Definition of ways of crystallization in two-component systems condition diagrams (definition of consequences of crystallization processes)
Example 5. In the diagram of condition of Al2O3 – SiO2 according Bowen and Greig (pic.3) the point m is responsible for the composition 20% Al2O3 and 80% SiO2.
To define the way of mixture crystallization, corresponding to the point m.
Solution. The point m corresponds to one phase – liquid one (fusion), as it is higher than the liquation line in the diagram. At withdrawing heat the system temperature decreases, the point m of the system slides down vertically, however fusion composition remains constant till achieving the point m´ in liquation. In the point m´ crystals of mullite 3Al2O3·2Sio2 begin to evolve from fusion. At this fusion begins to be enriched with silica (SiO2), as mullite is richer in Al2O3 than the mixture of m composition. If one draws connoda from the point m, finite points of connoda m´ and t´ will show the compositions of equilibrium phases m´ - liquid and t´ - solid (mullite A3S2)*.
For further A3S2 crystals sorting temperature decrease is required. Crystallization course according to cooling and release of A3S2 will be defined near connodas, located lower depending on temperature. One end of connodas slides down along liquation m´E, showing composition of liquid phase, and another end – composition of solid mullite A3S2. In the point E crystallization ends releasing simultaneously crystals A3S2 and a-cristobalite. The point E is called eutectic, and composition corresponding to it – eutectic one.
In the point E such states of systems are possible:
fusion of mixture m when cooled has reached the point E; at this there are two phases: solid A3S2 and liquid one of composition E;
eutectics crystallization flows, then
F = 2 + 1 – 3 = 0
since three phases are in equilibrium: solid A3S2, SiO2 and remained, still non crystallized liquid of composition E. According to withdrawing of heat (at eutectic temperature)
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amount of liquid decreases, quantity of solid phases increases;
eutectics crystallization has just finished;
F = 2 + 1 – 2 = 1
Pic.3. The diagram of the state of systems A3S2 - SiO2 (according to Bowen and Greig)
Two solid phases A3S2 and α- cristobalite are in equilibrium.
Further cooling and settling of representative point of mixture from m´´ to m´´´ will not cause any phase transformations, since this point will transfer in two-phase field of solid phases A3S2 and SiO2 of constant composition.
Example 6. In the diagram of the state of SiO2 – α – Al2O3 (see pic.3) representative point n, responsible for composition 70% of Al2O3 and 30% of SiO2 is taken.
To draw the way of fusion crystallization of this mixture.
Solution. The point n corresponds to liquid state. At displacement (cooling) of the point n to n´ one phase retains – fusion of constant composition. In the point n´ solid phase – A (corundum) occurs.
According to cooling and displacement of the point n´ to n´´ composition of liquid phase slides over curve liquation from n´ to peritectic H. In the point n´´ peritectic reaction between liquid phase of composition H and previously released crystals A (in this case of withdrawing heat) forming new solid phase A3S2.
However in consequence of excess of liquid phase (since the point n´´ is between the points of mullite compositions and N) its part retains even after A3S2 formation reaction. Crystallization of mixture of composition of the point n finishes, when it enters n´´´, that is when the composition of fusion corresponds to eutectics E. At this solid phases will be the same as for mixture of the point m´´.
Example 7. In the diagram of the state of systems SiO2 – α – Al2O3 (see pic.3) the point k, responsible for composition 80% of Al2O3 and 20% of SiO2 is taken.
To draw the way of fusion crystallization of this mixture.
Solution. In the point k there is one liquid phase of constant composition. At cooling and displacement of the point k to k´ phase changes do not take place. In the point k´ crystals of corundum occur, which are in equilibrium with fusion of composition of k´. At further cooling of mixture and displacement of the point k´ to k´´ corundum drops out, and composition of liquid phase changes along curve liquation k´H to peritectic point H.
In the point k´´ peritectic reaction between liquid phase H and previously released solid phase of corundum (at withdrawing heat) flows forming A3S2 mullite. In consequence of release of excess of corundum crystals all liquid phase will react (as the point k´ is between compositions of A and A3S2) and part of corundum will retain. Thus, in the point k´´ crystallization of mixture of k composition will finish forming two solid phases: corundum and mullite. At further cooling of mixture phase changes will not take place. The stated ways of crystallization are presented in the table on page 12.
Example 8. To compose a chart if ways of crystallization for fusion of mixture of composition 55% of SiO2 and 15% of CaO (the point g) in the diagram of the state of systems CaO-SiO2 (see pic.2).
Solution. Mixture of the point g, if it lies out of liquation field, consists of one liquid phase of definite composition. At displacement of the point g to any point of curve liquation of the liquation field (higher than kk´ is not shown) phase changes do not take place.
Points of initial mixture |
Points of the composition of liquid phase |
Solid phases |
Comment |
m m→m’ m m’→m” m”
Low m” n n→n’ n’ n’→n” n” |
m m→m’ m’ m’→E E disappear
no n n→n’ n’ n→H H |
- - A3S2 vegetation A3S2 A3S2+S vegetation A3S2+S - - A vegetation A A disappear A3S2 vegetation |
Reaction H + A → A3S2 In the product of reaction liquid phase is also excess Solid phase A disappears |
n” →n”’ n”’
Low n’” k k→k’ k’ k’→k” k”
Low k” |
H →E E disappear
- k k→k’ k’ k’→H H disappear
- |
A3S2 A3S2 +S vegetation A3S2 +S - - A vegetation A A+ A3S2 vegetation
A+A3S2 |
Reaction H + A → A3S2 In the product of reaction A is also excess Liquid phase H disappears |
Representative points of initial mixture |
Representative points of liquid phase |
Solid phases |
g g →g’ g’ g’ →gII
gII
gII→gIII gIII
gIII→gIV gIV
gIV → gV gV
gV →gVI gVI
gVI →gVII
gVII |
g g →g’ k+k’
A’+A disappear
A’ →B’ B’
B’ →E E disappear
- - - -
-
- |
- - -
α-crystoballite disappear α- crystoballite α- crystoballite disappear α-tridymite disappear
α- tridymite α- tridymite + α-СS vegetation α- tridymite + α-СS α- tridymite α-СS- disappear, +β-СS – vegetation α- tridymite + β-СS α-quartz vegetation, α- tridymite disappear+ β-СS α- quartz + β-СS α- quartz disappear |
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Pic.4. The diagram of the state of the systems anorthite – albite
B. Graphic methods of calculation. The lever rule and its employment to calculate the state diagrams
In the process of equilibrium changes of systems one has to deal with decomposition of one phase into two phases of another composition or, on the contrary, formation of new phase from two ones. Therefore it is important to establish dependence between concentrations and quantities of disappearing and appearing phases. This dependence is established according to lever rule.
Let’s suppose that initial phase of composition (concentration) m (pic.5) contains b% of component B and, taken in the amount of n g, decomposes into two phases:
phase of composition m1, containing a % of component B and mass k g;
phase of composition m2, containing c % of component B and mass l g.
Then
n = k + 1
The total number of substances doesn’t change at transformation. In this case the following rules are true.
Points, representing compositions of received phases m1 and m2, lie on one right line (axe of concentration x) on the both sides from the point of composition m of the initial phase.
This is applicable for formation of phase m from two phases m1 and m2. The point m must lie between the points m1 and m2 of initial phases.
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Pic.5. To the conclusion of the lever rule
The level rule establishes correlation between quantities of received phases m1 and m2 and distances between representative points of their compositions and representative point m of initial phase
The lever rule is employed to defined quantative correlation of phases in the diagram.
Let’s suppose that the initial liquid phase m1 is cooled to the temperature t2 (pic.6). m2t2 will correspond to this state of connoda and representative point of initial mixture is on it. At this temperature mixture p1 is decomposed into two phases: liquid composition m2 and solid B (the point t2 representing it).
Having employed the lever rule, one can calculate the quantity of both phases at the temperature t2:
To define percentage composition of any phase in mixture, for example liquid, one takes derivative proportion from the previous one: content of liquid phase p of composition m2 will be
where g – amount of initial mixture, weight. %
If we accept g = 100, we get the desired result in percentage terms at once. For example, we get percentage content of solid substance B for solid phase in a similar way.
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Pic.6. The diagram of the state of two-component system A – B with eutectics without solid solutions and chemical compounds
Example 10. To define percentage content of solid phase and fusion in the diagram Na2O-SiO2 for mixture of composition 55% of SiO2 and 45% of Na2O at the temperature 900º.
Solution. We draw connoda ab at 900º and plot the point c of initial mixture on this cannoda (55% of SiO2). The point c divides connoda into two cuts: cb up to crossing with liquation in the points b and ca up to crossing of connoda with vertical Na2O·SiO2 (the point a). The content of liquid phase p equals to correlation of cuts
The quantity of solid phase Na2O·SiO2 will be
Example 11. To define the composition of initial mixture, formed at the temperature 1100° 40% of fusion in the diagram of Na2O·SiO2 in the field of crystallization of α- tridymite.
Solution. We draw (see pic.7) connoda dl at 1100°. Having taken the length of all connoda for 100%, we divide it into 10 parts. We put four parts from the point l aside d and find the point k on connoda. Having let drop vertical from the point k to the axe of concentration, we define the desired composition of mixture: 90,5% of SiO2 and 9,5% of Na2O.
Example 12. The given composition of fusion contains 70% of Al2O3 and 30% of SiO2 (see pic.3). Fusion is cooled to peritectic temperature 1810º. To define amount of fusion and solid phase at the given temperature before peritectic reaction starts and after it finishes.
Solution.
Before the reaction starts on the horizontal Hf of peritectic reaction we plot the point nn, responsible for the given composition of initial mixture. Amount of fusion at this will be
Amount of solid phase α – Al2O3 will comprise
100 – 69,5 = 30,5%
After the reaction finishes mullite A3S2 will be as solid phase, since all the amount α – Al2O3 will be used up at the reaction with liquid H. On the same connoda Hf the point l is responsible for composition of mullite. Amount of fusion at this will be
Amount of solid phase A3S2 will comprise
100-17=83%
Example 13. To define the temperature for initial mixture of composition 8% of SiO2 and 92% of Al2O3, giving 40% of fusion at correlation of solid:liquid = 3:2 in the diagram of the state of systems Al2O3-SiO2 (pic.8).
Solution. We plot the point x of initial mixture to axe of concentration. Al2O3(A) will be solid phase. We divide the cut Ax, corresponding to 40% of fusion, into two different parts. We put three such parts along axe of concentration from the point x aside SiO2 and find the point b. From the point b we reset vertical to crossing with liquation in the point a. From the point a we draw connoda and find the desired temperature t.
Having drawn vertical xc from the point x, we convince that in the point c (at temperature t) mixture contains 40% of fusion in correlation to quantity of cuts tc and ca.
C. Analytical methods of calculation
Pic.8. The diagram of the state of the system Al2O3-SiO2 (according to Toropov and Galakhov)
Along with graphic method of calculation in state diagrams there is also analytic method. At using analytic method it is necessary to differentiate the following two cases:
In the system there is a component, which remains in liquid phase, then when another is partly crystallized and partly remains in liquid. For example, in the diagram of the state of systems Al2O3 – SiO2 (see pic.8) all fusions of compositions, lying between eutectics E1 and SiO2, will be crystallized (higher than eutectic temperature) releasing only SiO2, and Al2O3 fully remains in fusion.
In solid phase two components drop out. For example, in the same diagram of the state of systems Al2O3 – SiO2 fusions of compositions, lying between eutectics E2 and chemical compound 3Al2O3·2SiO2 (mullite), will release crystals of mullite at crystallization, that is simultaneously two components Al2O3 and SiO2.
To implement calculation in diagram of the state by analytical method the following designations are used: X – content of component in initial mixture, %, x – content of this component in fusion, %; x´ - content of this component in solid phase, %; p – amount of fusion, %. Then for the case, if there is a non-dropped out component
In formula (I, 5) for X and x we insert percentage content of a non-dropped out component.
In case a non-dropped out component is absent, formulas (I, 5) take the form
In final case the task can be solved concerning any component, since they are both drop out.
Example 14. At crystallization only one component drops out.
To define the composition of initial mixture, formed at 1100º 30% of fusion in diagram of systems Na2O – SiO2 (see pic.7) in the field of α-tridymite + liquid.
Solution. We draw connoda at´´ at temperature 1100º. We use the formula (I, 5) and solve concerning a non-dropped out component Na2O:
We find value of xNa2O in the point a, as the point a corresponds to the composition of fusion, equilibrium with SiO2 (α-tridymite): at the temperature 1100º xNa2O = 22º. Then we have
From here the content SiO2 in initial mixture will be
Example 15. Liquid phase at 1650º contains 89% of SiO2 and 11% of Al2O3 (see pic.8). To define the composition of liquid by the point on liquation. The composition of solid phase – mullite – corresponds to 71,8% of Al2O3 and 28,2% of SiO2; amount of the fusion – 40 weight %.
To define the composition of initial mixture.
Solution. Taking the mass of initial mixture equal to 100 g, the content of Al2O3 in liquid phase will be
The content of Al2O3 in solid phase will be
The total content of Al2O3 in initial mixture of XAl2O3 equals to the sum of these quantities
From here we calculate the content of SiO2 in initial mixture by difference
Example 16. To define percentage content of solid and liquid phase in the system CaO – SiO2 by the composition of initial mixture 45% of CaO and 55% of SiO2 and liquid phase 39% of CaO and 61% of SiO2.
As a non-dropped out component is absent, we employ the formula (I, 6). The amount of liquid phase will be
The amount of solid α - CaO·SiO2 will be
Example 17. To calculate the quantity of formed two liquid phases at 1720º in the system MgO - SiO2 (see pic.1) by the given composition of initial mixture 20% of MgO and 80% of SiO2.
Solution. We draw connoda kf and plot the point of initial mixture n, lying in the field of liquation on it. The points of crossing of connoda with curve of liquation show composition of equilibrium liquid phase; richer for SiO2 the liquid contains 2,5% of MgO and 97,5% of SiO2, less rich for SiO2 the liquid contains 32,0% of MgO and 68% of SiO2.
We define the amount of liquid, richer for SiO2, concerning the component of SiO2
The amount of liquid, less rich for SiO2 will be
Example 18. To define the quantity of solid and liquid phases at 1600º in the systems CaO - SiO2 (see pic.2) by the given composition of initial mixture 20% of CaO and 80% of SiO2.
Solution. At 1600º CaO will be a non-dropped out component for the given initial mixture. The composition of liquid phase is defined by the point, lying on the crossing of curve of liquation with connoda, drawn at 1600º; it corresponds to 33% of CaO. The amount of liquid phase will comprise
The quantity of solid phase of α-cristobalite
Example 19. T define the percentage content of solid solution and liquid phase in the system of helenite 2CaO·Al2O3·SiO2 – okermanite 2CaO·MgO·2SiO2 (pic.9) by the given composition of initial mixture: 40$ of okermanite and 60% of helenite at the temperature 1460º.
Solution. We draw connda at the temperature 1460°. By the ends of connoda we find the compositions of phases: liquid phase contains 51% of okermanite and 49% of helenite. The composition of solid solution: 22% of okermanite and 78% of helenite. The amount of liquid phase will be
The amount of solid phase –
Example 20. The system SiO2 – TiO2 (pic.10). To define the percentage content of solid solution and liquid phase by the given composition of initial mixture 60% of TiO2 and 40% of SiO2.
Pic.9. The diagram of the state of the system helenite – okermanite
Pic.10. The diagram of the state of the system TiO2 – SiO2 (according to Riker and Gummel)
Solution. We draw connoda at the temperature 1600°. We find the composition of liquid phase, equal to 27% of TiO2 and 73% of SiO2 by one end of connoda. By another end the composition of solid solution corresponds to 80% of TiO2 and 20% of SiO2.
The amount of liquid phase will be
The amount of solid phase –