Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations The Obturator on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Help with check ball design 4

Status
Not open for further replies.

nickjk

Mechanical
May 10, 2007
74
I am having problems with a Spring seated check ball used in a low volume low flow (130 cubic in./min max) hydraulic device.

The device uses a .250" steel ball, 4.4 lb spring with a 54 lb/in spring rate, 118 degree included angle seat with a .0147 sq. in. contact area.

The original design intent was to prevent the actuation of the device until a pressure of 300 psi has been reached.

My problem is there appears to be a large pressure drop in the device after the check ball.

With the check ball installed the device engages at approx 640 psi.
With the check ball removed the device engages at approx 420 psi.

What could be causing the large 33.4% pressure drop, could it be caused by the spring rate?

Any ideas would greatly be appreciated.

Thanks in advance,

Nick
 
Replies continue below

Recommended for you

"With the check ball installed the device engages at approx 640 psi.
With the check ball removed the device engages at approx 420 psi.
"

Not following. How does the device "engage" if there is no ball to seal the port?
 
btrueblood

With the check ball installed, once the hydraulic pressure overcomes the spring the ball is unseated and the hydraulic fluid flows through to the device. This is where I think the pressure drop is.

In the other case I removed the ball and spring so the fluid can traveld straight through to the hydraulic device without building pressure first.

I hope this helps

Thank you,

Nick






 
What cracking pressure did you design the spring and ball combination for? I.e. how much preload is in the spring when the ball is seated?

If your seat area is 0.015 in^2, or roughly a .14 inch diameter circle, and you put 4.4 lbs. of preload on the ball, then the cracking pressure should be roughly 300 psi. That sounds fairly close to your number of 400 psi. This is for a very low flow rate, when the ball has barely lifted from the seat.

With no spring or ball, the same orifice should produce a pressure drop of about 1 psi (using water as the reference fluid). I'm assuming there is little or no restriction between the ball and the valve housing.

The ball and orifice together with a 54 lb/in spring should produce a bit more drop. I usually use 20-30% of seat diameter as a rough opening distance for the ball, which gives a spring deflection of 0.2*.14 = .03 inch, or spring delta-force of .03*54 = 1.4 lbs, acting on the projected seat area of .015in^2 gives an additional 100 psi. This is an upper ballpark for the flowing check valve. There are published correlations for ball seats, but I've had trouble with the one listed in the Valve Designer's Handbook.

Hope that all helps.
 
btrueblood

""If your seat area is 0.015 in^2, or roughly a .14 inch diameter circle, and you put 4.4 lbs. of preload on the ball, then the cracking pressure should be roughly 300 psi.""

Your statement above is my original design intent.

And if I understand correctly, the ball will lift a distance that is 20-30% the seat diameter. Does this mean the distance traveled is independent of spring rate.
Because the ball is traveling that additional .03" the spring force goes up which in the example increases the pressure applied to raise the ball to a total of 400 psi. Will the pressure above the ball be 300 or 400 psi?

Where I am having a problem is based on if I apply approx. 650 psi to the check ball I would expect the check ball to unseat, pressure to build and equalize on the other side and the ball to reseat with maybe a slight 10 to 25 psi drop. It appears I am getting approx. a 230 psi drop.

I did not know about the 20-30% for ball travel.

Thank You for your time and help,

Nick
 
Well I think the light bulb finally came on.

Please anyone, help me to confirm my thoughts.

In the above example, The preload of the spring and seat area are designed for a 300 psi cracking pressure. Because of the addition .03 travel of the ball and spring this pressure is more likely 400 psi.
If I apply 650 psi to the inlet of the check ball the outlet pressure will be the 650 psi - (spring preload / ball seat area) = 350 psi, hence pressure drop 650 psi inlet 350 psi outlet
Because of low flow I neglected pressure drops caused by flow.

See enclosed attachment

Thank you,

Nick
 
 http://files.engineering.com/getfile.aspx?folder=43536ad8-8647-4158-9461-c663474f979c&file=checkball.pdf
nickjk

Your drawing describes the condition where the ball actually seals the flow. However, when the ball moves (float) the picture is no longer valid. The forces of fluid flow drag forces hold the ball against the spring and you can no longer say that the 650psi pressure operates on the 0.0147 seat area. The 20-30% travel is what needed to use the full flow capability of the orifice diameter, not the seat diameter.

What is not clear from your description is: Do you have a constant flow through the check valve or it opens and fills a chamber until the pressure in the chamber is such that the check valve closes and reseal the flow.
 
Israelkk

The check ball opens and fills a chamber until the pressure in the chamber is such that the check valve closes and reseal the flow. Also the volume is very small.

Thank you,

Nick

 
What is the resael pressure? How the pressure in the chamber behaves? Is it continuously increases and then then check valve seals, or it increases and then decreases before it seals? If it goes up and down how much?
 

The others above have already in my opinion given you the answer, but let m try anyway:

It seems to me that you are trying to look at the pressure/forces on the ball for the different situations related to the fluid pressure contra the spring force (wich is again related to ball/spring travel) in 'static' situations

For a check valve you have normally three situations:

Situation 1: valve closed until a certain pressure; = cracking pressure.

Situation 2: valve fully opened at a given flow at a given pressure. This is the minumum flow/pressure to keep the valve fully opened.

Situation 3 : flow (and pressure)is 'in between' situation 1 and 2. . The flow is too low for the given reaction force (spring force compressed) to keep the valve completely opened. The ball is riding on a too low flow, or 'gulping' between closed and (partly)opened.

It seems to me that you could perhaps be in situation 3. This means the spring force is too high or the flow too low. In addition the partly opened system may restrict the flow further.

If your situation reaches a'no flow', but open valve situation, the pressure on all walls in two interconnected closed chambers would be equal.


 
gerhardl

I agree, I feel this case would fall into situation 3.
My fluid flow is very low. During testing with a hand pump you will see cracking pressure being reached and then a slight drop in pressure then cracking pressure being reached again then a slight drop in pressure then it seems like you see the cracking pressure going up slightly then a slight drop in pressure. I would have to test again for what the difference in pressure is.
This probably agrees with what you describe as gulping between closed and partly opened.

Do you agree that situation 3 with a sealed outlet that will just build pressure by slowly opening and closing or gulping will act similar to the static observation?

Your right, the others have answered my questions. I am just trying to fine tune the design by understanding what is going on.

Thank you all,

NickJK
 
Back pressure on a check valve can create many flow related problems. One of my earlier expensive experiences was with a manifold that had a HydraForce CV08-21 check valve in it.
This worked fine but the flow was too low after the system ran for some time and caused motor shaft seal failure. I redesigned the manifold to accept the HydraForce CV10-21 check valve.
This valve has a different design and when the back pressure reached a certain level the valve would open wide open and cause a malfunction in the system. To trouble shoot the problem I had to build a data collection system with a gear type flow meter and do testing in real time on a machine during operation. In this case some of the back pressure was from a directional control valve 30 ft from the manifold. To resolve the problem it was less expensive to modify a RV10-22 valve than redesign and manufacture two different complex manifolds.
The current project is a 15,000 psi cartridge relief valve. It uses a design similar to your check valve but it has a stack of disc springs for pressure and an adjustment screw. I used Algor to do the CFD and found it necessary to restrict the flow (up to 60 GPM) into the face of the ball or the flow speed and pressure at the ball face would cause the oil to burn. We have replaced the disc springs with lighter coil springs and are able to use it as a check valve in a low flow pressure intensifier.

Ed Danzer
 
Nick,

Ed's 1st post is my concern also - the "chamber B" seems to be too small in diameter, you should verify that the flow area around the ball is larger (by ~2x or more) than the open orifice area.

The 20% to 30% is a ballpark for full flow thru a ball/poppet seat. This is a ball travel of .2 to .3 times the seat orifice diameter.

My use of seat projected area on the ball for the differential pressure of the flowing valve is also a ballpark, but it's a fairly good one when the diameter of ball is a lot larger than the seat, like yours is. You can do some CFD like Ed does, and when you have system dynamics and high flow rates/velocities like his, it's a good approach. But if you just want to tweak your cracking pressure, having a simple model to guide you in picking a spring is more helpful than hours spent behind the PC. One test data point from the lab is worth 1,000 runs of CFD code.
 

nickjk:

Your description seems self evident in explanation of what is happening.

I am not quite sure I understand your last question, but both EdDanze and btrueblood is giving valuable input and seems more familar with the design than myself.

I fully agree with trying a model in practice.

Theory will bring you to a certain point, but then it could be practically impossible to take into consideration all variables in the total system that could influence the valves behaviour.

A revised or different models could be cheaper and faster.

Good luck!

 
Status
Not open for further replies.

Part and Inventory Search

Sponsor