steel-entrained-air system.

A measurement of flow during the depressurization would give a good number for compliance vs pressure.

Here's an academic looking paper on fluid mechanics in wellbores.
Since this consultation is in petro field maybe it'll have some vocabulary words you could use to prime the conversation pumps.
http://petrowiki.org/Fluid_mechanics_for_drilling