This document discusses phase diagrams for binary systems with different types of solubility between liquid and solid phases. It provides examples of phase diagrams for systems with: (1) complete liquid and solid solubility, (2) complete liquid but zero solid solubility, and (3) complete liquid and limited solid solubility. It also works through examples of determining phase compositions and amounts for steel alloys based on their placement on an iron-carbon phase diagram.

1. PHASE DIAGRAMS-2
Eng. Kareem. H. Mokhtar

2. Phase diagrams

3. Q3) Draw schematic phase diagrams for
binary systems with
■ a) complete liquid and solid solubility
b) complete liquid but zero solid solubility,
c) complete liquid and limited solid solubility.

4. Fe
3
C
(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L
g
(austenite)
g +L
g + Fe3C
a + Fe3C
L+Fe3C
d
Co, wt% C
1148°C
T(°C)
727°C

5. Q4) For a 99.6 wt% Fe-0.40 wt% C at a
temperature just below the eutectoid,
determine the following knowing that
the maximum solubility of C in alpha is
0.022%
a) composition of Fe3C (cementite) and alpha
b) the amount of cementite in grams that forms
per 100 g of steel
c) the amount of austinite and alpha at pre-eutectoid
Fe
C
(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L
g
(austenite)
g+L
g + Fe3C
a+ Fe3C
L+Fe3C
d
Co, wt% C
1148°C
T(°C)
727°C

6. a) composition of Fe3C and alpha
CO = 0.40 wt% C
Ca = 0.022 wt% C
CFe C = 6.70 wt% C
3
Fe
3
C
(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L
g
(austenite)
g+L
g + Fe3C
a+ Fe3C
L+Fe3C
d
Co, wt% C
1148°C
T(°C)
727°C
CO
R S
CFe C
3
Ca
Solution

7. b)the amount of cementite in grams
that forms per 100 g of steel
b) Amount of cementite= mass of alloy x percent of cementite
Fe
3
C
(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L
g
(austenite)
g+L
g + Fe3C
a+ Fe3C
L+Fe3C
d
Co, wt% C
1148°C
T(°C)
727°C
R S
CFe C
3
Calpha

8. a) the amount of Austinite and alpha at pre-eutectoid
Co = 0.40 wt% C
Ca = 0.022 wt% C
Caus = 0.76 wt% C
austinite= 51.2 g
alpha= 48.8 g
Fe
3
C
(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L
g
(austenite)
g+L
g + Fe3C
a+ Fe3C
L+Fe3C
d
Co, wt% C
1148°C
T(°C)
727°C
CO
R S
Caus
Calpha

9. For Fe- C alloy at a temperature just
below the eutectic, determine the
following at the eutectic composition
a) What are the phases present
b) What is the composition of each phase
c) What is the amount of each phase
Fe
C
(cementite)
1600
1400
1200
1000
800
600
400
0 1 2 3 4 5 6 6.7
L
g
(austenite)
g+L
g + Fe3C
a+ Fe3C
L+Fe3C
d
Co, wt% C
1148°C
T(°C)
727°C

10. Solution
■ a) Austenite and Cementite
■ B) Austenite is composed of 97.8%Fe and 2.2%C
Cementite is composed of 93.3% Fe and 6.7% C
■ C) the amount of austenite is (6.7-4.3)/(6.7-2.2)
The amount of Cementite is (4.3-2.2)/(6.7-2.2)