How To Maintain Your Plural Component Spray System

Troubleshooting for a Graco Reactor 2 Integrated Elite E-XP2i ...
Plural-component spray equipment can be utilized for using spray polyurethane foam (SPF), polyurea, urethanes, and epoxies. If you supply spray used services, it is necessary to do every little thing you can to maintain your plural-component devices and sprayers in good condition whatsoever times to reduce downtime and upkeep costs. Follow these 5 standards to obtain one of the most out of your tools with every passing year.
1. Examine Your Power Supply Daily
Daily, examine the voltage on your device( s). Be sure to examine coolant and oil levels also. Beginning the engine up as well as allowing it run for around five minutes will help you identify any other concerns you might not notice once you leave the generator.
Power Supply | Pinnacle West
When beginning the engine, turn on the main circuit breaker and inspect the radiator as well as compressor colder for debris. If there is debris, utilize pressed air to cleanse the parts. Evaluate the engine, compressor, as well as radiator for leaks, as well as the fuel and also coolant tubes.

Make certain to inspect the hr meter for any kind of future service( s) that might be due. You can locate the service schedule for oil, fuel, and also air filter modifications in your tools manuals. This will certainly guarantee every little thing is functioning as effectively as possible, and also it decreases the danger of damage and also downtime down the line.

A couple of basic checks daily can save you a great deal of cash on repair work in the future, and also it virtually ensures much less stress and anxiety from economic losses due to being down on the work.


2. Care for Your Hoses
Delivering more to your bottom line - New Graco Series 2 E-XP2 ...Hoses are responsible for transporting plural-component products to where they need to go. As a result, maintaining them in superior conditions is crucial. To prevent your plural-component chemicals from freezing in cold weather, always bring your pipes right into the rig and hang them correctly on a hose rack. Be sure to maintain them kept and also change scuff jackets as needed. An alternative to typical scuff is making use of shielded or non-insulated items. These form-fitting items are made with a thick nylon that make them immune to tearing and also other environmental factors, while making use of Velcro to maintain them in position and also kind to the hose pipe.

When changing in between open- as well as closed-cell products, ensure you purge your B side with water so you don't infect and also lock up your whole system and the pipes, which can be pricey to repair. While you are not spraying, or if you are placing your gear away for lasting storage, be sure to flush both the An and B sides with an authorized solvent. Refraining so can force you to squander a lot of time on the following task trying to get every little thing prepped as well as prepared to go. If you have any type of questions, consult your chemical supplier or a tools professional on flushing processes, and also, as constantly, make sure to adhere to appropriate safety and security method when taking care of the products.

3. Purchase Annual Electric Motor Maintenance
If a motor is not in good working condition, your equipment can refrain from doing a lot. Having it serviced a minimum of when annually will certainly make a massive influence on your procedures. An expert in the field can take a detailed consider your system as well as identify whether new parts or repair work is needed. Be sure to utilize a devices technician who is qualified to deal with your brand name of device in order to guarantee appropriate upkeep.

Preventive maintenance checklist for centrifugal pumpsFor power sources, make sure to take your unit to a qualified solution center of the producer to ensure that all service warranties are promoted. You can go to the internet site for the producer to discover service facilities, and also your initial documentation might detail the terms of your warranty. This is a far better choice than waiting till your spray tools fails.

4. Maintain Spare Parts handy
The best option to prevent longer downtimes, in case a component of your spray machine is not functioning, is to maintain extra components available. This is a simple and also budget-friendly action to take, and also it is a much better alternative than getting what you believe to be the issue at the time of failure.

Your spray gun is the component in your spray system that is more than likely to have problems. Its internal components break from usage more quickly than any other component of the spray system. Taking care of your spray gun is a bit extra tedious, but it will certainly profit your work and also company. After every job, it's important to get rid of any excess chemical from within the gun to stop build-up and clogging. It's an excellent concept to break down the weapon every now and then and also soak all the pieces in a cleansing solvent, thus removing any debris as well as cleaning the components. When rebuilding the gun, make sure to grease all the O-rings to stop them from breaking down quicker.

If you are not keeping your spray gun daily after splashing, you can end up with a great deal of issues that can cause you downtime as well as economic loss, so you want to be sure to maintain usual spare components on your spray gear in case of gun failure. Buying a back-up weapon in case the parts required are out hand is likewise a perfect means to reduce downtime on a job and also minimize troubles-- particularly if you are not currently changing the components as needed when they are failing. Transfer pump parts are other excellent extra components to continue your spray rig to ensure your system is always running.

5. Follow Service and Upkeep Schedules
Many layers specialists can agree that, as a result of the nature of their business, regular maintenance on their spray gear is typically overlooked. Lengthy days and requiring timetables can leave specialists having little to no time at all to execute regular and also preventative maintenance on their devices. Preventive and also scheduled upkeep on all tools is key to maintaining the components of your spray system running longer and without failure.

310666H Xtreme Mix 185 Repair, Plural Component Proportioner, U.S. ...
Describe each equipment supplier's standards for maintenance schedules. Tools suppliers and service technicians can also give you with precautionary maintenance to assist keep your system up and running. You ought to always describe the producers' routines for upkeep, but the complying with sample timetable will certainly give you a suggestion of what to expect:


Compressor

Daily: Examine oil levels, clean air/oil external radiator, tidy cooler filter, and clean air filter.

1,000 hours: See guidebook.Air DryerDaily: Inspect that auto drainpipe is discharging.Regular monthly: Clean particles from ambient air filter, tidy with soap and water, as well as permit to dry before reinstalling.Annual: Change separator filter component.Spray Foam MakerDaily: Examine pump lubrication degree. Fill up as needed with throat seal fluid.Weekly: Grease circulation valves with suitable oil. Tidy particles out of electrical cupboard vent holes.Spray GunDaily: Grease weapon with ideal grease, drill blending chamber, and examine shutoff displays.As needed: Tidy blending chamber nozzle by involving piston security lock, usage right size drill to clean nozzle, make use of stiff bristle brush to tidy air cap, as well as clean liquid manifold, impingement components, and/or passages.Fresh Air System500 hours: Change inline outlet filter and inlet filter.DesiccantDaily: Look for shade change from blue to pink and also replace as required.




Verdict
By doing daily checks, by complying with manufacturers' timetables for servicing your source of power part and compressor, by caring for your hoses, as well as by maintaining spare parts accessible, you'll have the ability to keep your plural-component sprayer at its optimum efficiency. If you ever have concerns regarding the upkeep to be carried out weekly as well as yearly, connect to your tools provider, who can give you guidelines for each and every element.
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Thermodynamics, ideal solutions, entropy











There is an old adage, I don't know who came up with this, that the first time you learn thermodynamics, you don't understand it and the second time you learn thermodynamics, you think you understand it and the third time you learn thermodynamics, you're comfortable with not understanding it And I think that's the category where most professional scientists find themselves, because it is quite a complex topic, but we can understand what we we can understand the basics using a few simple ideas. What is what is the the quantity that must be minimized at equilibrium in any chemical process? Gibbs, free, energy G, so G is the. What is what is G? What is the Gibbs free energy Gibbs? Free energy is the thermodynamic potential that must be minimized at constant temperature and pressure in a chemical system. Okay, that tells us precisely nothing about what it actually is. You may have in first semester, physics or chemistry derived the expression for G and what you ultimately end up with is G which, just as a reminder, is at constant temperature and pressure equals something called H. Minus something called PS H is the enthalpy T. Is the temperature and s is the entropy? Where does or does H come from? Where do the HS come from? Where does it or does H? Come H is heat, so it's the heat of reaction or the heat that results from a particular chemical process. But where does it really come from bonds between atoms bonds between atoms? How about how about intermolecular forces as well? So if you have two molecules that are far away from each other in space and they're not charged and you bring them closer to each other, the and they're not they're not charged, or they have opposite charges. The potential energy over here compared to over here is the potential energy higher in one or two potential energy is higher in one. So we minimize the potential energy by bringing two uncharged molecules together by Vander Waals forces and by conservation of energy. Something has to have to be ejected, and that is Heat, incidentally, do uncharged molecules, attract or all uncharged molecules attract in a vacuum. Is that a true statement, so oil and oil they do they attract in a vacuum there. The only molecules in the universe do they attract; they do how about water and water. Do they attract? Yes, they do. How about oil and water do they attract if it's only one molecule of water and one molecule of oil they do attract, they do attract in a vacuum because they both have protons and electrons, and one has a permanent dipole moment which can induce a dipole moment In the other, they also have electron clouds that transiently deflect each other and create Vander, Waals bonds, which is called the dispersion or London interaction. So the answer is that all uncharged molecules attract pairwise in a vacuum, but a chemical system is not always pairwise or basically never pairwise, because you have lots of different molecules interacting in some environment, so, incidentally, like charged molecules or molecules that are constrained so that the Negative end, so that the same side of a dipole has to come closer if they're not allowed to rotate. That'S that's! That'S that's unfavorable and Delta Klee. So that gives you a positive H, but for uncharged or freely rotating dipoles Delta, H is always is always negative. You get it give off heat because the potential energy is being lowered in that interaction. So we know so so. If G at constant temperature and pressure equals H, minus TS for a h, minus tough for any given process, then Delta G for a process. The change in Gibbs free energy is Delta, H, minus T Delta s. Now. What we're going to be predicting in the next couple of classes is whether or not certain pairs of molecules so solvents when they come together as mixing favorable or unfavorable or polymers, when they come together as mixing favorable or unfavorable, and what we? What we get has a bearing on the Faye's behavior of a solid polymer material, which controls all of its thermal and mechanical properties that we're trying to engineer in a solid material. So everything comes down to to minimizing the the Gibbs free energy through these parameters. Delta. H and Delta s, so H is fundamentally derived from from from bonds, and I mean bonds in a very general sense. You have a chemist's bond, which is a covalent bond which involves formal sharing of electrons between atoms in a molecule, and we also have the physicists bond, which is like a van der Waals bond, which is just an electrostatic interaction that it that comprises components from dipole-dipole Interactions, dipole induced dipole interactions and dispersion interactions, so this is chemical bonds or covalent bonds and also van der Waals bonds and also will say, ionic bonds. These two are really electrostatic in nature. This one is a little bit more quantum mechanical in nature, because you have these atomic orbitals that hybridize to form molecular orbitals, but in terms of their effect on Delta, H, they're, similar Delta s Delta s can mean a lot of different it Delta s comprises a Wide range of entropic terms, but the one we're mostly we mostly care about the only one that we care about in in at least our discussion of this topic in the abstract is configurational entropy, and that is we picture the molecules as spheres as occupying points. On a lattice, the circles or spheres and two-dimensional to three-dimensional spaces, but they have no, they have total rotational symmetry they're, not they're, not rods, because if you have rods, you had a review of rotational degrees of freedom as well. We'Re only talking about configurational entropy. So, where are the the these circles or spheres on a lattice and are there more states available before a chemical process or after a chemical process? Yes, this simplifications. We can express them all in three characters rather than I, if, if you, if you want to in a real system, you have to break Delta s apart into the configurational term and the rotational term and other terms that might that might arise. So this is the change in the number of statistical microstates available and for our discussion this week and for the rest of the class except under specific circumstances. So the mathematical treatment will all be for configurational entropy. Only we're not going to worry about about rotation and a rotational entropy for now or mathematically we're not going to worry about it at all, but I may give you examples of cases where rotational entropy plays a plays. A significant role number is the statistical microstates available and this is configuration only in in our discussion. Okay, so let's modify this equation a little bit with a superscript M which means mixing because ultimately, if we want to learn something about the phase behavior of and the dissolution of polymers, in a solution to do solution, processing or polymers in polymers to do melt casting Or to predict that the phases that we ultimately get, what we care about is mixing, so the unmixed state before and the mixed state is after yes, yes, they can be measured, they can be measured, color, calorimetric, ly, Delta H's is the easiest because, because it just Just comes off as as heat, and you can measure the other quantities by various curve, fitting that we're not going to talk about in the class, but they are measurable. Okay, so Delta s of Delta s for a mixing process configuration only. So you have two separate species over here and then in the mixed state. They'Re jumbled up is Delta s positive or negative Delta. S is going to be positive for the configurational change in entropy upon mixing every time. Yes, the there are. So why do unlike solvents? Sometimes not mix so, if Delta, we know that for a spontaneous process, so something that that lowers the Delta G we have. If we have a positive Delta s term, then this term is all negative and contributes toward making this process happen because it gives us a negative Delta G. So why would say say: cyclohexane and a seed, a nitrile, don't mix sorry, so we're taking the mixed state. We'Re just saying that there is a mixed state, even though they don't mix the mixed state has all the molecules. Jumbled together, we can all agree that there are way more distinguishable microstates available when they're all mixed together versus when they're all when they're separate in like oil and water. The reason they don't mix is because delta h of mixing must be unfavorable, but i just told you that uncharged molecules attract in a vacuum and have a negative Delta H. So how is the enthalpy of mixing unfavorable yeah yeah? That'S halfway there. So you've all heard the phrase life dissolves like that. That gets you part of the way there, but it's still kind of wrong. The reason it's still kind of wrong is because a a molecule with a relatively low Vander, Waals coefficient. So because we have all these dipole terms in the Osito nitrile, it has a very high Vander volts coefficient. It likes interacting with other things and because you have two things: if you have two molecules of this, that both have a high vander waals coefficient, then that's a very favorable interaction. This - and this also has a favorable interaction, but it's, but it's less favorable than this end this. So, while it's really because the heceta nitrile is excluding the cyclohexane that they separate, because it's more favorable for these, these molecules to continue interacting with itself rather than mixing cyclohexane, as if cyclohexane had a brain and it had like crushes on things, it would say I Really wish to see de nitrile would like me because it has a really high vander waals coefficient and I would really like to interact with it. But heceta nitrile says no. I, like my own friends so so the result is not that like dissolves like, but that more the one of the greater intermolecular forces excludes the one with the with the less favorable intermolecular forces. That'S quite a simplified picture because there are some examples like the hydrophobic effect, for example, which has a best L on water, what we always think about with immiscible solvents. But in fact the Delta H is slightly favorable upon mixing. But it's actually entropically quite unfavorable to mix because of rotational terms that again we're not going to talk about, because if you we're not going to talk about mathematically I'll, tell you why the hydrophobic effect happens because you get solvent cages of of hydrogen bonds that are Tangential to dissolved hydrophobic species in water and that restricts the rotational freedom of water molecules that have to soluble eyes the the organic species and therefore, even though the Delta H is, is even a hair favorable, the entropy is very unfavorable, so this is. This is really our analysis depends on the fact that we're considering only configurational, entropy, okay, so like solvents, tend to to two like each other, like chemical species tend to dissolve in in each other. So let's take, let's take Applause ]. This is called toluene and this is styrene vinyl benzene. So would we predict, based on the argument that similar molecules tend to be soluble in each other, that these would be soluble in each other? They would form a miscible mixture, yeah yeah. They definitely would but polystyrene, where you have a long zigzag with a bunch of benzene rings that hang off of it. If you have really high molecular weight polyester in does it become less soluble or more soluble in the in the toluene less valuable and that that's a little bit intuitive from our everyday experience, because we have you know we have high molecular weight polyethylene that won't dissolve In solvents, because we have solvent bottles that we have in high molecular weight, polyethylene and but the reason is, is a little bit know counterintuitive! It'S not it's not that easy to understand just by just by you know, thinking because the toluene and the styrene, so the monomer is soluble. But if you have a certain number of repeat units, it becomes more and more insoluble, the more repeat units you add and even though the enthalpy of mixing isn't going to change to too much because you still have this interacting with this, which is pretty similar. The entropy term, the change in entropy, could be comes much and much much much much less, because what you're doing is by making a polymer you're confining the monomers you're telling them that they have to be next to each other. They can't be anywhere. They have to be next to each other, so the entropy of mixing is much less favored. So if you have infinite molecular weight, the entropy of mixing goes to goes to zero, and even the slightly the slightly unfavorable enthalpy of mixing takes over and makes them insoluble. In each other yep, you see an increased effect of that and continuously polymers, because there's so the question is: do you see more mixing in isotactic and syndiotactic of like polypropylene or polystyrene, because they're so chemically similar? You would certainly predict more mixing in that case, based on based on these arguments, but the. But if you take into account crystallization and the fact that one wants to crystallize and the other doesn't then you would predict a different kind of phase segregation that resulted from crystallization. But that's a good. That'S a good question: okay, okay, let's think about polymers in solution - and these are just a couple of sort of high-level points that I'm going to write down and by the end of the by the end of next week, we'll kind of understand why why this is In a good solvent - and this is favorable intermolecular forces - can I abbreviate that IMS: favorable intermolecular forces a polymer coil expands in order to increase interactions with the solvent relative to in a poor, solvent it contracts. If you see a homogeneous solution - and this is of a polymer or molecules taken together, what you can say about it, if you see some sprite, you know that that it's it's Delta G of mixing is less than zero and we got Delta G by considering the Delta G M equals G of of 1

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POLYMER SOLUTIONS: Flory-Huggins Theory










Welcome to this movie on statistical mechanics, my name is your station and I teach statistical mechanics at the Delft University of Technology in the Netherlands. This is the third movie in a series of three on elementary polymer and solution physics, and in particular, this movie deals with polymers in solution and the theory which describes those systems is the so called flory-huggins theory. The first movie, which I made dealt with free chains where the chain doesn't have any interaction. The beats of the chain do not have any interaction. The second one was a movie on polymers, where the beads cannot overlap, and then you get the so-called Florrie exponents. You have a self avoiding chain, and in this movie we are treating a much more complicated problem. Those are chains which cannot overlap, so the beats of change do not want to overlap, but in addition, there is a solution present and then you can have a search that either the polymers collapse into one part of the volume where they form a dense, polymer melt Or they will really dissolve in the solution and that's the problem I want to address in this movie, and I hope that you enjoyed this movie and learned something about it. On flory-huggins theory, in the flory-huggins theory for polymer solutions, we start with polymers, whose beats cannot overlap. That'S similar to the floor, a mean field theory where we had also beats that could not overlap. They had a hard core repulsion. They were moving in the continuum space here we put the polymers on a lattice and the solvent atoms of the solvent molecules are also put on lattice sites and on each letter site that can either be a polymer bead or there can be a solvent molecule, and We can envisage that in that case you have the following two situations: on the left-hand side, we see polymers, they are the colored sites. The color here only serves to distinguish to differ chains, and we have also great circles white insight and they are the solvent molecules and we see that of maalik solvent and the polymer they mix, and so this is a real solution which is characterized by a high Entropy, so we can say that here the entropy is high, it's certainly higher if we compare it to the right hand, side where we see that face has separated. So we have a region in space where the polymers all go, so they form a dense melt and we have, in addition, a solvent which separates from the polymers, and so this is called a separated phase. The polymers here in they form a melt and, on the left hand, side. We have a solution, although we can make a definite statement about the entropy which is higher on the left hand case in the solution than on the right hand, side in a separate two rated face. It'S difficult to tell what the energy in the two cases will be, because that depends on how much the sites of the polymer and the beats of the polymers do like each other and how much they like the solution and also on how much the molecules inside The solution like each other, so we have interaction parameters that we will come back to in the end, describing these three different energies of polymer, polymer, polymer, solvent and solvent, solvent interactions and, depending on the values that these parameters have, the energy can be higher or lower. On the right hand, side and, on the left hand, side. The face which dominates is the one which has the lowest free energy and therefore we need indeed to evaluate the entropy for the two phases. And we need to evaluate the energy. And we start by the entropy, but before we do that, we need to define several quantities that we will use in the calculation and the first one is, and sites and sites is just a number of sites in the system, so it's equivalent to the volume. The second number, which is important, is n and n is the number of beats for each polymer. The next number is n, P and n P is the number of polymers we have in the solution or in the melt. So the number of polymer beats in the system is just n times and P. Finally, we have the number Q and that's the lattice coordination number. That'S the number of nearest neighbor sites for each sites on the lattice. So now we start calculating the entropy by imagining that we build up the system polymer before polymer and we will use some mean field arguments in doing that before we continue. It should be pointed out that the grid that we have on here is a two-dimensional grid, but it could also be three dimensional or one dimensional or four dimensional. Whatever we now evaluate how many possibilities we have when we built up the J's polymer of the system, so we assume that there are already J minus one polymers in the system and we are going to build up now. The j'the polymer and we start with the first site for that first beat of the polymer. We can choose any site which is not yet occupied by the J minus one previous polymers. So in total we have how many sites at our disposal we had tons n sites in total and of these J minus 1 times the number of sites per polymer and that was n has been occupied by other polymers by other beats, and therefore we have this. Many probably possibilities to put our first beat for our second beat. We have in principle, we can choose any of the q neighbors of the previous one, but maybe that neighbor is already occupied by one of the previous polymers. So we have to multiply this by a probability that the site is unoccupied and for this probability we use a mean field assumption we just take Q times, then we just take the sites that are still unoccupied. So that's n sides minus J, minus 1 times n minus 1 because we have already placed the first beat and we have to divide that by n sites. This is a mean-field assumption, because we do not. We assume that the average probability of occupation is independent. Of that of the fact that we have placed a first beat here, so it's homogeneous everywhere in the lattice. That assumption is the mean field assumption in order to see how many possibilities we have for the third beat. Let us take first first beat and then we could put a second beat either here or there or there or there. If you are on a square lattice, so suppose we have put it here and then for the second beat we can again in principle. It has four neighbors, so Q is four in this case of a square lattice in the plane, but one has already been occupied. So instead of Q neighbors, we only have Q minus one neighbors. We can where we can place the next site, and we should multiply again by this probability that that site is unoccupied. It could be that some other polymer which started here as occupied, for example this or that or that site, and so what do we obtain? We have aq minus 1 times n sites, minus AJ, minus 1, times N, and now we have minus 2 because we have placed already the second bead. So there's one less site free and we carry on in the same fashion. So for the cave beat you have. The same expression as for the third beat very similar, except for the minus two which we had for the third beat. We have a minus k plus 1 for the cave beat and he carry on until we have exhausted all the N beats of Palawan and the J all in all for polymer J. In order to find the number of possibilities we have for building up that polymer, we should multiply all these numbers, so this number times that number etc etc, and we obtain the following expression so recall that for the first site we just had all the as many Possibilities as there were empty sites for the second one, we could take Q neighbors, and then we should multiply that by the probability that they are unoccupied. That'S this factor, then, for the next one. We have only Q minus 1 because the previous beat was already occupied and we have to multiply that by the unoccupied, probability, P unoccupied, and we carry on like that until we arrive at beat number n now note. When an enumerator, we have a product of terms which decrease. So here we have a certain number. This is that same number, minus 1, minus 2, minus 3, etc. Until we arrive at the last number - and there is a compact way of representing that - namely by the fraction of two factorials - and what we obtain then in the end is that we have n sides minus J, minus 1 times n factorial. So that's what we would have if this would continue all the way down to one, but that's not the case. So we have to divide out all the terms that are smaller than this and and that turns out to be n sites minus J n. So that takes care of all the integers in the denominator. Apart from Q, then we have one factor of Q and in addition, we have n minus two factors of Q minus 1. That'S because we have the terms of Q, they all correspond to links. If you have n sites, we have n minus 1 links the first one of those is Q, and so for the remaining n minus 2. We have Q minus 1 and then we have this factor of n sites in the denominator and that occurs for each beat. Except for the first one, so that should be raised to the power and minus 1. So this is the number of configurations that we have for polymer number J and in order to find the number of configurations for all the polymers, we need to multiply this quantity. Omega J for each polymer soda J runs from 1 up to the number of polymers and then, of course, we need to divide by an P factorial, because when we reorder the polymers, then we have in fact the same configuration and the end doesn't matter in which Order we built up the polymers and that's why we have to compensate here with one of our and P factorial. So when we worked that out, it's wise to start with the prefactor. That'S rather simple, because it's the same for each polymer. So we simply have to raise that to the power RNP, so we have first of all our one over n P factorial, and then we have Q to the power n P. Then we have Q, minus 1 raised to the power n minus 2 times n P, and we divide by n sites raised to the power n P times: n minus 1. So the remaining terms are a bit more complicated for polymer number one. We have here n sites factorial and we need to divide by n sites minus n factorial. So that's this expression for J is 1. So then we obtain the sequence of these terms and sites factorial divided by n science and factorial, and then we have the same term in the numerator and a new term in the denominator that same term in the numerator new term in the denominator, etc, etc. So we see that there are a lot of cancellations. This term cancels against this term. This term cancels against that term, this term councils against the next one, etc, etc. Finally, this term will also cancel and the only terms which remain our n sites factorial and this term n sites minus NP times n factorial. So finally, we have the same term as we had here and then we are left with n sites. Factorial / n sides, n P times n factorial, and so that's the number of ways we can put the polymers in the solution phase. We call that the solution phase was this phase that we had in the left picture. What we want to find out is whether the left or the right situation, so the solution or the separated face is stable, and so we can. We know that when we know the free energies of both - and we are dealing now with the entropy, which is an important part of that free energy, so for the entropies, we need to compare the entropy of this face with that of this, and we are focused Only on the entropy of the polymers and in the right hand, side we can use the the procedure that we are followed here, except that we just assume that there is no solvent present and that we have fewer sites. So here are all the sites that can be occupied by polymer, so it's just n times n P. So that's the number of beats we have available in the polymers and we have no solution for the separated phase that we use. Therefore, the same expression. But now we take n sides to be equal to n times n P. So, for example, this term this P R sub n from the expression and we then obtain 1 over and P factorial. And then we have Q and P Q, minus 1, n minus 2 and P, and now in the denominator, we have n times n P, raised to the power and P times n minus 1, and then we have instead of n sites and times and P factorial. In order to find the entropy difference, we take the K log of the Omega solvent and that of the separated face, and so that means that in the end we have to take the logarithm of the ratio of these two terms and then one thing is simple: We see that these terms in those are exactly the same, and so we then recover the expression only with the different terms and the result then becomes the following expression, which is rather cumbersome to work out and in order to make life a bit easy easier. We define a new parameter, which is the polymer fraction, so we define a parameter that we call Phi and that is n times. N, P and P is the number of polymers and is the number of beats per polymer. So this is the total number of beats divided by the total number of sites. So that's the polymer fraction and in terms of this expert parameter, we can express this entropy difference in a rather straightforward way. First, we recognized this parameter here in the first term. In the logarithm and working it out, we get an PE times, n minus 1 times the logarithm of Phi, and then we use the Stirling approximation for the other terms that occur here. All these factorials here is the result where we have used the following. This is Phi times n sites, and that rightly gives rise to the last term here and sites Phi times the violin and science Phi. Then we have here 1 minus Phi times n sites which gives rise to this term, and the term in the numerator is simple. It just gives me this, and then we realize that the last terms are all proportional to n sites, so we could put n sites here in the denominator. At first sight it doesn't. It seems that this term is not proportional to n sites but and P times n can be written as Phi times n sites and the minus NP can be written as Phi times n sites divided by N, and so we can then write the following for s. Delta s, divided by the number of sites undivided by KD, gives minus Phi over n log Phi minus 1 minus Phi algorithm of 1 minus Phi. This expression follows straight over formally from the previous one. You just need to work out all the terms. One important remark in working out the entropies for the solution and for the separated phase. I have neglected the multiplicity of the solvent itself, but that doesn't matter because that's the same in both case, so that doesn't affect the final difference. The next step is done to calculate the energy both of the solution and of the separated face, and we start by the solution. Now the energy obviously depends on the interaction between solvent molecules, interactions between the polymer beats interactions between polymer beats and the solution, and therefore we now introduce a few interaction parameters where we assume that there are only nearest neighbor parameter interactions and so for the solvent. We have the parameter J SS, which describes us as a solvent, solvent interaction which only interacts between nearest neighbor sites, both occupied by solvent molecules. So we ask ourselves how many such interactions will there be? Well, in total, we have n sites times Q, because the lattice coordination number is Q. So this is the total number of nearest neighbors, but we have come to them doubly. So we need to divide this by 2. We need to multiply by j SS, but now these are. This is the total number of sites and, of course, the we can only allow sites which are not occupied by the polymer and therefore provided. If we multiply now by the occupation probability that the first site is indeed not occupied by a polymer, we have 1 minus Phi. For that, because Phi is the polymer fraction and if we bond the second site to be empty as well. We need to multiply once more with by 1 minus Phi, and therefore this is the interaction energy of the solvent. Then we need to take into account of solvent polymer interactions and we assume that the interaction constant we take for that. The symbol JSP now here is a section of a polymer and that can interact with, in this case two neighbors, because the other two have been occupied by neighboring polymer beads and therefore we start by allowing each polymer beat to F neighbors, and there are n times And P polymer beads, but because the these two sites have already been occupied by other Peet's beats. We have here a Q minus two, because two sites have been occupied and then we need to multiply by the, by the probability that the remaining sites are unoccupied and that probability is 1 minus Phi, and then we multiplied by j SP. Obviously, you could argue that we should also consider the first and the last beat, but we assume that we are in the large and limit, so we have long polymers and then they give a negligible correction to this result. Next, we consider polymer polymer interactions and that's the case which is very similar to the previous one, except that we want to read cross sites now to be occupied by another polymer by another beat, and it's important to emphasize here that, with these polymer polymer interaction, we Do not include the interactions between beads that are neighboring on a chain they're always in the same number of neighbors on a chain, so that gives a constant contribution to the energy, and we don't consider that here. So we take the case where, at this site, the Red Cross or that site is occupied by a beat of another polymer or a bit further on on this polymer, and therefore we have something similar to the previous one. We have n times and P times. Q minus 1, but now we need to divide by 2 because we have both polymer sites. They are indistinguishable, so swapping them gives the same and we do not want to off count. And then we need to multiply by the probability that the neighboring site is indeed occupied, and that gives me a Phi and then I have a J Palmer Palmer and I can add up all these terms. But I've done something in addition and times NP can be written as n sites times Phi and I've used that here. I have also used that for this contribution and therefore, in the final expression, we have an extra factor of Phi included in these two terms and that enables us to write the solvent energy as n sites times a single energy. So we are interested in the end. In the energy per site, we need to compare that solvent energy to the energy of the separated phase, and in that case we have a fraction of the volume which is occupied by the polymer and we have a solvent and because the boundary is a lot smaller. It took it involves much fewer links than the bulky parts we neglect the boundary altogether, so we consider the polymer volume separately from the solvent volume. Then we realized that every polymer in the melt has Q minus 2 neighbors and we need to multiply this by the number of polymer sites and that's n sites times Phi and for the solvent. We have n sites times 1 minus Phi solvent sites. So we subtract polymer fraction from the total and then we have a queue. Each site can have Q neighbors. Now because there is no restriction, you know dealing with the solvent and not with the polymers. We need to divide by 2 for double-counting and we have a J solvent solvent and the entry are interested in the difference between the free energy. So here we focus on the difference between these two energies. All the terms are proportional to n sites, so we can divide by that and if we subtract these two expressions, then after some calculation we find the following expression and we see that we have extracted now all the terms which depend on the polymer concentration in front Of the square brackets and inside the square brackets, we have three terms solvent, solvent, solvent polymer and polymer polymer, and we see that the terms wherever particles interact among particles among them in themselves. So, with the particles of the same kind, they have a minus sign and four different particles. We have a plus sign. Obviously it did this parameter just as a fixed value, because the QD coordination number is fixed and all the interaction parameters they depend on the chemistry of the solvent and of the polymer. So all these terms are fixed and we lumped all those terms together into KBT. Chi, that's called the flory-huggins parameter guy and we see that, although the original expression here does not depend on temperature at all by this procedure, we have introduced the temperature dependence in Chi such that for low temperature, guys big and for high-temperature guy is small. Now, let us recall the difference between the entropy which we calculated before and which is this expression, and that enables us to find the free energy difference and using the Delta e and using the Delta s. We can find the Delta F just two Delta e minus T Delta s, and then we have a lot of terms that we can work out and the final result is then, this expression that we see here and this term arises from the energy difference. The other terms are just the Delta s, and so this is the energy free energy difference between the solution and the separated face, and this equation determines whether the solution is stable. That'S the case when Delta F is smaller than zero, because then the free energy of the solution is lower than that of this separated face and for positive values of Delta. F, we find that a separated face is stable. Here you see a plot of Delta F as a function of Chi and off I'd two parameters in the formulas given here on top and the color indicates whether the value of Delta F is positive. In that case, the separated face is stable or negative. That'S the blue region and then the solution is stable and then on the right picture. You see a few scans that were taken for fixed values of Chi, so those are cuts in this direction and you see that there are even values of Phi, for which the value for Delta F can be either positive or negative, depending on the concentration Phi. In summary, we have compared two different phases for a collection of polymers and solvent molecules. The left phase is the phase in which the polymers mix with the solute with the solvent molecules, and so they form a solution on the right hand, side they form a separated face, which one of the two is stable depends on the relative free energies and in Order to calculate that, if we are first analyzed the entropy and then the energy, both in a mean field approach - and it turns out that these quantities can be expressed in terms of two parameter. The first of these is the polymer fraction. That'S fine and that's the number of polymer beats divided by the total number of sites, so we have NP times and polymer. Beads and P is number of polymers and and mr. length of the polymer measured in beats and the remaining sites are occupied by the solvent. The second parameter is Chi, that's the flory-huggins parameter and it depends on the interaction constants. These are all nearest neighbor interaction, constants, the first one for two sites, neighboring sites being occupied by solvent molecules, then solvent polymer and then polymer polymer. The interactions were, the occupations of two neighboring sites are the same, set its SS or PP. They both have a negative minus sign and the one in which a solvent talks to a polymer has a positive sign. Furthermore, there is some temperature in introduced because we have divided by KBT. The stability of the face is determined by the sign of Delta F, which depends on those two parameters. Guy and Phi - and here you see exactly for which values of Chi and Phi 2 Delta F is positive, then it's red or when it's negative. In that case, it's blue

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