Chapter 13, Problem 13.13P

Interpretation Introduction

Interpretation:

The interference in $\text{1080}$ between ferrite and cementite and percent reduction in the surface needs to be determined.

Concept Introduction:

Pearlite is composed by adding layers of ferrite containing $\text{87}\text{.5wt\%}$, and cementite contain $\text{12}\text{.5wt\%}$ in contribution. Pearlite was identified by Sorby and named it sorbate, pearlite is formed by the cooling of austenite. Pearlite is one in the strongest structure.Stainless steels may be attracted into slender cables using a perlitic structure or a near-pearlitic microstructuring. Such cables, often packed into cords, have been used extensively for piano cables, suspension bridge cables, and tire reinforcement steel cable.

Answer to Problem 13.13P

Area of interference of pearlite is ${\text{11,000cm}}^{\text{2}}$ and percentage reduction is $\text{98}\text{.4\%}$.

Explanation of Solution

Given:

Thickness of cementite = ${\text{4×10}}^{\text{-5}}\text{cm}$

Ferrite = ${\text{14×10}}^{\text{-5}}\text{cm}$

Diameter of cementite sphere = ${\text{4×10}}^{\text{-3}}\text{cm}$

Density of ferrite = $\text{7}\text{.87}$ g/cm³

Density of cementite= $\text{7}\text{.66}$ g/cm³

Calculations:

The formula for the lever is given as

  $\text{Xa=}\frac{\text{c-b}}{\text{a-b}}\text{×100}$

Where,

  $\text{ c= 0}\text{.8=}$ Entire alloy

  $\text{a=6}\text{.67 =}$

  $\alpha$ Phase weight percentage

  $\text{b= 0}\text{.0218 =}$ $\theta$ Phase weight percentage

Therefore,

  $\begin{array}{l}\text{\%Fe3c=}\frac{\text{0}\text{.80-0}\text{.0218}}{\text{6}\text{.67-0}\text{.0218}}\text{×100}\\ \\ \text{=}\frac{\text{0}\text{.7782}}{\text{6}\text{.6482}}\text{×100}\\ \\ \text{=11}\text{.705\%}\end{array}$

In 1080 steel alloy weight percent of ferrite,

  $\begin{array}{l}\text{\%Fe=100-\%Fe3c}\\ \text{=100-11}\text{.705}\\ \text{=88}\text{.295\%}\end{array}$

Now, percent of volume fraction of $\text{Fe3c}$ is given by,

  $\text{\%volume fraction Fe3c }$ $\text{=}\frac{\text{[}\frac{\text{weight}\text{}\text{of}\text{}\text{Fe3c}}{\text{density}\text{}\text{of}\text{}\text{Fe3c}}\text{]}}{\text{[}\frac{\text{weight}\text{}\text{of}\text{}\text{Fe3c}}{\text{density}\text{}\text{of}\text{}\text{Fe}}\text{]+[}\frac{\text{weight}\text{}\text{of}\text{}\text{Fe3c}}{\text{density}\text{}\text{of}\text{}\text{Fe}}\text{]}}\text{×100}$ $\begin{array}{l}\text{Substitute, Weight of Fe3c= 11}\text{.705}\hfill \\ \text{Density of Fe3c= 7}\text{.66}\hfill \\ \text{Weight \% of Fe= 88}\text{.295}\hfill \\ \text{Density of Fe = 7}\text{.78}\hfill \end{array}$

Therefore $\text{\%}$ volume fraction becomes

  $\begin{array}{l}\text{=[}\frac{\frac{\text{11}\text{.705}}{\text{7}\text{.66}}}{\text{[}\frac{\text{11}\text{.705}}{\text{7}\text{.66}}\text{]+[}\frac{\text{88}\text{.295}}{\text{7}\text{.87}}\text{]}}\text{]×100}\\ \\ \text{=}\frac{\text{1}\text{.5281}}{\text{1}\text{.5281+11}\text{.219}}\text{×100}\\ \\ \text{=11}\text{.987}\end{array}$

Calculate the number of the interface in the thickness of ferrite and cementite

Number of interface per cm = $\frac{\text{Number of interface}}{\left[\text{thickness of ferrite + thickness of cementite}\right]}$

Where,

  $\begin{array}{l}\begin{array}{l}\text{Number of interface= 2}\hfill \\ {\text{Thickness of ferrite = 4×10}}^{\text{-5}}\text{cm}\hfill \\ {\text{Thickness of cementite =14×10}}^{\text{-5}}\text{cm}\hfill \end{array}\hfill \\ \hfill \end{array}$

Therefore,

Number of interface per cm=

  $\text{= }\frac{\text{2}}{{\text{4×10}}^{\text{-5}}{\text{cm+14×10}}^{\text{-5}}}$

  $\text{=}\frac{\text{2}}{{\text{18×10}}^{\text{-5}}}$

  $\text{=1}{\text{.1×10}}^{\text{4}}$

  $\text{interface /cm}$

Now for the area of interface (A) in pearlite

A= number of interface

  $\begin{array}{l}\text{= }\left(\text{1}{\text{.1×10}}^{\text{4}}\text{interface/cm}\right)\left({\text{1cm}}^{\text{3}}\right)\hfill \\ {\text{ =11,000cm}}^{\text{2}}\hfill \\ \hfill \end{array}$

Calculate the volume of cementite sphere with radius ${\text{2×10}}^{\text{-3cm}}$

  $\begin{array}{l}\text{v=}\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}\\ \\ \text{=}\frac{\text{4}}{\text{3}}{\text{π(2×10}}^{\text{-3}}{\text{cm)}}^{\text{3}}\\ \\ \text{=3}{\text{.35×10}}^{\text{-8}}{\text{cm}}^{\text{3}}\end{array}$

Now, the number of cementite in the sphere,

  $\begin{array}{l}\text{N=}\frac{\text{volume fraction of cementite}}{\begin{array}{l}\text{Volume fraction of cementite sphere}\hfill \\ \hfill \\ \hfill \end{array}}\\ \text{=}\frac{\text{0}{\text{.11987cm}}^{\text{3}}}{\text{3}{\text{.35×10}}^{\text{-8}}{\text{cm}}^{\text{3}}}\\ \\ \text{=3}{\text{.58×10}}^{\text{6}}\end{array}$

Calculate the total surface area of a sphere,

Sphere = ${\text{(4πr}}^{\text{2}}\text{)(N)}$

Where,

  $\text{N=3}{\text{.58×10}}^{\text{6}}$

Interface,

  $\text{\%}$

  ${\text{=180cm}}^{\text{2}}$

  ${\text{interface/cm}}^{\text{3}}$

Now, the percentage reduction in the surface during anodizing pearlite steel

%reduction in the interface area $\text{=}\frac{\text{A-sphere}}{\text{A}}\text{×100}$

Where,

  $\begin{array}{l}{\text{A=11000cm}}^{\text{2}}\\ \\ {\text{sphere=180cm}}^{\text{2}}\\ \end{array}$

  $\text{\%}$ Reduction

  $\begin{array}{l}\text{=}\frac{\text{11000-180}}{\text{11000}}\text{×100}\\ \\ \text{=}\frac{\text{10820}}{\text{11000}}\text{×100}\\ \\ \text{=98}\text{.4\%}\end{array}$

Conclusion

Area of interference of pearlite is ${\text{11,000cm}}^{\text{2}}$ and percentage reduction is $\text{98}\text{.4\%}$.