A hypothetical \(A-B\) alloy of composition \(40 \mathrm{wt} \% \mathrm{B}-60 \mathrm{wt} \% \mathrm{A}\) at some temperature is found to consist of mass fractions of 0.66 and 0.34 for the \(\alpha\) and \(\beta\) phases, respectively. If the composition of the \(\alpha\) phase is 13 wt\(\% $$\mathrm{B}-87 \mathrm{wt} \% \mathrm{A},\) what is the composition of the \(\beta\) phase?

Short Answer

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Question: Calculate the composition of the β phase for an A-B binary alloy system with an overall composition of 40 wt% B-60 wt% A, mass fractions of α and β phases as 0.66 and 0.34, respectively, and α phase composition of 13 wt% B and 87 wt% A. Answer: The composition of the β phase is 92.41 wt% B and 7.59 wt% A.

Step by step solution

01

List given information

We have been given the following information: 1. Overall alloy composition: \(40 \mathrm{wt} \% \mathrm{B}-60 \mathrm{wt} \% \mathrm{A}\) 2. Mass fractions of α and β phases: 0.66 and 0.34, respectively. 3. Composition of α phase: 13 wt\(\%\) B and 87 wt\(\%\) A.
02

Applying the Lever Rule

The Lever Rule states that for a given two-phase alloy system, the ratio of the mass fractions of the two phases is equal to the ratio of the difference in composition between the overall alloy composition and each of the phase compositions. Mathematically, this can be represented as follows: \(\frac{W_\alpha}{W_\beta} = \frac{C_\beta - C_{overall}}{C_{overall} - C_\alpha}\) Where \(W_\alpha\) and \(W_\beta\) are mass fractions of α and β phase, respectively, \(C_\alpha\) and \(C_\beta\) are compositions of α and β phase, respectively (weight percent B), \(C_{overall}\) is the overall alloy composition (weight percent B).
03

Substitute given values

We have: \(W_\alpha\) = 0.66 \(W_\beta\) = 0.34 \(C_\alpha\) = 13 wt\(\%\) B \(C_{overall}\) = 40 wt\(\%\) B Let's substitute these values in the equation and solve for \(C_\beta\): \(\frac{0.66}{0.34} = \frac{C_\beta - 40}{40 - 13}\)
04

Solve for \(C_\beta\)

Rearranging the equation and solving for \(C_\beta\): \(C_\beta - 40 = \frac{0.66}{0.34} (40 - 13)\) \(C_\beta - 40 = 1.941176 (27)\) \(C_\beta = 1.941176 (27) + 40\) \(C_\beta = 52.41176 + 40\) \(C_\beta = 92.41176\) Thus, the composition of the β phase is 92.41 wt\(\%\) B and \(7.59 \mathrm{wt} \% \mathrm{A}\).

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