A straight piston stem won't be air balancing?

[nohappy]

On "Regulator savvy book" section one page 31, the author described that a straight piston stem won't be actual air balanced because there exists annular area on the knife edge of a piston.

But that really confused me, I assume that when a piston first stage is in close position, there is no HP air between the piston knife edge and the HP seat. Hence, HP air should only exert vertical force on the piston outer diameter. I can't see any horizontal force by HP air that would make it unbalanced.
I drew an exaggerated figure to illustrate my question. Unless there is HP air at the orange area, there should be no horizontal force by HP air. How could the author comparing the diameter of the sealing edge on the end of the piston and the sealing diameter of the piston stem o'ring and then calculating annular area by them? Instead, I think he should only consider the outer diameter on each side. As long as the piston stem is straight, it should be air balanced. Isn't that right?
To sum up my question, does the author assume there is HP air in the orange area (HP seat) or did I miss anything? How is a straight piston not air balanced? Why does the inner annular area on the piston edge matter?

 

[axxel57]

Thanks for thread.

That page in the Savvy has bothered me for long time. I think I understand the basic mechanism, but until today I’m missing the complete explanation.

I will try to show how far I got.

I’m sure you are right. When the piston seals with the HP seat, there is only vertical force on the piston stem in the HP chamber.

The point is, that, to create the IP behind the piston, the piston has to move and open the seal.

That’s when the streaming HP force starts to exert some force on the piston edge and the piston stem outside and inside ( I guess some sort of friction). Together with the main spring it forms a ‘down stream force’ against the developing IP in the LP area (LP chamber and LP hoses) until the gas pressure behind the piston has reached IP level and closes again the seal.

Remember, balanced piston 1st means that, quite independent from the supply pressure, the mechanism will create the ‘same’ IP over the course of a full dive. It’s by ‘definition’ not a matter of any balance within the 1st as I understand it.

In school Physics has been always my worst subject, so I cannot refer to any formula or physical law (and my English is not too good), but in second stages one can balance completely the valve by designing the orifice diameter and the sealing o-ring to the same size. As divers we don't use that type of valve, we keep the mechanism always slightly unbalanced to avoid a dead lock in the LP hoses in case of a 1sts failure.

In barrel poppet or coaxial 2nds designs it is more clear where the gas forces will work, but there must be a similar force effecting the movement of the 1sts piston (maybe it is really about friction and the piston edge).

I hope Luis or someone of the ‘usual’ suspects can chime in and enlighten us……

 

[Bob DBF]

I think he is pointing out the difference between theoretical and actual. Theoretically the knife edge has no thickness, in actuality it does, and this difference makes it not perfectly balanced. Which is why a difference in IP is seen between a full and empty tank.


Bob
No expert

 

[nohappy]

Thanks for your reply @axxel57 . I think IP value is decded in close position instead of open position. First of all, I believe the pressure in LP and HP chamber should be the same in open position. Secondly, I think there is nothing to do about air firstion on the piston. But I'm still appreciated for your help. Yes, I hope expert like @Luis H or someone can help me out.

Thank you, @Bob DBF . Yes, I understood he is pointing the the difference makes it not perfectly balanced. But my problem is that I don't understand why is that difference makes it not perfectly balanced. Or I should say that I think that his explanation is not making sense to me.

 

[Nemrod]

Because it (knife edge) has area and HP air is acting against it in the direction of your little arrows. N

 

[Couv]

I was hoping @Luis H would jump right into this thread as his understanding of the physics involved plus his ability to articulate on the subject far exceed mine-but I'll take a stab at it until he shows up. Of course I welcome any corrections.

On "Regulator savvy book" section one page 31, the author described that a straight piston stem won't be actual air balanced because there exists annular area on the knife edge of a piston.

But that really confused me, I assume that when a piston first stage is in closed position, there is no HP air between the piston knife edge and the HP seat.

I've highlighted closed position, because your statement is exactly correct. It is when the valve is in the open position the knife edge exposes more surface area to the supply pressure. The surface area times the supply pressure provide an additional factor in the total opening force. Therefore, as the supply pressure diminishes so does the total force holding open the valve. Remember, the IP only builds when the valve is open, so less force holding it open allows less force (IP x piston head size) to close the valve.

 

[rsingler]

I think a more careful reading of what was written might yield the answer.

" The seating edge will always be minutely smaller than the full outside diameter of the stem."
This is the key to understanding the issue of balancing, in the closed position.
The IP x area of the closed knife edge disc must equal the area of the stem at the point where it passes thru the body, in order for balancing to be perfectly equal.
But that is not the case. Microscopically, it is not a knife edge. It is rounded (microscopically). Therefore the circle of contact on a flat seat is just inside the outer diameter. The circle of contact is slightly closer to the outside of the stem on a cone seat, but is still slightly inside. Therefore, by definition, the pressure x area holding the valve closed will be slightly less than the tank pressure x area trying to force the valve open. Therefore the valve will stay open longer, resulting in a slightly higher IP. As tank pressure drops, the difference will decrease, and IP will fall.

You can eliminate this issue, or even overcompensate (overbalance) for this issue, by having the knife edge of the piston formed from a larger diameter than the stem that passes thru the body.

It's easiest to visualize, if you look at a picture of a center-balanced valve.

Imagine two different diameters exiting the center chamber due to differing diameters of the stem component. Obviously the larger diameter will exert more force on its side, sliding the center balanced stem to the side of the larger diameter. The same thing takes place in a piston first stage.

 

[Luis H]

Thanks @rsingler for the reply. I saw my name, but I haven't had time and I would have needed to produce some drawing to illustrate. Your sketch and pictures show it very clearly. Thanks

In theory the knife edge would need to be sharp enough that would be very delicate and it would cut right into the seat. In reality is not only basically impossible to create such a sharp edge, it is impracticable.

You do need a thin sealing edge, but it does need a little radius so it doesn't "excessively" cut the seat.

The first diagram from @nohappy shows a flat sealing surface. It would be very hard and unreliable to create a good seal with so much contact area. They would have to be perfectly parallel, flat and even all the time.

The smaller the contact area between the seat and the orifice, the least amount of force needed to create a continues contact sealing ring. To create a seal you only need a continues contact ring.

 

[rsingler]

@Luis H , it would be really helpful if you could diagram the net forces in a balanced piston, to show how the smaller knife edge diameter contributes to differing IP at falling tank pressure. I just can't figure out how to portray it.

We've got tank pressure times stem cross-sectional area for a force pushing out on the piston (opening the valve), where the piston enters the HP compartment.
We've got spring pressure pushing out on the piston head.
We've got IP x piston head area (minus what amount?) pushing the valve closed in the IP compartment.
Where is the vector applied for the smaller area of contact right at the seat? Is it subtracted from Piston Head area?

I'm stumped on portraying two pictures side-by-side with differing tank pressures and differing IP's.

 

[Luis H]

The math is fairly straight forward.

According to Pete Wolfinger, he is assuming the piston head is one square inch, but you actually have to subtract the inside projected area of the piston stem, (and to be accurate you need to add the projected area of the head O-ring.

But just for a rough order of magnitude, lets say the piston area is about 1 in square (the projected area inside the piston stem is about 0.03 in square which probably cancels with the O-ring area).

Simple math pushing the piston closed 1 in^2 x 130 psi = 130 pounds

The only area that is affected by the tank pressure is that little exposed ring.

Area = ((0.25^2) x Pi/4) - ((0.246^2) x Pi/4) = 0.00156 in^2

If we multiply that area times the change in pressure: 2700 psi x 0.00156 in^2 = 4.21 pounds

The 4.21 pounds is about 3% of the 130 pound force needed to close the piston.

I know that a drawing would help, but that would take me a lot longer.