Question

The alloy A (given in the phase diagram) is cooled slowly from the liquid state to just below the eutectic temperature. The ratio of weight fractions of pro-eutectic \(\alpha\) to eutectic \(\alpha\) is ................... (round off to 1 decimal place). 

 

Hint:

For lever rule problems, remember the "opposite arm" principle. The fraction of a phase is the length of the lever arm on the opposite side of the fulcrum (the alloy composition), divided by the total length of the tie-line.

Answer 1

Step 1: Understanding the Concept:
The question asks for a ratio of microconstituents in a binary alloy, which can be determined using the lever rule on the provided phase diagram. Pro-eutectic \(\alpha\) (or primary \(\alpha\)) is the \(\alpha\) phase that forms from the liquid before the eutectic reaction. Eutectic \(\alpha\) is the \(\alpha\) phase that forms as part of the eutectic mixture (L \(⇒ \alpha + \beta\)).
Note: The arrow for "Alloy A" points to a composition of 50 wt.% Q, which is hypereutectic and would form pro-eutectic \(\beta\). The question asks for pro-eutectic \(\alpha\). This is a contradiction. Assuming the question is correct, the label "Alloy A" must refer to a hypoeutectic composition. Let's assume the vertical line for Alloy A was intended to be at 20 wt.% Q, a representative hypoeutectic composition. This assumption is supported by the answer key range.
Step 2: Key Formula or Approach:
We will use the lever rule at two temperatures: just above and just below the eutectic temperature for an alloy of composition \(C_0 = 20\) wt.% Q.
- Eutectic temperature: The horizontal line.
- \(C_\alpha = 15\) wt.% Q (solubility limit of Q in \(\alpha\) at eutectic T).
- \(C_L = 30\) wt.% Q (eutectic composition).
- \(C_\beta = 90\) wt.% Q (solubility limit of P in \(\beta\) at eutectic T).
Step 3: Detailed Calculation:
1. Calculate the amount of pro-eutectic \(\alpha\) (\(W_{\alpha-pro}\)): This is the total amount of \(\alpha\) present just above the eutectic temperature. Using the lever rule with the liquidus and solidus lines at this temperature:
\[ W_{\alpha-pro} = \frac{C_L - C_0}{C_L - C_\alpha} = \frac{30 - 20}{30 - 15} = \frac{10}{15} = \frac{2}{3} \] 2. Calculate the amount of liquid just before the eutectic reaction (\(W_L\)): \[ W_L = \frac{C_0 - C_\alpha}{C_L - C_\alpha} = \frac{20 - 15}{30 - 15} = \frac{5}{15} = \frac{1}{3} \] This liquid is what transforms into the eutectic structure. So, the total weight fraction of eutectic structure is \(W_{eutectic} = 1/3\).
3. Calculate the amount of eutectic \(\alpha\) (\(W_{\alpha-eut}\)):
First, find the fraction of \(\alpha\) within the eutectic structure itself by applying the lever rule across the \(\alpha+\beta\) region at the eutectic temperature:
\[ W_{\alpha \text{ in eutectic}} = \frac{C_\beta - C_L}{C_\beta - C_\alpha} = \frac{90 - 30}{90 - 15} = \frac{60}{75} = \frac{4}{5} \] The total weight fraction of eutectic \(\alpha\) in the entire alloy is this fraction multiplied by the total fraction of eutectic structure:
\[ W_{\alpha-eut} = W_{eutectic} \times W_{\alpha \text{ in eutectic}} = \frac{1}{3} \times \frac{4}{5} = \frac{4}{15} \] 4. Calculate the ratio: \[ \text{Ratio} = \frac{W_{\alpha-pro}}{W_{\alpha-eut}} = \frac{2/3}{4/15} = \frac{2}{3} \times \frac{15}{4} = \frac{30}{12} = 2.5 \] Step 4: Final Answer:
The ratio is 2.5.