The principle behind throttling valves

[tony_engin]

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?

 

[Clausius2]

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?

Don't rush. You'll heard about Bernoulli equation several times in your life. Just wait.

 

[Q_Goest]

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!

 

[tony_engin]

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

 

[Q_Goest]

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:
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.

 

[tony_engin]

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

 

[tony_engin]

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?

 

[Q_Goest]

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.

Hope that explains it a bit better.

*PdV is Pressure times differential volume

 

[tony_engin]

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...

 

[FredGarvin]

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

 

[minger]

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...

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.

 

[tony_engin]

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?

 

[Q_Goest]

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.

 

[tony_engin]

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.

 

[FredGarvin]

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."

 

[tony_engin]

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?

 

[Q_Goest]

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.

Microscopic Energy:
Internal energy involves energy on the microscopic scale. For an ideal monatomic gas, this is just the translational kinetic energy of the linear motion of the "hard sphere" type atoms , and the behavior of the system is well described by kinetic theory. However, for polyatomic gases there is rotational and vibrational kinetic energy as well. Then in liquids and solids there is potential energy associated with the intermolecular attractive forces. A simplified visualization of the contributions to internal energy can be helpful in understanding phase transitions and other phenomena which involve internal energy.

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

 

[sree]

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?

 

[sree]

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?

 

[Q_Goest]

Hi Sree, welcome to the board.

... does it mean that a throttle valve is used to reduce the pressure(or create a back pressure in front of the valve) but not the flow across it?

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.

 

[daneshjou]

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?

 

[Q_Goest]

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.

 

[Windnsails]

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).

 

[Hemanasher]

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?

 

[stevmg]

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.

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?

stevmg

 

[Q_Goest]

Hi stevmg, I'm not sure exactly what you're asking but will try and clear up what I can.

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?

When an ideal gas flows through a restriction from high pressure to low pressure, the molecule itself doesn’t pick up speed. The velocity increase is in the bulk velocity of the fluid. Molecular velocity doesn't change for an ideal gas.

The gas in this throttling process has different forms of energy including kinetic energy and enthalpy. Energy is conserved (per first law of thermodynamics) so given a control volume put around a valve which is throttling flow, the enthalpy in plus the kinetic energy into the control volume is equal to the enthalpy out plus the kinetic energy out. Note this assumes no heat or work is done on or by the fluid by the valve which is generally a pretty reasonable assumption. Normally, the bulk velocity increase is small so we neglect kinetic energy increases and consider this an isenthalpic process or isenthalpic expansion.

For a real gas, the temperature may increase or decrease following an isenthalpic expansion. Helium for example, can warm up under fairly 'normal' conditions. For your proposed example of there being attractive forces between molecules that drop off as a function of distance, note that as the gas goes from a location of high pressure where the mean distance between molecules is much closer than the mean distance between molecules after they go through the isenthalpic expansion, this attractive force you've proposed will also drop off. So the energy that increases the temperature of the gas downstream of this isenthalpic throttling process is already inside the gas due to the close proximity of the molecules.

 

[Toolbox13]

Some great answer by Q_Goest, if still online can you explain the following. Has me a bit stumpted.

For flow of an ideal gas across a throttle the enthalpy is constant and hence since enthalpy, Cp(DT), is constant there can be no temperature change across a throttle for a gas flow.

However the head loss across a throttle results in the conversion of mechanical energy to thermal energy (through frictional heating and such) and hence there would be a change in the internal energy, Cv(DT), therefore a change in internal energy requires a change in temperature and hence a change in enthalpy which contradicts my first point of constant enthalpy for a throttle device (as noted above several times).

I am particularily thinking of frictional heating in a flowing system across a throttle, this viscous heating will result in a temperature rise across the throttle but then how does this equal with a constant enthalpy defined as Dh = Cp(DT)?

Note again that this is for ideal gas so no joule-thomson effect.

 

[Chestermiller]

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...

The answer to this question is related to what Q_Goest said before. He indicated that, if you could solve the Navier Stokes equations, you would be able to predict the pressure drop. Now, you have to ask yourself "what is it about the Navier Stokes equations that gives you a different answer from the Bernoulli equation?" The Bernoulli equation is based on the differential force balance equation for an ideal fluid, one which has no viscosity. The Navier Stokes equations include the viscous stresses that are related to the rate of deformation of the fluid, and which are omitted from the flow equations for an ideal fluid. It is the effect of fluid viscosity that brings about the pressure drop and prevents recovery of the pressure. For some very viscous fluids, the Bernoulli (inertial) effects are negligible compared to the viscous effects. Even in the case of air (low viscosity), the viscous effect is typically significant.

 

[Toolbox13]

Thanks for answer, but nothing to do with question. You do not have to solve the Navier Stokes equation for this, there are plenty of correlations for pressure drops across all types of throttles including channels that are reasonable.

The Bernoulli equation has nothing to do with this, it is the energy equation (which is the Bernoulli equation with the Q/W terms dropped). As per my question there is no enthalpy term in the Bernoulli equation (correct term is the mechanical energy equation with head loss terms included) as it assumes constant temperature, no work done or heat transfer and a range of other things.

 

[Chestermiller]

Thanks for answer, but nothing to do with question. You do not have to solve the Navier Stokes equation for this, there are plenty of correlations for pressure drops across all types of throttles including channels that are reasonable.

The Bernoulli equation has nothing to do with this, it is the energy equation (which is the Bernoulli equation with the Q/W terms dropped). As per my question there is no enthalpy term in the Bernoulli equation (correct term is the mechanical energy equation with head loss terms included) as it assumes constant temperature, no work done or heat transfer and a range of other things.

You were asking why the Bernoulli equation doesn't work for predicting the pressure drop. I was just using the Navier Stokes equations to help identify the reason. It was never my intention to tell you that the only way to get what you wanted was to solve the Navier Stokes equations.

My answer was that all fluids and gases have viscosity which is responsible for the dissipation of mechanical energy to heat. The actual amount of heat generation is often very small (for gases) so the effect on temperature is small, but the effect on dissipating mechanical energy is significant. Those correlations you are referring to would not be necessary if the fluid viscosity were zero, and it is the viscosity effect that determines the oriface coefficients in your correlations.

Chet

 

[Chestermiller]

Some great answer by Q_Goest, if still online can you explain the following. Has me a bit stumpted.

For flow of an ideal gas across a throttle the enthalpy is constant and hence since enthalpy, Cp(DT), is constant there can be no temperature change across a throttle for a gas flow.

However the head loss across a throttle results in the conversion of mechanical energy to thermal energy (through frictional heating and such) and hence there would be a change in the internal energy, Cv(DT), therefore a change in internal energy requires a change in temperature and hence a change in enthalpy which contradicts my first point of constant enthalpy for a throttle device (as noted above several times).

I am particularily thinking of frictional heating in a flowing system across a throttle, this viscous heating will result in a temperature rise across the throttle but then how does this equal with a constant enthalpy defined as Dh = Cp(DT)?

Note again that this is for ideal gas so no joule-thomson effect.

There is a reason for this. The equation Enthalpy in = Enthalpy out is only an approximation, and omits a term in the energy balance. The term that is missing is the viscous heating. This term is usually very small for a gas, and, can be neglected. See the advanced treatise Transport Phenomena by Bird, Stewart, and Lightfoot, Table11.4-1, next-to-last equation.

Do a calculation for a specific example of the mechanical energy loss associated with the pressure drop using one of the oriface correlations, and convert this into a temperature rise to see how small it actually comes out to be. If, for an ideal gas, you also assume that its viscosity is zero, then the viscous heating is zero, and the Joule Thompson coefficient is exactly zero.

 

[Toolbox13]

Thanks, the above was close to my thinking also but could not find a textbook that explains it in detail. I will look up the one that you suggested. I am also not sure what you propose is fully correct.

My thoughts are that the Enthalpy in must equal Enthalpy out based on steady flow energy equation where KE/PE work out of control volume and heat out are zero. If this is the case then h_in must equal h_out, however it is plausible that while h_in = h_out there could be a temperature change due to transfer of energy from pressure term to internal energy term in enthalpy. If this is the case, then it is not h_in = h_out is not valid but rather the relation Dh = Cp(DT) is questionable in such instances. In fact this is derived on teh basis of constant pressure originally, must look it up again.

In short thanks for answer

 

[Toolbox13]

Chestermiller,
forgot to add the problem I am looking at has 300 to 8000W of viscous heating, hence hard to ignore it.
Microchannels with high speed flows.

 

[Chestermiller]

Thanks, the above was close to my thinking also but could not find a textbook that explains it in detail. I will look up the one that you suggested. I am also not sure what you propose is fully correct.

My thoughts are that the Enthalpy in must equal Enthalpy out based on steady flow energy equation where KE/PE work out of control volume and heat out are zero. If this is the case then h_in must equal h_out, however it is plausible that while h_in = h_out there could be a temperature change due to transfer of energy from pressure term to internal energy term in enthalpy. If this is the case, then it is not h_in = h_out is not valid but rather the relation Dh = Cp(DT) is questionable in such instances. In fact this is derived on teh basis of constant pressure originally, must look it up again.

In short thanks for answer

As you noted, there are really two issues here:
1. Is the relationship Hout = Hin a sufficiently accurate representation of the thermal energy balance (first law) for a steady flow system in which viscous heating is not neglected?
2. Are the thermodynamic property equations we use to calculate the change in enthalpy per unit mass of a given material correct?

I can tell you that the thermodynamic property equations are definitely solid, so that the answer to the second question is "yes," particularly if we use the following fundamental relationship to calculate the change in enthalpy of a material in terms of the changes in temperature and pressure:
dH=C_pdT+(V-T\frac{\partial V}{\partial T})dP
where H is the enthalpy per mole, V is the volume per mole, and C_p is the molar heat capacity at constant pressure. This equation applies to any material, including liquids, ideal gases, and non-ideal gases. Note that, in the case of an ideal gas, the second term is identically equal to zero.

The way that this equation is typically applied is to choose a reference state T₀, P₀ at which the enthalpy is taken to be zero. The equation is first integrated with respect to T at constant P = P₀ (such that only the first term is involved) from T=T₀ to T = T; then the equation is integrated with respect to P at constant T = T (such that only the second term is involved) from P = P₀ to P = P. If you know how the heat capacity varies with T at P = P₀, and you know the PVT behavior of your liquid or gas, then this equation is guaranteed to deliver the correct values of the enthalpy relative to the reference state. The derivation of the above equation is pretty much in every thermo book, and you should look it up if you are not yet familiar with it.

The first question is related to the general way that the first law is expressed in elementary thermo books. It typically neglects the contribution of viscous stresses to the overall stresses at the boundary of the system, and thus in calculating the rate of work done on the surroundings. It only includes the isotropic pressure portion of the stress. BSL provides a more rigorous development which includes the viscous contributions to the stresses. BSL first writes down the mechanical energy balance equation, which is the result of dotting the differential force balance equation (equation of motion) with the velocity vector. This essentially delivers the Bernoulli equation, with some extra terms included to account for viscous stresses. They then write down the overall energy balance equation in terms of both thermodynamic energy quantities and mechanical energy quantities, some of whose terms are the same as in the mechanical energy balance equation. They then subtract the mechanical energy balance equation from the overall energy balance equation to arrive at the thermal energy balance equation. This is basically a differential form of the first law that properly includes the contributions of the viscous stresses. I very strongly encourage you to read Chapter 11 of BSL, and to study the derivations carefully, especially the way they arrive at the thermal energy balance equation.

I want to commend you on identifying all these relevant and important questions that need to be addressed in order to solidify ones understanding.

Chet

 

[Toolbox13]

Hi
On what you term the first question, you are correct but it is generally not necessary to delve into the tensors of Reynolds and normal stresses when control volumes are considered. I am quite familiar with the dissipation term in the differential energy equation, actually spent quite a while working with it. The above result that propose from the textbook I expect is easily found from the first law Q-W=d(H)+d(KE)+d(PE). For throttle everything except d(H) is assumed to go to zero. You may then expand out d(H) and show that viscous dissipation results in a temperature change but then this would contradict the constant h across the throttle assumption (which must be true since it is simply energy conservation afterall). So I think one can view it from two perspectives,

Allow the frictional work to show up in the energy equation giving W(viscous) = d(H) but this implies the the control volume must be drawn within the throttle rather than around it.

Or allow h to be constant when control volume is drawn around the throttle and then split the enthalphy into two terms, u+PV, the u term can then be assigned to equal the viscous dissipation (Cv(dT)) and hence giving a temperature rise. This initially seems in conflict with the constant enthalpy requirement as it indicates a temperature rise but this temperature rise is compensation for by the (V−T∂V∂T)dP using your equation above. This way the enthalpy can remain constant even though there is a pressure rise.

Both ways to think about it are probably reasonable and I think correct.

 

[Chestermiller]

Hi
On what you term the first question, you are correct but it is generally not necessary to delve into the tensors of Reynolds and normal stresses when control volumes are considered. I am quite familiar with the dissipation term in the differential energy equation, actually spent quite a while working with it. The above result that propose from the textbook I expect is easily found from the first law Q-W=d(H)+d(KE)+d(PE). For throttle everything except d(H) is assumed to go to zero. You may then expand out d(H) and show that viscous dissipation results in a temperature change but then this would contradict the constant h across the throttle assumption (which must be true since it is simply energy conservation afterall). So I think one can view it from two perspectives,

Allow the frictional work to show up in the energy equation giving W(viscous) = d(H) but this implies the the control volume must be drawn within the throttle rather than around it.

Or allow h to be constant when control volume is drawn around the throttle and then split the enthalphy into two terms, u+PV, the u term can then be assigned to equal the viscous dissipation (Cv(dT)) and hence giving a temperature rise. This initially seems in conflict with the constant enthalpy requirement as it indicates a temperature rise but this temperature rise is compensation for by the (V−T∂V∂T)dP using your equation above. This way the enthalpy can remain constant even though there is a pressure rise.

Both ways to think about it are probably reasonable and I think correct.

I looked over Chapter 11 in Bird, Stewart, and Lightfoot last night, and I'm very confident you are going to find the answers to all your questions in that chapter. Unfortunately, I have not been able to articulate the essence of that development in a way that satisfies your doubts.

I also am sending you a private message in which I am proposing that we work together to apply the equations in BSL to solve a specific problem, in order to develop a practical understanding of exactly what is going on. I hope you are interested in participating.

Chet

 

[Chestermiller]

Several weeks ago, in posts to this thread, Toolbox13 and I had been puzzling over some of the thermodynamics fundamentals of throttle values, particularly with regard to the manner in which viscous heating comes into play, and its influence on the temperature of the fluid exiting from the valve. As a result of these interactions, in private messages, we resolved to collaborate in analyzing the system in much greater detail, with the objective of removing the gaps in our understanding. We are now reporting back to the thread on the results of our deliberations, with which we are now in full agreement.

In order to keep things simple, we decided that, rather than deal with the geometric complexities of a throttle valve, we would analyze the simpler situation involved in the thermodynamics of adiabatic laminar fluid flow through a long straight section of pipe. The qualitative results we obtained for this case are fully compatible with those for a throttle valve.

We considered three types of fluids: Ideal (compressible) gases, ideal incompressible liquids, and real compressible gases. In all three cases, viscous dissipation was included in our analyses.

We also included three versions of energy balance equations:
1. Mechanical energy balance equation (obtained by dotting the differential equation of motion with the velocity vector)
2. Overall energy balance equation, including both mechanical and thermal effects.
3. Thermal energy balance equation, obtained by subtracting the mechanical energy balance equation from the overall energy balance equation. This focuses mainly on how the thermodynamic functions vary with position and time.

We considered the differential forms of these energy balance equations as well as the forms of the equations obtained by integrating over a fixed control volume, such as a pipe or a throttle valve. We also considered the predictions for specific thermodynamic functions, such as internal energy or enthalpy.

The following is a summery of our findings:
1. The change in mixing cup average enthalpy between inlet and exit of the pipe section (or throttle valve) is zero.
2. Even though the change in mixing cup average enthalpy is zero, there are non-uniform enthalpy variations over the exit cross section of the pipe. In the case of an ideal gas, this translates into radial variations in temperature at the exit; however, the mixing cup average temperature of the gas at the exit is the same as at the entrance.
3. In the case of an ideal gas, there are two competing physical mechanisms affecting the internal energy change and the temperature change from inlet to exit. These are: viscous heating and expansion cooling. These two effects exactly cancel one another for an ideal gas, such that there is no change in mixing cup average temperature.
4. For incompressible liquids, even though the mixing cup enthalpy change is zero, there is no expansion cooling, and the viscous heating effect brings about a temperature rise.
5. For real gases, even though the mixing cup enthalpy change is zero, the effect of pressure on enthalpy leads to Joule-Thompson heating or cooling.

 

[Dennis C]

For Tony:
"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..."
You don't want it to recover, you are controlling the valve (opening and closing it) depending on the downstream load.

Tony, minus all the engineering, think of trying to regulate pressure downstream of a MANUAL valve with a varying downstream load.
You tie a 1" water line into a 6" main header that is at 100 psi. you put a manual gate valve on it and it is in the closed position.
On the downstream end of the manual gate valve you have a garden hose (for simplicity) and another valve, your "load" valve. Now you open your load valve wide to atmosphere, then you throttle open the manual gate valve 1/2 way, the upstream line will still be at 100 psi, and if you have a gauge installed downstream of your throttle valve it may well read zero because we are open to atmosphere and can move the water without line losses. It will never recover because the load is there (open to atmosphere). Now, close your load valve to 90 percent, (Your downstream load is reduced) your original throttle valve still open 50% will now not produce the drop you want, if fact it may very well "recover" because the load is so low and the valve is still at 50%. When they are talking about throttling, the downstream load has to be there for a pressure drop.

Let's say you want to drop pressure in a steam line from 200 psi to 80 psi
the control valve will vary its position depending on downstream load, that is why it won't "recover". It is not suppose to. If the downstream load on the steam main drops 90 percent (like when you closed the load valve 90% on the hose) because a process or a heat exchanger was shut down,the throttle valve (control valve) will also close off a huge percentage to keep the downstream press at your predetermined pressure, 80 psi. It is not suppose to recover!
Back to the original manual valve, you can not throttle pressure with a manual valve if the downstream load varies! if the downstream load is constant and the fluid and upstream pressure is constant, you can.
Hope this helps
Dennis

 

[PepperAngus]

Can anybody explain why the pressure decreases across a throttling valve using the equation U1 + P1V1 = U2 + P2V2? Furthermore, for gases flowing through a throttling valve, is it necessary for the outlet pipe to be greater than the inlet? If not, how can there be an "expansion" or increase in volume?

 

[Chestermiller]

Can anybody explain why the pressure decreases across a throttling valve using the equation U1 + P1V1 = U2 + P2V2?

The pressure decreases across a throttling valve, not because of conservation of energy between inlet and outlet, but because of viscous drag forces acting on the fluid in passing through the valve, which converts mechanical energy to thermal energy. If you had flow through a porous plug instead of a valve, would you really feel that the inlet and outlet pressures should be the same?

Furthermore, for gases flowing through a throttling valve, is it necessary for the outlet pipe to be greater than the inlet?

No.

If not, how can there be an "expansion" or increase in volume?

The velocity of the gas increases.

Chet

 

[piyush007]

so that means if i have a throtlling valve which decreases the pressure to 0.2 bar from 23 bar of steam, and i am planning to have an arrangement that will give 0.2 bar from 12 bar of steam, which is more beneficial and why, if possible prove by calculation.

 

[Chestermiller]

so that means if i have a throtlling valve which decreases the pressure to 0.2 bar from 23 bar of steam, and i am planning to have an arrangement that will give 0.2 bar from 12 bar of steam, which is more beneficial and why, if possible prove by calculation.

Exactly what do you mean by "more beneficial?"