Modelling very large pressure pressure drops in liquids 2

[ColourfulFigsnDiags]

I have a system where there is a near 100Barg pressure drop in a liquid stream over a valve. According to HYSYS this pressure drop results in an almost 1% increase in the absolute temperature of the liquid.

Is this correct? If so, what is the fundamental reason why this happens?

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[Latexman]

This is enough dP on a liquid to change the volume by about 1 %. I know 3000 psi will reduce water's volume by 1%. Throttling through a valve is usually isenthalpic.

h = u + pv = 0
du = Cv dT

pv decreased, so T had to increase.

Good luck,
Latexman

 

[sailoday28]

Latexman (Chemical)

du=CvdT- d/tau [p*tau]dV where the bracket term derivative is at const volume AND
tau=1/T T=absolute temp
p= pressure
V= specific vol.
Cv= spec ht at const specific vol
u= spec internal energy.

or
du= CvdT + Tdp/dT[@const v * dV - p dV

Sorry, I don't know how to type in the partial derivatives.

Regards

 

[Latexman]

sailoday,

The equations I pulled from memory were strictly for ideal gases and not correct for this case. Thanks for the correction.

Good luck,
Latexman

 

[UmeshMathur]

If you express the fundamental equation for enthalpies,

H = U + P*v

Then, dH = dU + p.dv + v.dP

At constant enthalpy, dH = 0. Also, the p.dv term in this case is very small compared to the v.dP term, since the specific volume change is negligible. Therefore, we can say that dU approximates -v.dp. The change in temperature, of course, is proportional to the dU term.

In Hysys, they solve the appropriate equation of state iteratively at the final pressure to obtain an enthalpy equal to the value at the inlet. Note that it is possible to end up with a two-phase mixture at the outlet condition, depending on how close the incoming liquid enthalpy is to the outlet liquid's enthalpy at the bubble point temperature. This check is made rigorously in Hysys.

 

[sailoday28]

Then, dH = dU + p.dv + v.dP

Refer to my previous posting on this thread for dU.
While specific vol change is negiligible, what about the temp change?

Yes an appropriate equation of state should be used.

For example

dH=CpdT d [ (v*Tau)/dtau at const press ] *dP

Where Cp is spec ht at const pressure,
v spec vol
Tau= 1/T
T= abs temp.
Assuming a const enthalpy process, the change in temp can
obtained from an equation of state.
If v=v(P, T} use the above eq.
If p=P(v,T) use my previous posted equation to get dU.

Regards.

Having the change in press

 

[UmeshMathur]

sailoday28:

Solving the equation of state (EOS) numerically at constant enthalpy is the proper way to get the new temperature after the valve. The solution method for the Soave or BWR equations of state is found in, e.g., T. F. Daubert: "Chemical Engineering Thermodynamics" (McGraw-Hill, 1985). The input data required are: critical temperature and pressure, Pc, acentric factor, and the ideal gas enthalpy coefficients, h0=f(T) for all components in the mixture.

ColourfulFigsnDiags's original question is best answered, I think, by reminding him that maintaining fluid enthalpy after pressure is reduced drastically cannot be done without a temperature increase for most fluids, unless you have the rare case of a negative Joule-Thompson coefficient:

(?T/?P)|H = 1/Cp*[-T*(?V/?T)|p + V] , where the ? operators denote partial derivatives.

The Joule-Thompson coefficient is negative only for real gases above reduced temperatures of about 6, and also for hydrogen and helium gases. For liquids, it is always negative.

 

[sailoday28]

Correction to my previous post.
dH=CpdT + d [ (v*Tau)/dtau at const press ] *dP

or
dH=CpdT +vdP - T?v/?T|p dP

or for JT coefficient that of posting
by UmeshMathur (Chemical)

"In Hysys, they solve the appropriate equation of state iteratively at the final pressure to obtain an enthalpy equal to the value at the inlet. Note that it is possible to end up with a two-phase mixture at the outlet condition, depending on how close the incoming liquid enthalpy is to the outlet liquid's enthalpy at the bubble point temperature. This check is made rigorously in Hysys"

Does Hysys check to see if flashing occurs within valve?

Regards

 

[UmeshMathur]

If the outlet stream from the valve will be a two-phase mixture, Hysys gives you the molar vapor fraction (i.e., vapor moles/feed moles) at outlet T and P.

 

[sailoday28]

UmeshMathur-My real question is== how does Hysys compensate for flashing within a valve?
Regards

 

[UmeshMathur]

The exact geometry of a valve is extremely irregular, and so an increasing amount of flashing occurs from the point where the local bubble point pressure is reached until the valve exit. If the fluid is already in a two-phase condition at the entrance to the valve, the percent vaporization simply increases as pressure drops.

I rather doubt that anyone is interested in the progression of flashing within the valve body. The pertinent engineering question is: "What is the condition at the end of the valve for a specified total pressure drop?". This calculation requires an isenthalpic flash to be performed at valve outlet T and P. This provides the final value of the V/F ratio. To make this calculation, Hysys uses the user-specified thermodynamic option, e.g., an equation of state such as PR, SRK, BWR, etc.

As always, the user remains responsible for the proper choice of thermodynamic specifications from the many options available in Hysys (or any other process simulator).

 

[ColourfulFigsnDiags]

Fascinating discussion guys, and thank you for the answers!

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[ColourfulFigsnDiags]

Ok, I guess (after reading this thread) it is clear that if the pressure drop across the valve occurs isenthalpicly, then there will (nee. must) be a corresponding temperature rise in the fluid.

The next question is, how valid is the assumption that the pressure drop across the valve is isenthalpic?

I mean, if you were out in the field, with a valve, would the temperature increase be a real thing that you would measure?


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[Aggie03]

If there isn't any vaporization of the liquid you will often see an increase in temperature, albeit small. This is especially the case for water and is sometimes referred to as a "reverse JT effect". JT systems are typically used to lower temperature by vaporizing a portion of the liquid. Check your college thermo boook, it should have a discussions of JT systems and times when the JT coefficient is negative (this is your case).

By definition a JT system is isenthalpic and is an accurate (and necessary) assumption here. Remember your first principles. Without heat flow, enthalpy must remain constant.

 

[sailoday28]

??By definition a JT system is isenthalpic and is an accurate (and necessary) assumption here. Remember your first principles. Without heat flow, enthalpy must remain constant.??
IF AND ONLY IF SPECIFIC VOL CHANGES ARE NEGLIGIBLE--ie change in KE is negligible. Which for liquids is a reasonable approx. Oh--I forgot to include change in elevation----

regards

 

[ColourfulFigsnDiags]

Sorry, perhaps I was not clear enough. I understand that any system without heat flow will be isenthalpic, however, I guess what I was asking is if there is any significant transfer of heat via friction within the valve, or via heat transfer / losses to the outside world.

It is now clear to me that most liquids will experience an inverse JT effect (albeit very small), and that HYSYS takes account of this in its calcs (which was my original question). But if we walk away from the cleanliness of theory, and look at the messiness of reality, the pressure inside the valve itself will clearly rapidly change and vary dependent on the valve geometry. In general, will this effect be observed in the plant, or will external losses / effects swamp it?

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[UmeshMathur]

I'm afraid we have some misconceptions regarding the fundamentals. The following comments are offered in an attempt to clarify the issues.

A negative Joule-Thompson coefficient is observed only rarely - hydrogen and helium gases, and very hot gases (above a reduced temperature of 6 or so). For those who have forgotten, reduced temperature is the ratio of the system temperature to the critical temperature of the fluid (absolute units). Thus, most fluids cool down when pressure is reduced isenthalpically.

The degree of flow turbulence and sudden changes in direction of flow across the valve are irrelevant to the issue here. The enthalpy is preserved, except for heat losses. These losses are negligible, considering that the total surface available for heat transfer to the outside world is barely a few square feet in most valves and, therefore, quite insufficient to allow significant heat transfer by natural convection to the outside air. Recall that convective heat transfer coefficients are also very small.

Finally, huge changes in velocity from inlet to outlet can have effects on outlet temperature arising from the kinetic energy term. Again, these are generally negligible for most real-world valves.

Therefore, the theoretical prediction of outlet temperatures is pretty accurate, as long as you choose the right thermodynamic options in Hysys. In fact, these changes are quite small for liquids that do not show a phase change.

If, on the other hand, the incoming fluid flashes to any significant extent, the valve outlet gets much cooler, as the latent heat of vaporization is being drawn from the enthalpy of the fluid itself. That is why, in the refrigeration business, such valves are called "JT valves".

 

[sailoday28]

If the internal characteristics of the valve are not considered, can't flashing within the valve affect the flowrate?
Regards

 

[Aggie03]

Colourfulfigs,

The answer is yes. But remember that steady state operation is something that is rarely seen in a plant, no matter what people say (depending on your definition of steady state that is). So you're probably not going to get the model to match to 1%, which is the change that you noted in the simulator.

Without any vaporization the volume isn't going to change appreciably, so there will be a small temperature increase. Almost every other effect for this system is typically negligible.

But whether the operator will actually see such an effect depends more on the accuracy of their instrumentation than anything. Such a change is typically not measurable in a plant (in a lab, yes). It's not really going to be seen in operation unless you start vaporizing something and get a large delta T.

UmeshMathur, you're right on for gases (H2 and He). We're talking a liquid here, and maybe in reality the JT coefficient term doesn't necessarily apply (I don't really know). But I do know (dt/dp)h is negative when the temperature increases during a throttling application.

 

[sailoday28]

(?T/?P)|H = 1/Cp*[-T*(?V/?T)|p + V] , where the ? operators denote partial derivatives.

The Joule-Thompson coefficient......... For liquids, it is always negative. MAYBE

The maximum density of water is at 4C at atm pressure. Therefore ?V/?T|p can be negative.

Regards