Material Science engineering, Section 6

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

2. Phase diagrams

3. From 0 % to 2% Sn

4. At 15% Sn

5. Question 1
■ Using the following diagram determine
■ (a) the solubility of tin in solid lead at 100°C
■ (b) the maximum solubility of lead in solid tin
■ (c) the amount of B that forms if a Pb-10% Sn alloy is cooled to 0°C
■ (d) the masses of tin contained in the ⍺ and B phases
■ (e) the mass of lead contained in the and phases.Assume that the total mass of the
Pb-10% Sn alloy is 100 grams.

6. a) the solubility of tin in solid lead at
100°C
Answer
6%

7. b) the maximum solubility of lead in solid tin
■ Answer
97.5% Sn or 2.5% Pb

8. c)The amount of B that forms if a Pb-10% Sn
alloy is cooled to 0°C
■ B =
10−2
100−2
= 8.2 %

9. d) the masses of tin contained in the ⍺ and B
phases at 0°C(alloy mass=100grams)
Mass of tin in ⍺ phase = percent of Sn x mass (⍺)
Mass of tin in B phase = percent of Sn x mass (B)
Percent of Sn at ⍺ = 2%
Percent of Sn at B = 100%
From (C) we found that
B= 8.2% and ⍺ = 91.8%
As the total alloy mass is 100g
B= 8.2 g and ⍺ = 91.8 g
Mass of tin in ⍺ = 1.836 g
Mass of tin in B = 8.2 g

10. d) the masses of tin contained in the ⍺ and B
phases at 0°C(alloy mass=100grams)
■ To be more accurate
■ For Pb-10% Sn
■ And given that the alloy mass is 100 grams
■ Pb= 90g
■ Sn= 10 g
■ Calculated that Mass of tin in ⍺ = 1.836 g
■ Then the mass of tin in B= 10 - 1.836
= 8.164 g

11. e) the mass of lead contained in the ⍺ and B
phases. Assume that the total mass of the
Pb-10% Sn alloy is 100 grams.
■ Mass of Pb in ⍺ phase = percent of Pb x mass (⍺)
■ = 0.98 x 91.8= 89.964 g
■ Mass of Pb in B phase =
■ Total mass of pb – mass of Pb in ⍺ phase
■ 90 - 89.964 = 0.036 g

12. Question 2 (assignment)
■ 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.