Could this be?

[SteveWag]

I have a simple weighted arm swing check, 8-inch, horizontal installation, resilient seated, NOT installed as a check (not intended to prevent backflow). The valve opens in the direction of flow and is bypassed by a two-inch connection (18-feet sched. 40, two ells, one magnetic flow meter and one injection tee, all 2-inch). The arm and weight have been modified so the valve opens when 1.7 PSI of differential is developed - by flow through the 2-inch.
When flowing more than 50 GPM or when not pressurized, the valve arm, and valve disc, opens freely, with manual lifting of the weighted arm. When only pressurized (90 PSI) and not flowing, lifting the arm becomes very difficult, initially. A coworker suggests, and I agree, that the 90 PSI is acting on different upstream and downstream areas. That is, the upstream pressurized area (tending to open the valve) is the inside diameter of the seat opening, while the downstream pressure is acting on the area of the opening plus the seating area of the resilient seat. This makes the upstream area, D=8” while the downstream area is D=8 1/8” or 8 1/4 “. The difference in area is 1.9 to 3.2 square inches and the difference in force is, at 90 PSI, 171 to 280 pounds. Does this seem reasonable?
Thanks
Steve

 

[BigInch]

Yes, but the 1.7 psi differential x the upstream area would also seem to answer the how much force is required and that is much less than the 171 lbs.

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek

 

[katmar]

The 1.7 psi fits very well with the expected pressure drop for a flow of 50 GPM through the bypass. If the data at 90 psi is an arbitrary point then there is no necessary link to the opening force required for the valve at normal operating conditions. The 1.7 psi on the upstream area inside the seat is only about 85 pounds, suggesting that your normal operating pressure is about 45 psi. If the normal operating pressure is 90 psi then I agree with BigInch that there seems to be a bit of an anomaly. It all boils down to a question of what your normal operating pressure is.

Katmar Software - Uconeer 3.0

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[BigInch]

And if you have a pressure gauge on both sides of the valve. When it is closed, pressure on one side is probably not the same as on the other.

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek

 

[SteveWag]

Thanks all for your comments! I guess I did not describe the situation so well. The whole system seems to work well. First, the nominal pressure is 90 PSIG. The upstream and downstream sides of the swing valve are connected with 18-feet of 2-inch pipe. I cannot see, at zero flow, how a differential water pressure can exist. I can verify zero flow by use of the flow meter in the 2-inch. The problem is that the Operator tries to lift the arm, and weight, and encounters a “hard spot” where the lifting of the arm is almost impossible. He interprets this as a valve problem. Says something is stuck in the valve. We have torn apart the valve many times and found nothing. During reassembly the valve opens manually very easily. After reassembly the valve manually opens very easily. After pressurizing, the valve is very hard to manually operate. When any significant flow exists, the valve opens automatically as expected.

I think that the 90 PSIG (same on both sides) is operating on unequal areas, with the downstream area being a little larger. The initial manual lifting of the arm removes some slop in the keyway and the encounters the breaking force required to overcome the differential force.

Is this possible?
Thanks
Steve

 

[katmar]

Let's turn it around and instead of asking "what is the force on the flap", rather ask "what is the outside diameter of the seal" because that seems to be a bit of an unknown. Assuming that the inside diameter of the seal is exactly 8" and that the differential pressure to open the flap is 1.7 psi a simple force balance tells us that the outside diameter of the seal is 8.075" (or 8 and 5/64"). This is just a little bit less than the lower limit of 8 1/8" you originally specified. If the seal is elastic to some extent, as the pressure across the check valve increases the seal will revert to its original shape (presumably a circular cross section). When it gets to the point where the contact area with the flap has decreased to a diameter of 8 5/64" the flap pops open. Sounds possible and reasonable to me.

When there is no differential pressure across the valve and you have 90 psi both sides, then the seal is compressed and the diameter of the contact area could increase to 8 1/8" or more, giving the 171+ pounds you originally calculated

Katmar Software - Uconeer 3.0

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[StoneCold]

SteveWag
I don't have an answer but I do have two more questions.
Is the piping horizontal or vertical? (can there air trapped under the check valve?)
Is there a check valve in the 2" branch piping that stops water from flowing backwards during the "no forward flow" situation?

Thanks
Stonecold

 

[SteveWag]

katmar:

I think you and I agree. I am not interested in the cracking pressure during normal flow conditions, just now only in the opening force required to open the valve externally.

StoneCold:
Horizontal, 2-inch taps the 8-inch on top, rises vertical, about a 4-foot horiz. of 2-inch w/ a 2-inch air release on top :No check in the 2-inch bypass.

This is a domestic water distribution system, about 7K feet of down steam 8 & 6 inch and 25 houses, no storage. The idea is to force low flows through the 2-inch and add chlorine based on flow (in the 2-inch), high flows open the check and go on w/o extra chlorine.
Thanks
Steve
P/S additional comments welcomed!