Chapter 17, Problem 17.18P

Interpretation Introduction

(a)

Interpretation:

The volume fraction of silicon carbide for given grinding wheel should be calculated.

Concept introduction:

According to the rules of mixtures for particulate composites:

  $\begin{array}{l}{\rho }_c={\displaystyle \sum \left(f_i{\rho }_i\right)}\text{for}\left\{\text{i}\text{=}\text{1}\mathrm{..........}\text{n}\right\}\\ {\rho }_c=f_1{\rho }_1+f_2{\rho }_2\mathrm{...................}{f}_n{\rho }_n\end{array}$

Where

  ${\rho }_c=$ density of composite

  $f_i=$ volume fraction of i component

  ${\rho }_i=$ density of i component

Volume fraction is defined as:

  $\begin{array}{l}f=\frac{V_1}{V_1+V_2}\\ f=\text{volume}\text{fraction}\\ V_1=\text{volume}\text{of}\text{first}\text{component}\\ V_2=\text{volume}\text{of}\text{second}\text{component}\\ \text{Where}{\text{V}}_{\text{1}}\text{and}{\text{V}}_{\text{2}}\\ {\text{V}}_{\text{1}}\text{=}\frac{\text{weight}\text{of}\text{first}\text{component}}{\text{density}\text{of}\text{first}\text{component}}\\ {\text{V}}_{\text{2}}\text{=}\frac{\text{weight}\text{of}\text{second}\text{component}}{\text{density}\text{of}\text{second}\text{component}}\\ \text{in}\text{terms}\text{of}\text{weight}\text{and}\text{density}\end{array}$

  $f=\frac{{\text{V}}_{\text{1}}\text{=}\frac{\text{weight}\text{of}\text{first}\text{component}}{\text{density}\text{of}\text{first}\text{component}}}{{\text{V}}_{\text{1}}\text{=}\frac{\text{weight}\text{of}\text{first}\text{component}}{\text{density}\text{of}\text{first}\text{component}}+V_2=\frac{\text{weight}\text{of}\text{second}\text{component}}{\text{density}\text{of}\text{second}\text{component}}}$

Answer to Problem 17.18P

The volume fraction of silicon carbide is $0.336571$.

Explanation of Solution

Given information:

Diameter of grinding wheel = $9\text{inch}$

Thickness of grinding wheel = $1\text{inch}$

Weight of grinding wheel = $6\text{lb}$

Density of silicon carbide = $3.2\text{g}{\text{/cm}}^{\text{3}}$

Wheel is $5\%$ porous

Cubes of silicon carbide = $\text{0}\text{.04 cm}$

Based on given information:

Applying rule of mixing,

  ${\rho }_c=f_1{\rho }_1+f_2{\rho }_2+f_3{\rho }_3\mathrm{..............}\left(1\right)$

  ${\rho }_c$ = density of wheel

  $f_1$ = the volume fraction based on porosity

  $f_2$ = the volume fraction of silicon carbide

  $f_3$ = the volume fraction of silica glass

  ${\rho }_1$ = density of porous material is $0$

  ${\rho }_2$ = density of silicon carbide is $3.2{\text{g/cm}}^{\text{3}}$

  ${\rho }_3$ = density of vulcanized rubber is $2.5{\text{g/cm}}^{\text{3}}$

Calculation of volume fraction of wheel with the use of porosity:

  $\begin{array}{l}{f}_1=\left(\frac{5}{100}\right)\\ f_1=0.05\end{array}$

Let the volume of grinding wheel be V:

  $\begin{array}{l}{V}_{\text{Wheel}}=\frac{\pi }{4}{d}^{2}t\\ \text{d}=9\text{inch}\\ 1\text{inch}=2.54cm\\ \text{t}=1\text{inch}\\ V_{\text{Wheel}}=\frac{\pi }{4}{\left(\text{9 inch}\times \frac{\text{2}\text{.54}\text{cm}}{\text{1}\text{inch}}\right)}^2\times 1\text{ inch}\times \frac{\text{2}\text{.54}\text{cm}}{\text{1}\text{inch}}\\ V_{\text{Wheel}}=1402.499{\text{cm}}^3\end{array}$

Calculation of density using weight of grinding wheel and volume of grinding wheel

  $\begin{array}{l}{\text{ρ}}_{\text{grinding}\text{wheel}}\text{=}\frac{\text{weight}\text{of}\text{grinding}\text{wheel}}{\text{volume}\text{of}\text{grinding}\text{wheel}}\\ {\text{ρ}}_{\text{grinding}\text{wheel}}\text{=}\frac{\text{6}\text{lb}}{\text{1042}\text{.499}{\text{cm}}^{\text{3}}}\end{array}$

Conversion of lb/cm³ to g/cm³:

  $\begin{array}{l}\text{1}\text{lb}\text{=}\text{0}\text{.45359}\text{kg}\\ \text{1}\text{kg}\text{=}\text{1000}\text{gram}\end{array}$

  ${\text{ρ}}_{\text{grinding}\text{wheel}}\text{=}\frac{\text{6}\text{lb×}\frac{\text{0}\text{.45359}\text{kg}}{\text{1}\text{lb}}\text{×}\frac{\text{1000}\text{gram}}{\text{1}\text{kg}}}{\text{1042}\text{.499}{\text{cm}}^{\text{3}}}$

  ${\rho }_{grindingwheel}=2.6106\text{g}\text{/}{\text{cm}}^{\text{3}}$

The density of grinding wheel is $2.6106\text{g}\text{/}{\text{cm}}^{\text{3}}$.

Applying rule of mixing from equation (1):

  ${\rho }_c=f_1{\rho }_1+f_2{\rho }_2+f_3{\rho }_3$

The value of volume fraction of silicon carbide and silica glass is unknown.

From the rule of mixing,

  $\begin{array}{l}{f}_1+f_2+f_3=1\\ f_3=1-f_1-f_2\\ f_3=1-0.05-f_2\end{array}$

Substituting the volume fraction of silica glass in terms of volume fraction of silicon carbide and porosity:

  $2.6106=\left(0.05\times 0\right)+\left(3.2\times f_2\right)+\left(2.5\right)\times \left(1-0.05-f_2\right)$

  $f_2=0.366571$

The value of volume fraction of silicon carbide = $0.366571$

Interpretation Introduction

(b)

Interpretation:

The number of particles of silicon carbide lost by wheel when diameter is worn to 8 inch should be calculated

Concept introduction:

Volume fraction is defined as the ratio of volume of specified component to the sum of total volume of components.

  $\begin{array}{l}f=\frac{V_1}{V_1+V_2}\\ f=\text{volume}\text{fraction}\\ V_1=\text{volume}\text{of}\text{first}\text{component}\\ V_2=\text{volume}\text{of}\text{second}\text{component}\end{array}$

Volume of cubic particle is defined as:

  $V_{cubicparticle}={\left(a\right)}^3$

Answer to Problem 17.18P

The number of particles of silicon carbide = $1150646.830$.

Explanation of Solution

Given information:

Diameter of grinding wheel = $9\text{inch}$

Thickness of grinding wheel = $1\text{inch}$

Weight of grinding wheel = $6\text{lb}$

Density of silicon carbide = $3.2\text{g}{\text{/cm}}^{\text{3}}$

Density of silicon glass = $2.5\text{g}{\text{/cm}}^{\text{3}}$

Wheel is $5\%$ porous

Cubes of silicon carbide = $0.04$

Number of particles per cubic centimeter is defined as the ratio of number of silicon carbide particle to volume of silicon carbide particle:

  $n=\frac{numberofsiliconcarbideparticle}{volumeofsiliconcarbideparticle}$

Volume of silicon carbide particle = $V_{initialofSiCparticle}-V_{finalofSiCparticle}$

Calculation of initial volume of particle:

  $\begin{array}{l}{V}_{initialvolumeofSiCparticle}=\frac{\pi }{4}{\left(9\right)}^2\times 1\times \frac{\text{2}\text{.54}{\text{cm}}^{\text{3}}}{\text{1}\text{inch}}\\ V_{initialvolumeofSiCparticle}=1402.499{\text{cm}}^3\end{array}$

Calculation of final volume of particle:

  $\begin{array}{l}{V}_{finalofSiCparticle}=\frac{\pi }{4}\times d^2\times t\\ V_{finalofSiCparticle}=\frac{\pi }{4}\times {\left(8\right)}^2\times 1\times \frac{2.54\text{cm}}{1}\\ V_{finalofSiCparticle}=823.7036{\text{cm}}^3\end{array}$

Volume of silicon carbide = $\text{1042}\text{.499-823}\text{.7}\text{=}\text{218}\text{.799}{\text{cm}}^{\text{3}}$

For cubic particle of 0.04 cubes of silicon carbide:

  $\begin{array}{l}{V}_{cubicparticle}={\left(a\right)}^3\\ V_{cubicparticle}={\left(0.04\right)}^3\\ V_{cubicparticle}=6.4\times {10}^{-5}{\text{cm}}^3\end{array}$

Number of silica particle lost is defined as,

  $n_{lost}=\text{}$ (Number of silicon carbide particle per cubic centimeter) $\times$ (Volume of wheel lost in wear cubic centimeter)

For 1 cubic centimeter of particle of silicon carbide:

  $\begin{array}{l}{n}_{SiC}=f_2\times 1c{m}^3\\ n_{SiC}=0.336571c{m}^3\end{array}$

Number of silicon carbide particle per cubic centimeter = $\frac{0.336571}{6.4\times {10}^{-5}}=5258.9218$

  $\begin{array}{l}{n}_{lost}=\text{}5258.9218\times 218.799\\ n_{lost}=\text{}1150646.830\end{array}$

The number of silicon carbide lost particle = $1150646.830$