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So I have an equilibrium phase diagram of steel and I am asked to 'Calculate the proportion of pearlite in the microstructure of 0.4 wt% C steel just below the eutectoid temperature (727 °C).'

I have been using the forumlas

$Wa=\frac{Cb-Co}{Cb-Ca}$%

$Wb=100-Wa$%

$Ca=weight$% of (Carbon in this example) at the first intersection point on the eutectoid line.

$Cb=weight$% of (Carbon in this example) at the second intersection point on the eutectoid line or the bottom of the 'V'. I don't know how to explain any better sorry.

So looking at my diagram, I see that $Ca=0.022wt$% Carbon and $Cb=0.76wt$% Carbon, and $Co$ is just 0.4%.

So $Wa=\frac{0.76-0.4}{0.76-0.022}$%

$=0.487805%$%

This is where I am confused. What exactly is Wa? Is that how much ferrite/austenite/cementite that is in the pearlite then? Or how much pearlite is at that point compared to ferrite? And in any case, what does that mean Wb is?

Any help would be much appreciated. Thanks

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Assuming you mean this phase diagram, it looks like your calculations are correct.

Based on my understanding of the diagram, at $727\ ^\circ \text{C}$, you have 100% pearlite at $C_b = 0.76\ \%C$ and 0% pearlite at $C_a=0.022\ \%C$. $W_a$ then appears to be how far $C_o$ is between $C_a$ and $C_b$, so $W_a$ should be the amount of pearlite.

The units of $C_a$, $C_b$, and $C_o$ are $\% C\ by\ mass$. That means when you calculate $W_a$ initially, you are calculating the fraction of pearlite and not the percent of pearlite. All of the $\% C\ by\ mass$ units cancel:

$$W_a=\frac{0.76\ \%C-0.4\ \%C}{0.76\ \%C - 0.022\ \%C}=\frac{{\%C}(0.76-0.022)}{{\%C}(0.76-0.022)}=0.487805\implies48.7805\%$$

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  • $\begingroup$ So Wa is my pearlite amount? And what happens to the amount of pearlite after point Cb? Does it gradually go down until the Cementite area? $\endgroup$ – user88720 Nov 10 '13 at 4:51
  • $\begingroup$ @user88720 That is exactly my question..Tell me something why did you choose $0.76 % C$ on the other end(I mean why did we not take the complete end that is $6.67%C$). $\endgroup$ – user586228 Sep 23 at 18:15
  • $\begingroup$ @user586228 - We use 0.76%, because that is the 100% pearlite line. To the left is a mix of pearlite and ferrite $\alpha$. To the right is a mix of pearlite and ledeburite. At 6.67% you have a crossover from a mix of $\ce{F3C}$ (cementite) and ledeburite to 100% cementite. There is no pearlite at this line. Since the goal was to determine the proportion of pearlite at 0.4%, we need a limit of 100% pearlite on one side (happens at 0.76%) and 0% pearlite on the other (0% C). $\endgroup$ – Ben Norris Sep 23 at 23:42
  • $\begingroup$ @BenNorris How did you predict the crossover at 6.67%?We are generally taught to draw the phase diagrams particularly upto 6.67%.Is there any intuition that helps us predict this?Or is it purely by knowledge? $\endgroup$ – user586228 Sep 24 at 4:47

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