Jump to content

  • Log in with Facebook Log In with Google      Sign In   
  • Create Account


ChExpress Blog - 11/12/14

Read the latest news from the chemical industry in Christa's blog.

Featured Articles

Check out the latest featured articles.

File Library

Check out the latest downloads available in the File Library.

New Article

Product Viscosity vs. Shear

Featured File

Air Vessel Sizing for Water Hammer Prevention

Isentropic Efficiency And Its Implications


This topic has been archived. This means that you cannot reply to this topic.
30 replies to this topic
Share this topic:

#1 JoeWong

JoeWong

    Gold Member

  • ChE Plus Subscriber
  • 1,223 posts

Posted 22 February 2007 - 11:31 AM

Dear all,

Low temperature embrittlement due to system blowdown shall always be considered during material selection study. During Basic Engineering phase, normally depressuring unit in HYSYS and / or PRO-II will be used to carry out blowdown and it low temperature effect. One of the parameter seriously affect the results especially temperaure is ISENTROPIC EFFICIENCY.

The higher the isentropic efficiency, the lower the final temperature. In some case , when we use Isen. Eff = 80%, its final temperature is lower than the pre-selected material (LTCS) low temperature limit (-46 degC) and a much expensive material (SS) is required. If we apply lower isen. eff=50%, its final temperature is still higher than the CS LT limit.

I have gone through several project. Some project use isen. eff= 50%, some use 80%, etc. When looks for engineering judgement / supporting doc., most of the time having the following answers :

- COMPANY practice
- Past project experience
- Project Design Basis

I really would like to take this oppurtunity to gather some informations, thaughts and advices from all of you.


Looks forward your advice.

JoeWong

#2 vicini

vicini

    Junior Member

  • Members
  • 27 posts

Posted 23 February 2007 - 10:14 PM

First off, don't worry about the low temperatures during blowdown or depressering. As the temperature drops, the stress drop and there no problems. Suggest reading hydrocarbon processing July 2004 Material selection for low temperature applications. ASME allows sub - 40 C for carbon steel at reduced stress.

#3 JoeWong

JoeWong

    Gold Member

  • ChE Plus Subscriber
  • 1,223 posts

Posted 27 February 2007 - 05:57 AM

Dear vicini,

Thanks for your advice.

I understood your advice you were based on the ASME fig. 323.2.2B, Reduction in minimum design metal temperature without impact test. I have applied this allowance for FLARE piping in one of my previous project. However, one shall understand that the condition (2) stated this allowance will limit the internal pressure to a very low pressure.

In our case, the low temperature occured at the vessel instead of BDV downstream piping. During blowdown, the internal pressure will exceed the limitation per condition (2). I doubt we can apply above allowance in this case.

JoeWong

#4 markk

markk

    Brand New Member

  • Members
  • 3 posts

Posted 12 September 2007 - 01:59 AM

JoeWong & Vicini,

You need to be careful when taking credit for the fact that the pressure reduces as the temperature falls.
This is because after the depressurisation, the vessel could be repressurised (before it has been warmed up) due to operator error or ignorance, leading to a high pressure at the low temperatures.

The likelihood of this case should be taken into account.

#5 JoeWong

JoeWong

    Gold Member

  • ChE Plus Subscriber
  • 1,223 posts

Posted 13 September 2007 - 12:27 AM

markk,

You are right that immediate repressurization after depressuring potentially introduce stress on the system again while it still cold. Thus generally we do not consider reduced stress at low temperature for vessel (upstream of BDV/RO). But we may take credit on the BDV/RO downstream where pressurization of BDV/RO piping is very unlikely (subject to scenario analysis).

Anyway, good point to highlight.

JoeWong

#6 Ariel

Ariel

    Junior Member

  • Members
  • 26 posts

Posted 29 February 2008 - 05:05 AM

Dear Joe,

I don’t have a lot experience at Blowdown systems but I don’t understand why you are concern about low temperatures at the vessel. I always have been worry about low temperatures after BDV/RO system. Could you explain why would you have low temperatures at the vessel from blowdown o depressurization? Or you are talking about KOD vessel?
I have a home made tutorial (not by me) for using Depressurization Utility in hysys but I don´t have it now. I´ll look if there is a recommendation for isentropic efficiency there.

Bye,
Ariel

#7 rxnarang

rxnarang

    Gold Member

  • Members
  • 132 posts

Posted 03 March 2008 - 03:56 AM

Joe,

My understanding is as follows: The isentropic efficiency in a depressuring calculation refers to the vapour expansion in the vessel as the vessel depressurises. The expansion can be 100% isentropic, which will lead to lowest temps or 0% isentropic which means a isoenthalpic expansion. HYSYS gives some guidelines on the efficiency, whcih you can read on the Design/opions page; 87-98% for vapors.

Now to use this meaningfully one has to simulate the depressuring situation as authentically as possible. This means one also has to put in the the metal mass in contact with the vapor (or liquid). As the metal mass has some heat capacity, I found that the drop in temperature is moderated significantly.

To further make matters interesting, one should also simulate the heat gain from the ambient ( and fire, if that is the assumption), including any insulation. This can moderate low temperatures, depending on the details.

For MDMT calculation, ASME Section VIII mentions auto refrigeration as one of the plausible considerations. Many company procedures calculate this by depressuring the fluids ( liquid or vapor- in case of light liquids, the atmospheric boiling point can be used straightaway) isoenthalpically to atmospheric pressure. This means a simple simulation using a valve should do the trick.

I would calculate the design temperstures across the plant as a separate excercise, and then authenticate the MDMT when doing the depressuring. In case I get lower temperatures than isoenthalpic expansion, I would further build in metal mass and heat loss in the depressuring module and rerun the calcs. I anticipate you will find that the isoenthalpic expansion to atmospheric pressure will suffice.

Regards
Rajiv

#8 Art Montemayor

Art Montemayor

    Gold Member

  • Admin
  • 4,880 posts

Posted 03 March 2008 - 08:55 AM


Joe / Rajiv:

I have been trying to follow this thread because I consider it as one that is of important and practical interest to all process engineers. However, I have trouble understanding what the basic issues and basis are for the specific problem that has been presented as an example:

De-pressurization of a process vessel (especially one filled with hydrocarbons) is an important operation that must be recognized and dominated by process engineers. However, we must have a basic definition of whether the vessel contains 100% gas (or vapor), 100% liquid filled, or is partially filled with liquid in equilibrium with its vapor.

Any free, adiabatic expansion (irreversible, where no work is produced) of a gas or liquid follows an isenthalpic process (enthalpy is held constant). This is what occurs when a pressure vessel is de-pressurized.

Any expansion of a gas where work is done (as in a turboexpander, steam turbine, expansion engine, gas turbine, etc.) takes place with the fluid’s entropy held constant. This is called an isentropic process. This is not what ordinarily happens when a depressurization valve is opened on a pressurized process vessel – whether it contains gas or liquid.

When an adiabatic, isenthalpic expansion of a liquid or of a gas takes place, there is usually a cooling effect that follows (in the product, expanded fluid – if the source fluid pressure is presumed to be held constant, as in a steady state). The Joule-Thomson effect and the refrigeration effect of an expanded liquid refrigerant are typical examples of this.

However, in the case of a shut-in, pressurized vessel subjected to a “blow-down” effect due to de-pressurization, the process is basically a batch process that involves free, adiabatic expansion (isenthalpic) with no work done. But since the process is also batch, the vessel itself undergoes a decrease in pressure (gas expansion) as well as the released gas. Both fluids – the gas released and the gas remaining in the vessel – undergo free, isenthalpic expansion and should be expected to also cool down - normally.

The problem I have is understanding where the term “isentropic” enters the picture here. Please see the attached definition of isentropic efficiency and correct me if I am wrong or if I am mis-interpreting this dialogue. To my understanding, the moment we recognize that no work is being done on the system or by the system, the term isentropic is not applicable. Am I missing something?

I also think that the thread has failed to identify WHICH fluid temperature is the one of primary interest. I know that the gas that remains behind (in the vessel) and is being cooled is the culprit that could contribute to brittleness in the vessel material and cause early failure - but that hasn't been specifically stated and identified. I think it should be cleared up because the exiting cool gas is also affecting the vent piping itself - which is, I think, a lesser problem.

Also, what is "isoenthalpic" expansion?
Attached File  Isentropic_Efficiency.doc   27.5KB   521 downloads


#9 Zauberberg

Zauberberg

    Gold Member

  • Store Customers
  • 2,193 posts

Posted 03 March 2008 - 10:38 AM

Good evening,

Bringing the term "isentropic efficiency" in relationship with blowdown process - which doesn't have any work extracted from a process fluid (W=0), is a modeling software issue - Aspen HYSYS in particular.

According to software developers, applying Joule-Thomson effect in blowdown calculations of 100% noncondensible vapor filled vessel, does not match with low temperatures downstream of restriction orifice which are measured in practice. From their point of view, isentropic efficiency is just a tuning parameter in order to achive compliance of software outputs versus field measured data. They emphasize that each system is different and should be submited to rigorous calculations for obtaining correct results. This can be considered OK - in my opinion - otherwise there are no ways to make any influence on temperature profile (both in the vessel and vent piping downstream of orifice). We all agree that isentropic expansion leads to the lowest possible temperature of fluid undergoing expansion.

However, the strange thing is that recommended values for isentropic efficiency in blowdown simulations of 100% gas filled vessel are approaching 100%, while for liquid and 2-phase fluids this parameter should be lower, without any further explanation. I consider this to be very poor and insufficient information for accurate modeling of blowdowns, but I guess we have to live with it. In designing completely new ("grasroot") plant, there is no other way to estimate temperature profile during system blowdown, except on relying on Joule-Thomson effect, simulation software and similar industrial services and system conditions.

Nevertheless, if I would have to define what could "isentropic efficiency" present in a blowdown process, I would say the following:

1. Literally, blowdown work is not zero. Expanded fluid causes compression of the fluid present in the vent system (if closed), or compression of atmospheric air (if open type venting). Another issue is that "work" is not used for anything. As I have said in one similar post, this reminds me on R. Meyer's experiment of determination of Cp and Cv.

2. Isentropic expansion is sometimes called "frictionless expansion". Comparing blowdown process with true expansion (steam turbine, turboexpander etc.), one can use this relative amount of fluid energy during depressuring which has not been lost in friction, shock waves, vibration, heat transfer with surroundings etc., to serve as a measure of how close blowdown can be to the true isentropic expansion.

This is my 2 cents opinion. I have heard people working for decades in plant operations, saying there is no way for gas to reach temperatures lower than the one calculated based on Joule-Thomson effect. And liquid will reach boiling temperature at final blowdown pressure, taking into account final liquid composition. And I cannot say anything against that, it is a good engineering logic. This was my way to combat against various software limitations.

Best regards,

#10 JoeWong

JoeWong

    Gold Member

  • ChE Plus Subscriber
  • 1,223 posts

Posted 16 March 2008 - 05:30 AM

QUOTE (rxnarang @ Mar 3 2008, 04:56 AM) <{POST_SNAPBACK}>
Joe,

My understanding is as follows: The isentropic efficiency in a depressuring calculation refers to the vapour expansion in the vessel as the vessel depressurises. The expansion can be 100% isentropic, which will lead to lowest temps or 0% isentropic which means a isoenthalpic expansion. HYSYS gives some guidelines on the efficiency, whcih you can read on the Design/opions page; 87-98% for vapors.

Now to use this meaningfully one has to simulate the depressuring situation as authentically as possible. This means one also has to put in the the metal mass in contact with the vapor (or liquid). As the metal mass has some heat capacity, I found that the drop in temperature is moderated significantly.

To further make matters interesting, one should also simulate the heat gain from the ambient ( and fire, if that is the assumption), including any insulation. This can moderate low temperatures, depending on the details.

For MDMT calculation, ASME Section VIII mentions auto refrigeration as one of the plausible considerations. Many company procedures calculate this by depressuring the fluids ( liquid or vapor- in case of light liquids, the atmospheric boiling point can be used straightaway) isoenthalpically to atmospheric pressure. This means a simple simulation using a valve should do the trick.

I would calculate the design temperstures across the plant as a separate excercise, and then authenticate the MDMT when doing the depressuring. In case I get lower temperatures than isoenthalpic expansion, I would further build in metal mass and heat loss in the depressuring module and rerun the calcs. I anticipate you will find that the isoenthalpic expansion to atmospheric pressure will suffice.

Regards
Rajiv


Rajiv,

Thanks for your response.

This is an interesting topic and it has been discussed again and again. This topic discussed seriously in other forum. I would like to invite you to read it (CLICK HERE) and get some outcome from the discussion.

I guess "isoenthalpic" is "Isenthalpic". Am i right ?

#11 JoeWong

JoeWong

    Gold Member

  • ChE Plus Subscriber
  • 1,223 posts

Posted 16 March 2008 - 06:33 AM

Mr. Montemayor,

Thanks for your interest in this old topic.
This topic has seriously discussed in other forum. It has invited more inputs and though in this topic. Click HERE to read more (if you have not).

QUOTE (Art Montemayor @ Mar 3 2008, 09:55 AM) <{POST_SNAPBACK}>
[size=3][font="Times New Roman"]
De-pressurization of a process vessel (especially one filled with hydrocarbons) is an important operation that must be recognized and dominated by process engineers. However, we must have a basic definition of whether the vessel contains 100% gas (or vapor), 100% liquid filled, or is partially filled with liquid in equilibrium with its vapor.


I would say depressuring are applicable to all conditions where vessel contains 100% gas (or vapor), 100% liquid filled, or is partially filled with liquid in equilibrium with its vapor.


QUOTE
Any free, adiabatic expansion (irreversible, where no work is produced) of a gas or liquid follows an isenthalpic process (enthalpy is held constant). This is what occurs when a pressure vessel is de-pressurized.


In depressuring context, fluid (vapor "expansion" only, liquid flashing plus vapor "expansion" for vapor/liquid or liquid filled system) in the vessel is "moved" from vessel to "sink" (i.e. atmosphere), there is merely "no work" if the boundary is within the vessel itself. However, the depressuring process involve bringing a fluid at one state (pressurised fluid in vessel boundary) to another state (to low pressure region where fluid is expanded and move the boundary to atmosphere), the "work" is done to move the boundary. However, this "work" is not mechanical work that always in our mind.

What's your opinion ?

QUOTE
Any expansion of a gas where work is done (as in a turboexpander, steam turbine, expansion engine, gas turbine, etc.) takes place with the fluid’s entropy held constant. This is called an isentropic process. This is not what ordinarily happens when a depressurization valve is opened on a pressurized process vessel – whether it contains gas or liquid.


Expansion in machinery (e.g. turboexpander, steam turbine, expansion engine, gas turbine, etc." involve mechanical work by keeping entropy constant (IDEAL). Introduction of Isentropic Efficiency to derive a real mechanical work done by shifting the entropy (NON-IDEAL). I am fully in agreement with it.

The main problem with depressuring is fluid expansion involve boundary movement. Is "work" involved by moving boundary ?

I must make the depressuring context clear here. It is related to fluid expansion from pressurized vessel to atmosphere but not related to depressuring valve (BDV/RO).


QUOTE
The problem I have is understanding where the term “isentropic” enters the picture here. Please see the attached definition of isentropic efficiency and correct me if I am wrong or if I am mis-interpreting this dialogue. To my understanding, the moment we recognize that no work is being done on the system or by the system, the term isentropic is not applicable. Am I missing something?


You are right that it is hard to imagine how the term "isentropic" enters into the picture if we consider there is NO work involve in depressuring process. I believe the "work" is real mechanical work.

If i view the moving of boundary from one state to another state is somewhat "a work (not mechanical work)", then there shall be some term to related this IDEAL "work" and NON-IDEAL "work". That's where the term "isentropic" come in. However, i may be absolutely wrong in this concept and many experienced engineers (especially those who has strong mechanical background) may not be accept this concept. If you read the thread in other forum, you will understand.

QUOTE
I also think that the thread has failed to identify WHICH fluid temperature is the one of primary interest. I know that the gas that remains behind (in the vessel) and is being cooled is the culprit that could contribute to brittleness in the vessel material and cause early failure - but that hasn't been specifically stated and identified. I think it should be cleared up because the exiting cool gas is also affecting the vent piping itself - which is, I think, a lesser problem.


I would apologize for not making this clear. As mentioned earlier, it is related to fluid expansion from pressurized vessel to atmosphere but not related to depressuring valve (BDV/RO). Thus, the primary interest in fluid behind the vessel and associated metal temperature.

QUOTE
Also, what is "isoenthalpic" expansion?


I guess it is isenthalpic expansion. Let Rajiv to confirm my guess.

#12 JoeWong

JoeWong

    Gold Member

  • ChE Plus Subscriber
  • 1,223 posts

Posted 16 March 2008 - 07:16 AM

Zauberberg,
Thanks for you input again. I believe you have participated in similar discussion in other forum.

QUOTE (Zauberberg @ Mar 3 2008, 11:38 AM) <{POST_SNAPBACK}>
Bringing the term "isentropic efficiency" in relationship with blowdown process - which doesn't have any work extracted from a process fluid (W=0), is a modeling software issue - Aspen HYSYS in particular.


If you read the response that in earlier post by Mr. Montemayor, i have brought up "work" (but not mechanical) concept by viewing moving of fluid from one state to another state. This is close to the experiment by R. Meyer's in determining Cp and Cv.


QUOTE
According to software developers, applying Joule-Thomson effect in blowdown calculations of 100% noncondensible vapor filled vessel, does not match with low temperatures downstream of restriction orifice which are measured in practice. From their point of view, isentropic efficiency is just a tuning parameter in order to achive compliance of software outputs versus field measured data. They emphasize that each system is different and should be submited to rigorous calculations for obtaining correct results. This can be considered OK - in my opinion - otherwise there are no ways to make any influence on temperature profile (both in the vessel and vent piping downstream of orifice). We all agree that isentropic expansion leads to the lowest possible temperature of fluid undergoing expansion.

However, the strange thing is that recommended values for isentropic efficiency in blowdown simulations of 100% gas filled vessel are approaching 100%, while for liquid and 2-phase fluids this parameter should be lower, without any further explanation. I consider this to be very poor and insufficient information for accurate modeling of blowdowns, but I guess we have to live with it. In designing completely new ("grasroot") plant, there is no other way to estimate temperature profile during system blowdown, except on relying on Joule-Thomson effect, simulation software and similar industrial services and system conditions.


The quote in HYSYS "use 87-98% for vapors, 40-70% for liquid...". Personally, i don't buy this idea as i have not received / read any supporting facts behind or published report by HYSYS or others to prove the data is acceptable to certain confident level.

Because of this, i brought this issue up here to gain more insights.


QUOTE
Nevertheless, if I would have to define what could "isentropic efficiency" present in a blowdown process, I would say the following:

1. Literally, blowdown work is not zero. Expanded fluid causes compression of the fluid present in the vent system (if closed), or compression of atmospheric air (if open type venting). Another issue is that "work" is not used for anything. As I have said in one similar post, this reminds me on R. Meyer's experiment of determination of Cp and Cv.

2. Isentropic expansion is sometimes called "frictionless expansion". Comparing blowdown process with true expansion (steam turbine, turboexpander etc.), one can use this relative amount of fluid energy during depressuring which has not been lost in friction, shock waves, vibration, heat transfer with surroundings etc., to serve as a measure of how close blowdown can be to the true isentropic expansion.


I am pretty inline with this though. Depressuring involve energy lost for fluid movement in frictional aspect, heat transfer barrier, shock wave limitation at inlet nozzle/pipe, movement of liquid due KH effect, etc. Somehow an efficiency factor shall be in to cater for all this uncertainties. We can not just ignore them.

I am in agreement with HYSYS where it brought this deviation into the depressuring unit. Similarly it was in PRO-II as well. However, the PV term as how HYSYS described has weak thermodynamic meaning.

PV term definition in HYSYS for those who are interested...
QUOTE
The PV work term (or isentropic efficiency) represents the work done on the surroundings by the expanding fluid. It is part of the dynamic energy balance [1]:

F0 Cp0 (T0 -Tref) + UA(Tbulk - Twall) + d(PV)/dt = (M Cp)holdup d(Tbulk)/dt [1]

Energy Flow Out of Vessel + Wall Effects + PV Work of Expanding Fluid = Rate of change of energy of fluid in vessel.

In HYSYS the d(PV)/dt term above is actually calculated as:

d(PV)dt ~ (VesselHoldupPressure@CurrentTimeStep - VesselPressure@LastTimeStep) / Density * fluidWorkFactor [2]

where:

fluidWorkFactor = PV Work Term Contribution/100

This correction is included so that the user can "tune" the depressuring utility to match plant or measured data.


Prof. S.M. Richardson and Prof. Haroun Mahgerefteh have spend their time and effort in understanding this issue. The spirit behind is really to understand how it behave and provide sufficient confident level to define a good approach in this depressuring aspect.

#13 JoeWong

JoeWong

    Gold Member

  • ChE Plus Subscriber
  • 1,223 posts

Posted 16 March 2008 - 10:10 AM

QUOTE (Ariel @ Feb 29 2008, 05:05 AM) <{POST_SNAPBACK}>
Dear Joe,

I don’t have a lot experience at Blowdown systems but I don’t understand why you are concern about low temperatures at the vessel. I always have been worry about low temperatures after BDV/RO system. Could you explain why would you have low temperatures at the vessel from blowdown o depressurization? Or you are talking about KOD vessel?
I have a home made tutorial (not by me) for using Depressurization Utility in hysys but I don´t have it now. I´ll look if there is a recommendation for isentropic efficiency there.

Bye,
Ariel


Airel,

Depressuring involve partial removal of inventory from vessel to receiver (i.e flare disposal system) by pressure different. This process will carry away energy from the fluid in the vessel and temperature of fluid will drop...(like sweating process...water from our skin and evaporate to atmosphere. The evaporated water will carry away heat from our skin).

Every material has it own minimum allowable temperature. Exposing material below this minimum allowable temperature would potential lead to low temperature embrittlement failure (failure similar to glass break), a complicated phenomenon.

#14 djack77494

djack77494

    Gold Member

  • ChE Plus Subscriber
  • 1,282 posts

Posted 02 September 2008 - 03:48 PM

To all,
Thanks for a very interesting discussion on an interesting topic. I, for one, am still more than a bit confused. I can see that work (specifically PV work) might be done on the surroundings, yet on the other hand I am accustomed to thinking of expansion across an expansion valve as being isenthalpic. Furthermore, not even thinking about whether or not work is being done, if there is no heat transfered across the boundary, then the process should be isenthalpic, no?? So I hope someone can explain if the process should be considered isenthalpic, isentropic, neither, or "it depends". I'd also be interested in seeing a bit of explanation as to their reasoning.
Thanks,
Doug

#15 kheong83

kheong83

    Brand New Member

  • Members
  • 4 posts

Posted 02 September 2008 - 11:50 PM

Djack,

I think there is no doubt that the expansion across an expansion valve as being isenthalpic as where entropy will increase while mantaining enthalpy as irreversible work is done. The PV work term in Hysys actually is considering the work done to expand the fluid in the vessel itself to transfer it to the valve upstream. Therefore, if we consider pv work term, the actual temperature at the valve upstream is actually lower than the initial valve upstream temperature as the vessel is a batch process where the pressure in the vessel will drop with time unlike our normal valve that is in continous process. Thus, due to lower upstream temperature at the valve, the depressuring dynamics minimum downstream temperature will be lower if PV work term is consider (set > 0%)

Thus, if we consider the valve itself, it is still an isenthalpic process but if we consider the whole system that includes transfering the fluid from the vessel to the valve, its not an isenthalpic process but it will be closer to isentropic process as the increase in entropy downstream of the valve is compensated by the decrease in entropy cause by losses in the form of work done to expand the fluid from vessel to upstream of the valve.

Regards,
kheong83

#16 shan

shan

    Gold Member

  • ChE Plus Subscriber
  • 413 posts

Posted 03 September 2008 - 08:26 AM

Let me try to use an example to make the issues simpler.

Use Fig 24-23 Methane P-H Diagram on GPSA as example.

Initial Conditions: P=1000 psia, T=100 F, H=-1540 Btu/Lb and S=2.24 Btu/Lb-R

Isenthalpic Expansion to 100 psia, Vertical along H-1540 Btu/Lb Line Downward
Equivalent to Hysys PV Term Work Contribution = 0%
Finial Condition: P=100 psia, T=52 F, H=-1540 Btu/Lb and S=2.50 Btu/Lb-R

Isentropic Expansion to 100 psia, along S=2.24 Btu/Lb-R Line Downward
Equivalent to Hysys PV Term Work Contribution = 100%
Finial Condition: P=100 psia, T=-150 F, H=-1650 Btu/Lb and S=2.24 Btu/Lb-R

In real life, there is neither Isenthalpic Expansion nor Isentropic Expansion. There is always work such as heat transfer between different temperature objects and friction existed. Also, Based on the Second Law of Thermodynamics, entropy always increases in all the processes.

You may check with Hysys help desk to ask why their recommendation is 87%-98% PV Term Work and follow the go-bys to ensure if 50% PV Term Work is save assumption.

Regards

Shan

#17 kheong83

kheong83

    Brand New Member

  • Members
  • 4 posts

Posted 03 September 2008 - 08:01 PM

In addition to Shan explanation, isentropic process according to a book, "Thermodynamic: An Engineering Approach" by McGraw Hill can happen in 2 situations:
1. Adiabatic and reversible process (turbo expander, etc.)
2. Irreversible process (valve) but with losses (in terms of heat, friction,etc)

The PV work term in HYSYS makes the process in the depressuring valve closer to the second isentropic process situation mention above where losses occurs to transfer the fluid from vessel to the valve makes the whole process closer to isentropic process as we increase PV work %.

Regards,
Kheong83

#18 shan

shan

    Gold Member

  • ChE Plus Subscriber
  • 413 posts

Posted 04 September 2008 - 07:14 AM

Thanks Kheong83 for the addition of explanation.

Hysys PV work term contribution is 100% in the ideal turbine case (100% Efficiency, S=constant) and Hysys PV work term contribution is 0% in ideal valve case ( 0% Efficiency, H=constant).

Therefore, PV work term contribution is something between 0% and 100% in the real life. If you have no idea what the value is, just use 50% because the biggest inaccuracy that you may have is 50% either over or under. Just Kidding! I believe that Hysys have some reasons to recommend 87% to 98% but I don’t know what reasons are.

Regards

Shan

#19 djack77494

djack77494

    Gold Member

  • ChE Plus Subscriber
  • 1,282 posts

Posted 04 September 2008 - 02:04 PM

QUOTE (shan @ Sep 4 2008, 04:14 AM) <{POST_SNAPBACK}>
I believe that Hysys have some reasons to recommend 87% to 98% but I don’t know what reasons are.


I wrongly thought I followed the explanation up to this point. First let me present a new question - We are talking about the remaining fluid (residual) in the vessel when we do this calculation, right? I do understand that this is an unsteady state or batch process and I think we're mainly interested in knowing how the properties (T & P) of the fluid in the vessel change during depressuring. So, can we continue? Please explain how we can have up to 98% efficiency in a process with a useless, wasted, near free expansion across a valve. Thanks.

#20 shan

shan

    Gold Member

  • ChE Plus Subscriber
  • 413 posts

Posted 04 September 2008 - 04:46 PM

Hi Djack77494,

We are not talking about residual fluid or discharge fluid. We are discussing the depressuring process [from state A (higher pressure) to state B (lower pressure)].

My point is that Hysys PV work term contribution is a measurement of the gas depressuring process close to isenthalpic process (H=constant) or isentropic process (S=constant). If the expansion process isenthalpic process, PV work term contribution is 0% and if the expansion process is isentropic process, PV work term contribution is 100%.

You have the freedom to input any PV work term contribution value based on your own engineering judgment. Could you please disclose how many case Hysys depressuring you have simulated and what were PV work term contribution values you applied? Do you have any evidence to prove that your PV work term contributions match with real operating case?

Again, I am not in the position to defend Hysys recommendation (87% to 98%). However, I think this is a reasonable range. Please image the released gas front as a piston. The work is done by pushing backpressure resistance until the force exhausted. This is the same thing as wind always contributes work no mater windmill power generators are installed or not.

Regards

Shan

#21 shan

shan

    Gold Member

  • ChE Plus Subscriber
  • 413 posts

Posted 05 September 2008 - 09:32 AM

For everybody’s convenience, I am posting the example diagram as the attachment. Please see my 9/3/08 thread for the explanation.

If you understand why point D and point E are impossible, you got what I was talking about.

Attached Files



#22 djack77494

djack77494

    Gold Member

  • ChE Plus Subscriber
  • 1,282 posts

Posted 05 September 2008 - 04:36 PM

QUOTE (shan @ Sep 5 2008, 05:32 AM) <{POST_SNAPBACK}>
If you understand why point D and point E are impossible, you got what I was talking about.


shan,
Thanks for the diagram and your attempts to "drive these concepts home". I don't think you've succeeded, however, in getting full understanding through my own (thick) cranial structure.

As I look at your CH4 Mollier Diagram, I fully understand that point "D" is impossible. It would violate the second law of thermo. Point "E" is also impossible unless external heat were added to our system. So really the whole issue comes down to where on the P=100 psia line will we land. Obviously it's between points "B" and "C".

Please correct any misconceptions I may have in the following.
1) The system consists of the vessel and a portion of a flare system that we will blowdown into.
2) Let me imagine the vessel is originally at 1000 psia and it will be rapidly depressured through an expansion valve to 100 psia (like your diagram).
3) Furthermore, let us imagine that the flare system is at 14.7 psia (atmospheric pressure) throughout this operation.

Now, please allow me to imagine that there is an empty balloon attached to the valve discharge as a useful tool to help me visualize what happens. Furthermore, let's say the vessel has a volume of 1 cubic foot.

Other than PV work, there is no work and no heat transfered into/out of the vessel plus the balloon during depressuring.

I envision that the vessel originally had a PV term =
1000 psia * 1 ft^3 / 144 in^2/ft^2 = 6.94 ft*#
The vessel's final PV term =
100 psia * 1 ft^3 / 144 in^2/ft^2 = 0.694 ft*#
PV work was done on the surrounding by virtue of the fact that a mass of gas "pushed back" the atmosphere. Intuitively, the balloon's volume would have changed from zero initially to some final volume at an unknown temperature.

So, am I correct in thinking that the vessel's final temperature is not relevant to this calculation? (I know that it may be of interest for other reasons.) If we have an ideal gas, is this process isothermal? I'm thinking that if the process is isothermal, then the initial and final vessel and balloon temperatures are all the same, and that the final balloon volume would be 9 ft^3. Then the PV work done to the balloon (surroundings) would be
14.7 psia * 9 ft^3 / 144 in^2/ft^2 = 0.919 ft*#.
The PV work that was "lost" be the fluid in the vessel would be 6.94 - 0.694 = 6.246 ft*#.

For those who have not yet dozed off (probably very few), would not the overall efficiency be 0.919/6.246 * 100 = 14.7%??

#23 djack77494

djack77494

    Gold Member

  • ChE Plus Subscriber
  • 1,282 posts

Posted 05 September 2008 - 04:58 PM

In thinking about this a bit more, another thought occured to me. First let me state that I had been considering a vessel that is 100% vapor filled.

If we consider just the vessel itself, then as gas is released and the pressure drops, the temperature of the gas in the vessel will also drop. But it does not matter whether we have free expansion into a vacuum or if we are extracting 100% of all possible work from the gas. The vessel temperature should be the same in both cases, no?

The temperature of the gas downstream of the expansion WILL be vastly different depending on the nature of the expansion process. I'd like to get a handle on both temperatures.

#24 shan

shan

    Gold Member

  • ChE Plus Subscriber
  • 413 posts

Posted 08 September 2008 - 09:39 AM

Hi djack77494,

The “success” is so attractive to me. Let me put some more efforts.

Actually, since you have agreed that the finial point should be some place between B and C, which is whole point what I wanted to make. I think I am able to declare 99% success. It is not important at all whether the efficiency is 17.4% or it is 98%. Anyway, let me try to make the issue a little bit more clear.

First, your efficiency calculation is incorrect. As you assumed, the ideal gas law (P1*V1)/T1 = (P2*V2)/T2. Please keep it in your mind that this is under the conditions of same amount of gas moles. This means that the depressured vessel and “balloon” have same pressure P2 at the end.
As you assumed T1=T2, therefore: P1*V1=P2*V2
If P2 = 100 psia, V2=P1*V1/P2=1000*1/100=10 ft3, Work = (10 ft3 – 1 ft3) * 100 psi / (144 in^2/ft^2) = 6.25 ft*# PV Efficiency = 6.25/6.94 = 90.0% (based on your efficiency definition)

If P2 = 14.7 psia, V2=P1*V1/P2=1000*1/14.7=68 ft3, Work = (68 ft3 – 1 ft3) * 14.7 psi / (144 in^2/ft^2) = 6.84 ft*# PV Efficiency = 6.84/6.94 = 98.6% (based on your efficiency definition)

Actually, the efficiencies of both cases are 100%. All the available energy is recovered. Otherwise, could you explain where your 85.3% energy (100%-14.7% based on your calculation) goes? Please remember the first law to thermo (no energy is created or destroyed).

Hopefully, you know why you were wrong now.

Secondly, if you persist on your opinion 14.7% depressing efficiency, it is OK to me. Sure, it is a possiblility under certain conditions. However, I just warn you that your depressuring temperature is in the high side (0% is the highest temperature), which may result underestimate low temperature impact on your piping and equipment.

Finally, let me use two valves (now comparing apple to apple instead of a valve and a turbine that may create confusion) as an example to clarify the implication isentropic process and isenthalpic process.

Isentropic process: Gas depressures through an infinite large valve and contributes 100% work around (no friction and no heat transfer) so that gas temperature is lowest and contains the least H at the end.

Isenthalpic process: Gas depressures through an infinite small valve and contributes 0% work around (you may image 100% work turn into100% friction heat and 100% friction transfer back to the gas to raise the gas temperature) so that gas temperature is highest and contains the most H at the end.

Now somebody may argue that no work was seen in any depressuring process. Oh, my god, how we can jump out of this kind of loop questions. OK, let me borrow djack77494’s giant balloon and collect all relieved gas from the vent tip to expand the balloon (do work). I will ride the giant CH4 balloon to fly over head of your guys to show the work from lowdown gas and to ask your grant of success.

Regards,

Shan

#25 djack77494

djack77494

    Gold Member

  • ChE Plus Subscriber
  • 1,282 posts

Posted 10 September 2008 - 08:52 AM

QUOTE (shan @ Sep 8 2008, 05:39 AM) <{POST_SNAPBACK}>
The "success" is so attractive to me. Let me put some more efforts.


shan,
Thank you for your additional efforts, and please feel free to borrow my giant balloon.

To complete my understanding, addressing our example I suggest that the pressure temperature relationship of the vessel's contents is independent on the valve/turbine through which expansion occurs. Would that be correct? If so, then the efficiency of the expansion device ONLY affects the gas on its downstream side. We are on the verge of having the grant of success!




Similar Topics