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Lever rule doubt

by jmex
Tags: doubt, lever, rule
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jmex
#1
Jul5-14, 04:39 PM
P: 23
Hello,

From the figure how is it possible to say that this line shows amount of liquid while other is amount of solid. In the figure they showed MP is amount of liquid while OM is amount of solid. How?Click image for larger version

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256bits
#2
Jul6-14, 08:17 AM
P: 1,468
Seems backwards, doesn't it.

But, if you read the explanation regarding Fig 3 on this site, as a mixture cools,
http://www.tulane.edu/~sanelson/eens...pphasdiag.html
your understanding will be enhanced.

As an example wuich may be more intuitive to understand, for your picture as the composition K cools it will reach the liquidus line at N. Taking a horizontal line at this point N, and using the point similar to that for the horizontal line through N, we have O'N and NP' where the line meets the liquidus and solidus. You can see that O'N = 0 so there should not be any solid forming yet, after all everything started as a liquid. All of the commposition should still be liquid represented by length NP'. The first solid though, as we cool just a tiny bit more will be of the composition where a vertical line from P' on the solidus meets the horizontal x-axis.

Hope that helps.
Chestermiller
#3
Jul6-14, 06:42 PM
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This is usually done algebraically. Let L be the fraction liquid, and S be the fraction solid, so that L+S=1. Let [itex]\bar{x}_B[/itex] be the overall concentration of species B, corresponding to the vertical line through point M, let [itex]x_{Bliq}[/itex] correspond to the concentration of B in the liquid at point O (along the liquid equilibrium line), and let [itex]x_{Bsol}[/itex] correspond to the concentration of B in the solid at point P (along the solid equilibrium line). Then,
[tex]Lx_{Bliq}+Sx_{Bsol}=\bar{x}_B[/tex]
This can be combined with the equation L+S=1 to solve for L and S. From this, you will see that that the lever rule follows.

Chet


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