Chapter 17, Problem 17.20P

Interpretation Introduction

Interpretation:

The volume fraction of original compact and volume fraction of closed porosity 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 17.20P

The volume fraction of original compact = $0.6006$

The volume fraction of closed porosity = $0.0097$

Explanation of Solution

Given information:

Weight of tungsten compact = $125\text{gram}$

Amount of silver infiltered= $105\text{gram}$

Density of composite = $13.8\text{g}{\text{/cm}}^{\text{3}}$

Based on given information,

Applying rule of mixing:

  ${\rho }_c=f_{pore}{\rho }_{pore}+f_{tungsten}{\rho }_{tungsten}+f_{silver}{\rho }_{silver}\mathrm{..............}\left(A\right)$

  ${\rho }_c$ = density of composite is $13.8\text{g}{\text{/cm}}^{\text{3}}$

  $f_{pore}$ = the volume fraction of pore

  $f_{tungsten}$ = the volume fraction of tungsten

  $f_{silver}$ = the volume fraction of silver

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

  ${\rho }_{tungsten}$ = density of tungsten

  ${\rho }_{silver}$ = density of silver

Calculation of volume fraction of tungsten:

  $f_{tungsten}=\frac{V_{tungsten}}{V_{tungsten}+V_{silver}+V_{pore}}$

Volume of tungsten is calculated as:

  $\begin{array}{l}{V}_{tungsten}=\frac{massoftungsten}{densityoftungsten}\\ V_{tungsten}=\frac{\text{125}\text{gram}}{\text{19}\text{.25}{\text{gram/cm}}^{\text{3}}}\\ V_{tungsten}=6.493{\text{cm}}^3\end{array}$

Volume of silver is calculated as:

  $\begin{array}{l}{V}_{silver}=\frac{massofsilver}{densityofsilver}\\ V_{silver}=\frac{\text{105}\text{gram}}{10.49{\text{gram/cm}}^{\text{3}}}\\ V_{silver}=10.009{\text{cm}}^3\end{array}$

Substituting the values of volume of silver and volume of tungsten:

  $\begin{array}{l}{f}_{tungsten}=\frac{V_{tungsten}}{V_{tungsten}+V_{silver}+V_{pore}}\\ f_{tungsten}=\frac{6.493}{6.493+10.009+V_{pore}}\\ f_{tungsten}=\frac{6.493}{16.502+V_{pore}}\mathrm{...................}\left(1\right)\end{array}$

Calculating volume fraction of silver:

  $\begin{array}{l}{f}_{silver}=\frac{V_{silver}}{V_{tungsten}+V_{silver}+V_{pore}}\\ f_{silver}=\frac{10.009}{6.493+10.009+V_{pore}}\\ f_{silver}=\frac{10.009}{16.502+V_{pore}}\mathrm{.......................}\left(2\right)\end{array}$

Calculation of volume fraction for pore:

  $f_{pore}=\frac{V_{pore}}{V_{tungsten}+V_{silver}+V_{pore}}$

  $f_{pore}=\frac{V_{pore}}{6.493+10.009+V_{pore}}$

  $f_{pore}=\frac{V_{pore}}{16.502+V_{pore}}\mathrm{...........................}\left(3\right)$

On substituting the values in equation (A),

Values taken are

Density of pore = 0

Density of tungsten = $19.25g/c{m}^3$

Density of silver = $10.49g/c{m}^3$

  $\begin{array}{l}13.8=\left(\frac{V_{pore}}{16.502+V_{pore}}\right)\times \left(0\right)+\left(\frac{6.493}{16.502+V_{pore}}\right)\times \left(19.25\right)+\left(\frac{10.009}{16.502+V_{pore}}\right)\times \left(10.49\right)\\ V_{pore}=0.163c{m}^3\end{array}$

Calculation of volume fraction of original composite:

  $\begin{array}{l}{f}_{originalcompact}=\frac{V_{silver}}{V_{total}}\\ V_{total}=V_{silver}+V_{tungsten}+V_{pore}\\ V_{total}=10.009+6.493+0.163\\ V_{total}=16.665c{m}^3\end{array}$

Substituting the value:

  $\begin{array}{l}{f}_{originalcompact}=\frac{10.009}{16.665}\\ f_{originalcompact}=0.6006\end{array}$

The required value of volume fraction of original compact is 0.6006.

Calculation of volume fraction of closed porosity:

  $\begin{array}{l}{f}_{closedporosity}=\frac{V_{pore}}{V_{total}}\\ f_{closedporosity}=\frac{0.163}{16.664}\\ f_{closedporosity}=0.0097\end{array}$

The value of volume fraction for closed porosity = 0.0097.

Conclusion

The volume fraction of original compact = $0.6006$

The volume fraction of closed porosity = $0.0097$