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Good morning, we continue our lecture on solidification of binary alloys. Now we have, so far lookedat isomorphous system, it is cooling curve; we also looked at the lever rule, which canbe applied to find out the percentage phases, which are present in a particular case, Imean wherever you have two phase is coexisting. We also looked at the difference between anideal and real solid solutions. We talked about free energy composition diagram in thecase of isomorphous system. Then, we moved over to cases, where there is solubility limit,that means, two metals, they are not soluble in solid state in all proportion; in thosecases, there will be a situation, where three phases can coexist. And we looked at two particularcases; eutectic and peritectic system. And and we looked at how this structure evolvesin detail in eutectic and peritectic alloy. Now we will continue for here onwardsAnd basically, what we have seen that the transformation that takes place during solidificationof a binary alloy is represented in the form of a diagram, which is called binary phasediagram. This is a graphical representation of phase compositions and and their amountat a given temperature. And often we do it at constant pressure, and usually the pressureis one atmosphere. And one important condition here if we assume that the cooling is veryslow, such that at every stage, there exists an equilibrium between liquid and solid.And these binary diagrams, they are well useful tool for quantitative evaluation of the microstructure of the alloy; and hence, it can also in for some thing about its mechanicaland other physical behavior of the alloy. Now, in the case of an isomorphous system,where there is unlimited solubility, and both liquid and solid state, the phasing takesplace over a range of temperature, we looked at this. We looked at eutectic system, wherethere is a partial solid solubility, and here there will be a some stage, a liquid, whichdissociates into a mixture of two solids - alpha and beta; in phase diagram we always representliquid using an this roman alpha that L represents liquid, and Greek alphabets are used to berepresent phases. Now, we also looked at a peritectic system, where… And this is alsoa case, where there is a partial solid solubility; there is a limited solubility in the solidstate, but unlimited solubility in liquid state.And here, on the second conditions over a range of competition, the liquid reacts withsolid, which has precipitated out to give a different solid; like in this particularcase, liquid reacts with alpha to give beta, where beta has entirely different physicalcharacteristics and structures from that of alpha. Now these two cases - eutectic andperitectic, these are examples of invariant reactions; and in fact, at 1 atmosphere pressure,if three phases are coexisting in a binary system, that in binary system, where numberof component is two; in that case, if you applied this phase rule, you will find thedegree of freedom is 0. So, that means this type of equilibrium can coexist at a fixedtemperature and between a fixed composition of liquid and to solids. So, there is no variablethat can be altered. So, in fact, in principle from the thermodynamic characteristics, itis possible to calculate these temperatures. And we will see today, one such example.Now, there will be several order invariant reactions as well, which are possible apartfrom this, but basically similar to that, that one phase is separating into two phaseor two phases reacting to give a third phase. So, this type there are certain other possibilitiesas well, and we will look at them in subsequent part of the lecture. And also we will lookat some different variants of isomorphous system.First start with this isomorphous system in the phase diagram, you know just to recollectthat, this is that phase diagram of an isomorphous system. And in the two phase region, one canapply phase rule, that lever rule; one can apply the lever rule to find out proportionof alpha and liquid, which are coexisting; like over here, the alpha of this compositiongiven by that point p, that composition is X 1. So, this can coexist with a liquid ofthis composition X 2; and in this case, in using lever rule you can say that alpha, amountof alpha is proportional to q r q r, and amount of liquid is proportional to p q. And thereare certain alloys, which exceed it this kind of phase diagram, which are listed here, likeone of the common alloy, is copper nickel. And here, if you look at both copper and nickel,they have identical crystal structure, they have nearly similar lattice parameter, differenceis less. So therefore, atomic size’s difference is less, so there exhibit unlimited solubility.And same, I think will be valid, and for you to check up what are the crystal structures,and lattice parameters of these cases germanium-silicon, antimony-bismuth, they also exhibit a similarisomorphous behavior; this kind of table will have at least in the half; until solidificationis complete, they will exhibit this type of phase diagramNow, there are cases where this type of isomorphous system can exhibit maxima or minima; so thatmeans, in between an alloy, there can be an alloy, which has in this particular case,higher melting point than both A and B; and which is shown in this diagram. And we willlook at it here, say suppose this is actually T over A, this is B, you have percentage ofB with percentage plotted along this axis, this is melting point of A, here it is liquid,and here it is alpha. Now, here you have a maxima, at so that means, you can look atit as if it is two isomorphous system, one this side, another this side.You can also have a case something like this, exactly similar; here you have liquid, hereyou have alpha, and here this is T over A, this is T over B, this is melting point ofA, melting point of B; at in between, you have an alloy, which melts in exactly sameway as that of pure metal. If you try and plot, it is cooling curve of this alloy, itwill actually be a temperature and time; its plot will be more or less exactly similarto that of pure metal. So, this is the melting point of this law, let us say, this is themelting point; and it will exactly, I mean, its solidification behavior will be similarto that of pure metal. So, this kind of deviations from an ideal isomorphous system, they areresult of deviations from ideal deviations from ideal solid solutions. And we will lateron, we will see how we can express this deviations in a more quantitative fashion.Now, binary eutectic will looked at eutectics, an in this particular cases, where where wehave two fields, alpha, beta these are terminal solid solution, this is liquid; here you havealpha plus eutectic, here you have eutectic plus beta. And the two cases, you know, thereare certain binary alloy system will exhibit this kind of behavior like silver-copper,aluminum-silicon, led-tin, they have this type of phase diagram; and this is a commonsolider alloy, this is used as led-tin is is a solder alloy, it is a low melting material.And the type of structure that you can get in eutectic is shown here; and we can lookat this, say one type of structure in eutectics, say this alloy, here it is totally liquid;and once it solidifies, solidification is complete, you will have a structure, whichwill give you really an intimate mixture of two phases alpha and beta; here, suppose ifsee alpha solidifies, surrounding liquid becomes rich in beta, so automatically, the beta solidifies;so in fact, one after the other, it will start farming as so, you have a layer of alpha andthat a layer of beta will form; say something like this, you you have a layer of alpha forming,then within that, you will have beta, then again alpha. So, this is be a lamella structurecan form. Similarly, you can also have this type ofstructure; if alpha is precipitates out, surrounding if alpha precipitates out, surrounding isbeta, again alpha precipitates out. So, you can have this kind of a very intimate mixtureof this kind of a rod like this precipitate, which is chromatically shown in this diagrammixed one here. You can also have certain irregular structures are also possible, wheremaybe one phase is like this; regular structures are also possible; but what it means thateutectic will be an intimate mixture of two phases. So, this is the say beta, this isalpha, and this pattern can be regular or irregular, and this depends on certain physicalcharacteristics of the two metric A and B.Now, you can it is also possible to have one extreme case, where the solids are totallyinvisible like here; this is a level diagram of an binary eutectic, where there is no solubilityin the solid state. There are certain clear examples like cadmium-bismuth, antimony-led,which exhibit this kind of phase diagram. And here, the eutectic will be an intimatemixture; the hypoeutectic, you will have primary alpha, primary that means, A crystal of Aand eutectics; and here in the hypereutectic alloy, you will have primary B crystals ofB and then eutectic. Now, use in thermodynamics, it will also possibleto calculate if suppose, we take the case of cadmium-bismuth; if you know the meltingpoint of cadmium and bismuth unknown, the in a latent heat of fusion of cadmium andbismuth is known; in that case, if we assume that liquid is an ideal solid solution, itis possible to calculate from this thermodynamic data that means, melting point and heat offusion. The composition as well as the temperature, that eutectic temperature and eutectic composition,it is also possible to plot this liquid as line, this liquid as line as well as this.And we will see look at one such example of calculating this type of phase diagram; infact, we have seen one such example in the case of isomorphous system, where both liquidand solid are assume to be Raoult’s law; that means, both liquid and solid are assumeto be ideal solution.So, let us look at the determination of eutectic diagram from thermodynamic properties. Nowthis is a problem bismuth-cadmium, they are soluble in liquid state, but insoluble insolid state; estimates it eutectic composition and temperature; the melting points and latentheats of fusion of bismuth and cadmium, they are given below. Assume the liquid to be anideal solution; and since pure bismuth and pure cadmium are precipitating out, we canassume that their activity, you will be equal to 1. And these are given over here; thisis the melting point of bismuth is 271 degree centigrade, its heats of fusion that is latentheat of fusion is 2.6 kilo calorie per mole; cadmium melting point 321 degree centigrade;and its latent heat of fusion is 1.53 kilo calorie per mole. And let us see, how we proceedwith the calculation. So, here the main principle is that you have to calculate the free energyof this transformation.And look at a case in the phase diagram, consider a temperature over here. If you are in thisregion, you have pure A a equilibrium, pure A is in equilibrium with solution; that liquidconsisting of liquid or as a liquid solution of B in A. Now, how do we calculate the freeenergy of transformation,
which is illustrated here; say suppose in this particular case,you take that pure A as standard state; in free energy calculation, definition of standardstate is quite important; and we take the standard state that is pure A at T, temperatureT. Now, in this case in that case, what is thefree energy of pure A? Say since it is a mixture, it is RT ln activity of A; so we write itmole fraction of A and activity is equal to mole fraction. So, in that case, this is 1,because it is pure A, so this is 1, so therefore, for pure A, this is 0. Now how do you calculatethe free energy of that liquid or let us say partial we want to find out partial free energyof A in liquid; how will you find out? In that case, you take first pure A - solid,it transforms into liquid; and then we assume that this A liquid goes into solution. So,in that case, you can easily write it down, you can check up your earlier notes.And this particular case, here that free energy will be delta H of that melting of A times1 minus T over melting point of A; whereas, this is RT ln we assume ideal solution; solutionto be ideal, then this is the atom fraction A in liquid. So, you add the two, this isthe free energy, so this is the partial model free energy of A in liquid, and this shouldbe equal to this. So, that is the chemical that partial molar free energy is also knownas the chemical potential; and this two potential should be 0. So that means, over this, ifyou sum these together, so you assume that this is some kind of a chemical reaction,and you add the two, and they are in equilibrium. So, therefore, these two should balance; andwith this, what you get is listed here.In this particular case, you can see that atom fraction, the large atom fraction A inliquid will be a function of which is shown over here. So, this you can just algebraicsimplification, this is equal to this; and you can apply the same concept to this regionof the diagram, and then you get similar expression for this. Now, if here, so all these are known,this is known, this is known, this is known, temperature is fixed. So, it is possible tocalculate N A that is composition of the liquid as a function at any given temperature T.And if you do that, you will be able to generate this plot.In this same way, if you try and solve this, then you will be able to generate this plot;and wherever these two meet, so this is where composition of the liquid, which is in equilibriumwith A is exactly same as the composition of the liquid, which is in equilibrium withB. So, in this particular case, the sum total of this will be 1. So, in this case that thatthat you have to possible to find out the eutectic temperature as well as eutectic composition,and for this you can easily set it up in a spreadsheet, and then and these results whichare shown here, on the next one.So, these are the different temperature; this is the melting point of 1 or the 2, whichis the higher; and this is in kelvin, degree kelvin; and you calculate that N B - atomfraction B; and this is the other one.And if you plot the two, then you get this is the diagram you get, and this eutecticcomposition comes out to be close to around let us say 0.55 atom fraction, and this temperaturecomes out to be 408 degree kelvin.So in fact, if you check up this diagram, you will find you will find that althoughthis, there is a good match between this atom fraction, this composition is exactly seenas as given in the phase diagram, which is experimentally found out. But this temperatureis possibly, is a slightly lower; what is there we have found out by this. So, in thatcase, why is this deviation, why it is not matching? So, one possible explanation couldbe we have made on assumption that the liquid is ideal, second is we are assumed that thereis no solubility. So now, it is a question of is there a slight,which is extremely small amount of A is soluble in B or B in A, these are some of the possibilities,we can think about; so that means, you have to question this assumptions only, but inthe procedure, this is a procedure which is applied; and in one way, you can see thatthermodynamic properties are also evaluated from the phase diagram. So, actual experimentallyyou determine, and then you try to conclude whether that liquid, a solution is an idealor not; if it is not ideal, from this it is possible to calculate, how much is its deviationfrom ideality?Now, next we also looked at the binary peritectic system. Now, binary peritectic system youknow, this is one way, you can when you look at the phase diagram, you said this is a veryuseful tool, you know by which you can know about the phase percentages, you make somequantitative estimates, like say in this particular case, say this alpha beta region, you knowhow well its property depend? Suppose we assume that alpha is softer, beta is harder. So,there is a possibility, what we can see? You can apply the rule of mixture to find outis properties. So, if you know percentage alpha, it is possibleto… And to find out the property of the mixture; say suppose we are we say that aparticular composition say say and this, we try and find out that on this axis, we superimpose tensile strength or hardness. Say suppose beta has a certain hardness, say of beta ofa particular composition, this composition has certain hardness, say let us say, thisis the value of this hardness; and then alpha has certain hardness, in that case in thisregion, as amount of beta increases, if you go along this, it will follow is some somethinglike this, a linear dependent. So, if you know percentage beta, you will be able tofind out such properties. And also, if you look at the micro structure;say suppose of this diagram is not known, it is also possible to find out or locatethis composition these compositions from micro structure, analysis of the micro structure.So, what you have to do? Look at two micro structure or two compositions, look at thismicro structure, find out percentage alpha here; find out percentage alpha here. Andin both cases, both the two phases alpha and beta, say here as it is shown here, they willhave a composition, alpha will have this composition, beta will have this composition. So, usingfew a lever rules, you can set up one equation with these compositions.And secondly, follow the second alloy, you can write a similar expression; in that case,and once you know have two equation, two unknown, you will be able to find out these compositions.So, it is both way possible; if the phase diagram is known, you can find out its propertiesfrom phase percentages if or else if phase diagram is not known that you find that, thisis the region you have two phase structure; then you find out at least make two alloyin the similar in the region, and then find out percentage alpha and beta. And from that,you will be able to find out the compositions that is of this solvers, that is solubilitylimit of alpha and beta.And now, let us look at we have looked at two types of invariant reaction; see one isa eutectic, where liquid dissociate into a mixture of two different phases, two distinctphases having different physical and properties physical mechanical properties; a crystalstructure like alpha and beta. And we also have looked at a peritectic system, wheretwo phases say liquid reacts with alpha giving giving a totally different crystal say beta,a different phase called beta. So, these two cases, we had looked at.But real phase diagrams are not see basically, not not simple solid solution or eutecticor peritectic, they often consist of several two phase regions, and invariant reactionsinvolving three phases in a binary system. Now, these commonly invariant reactions, youhave some invariant reactions other than eutectic and peritectic, and we will look at them.And we can broadly classified this invariants into two groups in one case, one phase oncooling separates into two different phases like that in eutectic; and another case, twophases on cooling reacts to form a different phase like in peritectic.Now, let us look at the first that is one phase dissociating into two phases; now thisexample we have looked at in detail, say that is eutectic; and here this shows a small partof that eutectic phase diagram we eutectic isotherm. So, like here this is eutectic,you have two solid solution alpha and beta, and this is the liquid, and this dissociateshere. So, here it is totally liquid, and here it is a mixture of alpha and beta; and theyare intimate mixture. And now we we can esteem this, in cases where the liquids also canhave solubility limit, there are several examples in liquid, which you can think about try andmakes all and water, they do not makes. So, study the possibility that there can bein these two phases, if you are trying to makes in the liquid state, they may not besoluble in each other; they may, and there or they may be partly soluble in each other.So, there can be several such examples. And let us generalize that we have a liquid, andit dissociates into two different liquids having different composition; one we representas L 1 and another we represent as a L 2. And if we have a reaction say one liquid L1 dissociates into a solid another different liquid, and which is shown over here; herethis is alpha, this is L 1, and this side, it is liquid is L 2, and this is the miscibilitygap; so that means, liquid here, in this particular composition liquid, if it has this compositionat this temperature, they will not mix and they are not totally miscible, they are not…They will dissociate into two liquid L 1 and L 2.So, if you extend this, there may be a possibility that higher temperature chance of miscibilityis higher. So, here they are miscible, if you go down this temperature, there is a miscibilitygap, it is dissociates into two liquids L 1 and L 2, this is a miscibility gap. So,in this particular case, this liquid is here, it is reacting, it is it is dissociating asyou cool this temperature is dissociating into alpha, another liquid; and this typeof reaction is called monotectic reaction or this is also an invariant, because youhave three phases coexisting in a binary system. And you can have amount of variant, it isnot bad, it is only liquid can dissociate on cooling into a mixture of two phases, youcan also have a solid, say gamma; on cooling, it dissociates into two phases, two othersolid different solid phases, alpha and beta; and this diagram is shown is here. So, gammais dissociating into alpha and beta, so here. So,
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