Chapter 8.15, Problem 66SEP

(a) In the Ti–Al phase diagram. Figure P8.42, what phases ate available at an overall alloy composition of Ti-63 wt% A1 at temperatures below 1300°C? (b) What is the significance of the vertical line at that alloy composition? (c) Verify the formula next to the vertical line, (d) Compare the melt temperature of this compound to that of Ti and Al. What is your conclusion?

To determine

In the Ti–Al phase diagram, Figure P8.42, what phases are available at an overall alloy composition of Ti–63 wt% Al at temperatures below 1300°C.

Explanation of Solution

Refer Figure P8.42, "Ti–Al phase diagram", is presenting the formation or development of an intermetallic compound between aluminum and titanium as Al–63wt% Al indicates some substitution between them.

The available phase between aluminum and titanium is ${\text{TiAl}}_{\text{3}}$.

To determine

What is the significance of the vertical line at that alloy composition.

Explanation of Solution

The vertical line signifies the formation or development of an intermetallic compound as there are two elements of metals and the intermetallic compound is stoichiometric.

To determine

Verify the formula next to the vertical line.

Answer to Problem 66SEP

The formula is ${\text{Al}}_{\text{3}}\text{Ti}$.

Explanation of Solution

Express the number of moles in titanium.

$N_{\text{Ti}}=w{t}_{\text{Ti}}\left[\frac{\text{1}\text{mol}}{{\text{M}}_{\text{Ti}}}\right]$ (I)

Here, number of moles in titanium is $N_{\text{Ti}}$, weight percentage of titanium in compound is $w_{\text{Ti}}$ and atomic mass of titanium is ${\text{M}}_{\text{Ti}}$.

Express the number of moles in aluminum.

$N_{\text{Al}}=w{t}_{\text{Al}}\left[\frac{\text{1}\text{mol}}{{\text{M}}_{\text{Al}}}\right]$ (II)

Here, number of moles in aluminum is $N_{\text{Al}}$, weight percentage of aluminum in compound is $w_{\text{Al}}$ and atomic mass of aluminum is ${\text{M}}_{\text{Al}}$.

Express mole percentage of aluminum.

$\text{mol\%}\text{Al}=\frac{N_{\text{Al}}}{N_{\text{Al}}+N_{\text{Ti}}}$ (III)

Express mole percentage of titanium.

$\text{mol\%}\text{Ti}=\frac{N_{\text{Ti}}}{N_{\text{Al}}+N_{\text{Ti}}}$ (IV)

Conclusion:

To verfify the formula for the compound, let the mass of compound be $\text{100}\text{g}$. Thus, the weight percentage of of titanium is 37wt% and that of alumium is 63wt%.

$\begin{array}{l}{w}_{\text{Ti}}=37\text{wt\%}\\ w_{\text{Al}}=63\text{wt\%}\end{array}$

Write the atomic mass of titanium and aluminum as below:

$\begin{array}{l}{\text{M}}_{\text{Ti}}=47.88\text{g}/\text{mol}\\ {\text{M}}_{\text{Al}}=26.98\text{g}/\text{mol}\end{array}$

Substitute $\text{37}\text{g}$ for $w_{\text{Ti}}$ and $47.88\text{g}/\text{mol}$ for ${\text{M}}_{\text{Ti}}$ in Equation (I).

$\begin{array}{c}{N}_{\text{Ti}}=\left(\text{37}\text{g}\right)\left[\frac{\text{1}\text{mol}}{47.88\text{g}/\text{mol}}\right]\\ =0.77\end{array}$

Substitute $63\text{g}$ for $w_{\text{Al}}$ and $26.98\text{g}/\text{mol}$ for ${\text{M}}_{\text{Al}}$ in Equation (II).

$\begin{array}{c}{N}_{\text{Al}}=\left(63\text{g}\right)\left[\frac{\text{1}\text{mol}}{26.98\text{g}/\text{mol}}\right]\\ =2.36\end{array}$

Substitute $0.77$ for $N_{\text{Ti}}$ and $2.36$ for $N_{\text{Al}}$ in Equation (III).

$\begin{array}{c}\text{mol\%}\text{Al}=\frac{2.36}{2.36+0.77}\\ =0.75\\ =75\%\end{array}$

Substitute $0.77$ for $N_{\text{Ti}}$ and $2.36$ for $N_{\text{Al}}$ in Equation (IV).

$\begin{array}{c}\text{mol\%}\text{Ti}=\frac{0.77}{2.11+0.77}\\ =0.25\\ =25\%\end{array}$

For each titanium atom, there are 3 atoms of aluminum, hence the formula is ${\text{Al}}_{\text{3}}\text{Ti}$.

To determine

Compare the melt temperature of this compound to that of Ti and Al. What is your conclusion.

Explanation of Solution

The melting temperature of ${\text{Al}}_{\text{3}}\text{Ti}$ is just below the temperature of $\text{1400°C}$, which is considerably greater than that of aluminum $\text{660°C}$ but lesser than that of titanium $\left(\text{1670°C}\right)$.

Hence, titanium aluminide can substitute all aluminum in applications that require high temperature requirement with appreciably higher strength.