CHAPTER 7

7.6)

Let A and B be the two components constituting the binary phase diagram. Consider co-existing phases  (with composition C1) and  (with composition C2) with a mean composition C0. ‘f’ denotes the fraction of a particular phase. Let be the compositions expressed in terms of the percentage of B.

Using lever rule: ,

Amount of B in C0 (L) = Amount of B in  + Amount of B in 

==

7.9)

Given: 88 vol.% eutectic mixture, 12 vol.% proeutectic  phase

Assumption: area fraction = volume fraction

Using the fact that there is 12 vol%  phase:

(7.9-1)

Where,

Substituting in equation (7.9-1):

Using lever rule: , C0 = 0.6562 = 65.62%

7.12)

Given: 0.2 % C steel, temperatures are just above or below the indicated temperatures (as tie lines can be drawn only in the two phase fields).

(i) T > 1493C  L +  equilibrium

,

(ii) T < 1493C  L +  equilibrium

,

(iii) T > 725C  +  equilibrium

,

(iii) T < 725C  + Fe3C equilibrium

,

7.14)

Given: Case-1: 5% ferrite () at grain boundaries (GB),

Case-2: 5% cementite (Fe3C) at grain boundaries (GB)

Case-1: referring the Fe-C diagram , C1 = 0.76% C

Case-2: referring the Fe-C diagram , C2 = 1.09% C

7.24)

Data: mCo = 58.93 g/mole, mTi = 47.9 g/mole

Wt. % Co in Ti2Co =

First the data is condensed into the table and diagrams below:

Phase 
Temperature
 /  /  / L / Ti2Co / (ii) Invariant reaction
1020C / 17% Co / 27% Co / 38 % Co /
Eutectic reaction
685C / 1% Co / 8% Co / 38 % Co /
Eutectoid reaction
  • Ti2Co stable upto 1058C

  • Phase diagram for Ti:

L / 1670C
 (BCC)
882C
 (HCP)

Next we follow the following steps:

  • Mark the relevant temperatures (T) and compositions (C)
  • Mark the phases present at various T-C coordinates
  • Join the lines connecting the stability of the phases
  • Label the phase diagram

(iii)

,

7.11)

1