1. Building Binary Diagrams An aid in the construction of binary diagrams from experimental data. Click to continue

2. Given Information: Aluminum (melting point 658°C) and Cobalt (melting point 1480°) form three compounds, of which AlCo melts at 1630°C, Al 5 Co 2 decomposes at 1175°C, and Al 4 Co decomposes at 943°C. A complete series of solid solutions forms from AlCo to pure cobalt. Step 1: Draw the outline of the binary diagram with the proper scale. Label the two end components. 1700 1500 1300 1100 900 700 AlCo mol% 20406080 Step 2: Draw lightly some guide lines showing the temperature and compositions in the problem. Step 2 Calculations: AlCo - 1/(1+1) = 50 mol%Co Al 5 Co 2 - 2/(5+2) = 28.6 mol%Co Al 4 Co - 1/(4+1) = 20 mol%Co Click to continue

3. Do Not Mark Here at This Time Solid Solution Region Given Information: Aluminum (melting point 658°C) and Cobalt (melting point 1480°) form three compounds, of which AlCo melts at 1630°C, Al 5 Co 2 decomposes at 1175°C, and Al 4 Co decomposes at 943°C. A complete series of solid solutions forms from AlCo to pure cobalt. Click to continue 1700 1500 1300 1100 900 700 AlCo mol% 20406080 Step 3: Mark the congruent, incongruent, and dissociation points (if given). Do Not Mark In Regions Where Solid Solutions Exist!!!! Congruent MeltingIncongruent Melting Dissociation (Solids only or Liquids only) Step 4: Draw intermediate compound and end component lines. Erase guide lines that have been used.

4. Given Information: Aluminum (melting point 658°C) and Cobalt (melting point 1480°) form three compounds, of which AlCo melts at 1630°C, Al 5 Co 2 decomposes at 1175°C, and Al 4 Co decomposes at 943°C. A complete series of solid solutions forms from AlCo to pure cobalt. Click to continue Step 5: Determine the peritectic liquid composition for each incongruent melting point using the “Peritectic Reaction.” Example: Consider the incongruent melting point of Al 5 Co 2,which decomposes into a solid and a liquid (from the diagram, the solid is AlCo). Write the reaction: Al5Co2  AlCo + Liq (x # of Al atoms, y # of Co atoms) Balance the reaction: Al5Co2  AlCo + Liq (4 Al, 1 Co) Determine liquid composition: Liq(4Al, 1 Co)  1/(4+1) = 20 mol%Co The same process is used to determine the peritectic reaction and peritectic liquid composition for the incongruent melting point of Al 4 Co. The resulting peritectic liquid composition is 12.5 mol%Co

5. Click to continue Given Information: Aluminum (melting point 658°C) and Cobalt (melting point 1480°) form three compounds, of which AlCo melts at 1630°C, Al 5 Co 2 decomposes at 1175°C, and Al 4 Co decomposes at 943°C. A complete series of solid solutions forms from AlCo to pure cobalt. Step 6: Mark the location of the peritectic liquids on the diagram. Draw the associated tie-lines. Do Not Mark Here at This Time Solid Solution Region 1700 1500 1300 1100 900 700 AlCo mol% 20406080 Use these symbols: Liquid Solid

6. Given Information: Aluminum (melting point 658°C) and Cobalt (melting point 1480°) form three compounds, of which AlCo melts at 1630°C, Al 5 Co 2 decomposes at 1175°C, and Al 4 Co decomposes at 943°C. A complete series of solid solutions forms from AlCo to pure cobalt. Click to continue Step 7: Start to draw the liquidus line. Do Not Draw in the Solid Solution Region!!! Notice that the liquidus line at 12 mol%Co has a positive slope. You may not connect the point to the edge of the diagram as is. You must add an intermediate point which will change the slope of the liquidus line. The only point that does that is an eutectic point. We are not told what the eutectic temperature is and we can not determine its composition. The point must be placed on the diagram somewhere which allows the liquidus line to be drawn. Draw the associated tie-line. Do Not Mark Here at This Time Solid Solution Region 1700 1500 1300 1100 900 700 AlCo mol% 20406080 Use this symbol:

7. Given Information: Aluminum (melting point 658°C) and Cobalt (melting point 1480°) form three compounds, of which AlCo melts at 1630°C, Al 5 Co 2 decomposes at 1175°C, and Al 4 Co decomposes at 943°C. A complete series of solid solutions forms from AlCo to pure cobalt. Click to continue 1700 1500 1300 1100 900 700 AlCo mol% 20406080 Step 8: Solid Solutions: “Complete series” means “continuous,” no other information has been given (no max. or min.) Refer to Figure 3.11 in the text for an example. Now draw the liquidus and solidus lines in the continuous solid solution region on the diagram

8. Click to repeat 1700 1500 1300 1100 900 700 AlCo mol% 20406080 Step 9: Label diagram. Al 5 Co 2 + AlCo Continuous Solid Solution (Co, AlCo) Liq. + Al 4 Co Liq. + Al 5 Co 2 Liq. + AlCo Liq. + ss Liq.+ Al Al 5 Co 2 + Al 4 Co Al + Al 4 Co Liquid Al 4 Co MENU