# Example ```Type -3- Two metals, completely soluble in all proportion in the
liquid state, are only partially soluble in the solid state
Since most metals show some solubility each other in the solid
state, this type is the most common and therefore, the most
important alloy system
Fig -10- Phases diagram illustrating partial solid solubility
The melting points of the two pure metals are indicated at points
TA and TB, respectively. The liquidus line is TA E TB, and the
solidus line is TA F E G TB. Above the liquidus line, there is
only a single phase liquid solution. These metals (metal A and
metal B) are partly soluble in the solid state, a solid solution
must be formed. Alloys in this system never solidify crystal of
pure A or pure B but always a solid solution or mixture of solid
solutions. At TE, the
solid solution dissolves a maximum of
20 % B as shown by point F, and the B solid solution a
maximum of 10 % A as shown by point G. With decreasing
temperature, the maximum amount of solute that can be
dissolved decreasing, as indicated by lines FH and GJ. These
lines are called solvus lines and indicate the maximum solubility
(saturated solution) of B in A ( - solution) or A in B (B
solution) as a function of temperature point E, where the
liquidus lines meet at a minimum is known as the
(eutectic point).
Example
Draw cooling curve of alloy (1) (Fig. 10)
Fig -11- The cooling curve and microstructure at various
temperature during solidification of 95 % A – 5 % B alloy
As example:
Example
Determine the relative amounts of
mixture (Fig. 10)
and B in the eutectic
Solution
The relative amounts of
and B in the eutectic mixture may be
determined by applying (lever rule)
EG
 100% 
FG
EF
%B 
 100% 
FG
% 
20
 100  28.6%
70
50
 100%  71.4%
70
Example
Determine the relative amounts of
point K. (Fig. 10)
and β at room temperature,
Solution
KJ
95  70
 100% 
 100  29.4%
HJ
95  10
HK
70  10
B% 
 100% 
 100  70.6%
HJ
95  10
% 
Example
Write the eutectic reaction (from Fig. 10)
Solution
This reaction is the same as the one which occurred in the type
(2) of thermal equilibrium diagram, except for the substitution
of solid solution for pure metals.
Because of the change in solubility of metal B in metal A, line
FH, and of metal A in metal B, line GJ, there will be a slight
change in the relative amounts of
and B as the alloy is cooled
to room temperature.
The relative amounts of
and B at room temperature are.
KJ
25
100%  100  29.4%
HJ
85
HK
60
B% 
100  100  70.6%
HJ
85
% 
Example
Draw cooling curve of alloy 3 (Fig. 10)
Fig -12- The cooling curve and microstructure at various
temperature during solidification of alloy 3 (60 % A + 40 % B)
```