Characterization of a valve

[magicfrog]

Hi everyone, I hope this is the right section to post this problem I'm getting lost in. In theory it seems trivial but in the end the reasoning doesn't make sense to me.

I refer to a real case study, a check valve (made internally with a spring ball system) in a piston that moves inside a viscous fluid.

Obviously the geometry and the inlet velocity of the fluid are known (vi). The pressure in the system is assumed to be atmospheric pressure (pi).

To activate the system, the hydraulic force must exceed the force of the spring. The question I don't know how to answer is to understand which are the dynamic laws that govern this system since the fluid is viscous and Bernoulli can't help me. Nor can Poiseuille's law because it is applicable for sufficiently long straight sections.

It would be interesting to understand the pressure variation in the first section (d2) and the subsequent one between the sphere housings and the sphere itself (d3-ds).

I'm probably lost in a glass of water but I can't figure it out.
obviously the characteristics of the spring and the fluid are assumed to be known.

Thanks to whoever answers!

 

[russ_watters]

To activate the system, the hydraulic force must exceed the force of the spring. The question I don't know how to answer is to understand which are the dynamic laws that govern this system since the fluid is viscous and Bernoulli can't help me.

When the valve is closed, the velocity of the fluid is zero. Are you really asking the pressure to open it or pressure drop to maintain a certain velocity?

 

[magicfrog]

When the valve is closed, the velocity of the fluid is zero. Are you really asking the pressure to open it or pressure drop to maintain a certain velocity?

The piston speed is known and therefore also the fluid velocity. I'm asking for pressure drop when the fluid has a certain velocity in the two stations (first orifice and meatus between seat and sphere) knowing however that the fluid is incompressible but viscous.

 

[magicfrog]

The only thing I can say for sure is the conservation of mass as an incompressible fluid. Knowing the velocity that insists on the section of diameter d1 I can derive the velocity in the various sections. But how can I relate this result to pressure drops? Should I go back to the Navier-Stokes equations (of which a simplification is precisely Poiseuille's law) ?

 

[erobz]

Mass, Momentum, and Energy effects need to be applied to the control volume, so if you don't know the appropriate velocity, pressure, temperature distributions then I suppose its Navier Stokes.

If you can make reasonable assumptions about the distributions you can get away with the integral forms:

$$ 0 = \frac{ d}{dt} \int_{cv} \rho d V\llap{-} + \int_{cs} \rho ( \boldsymbol {V}\cdot d \boldsymbol{A} ) \tag{mass}$$

$$ \sum \boldsymbol{F} = \frac{ d}{dt} \int_{cv} \rho \boldsymbol{v} d V\llap{-} + \int_{cs} \boldsymbol{v}\rho ( \boldsymbol {V}\cdot d \boldsymbol{A} ) \tag{Momentum}$$

$$ \dot Q - \dot W_s = \frac{d}{dt} \int_{cv} \rho \left( \frac{V^2}{2} + gz + u \right) d V\llap{-} + \int_{cs} \left( \frac{V^2}{2} + gz + u + \frac{p}{\rho} \right) \rho \boldsymbol {V}\cdot d \boldsymbol{A} \tag{Energy}$$

 

[jrmichler]

There is no simple hand calculation for check valve pressure drop. Spring force and orifice area control the cracking pressure. At low flow, the primary restriction is the annular area between the ball and the orifice. At larger flow, both the orifice area and area between the ball and barrel control the flow.

Check valve manufacturers publish pressure drop curves for their check valves. The figure below is an example from one such manufacturer: https://www.sunhydraulics.com/model/CXBA.

I used search term hydraulic check valve to find that graph. There were other good hits from that search. The page that lead to the image above, https://www.sunhydraulics.com/models/Cartridges/directional/check, shows the range of cracking pressures and flow rates for the check valves from one manufacturer.

A good exercise is to attempt to calculate the pressure drop vs flow rate for a check valve, then compare to the published curve for that check valve. You will then get an excellent understanding why these curves are measured experimentally.

 

[erobz]

Can we relate the pressure in the separation zone ( directly behind the sphere) to the drag force? I can’t seem to find a straight answer on this question, which is probably important for the “tinker model” I would like to develop for this.

 

[magicfrog]

Excuse my ignorance but what do you mean by ‘tinker model’?

Drag force? Can you explain better what you mean?
Thank you very much!

 

[erobz]

I just meant a model to play around with that attempts to quantify some effects.

I would like to just play around ("tinker") with the equations in post 5 to come up with some plausible uniform magnitudes pressure/velocity distributions acting over and passing through the control surface enclosing the sphere. The drag force acting on the sphere inside the control volume is partially responsible for the pressure distribution $P_3$, as well as mass, and energy requirements.

These taken together will output a force that the spring must balance if the system is in steady state ( for starters). I was asking if anyone knew how to quantify the pressure distribution more or less based on the drag coefficient of the sphere?

These equations are the basis for the Navier Stokes Equations (as far as I know) - the integral version.

 

[erobz]

ChatGPT tells me it could be approximately like $p_b \approx p_{\infty} +\frac{1}{2}C_{p_b} \rho U_{\infty}^2$. with $-0.4<C_{p_b}<-0.2$. It bases this of empirical results. I don't know how bad it is, but it seems plausible. Any takers?

 

[magicfrog]

Thanks for the explanation!

I don't want to say something stupid, but aren't there edge conditions required to apply this relationship? From what I remember it refers to the generic formula of the lift or resistance of a body .I seemed to remember that there were unperturbed flow or boundary layer conditions to be met and I honestly don't think this is the case with an overpressure valve.

Maybe it can be a rough approximation.

 

[erobz]

Thanks for the explanation!

I don't want to say something stupid, but aren't there edge conditions required to apply this relationship? From what I remember it refers to the generic formula of the lift or resistance of a body .I seemed to remember that there were unperturbed flow or boundary layer conditions to be met and I honestly don't think this is the case with an overpressure valve.

Maybe it can be a rough approximation.

I'm going to be frank, it's likely that you have forgotten more than I ever knew (I don't have any special training in it ). Which is why I'm asking about it - hence the idea of just tinkering around.

About the only thing I do know is there is a fine line between getting something down on paper and hell in a hand basket...and I don't know where that line is! So if you are foggy on the concepts you mention and I, uniformed, my feeling is to just move forward to see if what pops out of the equations can be rationally interpreted while waiting for an expert to weigh in on that question. If you are ok with accepting the inherent complexity as stated by @jrmichler, that's fine too. I don't want to hold you hostage while I play around with the equations; I find that hardly anyone ever joins in or comments ( gives feedback) about them anyhow on this site.

 

[magicfrog]

Let's play with these equations and see what we come up with. Somehow we will come out of it!

 

[erobz]

First Conservation of Mass-steady flow:

$$ 0 = \int_{(1)} \rho \boldsymbol V \cdot d \boldsymbol A + \int_{(2)} \rho \boldsymbol V \cdot d \boldsymbol A + \int_{(3)} \rho \boldsymbol V \cdot d \boldsymbol A $$

With constant density and uniform velocity distributions this becomes (bypassing the vector formalities - and dividing out constants):

$$ 0 = -V_1 A_1 + V_2 A_2 + V_3 A_3 $$

$$ V_1 d^2 = V_2 D^2 - V_2 b^2 + V_3b^2 $$

So $V_1, d ,D, b$ are specified leaving two variables $V_2, V_3$.

The next equation to naturally go to is Energy to work on those pressure magnitudes. Any Ideas?

 

[magicfrog]

I don't want to ruin the party but, shouldn't viscosity also enter into the first law of NS?

Turning to the energy equation I would be of the opinion to neglect the variation of heat as well as all variations of potential energy. I have doubts about work.

 

[erobz]

I don't want to ruin the party but, shouldn't viscosity also enter into the first law of NS?

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Turning to the energy equation I would be of the opinion to neglect the variation of heat as well as all variations of potential energy. I have doubts about work.

The viscous effects you would be citing in the differential equations are synonymous with thermal quantities $Q, u$ in the integrals. The way I imagine it is if inviscid flow was in thermal equilibrium with its surroundings and there was a switch that could turn on viscosity; two things are going to happen when you flip it( 3 if you count the dependent effects on pressure), the flow is going to warm downstream i.e. the internal energy ( per unit mass) $u$ is going to rise, and due to this warming $\dot Q$ is going to transfer out of the system as irrecoverable losses to the surroundings.

When they later develop the "Energy Equation" ( you have probably seen this version that does all this math with average values assuming steady flow giving you something that looks like Bernoulli's, but with additional terms for losses $h_l$ and shaft work ( pumps and turbines $h_p,h_t$) in terms of "head" and kinetic head correction factors that we use in plumbing more or less ) my authors group the terms and say (basically for this situation we are discussing) that the quantity $\left[ \int_{(1)} u \rho ( \boldsymbol V \cdot d \boldsymbol A ) + \int_{(2)} u \rho ( \boldsymbol V \cdot d \boldsymbol A ) - \dot Q \right] > 0$ as a consequence of the Second Law. Since it’s always positive they give it a proportionality to $\frac{V^2}{2}$ for convenience as far as I can tell. Still later in the book after developing Darcy-Weisbach they empirically measure different loss scenarios and make charts for us to follow. So basically, it’s usefulness and empiricism at the sacrifice of precision and ability to predict novel situations that you could find with the Navier Stokes equations. So I think the viscous effects are in there.

Does that sound more or less reasonable?

 

[magicfrog]

This only means one thing, considering thermal effects negligible implies considering a fluid non-viscous.

Correct?

 

[erobz]

This only means one thing, considering thermal effects negligible implies considering a fluid non-viscous.

Correct?

Yeah, that sounds correct, but only for pipe flow. So this is specific, not general result.