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Lever rule doubt 
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#1
Jul514, 04:39 PM

#2
Jul614, 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 xaxis. Hope that helps. 


#3
Jul614, 06:42 PM

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P: 5,172

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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