Valve CV vs thrust loads on Piping

[invent11]

In a water piping system we are replacing a 6" check valve with a gate valve (isolation valve) with much higher CV. What would be effect on thrust loads when valve is open? Is there any rule of thumb to have a rough estimate of its impact? Thanls for sharing your thoughts

 

[RVAmeche]

A high cV just allows a higher flow rate, if your system can handle it.

The valve itself doesn't generate thrust.

 

[invent11]

What about thrust on piping as a result of higher flow rate?

 

[davefitz]

The thrust can occur under 2 scenarios. First, if there is a change in momentum, such as a change in the rate of flow over a short period of time. If the valve were to slam closed suddently, the change in water momentum leads to "water hammer", normally worse at the nealrly closed position. The second situation is if the fluid flashes from liquid to steam or vapor and the velocity and density changes at the valve's throat due to throttling. This second effect is not relevant for the cold a water situation you described.

"...when logic, and proportion, have fallen, sloppy dead..." Grace Slick

 

[LittleInch]

How about you tell us some more details.

Which direction was this NRV?

Is the valve within a piping system or free to the air / open end?

A drawing, sketch or some idea about flow rates/ velocities, fluids etc all help.

There might be a bit more load but usually it's not worth talking about and is balanced by forces in the downstream pipework.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[bimr]

Whether the thrust load is a concern or not, will depend on your piping system which you have not described. If the system is constructed of steel piping, the piping system will handle the thrust load because the piping joints are welded with mechanical joints. If the system is constructed of push on joints, then you need to analyze the thrust.

 

[fel3]

The net thrust across an open gate valve will be tiny. So tiny that we can ignore it. Since the gate valve has a larger Cv than the check valve, the net thrust here will decrease from a small value to an even smaller value.

Take your data for maximum flow rate in the 6" line (Q, gpm), valve coefficient for the replacement gate valve (Cv), and specific gravity for your liquid (SG), and you can easily calculate the pressure drop across the valve (dP, psi). Then, multiply dP by the flow area of your 6" pipe (A, in^2) and you will get the net thrust across the open gate valve (T, lbf).

Cv=Q*SQRT(SG/dP) => dP=SG*(Q/Cv)^2

I found a table from Nibco that gave Cv=2,250 for their 6" gate valve, fully open. Assuming water (SG=1) and a flow rate Q=500 gpm (V=5.7 fps, which is reasonable), I get dP=0.049 psi and T=1.4 lbf. Tiny. This is barely more than my main camera body weighs.

Now compare this net thrust with the thrust on a 6" 90° elbow in a system that operates at 100 psi. The thrust equation for elbows is T[elbow]=2*P*A*sin(theta/2). For my assumptions, T[elbow]=3,999 lbf. The reason we can ignore the net thrust across the gate valve is that it is equivalent to a rounding error in the thrust for an elbow.

Fred

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"Is it the only lesson of history that mankind is unteachable?"
--Winston S. Churchill

 

[invent11]

Thanks for feedbacks

 

[1503-44]

Forces can be generated on the pipe/valve free body, or control volume, by both fluid momentum vectors and internal pressure.
In the absense of changes in pipe direction, high flows and significant surge pressures, the greatest net force on a valve is generated when the valve is closed.

Momentum: A net force will be generated on pipe or valve when there is a change in momentum of fluid within, ie. when mass, velocity or direction of fluid entering the control volume differs from that leaving, such as at an elbow location, with or without a valve, where a simple change of direction of the fluid's momentum, even during steady state flow, causes an opposite reaction on the elbow. When there is a change of direction, lateral forces on the pipe and valve can be generated. In the absense of a change in direction of fluid flow and the case of a fully open valve, it is not possible to generate a net resultant axial force on the valve as there is no colinear surface on which such forces can act. Cv versus valve open/closed position: As the valve closes, presumably surfaces on which the forces aligned with the pipe can act are created and the axial "thrust" on the valve increases as the valve closes. For a time steady state flow may be maintained and developed force may remain more or less constant, because as mass flow remains equal, the flow area is reduced and velocity increases, which causes no net change in momentum. Eventually as the valve closes more and more, undoubtedly steady state flow will be inturrupted, rapidly decreasing mass transfer to zero and eventually losing all momentum as the valve closes. That may or may not cause surge pressures to act on the closed flow surfaces within the valve, which are responsible for "surge forces" on the valve. Force on the valve then becomes differential_surge_pressure x area of flow. When surge pressure eventually come too rest, only internal pressure remains.

Internal Pressure: A net force on a control volume can also be generated from any internal differential pressure. The greatest net force on a valve is often generated when the valve is closed, similar to the end cap force in a closed pressure vessel. F= (Upstream pressure-Downstream Pressure) * cros-sectional area of flow. In the case of a pipeline block valve, that net force is resisted by the by axial stress within the pipe wall, which is typically transferred to the soil through friction over a length of pipe from the valve to virtual soil anchors on each side, or by more immediate means, directly through an anchor/support to which the valve is firmly attached.

“What I told you was true ... from a certain point of view.” - Obi-Wan Kenobi, "Return of the Jedi"