Derivation of Lever Rule

In two-phase region the composition of two phases is different and is NOT the nominal composition of the alloy.

Lever rule is used to find the amount of phases present in only the two-phase regions of binary phase diagrams.

Let's consider an isomorphous system having components A and B.
Say we have to find the amount of two phases at T temperature and for C_o .

Draw a horizontal temperature line for T temperature as shown below.

Fig.1

The temperature line ties both liquidus and solidus lines together, that is why it's call tie line.

Note : Tie lines can be only drawn in two-phase coexistence region.

C_L = Composition of liquid
C_O = Composition of alloy
C_S = Composition of solid

Derivation of Lever Rule 

Consider,
M_B,total = Total mass of B in system
M_B,liquid = Mass os B in liquid
M_B,solid = Mass of B in solid
M_L = Total mass of the liquid
M_total = Total mass of the system

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Let's do mass balance for B, as per law of conservation on mass. \[M_{B,total}\;=\;M_{B,liquid}\;+\;M_{B,solid}\] We can find the value of M_B,liquid and M_B,solid by multiplying M_L with the respective compositions C_L and C_S . Thus equation becomes, \[M_{B,total}\;=\;\left(M_{L}\cdot C_{L}\right)\;+\;\left(M_{L}\cdot C_{S}\right)\] Let's divide both sides with M_total thus equation becomes, \[\frac{M_{B,total}}{M_{total}} = \frac{\left (M_{L}\cdot C_{L} \right )}{M_{total}} + \frac{\left (M_{S}\cdot C_{S} \right )}{M_{total}}\] \[\therefore \left (\frac{M_{B,total}}{M_{total}}\right ) = \left (\frac{M_{L}}{M_{total}}\right)\cdot C_{L} + \left (\frac{M_{S}}{M_{total}}\right)\cdot C_{S}\] But we know that \[ \frac{M}{M_{total}}\;=\;C\;=\;Overall\;composition,\] \[\frac{M_{L}}{M_{total}}\;=\;W_{L}\,=\,Weight\;fraction\;in\;liquid,\] \[\frac{M_{S}}{M_{total}}\;=\;W_{S}\,=\,Weight\;fraction\;in\;solid\] therefor the equation becomes \[\left ( C_{O} \right ) = \left (W_{L}\cdot C_{L}\right ) + \left (W_{S}\cdot C_{S}\right )\;........(1)\] But since the system only have two phases ( i.e. solid & liquid ), summation of the weight fractions will be one. \[W_{S}\;+\;W_{L}\;=\;1\] \[\therefore\;W_{L}\;=\;1\;-\;W_{S}\] If we substitute the value of W_L in eq-1 then we will get \[\left ( C_{O} \right )\;=\;\left(1\;-\;W_{S}\right)\cdot C_{L}\;+\;\left (W_{S}\cdot C_{S}\right )\] \[\therefore C_{O}\;=\;C_{L}\;-\;W_{S}\cdot C_{L}\;+\;W_{S}\cdot C_{S}\] \[\therefore C_{O}\;=\;C_{L}\;-\;W_{S}\left(C_{L}\;-\;C_{S}\right )\] \[\therefore W_{S}\left(C_{L}\;-\;C_{S}\right )\;=\;C_{L}\;-\;C_{O}\] thus we will get \[\therefore W_{S}\;=\;\frac{C_{L}\;-\;C_{O}}{C_{L}\;-\;C_{S}}\;\;\;\;\;and\;\;\;\;\;W_{L}\;=\;\frac{C_{S}\;-\;C_{O}}{C_{S}\;-\;C_{L}}\]


Numerical

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