Pressure drop across valves

[Ask1122]

Hi guys, i want to ask why is there pressure drop across valves? Because if you look at bernoulli's equation, the head will be the same in 3 points, before the valve, at the valve and after the valve. The only thing changes when a fluid flow through a valve is its pressure and velocity at the valve, due to the change of area. But before and after the valve, pressure should be the same, because the area of the pipe goes back to the orginal size.
So why is there a pressure drop across a valve?

The best i can come up with is maybe because some of the energy in the fluid flow has been converted to vibration energy (i think there is vibrations in the valve disks but i am not sure how/why the convertion takes place). Is this assumption true?

And is there a way to predict how much pressure loss you will have across a valve? Because without some kind of predictions, how would you design/select a pump for a pipe line with valves (since your pressure drop across a valve, you might not have enough NPSHr for the pump)? What do people in the industry do? Do they size the pump with NPSHr value lower than the pressure after the valve?

Thank you for the helps!

 

[tiny-tim]

… So why is there a pressure drop across a valve?

The best i can come up with is maybe because some of the energy in the fluid flow has been converted to vibration energy (i think there is vibrations in the valve disks but i am not sure how/why the convertion takes place). Is this assumption true?

And is there a way to predict how much pressure loss you will have across a valve? Because without some kind of predictions, how would you design/select a pump for a pipe line with valves (since your pressure drop across a valve, you might not have enough NPSHr for the pump)? What do people in the industry do? Do they size the pump with NPSHr value lower than the pressure after the valve?

Hi Ask1122!

Yes, it's because of "unrecoverable" energy loss.

See the whole of https://www.physicsforums.com/showthread.php?t=67757" for fascinating details. Here's some extracts …

Well Tony, you've picked up on why it's so difficult to explain things in terms of Bernoulli's equation. When in school, it seems as if they explain way too much in those terms, as if to say Bernoulli's equation is all you need to determine pressure drop through a fluid system. That's totally false. Real fluid systems don't abide by Bernoulli's law, only in part. Real fluid systems must include thermal affects on the fluid which Bernoulli's does not. Bernoulli's only looks at the conservation of potential and kinetic energy, totally ignoring thermal energy. Pressure drop through a valve can't be determined solely on changes in potential and kinetic energy terms.

Even for an "incompressible fluid" such as water, there is some compressibility. The first law of thermodynamics applies to a liquid every bit as much as it applies to a gas. The PdV energy in the liquid must be exchanged with the internal energy.
…
Once you calculate the amount of pressure drop through a valve using these equations, you've essentially calculated what is commonly referred to as "unrecoverable" pressure drop. Bernoulli's does not allow for this, it assumes all pressure is 'recovered'. So the unrecoverable pressure drop is a change of internal energy (thermal energy of the molecules) into PdV energy.

Note that enthalpy is constant across the valve. If you want to calculate the real pressure drops in a system (without performing a CFD analysis) you'll need to use equations such as these at each step to determine unrecoverable pressure drop, along with using the first law along with Bernoulli's for changes in potential and kinetic energy. There's a lot more to calculating pressure drop in a real system than simply applying Bernoulli's.

Tony, the question you're asking is actually a very difficult one to understand in my opinion, but a very good question.

Most gasses when expanding through a throttling valve or other restriction, will decrease in temperature while a small number of gasses such as helium for example, actually increase in temperature after expanding through a restriction.

Regardless of if the gas cools or heats up, two er... three things are true:
1) The gas decreases to a lower pressure.
2) The gas will increase in volume.
3) Enthalpy will stay the same (one caveat regards overall fluid velocity, but for your typical fluid system, this contribution to enthalpy can be neglected).

Enthalpy is:
H = U + PdV
Where H = enthalpy
U = internal energy
PdV is a pressure energy term.
…
Internal energy is a combination of a large number of factors.
Ref: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/inteng.html#c3

So internal energy may increase, or it may decrease.
1) If internal energy DECREASES, the temperature will be LOWER, and the PV term will be HIGHER after expansion. This is the most common case.
2) If internal energy INCREASES, the temperature will be HIGHER, and the PV term will be LOWER after expansion. This is more unusual.
…
So the temperature after expansion is dependant on how the atoms or molecules rearrange themselves after expanding, which is dependant on the amount of different types of 'microscopic' energy available to the atom or molecule.

Why does the pressure not recover? Bernoulli's does not account for all types of energy in a system, only the recoverable mechanical ones. I've seen it said that energy is lost as the fluid expands through the valve, but that's not exactly correct. Only the recoverable mechanical energy is lost. Conservation of energy still applies. The energy is simply changing or moving from the thermal energy of the fluid (internal energy) by expanding, resulting in PdV* energy which can not be recovered.

Here's a decent link on Enthalpy:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/firlaw.html#c2

Here's another good link on Internal Energy:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/inteng.html#c3

Real fluids are not incompressible, even liquids. They undergo an expansion (PdV). This is mechanical energy which can't be recovered.

Another way of looking at it is to say the entropy increases as the fluid expands, and the entropy can't decrease without work being done on the fluid. Note that entropy would not decrease if we applied Bernoulli's equation in which only kinetic and potential energy is exchanged.

The pressure recovers to a degree. That amount of recovery is going to be greatly dependent on the geometry of the valve, piping and the flow conditions. Simply put, some of the flow stream energy is going to be used when going through the valve when things like turbulence, noise, heat and god forbid, cavitation are created. There's no way around it. That is one of the tradeoffs engineers make when designing a fluid system. If you look at the recovery characteristics of a venturi vs. those of a gate or butterfly valve, you can readily see the effects.

… due to an irrecoverable loss of energy due to friction, heat transfer and noise, the pressure of the fluid does not increase back to it's original value before the valve.

 

[Ask1122]

Thanks tiny-tim ;)
It is a very informative thread, but some parts are above my head though, i think i will need to get my thermodynamics up to scratch before i go back to that thread ;P

 

[Brian_C]

Bernoulli's equation does not apply to a viscous fluid, or a fluid that experiences nonconservative body forces.

 

[banerjeerupak]

well which equation is used to calculate pressure drop in viscous flows??

 

[FredGarvin]

It is usually referred to as the modified Bernoulli equation. You know, the Moody Diagram and all that good stuff.

https://www.physicsforums.com/showpost.php?p=1588952&postcount=4