Lever Rule: Determining Amount of Liquid & Solid

In summary, according to the figure, it is possible to say that the line through point M shows the amount of liquid while the other line through point P shows the amount of solid. This is done by taking a horizontal line at point M and using the point similar to that for the horizontal line through point N, which will give you O'N and NP' where the line meets the liquidus and solidus. 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.
  • #1
jmex
38
1
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?
uploadfromtaptalk1404596350450.jpg
 
Engineering news on Phys.org
  • #2
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/eens211/2compphasdiag.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.
 
  • #3
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
 

1. How does the Lever Rule determine the amount of liquid and solid in a mixture?

The Lever Rule is a method used to calculate the amount of liquid and solid in a two-phase mixture based on their proportions and densities. It uses the principle of equal lever arms, where the weight of the liquid and solid on one side of the lever is equal to the weight of the mixture on the other side.

2. What is the formula for the Lever Rule?

The formula for the Lever Rule is:
Amount of Liquid = (Distance from the Solid end / Total Length of Lever) x Total Amount of Mixture

3. When is the Lever Rule most commonly used?

The Lever Rule is most commonly used in materials science and engineering to determine the amount of liquid and solid in a solidification process, such as in the production of alloys or the solidification of polymers.

4. How does the Lever Rule account for changes in temperature and density?

The Lever Rule assumes that the temperature and density of the liquid and solid phases are constant throughout the mixture. However, if there are significant changes in temperature or density, the Lever Rule may not accurately determine the amount of liquid and solid present.

5. Are there any limitations to using the Lever Rule?

Yes, the Lever Rule is only applicable to two-phase mixtures where the liquid and solid phases are completely miscible and have distinct densities. It also assumes that the mixture is in thermal equilibrium and that there are no chemical reactions taking place. Additionally, the Lever Rule does not account for changes in volume or mass during solidification.

Similar threads

  • Materials and Chemical Engineering
Replies
21
Views
2K
  • Materials and Chemical Engineering
Replies
1
Views
2K
  • Classical Physics
Replies
20
Views
921
  • Materials and Chemical Engineering
Replies
6
Views
2K
  • Materials and Chemical Engineering
Replies
4
Views
3K
Replies
8
Views
1K
Replies
11
Views
981
  • Introductory Physics Homework Help
Replies
12
Views
937
  • Astronomy and Astrophysics
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
895
Back
Top