Question

A steel sample with 1.5 wt.% carbon (no other alloying elements present) is slowly cooled from 1100 °C to just below the eutectoid temperature (723 °C). A part of the iron-cementite phase diagram is shown in the figure. The ratio of the pro-eutectoid cementite content to the total cementite content in the microstructure that develops just below the eutectoid temperature is ................ (Rounded off to two decimal places)

Hint:

Remember the distinction between phases and microconstituents. Pro-eutectoid cementite is both a phase and a microconstituent. Pearlite is a microconstituent made of two phases (\(\alpha\) and Fe3C). Total cementite is a phase, calculated on the final \(\alpha\)-Fe3C tie-line.

Answer 1

Step 1: Understanding the Concept:
This problem requires the application of the lever rule on the Fe-Fe3C phase diagram to determine the weight fractions of different microconstituents in a hypereutectoid steel (Carbon %>0.8).
- Pro-eutectoid cementite is the cementite that forms from austenite when the steel is cooled through the austenite + cementite (\(\gamma\) + Fe3C) phase region, before the eutectoid reaction occurs.
- Total cementite is the sum of the pro-eutectoid cementite and the eutectoid cementite (which forms as part of the pearlite structure).
Step 2: Key Formula or Approach:
The lever rule is used to find the weight fraction of a phase or microconstituent. \[ \text{Weight Fraction} = \frac{\text{Length of opposite lever arm}}{\text{Total length of tie-line}} \] 1. Calculate the weight fraction of pro-eutectoid cementite (\(W_{\text{Fe3C, pro}}\)) using a tie-line just above the eutectoid temperature (723 °C).
2. Calculate the weight fraction of total cementite (\(W_{\text{Fe3C, total}}\)) using a tie-line just below the eutectoid temperature.
3. Find the ratio of the two quantities.
Step 3: Detailed Calculation:
From the phase diagram:
- Overall carbon concentration, \(C_0 = 1.5\) wt.%.
- Eutectoid composition (austenite to pearlite), \(C_\gamma = 0.8\) wt.%.
- Carbon concentration in cementite (Fe3C), \(C_{\text{Fe3C}} = 6.7\) wt.%.
- Carbon concentration in ferrite (\(\alpha\)) at 723°C, \(C_\alpha = 0.035\) wt.%.
1. Calculate Pro-eutectoid Cementite Content:
Apply the lever rule just above 723°C. The phases are austenite (\(\gamma\)) and pro-eutectoid cementite (Fe3C). The tie-line runs from 0.8% to 6.7%. \[ W_{\text{Fe3C, pro}} = \frac{C_0 - C_\gamma}{C_{\text{Fe3C}} - C_\gamma} = \frac{1.5 - 0.8}{6.7 - 0.8} = \frac{0.7}{5.9} \approx 0.1186 \] 2. Calculate Total Cementite Content:
Apply the lever rule just below 723°C. The final phases are ferrite (\(\alpha\)) and cementite (Fe3C). The tie-line runs from 0.035% to 6.7%. \[ W_{\text{Fe3C, total}} = \frac{C_0 - C_\alpha}{C_{\text{Fe3C}} - C_\alpha} = \frac{1.5 - 0.035}{6.7 - 0.035} = \frac{1.465}{6.665} \approx 0.2198 \] 3. Calculate the Ratio: \[ \text{Ratio} = \frac{W_{\text{Fe3C, pro}}}{W_{\text{Fe3C, total}}} = \frac{0.1186}{0.2198} \approx 0.53958 \] 4. Round to two decimal places: \[ \text{Ratio} \approx 0.54 \] Step 4: Final Answer:
The required ratio is 0.54.
Step 5: Why This is Correct:
The solution correctly identifies the microconstituents and applies the lever rule at the correct temperatures (just above and just below the eutectoid line) to find the weight fractions of pro-eutectoid and total cementite, respectively. The final ratio is calculated from these values.