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  1. #101
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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq



    Quote Originally Posted by Chef View Post
    I also make it 9% and we still have usefull work that the gas can do.

    So next how much energy do we have in the 3 streams. Lets start with the theoretical amount and then we can reduce it by an efficiency dependant on which method id used to gain that energy.

    Chef
    I will work out, using my chosen method, but will wait till others show their method (not a cop out). I do not want to influence how the following calculations are made. I feel that my method maybe flawed, even though the method does seem plausable to me?



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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    So I was right then, Jedi Master

    R's chillerman
    Last edited by chillerman2006; 24-08-2011 at 08:28 PM.
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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Chef,

    I hope you dont mind, a small contribution to make this a little easier for 'yous' to quantify



    Originally Posted by Chef
    Looking at a simple TXV system. Lets assume after the TXV we seperate the gas down one pipe and the liquid down another. The question is how do we assign the correct thermodynamic properties to each fluid when it gets to end of its pipe.

    The gas we want to do work on like heat it or compress it but before we can see where it ends up and predict its temperature and pressure we need to know from where we started? So do we move the gas out of the middle of the bell on a PH diagram to its edge and similarly do we move the liquid back to the saturated liquid line to gets its new properties?

    Simply jumping out of the bell interior means that the enthalpy and entropy both make huge changes without any process being done. Hmmm

    Your thoughts welcome



    Originally Posted by Chef
    So I split the gas and liquid but what h and s values are then given to each seperate flow. In a simple example lets have the evap at 1 bar and x=0.3
    The gas/liquid is at h=231 and s=1.13 but now rip away the gas and suddenly it has a value of h=384 and an s=1.75 but we have not done anything except seperate the flows and we see a huge change in properties?

    This means the gas/liquid together has one set of properties but when seperated they both have differant set of properties. What are the rules for reassigning the properties to each of the seperate streams.

    Simply I want to know where to plot the 2 seperate streams on a PH diagram.

    Chef





    Originally Posted by chillerman2006

    chef

    As far as plotting the cycle on a chart, your vapour is to the right, your saturated vapour is in the middle & the liquid is to the left

    but I cant see how this can be plotted without combining temperatures with pressure or why you would want to do this



    Originally Posted by mad fridgie
    Hi Chef, good to see you are looking into our little problem.

    As far as enthalpy goes, if the vapor is just that vapour (full seperation) then the enthalpy will be on the saturation point of the curve. But how is entropy effected.???????





    Originally Posted by mad fridgie
    The question is not based upon what you would expect!



    Originally Posted by mad fridgie
    Firstly read the first question, there is no mention of an evaporator.

    So without thinking any further than what I say next

    We have an open flash economiser vessel, liquid in, liquid out vapour out. what are the properties of all. Are there any strange things happening???



    Originally Posted by mad fridgie
    Yes seperation is an issue, but not the issue in hand (it must happen in an open flash vessel anyway) We are onlt talking about internal energy, (law of conservation)


    Originally Posted by mad fridgie
    high pressre liquid in (from cond) reduce pressure, what happens?



    Originally Posted by mad fridgie
    Not quite, apart from liquid head there is no sub cooling, and you are converting some liquid into vapour to keep the energy balanced. What then are actual properties of each exit stream. This is the question in hand?



    Originally Posted by mad fridgie
    You are cooling the liquid by vapourizing, the pressure has dropped, so the liquid is at saturation, hence is not sub cooled.


    Originally Posted by chillerman2006
    Liquid in - saturated liquid

    liquid out - liquid only

    vapour out - vapour only

    neither are superheated or subcooled

    & therefore

    to go back to the question from Chef - #5

    They would both be plotted on a ph chart along their respective lines at the point of saturation

    ???????????



    Originally Posted by mad fridgie
    Yes one would think so?????, unless someone can show something else??? and this is all chef was asking


    Originally Posted by desA
    The methodology for solution will very much depend on the two-phase model you select. Flow regime for two-phase flow can be rather interesting & the theory is far from settled.

    With different flow regimes, comes different levels of phase interaction.

    So, pick a suitable model, or models, predict the flow regime & determine the appropriate properties.

    I have a few very useful references to hand. If you pm me, with your current e-mail address, I'll get these onto you.





    Originally Posted by Chef
    Chillerman, Mad is right and the whole question is just about how to get the properties of the split streams. If I read the previous posts correctly it seems it should be possible to simply move the 2 streams to their respective lines on the bubble but does that meet the no work and no heat input criteria is met. So I propose this:- (which is way too complicated for a Sunday morning I should add)

    As no work or heat has been applied to either of the separated streams we can write the state function as

    dS=dQ-dW

    and the process as dh=dQ-dW+pdv+vdp

    As dQ the heat input is zero and dW the work done is zero the dS must be zero

    So x*S4-S3+S5-x*S5 = 0

    And as both p the pressure and v the volume are constant dh can be set to zero so there is no gain or loss of enthalpy

    So m*(h4-h3)-m*(1-x)*(h4-h5) = 0

    Where h3 is the enthalpy at a quality of x
    And h4 is the enthalpy at the gas line
    And h5 is the enthalpy at the liquid line
    Also S3, S4 and S5 are the entropy at the same locations.

    This boils down to the fact if these state functions are equal to zero then we can indeed split the gas out and position directly on the gas line whilst the liquid must then be plotted on the liquid line with the caveat that both streams must have their respective masses carried with them and new processes can then be started.

    But I am still mystified why the gas can have its entropy changed without work or heat input?

    Chef




    Originally Posted by Chef
    Your were right Mad and so we went to the sun deck and had a couple of Fiji Golds and the watched the world cup final and saw Brazil win 3 - 2. Exemplorary. Good input.

    So the last test seems to be that h5-h3-x(h5+h4)+2xh3 must tend to zero to make

    dh=d(S+pv) true.

    But for this to happen it makes it an isochoric process and that then means that the gas portion at the outlet of the TXV already has the properties of the gas at the gas line. It is therefore only the liquid that holds the core thermodynamic properties. ie it does not matter what you do with the gas as it cannot affect the liquid except for normalising the mass if one removes it. Interesting!

    How to calculate this on the fly and use it is a distant goal at the moment. But without it how can we go to stage 2.

    Chef





    Originally Posted by mad fridgie

    So it would seem that all we need to know at this point, is the % by mass of the lquid and vapour streams at the relevent saturation points?





    Originally Posted by Peter_1
    Just noticed now this post. I thought if Chef's asking a question and scrolling down, Mad Fridgie replies to it, then I have to read ever comma of this thread because both of them will dig very deep in the Thermodynamic laws again. Which I will do right now...reading, not digging.





    Originally Posted by desA
    Annular flow, slug flow, plug flow - all have vapour split from liquid. Take your pick.

    You will need to understand the flow regime at some point, I fear. Thermodynamics will have assumptions about the fluid under consideration. Your final system will be somewhere between two extreme positions, due to liquid-vapour interaction.





    Originally Posted by mad fridgie
    Would the flow regime relate directly to the exit of the expansion valve and its effects reduce if the bell (liquid vapour seperator) is a reduced velocity and impegement type. (basically allowing the heavy liquid to fall to the bottom and the lighter vapour to rise to the top)





    Originally Posted by mad fridgie
    I am certainly not at the level chef, DesA or Peter, just a different level. I am always learning from these guys,(thank you), If you are unsure of what is being said. ask!, and no doubt we will try to answer in a different termonology.





    Originally Posted by desA
    Basically separating & then forcing flow in a certain pattern. An interesting idea. Perhaps pipe diameter would play a useful part here?

    Keep the liquid on the bottom part, with vapour sliding along the top part. There will be a critical vapour velocity at which surface waves will occur in the liquid - trying to re-introduce another more complicated flow pattern.

    Keep below the critical velocity, or alter pipe diameter & we should be able to convey both separately. Thereafter, we can then treat Chef's problem from a thermodynamic point of view.





    Originally Posted by desA
    Ok, Chef. I've been re-reading your post.

    Would you want to re-combine the separate flows, or keep them separate? If separate, then the thermo should be a little easier to resolve, I'd think. If re-combined, then flow regime will impact the final solution.

    Can I suggest that we introduce some diagrams, then superimpose their thermodynamic states on the log(p)-h diagram?

    If the two fluids remain at same pressure, then their properties ?will? lie on the same pressure line on the log(p)-h diagram - no? Their individual states should surely then move to liquid sat line & vapour sat lines respectively?

    Interesting.





    Originally Posted by Peter_1
    To be honest, I explain this each year to my students. I take the fully expanded point on the refrigerant cycle where evaporation starts.
    For me, the enthalpy of the gas and the liquid are there the same. Enthalpy can be found in charts and saturated tables (ST)
    I then took the specific volume in the chart because you can't find those in the ST. For gas I took the value of both 'v' value on both sides of the saturated and multiplied this with the 'x' value, the lather found again in the chart.
    For the liquid same but multiplying with 1-x.

    DesA, isn't this a little bit the same conversation we once had via PM about something I didn't understand in the Dossat book Principles of Refrigeration?

    I'm struggling also with the fact that a chart is a chart for a standstill refrigerant. But in a real life cycle, the refrigerant is flowing and then Bernouilli is coming around the corner. Then my brain starts to twist because I never studied thermodynamics really into its deep aspects and therefore, I can't combine a static situation with a flowing situation.

    The compression process.. I had to study this the last in 1999 when I took my ATPL exams and we there had a complete course about jet engines, going really very deep into the theoretical side of a jet. I still remember I struggled with this one but anyhow, I made this one. There were +/- 130 out of 180 who had to go back a second time. I still have my exams about this course somewhere and if I re-read those now, it's far, so far away.

    Chef, perhaps we should ask this once to real authorities in this field like Cengel , Boles, White,...





    Originally Posted by desA
    Good memory, Peter.



    Would be lovely to invite them.





    Originally Posted by chillerman2006
    Evening Peter

    Heisenberg's Uncertainty Principle, is also rearing his head (for me in more ways than one)

    momentum = mass x velocity, the refrigerant has momentum p = mv, where 'p' is momentum, 'm' is mass and 'v' is velocity.

    'Mass' can be considered as the amount of 'stuff' an object is made of. 'Velocity' is the component of the particle's speed acting in a specific, defined direction.

    Basically the constant changing flow in a system, makes so many changes to your calculations (If I am understanding correctly so far) ??????


    R's chillerman





    Originally Posted by Chef
    Interetsing question and an even more interesting outcome.

    So the state equations seem to show the gas has no effect on the process of a 2 phase flow and therefore the corollary to this discussion so far is that the internal properties of say an evaporator can be described in fixed enthalpy values rather than a variable enthalpy.

    Let me try and explain. After the TXV the value of h3 is known but as we move along the evaporator and more liquid evaporates to a gas the value of h changes and to define the properties at a new location we need to re-evaluate h from say a PH diagram.

    But if we use the new equation

    x=[dQ/dy*m-h4+h5]/(h4-h5)

    we can now use fixed values of h namely those of the liquid line and gas line at the respective pressure. Now we can relate the fluid properties at any location with just heat input and do not need to revisit the PH diagram. The properties are defined by heat per unit length ie dQ/dy.

    That would make calculations rather simpler in a 2 phase evaporator.

    But this digresses from the original thread and just an interesting aside but should answer Mad's question? I think?

    Chef





    Originally Posted by Chef
    Mad – I am not sure that putting up widget #1 is your best idea. Better to solve the process issues which will then be public and other widgets may evolve along the way.

    Widget #2 is not even close to being resolved yet.

    Peter – very nice to see you here and I am not sure who Boles is but he sounds interesting and we await his thoughts.

    There seems to be a lot of stuff needs to be done to fully understand the bell and the consequences of messing with it can be fruitful or a failure. So far the out come as noted in my last post is to relate the properties to fixed h points and that is significant from a computational point of view.

    Chef





    Originally Posted by mad fridgie
    Hi chef, I did have to think hard about what i was going to give, but after the last few attempts at commercialization, on other ideas, (which were somewhat easier, to prove, build and test) i have decided to cut my losses, enjoy the ride. i have been in contact with one of the big multi-national component manufactures, who showed interest, only if i could front with a full working prototype and the theorectical calculations to back other selections. They would then look at the manufacture and marketting
    At this time more than i am willing and can afford to risk.
    To be truthful it is not an exiting enough product/idea for venture capalists





    Originally Posted by desA
    I'd like to introduce the basic laws of thermodynamics, to clear up the discussion a little:

    http://en.wikipedia.org/wiki/Thermodynamics



    So, the moment the vapour & liquid are 'physically separated e.g. by a wall, then they can be considered in isolation, with their own states. Once separated, thermodynamic theory would then instruct you to place the new state point at its pressure/temp/enthalpy/entropy value. In some cases, the point seems allowed to 'jump' to its new position.

    Perhaps we can now explore this via thermodynamic diagrams, Chef. (Very interesting thread. Making me think a fair bit... )







    Originally Posted by Peter_1
    This will clarify
    http://books.google.com/books/about/...d=5-hSAAAAMAAJ





    Originally Posted by Chef
    OK we are both on the exact same page and 'jumping' to a new position with new properties seems to be possible but there is no precedence or rules as of yet.

    The thermodynamic relations I proposed earlier seem to show this jump is supported for isobaric and isochoric processes but may only work for specific circumstances. What is the general form of the equations that allow us to jump to any part of the bell still keeping p and T constant?

    So what have you found for a process at the moment that may take us forward?

    Thanks

    Chef





    Originally Posted by Chef
    So here is a PH diagram of a 4 stage drop with phase seperation. To keep it simple I have started at h2 at the outlet of the condenser and ended up at h3-4 at the inlet to the evaporator. At each location after a stage drop the phases are split into their gas and liquid forms and repositioned on the PH diagram to relevant gas or liquid line.

    This is a very basic start at suggesting a process to discuss further and obvious errors include gas that has been removed at stage 1, 2 and 3 are cooled to the final evap temperature and the process should follow an isenthalpic drop but maybe we can include that at version 2 of the diagram.

    Now we have the same (actually more cooling effect) cooling as a one stage drop but several gas flows at intermedite pressures that can do work. Enter the Carter Widget.

    Any comments so far?

    ChefAttachment 7134





    Originally Posted by mad fridgie
    The question then begs, when we have our various vapour streams, at different pressures (which has very little practical use in the net cooling effect) is how we can utalise the ? (being vague on purpose) to increase the pressure at the compressor suction without restricting flow in the evaporator.
    What is the best possible theorectrical out come





    Originally Posted by Chef
    Well I presume a multi ring expander with each concentric ring sized correctly for the pressure input and flow rate to combine the flows and then use the suction provided to reduce the last section of flash gas and lower its pressure and temperature and hence get more cooling. then the final gas goes to the compressor and combines with the evap gas. Seems logical.





    Originally Posted by mad fridgie
    that is the problem, "it just seems logical", but somewhere in the back of mind, i feel the "Murphy effect" maybe present, the guy who trips you up at the end!

    Because it is so logical, surely someone else must of been down this track?





    Originally Posted by Chef
    Maybe you missed my point so I will try again. We have removed gas at the 8 bar line and so this gas is no longer cooled by flashing liquid, similarly the gas at 4 and 2 bar are seperated and they also need no further cooling. In a standard TXV system all the gas is cooled to the evap temp so we have saved some energy here. The question is - is it enough to help and then we have the side streams of gas.

    Got to replace the mainsail so the diagram will come later.

    Chef





    Originally Posted by mad fridgie
    I understand, each proportion that is removed no longer requires the sensible energy removed. to the next stage.
    Each stage then will have a greater amount of potential energy to be used for work. Whilst increasing the actual liquid mass flow to the evap (less total flash gas produced). Struggling in how to manufacture cheaply for small to medium systems, larger becomes much easier. (maybe this where it should be?)





    Originally Posted by Chef
    The savings in liquid flashing off to cool the gas is the first part. You seem to like that bit and then something like a deLaval nozzle, they can be multi ringed concentric and have nice thermo properties. Small system is a financial problem but medium would be OK. Large systems would definetly use a compander to extract the energy sucked out of the flash gas but that is known technology for decades in the cryogenic industry so not new by an means.

    To get the theory of this nailed is huge - really huge so may be a while till this thread gets a final answer unless someone pops up with a neat idea!!! Still I love a challenge.

    Chef





    Originally Posted by desA
    Ok, Chef. So you're looking for the 'jump' condition inherent within the equations - rather than simply taking for granted that the fluid 'has to' jump when split into vapour & liquid.

    Let me look a little more into the equations for the 'jump' condition. Never so easy, but it should surely be there to justify the change. (I love jump conditions...





    Originally Posted by Chef
    So a small experiment is due here I feel. I pour lighter fuel on my hand and as it evaporates it gets cold, if I blow on it it gets even colder as more evaporation takes place. So you are telling me that the gas after evaporation is cold and in the time available before it disperses in the atmosphere it has time to transfer it coldness to the liquid. Yeah right I beleive that. If you wish you could provide a useful referance to support your comments.

    In the PH diagram I posted there are 3 places where we propose to strip the gas away and pipe it off for its own process. The first split is at 8 bar which is 31.4C so the gas remains at that temp in its side pipe.

    If we had left that gas with the rest of the refrigerant as in a standard TXV system it would need to be cooled down to -26C which is the evap temp in this case.

    So we are going to need to cool that gas and the only source of 'cold' is the liquid hence my comment the liquid will cool the gas. The JT effect is not enough on its own. It is the phase transition that does the cooling.

    But you say the gas cools the liquid so by dropping some gas from 8 bar to 1 bar it will cool in a process that not not only cools itself but has enough (negative) energy to also cool some of the liquid. ????

    Chef





    Originally Posted by mad fridgie
    Actually, the liquid cools the liquid.
    Liquid at saturation, "no gas/vapour" so how can the gas cool he liquid.
    Part of the liquid stream absorbs energy for the remainder of the liquid stream, for the correct amount of energy to be absorbed, (to reach balance) then some of the liquid has to convert into vapour(gas). The gas/vapour is a "by product". of the process. "It is the conversion" of state
    I would say that the choice of word "cool" is incorrect.





    Originally Posted by mad fridgie
    If the vapour does not exist, then it can not do any work. Yes it is the transformation, the new formed vapour holds the energy. No energy is transfered to the vapour after the vapour is formed, so the vapour can not cool the liquid.





    Originally Posted by mad fridgie
    Yes the vapour hold the energy of transformation which gives the "impression" that the vapour must of cooled the liquid. For the vapour to cool the liquid, then for heat transfer to occur the liquid has to be warmer than the vapour (vapour has to be colder than the liquid).

    I supose to be factual; it is the change in pressure that cools the liquid by method of liquid to vapour transformation





    Originally Posted by Chef
    Good so we all agree the cooling effect comes from the latent heat of vapourisation.

    So back to the point how much energy do we save by not cooling the seperated gas flows?





    Originally Posted by mad fridgie
    On R134a 50SCT and -30SST no sub cooling

    3 takes off

    30C
    10C
    -10C

    between 9-10% extra net cooling for the same liquid mass flow





    Originally Posted by Chef
    I also make it 9% and we still have usefull work that the gas can do.

    So next how much energy do we have in the 3 streams. Lets start with the theoretical amount and then we can reduce it by an efficiency dependant on which method id used to gain that energy.

    Chef

    Originally Posted by Chef
    If you expand the gas at 8bar and 31C down to 4bar its temperature will end up being 26C. (Isentropic expansion)

    The saturated temperature at 4 bar is 9C.

    Somehow the expanded just is just not getting cold enough. It needs further cooling from the latent heat.

    You have flogged this dead horse so hard even the animal rights groups are looking for you. Bury the old nag in the garden and get a new pony and call it 'Latent Heat'.


    Originally Posted by mad fridgie
    I think this explains all, sometimes the numbers just do not lie.

    and for the humour, I like it!





    Originally Posted by Chef
    I have not gone quiet but at each stage I am coding it so just reset the input parameters and the answer drops out.

    It takes time code and check but so far just about keeping up.

    Of course I do have a distraction, well several actually as marinas are a great source of talent, sailing talent that is.

    I am going for a concentric de Laval if you want to know what the widget is.





    Originally Posted by mad fridgie
    I may have this completly wrong, "de lavel" does the widget only do work on itself!


    Originally posted by desA
    I have developed some food-for-thought, for the brave, based on flow & energy principles (I trust there are no silly typing errors - please advise if so found):



    Attached Files Attached Files
    Last edited by chillerman2006; 25-08-2011 at 05:58 PM.
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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    If heat is added to vapor in compressing it, then surely heat is removed from vapor in de-compressing it (stepping down it's pressure)... in which case there could be sensible heat transfer... the fully formed reduced heat content vapor in fact cooling the liquid.

    How's that for beating the dead horse?

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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Quote Originally Posted by Gary View Post
    If heat is added to vapor in compressing it, then surely heat is removed from vapor in de-compressing it (stepping down it's pressure)... in which case there could be sensible heat transfer... the fully formed reduced heat content vapor in fact cooling the liquid.

    How's that for beating the dead horse?
    Argh Gary,
    I see you have been to MF's Training school too, who says you cant teach an old dog new tricks
    You keep kicking a dog & it will bite, Basically I can either keep trying to learn or can spend all my time, snapping at the heels, easy either way !
    Last edited by chillerman2006; 25-08-2011 at 04:58 PM.
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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Quote Originally Posted by Gary View Post
    If heat is added to vapor in compressing it, then surely heat is removed from vapor in de-compressing it (stepping down it's pressure)... in which case there could be sensible heat transfer... the fully formed reduced heat content vapor in fact cooling the liquid.

    How's that for beating the dead horse?
    If you expand the gas at 8bar and 31C down to 4bar its temperature will end up being 26C. (Isentropic expansion)

    The saturated temperature at 4 bar is 9C.

    Somehow the expanded just is just not getting cold enough. It needs further cooling from the latent heat.

    You have flogged this dead horse so hard even the animal rights groups are looking for you. Bury the old nag in the garden and get a new pony and call it 'Latent Heat'.

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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Quote Originally Posted by Gary View Post
    If heat is added to vapor in compressing it, then surely heat is removed from vapor in de-compressing it (stepping down it's pressure)... in which case there could be sensible heat transfer... the fully formed reduced heat content vapor in fact cooling the liquid.

    How's that for beating the dead horse?
    I think the hourse is not just dead, it is rottern.

    For your statement to be correct, the vapour has to be formed first at the high pressure then expanded, to cool the liquid. If we could imagine that this process took a larger time scale, the you would have some flash gas to start with a slightly higher pressure, as this is dropped, then it would actually cool the new vapour being produced not the liquid. But this just semantics.

    gas expanding does require energy "joule thompson"

    You could say a transcritical refrigeration system uses this principle.
    Last edited by mad fridgie; 24-08-2011 at 10:42 PM.

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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Gents,

    The step down in temperature you's propose from h2 through h3-1, h3-2, h3-3 to h3-4 (chefs ph chart)

    How is this being achieved ???

    Does your proposed vessel have multi-stages & as you remove vapour at each stage your lowering pressure each time, lowering the liquid temp ???

    I am trying to follow closely & quietly, just I am getting a bit confused

    Also what is the correct way to write the part in bold ???

    R's chillerman
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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Quote Originally Posted by Chef View Post
    If you expand the gas at 8bar and 31C down to 4bar its temperature will end up being 26C. (Isentropic expansion)

    The saturated temperature at 4 bar is 9C.

    Somehow the expanded just is just not getting cold enough. It needs further cooling from the latent heat.

    You have flogged this dead horse so hard even the animal rights groups are looking for you. Bury the old nag in the garden and get a new pony and call it 'Latent Heat'.
    I think this explains all, sometimes the numbers just do not lie.

    and for the humour, I like it!

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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Quote Originally Posted by chillerman2006 View Post
    Gents,

    The step down in temperature you's propose from h2 through h3-1, h3-2, h3-3 to h3-4 (chefs ph chart)

    How is this being achieved ???

    Does your proposed vessel have multi-stages & as you remove vapour at each stage your lowering pressure each time, lowering the liquid temp ???

    I am trying to follow closely & quietly, just I am getting a bit confused

    Also what is the correct way to write the part in bold ???

    R's chillerman
    At this stage in the thought pattern, just thing of it as 3 vessels each with there own control.

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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    This Quasistatic process really makes this whole process quite simplified

    http://www.youtube.com/watch?v=WLKEVfLFau4&NR=1

    Enjoy, chillerman
    Last edited by chillerman2006; 25-08-2011 at 03:07 AM.
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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Last edited by chillerman2006; 25-08-2011 at 02:54 AM.
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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Right I am nearly there,

    I can see what happens on the ph chart when I strip away the vapour

    & return to the saturated liquid curve of the bell

    My calculations are

    https://docs.google.com/viewer?a=v&p...4OThh&hl=en_GB

    h3-1 = 8.0bar @ 31c
    h3-2 = 4.0bar @ 10c
    h3-3 = 1.9bar @ -9c

    R's chillerman
    Last edited by chillerman2006; 25-08-2011 at 03:37 AM.
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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Nice Video, Explains as all above !

    http://www.youtube.com/watch?v=Xb05CaG7TsQ

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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Quote Originally Posted by Chef View Post
    If you expand the gas at 8bar and 31C down to 4bar its temperature will end up being 26C. (Isentropic expansion)

    The saturated temperature at 4 bar is 9C.

    Somehow the expanded just is just not getting cold enough. It needs further cooling from the latent heat.

    .
    http://www.youtube.com/watch?v=iiFWoXQPOJc&NR=1
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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Chillerman - many thanks indeed for the synopsis of the thread as it strips away nicely the irrelevant stuff and keeps the thrust flowing well, certainly makes the whole thing much more understansable. You must have a really fast connection to get all that stuff collated. Well done.

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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Quote Originally Posted by Gary View Post
    If heat is added to vapor in compressing it, then surely heat is removed from vapor in de-compressing it (stepping down it's pressure)... in which case there could be sensible heat transfer... the fully formed reduced heat content vapor in fact cooling the liquid.

    How's that for beating the dead horse?
    https://rapidshare.com/files/3020151...yles_s_law.gif
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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Quote Originally Posted by Chef View Post
    I also make it 9% and we still have usefull work that the gas can do.

    So next how much energy do we have in the 3 streams. Lets start with the theoretical amount and then we can reduce it by an efficiency dependant on which method id used to gain that energy.

    Chef
    All gone quiet on this question.
    I am going to change the question slightly, "what is the potential energy available for work in the 3 vapour streams"

    A bit of assistance to get you on your way.

    First stage 50C liquid to 30C converted approx 17.5% of the original liquid flow into vapour, the vapour has "h" of 413.5 Kj/kg.

    Remember that the mass (amount in weight) of liquid in each stage, is reduced by the % of vapour that is being removed.

    hope that gets you thinking?
    Last edited by mad fridgie; 25-08-2011 at 05:20 AM.

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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    I have not gone quiet but at each stage I am coding it so just reset the input parameters and the answer drops out.

    It takes time code and check but so far just about keeping up.

    Of course I do have a distraction, well several actually as marinas are a great source of talent, sailing talent that is.

    I am going for a concentric de Laval if you want to know what the widget is.

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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    I may have this completly wrong, "de lavel" does the widget only do work on itself!

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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    got a pic of a "concentric de laval". It's all greek to me.

    I wish I had the math for this. most of what I do is by understanding concepts and intuition.

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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    No worries, found it. I thought it would be something from the dark reaches of engineering but it is pretty common.

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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    I have developed some food-for-thought, for the brave, based on flow & energy principles (I trust there are no silly typing errors - please advise if so found):



    Last edited by desA; 25-08-2011 at 04:06 PM.
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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Hi DesA,

    Too much bravery gets you shot, & I am not ready to become cannon fodder,

    There will be less takers to this than Chef's first post ! Unless there are some mercenaries amongst the members

    R's chillerman
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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Gents

    I have found a widget type thingamegig, is this like your proposal

    R's chillerman
    Last edited by chillerman2006; 26-08-2011 at 02:10 AM.
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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Quote Originally Posted by chillerman2006 View Post
    Gents & Mr G

    I have found a widget type thingamegig, is this like your proposal

    R's chillerman
    Nicely done. Maybe you should change your avatar to the Pink Panther.

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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Quote Originally Posted by desA View Post
    I have developed some food-for-thought, for the brave, based on flow & energy principles (I trust there are no silly typing errors - please advise if so found):



    Somehow the url's were missing on my screen but there in the 'reply with quote'?

    I will study it carefully - thanks

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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Des

    I agree and your method was differant to mine but just as applicable and your equation 4

    m1*(h2v-h1)=m2l*(h2v-h2l)

    is the same as i had but you have used a differant format. Mine was an equality that needed to tend to zero to make it true and so a proof and yours is the final conditions so I like that as it becomes useable in future calculations.

    Using your method we can easily see how we can go through 4 expansion and seperation processes and easily calculate the end result. I see you had to multiply both sides by -1 so the enthalpies were more understandable and I assume that is purely from a definition of W being either positive or negative but either solution is correct and so makes little differance.

    I also agree gZ is small enough to be ignored and C can be treated similarly as it would be low in practical cases.

    It is an elogant solution and very nicely presented and glad to see your on the case. Many thanks indeed.

    Using Equ 4 I will work up a value for the 4 stage process and see what it looks like.

    Chef
    Last edited by Chef; 26-08-2011 at 02:20 AM.

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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Hi Des and chef,

    even though I think I understand the final equation, I am no longer up with mathmatical and scientific termonolgy.

    can we please have legend!

    I presume the little dots mean (square) and multiply

    I can not see that the -1 calcultation, or is this just assumed and a known mathmatical method

    sorry, but if i do not ask, i can not make an opinion.

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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    The dot over the m means mass flow rate and the dot between items means multiply.

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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Thanks for your comments, Chef. An absolute pleasure to contribute.
    Engineering Specialist - Cuprobraze, Nocolok, CD Technology
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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    You guys astound me. I have no idea how you find the time unless Chef is in port all the time or or he has auto-furl and powered winches and a good auto-helm.

    CM, don't you need to work and is your wife starved for attention.

    I can't keep up but I will try. Cheers

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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Quote Originally Posted by MikeHolm View Post
    You guys astound me. I have no idea how you find the time unless Chef is in port all the time or or he has auto-furl and powered winches and a good auto-helm.
    CM, don't you need to work and is your wife starved for attention.
    I can't keep up but I will try. Cheers
    Mike - Only been in a Marina 3 times in the last 20 years and no longer than a few weeks and no auto furl or power winches I lament, they are powered by cheese sandwiches and granola bars but keeps you fit I suppose. Just think about it as an office with a changing scenic view.

    Chef

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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Here are some results condensed a bit.

    If we use a standard cycle and drop from 13.3 bar down to 1 bar with a standard TXV and assume 1 for the mass flow we get a cooling effect of 111.6 Kj

    If we go through a 4 stage let down as shown on the PH diagram posted earlier and useing the formulae from DesA then we have a cooling effect of 120.6 Kj
    Each of the 4 stages was anaylsed and the results added together.

    This extra cooling effect comes from the fact the gas is seperated and does not require any further cooling. So now we have 3 streams of seperated gas at 8, 4 and 2 bar and these streams will contain the missing energy so there is 9Kj available if we can use the gas efficiently. Even if we only get 50% usage then the total energy added to the cycle will be 13.5Kj

    Chef

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    Re: How to assign thermodynamic properties after splitting two phase into gas and liq

    Very interesting results & feedback, Chef. Thank you.
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