Chapter 12, Problem 12.72P

Interpretation Introduction

(a)

Interpretation:

To estimate the number of phases.

Concept Introduction:

The graphical representation of the physical state of a substance which is under various conditions of pressure and temperature is known as phase diagram. A phase diagram has temperature on the x-axis and pressure on the y-axis. It is a type of chart which is used to demonstrate conditions at which thermodynamically different phases takes place and coexist at equilibrium.

Answer to Problem 12.72P

The phases which are present in the steel are ferrite, pearlite, cementite, Primary ferrite, Primary cementite with the amount of $\text{97\%, 3\%, 23}\text{.8\%, 76}\text{.2\%}$, respectively.

Explanation of Solution

Given information:

Carbon%	Yield strength
0.2	295
0.4	353
0.6	372
0.8	376
0.95	379

To calculate the number of phases and microconstituents present at a given percentage of carbon (%C), apply the lever rule,

To calculate the percentage of ferrite (a)

  $\text{\% α=}\frac{\text{6}\text{.67- \%C}}{\text{6}\text{.67-0}}\text{×100}$  ......(1)

  $\text{\% α=}\frac{\text{6}\text{.67-\%C}}{\text{6}\text{.67}}\text{×100}\text{\%C= 0}\text{.2}$

Where,

  $\text{\% α=}\frac{\text{6}\text{.67-0}\text{.2}}{\text{6}\text{.67}}\text{×100}$

  $\text{\% α}$

  $\text{=97\%}$

To calculate the percent of ferrite in $\text{0}\text{.4\%c}$, substitute this value in equation (1)

  $\text{\% α=}\frac{\text{6}\text{.67-0}\text{.4}}{\text{6}\text{.67}}\text{×100}$

  $\text{\% α}$ = $\text{94\%}$

To calculate the percent of ferrite in $\text{0}\text{.6 \%C}$, substitute this value in equation (1)

  $\text{\% α=}\frac{\text{6}\text{.67-0}\text{.6}}{\text{6}\text{.67}}\text{×100}$

  $\text{\% α}$ = $\text{91\%}$

To calculate the percent of ferrite in $\text{0}\text{.8 \%C}$, substitute this value in equation (1)

  $\text{\% α=}\frac{\text{6}\text{.67-0}\text{.8}}{\text{6}\text{.67}}\text{×100}$

  $\text{\% α}$ =88%

To calculate the percent of ferrite in $\text{0}\text{.95 \%C}$, substitute this value in equation (1)

  $\text{\% α=}\frac{\text{6}\text{.67-0}\text{.95}}{\text{6}\text{.67}}\text{×100}$

  $\text{\% α}$ = $\text{85}\text{.8\%}$

Now,

Applying the lever rule to calculate the percentage of cementite (Fe₃C) available,

  ${\text{\% Fe}}_{\text{3}}\text{C=}\frac{\text{\%C-0}}{\text{6}\text{.67-0}}\text{×100}$  ......(2)

Percentage of cementite for $\text{0}\text{.2 \%C}$,

  ${\text{\% Fe}}_{\text{3}}\text{C=}\frac{\text{0}\text{.2}}{\text{6}\text{.67}}\text{×100}$

  ${\text{\% Fe}}_{\text{3}}\text{C}$

  $\text{=3\%}$

Percentage of cementite for $\text{0}\text{.4 \%C,}$

  ${\text{\% Fe}}_{\text{3}}\text{C=}\frac{\text{0}\text{.4}}{\text{6}\text{.67}}\text{×100}$

  ${\text{\% Fe}}_{\text{3}}\text{C}$ = $\text{6\%}$

Percentage of cementite for $\text{0}\text{.6 \%C}$ ,

  ${\text{\% Fe}}_{\text{3}}\text{C=}\frac{\text{0}\text{.6}}{\text{6}\text{.67}}\text{×100}$

  ${\text{\% Fe}}_{\text{3}}\text{C}$ =

Percentage of cementite for 0.8 %C,

  ${\text{\% Fe}}_{\text{3}}\text{C=}\frac{\text{0}\text{.8}}{\text{6}\text{.67}}\text{×100}$

  ${\text{\% Fe}}_{\text{3}}\text{C}$ =12%

Percentage of cementite for 0.95 %C,

  ${\text{\% Fe}}_{\text{3}}\text{C=}\frac{\text{0}\text{.95}}{\text{6}\text{.67}}\text{×100}$

  ${\text{\% Fe}}_{\text{3}}\text{C}$ = $\text{14}\text{.2\%}$

The composition which is present when the carbon content is less than $\text{0}\text{.77\%}$ are as follows:

1. Pearlite
2. Primary Ferrite

The lever rule to determine the percentage of pearlite,

  $\text{\% Pearlite=}\frac{\text{\%C-0}\text{.0218}}{\text{0}\text{.77-0}\text{.0218}}\text{×100}$  ......(3)

Percentage of pearlite containing $\text{0}\text{.2 \%C}$,

  $\text{\% Pearlite=}\frac{\text{0}\text{.2-0}\text{.0218}}{\text{0}\text{.77-0}\text{.0218}}\text{×100}$

  $\text{\% Pearlite}$ = $\text{=0}\text{.238×100}$

  $\text{\% Pearlite}$ = $\text{= 23}\text{.8\%}$

Percentage of pearlite containing $\text{0}\text{.4 \%C}$,

  $\text{\% Pearlite=}\frac{\text{0}\text{.4-0}\text{.0218}}{\text{0}\text{.77-0}\text{.0218}}\text{×100}$

  $\text{\% Pearlite}$ = $\text{77}\text{.3\%}$

Percentage of pearlite containing $\text{0}\text{.8 \%C}$ ,

  $\text{\% Pearlite=}\frac{\text{0}\text{.8-0}\text{.0218}}{\text{0}\text{.77-0}\text{.0218}}\text{×100}$

  $\text{\% Pearlite}$ = $\text{99}\text{.5\%}$

Percentage of pearlite containing $\text{0}\text{.95 \%C}$ ,

  $\text{\% Pearlite=}\frac{\text{0}\text{.95-0}\text{.0218}}{\text{0}\text{.77-0}\text{.0218}}\text{×100}$

  $\text{\% Pearlite}$ = $\text{96}\text{.9\%}$

Apply the lever rule to calculate the percentage of primary ferrite.

  $\text{\% Primary Ferrite=}\frac{0.77-\%C}{0.7482}\times 100$  ......(4)

Percentage of primary ferrite containing 0.2 %C,

  $\text{\% Primary Ferrite=}\frac{0.77-0.2}{0.7482}\times 100$

  $\text{\% Primary Ferrite}$ = $\text{76}\text{.2\%}$

Percentage of primary ferrite containing $\text{0}\text{.4 \%C}$,

  $\text{\% Primary Ferrite=}\frac{0.77-0.4}{0.7482}\times 100$

  $\text{\% Primary Ferrite}$ = $\text{49}\text{.5\%}$

Percentage of primary ferrite containing $\text{0}\text{.6 \%C}$ ,

  $\text{\% Primary Ferrite=}\frac{0.77-0.6}{0.7482}\times 100$

  $\text{\% Primary Ferrite}$ = $\text{22}\text{.7\%}$

The composition which is present in steel for carbon content more than 0.77% are as follows:

1. Pearlite
2. Primary Cementite

Applying the lever rule to find the percentage of primary cementite,

  $\text{\% Primary Cementite=}\frac{\%C-0.77}{5.9}\times 100$  ......(5)

Percentage of Primary cementite for $\text{0}\text{.8 \%C}$,

  $\text{\% Primary Cementite=}\frac{0.8-0.77}{5.9}\times 100$

  $\text{\% Primary Cementite}$ = $\text{0}\text{.5\%}$

Percentage of Primary cementite for $\text{0}\text{.95 \%C}$ ,

  $\text{\% Primary Cementite=}\frac{0.95-0.77}{5.9}\times 100$

  $\text{\% Primary Cementite}$ = $\text{3}\text{.1\%}$

Interpretation Introduction

(b)

Interpretation:

To determine the carbon content in each phase.

Concept Introduction:

Usually, carbon is the most important commercial steel alloy. By increasing the carbon content, one can increase the hardness and strength. The hardenability can also be increased by increasing the carbon content. Steel is known as an alloy of carbon and iron, and sometimes of other elements. Due to its high tensile strength and low cost, it is a major component used in infrastructure, buildings, ships, tools etc. Increasing the carbon content of steel is disadvantageous too. It increases the brittleness and reduces the weldability as it has the tendency to form martensite.

Answer to Problem 12.72P

The carbon content present is $\text{0}\text{.2 wt\%}$ and 6.67 wt%.

Explanation of Solution

  

The value of the amount of carbon content in each phase is as follows:

1. Ferrite (a) - The ferrite contains pure iron (Fe) at the room temperature and consists of $\text{0}\text{.2\%}$ the weight of carbon.
2. Cementite (Fe₃C) − The cementite contains $\text{6}\text{.67\%}$ weight of carbon,

Interpretation Introduction

(c)

Interpretation:

To plot the carbon content of pearlite, Cementite versus yield strength.

Concept Introduction:

Usually, carbon is the most important commercial steel alloy. By increasing the carbon content, one can increase the hardness and strength. The hardenability can also be increased by increasing the carbon content. Steel is known as an alloy of carbon and iron, and sometimes of other elements. Due to its high tensile strength and low cost, it is a major component used in infrastructure, buildings, ships, tools etc. Increasing the carbon content of steel is disadvantageous too. It increases the brittleness and reduces the weldability as it has the tendency to form martensite.

Answer to Problem 12.72P

  

Explanation of Solution

  

From the above-mentioned figure, the amount of hard cementite increases as the percentage of carbon content increases. But the increase in the percentage of carbon decreases the amount of ferrite present.

Hence it leads to an increase in yield strength. Eventually, this will increase the amount of pearlite present in the steel.