|Tie Lines and the Lever Rule||page 1 of 9|
|Consider a cooling alloy at the composition and temperature marked on the diagram. As shown on the phase diagram, the alloy is, at the given temperature, a mixture of alpha and liquid phases - but what are their exact compositions at this temperature?|
|Tie Lines and the Lever Rule||page 2 of 9|
(constant temperature) line through the alloy's position
on the phase diagram when it is in a two phase field,
intersecting the two adjacent solubility
curves, is called a tie line
(yes, that's the horizontal yellow line on the diagram).
The ends of the tie lines show the compositions of the two phases that exist in equilibrium with each other at this temperature. From the diagram we know that alpha and liquid phases will exist. The tie line shows that the alpha phase is 5.2%B and the liquid phase is 34.5%B at this temperature.
Remember, though, that the overall composition of the sample is unchanged - we are only discovering the compositions of the constituent phases within the sample.
|Tie Lines and the Lever Rule||page 3 of 9|
|For a cooling alloy at
and temperature Tx
, tie lines may be used to answer questions such
The answer to "what phases are present ?" is easy. Composition Co and temperature Tx meet in the beta + liquid phase field, so these are the two phases present.
|Tie Lines and the Lever Rule||page 4 of 9|
|To answer "what are
their compositions ?" we must draw a horizontal
tie line from the point to the nearest phase diagram
boundaries. The tie line shows us that the compositions
|Tie Lines and the Lever Rule||page 5 of 9|
|To answer the last question "if
the temperature is reduced to Ty,
how do the compositions of the two phases vary?" consider
the new tie-line, shown in yellow on the diagram.
The compositions of liquid and beta phase have both decreased in wt%B to:
Thus, both the liquid and the beta phases are getting richer in A as the sample is cooled.
|Tie Lines and the Lever Rule||page 6 of 9|
|Now that we know the compositions
of the two phases, we need to find how much
of each phase exists at the given temperature. The ratio
of the two phases present can be found by using the lever rule.
At first sight the lever rule can appear confusing. It is really invoking the conservation of mass, and can be proved mathematically, as shown below the diagram.
Essentially, we start off with an overall composition of our alloy - Co. From the tie-line we know that the two phases at a given temperature have two different compositions, but overall the amounts of these two compositions must add up to the alloy's overall composition, Co.
This is the basis for the lever rule. Using the lever rule itself is very simple, we'll show you with a diagram.....
|Tie Lines and the Lever Rule||page 7 of 9|
|Basically, the proportions
of the phases present are given by the relative
lengths of the tie line. So, the
proportions of alpha and liquid present on the diagram
(showing a portion of the whole phase diagram) are:
Simple, isn't it ?
But... which equation corresponds to which phase ?
|Tie Lines and the Lever Rule||page 8 of 9|
|Now, consider the same alloy
as it crosses the liquidus line.
It seems reasonable to assume that, at this point, the
alloy will be nearly all liquid. Looking at the diagram
it can be seen that Y1 is very
small here and so must be the proportion of
alpha present. Similarly X1 is
relatively large and so it
corresponds to the amount of liquid.
So, the left side of the tie line gives the proportion of the liquid phase (the phase to the right), and the right side of the tie line gives the proportion of the alpha phase (the phase on the left).
|Tie Lines and the Lever Rule||page 9 of 9|
|Distances along the tie line
can be found very simply by using a ruler on an accurate
phase diagram or, more correctly, by using data from the
composition axis (the x-axis).
For example, on the diagram shown, the percentage of alpha present can be calculated from the three pieces of composition data given:
Fraction of alpha = (34.5 - 23.7) / (34.5 - 5.2) = 0.3686
Thus, percentage of alpha = 0.3686 x 100 = 36.86%
and, as the alpha and the liquid make up 100% of the alloy's composition:
Percentage of liquid = 100 - 36.86 = 63.14%