CALCULATING TEMPERATURE DECREASE IN EXPANSION VALVE 2

Thermodynamic theory does not say that.

While expansion through a control valve is isenthalphic that does NOT mean there is no temperature change.

If you look at a Mollier diagram for a fluid and expand the gas at constant enthalphy, you'll see the temperature does decrease.

Be careful, in the Joule-Thompson expansion, temperature increases or decreases during the throttling depending whether the Joule-Thompson coeficient is negative or positive. I know it is rare for the temperature to increase, but it can bite you as I have found out.

devaxrayz:

Like TD2K, I also am wondering where you have learned such dysfunctional Thermodynamics. Nowhere does it state in Thermodynamics that there is no temperature decrease (or its negligible) in an isenthalpic expansion process. I just want to make sure that everyone agrees that Thermo is not being re-written with new "laws"!

TD2K has indicated the fastest and most practical way to find out the outlet temperature of a fluid exiting an expansion device: The Mollier diagram. You can also use my favorite, the Temperature - Entropy diagram (T-S). Both will indicate the final conditions of the exit fluid if you follow the constant enthalpy line down to the final pressure. True, there are some fluids, Hydrogen for one, which will exhibit an increasing temperature under certain conditions - but these are rare. And you can't go wrong using the corresponding Mollier diagram for the subject fluid. You can find these diagrams in textbooks or in the GPSA Databook. You can also use NIST Thermo tables and data.

The Joule-Thomson coefficient is a measure of the the temperature change you are looking for and it is a partial derivative, (dT/dp), @ constant enthalpy. It's hard-core Thermo and altogether another subject. Suffice it to say that you're much better off with a representative Mollier diagram.
Hope this helps you out.


Art Montemayor
Spring, TX

Gentlemen,

The confusion comes from the thermodynamic assumption for an ideal gas: not only does PV=nRT, but also dH=Cp*dT. This is what people remember from their thermo, the rest is often forgotten.

I get this in reverse all the time from people who do process simulations and chose a property method which models the vapor as an ideal gas. They want to know why they don't get a temperature change when they drop the gas pressure. A real gas model must be used to capture that sort of behavior.

thanx for all of the replies, its true that there is temperature decrease for isenthalpic expansion.

But i have a trouble in looking for the appropriate Mollier or T-S diagram for my stream. For your information i want to expand a liquid oxygen-nitrogen mixture from P1 =6 bar to P2 = 1 bar, T1 = -178 C. And the fraction of nitrogen is 54%

The composition of the stream affect to the temperature decrease(special composition need special diagram), don't they??

PS : i don't have any of GPSA Databook or NIST Thermo tables and data . So i will very appreciate if you can give me another clue

devaxrayz:

All of a sudden, when we get the rest of the basic data, we now have a LIQUID stream expanding adiabatically. Most of us, including myself had assumed your stream was a gas. Now, the Joule-Thomson coefficient does not apply. It is only valid for gases. We are now talking of a liquid freely expanding and vaporizing. Yes, you can find the resultant temperature on a Mollier, but you have to have the specific composition(s). You haven't given much basic data, so that's all I can do except refer you to a convenient simulation program like Aspen and others. You're going to need a lot more basic data than just your pressures.

Art Montemayor
Spring, TX

Well then, due to restriction of data resources that i had, can i make a rough estimation using both Nitrogen and Oxygen T-S diagram?

First i find the temperature of nitrogen (T'1)using N2 T-S diagram if i expand it isenthalpicly from 6 bar -174 C to 1 bar

Then similarly i find temperature of oxygen (T'2)

From the composition data (nitrogen 54% and oxygen 46%) I extrapolate that two temperature found (T'1 and T'2) to get the stream resultant temperature.

What is your comment on this method?? Is this valid enough ?? I appreciated all of your replies.

Try this link. The Daemon can handle non ideal gases, and mixtures of gases.

HAZOP at

FYI, I calculated -191.5C via a flash calc (assumes pressure in absolute units, molefrac composition). Your upstream condition looks to be only a few degrees subcooled. A general flash calc uses heat of vaporization and heat capacity data to converge on the outlet vapor fraction (which was 20%).

Because you are calculating by hand, you will get within a few degrees with minimal effort by just calculating the bubble point temp of your feed at 1 bar. Do this by itteratively guessing a temp and calculating until: PvapN2 + PvapO2 = 1bar.

At the reduced temperatures and pressures indicated by devaxrayz for both oxygen and nitrogen, the equilibrium vapour would be richer in nitrogen. Even then, the Miller equation for the inversion curve based on their critical properties would predict warming rather than cooling on isoenthalpic expansion. See Perry VI, table 3-150.

sshep :
thanx, i've follow montemayor sugestion on do it on simulation program (i use HYSIS) and got the result is
-191,6 C.

and then came along your suggestion with general flash calculation (u got -191.5 C as the result).

so i recalculate following your sugestion (make a flash calculation while maintain the sum of vapour enthalpy and liquid enthalpi equal to feed enthalpy).

I got the result of -191,46 C. Seem that this method quite work allright. Thanx man

I really appreciated if there is any comments or correction or if u think something else.

To devaxrayz, kindly let us have the initial and final mass fractions of oxygen and nitrogen in liquid and vapor form. Kindly confirm no argon is present. Thanks.

25362 :
I want to expand stream A to stream B

Stream A : (liquid)
TA = -174 C
PA = 5.7 bar
composition : Z_N2 = 54.25%; Z_O2 = 45.75% (i assume there is no argon)

Stream B vapor-liquid)
PB = 1 bar
Liquid (83.36%) composition : X_N2= 49%; X_O2= 51%
Vapor (14.64%)composition : Y_N2= 80.8%; Y_O2= 19.2%


The result :
TB = -191.6 (by HYSIS)
= -191.4 (my calculation using isenthalpic general flash calculation)
= -191,5 (sshep calculation)

hope that fulfill your request. thanx for your curiosity

devaxrayz

devaxrayz:

Congratulations on getting satisfactory and very consistent results with HySis (HySim?). I think you've got a handle on the process and can appreciate what is going on in the adiabatic free expansion.

I got very similar results in 1965 when I did a manual heat and material balance of my Supairco air separation column in Lima, Peru. I remembered the hand calcs and found my old files. Your numbers reflect my analysis of the crude Oxygen stream exiting the HP column and expanding into the LP column. As I said, Thermodynamics has definitely not changed. I did my old calcs manually and laboriously with a slide rule, but the process and the Thermo were exactly the same. Congrats.




Art Montemayor
Spring, TX