Chapter 5, Problem 5.18P

(1)

To determine

The transformation product after quenching and then after cool at room temperature.

Explanation of Solution

Introduction:

The process of rapid cooling of a metal in water, or air or oil in order to alter the mechanical properties of the metal is called quenching. The process of quenching helps in the prevention of undesired low-temperature processes like phase transformation. Under normal conditions quenching is performed for few seconds.

(a)

The steel has eutectoid composition that is initially equilibrated at $730°\text{C}$ . When it is quenched to the temperature of $650°\text{C}$ and then hold for about $100$ seconds, then the transformation product formed is the $100\%$ pearlite.

(b)

When the product formed in part (a) is cooled to room temperature, then the transformation product that is formed will be the $100\%$ pearlite. So, the final product after both the process will be $100\%$ pearlite.

Conclusion:

Thus, the product formed after quenching is $100\%$ pearlite and remains $100\%$ pearlite when cooled at room temperature.

(2)

To determine

The transformation product after quenching and then after cool at room temperature.

Explanation of Solution

Introduction:

The process of rapid cooling of a metal in water, or air or oil in order to alter the mechanical properties of the metal is called quenching. The process of quenching helps in the prevention of undesired low-temperature processes like phase transformation. Under normal conditions quenching is performed for few seconds.

(a)

The steel has eutectoid composition that is initially equilibrated at $730°\text{C}$ . When it is quenched to the temperature of $650°\text{C}$ and then holds for about $2$ seconds, then the transformation product formed is the $100\%$ austentite but it is unstable in nature.

(b)

When the product formed in part (a) is cooled to room temperature, then the transformation product that is formed will be the $100\%$ martensite. It is stable in nature, so, the final transformation product will be $100\%$ martensite.

Conclusion:

Thus, the product formed after quenching is $100\%$ unstable austentite and product after cool at room temperature is $100\%$ martensite.

(3)

To determine

The transformation product after quenching and then after cool at room temperature.

Explanation of Solution

Introduction:

The process of rapid cooling of a metal in water, or air or oil in order to alter the mechanical properties of the metal is called quenching. The process of quenching helps in the prevention of undesired low-temperature processes like phase transformation. Under normal conditions quenching is performed for few seconds.

(a)

The steel has eutectoid composition that is initially equilibrated at $730°\text{C}$ . When it is quenched to the temperature of $650°\text{C}$ and then holds for about $10$ seconds, then the transformation product formed is the $50\%$ pearlite and $50\%$ unstable austentite.

(b)

When the product formed in part (a) is cooled to room temperature, then the transformation product that is formed will be the $50\%$ pearlite and $50\%$ martensite. So, the final transformation product will be $50\%$ pearlite and $50\%$ martensite.

Conclusion:

Thus, the product formed after quenching is $50\%$ pearlite and $50\%$ unstable austentite and product after cool at room temperature is $50\%$ pearlite and $50\%$ martensite.

(4)

To determine

The transformation product after quenching and then after cool at room temperature.

Explanation of Solution

Introduction:

The process of rapid cooling of a metal in water, or air or oil in order to alter the mechanical properties of the metal is called quenching. The process of quenching helps in the prevention of undesired low-temperature processes like phase transformation. Under normal conditions quenching is performed for few seconds.

(a)

The steel has eutectoid composition that is initially equilibrated at $730°\text{C}$ . When it is quenched to the temperature of $400°\text{C}$ and then holds for about $3.16$ seconds, then the transformation product formed is the $100\%$ unstable austentite.

(b)

When the product formed in part (a) is cooled to room temperature, then the transformation product that is formed will be the $100\%$ martensite. So, the final transformation product will be $100\%$ martensite.

Conclusion:

Thus, the product formed after quenching is $100\%$ unstable austentite and product after cool at room temperature is $100\%$ martensite.

(5)

To determine

The transformation product after quenching and then after cool at room temperature.

Explanation of Solution

Introduction:

The process of rapid cooling of a metal in water, or air or oil in order to alter the mechanical properties of the metal is called quenching. The process of quenching helps in the prevention of undesired low-temperature processes like phase transformation. Under normal conditions quenching is performed for few seconds.

(a)

The steel has eutectoid composition that is initially equilibrated at $730°\text{C}$ . When it is quenched to the temperature of $400°\text{C}$ and then holds for about $25$ seconds, then the transformation product formed is the $50\%$ bainite and $50\%$ unstable austentite.

(b)

When the product formed in part (a) is cooled to room temperature, then the transformation product that is formed will be the $50\%$ bainite and $50\%$ martensite. So, the final transformation product will be $50\%$ bainite and $50\%$ martensite.

Conclusion:

Thus, the product formed after quenching is $50\%$ bainite and $50\%$ unstable austentite, and product after cool at room temperature is $50\%$ bainite and $50\%$ martensite.

(6)

To determine

The transformation product after quenching and then after cool at room temperature.

Explanation of Solution

Introduction:

The process of rapid cooling of a metal in water, or air or oil in order to alter the mechanical properties of the metal is called quenching. The process of quenching helps in the prevention of undesired low-temperature processes like phase transformation. Under normal conditions quenching is performed for few seconds.

(a)

The steel has eutectoid composition that is initially equilibrated at $730°\text{C}$ . When it is quenched to the temperature of $400°\text{C}$ and then holds for about $200$ seconds, then the transformation product formed is the $100\%$ bainite.

(b)

When the product formed in part (a) is cooled to room temperature, then the transformation product that is formed will be the $100\%$ martensite. So, the final transformation product will be $100\%$ martensite.

Conclusion:

Thus, the product formed after quenching is $100\%$ bainite, and product after cool at room temperature is $100\%$ martensite.

(7)

To determine

The transformation product after quenching and then after heating.

Explanation of Solution

Introduction:

The process of rapid cooling of a metal in water, or air or oil in order to alter the mechanical properties of the metal is called quenching. The process of quenching helps in the prevention of undesired low-temperature processes like phase transformation. Under normal conditions quenching is performed for few seconds.

(a)

The steel has eutectoid composition that is initially equilibrated at $730°\text{C}$ . When it is quenched to the temperature of $0°\text{C}$ and then holds for about $10$ seconds, then the transformation product formed is the $100\%$ bainite.

(b)

When the product formed in part (a) is heated at the temperature of $600°\text{C}$ and holds for $1000$ seconds then, the final product formed will be $100\%$ tempered martensite.

Conclusion:

Thus, the product formed after quenching is $100\%$ bainite, and product after heating is $100\%$ tempered martensite.