Chapter 5, Problem 5.13P

(1)

To determine

The phases present in chemical composition.

Explanation of Solution

Introduction:

A phaseis a homogeneous portion of a material that has uniform physical and chemical characteristics. A phase transformation occurs when a material changes from one phase to another. Phase transformations enter into many material processes, such as when parts are soldered or welded, or when metal is melted and cast into a shape that is then cooled to form a solid part.

(a)

The composition of $95$ atomic % Ni and temperature $1200°\text{C}$ occurs in the single phase region of Ni. Thus, Ni present in the phase diagram is the FCC (Face-Centered-Cubic) Ni crystal structure containing additional Al atoms. So, the number of phase available is $1$.

(b)

The composition of $80$ atomic % Ni and temperature $1000°\text{C}$ occurs in the double phase region of $\text{Ni}$ and ${\text{AlNi}}_3$. Thus, $\text{Ni}$ and ${\text{AlNi}}_3$ present in the phase diagram is the FCC (Face-Centered-Cubic) Ni crystal structure containing additional Al atoms. So, the number of phase available is $2$.

(c)

The composition of $40$ atomic % Ni and temperature $1134°\text{C}$ occurs in the double phase region of $\text{Ni}$ and ${\text{AlNi}}_3$. Thus, $\text{Ni}$ and ${\text{AlNi}}_3$ present in the phase diagram is the FCC (Face-Centered-Cubic) Ni crystal structure containing additional Al atoms. So, the number of phase available is $2$.

(d)

The composition of $40$ atomic % Ni and temperature $1133°\text{C}$ occurs in the three phase region of $\text{Ni}$ and ${\text{AlNi}}_3$. Thus, $\text{Ni}$ and ${\text{AlNi}}_3$ present in the phase diagram is the FCC (Face-Centered-Cubic) Ni crystal structure containing additional Al atoms. So, the number of phase available is $3$.

(e)

The composition of $40$ atomic % Ni and temperature $1133°\text{C}$ occurs in the single phase region of $\text{Ni}$ and ${\text{AlNi}}_3$. Thus, $\text{Ni}$ and ${\text{AlNi}}_3$ present in the phase diagram is the FCC (Face-Centered-Cubic) Ni crystal structure containing additional Al atoms. So, the number of phase available is $1$.

Tabulate the obtained result.

S.No	Atomic Percent Nickel	Temperature	Phase
1	$95\text{atomic\%}\text{Ni}$	$1200°\text{C}$	$1$
2	$\text{80}\text{atomic\%}\text{Ni}$	$1000°\text{C}$	$2$
3	$40\text{atomic\%}\text{Ni}$	$1134°\text{C}$	$2$
4	$40\text{atomic\%}\text{Ni}$	$1133°\text{C}$	$3$
5	$40\text{atomic\%}\text{Ni}$	$1132°\text{C}$	$1$

Table (1)

Conclusion:

Therefore, the phases present in chemical composition is shown in Table (1).

(2)

To determine

The degree of freedom in the material from Gibbs phase rule.

Answer to Problem 5.13P

The degree of freedom is shown in Table (2) .

Explanation of Solution

Formula Used:

Write the expression for the Gibbs phase rule as:

  $P+F=C+1$...... (I)

Here, $P$ is the number of phase, $F$ is the degree of freedom and $C$ is the composition.

Calculation:

(a)

  $95$ atom percent Ni and $1200°\text{C}$ :

Substitute $1$ for $P$ and $2$ for $C$ in equation (I)

  $\begin{array}{c}1+F=2+1\\ F=2\end{array}$

(b)

  $80$ atom percent Ni and $1000°\text{C}$ :

Substitute $2$ for $P$ and $2$ for $C$ in equation (I)

  $\begin{array}{c}2+F=2+1\\ F=1\end{array}$

(c)

  $40$ atom percent Ni and $1134°\text{C}$ :

Substitute $2$ for $P$ and $2$ for $C$ in equation (I)

  $\begin{array}{c}2+F=2+1\\ F=1\end{array}$

(d)

  $40$ atom percent Ni and $1133°\text{C}$ :

Substitute $3$ for $P$ and $2$ for $C$ in equation (I)

  $\begin{array}{c}3+F=2+1\\ F=0\end{array}$

(e)

  $40$ atom percent Ni and $1132°\text{C}$ :

Substitute $1$ for $P$ and $2$ for $C$ in equation (I)

  $\begin{array}{c}1+F=2+1\\ F=2\end{array}$

Tabulate the obtained result.

S.No	Atomic Percent Nickel	Temperature	Degree of freedom
1	$95\text{atomic\%}\text{Ni}$	$1200°\text{C}$	$2$
2	$\text{80}\text{atomic\%}\text{Ni}$	$1000°\text{C}$	$1$
3	$40\text{atomic\%}\text{Ni}$	$1134°\text{C}$	$1$
4	$40\text{atomic\%}\text{Ni}$	$1133°\text{C}$	$0$
5	$40\text{atomic\%}\text{Ni}$	$1132°\text{C}$	$2$

Table (2)

Conclusion:

Therefore, the degree of freedom is shown in Table (2) .

(3)

To determine

The chemical composition of phase.

Explanation of Solution

Introduction:

Many two-component materials have two phases in equilibrium that have different chemical compositions. Low-carbon steel has the two components iron and carbon, and at room temperature it has two phases. The first phase is $\alpha$ iron, which is BCC iron with a few carbon interstitial atoms, and the second phase is

iron carbide (Fe3C).

(a)

The chemical composition present at the temperature of $1200°\text{C}$ is $95$ atomic % of Ni and many two-component materials have two phases in equilibrium that have different chemical compositions. Thus,

  $C_{\text{Ni}}=95\text{atomic\%Ni}$.

(b)

The chemical composition present at the temperature of $1000°\text{C}$ is $80$ atomic % of Ni in $\text{Ni}/\text{atoms}$ in $\text{Ni}$.Thus, $\text{Ni}$ phase has $80$ atomic % $\text{Ni}$ and $20$ atomic % $\text{Al}$.Also, $75$ atomic % $\text{Ni}$ in ${\text{AlNi}}_3/\text{Total}\text{atoms}$ in ${\text{AlNi}}_3$ which is the stoichiometric composition of ${\text{AlNi}}_3$.

Thus,

  

(c)

The chemical composition present at the temperature of $1134°\text{C}$ is $40$ atomic % of Ni in $\text{Ni}/\text{atoms}$ in $\text{Ni}$.Thus, $\text{Ni}$ phase has $40$ atomic % $\text{Ni}$ and $27$ atomic % $\text{Al}$.Also, $43$ atomic % $\text{Ni}$ in ${\text{AlNi}}_3/\text{Total}\text{atoms}$ in $\text{L}$ which is the stoichiometric composition of ${\text{AlNi}}_3$.

Thus,

  $\begin{array}{c}{C}_{\text{Ni}}^{\text{L}}=27\text{atomic\%Ni}\\ C_{\text{Ni}}^{{\text{AlNi}}_3}=43\text{atomic\%Ni}\end{array}$

(d)

The chemical composition present at the temperature of $1133°\text{C}$ is obtained by the phase with $1133°\text{C}$ temperature line.

Thus,

  $\begin{array}{c}{C}_{\text{Ni}}^{\text{L}}=27\text{atomic\%Ni}\text{in}\text{L}/\text{total}\text{atoms}\text{in}\text{L}\\ C_{\text{Ni}}^{{\text{Al}}_2{\text{Ni}}_3}=40\text{atomic\%Ni}\text{in}{\text{Al}}_2{\text{Ni}}_3/\text{total}\text{atoms}\text{in}{\text{Al}}_2{\text{Ni}}_3\\ C_{\text{Ni}}^{{\text{AlNi}}_3}=4240\text{atomic\%Ni}\text{in}{\text{Al}}_2{\text{Ni}}_3/\text{total}\text{atoms}\text{in}{\text{Al}}_2{\text{Ni}}_3\end{array}$

(e)

The chemical composition present at the temperature of $1132°\text{C}$ is obtained by the phase with $1132°\text{C}$ temperature line.

Thus,

  $C_{\text{Ni}}^{{\text{Al}}_2{\text{Ni}}_3}=40\text{atomic\%Ni}$

Tabulate the obtained result.

S.No	Atomic Percent Nickel	Temperature	Chemical composition
1	$95\text{atomic\%}\text{Ni}$	$1200°\text{C}$	$C_{\text{Ni}}=95\text{atomic\%Ni}$
2	$\text{80}\text{atomic\%}\text{Ni}$	$1000°\text{C}$
3	$40\text{atomic\%}\text{Ni}$	$1134°\text{C}$	$\begin{array}{l}{C}_{\text{Ni}}^{\text{L}}=27\text{atomic\%Ni}\\ C_{\text{Ni}}^{{\text{AlNi}}_3}=43\text{atomic\%Ni}\end{array}$
4	$40\text{atomic\%}\text{Ni}$	$1133°\text{C}$	$\begin{array}{l}{C}_{\text{Ni}}^{\text{L}}=27\text{atomic\%Ni}\text{in}\text{L}/\text{total}\text{atoms}\text{in}\text{L}\\ C_{\text{Ni}}^{{\text{Al}}_2{\text{Ni}}_3}=40\text{atomic\%Ni}\text{in}{\text{Al}}_2{\text{Ni}}_3/\text{total}\text{atoms}\text{in}{\text{Al}}_2{\text{Ni}}_3\\ C_{\text{Ni}}^{{\text{AlNi}}_3}=4240\text{atomic\%Ni}\text{in}{\text{Al}}_2{\text{Ni}}_3/\text{total}\text{atoms}\text{in}{\text{Al}}_2{\text{Ni}}_3\end{array}$
5	$40\text{atomic\%}\text{Ni}$	$1132°\text{C}$	$C_{\text{Ni}}^{{\text{Al}}_2{\text{Ni}}_3}=40\text{atomic\%Ni}$

Table (3)

Conclusion:

Thus, the chemical composition of phase is shown in Table (3).

(4)

To determine

The atomic fraction of each phase.

Answer to Problem 5.13P

The atomic fraction for each phase is shown in Table (4) .

Explanation of Solution

Formula Used:

For $\text{Ni}$ phase:

Write the expression for the atomic fraction of the $\alpha$ phase as:

  $f^{\alpha }=\frac{C_0-C_{\text{Ni}}^{{\text{AlNi}}_3}}{C_{\text{Ni}}^{\text{Ni}}-C_{\text{Ni}}^{{\text{AlNi}}_3}}$...... (II)

Here, $C_0$ is the original chemical composition.

For $\text{Al}$ phase:

Write the expression for the atomic fraction of the ${\text{AlNi}}_3$ phase as:

  $f^{{\text{AlNi}}_3}=1-f^{\text{Ni}}$...... (III)

For $\text{L}$ phase:

Write the expression for the atomic fraction of the $\alpha$ phase as:

  $f^{\alpha }=\frac{C_0-C_{\text{Ni}}^{{\text{AlNi}}_3}}{C_{\text{Ni}}^{\text{L}}-C_{\text{Ni}}^{{\text{AlNi}}_3}}$...... (IV)

Here, $C_0$ is the original chemical composition.

For ${\text{AlNi}}_3$ phase:

Write the expression for the atomic fraction of the ${\text{AlNi}}_3$ phase as:

  $f^{{\text{AlNi}}_3}=1-f^{\text{L}}$...... (V)

Calculation:

(a)

The atomic fraction of the Ni phase at the temperature of $1200°\text{C}$ is $1$.As the yield strength of iron is linear with respect to the atomic fraction of solid-solution carbon atoms to the one-half power. An upper yield point and a lower yield point are present in low-carbon steel. The upper yield point is the stress necessary to force the dislocations away from the carbon solid-solution atoms.

(b)

Substitute $0.85$ for $C_0$ , $0.75$ for $C_{\text{Ni}}^{{\text{AlNi}}_3}$ and $0.87$ for $C_{\text{Ni}}^{\text{Ni}}$ in equation (II)

  $\begin{array}{c}{f}^{\alpha }=\frac{0.80-0.75}{0.87-0.75}\\ =0.42\end{array}$

Substitute $0.42$ for $f^{\text{Ni}}$ in equation (III)

  $\begin{array}{c}{f}^{{\text{AlNi}}_3}=1-0.42\\ =0.58\end{array}$

(c)

Substitute $0.40$ for $C_0$ , $0.43$ for $C_{\text{Ni}}^{{\text{AlNi}}_3}$ and $0.27$ for $C_{\text{Ni}}^{\text{Ni}}$ in equation (IV)

  $\begin{array}{c}{f}^{\text{L}}=\frac{0.40-0.43}{0.27-0.43}\\ =0.19\end{array}$

Substitute $0.42$ for $f^{\text{Ni}}$ in equation (V)

  $\begin{array}{c}{f}^{{\text{AlNi}}_3}=1-0.19\\ =0.81\end{array}$

(d)

The Peritectic reaction occurs at the temperature of $1133°\text{C}$ , so it is not possible to calculate the atomic fraction of the phase.

(e)

The atomic fraction of ${\text{AlNi}}_3$ in ${\text{AlNi}}_3/\text{total}\text{atoms}$ in ${\text{AlNi}}_3$ phase at the temperature of $1132°\text{C}$ is $1$.

Tabulate the obtained result.

S.No	Atomic Percent Nickel	Temperature	Atomic fraction
1	$95\text{atomic\%}\text{Ni}$	$1200°\text{C}$	$1$
2	$\text{80}\text{atomic\%}\text{Ni}$	$1000°\text{C}$	$0.42$ and $0.58$
3	$40\text{atomic\%}\text{Ni}$	$1134°\text{C}$	$0.19$ and $0.81$
4	$40\text{atomic\%}\text{Ni}$	$1133°\text{C}$	Not possible
5	$40\text{atomic\%}\text{Ni}$	$1132°\text{C}$	$1$

Table (4)

Conclusion:

Thus, the atomic fraction for each phase is shown in Table (4) .