Determine the Rate of Change of Pressure Across a Valve

[russ_watters]

You can't find the pressure drop across the valve unless you know the flow rate. So if you are changing the valve position, which changes both the pressure drop and flow rate, you need to know the characteristics of the entire system to find what the new pressure will be at the next increment of valve position.

 

[Purple_Dan]

You can't find the pressure drop across the valve unless you know the flow rate. So if you are changing the valve position, which changes both the pressure drop and flow rate, you need to know the characteristics of the entire system to find what the new pressure will be at the next increment of valve position.

I already know the pressure drop and the flow coefficient, from that I can determine the flow rate, but these are just instantaneous values. I want to know what the pressure drop will be after 1 second while I'm changing the valve position, which changes the flow coefficient. If I can determine that from the flow rate, how would I calculate what the flow rate will be in 1 second? I'm assuming I'll need the volume of the system...

It doesn't even have to be too exact, just a clue as to what a good position to put the valve in would be good. Then I can implement a control system based off that if I need to.
At the moment I am essentially completely blind as to how to implement this.

 

[HuskyNamedNala]

Like I told you, you won't find any analytical solution. You need to implement a predictor corrector algorithm, where based on the change in flow rate in the recent past, you estimate the derivative, then adjust the valve, re-measure the flow rate and correct. Repeat until you achieve steady state.
If you haven't taken a numerical methods class, you might be in trouble...

But from your posts, I gather what you are trying to do is use existing data to pre-calculate what the valve will be at steady state, rather than do a real time computation. Simply put, you won't be able to do this. Yes, theoretically there might be an exact solution to your problem, but practically the effort expended in looking for this solution will not yield a good return on your investment... unless you happen to solve the N-S equations, in which case you'll become a millionaire and world famous and won't have to work ever again.

The steady state valve curves will give you an approximate starting point. You will have to compute the derivative of the pressure drop in real time and estimate. This is the best solution I can think of.

 

[HuskyNamedNala]

Or... what you could do is open the valve, wait 10-20 seconds for steady state to be achieved, then readjust, repeat, etc. This method is not very elegant and is very slow, but it will work too.

 

[Chestermiller]

Are you trying to control the pressure drop at constant flow rate, the flow rate at constant pressure drop, or a combination of the two? If you're trying to control a combination of the two, then you need to quantify the interaction of the valve with the process equipment upstream of the valve (in terms of the fluid mechanics interactions). So far, aside from the control equipment, you haven't told us anything about the process equipment upstream.

Chet

 

[Purple_Dan]

Are you trying to control the pressure drop at constant flow rate, the flow rate at constant pressure drop, or a combination of the two? If you're trying to control a combination of the two, then you need to quantify the interaction of the valve with the process equipment upstream of the valve (in terms of the fluid mechanics interactions). So far, aside from the control equipment, you haven't told us anything about the process equipment upstream.

Chet

Other than that P&ID, I know relatively little about the equipment, I don't know lengths of pipes, diameters, volume or anything. I know that there's 0psi to the left of the isolation valve, 0psi below the control valve and the space inbetween can be between 50 and 23000psi.
If we take out all the unnecessary gumpf like gravity and pressure loss due to friction, what equation that will give me a prediction of the pressure? What variables will I need to know?

Thanks, I know it must be annoying not having all the data, but I don't have it either!

 

[russ_watters]

I already know the pressure drop and the flow coefficient, from that I can determine the flow rate, but these are just instantaneous values. I want to know what the pressure drop will be after 1 second while I'm changing the valve position, which changes the flow coefficient.

That doesn't make sense: If you already knew the pressure drop at each valve position, all you would need to do is divide the difference by how long it takes the valve to get to its next position.

 

[russ_watters]

If we take out all the unnecessary gumpf like gravity and pressure loss due to friction...

Pressure drop due to friction is the core issue you need to address!

 

[HuskyNamedNala]

You don't seem to understand, you're not going to get this analytically. Neglecting viscosity will get you Euler equations:
http://en.wikipedia.org/wiki/Euler_equations_(fluid_dynamics )

You can't even simplify the equations for 1-D because to conserve mass you would need to include 2 velocity components; presuming the velocity along the length of the pipe will vary.

Why are you so adverse to implementing the method I told you about?

 

[Purple_Dan]

Pressure drop due to friction is the core issue you need to address!

Yes, but I only care about the friction at the valve, not in the pipes.

 

[Purple_Dan]

Why are you so adverse to implementing the method I told you about?

Because, while I agree with you and would love for it to be that simple, I don't think the customer will allow me to wait a few minutes until it's got it right. Also, I know there's a better way to do it, and I'm a bit of a perfectionist. Sorry.

Also, I don't see what's wrong with Euler equations, with enough iterations, you get pretty close to the right answer. At least, I did in Uni...

 

[Purple_Dan]

That doesn't make sense: If you already knew the pressure drop at each valve position, all you would need to do is divide the difference by how long it takes the valve to get to its next position.

I know the psi at each position, what I don't know is what the psi will be after 1 second if the valve is open at 50%, I don't even have a clue what it would be. Just a nice guess would be good.

 

[HuskyNamedNala]

Because, while I agree with you and would love for it to be that simple, I don't think the customer will allow me to wait a few minutes until it's got it right. Also, I know there's a better way to do it, and I'm a bit of a perfectionist. Sorry.

Also, I don't see what's wrong with Euler equations, with enough iterations, you get pretty close to the right answer. At least, I did in Uni...

It is ironic that you believe you know a better way to solve this problem, yet with more information than any of us on this forum you are asking around for solutions. Good luck with whatever method you chose. The analytical route won't be fruitful.

 

[montoyas7940]

I know the psi at each position, what I don't know is what the psi will be after 1 second if the valve is open at 50%, I don't even have a clue what it would be. Just a nice guess would be good.

Because you only know the valve flow vs vessel pressure, you can't know how much static pressure will change in the vessel over time without first discovering how much volume you have to remove to achieve the change.

 

[jim hardy]

Scientific Wild Guess here, a SWIG:Seems to me that with 23,000 psi and a needle valve this must be some sort of leakage test rig.

At those kinds of pressures we must think differently about steel pipes and valves. Instead of rigid bodies they become elastic like balloons, and compressibility of water becomes noticeable.
A steel cylinder expands when you pressurize it. So it holds more water. As you let water out through your valve pressure decreases as the walls relax.
30,000 psi in steel gives ~ 1 part in a thousand elongation.
If you know the ratio of wetted area to wall cross section of your pressure vessel you can estimate its pressure versus volume curve, .

Without knowing the elasticity of that Christmas Tree Valve, which is a function of its dimensions, ... it's a Un-SWIG.

I could be way off base, frequently am...
Do you know for sure what parameter it is of that "Christmas Tree" they're wanting to measure?

here's a lead for compressibility of water.

http://ipims.com/data/fe31/E3131.asp?UserID=&Code=3298

i hope I'm on track here. feel free to correct.

 

[Purple_Dan]

Scientific Wild Guess here, a SWIG:Seems to me that with 23,000 psi and a needle valve this must be some sort of leakage test rig.

At those kinds of pressures we must think differently about steel pipes and valves. Instead of rigid bodies they become elastic like balloons, and compressibility of water becomes noticeable.
A steel cylinder expands when you pressurize it. So it holds more water. As you let water out through your valve pressure decreases as the walls relax.
30,000 psi in steel gives ~ 1 part in a thousand elongation.
If you know the ratio of wetted area to wall cross section of your pressure vessel you can estimate its pressure versus volume curve, .

Without knowing the elasticity of that Christmas Tree Valve, which is a function of its dimensions, ... it's a Un-SWIG.

I could be way off base, frequently am...
Do you know for sure what parameter it is of that "Christmas Tree" they're wanting to measure?

here's a lead for compressibility of water.

http://ipims.com/data/fe31/E3131.asp?UserID=&Code=3298

i hope I'm on track here. feel free to correct.

Well colour me impressed, I hadn't taken that into consideration. Unfortunately, like many of the questions I have been unable to answer, they test all different sizes of Christmas Tree Valves!
In your expert opinion, would the water compress enough to drastically affect my calculations? And would the volume change much at 23000 psi? I suppose that would depend on the specifics of the pipes and valves, seeing as hot water bottles are harder to inflate than balloons...

I hadn't thought that the water might compress, as my A-Level Physics knowledge didn't cover that!

 

[jim hardy]

I hadn't taken that into consideration.

i think that's what the guys have been trying to say when they ask for system details and what volume you need to move.
If i hit on the right words, well, it was just serendipity...

In your expert make that very amateurish, jh opinion, would the water compress enough to drastically affect my calculations? And would the volume change much at 23000 psi? I suppose that would depend on the specifics of the pipes and valves, seeing as hot water bottles are harder to inflate than balloons...

Well, keep in mind I'm just an old electronics instrument guy from a power plant . I have watched over the shoulders of genuine mechanicals and hope some will chime in here.

hopefully the system shape and size are among those things you know but can't divulge.

Let's just oversimplify and SWIG that the steel is loaded to 30,000 psi in the steel not in the water.
Ordinary steel can take that stress without yielding, exotics several times more.
So the designers would set wall and fastener thicknesses to limit stress to a fraction (maybe 70%?) of yield at design pressure.
so every dimension would increase by 0.1%, (look up Young's Modulus)
hence volume at design pressure would go up by ~ 1.001^3 = 1.003 .

I used to take care of an acoustic system attached to our reactor vessel, made from 4 inch and more thick steel..
It emits creaks and groans and snaps as you pressurize it and the metal stretches.Compressibility of water is usually ignored.
Per that link it's in the range of 2 to 4 X 10-6 per psi and gets smaller as pressure increases.
If we just use 3E-6
then at 20,000 psi it might compress say 3E-6 X 2E4 = 6E-2 = 6%.
That's probably a high number because as you see from that graph its compressibility gets smaller as pressure increases.
And dissolved gas in the water will have an effect.
But for the first few thousand psi water might compress 3E-6 X 1E3= 3E-3 , 0.3% per thousand psi.

At least that's what it looks like to this old instrument guy,
>>>and i'd appreciate corrections by a genuine mechanical engineer.

Maybe you want to ask the client some questions related to these effects. At least he'll know you're thinking about them..
You might get laughed off the stage,
or he might exclaim "At Last, somebody who asks the right questions".Good luck and keep us posted. I look forward to learning.

old jim

 

[russ_watters]

Yes, but I only care about the friction at the valve, not in the pipes.

You can't know one without the other.

I know the psi at each position, what I don't know is what the psi will be after 1 second if the valve is open at 50%...

Maybe I'm just being thick, but it seems to me that the second part contradicts the first part.

Or it could be that since we don't know some critical details about the system, we are speculating based on guesses.

In the beginning of the thread, it seemed like you are saying you are trying to calculate how pressure drop varies with valve position, as the valve position varies with time. But now it seems like you are asking about how pressure drop varies with flow, at a fixed valve position, as the system changes. So which is it? And if it is the system that is changing, how is it changing?

My first guess about the system was based on systems I see every day (trying to fit it into everyday experience): Take a heating hot water system, close the isolation valve at the discharge of the pump, and turn the pump on. Then open the valve slowly and see how flow rate and pressure drop across the valve and pump change with valve position.

Or is Jim right that this is some sort of leak test? Are you opening the valve and seeing how long it takes for the system to drain and pressure to drop to zero? Also, is this valve really holding back 23,000 psi on its own, giving it an initial pressure drop of 23,000 psi when you open it? There are additional problems with that, because the water flow will be supersonic, some will flash to steam as it goes through the valve due to the pressure drop alone (actually, you might get steam and snow at the same time), and it will heat-up tremendously due to the extraordinarily high friction of the extraodinarily high flow velocity. Of course, if 23,000 psi is just the static gravitational head and the entire system is a loop and the pressure drop is really only 10 or 20 or 50 psi across the valve, that issue won't exist.

 

[Purple_Dan]

You can't know one without the other.

Maybe I'm just being thick, but it seems to me that the second part contradicts the first part.

Or it could be that since we don't know some critical details about the system, we are speculating based on guesses.

In the beginning of the thread, it seemed like you are saying you are trying to calculate how pressure drop varies with valve position, as the valve position varies with time. But now it seems like you are asking about how pressure drop varies with flow, at a fixed valve position, as the system changes. So which is it? And if it is the system that is changing, how is it changing?

My first guess about the system was based on systems I see every day (trying to fit it into everyday experience): Take a heating hot water system, close the isolation valve at the discharge of the pump, and turn the pump on. Then open the valve slowly and see how flow rate and pressure drop across the valve and pump change with valve position.

Or is Jim right that this is some sort of leak test? Are you opening the valve and seeing how long it takes for the system to drain and pressure to drop to zero? Also, is this valve really holding back 23,000 psi on its own, giving it an initial pressure drop of 23,000 psi when you open it? There are additional problems with that, because the water flow will be supersonic, some will flash to steam as it goes through the valve due to the pressure drop alone (actually, you might get steam and snow at the same time), and it will heat-up tremendously due to the extraordinarily high friction of the extraodinarily high flow velocity. Of course, if 23,000 psi is just the static gravitational head and the entire system is a loop and the pressure drop is really only 10 or 20 or 50 psi across the valve, that issue won't exist.

Okay, so what happens is this:

• I build the pressure in the system to the desired target pressure via a pump
• I isolate the test area from the pump pressure
• Drop the pump pressure to zero (not via a control valve)
• The test runs and the test valve has to pass certain criteria
• I open the hold valve after the control valve
• I start opening the control valve to maintain a desired psi/sec drop

That last bit is what we care about.

I am only changing the valve
position, but from the specification documents of the valve I know the flow coefficient for the % that it is open.
Using this, I know the flow rate (or thereabouts).

So if a gallon of water has just burst out in 1 second, how do I calculate the psi drop.

If we ignore the compressibility of water and the change in volume of the system, which appears to be negligible for the accuracy I require.

And there really could be 23000psi across that control valve. I look forward to seeing snow steam, it sounds awesome...

And I'd like to apologise to everyone for being insufferable thus far!

 

[montoyas7940]

So if a gallon of water has just burst out in 1 second, how do I calculate the psi drop.

If we ignore the compressibility of water and the change in volume of the system, which appears to be negligible for the accuracy I require.

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.

 

[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.