[Unicorn91]

I tried to calculate the mass flow over a valve. First made a calculation by hand with Bernoulli:

(See image "Bernoulli.png")

Okay that was easy. Now I want to insert e pressure loss because of the valve geometry. In Literature it ist described this way:

(See image "Druckverlustbeiwert.png")

But now if zeta is < 1 my volume flow (bzw. mass flow) grows.
Is that possible?

Pipe contractions can have e zeta term of 0.5. Would that rise the mass flow?
In which situations is that possible?

Thanks for your help!

[FMDenaro]

As you know, Bernoulli cannot be applied when a dissipative process is acting, as happens for the viscous flow around a valve. Correction due to friction that casues pressure loss can be found in fluid mechanics textbooks, such as White. You find many exercises

[Unicorn91]

I was a bit confused as I realized that the definition of the pressure loss is based on Bernoulli as I described above.

https://de.wikipedia.org/wiki/Druckverlustbeiwert

First I startetd with the "first law of thermodynamics":
(See "first law of thermodynamics.png")

In my opinion I have to respect the pressure loss separately as I described. But I was not sure when I saw the definition of the pressure loss. And what happens when zeta is < 1?

@ FMDenaro: I've found the testbook from White. Wow almost 900 pages! Can you tell me which chapter or which exercise you mean?

[FMDenaro]

Originally Posted by Unicorn91

I was a bit confused as I realized that the definition of the pressure loss is based on Bernoulli as I described above.

https://de.wikipedia.org/wiki/Druckverlustbeiwert

First I startetd with the "first law of thermodynamics":
(See "first law of thermodynamics.png")

In my opinion I have to respect the pressure loss separately as I described. But I was not sure when I saw the definition of the pressure loss. And what happens when zeta is < 1?

@ FMDenaro: I've found the testbook from White. Wow almost 900 pages! Can you tell me which chapter or which exercise you mean?

Bernoulli requires the flow to be incompressible and the absence of any dissipation of energy due to viscosity and conducibility. This way the total pressure is invariant. Thus, you can compute the difference between the total pressure and the static pressure.


On the other hand, consider the example of the exact viscous solution in a channel (Poiseuille). You will find the pressure difference between two locations along the streamwise direction. This "pressure loss" is due to the friction.


In your case the flow around a valve produces separation due to viscous effect.The flow is not isentropic and the total pressure changes along the streamwise direction.



Search Bernoulli in the White textbook and compare to the section of viscous flows.

[JBeilke]

https://de.wikipedia.org/wiki/Druckverlust


There you find the bernoulli equation with pressure loss.

[FMDenaro]

Yes, the correction in the Bernoulli integral due to the viscous action highlights that the total pressure is no longer invariant in a non-isoentropic flow.

[Unicorn91]

Okay I forgot to say, in my situation I have a turbulent flow (RE = 10'000).

I tried an example from the textbook "White" for viscouse flow: see "Example Viscouse Flow Turbulent".
The resulting velocity (and massflow) is much too high.

So I think the main input of the pressure loss comes from the valve orifice.

@JBeilke: Right, that's how I calculated my second equation (see "first law of thermodynamics.png") So I think thats the right way.

But I was a bit confused about the definition of the flow coefficient Kv (see "Kv-Wert"). The flow coefficient is normaly given, when you buy a valve.

Maybe this is a simplification?