Leaky Check Valve

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blockmanjohn

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Hi,

I built a small check valve as part of a Jan Ridders designed engine. The valve seat was cut with a small boring bar at 45 degrees. It then uses a 3/16" steel ball to make the seal. There is no spring pressure on the ball. In any event, the valve leaks. I noticed that there are tiny, concentric grooves on the seat which were left by the boring bar. I think that if I were to some how polish the seat all would be well.

My plan is to turn an aluminum rod with a 45 degree chamfer to size, chuck the seat in the lathe, charge the rod with 1200 grit compound, use low speed and gently feed the rod into the seat with very little pressure until the seat is smooth. Does this sound like it would work? Is there a better way to do this? The valve is made from brass.

Thanks in advance, John.
 
I built a Jan Ridders Debbie 2 stroke. All you need to do is put the check ball in and using a small punch, tap it with a hammer to seat it. It worked great for me. It's best to use a new ball after seating it.
If you need any more info on this engine, feel free to ask. I have sorted out a lot of issues with mine-now it runs great.
 
Thanks for the tip Cheepo. I'll give it a try. This is for Jan's Scuderi Cycle engine. I have it completed but can not get it to run yet. I'm hoping this will do the trick.
 
Many will argue that a check valve ball should sit on an edge, not in a cone. I think you would do better to cut the seat square. Using a slot drill or end mill will leave it slightly high towards the centre, which is even better. Then you pop a ball in, and do what cheepo45 said. If the through hole is roughly 0.7 times the ball diameter, the ball will sit on the seat with a 45° tangent.
 
I'm sure you guys know more about check-valves than me, but consider this:-
By tapping the ball with a hammer, you are increasing the area of contact of the ball. This means that for a given force, from the spring, there will be less PRESSURE between the ball and the seat.
Jack
 
Instead of tapping the ball, another way is to glue a ball to a rod that has been centre drilled and twiddle the ball back and forward on the seat to burnish it. Just a few twiddles usually does the job. The rod should fit the top part of the valve to keep everything normal.
Obviously a new ball will be required for the valve afterwards.

Dave
The Emerald Isle
 
The other idea is to use a silicon nitride ball, instead of giving it a belt put it in Valve & both in the vice using a rod against the ball & give it squeeze to form the seat. How much squeeze is up to you... A couple of advantages of using the nitride ball is they are better dimensionally round & far harder, after using it for the seat you can use it in the valve. A stainless ball you need to throw away as it’s likely you will deform it.
Hope that’s is useful

Cheers Kerrin
 
I'm sure you guys know more about check-valves than me, but consider this:-
By tapping the ball with a hammer, you are increasing the area of contact of the ball. This means that for a given force, from the spring, there will be less PRESSURE between the ball and the seat.
Jack

The pressure between the ball and the seat isn't the most important thing.

The most important thing is that there must be 100% contact between the ball and the seat. A knife-edge contact is nice because it's less likely to have bumps that prevent complete seating. But a burnished high-area seat will seal better than a lumpy low-area seat.

The relevant pressure for opening the valve is the force of the spring (or gravity, if there is no spring) divided by the area of the ball exposed to the fluid.

Carl
 
Carl,
You're probably right about the pressure, but I wouldn't like to even begin calculating the area of the ball exposed to the fluid.
Any ideas please?
Jack
 
Carl,
You're probably right about the pressure, but I wouldn't like to even begin calculating the area of the ball exposed to the fluid.
Any ideas please?
Jack

Calculating the area of a sphere is simple math (4*Pi*radius^2) then you just need to estimate the proportion of the sphere exposed to the fluid itself. Alternatively, I would think it'd be approximately close enough to just use the area of the inlet pipe and treat the sphere as flat (area of a circle = Pi*radius^2).
 
Alternatively, I would think it'd be approximately close enough to just use the area of the inlet pipe and treat the sphere as flat (area of a circle = Pi*radius^2).

Not "close enough". Using the Flat projection of the sphere portion obstructing the inlet pipe is the EXACT area to use.

Any area that is not perpendicular to the pipe axis is symmetrically located around the axis and any radial force is balanced out.
 
Not "close enough". Using the Flat projection of the sphere portion obstructing the inlet pipe is the EXACT area to use.

Any area that is not perpendicular to the pipe axis is symmetrically located around the axis and any radial force is balanced out.

I think for it to be exact we would need to integrate over the entire curved surface but at the scale we're discussing it's overkill.
 
The area exposed to the fluid is A=pi(rx)^2

rx is the smaller of the pipe diameter and the (ball diameter)*sin(theta)/2, where theta is the angle between the centerline of the valve and the conical ball seat.

If one wishes to integrate over the entire ball surface, one can do so. But one must only integrate the component of the pressure parallel to the spring force; the component perpendicular to the spring force doesn't lift the ball.

The result of this integration is the flat projection of the sphere, as Mauro has said.

Carl
Provo, UT
USA
 
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In my previous life, we made tens of thousands of round balls in conical seats by coining (tapping) the ball into the seat. It was, most often, a carbide ball into a conical brass seat, but an ordinary ball bearing ball will work as well if it's harder than the seat material. One big advantage of this arrangement is that the ball will always seal regardless of the angle (within reason) of the stem it's attached to. A cone in a cone type of valve can be very sensitive to angularly off-axis articulation since, once off axis, the contact area wants to be an ellipse and it isn't, so it will leak.
 
I'm sure you guys know more about check-valves than me, but consider this:-
By tapping the ball with a hammer, you are increasing the area of contact of the ball. This means that for a given force, from the spring, there will be less PRESSURE between the ball and the seat.
Jack
Now wait up, this seems obscure to me. Do you mean less pressure per unit area of contact between the ball and seat? Or something else?
 
Just use a nitrile ball saves all the hassle
First use a steel ball of the same size give it a biff with a guided punch or just let a centre drill kiss the edge. The through hole is better reamed not drilled
Ideally the seat is coned upwards so that in use any dirt falls away from the seat
Iv never had a problem with Nitrile but had a s-it load of tears using stainless
 
Now wait up, this seems obscure to me. Do you mean less pressure per unit area of contact between the ball and seat? Or something else?
Pressure is force divided by area so if you increase the area of the seat you are reducing pressure on the seat. It's why someone stepping on your hand with a stiletto heel hurts more than when someone wearing sneakers steps on your hand (I really need to spend less time sprawled on the floor at parties).
 
The area exposed to the fluid is A=pi(rx)^2

rx is the smaller of the pipe diameter and the (ball diameter)*sin(theta)/2, where theta is the angle between the centerline of the valve and the conical ball seat.

If one wishes to integrate over the entire ball surface, one can do so. But one must only integrate the component of the pressure parallel to the spring force; the component perpendicular to the spring force doesn't lift the ball.

The result of this integration is the flat projection of the sphere, as Mauro has said.

Carl
Provo, UT
USA

Have you got a reference or at least a name for the equations you're drawing these from so I can look them up? I'm stuck on hydro-static pressure where we certainly need to integrate over the surface given the change in head pressure with depth. It's going to be minuscule in something this size, which is why I suggested we could neglect it, but my intuition says your equations are not truly exact. I would like to read up on it some more though - I haven't had much experience in fluid dynamics.
 
Pressure is force divided by area so if you increase the area of the seat you are reducing pressure on the seat. It's why someone stepping on your hand with a stiletto heel hurts more than when someone wearing sneakers steps on your hand (I really need to spend less time sprawled on the floor at parties).
yES, Of course. What I am asking, is that the little ring of contact between the ball and seat gets smaller the cleaner everything is, (that is, the cuts) which would increase the pressure per unit area of contact--BECAUSE the total pressure in the check valve has stayed the same, just the area of contact has reduced (or increased) thus the pressure PER unit area goes up or down inversely. So the original post was very unclear.
 
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