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The principle behind throttling valves

  1. Mar 18, 2005 #1
    Hi all!
    I am learning basic thermodynamics, and I find it difficult to undestand the principle behind throttling valves, which are used to reduce the pressure of the flow, right? How could this be done? Why velocity being unchanged after passing through the valve?
    Could anyone please help?
    Last edited: Mar 18, 2005
  2. jcsd
  3. Mar 18, 2005 #2


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    Don't rush. You'll heard about Bernoulli equation several times in your life. Just wait.
  4. Mar 18, 2005 #3


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    Hi Tony,
    The term "throttling valve" makes things sound very technical doesn't it? The difference between a "throttling valve" and any other valve is minimal. One might suggest a throttling valve has a better ability to vary it's restriction when compared to a fast opening valve such as a flat faced globe valve, a ball valve or a gate valve, but from a thermodynamic perspective, the first law can be applied to all these valves in exactly the same way. So from a thermodynamic perspective there is no significant difference.

    If you draw a control volume around the valve which is flowing at steady state, you can make the following observations:
    1) There is no stored mass or energy inside the control volume (ie: dU = 0)
    2) There is no work done by or on the control volume.
    3) Generally, the control volume is relatively small and the flow fast enough such that heat transfer can be neglected. This isn't always the case, but let's make a simplifying assumption that this is the case for now.
    4) There is a mass flow into the control volume, and a mass flow out of the control volume. These mass flows are equal.

    So the first law of thermodynamics applied to a valve results in a very simplified analysis. The first law reduces to - enthalpy in is equal to the enthalpy out.

    For example, if you know the pressure and temperature of the gas or liquid going in, you know the enthalpy entering the control volume (ie: you know the enthalpy of the fluid upstream of the valve). And if you know the enthalpy in, then you know the enthalpy out because it's the same. Pressure drop across a valve is an "isenthalpic" process.

    For most fluids, flow through a valve results in a dropping of both pressure and temperature. Some fluids such as helium have a "reverse Joule-Thompson" effect, which simply means that instead of the temperature dropping along with the pressure, the temperature actually increases.

    If you're asking why the velocity stays the same, I'm afraid it doesn't necessarily stay the same. The velocity downstream of a throttling valve is often higher than the upstream, simply because it's at a lower density, but velocity has little to do with anything here. The results of the first law analysis are identical regardless of velocity.

    If you're asking why the pressure drops, it's because there is no conservation of mechanical energy. Thermal energy plus mechanical energy is conserved, but Bernoulli's for example, does not address thermal energy. Enthalpy is the internal energy (ie: thermal energy) of the fluid PLUS the PdV energy. The fluid is essentially expanding across the valve which means the fluid's internal energy is changing by the amount PdV. If you need more detail than that, let me know.

    Best of luck in Hong Kong!
  5. Mar 19, 2005 #4
    Hi Q_Goest!
    Really thank you for answering me. I still have something not clear.
    For example, an incompressible fluid pass through a valve in a canal (the canal is of uniform cross sectional area), at steady state, the velocity of it before entering the value should be equal to the velocity after leaving the valve(Av=constant), right? So can I apply Bernoulli's equation on this flow? If so, isn't the pressure should stay unchanged? I am confused.

    I still don't really catch the main reason for the drop of pressure before the fluid entering the valve and leaving the valve in a uniform cross sectional area canal. Please help
  6. Mar 19, 2005 #5


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    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. The problem is we can't determine pressure drop given the first law. Perhaps someone with lots of experience with Navier Stokes equations and a computer that can perform a CFD analysis can do so, but for the rest of us mere mortals, we're stuck using empirical relationships such as the Darcy-Weisbach equation and equations for pressure drop through valves and orifices, etc… If you want to determine pressure drop through a valve or pipe, you can't use Bernoulli's and the Navier Stokes equations are far too complex for everyday use.

    Here's a web page that gives you the equation for pressure drop through a valve:

    Basically, a valve has a characteristic restriction given by the value Cv. If you're working in metric units though, you might try this web page:


    Same equations, different units. This page uses Kv.

    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.
  7. Mar 19, 2005 #6
    So, the main reason for a pressure drop across the valve is?
  8. Mar 19, 2005 #7
    Do you mean when a fluid pass through a valve, some internal energy will transfer to PdV energy, and since internal energy is reduced so as the pressure of the fluid. What actually is the PdV energy? V is referring the volume?
  9. Mar 20, 2005 #8


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    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:

    Here's another good link on Internal Energy:

    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.

    Hope that explains it a bit better.

    *PdV is Pressure times differential volume
  10. Mar 22, 2005 #9
    I don't understand why pressure will drop when the fluid pass through the valve but not "Why does the pressure not recover?"
    Could you please help again? I'm such an idiot...
  11. Mar 22, 2005 #10


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    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.

    The following is a good side by side compare of some different types of valves and their respective recoveries. The lowest pressure points are the pressure measured at the vena contracta.

    http://www.maintenanceresources.com/ReferenceLibrary/ControlValves/Images/cashcopressure1.gif [Broken]
    Last edited by a moderator: May 1, 2017
  12. Mar 23, 2005 #11


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    I bet you're thinking that as it goes through the valve, it should speed up and the pressure will drop, but after it is through, it slows back down and the pressure goes back up, basically equating Bernoulli's....

    I think the easist way to describe is through head loss. Head loss can be thought of as just lost engergy, typically though friction. It is a function of V², so as you close a valve which is able to throttle more and more, the fluid needs to speed up more and more to maintain continuity. As the speed increases, head loss increases, and head loss is irreversible.

    I'm not sure if what I said is correct, but it's the way I think of it.
  13. Mar 24, 2005 #12
    Hi minger !
    Yes, that's exactly what I think. And I also think of some kind of energy loss...but is this really the case?
  14. Mar 24, 2005 #13


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    There's a loss of mechanical energy, which is converted to thermal energy in the fluid. This conversion of energy from one form to another is irreversible, and entropy increases.

    Conservation of energy however, still applies. Even for phenomena of friction such as a block sliding across a surface. For the sliding block, the kinetic energy is converted to heat.
  15. Mar 25, 2005 #14
    So, is this interpretaton correct:

    When a fluid pass through a valve, the cross-sectional area is reduced, so velocity increase.(Cross-sectional aread X velocity = constant)
    And by Bernoulli's equation, the pressure will decrease in the valve.
    When leaving the valve, velocity should decrease due to increase in cross-sectional area, the energy should be used to increase the pressure back, but due to loss of mechanical energy into thermal energy of the fluid, the pressure of the fluid does not increase back.
  16. Mar 25, 2005 #15


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    I would modify what you said to say:

    "...but 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."
  17. Mar 25, 2005 #16
    if what said above is true, the temperature of the fluid after passing through the valve should be increased? But why textbook say the temperature as well as the pressure decreased drastically?
  18. Mar 25, 2005 #17


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    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.

    For example, if we have helium at 100 psia, 0 F, we have a specific volume of 12.3736 ft3/lbm. From this information we can determine the PdV term. Since this is a state, PdV is simply the pressure times the volume or:

    PV = (100 lb/in2) * (12.3736 ft2/lbm) * (0.001286 Btu/lb ft2) * (144 in2/ft2) = 229.1 Btu/lbm

    So if we knew internal energy, we can add PV to get enthalpy.

    If we had isenthalpic throttling across a valve, the difference in internal energy would be the difference of the two PV terms. Assuming we expanded helium from 100 psia and 0 F down to 10 psia, the new temperature would be very slightly warmer (about 0.7 F). Recalculating PV for this new state we have:

    PV = (10 lb/in2) * (123.475 ft2/lbm) * (0.001286 Btu/lb ft2) * (144 in2/ft2) = 228.7 Btu/lbm

    Note that these two values of PV are almost identical. Despite the very large drop in pressure, the helium expands to a very large volume. In this case the helium warms very slightly. Note also this is the "irreversible free expansion" which can not be recovered that we talked about earlier.

    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.

    Note also the internal energy may not change significantly, the PV term will remain the same, and the temperature will also remain the same.

    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.

    Edit: I can't count to 3... lol
    Last edited: Mar 25, 2005
  19. Aug 6, 2005 #18
    if the control volume(say, the valve) has only one inlet and outlet
    flow across it does not change but pressure decreases.with ref. to hydraulic systems does it mean that a throttle valve is used to reduce the pressure(or create a back pressure infront of the valve) but not the flow across it?
  20. Aug 6, 2005 #19
    pressure reducing valve

    if the control volume(say, the valve) has only one inlet and outlet
    flow across it does not change but pressure decreases.with ref. to hydraulic systems does it mean that a throttle valve is used to reduce the pressure(or create a back pressure infront of the valve) but not the flow across it?
  21. Aug 7, 2005 #20


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    Hi Sree, welcome to the board.
    The term "throttle valve" is used in various different industries to mean slightly different things. For example, a throttle valve in the refrigeration industry indicates a valve used to restrict flow between the condenser and evaporator in a closed loop system which creates the desired pressure drop. The flow rate is dependant on the compressor flow, so if one were to adjust the valve by turning the handle and adjusting the distance the plug is away from the seat, the valve will create more restriction. If the flow is constant because the compressor puts out a constant flow, the pressure changes. So an engineer in the refrigeration industry may look at a throttle valve as being one that controls pressure.

    A different example might be the chemical industry where the throttle valve is put between two pressure vessels. In this case, the vessels are generally at a constant pressure, something that can't be easily changed because they are so large. Now if you close down the valve a bit, the pressure upstream and downstream of the valve don't change, the flow changes instead. An engineer in the chemical industry may view a throttle valve as controlling flow.

    I guess what I'm getting at in my longwinded way is that the term "throttle valve" is a bit misleading. It is a valve that creates a restriction to fluid flow, but it doesn't necessarily always change either flow or pressure. It is often used for one or the other, and people in different industries may have a slightly different use for the valve, so you have to take into consideration the context in which the term is used.
  22. May 14, 2011 #21
    Hi Q_Goest,

    In a refrigeration system, what causes the pressure drop, expansion valve or the compressor? It is the compressor who makes a low pressure part in the system and liquid enters this low pressure area so naturally its pressure will be low so it will start to evaporate and eventually compressor maintains the low pressure; so what is the function of expansion valve? Is it only metering/feeding refrigerant to evaporator or it drops the pressure of the liquid as well? If it reduces the pressure who it does it to exact evaporation pressure set by compressor? Can we replace the expansion valve with an ordinary full bore globe valve which will not create much pressure drop when 100% open? Is the function of the compressor is just to reuse the refrigerant and continue the cycle?
  23. May 14, 2011 #22


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    We generally say the valve causes the pressure drop, but as you say, it obviously can't create a pressure drop in a closed loop system if there's nothing else in the system that creates a pressure rise. :smile:
  24. May 16, 2011 #23
    Bottom line is that the pressure loss is due mostly to turbulance.

    Think of a venturi - full line size squeezes to something less and then expandes gradually to keep stream lines as close to parallel as "practical".

    Now squeeze the throat of the venturi to something very small. The discharge becomes very long with nice gradually seperating streamlines. The velocity at the throat could be very high but friction effects would be almost the only energy lost.

    Now imagine the venturi with an abruipt expansion - you now have an orifice, with a tremendous velocity difference between the center and the wall just down stream.

    All that turbulence consumes energy via the several loss routes mentioned earlier (noise, vibration, cavitation, heat etc.). As the flow leaves the orifice, the flow pattern re-establishes a turbulent flow regime several pipe diameters downstream, but the pressure there will be missing the energy lost due to the abrupt expansion.

    Valves, (control valves included) are orifices with variable opening sizes (and shapes).
  25. Nov 28, 2011 #24
    I wanted to know how much pressuredrop can be their in joule thomson valve, my prof. gave an eg. in which inlet temp. is -84F and pressure is 990 PSIG and outlet temp is -88F and pressure is 330 PSIG, can we really get so much pressure drop?
  26. Jan 30, 2012 #25
    Q_Goest. Are you still on line about this subject? If so, please address this point:

    Suppose we have a non-ideal gas in which the molecules do have attractive forces on each other inversely proportional to the square of the distance between them (this not a new concept...) In a freely expanding situation, as the molecules "separate," the attractive forces will diminish. The attractive forces have the propensity to slow down each molecule's speed. As the forces diminish, the "slowing down" effect would tend to diminish. This, however, would not lead to an increase in the speed of the dispersing molecules but would merely reduce the "slowing down" effect. So, how the Dickens would a molecule pick up speed (and consequently temperature) in such a situation? In order for a molecule to pick up speed one would need an accelerating force on it. Where would that come from? The decrease in a decelerating force is not equivalent to an accelerating force.

    Is this only true in a "throttling situation" in which the molecules "speed up" as they pass through the throttle (and potential energy is converted into kinetic energy as portrayed, in some extent, by Bernoulli.) Is the throttling scenario necessary for some gases to actually increase in temperature?

    You know, there is another "paradox" known as the "water hammer" in which lateral motion of a fluid is converted to a higher potential energy than accounted for by Bernoulli. However, the volume of water elevated to this new "high" is much, much less than the initial moving volume of water. The analogy here would imply that if a non-ideal gas did pick up speed and temperature while going through a throttle, the volume of such a gas "heated up" would be far, far less than the original volume trying to get through the throttle.

    Am I making any sense?

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