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The principle behind throttling valves 
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#1
Mar1805, 07:48 AM

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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? 


#2
Mar1805, 11:51 AM

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#3
Mar1805, 06:54 PM

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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 JouleThompson" 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! 


#4
Mar1905, 01:44 AM

P: 45

The principle behind throttling valves
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 


#5
Mar1905, 08:16 AM

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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 DarcyWeisbach 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: http://www.cheresources.com/valvezz.shtml 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: http://www.thermexcel.com/english/ressourc/valves.htm 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. 


#6
Mar1905, 11:23 AM

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So, the main reason for a pressure drop across the valve is?



#7
Mar1905, 11:37 AM

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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?



#8
Mar2005, 05:56 AM

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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: http://hyperphysics.phyastr.gsu.edu...firlaw.html#c2 Here's another good link on Internal Energy: http://hyperphysics.phyastr.gsu.edu...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. Hope that explains it a bit better. *PdV is Pressure times differential volume 


#9
Mar2205, 10:37 AM

P: 45

Hi,Q_Goest!
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... 


#10
Mar2205, 11:27 AM

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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. 


#11
Mar2305, 08:17 AM

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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. 


#12
Mar2405, 10:53 AM

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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? 


#13
Mar2405, 12:32 PM

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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. 


#14
Mar2505, 01:21 AM

P: 45

So, is this interpretaton correct:
When a fluid pass through a valve, the crosssectional area is reduced, so velocity increase.(Crosssectional 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 crosssectional 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. 


#15
Mar2505, 09:08 AM

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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." 


#16
Mar2505, 10:53 AM

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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?



#17
Mar2505, 12:39 PM

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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. 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 


#18
Aug605, 05:17 AM

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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? 


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