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# Pressure Reducer Regulator Energy And Temprature Calcualtion

15 replies to this topic

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Posted 05 February 2012 - 03:02 AM

Hi Dear,
I have an N2 Regulator for reducing pressure from 200 Bar(g) To 30 Bar(g).When this proccess start regulator body temprature decrease.(@ 20 degree C)

I want know formula for the rate of energy absorb from ambience and body temprature?
Which formula can be explain this event.

Kind Regard,

### #2 marchem

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Posted 06 February 2012 - 05:05 AM

your outlet temperature is quite in agreement with that predicted from a EOS for the gas expansion process from 200 Bar to 30 Bar.
You can estimate the final temperature with Joule-Thomson or model the operation as adiabatic flash, in this case results are quite different because Joule Thomson is not constant over that range of pressures.
I calculated Joule Thomson for N2 at 200 Bar @ 293.15 K , model Soave Redlich Kwong with Prode Properties , you can download a copy, which is free for non commercial use, from
http://www.prode.com/en/properties.htm
the value is 0.06371 K/Bar
then the estimated final temperature is
293.15 - 0.06371*(200-30)
the result being 282.319 K
with Prode Properties you can model the reducer as adiabatic operation,
in that case the final temperature is 270.644 K
which is quite in agreement with your measure (293.15-270.644 = 22.5.

Generally for fluids distributed at ambient temperature there is little heat exchanged and you can assume an adiabatic operation.

Edited by marchem, 06 February 2012 - 05:07 AM.

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Posted 06 February 2012 - 05:26 AM

Hi Dear marchem

Kind Regard,

### #4 kkala

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Posted 06 February 2012 - 06:17 AM

I agree with marchem, having estimated more or less similar results when nitrogen was introduced from a metallic cylinder (~ 200 Atma, ambient temperature) to a fuel oil accumulator at a pressure of ~ 15 Atma. A simulator (I used design II at that case) showed a temperature drop, indicating that N2 is not quite an ideal gas at 200 Atma pressure.
Resulting temperature drop is due to throttling, and it may be estimated using an isenthalpic process rather than a formula. For instance, rough indications can be:
1. estimates (*) enthalpy versus pressure of N2 (number 16) for several temperatures. For 200 Bar a and 300 oK enthalpy is ~300 kJ/kg. This enthalpy corresponds to ~280 oK (~20 oC lower) at 30 Bara.
2. Perry, 7th edition, Section 2 (Physical and Chemical data), Fig 2-13 (Temperature entropy diagram for nitrogen) indicates for 300 oK and 200 Atma enthalpy=2700 kcal/kmol. Same enthalpy for 1 Atma corresponds to ~270 oK (30 oC reduction).
3. Printing the diagram per para 1 and using a ruler to measure distances in paras 1 and 2 may give more precise results instead of "rough indications".

Note (*): Go to Thermodynamic diagrams page

Edited by kkala, 06 February 2012 - 06:29 AM.

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Posted 06 February 2012 - 07:47 AM

Dear kkala
In this post i explain 2 way but dear ankur believe solution 1 is true when im reading Cryogenic Engineering book page 121 - 123 problem 3-2 know isenthalpic proccess is true.I want know what is diffrence between Solution1 and 2 in told post?
Kind Regard,

### #6 marchem

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Posted 06 February 2012 - 09:31 AM

I think you'll find the answer in a good textbook discussing thermodynamics,
if you model the valve expansion with a ideal gas as fluid then Joule Thomson = 0,
see this paper for the details

http://www.chemistry.../labs/2PA36.pdf

for a ideal gas (P*V)^(cp/cv) = constant

for real gases cp and cv ARE NOT constant,
and the above is not true.

### #7 kkala

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Posted 06 February 2012 - 10:49 AM

1. I think adiabatic means just isolated, not necessarily reversible (meaning isentropic). If isentropic, P1*V1^k=P2*V2^k (M Zemansky, Heat and Thermodynamics, 6.7 Quasistatic adiabatic process, McGraw-Hill 1951). The formula is valid for ideal gases.
2. In this case for P1=30 barg, T1=300 oK and P2=0.1 barg, T2=T1*(P1/P2)^(-0.4/1.4)= 300*(31/1.1)^-0.286 oK= 115.5 oK =-157.5 oC. A compressor stage could approach (but not reach) this situation, of course in the opposite sense (raising gas temperature).
3. Let us now consider a throttling process, like the pressure reduction of nitrogen through a valve. This is not reversible, thus not isentropic. But can be considered as isenthalpic, if kinetic energies (per fluid mass) upstream and downstream are negligible (common case). Velocities are high along the restriction, but quite low upstream and downstream, at some distance to eliminate eddies.
4. For an ideal gas, throttling does not result in temperature difference (internal energy and P*V depend on temperature only). But real gases do present a temperature difference during throttling, which can be estimated through the isenthalpic principle. This is often applied in pressure reduction of steam.

Edited by kkala, 06 February 2012 - 11:16 AM.

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Posted 06 February 2012 - 06:07 PM

Dear Kkala
I read your post and i knew your opinion is to use enthalpy equation for calculate the temp and only for compressor can use adiabatic formula? is this true?

With all this talk just a way for the calculation remains and this is use isenthalpic proccess with T-S Chart.

Kind Regard

### #9 Art Montemayor

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Posted 06 February 2012 - 07:26 PM

Babak:

In my opinion, when you have mentioned “T-S Chart”, you have hit the nail on the head. When you have a simple, straight-forward need to know the resultant tempeature of an isenthalpic (non-reversible adiabatic) expansion of a compressed gas, all you need to know is the initial and the final enthalpies. A simulator is fine when you are simulating a process and trying to optimize it. However, one glance at a T-S chart (or a thermodynamic database) and you quickly have the solution.

Refer to the attached workbook where I show how I have always answered these types of questions quickly – by using the free NIST thermodynamic data base for pure fluids found on the Internet.

Note that I assume you are expanding the nitrogen from an ambient temperature of 25 oC and that the database answer is approximately 1 oC – for a drop in temperature of 24 oC. Always state what your initial conditions are. The drop in temperature is different for different initial conditions. For these type of simple problems with pure fluids, I trust the accuracy of the NIST more than a simulator. Note that the NIST also gives you the Joule-Thomson coefficient for each individual state of temperature.

From your prior posts in our Forums, I get the impression that you are having problems differentiating between a reversible and a non-reversible process in thermodynamics. I can understand your struggle; I also labored early in my thermo courses because of some bad or faulty teaching of the processes. My solution was to read through several books until I discovered an author that explained the differences with clarity and logic. Remember: the Joule-Thomson expansion only applies to GASES – not to liquids.

### #10 kkala

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Posted 07 February 2012 - 09:15 AM

Diagrams with enthalpy as one main parameter (x or y), like Mollier diagrams (enthalpy versus entropy), are practical in isenthalpic calculations. Nevertheless two isobaric tables (constant pressure at each table) of physical properties for several temperatures can be an easier and practical way out.
One question, how can you obtain numerical values on the spreadsheet? This is easier than diagrams. Having tried NIST at http://webbook.nist....hemistry/fluid/, I obtain diagrams. We can then press "view data in HTML Table" and obtain tabulated numerical values. But probably the "download data as tab-delimited text file" is used to insert the data into excel, after some treatment.

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Posted 07 February 2012 - 02:56 PM

1- I search for PC based software for thermodynamic properties and diagram?
I used catt2 but its not complete.

2- Due to all these discussions can be concluded that for calculate the expansion temperature
must be use Mollier diagram and not use adiabatic formula.

3- this discussions for expansion if i want calculate compress temperature which method is tru?

Kind Regard

### #12 Art Montemayor

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Posted 07 February 2012 - 03:44 PM

KKala:

You and I are on "the same page" in resolving this simple isenthalpic problem. As you have noted, the basic Thermo data is quickly accessed in NIST (for free). What you are contemplating is not what I do to place the data in an Excel spreadsheet. Like you, I use the "view data in HTML Table"; but I do not resort to a tab-delimited text file. I simply (and quickly) copy the table exhibited on the screen and paste it into a spreadsheet directly. Everything comes out exactly as exhibited on the screen - except that now you have actual numbers in their respective cells and can quickly use this data in equations or other operations. I have used this technique for years in calculating many refrigeration cycles such as the Freons, Ammonia, Propane, Ethylene, and others and obtained good results.

Babak:

You seem to be missing the point here. I am sure that Ankur was trying to do the same thing that I have attempted to show you: the mechanical operation of a Thermo problem is extremely simple and direct - if you know and thoroughly understand what is actually ocurring in a thermdynamic process. You seem to want a simple rule that you can follow by using the word "adiabatic". And this is what gives away your lack of Thermodynamic knowledge. As a student, it is good that you learn this weakness now, and work to correct it as quickly and efficiently as you can. Read my first post carefully because you have missed the point I tried to help you with: "you are having problems differentiating between a reversible and a non-reversible process in thermodynamics."

I make this point because of your use of the word "adiabatic" without qualifying if the process is reversible or not. If the process is adiabatic and non-reversible, such as the free expansion of a compressed gas (a throttling of the gas), then is is ISENTHALPIC (the enthalpy remains constant). If the process is adiabatic and reversible, then it is ISENTROPIC (the entropy remains constant). An isentropic process is many times more efficient than an isenthalpic one. The big point here is identifying your process as reversible or not - and then proceeding to apply the equations or thermo operation that applies.

By applying the above knowledge, you can answer your own query #3: since the compression of gas is a reversible process, it is isentropic in nature - this is not perfectly true in real life, but close enough when carried out in a reciprocating compressor. I think you have your work cut out for you in Thermodynamics and that is why I strongly recommend that you especially study this subject with considerable vigor and hard work. You will never regret having done so after you graduate.

I hope this helps you out.

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Posted 08 February 2012 - 04:59 AM

Dear Montemayor
Thank you for your attention.Im confused because on the internet different ways for calcualtion is available.

1.I know the expansion gas by neddle valve is irreversible.(From 200 Bar @15 Degree C To 30 Bar)

2.I know Adiabatic expansion formula T2=T1*(P1/P2)^(-0.4/1.4) then T2= -99.5 Degree C

4.In this discussion what is Joule-Thomson and formula (Old post)

Simply i want know this proccess follow from which formula.

Edited by babakfarhadi, 08 February 2012 - 05:01 AM.

### #14 Art Montemayor

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Posted 08 February 2012 - 09:19 AM

Babak:

I am trying hard to explain to you what I suspect you are missing because of the manner in which you use the term “adiabatic”. All “adiabatic” means is that the process has NO HEAT TRANSFER between the system and the exterior. Most processes carried out are adiabatic. The important term to remember is not adiabatic but reversible or non-reversible – as I repeat my answer to your query: “Simply I want to know which formula this proccess follows.” (here, I correct your English). Please read my response again:

“I make this point because of your use of the word “adiabatic” without qualifying if the process is reversible or not. If the process is adiabatic and non-reversible, such as the free expansion of a compressed gas (a throttling of the gas), then is is ISENTHALPIC (the enthalpy remains constant). If the process is adiabatic and reversible, then it is ISENTROPIC (the entropy remains constant). An isentropic process is many times more efficient than an isenthalpic one. The big point here is identifying your process as reversible or not - and then proceeding to apply the equations or thermo operation that applies.”

Stated another way:
When the process is adiabatic and reversible (such as an expansion turbine), the equation to be used is (S2 - S1) = 0 (constant entropy).
When the process is adiabatic and non-reversible (such as an expansion-throttling valve), the quation to be used is (H2 - H1) = 0 (constant enthalpy).

In your 4th query, you ask, “In this discussion what is Joule-Thomson and formula (Old post)?” I simply mentioned the Joule-Thomson coefficient because of the comment made by Marchem in their post: “I think you'll find the answer in a good textbook discussing thermodynamics, if you model the valve expansion with a ideal gas as fluid then Joule Thomson = 0.” You can use a simulator on the Joule-Thomson relationship, but I believe (as I stated) this is an over-kill.

As a matter of personal interest, how many chemical engineering courses have you finished that expounded Thermodynamics – such as: Physics, Physical Chemistry, Chemical Engineering Thermodynamics, Mechanical Engineering Thermodynamics, Unit Operations, etc., etc.? I took all these courses in my undergraduate studies 52 years ago and I hope that they are still being taught in the same quantity and quality.

I hope that I succeed in helping you understand how to identify the process and apply the correct Thermodynamic relationship.

### #15 kkala

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Posted 08 February 2012 - 04:41 PM

The assumption of reversibility may need some time and effort to be assimilated. Believing I do not have a very clear idea (never have dealt with it at work), following notes may be useful for the topic. Comments are welcomed.
1. Books of Physics explain reversible and irreversible concept in the Chapter of Entropy. Looking at some old local ones, they generally report constant P*V^k (k=Cp/Cv) for an ideal gas adiabatic process, without clarifying it is true for reversible conditions (M. Zemansky, see post No 7). This could create the confusing impression that any adiabatic expansion (including throttling) follows this low, which is not true.
2. The "adiabatic formula" T2=T1*(P1/P2)^((1-k)/k) = T1*(P2/P1)^(1 - 1/k) derives from above P*V^k = constant, plus the ideal gas law. So it also concerns reversible conditions and ideal gas.
Under these conditions, work produced w=(P1*V1-P2*V2)/(k-1). Concerning entropy S1 = S2 (due to reversibility and adiabatic conditions, ie Q = 0).
3. Throttling is not reversible, since you cannot bring the total quantity of gas back to P1 without spending work (compression).
During throttling, work produced = 0, adiabatic conditions (Q=0), entropy S1 < S2.
4. Let us see examples of adiabatic expansion of steam from 70 bara, 450 oC to 40 bara.
4.1 If steam passes through a back pressure turbine, reversibility can be assumed for a gross theoretical estimate. S1=6.6368 kJ/kg/oK, S2=6.6368 kJ/kg/oK at 40 bara, so t2=363 oC. Theoretical work produced by steam w = 3289 - 3126 = 163 kJ/kg (difference in enthalpies). These according to steam tables ("corrections" could be applied to estimate actual turbine performance).
According to "adiabatic formula" T2=723*(70/40)^(-0.41/1.41)=614 oK, that is t2=341 oC. Also w = (70*0.04413 - 40*0.06825)*1E+5/(1.41-1) J/kg = 88 kJ /kg. Deviations could be due to the fact that steam is not an ideal gas.
4.2 If steam is throttled, enthalpies before and after are equal, H1 = H2 = 3289.1 kJ/kg; hence t2=432 oC. Work produced by steam = 0, entropy increase S2-S1 = 6.8797-6.6368 = 0.2429 kJ/kg/oK.
If steam were an ideal gas, t2=450 oC
4.3 Above could be easily estimated by use of a Mollier (H-S) diagram for steam too.
5. The Joule-Thomson coefficient concerns throttling; an explanation can be found in http://www.eng-tips....cfm?qid=86992 , post in 12 Feb 2004. Enthalpies seem more practical to use, if available.

Edited by kkala, 08 February 2012 - 05:06 PM.

### #16 kkala

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Posted 08 February 2012 - 06:18 PM

As you have noted, the basic Thermo data is quickly accessed in NIST (for free). What you are contemplating is not what I do to place the data in an Excel spreadsheet. Like you, I use the "view data in HTML Table"; but I do not resort to a tab-delimited text file. I simply (and quickly) copy the table exhibited on the screen and paste it into a spreadsheet directly. Everything comes out exactly as exhibited on the screen - except that now you have actual numbers in their respective cells and can quickly use this data in equations or other operations. I have used this technique for years in calculating many refrigeration cycles such as the Freons, Ammonia, Propane, Ethylene, and others and obtained good results

.
Thanks Art, it works satisfactorily and this way may be the easiest.
Problems were encountered when the delimited file was tried to paste.
Alternatively one can convert the delimited file into an excel file by following the "open an entire text file in Microsoft excel", from Excel Help (text file may not be ending to .txt).