MatE271
Answer HW # 5
10/26/2001
[9.9]
The system remain 100% liquid until nearly the eutectic temperature (577°C), at which
the entire system solidifies leaning a two-phase microstructure of solid solution α and β.
[9.21]
(a) m L = 1.0 * 50kg = 50kg
36.8 − 35
mα =
(50kg ) = 20.9kg
36.8 − 32.5
(b)
35 − 32.5
mβ =
(50kg ) = 29.1kg
36.8 − 32.5
(c)
xα ≈ 34.1
and
x β ≈ 46.5
giving
39.6 − 35
(50kg ) = 41.8kg
39.6 − 34.1
35 − 34.1
mβ =
(50kg ) = 8.2kg
39.6 − 34.1
mα =
(d) mα = 1.0 * 50kg = 50kg
(e)
xα ≈ 32.1
and
x β ≈ 46.5
giving
46.5 − 35
(50kg ) = 39.9kg
46.5 − 32.1
35 − 34.1
mβ =
(50kg ) = 10.2kg
46.5 − 32.1
(f)
mα =
xα ≈ 29.2
and
x β ≈ 46.5
giving
46.5 − 35
(50kg ) = 33.2kg
46.5 − 29.2
35 − 29.5
mβ =
(50kg ) = 16.8kg
46.5 − 29.2
[9.56]
(a) ≈ 910°C
(b) α , ≈ 30 wt% Zn
(c) ≈ 920°C
(d) From ≈ 920°C to ≈ 40°C
mα =
[10.10]
(a) This is comparable to the tempering history of figure 10-11 except that some
initial fine pearlite formation is allowed at 500 °C (see sample problem 10.3(c)
for the initial history). The final microstructure will be :
70% fine pearlite + 30% tempered martensite
(b) Figure 10-15 indicate that complete transformation to final pearlite occurs often
5 sec at 500 °C . Subsequent thermal history does not change that. The final
microstructure is then: 100% fine pearlie.
[10.35 & 10.36]
Taking xθ = 53wt %Cu and using Fig 9-27 at various temperatures:
5−0
× 100% = 9.45%
At 100 °C , wt % θ =
53 − 0
5 −1
× 100% = 7.69%
At 300wt % θ =
53 − 0
5 − 2 .5
× 100% = 4.95%
At 400, wt % θ =
53 − 2.5
At 500, wt % θ = 0 × 100% = 0% by inspection
Taking x β = 36.1wt %Cu and using Fig 9-28at various temperature:
5−2
× 100% = 8.8%
36.1 − 2
5−3
× 100% = 6.0%
At 200 °C , wt% β =
36.1 − 3
5 − 4 .5
× 100% = 1.6%
At 250 °C , wt % β =
36.1 − 4.5
At 100 °C , wt % β =