Chapter 8.15, Problem 29AAP

Consider the binary peritectic iridium–osmium phase diagram of Figure P8.28. Make phase analyses of a 40 wt% Ir–60 wt% Os at the temperatures (a) 2600°C, (b) 2665°C + ΔT, (c) 2665°C − ΔT, and (d) 2800°C. Include in the phase analyses the four items listed in Problem 8.28.

To determine

Refer Fig P8.28, and perform a phase analysis of a 40 wt% Ir and 60 wt% Os at temperature of $\text{2600°C}$.

Include the following in the phase analysis:

1. i. The phase present.
2. ii. The chemical composition of each phase.
3. iii. The amount of every phase is available.
4. iv. Draw the microstructure by the help of 2-cm-diameter circular fields.

Explanation of Solution

Refer Figure P8.28, and make a phase analysis at a temperature of $\text{2600°C}$,

1. i. Phases present or available are beta and alpha, that is $\beta +\alpha$.
2. ii. The chemical composition of Os in alpha phase is 42 wt% and in beta phase is 62.5 wt%.
3. iii. Express the weight percentage of beta and alpha:

   $\begin{array}{c}\text{wt\%}\text{alpha}=\frac{62.5-60}{62.5-42}\times 100\%\\ =12.2\%\end{array}$

   $\begin{array}{c}\text{wt\%}\text{beta}=\frac{60-42}{62.5-42}\times 100\%\\ =87.8\%\end{array}$

Thus, the weight percentage of alpha and beta is $\text{12}\text{.2}\text{wt\%}\text{and}\text{87}\text{.8}\text{wt\%}$ respectively.

1. iv. Show the microstructure as in Figure (1).

To determine

Repeat the same at $2665\text{°C}+\Delta T$.

Explanation of Solution

Refer Figure P8.28, and make a phase analysis at $2665\text{°C}+\Delta T$,

1. i. Phases present or available are liquid and beta, that is $L+\beta$.
2. ii. The chemical composition of Os in liquid phase is 23 wt% and in beta phase is 61.5 wt%.
3. iii. Express the weight percentage of liquid and beta:

   $\begin{array}{c}\text{wt\%}\text{beta}=\frac{60-23}{61.5-23}\times 100\%\\ =96.1\%\end{array}$

   $\begin{array}{c}\text{wt\%}\text{liquid}=\frac{61.5-60}{61.5-23}\times 100\%\\ =3.9\%\end{array}$

Thus, the weight percentage of beta and liquid is $\text{96}\text{.1}\text{wt\%}\text{and}\text{3}\text{.9}\text{wt\%}$ respectively.

1. iv. Show the microstructure as in Figure (2).

To determine

Repeat the same at $2665\text{°C}-\Delta T$.

Explanation of Solution

Refer Figure P8.28, and make a phase analysis at $2665\text{°C}-\Delta T$,

1. i. Phases present or available are beta and alpha, that is $\beta +\alpha$.
2. ii. The chemical composition of Os in alpha phase is 43 wt% and in beta phase is 61.5 wt%.
3. iii. Express the weight percentage of beta and alpha:

   $\begin{array}{c}\text{wt\%}\text{alpha}=\frac{61.5-60}{61.5-43}\times 100\%\\ =8.1\%\end{array}$

   $\begin{array}{c}\text{wt\%}\text{beta}=\frac{60-43}{61.5-43}\times 100\%\\ =91.9\%\end{array}$

Thus, the weight percentage of alpha and beta is $\text{8}\text{.1}\text{wt\%}\text{and}\text{91}\text{.9}\text{wt\%}$ respectively.

1. iv. Show the microstructure as in Figure (3).

To determine

Repeat the same at $2800\text{°C}$.

Explanation of Solution

Refer Figure P8.28, and make a phase analysis at a temperature of $\text{2800°C}$,

1. i. Phases present or available are beta and liquid, that is $\beta +L$.
2. ii. The chemical composition of Os in liquid phase is 45 wt% and in beta phase is 85 wt%.
3. iii. Express the weight percentage of liquid and beta:

   $\begin{array}{c}\text{wt\%}\text{liquid}=\frac{85-60}{85-45}\times 100\%\\ =62.5\%\end{array}$

   $\begin{array}{c}\text{wt\%}\text{beta}=\frac{60-45}{85-45}\times 100\%\\ =37.5\%\end{array}$

Thus, the weight percentage of liquid and beta is $\text{62}\text{.5}\text{wt\%}\text{and}37.5\text{wt\%}$ respectively.

1. iv. Show the microstructure as in Figure (4).