The document discusses the composition and phases of a lead-tin alloy at eutectic composition after the eutectic reaction. It determines that the alpha phase comprises 45% of the alloy and is 19% tin, while the beta phase comprises 55% of the alloy and is 97% tin. It then calculates that for 200g of alloy, the alpha phase mass is 90g and the beta phase mass is 110g. It further calculates the masses of lead and tin in each phase. The document also contains information about determining the phases and amounts of cementite, austenite, and alpha in a steel alloy.

1. SHEET 3
ENG. KAREEM. H. MOKHTAR

2. QUESTION 2
• Determine the amount and composition
of each phase in 200 g of a lead-tin
alloy of eutectic composition immediately
after the eutectic reaction has been
completed.
(b) Calculate the mass of phases present.
(c) Calculate the masses of lead and tin
in each phase.

3. DETERMINE THE AMOUNT AND COMPOSITION
OF EACH PHASE
• Composition
• ⍺ = 19% Sn
• B = 97% Sn
• Amount
• ⍺ =
97−62
97−19
= 45 %
• B =
62−19
97−19
= 55%

4. B) CALCULATE THE MASS OF PHASES PRESENT
• Mass of ⍺ = total mass x percent of ⍺
=200 x 0.45= 90 g
• Mass of B = total mass x percent of B
=200 x 0.55= 110 g

5. CALCULATE THE MASSES OF LEAD AND TIN
IN EACH PHASE.
• Mass of Sn in ⍺ = 0.19 x 90 = 17.1 g
• Mass of Sn in B = 0.97 x 110 = 106.7 g
• Mass of Pb in ⍺ = 90 - 17.1 = 72.9 g
• Mass of Pb in B = 110 – 106.7 =3.3 g
What we have calculated in (a)
Composition
⍺ = 19% Sn
B = 97% Sn
Amount
⍺ =
97−62
97−19
= 45 %
B =
62−19
97−19
= 55%
What we have calculated in (B)
Mass of ⍺ = total mass x percent of ⍺
=200 x 0.45= 90 g
Mass of B = total mass x percent of B =200 x 0.55= 110

6. 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.

7. Q4) FOR A 99.6 WT% FE-0.40 WT% C AT A
TEMPERATURE JUST BELOW THE EUTECTOID, DETERMINE
THE FOLLOWING
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 proeutectoid
Fe3C(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

8. a) composition of Fe3C and alpha
CO = 0.40 wt% C
Ca = 0.022 wt% C
CFe C = 6.70 wt% C
3
Fe3C(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

9. 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
Fe3C(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

10. a) the amount of Austinite and alpha at proeutectoid
Co = 0.40 wt% C
Ca = 0.022 wt% C
Caus = 0.76 wt% C
austinite= 51.2 g
alpha= 48.8 g
Fe3C(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
CausCalpha

11. TERM PAPER

12. TERM PAPER
• Aim
• To use the knowledge you have learned to solve real life problems
• Search for certain information online and in books

13. TERM PAPER
• Your duty
• Submit the term paper before the due date (Hardcopy- 4/12)
• Late submission but at the same week looses 30%

14. GRADING CRITERIA
• Technical writing 3 marks
• Correct procedure 1.5 marks
• Correct assumptions and correct references 1.5 marks
• Correct answers 4 marks