Suppose the alloy in Example 9.3 in your book is reprocessed to a temperature at which the liquid concentrate is 45 wt% B and the solid-solution composition is 85 wt% B. Calculate the amount of each phase.
Suppose the alloy in Example 9.3 in your book is reprocessed to a temperature at which the liquid concentrate is 45 wt% B and the solid-solution composition is 85 wt% B. Calculate the amount of each phase.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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I think I am just overthinking this but can someone double-check me and correct me?
Suppose the alloy in Example 9.3 in your book is reprocessed to a temperature at which the liquid concentrate is 45 wt% B and the solid-solution composition is 85 wt% B. Calculate the amount of each phase.
![EXAMPLE 9.3
The temperature of 1 kg of the alloy shown in Figure 9.30 is lowered
slowly until the liquid-solution composition is 18 wt % B and the solid-
solution composition is 66 wt % B. Calculate the amount of each phase.
SOLUTION
Using Equations 9.9 and 9.10, we obtain
Xss - x
66
50
(1 kg)
XL
(1 kg)
18
ML
Xss
66
0.333 kg
333 g
and
х — XL
50
18
(1 kg)
XL
(1 kg)
18
Mss
Xss
66
= 0.667 kg = 667 g.
Note. We can also calculate mss more swiftly by simply noting that
1,000 g
- m = (1,000 – 333) g = 667 g. However, we shall
Mss =
continue to use both Equations 9.9 and 9.10 in the example problems
in this chapter for the sake of practice and as a cross-check.
Temperature
FIGURE 9.30 A more quantitative
treatment of the tie line introduced in
Figure 9.6 allows the amount of each
phase (L and SS) to be calculated by
means of a mass balance (Equations 9.6
and 9.7).
L+ SS
T1
SS
30
50
80
100
A
Composition (wt % B)
В
ML +mss=Mtotal
0.30 mL +0.80 mss = 0.50 mtotal
→mL=0.60 mtotal
mss = 0.40 mtotal](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdb96b686-dbb4-403e-80c1-3924d9fbe7b4%2F6a095c63-c475-418a-b189-fad77cd05dfe%2F4bvwiz_processed.png&w=3840&q=75)
Transcribed Image Text:EXAMPLE 9.3
The temperature of 1 kg of the alloy shown in Figure 9.30 is lowered
slowly until the liquid-solution composition is 18 wt % B and the solid-
solution composition is 66 wt % B. Calculate the amount of each phase.
SOLUTION
Using Equations 9.9 and 9.10, we obtain
Xss - x
66
50
(1 kg)
XL
(1 kg)
18
ML
Xss
66
0.333 kg
333 g
and
х — XL
50
18
(1 kg)
XL
(1 kg)
18
Mss
Xss
66
= 0.667 kg = 667 g.
Note. We can also calculate mss more swiftly by simply noting that
1,000 g
- m = (1,000 – 333) g = 667 g. However, we shall
Mss =
continue to use both Equations 9.9 and 9.10 in the example problems
in this chapter for the sake of practice and as a cross-check.
Temperature
FIGURE 9.30 A more quantitative
treatment of the tie line introduced in
Figure 9.6 allows the amount of each
phase (L and SS) to be calculated by
means of a mass balance (Equations 9.6
and 9.7).
L+ SS
T1
SS
30
50
80
100
A
Composition (wt % B)
В
ML +mss=Mtotal
0.30 mL +0.80 mss = 0.50 mtotal
→mL=0.60 mtotal
mss = 0.40 mtotal
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