Material science

1. MATERIAL SCIENCE-4
PHASE DIAGRAMS (1)
ENG. KAREEM. H. MOKHTAR

2. AMOUNT OF EACH PHASE

3. LEVER ARM RULE
• The lever rule in general can be written as
• Phase percent =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎𝑟𝑚 𝑜𝑓𝑙𝑒𝑣𝑒𝑟
total length of tie line
x 100

4. PHASE PERCENT AT 1250 (OC)
• Phase percent =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎𝑟𝑚 𝑜𝑓𝑙𝑒𝑣𝑒𝑟
total length of tie line
x 100
• Ni in alpha= 45%
• Ni in Liquid= 32%
• Ni in alloy= 40%
• Alpha phase percent=
40−32
45 − 32
x 100 = 62%

5. EXAMPLE 3
• Regarding the cupper nickel phase diagram
• Find the following for 60%Cu-Ni alloy at 1250
a) while freezing, The first temperature at which
the first solid particle form?
b) while freezing, The first temperature at which
the alloy become totally solid?

6. EXAMPLE 2
• Regarding the cupper nickel phase diagram
• Find the following for Cu-50% Ni alloy at 1280oC
a) The phases present
b) The composition of each phase
c) the amount of each
phase.

7. EXAMPLE 2
• A) Solid and liquid
• B) Liquid  40% Ni and 60% Cu
Solid  55% Ni and 45% Cu
• C) Liquid phase = (55-50)/(55-40)= 33.33% and the rest is solid

8. • If we have Cu-20% Ni, and Ni and we want to produce 100Kg of alloy that has
40% Cu, find the amounts needed.
• Solution
• Total amount of Ni = amount of Ni in (Cu-20%Ni) + amount of Ni
• 60 = 0.2 x + (100-x)
• X = 50KG
• So we need 50Kg of alloy Cu-20% Ni and 50 kg of pure Ni

9. PHASE DIAGRAMS

10. FROM 0 % TO 2% SN

11. AT 15% SN

12. THREE PHASE REACTIONS IN BINARY PHASE
DIAGRAM

13. 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.

14. A) THE SOLUBILITY OF TIN IN SOLID LEAD AT 100°C
Answer
6%

15. B) THE MAXIMUM SOLUBILITY OF LEAD IN SOLID TIN
• Answer
97.5% Sn or 2.5% Pb

16. 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 %

17. 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

18. 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

19. 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.

20. 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%

21. 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

22. 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