Design of a Valve stem cross section in a Steam turbine emergency valve

[SurajS]

TL;DR

I want to design cross section area required for valve stem when valve is closed under spring force. Valve stem should be designed for buckling load but I am unable to calculate buckling load coming due to impact force coming when valve bangs on valve seat.
Is there any reference book on spring loaded valve design?

I want to design cross section area required for valve stem when valve is closed under spring force. Valve stem should be designed for buckling load but I am unable to calculate buckling load coming due to impact force coming when valve bangs on valve seat. Valve is having concentric spring.
Is there any reference book on spring loaded valve design?

 

[bobob]

I don't know for certain, but the best place to look and the place you would most likely get the most information about the requirements to meet engineering standards are the ASME manuals.

 

[AZFIREBALL]

Can you tell us what the word ' buckling ' means to you?
Also, are you trying to determine the smallest diameter for the valve stem, or the geometric configuration of the stem's cross section?

 

[JBA]

We need to see the configuration of the valve assembly, it is clearly not a conventional ASME type safety/pressure relief valve design.

 

[SurajS]

I am trying to design a cross section of valve stem and check its safety against buckling load.
This is schematic representation of emergency trip linkage from US patent:US4379544.

 

[JBA]

The spring is not the primary source of the valve closing it is the ΔP across the valve disc as approaches the valve seat that actually drives the valve closing and sealing. As a result stem load due to the spring load will be relatively low because the spring is only required to overcome the differential lifting force of the stem area x the system pressure to start the valve closing.
The primary source of the valve closing is the ΔP across the valve disc as it approaches the valve seat, which will reach a maximum force of F = ΔP x (the valve disc diameter) + the kinetic energy of the closing stroke. Your main focus should be on the stress on the valve disc upon closing because it can be very high depending upon the ΔP between the valve inlet pressure and outlet pressure at valve closing.
I am speaking from experience from developing an ASME Section 8 Pressure Relief Valve with a metal seat and disc and similar "bathtub plug" type of closing action. It took almost 6 months of engineering analysis, including FEA, and cycle testing focused singularly on incorporating the necessary closing cushioning to provide a design that would allow the valve to close and still have a reasonably long cycle life without damaging the disc and the seat sealing.
Admittedly, that valve is designed for very high pressure and temperature steam service, i.e. 900 psig @ 1000°F; but, point is that in this type of valve the seat disc will be very highly stressed and must be carefully designed to prevent disc overstressing upon closing. For reference, see US Patent No 5,011,116.

Just as another comment on the use of your valve, due to its design, once that valve is closed it will not possible to reopen it until the system inlet pressure is reduced to that of the outlet pressure with the valve closed.

 

[SurajS]

The primary source of the valve closing is the ΔP across the valve disc as it approaches the valve seat, which will reach a maximum force of F

Thanks for the insights. It is not a pressure relief valve. It is emergency valve of steam turbine which will required to be close the valve as soon as there is application of actuation force as shown in figure 1.
And steam is trying to open the valve. So only force which is closing the valve is Spring.
Now, I am trying to design the valve stem since it will be acted upon by impulse generated when valve closes.
As already asked in other thread, I was unable to calculate the force of impact analytically, the same has lead me to ask this question again.
Please let me know quantification of force estimation unlike qualitative way or how can I analyse this in FEA.
And It is pilot operated valve though figure doesn't show it explicitly.

 

[JBA]

Please Note: I have one concern about your above post, your valve drawing states that the "steam inlet" is the right side connection, which indicates to me that when the valve is closing or closed it is stopping flow from that inlet to the the lower pipe. So, unless you are trying to use your valve to block a reverse flow form the bottom pipe to the side "steam inlet"inlet, then all of my above still applies because it is the closing action, not the opening of my referenced valve, that creates the high impact disc stress.

If the reverse flow blocking I just described is the case, the only impact stress will be from the spring's stored energy with the valve open plus any spring's designed residual load at the valve's closed position.
To determine the impact force on your valve stem and bottom disc center, start by calculating the spring's potential energy in its stretched condition by PE spring = 1/2*k*h^2; and, that is the energy that will be delivered to the top of the stem and the disc at the bottom when the valve closes.
Then by using FEA, or a classic analysis (See @ http://www.roymech.co.uk/Useful_Tables/Mechanics/Plates.html for a sample "disc with a simply supported O.D. and a center load" analysis) determine the k disc for your disc. Then the PE applied to your stem and disc center will be equal to the above PE spring.
Next, using your established k for the disc loading; then, your: impact h disc deflection = sqrt(2 * PE spring / k disc); and, from that F impact = h disc deflection * k disc.
Next add to the above the force on the stem due to any residual spring loading after the valve closing; and, the sum of those two loads will be the total load on your stem top, as well as, on the bottom stem connection to your disc center.

Note: There will be some additional damping to reduce the above impact force due the compression stress on your stem that I have not included in this analysis; but; the above procedure for determining the disc deflection can be applied to the stem and then the total deflection used in the final F impact equation will be the sum of the stem and disc deflections.

 

[SurajS]

which indicates to me that when the valve is closing or closed it is stopping flow from that inlet to the the lower pipe. So, unless you are trying to use your valve to block a reverse flow form the bottom pipe to the side "steam inlet"inlet,

Thanks a lot for detailed analysis.
Here, please note, steam is present on above and below the valve disc in running condition. Valve needs to be closed against the effective unbalanced area (which is area of stem) on which steam force is acting in upward direction and spring closes against this force.

For worst case design, when no steam is admitted, valve will be impacting on valve seat due to spring energy.

I have considered the case of simply supported circular plate with load at center (Case 1 in the link you had provided).
The dimensions of valve disc considered. t=14 mm, r=38 mm, K_disc= Et^3/(0.552r^2)= 688506 N/mm
Spring stiffness considered= 17000N/m, and elongation in spring= 60 mm.
deflection of disc = 0.298 mm
Impulse force= 6888506* 0.298= 205271.93 N. It is very large force to be handled. I understand that we have ignored energy dissipation in valve seat and casting which are difficult to quantify.
What percent of energy of spring that will be directly transferred to valve is another big question.
Do I need to provide damper or disc spring on top of valve stem like a shock absorber?I had calculated the force of impact with with classical mechanics as F= change in momentum/ time

Time of impact of valve and valve seat , approximately considered as 0.001 sec since valve closing action completes in about 40 milliseconds.

mass of valve= 3kg
moment of inertia of lever system= 2.5 kg-m^2 about pivot.

Energy stored in extension spring= KE of lever system
0.5* 17000(N/m)* (60mm)^2 = 0.5* moment of inertia of lever system * angular velocity ^2

angular velocity of lever (to which valve and spring is attached)= 5 rad/sec
Linear velocity of valve = angular velocity * distance of pivot from valve = 5*0.14 = 0.7 m/s,

Force of impact = (mass of valve *(velocity before collision - velocity after collision))/ Time of collision
= 3* (0.7-0)/0.001= 2100 N

Conceptually this approach is correct but only glitch is time of collision!
How do I proceed further? Which is realistic value? The one suggested by former approach or the classical mechanics approach?

Please guide me. Thanks a lot in advance.

 

[JBA]

First, thanks for the update on what valve action you are actually addressing, that helps clarify my understanding of your problem. Ironically, because all of our pilot actuated POPRV's are pressure operated, I never considered the possibility of a zero pressure valve closing. Also, since I did not have your valve diagram in front of me while developing my above analysis I didn't even think to consider the operating arm in the analysis; nor, the force multiplication due to the lever effect of the spring connection offset from the top of the stem. I apologize for that and have now printed a copy of the of the drawing to have in front of me for future reference.

What percent of energy of spring that will be directly transferred to valve is another big question.
Do I need to provide damper or disc spring on top of valve stem like a shock absorber?

The amount of force the that is applied to the stem/disc center is what should be considered as the force delivered to the body valve seat since it is the item providing the opposing force to the impact. If you are concerned about the stress levels due to the current impact force then any added deflection between the spring and the valve stem, such as a disc spring, is one way to reduce the impact force and stem, etc stresses.

Conceptually this approach is correct but only glitch is time of collision!
How do I proceed further? Which is realistic value? The one suggested by former approach or the classical mechanics approach?

My best suggestion to helping resolve your conflict is to try performing the following third alternative impact force analysis to see how its result compares to your two prior analyses.

Use the average driving force of the spring (i.e. the average of the spring force with the valve open and the spring force after valve closing) as applied to the top of the stem; and, a consolidation of all of the lever assembly components placed at the top of the stem location; then use, a=F/m to establish the acceleration of that mass; and, since you know the valves closing distance: t = sqrt(2*s/a) and consequentially v = 1/2 a*t, then the total PE = KE = 1/2mv^2 applied at the top of the stem is established; and, as before:
Using your established k for the disc loading; then, h disc deflection = sqrt(2 * PE / k disc); and, from that F impact = h disc deflection * k disc.

 

[JBA]

Error Alert on above analysis calculation:

. . . then use, a=F/m to establish the acceleration of that mass; . . .

Because the valve stem and disc are included with the top mechanism their mass needs to be included with the mass of the lever etc in the above a = F/m calculation.

I can't believe I didn't see that originally. I think I must be getting to the age that I am losing some brain cells.

 

[SurajS]

t = sqrt(2*s/a) and consequentially v = 1/2 a*t, then the total PE = KE = 1/2mv^2 applied at the top of the stem is established; and, as before:
Using your established k for the disc loading; then, h disc deflection = sqrt(2 * PE / k disc); and, from that F impact = h disc deflection * k disc.

Thanks for your help.
I already estimated valve closure timing by writing differential equation of lever system which is second order DE. From that I had estimated time to close about 40 ms. This is not duration of collision. That is why i had assumed time of impact as 0.001 sec

and yes mass considered must be total mass (M lever + Mvalve)= 6kg (not 3 kg which I had wrongly considered)
So the force of impact from f= dp/dt = 4200N not 2100N.

And KE has to be calculated 0.5 *I*w^2. And it does not change answer much. Since impact force is mainly governed by Kdisc which is very large anyway.
But again which is correct value?

 

[JBA]

I have duplicated your No. 9 Post calculation procedure and my calculated impact force is 6886 N. A careful comparison with yours indicates that the apparent error in your calculation was the failure to convert your calculated disc k from N/m to N/mm in the calculation. Unfortunately, the 6886 N result is still not exactly equal to any of the prior alternate calculation methods but at least it is in their range.

One other item for the spring's PE calculation the actual spring travel distance should be used rather than the full spring extension of 60 mm. If there any residual spring extension and load when the valve is in its closed position, that residual spring load can be added to the calculated impact load but its residual spring extension length should not be included in the spring's PE calculation.

(Attached below is a copy of my Excel analysis sheet for your review)

 

[SurajS]

Hello, Thanks for excel calculation.

Please note the unit of E =202 GPa= 202 *10^3 MPa= 202 *10^3 N/mm^2

The dimensions of valve disc considered. t=14 mm, r=38 mm, K_disc= Et^3/(0.552r^2)

K_disc = 202 *10^3 (N/mm^2)* (14 mm)^3/(0.552* (38 mm)^2) = 774803 N/mm (I had Considered E =200 GPa earlier, so value is different)

About spring preload, it is 420 N. (At zero extension).
Hope it clarifies the things.
On google, I found the data of steel ball rebound force calculation. Have a look. Their PE is about 44 J. In my case it is 30 J. Can I approximate the force based on these data. Let me know. Thanks.

http://www.pcb.com/Contentstore/mktgcontent/WhitePapers/WPL_5_Impact.pdf

 

[JBA]

Edit: I have a completely new calculation in my US units and have gotten a completely different and lower impact force; but, I would like your input my below questions before sending it to you for your review. A main element is the US units difference between weight and mass.

I don't see any usable correlation between the referenced paper and our analysis. Please explain how you think you might apply the data in the above referenced paper.

Thanks for the clarification on the required conversion from MPa to N/mm^2. For the entirety of my life and profession in the USA I have used inches & lbs, so, metric is a bit of a foreign language to me, particularly Newtons and Pascals. I normally convert metric to in & lb when I have to solve a metric system problem; and, I started by doing that on this calculation, but then decided I could skip that step and just stay with metric. My mistake! So we are back to the high value result again; and, at this point I have no explanation as to why we are getting such a difference between its result and our other analysis result.

I need some clarification on the below items:

I have considered the case of simply supported circular plate with load at center (Case 1 in the link you had provided).
The dimensions of valve disc considered. t=14 mm, r=38 mm, K_disc= Et^3/(0.552r^2)= 688506 N/mm
Spring stiffness considered= 17000N/m, and elongation in spring= 60 mm.
deflection of disc = 0.298 mm
Impulse force= 6888506* 0.298= 205271.93 N.

Where does the above disc deflection = .298 mm come from?

Force of impact = (mass of valve *(velocity before collision - velocity after collision))/ Time of collision
= 3* (0.7-0)/0.001= 2100 N

You are stating that m * V / t = F, but the units conversion is: (kgm * mm/sec) / sec = (kgm * mm) / sec^2 how does that = N?

I also need some clarification about the spring extension, you said the spring's rate is 17 N/m and for that rate the spring must be extended 24.7 mm to provide 420 N of force at the valve's closed position, that indicates the total extension of the spring from its zero load length is 25 + 60 = 85 mm, is that correct; or is its valve open to valve closed deflection actually 35 mm?

 

[SurajS]

Hello,
I can understand the unit convenience. You can do the calculation in lb in system. I can convert comfortably. I am using SI units as here in India it is predominantly used.

I don't see any usable correlation between the referenced paper and our analysis. Please explain how you think you might apply the data in the above referenced paper.

I was thinking can my valve be treated as ball falling from the height mentioned (Especially for impact time) I am not so sure if it is right approximation since area of contact in ball and valve are significantly different. I was trying to use the experimental data which I don't have and neither I can have in future.

Where does the above disc deflection = .298 mm come from?

Next, using your established k for the disc loading; then, your: impact h disc deflection = sqrt(2 * PE spring / k disc); and, from that F impact = h disc deflection * k disc

You are stating that m * V / t = F, but the units conversion is: (kgm * mm/sec) / sec = (kgm * mm) / sec^2 how does that = N?

P= mV/t = kg* (m/sec) /sec= kg-m/s^2 = N

I also need some clarification about the spring extension, you said the spring's rate is 17 N/m and for that rate the spring must be extended 24.7 mm to provide 420 N of force at the valve's closed position,

It is an extension spring. There will be preload of 420 N which has to be induced at the time of manufacturing.
All I want is spring force should vary from 420N (94.4 lbf) from valve close condition to 1150N (258.5lbf) in valve open condition.
We can induce the pretension by the way you mentioned just maximum spring force should be 258.5 lbf at 85mm elongation so corresponding spring stiffeness will be 13.52 N/mm (77 lbf/in).

Thanks.

 

[JBA]

Below you will find new disc/stem impact calculation using my units system; and, includes both the Spring PE based solution and an Average Spring Force based solution for comparison.

Thanks for your responses and I apologize for my continuing string of questions; but, I am just trying to insure that I fully understand all of the elements of your problem.

With regard to the V/t = N question: I failed understand the F kg = kg*m / a = N relationship, i.e. just another SI units confusion issue on my part. For our system Mass = lbs / g = slugs; and, apparently we have a British physicist to blame for the "slugs" name choice.

I was thinking can my valve be treated as ball falling from the height mentioned (Especially for impact time) I am not so sure if it is right approximation since area of contact in ball and valve are significantly different. I was trying to use the experimental data which I don't have and neither I can have in future.

The ball drop really gave them no actual usable data; and, by performing the last suggested calculation, you have an equivalent of their drop test by calculating an acceleration using your average F spring and closing assembly mass (instead of their g acceleration), your valve's closing travel (instead of their drop height) and your calculated disc k factor and to find your F impact force (instead of their averaging of their impact test graph's recorded impact force profile to obtain their average impact force).

 

[SurajS]

Edit: I have a completely new calculation in my US units and have gotten a completely different and lower impact force; but, I would like your input my below questions before sending it to you for your review. A main element is the US units difference between weight and mass

Hello sir, please let me know. Thank you.

 

[JBA]

In your Post #16 you have already replied to my Post #15 requests; and, my new Excel analysis is the one attached in my above Post #17.

 

[SurajS]

hanks for your responses and I apologize for my continuing string of questions; but, I am just trying to insure that I fully understand all of the elements of your problem.

Thank you sir, you are really helping me by all means.

About the calculation you have provided,
Just one glitch. with constant acceleration problem v= u+ at , a= acceleration, u= initial velocity, v= velocity at time t. You have considered 0.5*a*t.

Nonetheless, force value is pretty much high. of the same order which we had calculated by considering valve as simply supported plate. Force is governed by stiffness value of plate and Kdisc is high. It is the same approach.

 

[JBA]

I have used v = 1/2 a*t for the average velocity in line with s = 1/2 *a *t^2; as opposed to v = a*t which is the terminal velocity and does not correctly calculate the time of travel.

 

[SurajS]

I have used v = 1/2 a*t for the average velocity in line with s = 1/2 *a *t^2; as opposed to v = a*t which is the terminal velocity.

Oh ok , understood. Can such high value of force be practical, because against this value valve stem has to designed for buckling.

Whereas with F=dP/dt , value is much less.

 

[JBA]

If you are referring to the average spring force calculation method, then yes it is a bit less and to be honest I would be inclined to use that; but higher the PE method result is a bit safer, barring an actual instrumented test.

I would not consider either calculated load value out of reason; so I suggest that you do the stem buckling, stem compressive stress and disc connection shearing stress and disc deflection stress calculations for both and forces to see their results; and, remember that for all of those analyses you must add the closed valve spring residual load to the impact load.
Keep in mind that because the stem has a buckling restraint just above its midpoint at the body seal there is a bit of an issue for the classic buckling analysis using the stem top connection (I assume you would treat the stem top connection as a guided end and the bottom disc connection as a rigid end connection but I am not sure how to treat the guiding restraint issue).

PS It is now 1:00 AM where I am so my responses to your further post(s) will have to wait until after I rise in the morning.

 

[SurajS]

If you are referring to the average spring force calculation method, then yes it is a bit less and to be honest I would be inclined to use that; but higher the PE method result is a bit safer, barring an actual instrumented test.

Thanks for the suggestions. And yes boundary conditions are bit of a challenge but force is critical issue.
One more thing, to absorb the valve shock, can we put disc spring on top of valve stem?

Let me know about your view.

PS It is now 1:00 AM where I am so my responses to your further post(s) will have to wait until after I rise in the morning.

Thank you for support.

 

[JBA]

Since protecting the stem is your primary concern placing a shock absorbing feature at the bottom of the stem is definitely worth considering; but, if your valve cap to disc space is very similar to that shown in the valve drawing you placed in this thread, it appears you are going to have a limited space for that modification. Placing the damping part feature there is easy but then adding the required disc to stem retaining feature without compromising the stem strength might be more difficult.

You can use the current calculations by substituting increasing values for the current calculated value to determine the required bottom deflection to protect your stem. In that respect, do mind telling me your stem diameter and the maximum expected ΔP across the valve disc when it closes under operating conditions.

PS From looking at our apparent time difference, if you need a quick turnaround post exchange on some issue, then your 8:30 - 9:30AM will be my 10:00 - 11:00PM so, depending upon your circumstances during that morning time, it might be a good time frame for starting one.

 

[SurajS]

Since protecting the stem is your primary concern placing a shock absorbing feature at the bottom of the stem is definitely worth considering; but, if your valve cap to disc space is very similar to that shown in the valve drawing you placed in this thread, it appears you are going to have a limited space for that modification.

Like you pointed out about the space, rather than bottom, I want to put the spring on top of stem. Stem diameter is about 16mm (0.63in).

I want to know: will the valve force of impact be directly transferred to resetting lever through valve stem which needs to be absorbed by spring? Do i need to put the spring? Since energy will be dissipated in various ways like energy lost in casting of turbine, sound etc. How to I know, what percent of energy (rather force as I have to design a spring against this load) is reaching top of valve stem? Will the valve rebound? if so, putting spring at top of valve stem could be the solution. Let me know.

Approximating valve stem at resetting lever as fixed and valve as pin when it hits the valve seat, Buckling load I am getting as 40000N (approx). P= 2*pi^2*EI/L^2, l=400mm, E= 100GPa (At 400 degree C. Steam inlet conditions are 40bara/400degree Celsius) . Since the moment valve closes, pressure difference across valve is approximately null.

PS From looking at our apparent time difference, if you need a quick turnaround post exchange on some issue, then your 8:30 - 9:30AM will be my 10:00 - 11:00PM so, depending upon your circumstances during that morning time, it might be a good time frame for starting one.

Sure, I am happy to know. Thanks.

 

[JBA]

First a some general items that I would like clarified before I respond to the stem spring design and valve rebound issues.
1.

Since the moment valve closes, pressure difference across valve is approximately null.

I assumed the purpose of the valve is to block the steam flow to the turbine(s) in the event of a turbine system problem of some kind; if so, then there will be a pressure force on the disc = P inlet x (disc area - stem area). If what I assume is true then the steam flow, which has flowing inertia, will be able to deliver the full inlet pressure to the disc at the valve's closing point.
If the above is not the purpose of the valve being closed, then please explain its alternative purpose to me. The pressure force will not directly affect the stem loading; but it may increase the disc's deflection at the closing point and thereby reduce the impact force for the stem and actuator.

2. I have assumed that the mass of the disc has been included with the stem in the impact calculation. If not, then the closing assembly's currently calculated acceleration and therefore the impact force are higher than they will be if the disc mass is included.

3. Based upon your above operating pressure and stem diameter, the max lifting force due to the stem is 180lbs (800N) and that is at the top of the stem and if your lever design is similar to the Valve Schematic posted then the required spring is even lower due to it lever advantage over the stem. So, apart from the stem seal friction, why is your designated valve open spring force 258lbs (1147.6N) so much higher?

Now for your items I can address regardless of the above:

Since energy will be dissipated in various ways like energy lost in casting of turbine, sound etc. How to I know, what percent of energy (rather force as I have to design a spring against this load) is reaching top of valve stem?

Any effect in reducing the impact force from the valve body will be minuscule because of its thick walls and therefore low stress and resulting strain from the impact force; so there is no help there and should not be considered in your analysis.
A Note of Caution: Be very careful about attempting to solve or reduce a problem with concepts such as the above without any verified calculations or data, because doing so has the potential to end up causing you real problems that never should have occurred. A bit of over-design is much better than the risk of a potentially under-designed one.

 

[SurajS]

I assumed the purpose of the valve is to block the steam flow to the turbine(s) in the event of a turbine system problem of some kind; if so, then there will be a pressure force on the disc = P inlet x (disc area - stem area). If what I assume is true then the steam flow, which has flowing inertia, will be able to deliver the full inlet pressure to the disc at the valve's closing point.
If the above is not the purpose of the valve being closed, then please explain its alternative purpose to me. The pressure force will not directly affect the stem loading; but it may increase the disc's deflection at the closing point and thereby reduce the impact force for the stem and actuator.

Yes , this is stop valve, so pressure difference has to be there across valve disc. I told null pressure difference just at the time of closure. I want to design for the worst case where to test the trip system, valve will be tested in absence of any steam. In presence of steam, impact will be lesser.

2. I have assumed that the mass of the disc has been included with the stem in the impact calculation. If not, then the closing assembly's currently calculated acceleration and therefore the impact force are higher than they will be if the disc mass is included.

Yes i have considered mass of disc and lever too.

3. Based upon your above operating pressure and stem diameter, the max lifting force due to the stem is 180lbs (800N) and that is at the top of the stem and if your lever design is similar to the Valve Schematic posted then the required spring is even lower due to it lever advantage over the stem. So, apart from the stem seal friction, why is your designated valve open spring force 258lbs (1147.6N) so much higher?

Well, from the data observed, Ratio of moment of spring force and moment of steam force about pivot, I have kept as 3. Closure time should be less than 0.5 sec or so (I had posted the question related to this too). I could not arrive at the stiffness of the spring required so I checked previous data and found this relation.
In static condition, just to close valve minimum spring force required = 360N with leverage.