Chapter 17, Problem 17.56P

Interpretation Introduction

Interpretation:

The comparison of cost of material in the composite quarter with the quarter made up of nickel should be determined.

Concept introduction:

The rule of mixtures for particulate composites is:

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

Where

  ${\rho }_c=$ density of composite

  $f_i=$ volume fraction of i component

  ${\rho }_i=$ density of i component

Volume fraction is defined as

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

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

Density is defined as the ratio of mass per unit volume expressed as

  $\rho =\frac{m}{V}$

Where,

m is the mass of material in kg.

V is the volume of material in cubic meter.

The unit of density is kg/cubic meter.

Answer to Problem 17.56P

The cost of quarter made up of copper and nickel is ${\left(\text{Cost}\right)}_{quarter}=\$0.02984$.

The cost of quarter made up of nickel is ${\left(\text{Cost}\right)}_{Ni}=\$0.0568$

Thus, the cost of quarter made up of copper and nickel is half of the cost of quarter made up of nickel.

Explanation of Solution

Given information:

Diameter of US quarter = $\frac{15}{16}\text{inch}$.

Thickness of US quarter = $\frac{1}{16}\text{inch}$.

Cost of copper = $\$1.10\text{per}\text{pound}$.

Cost of nickel = $\$4.10\text{per}\text{pound}$.

From given information, the volume of copper in US quarter = $\frac{2}{3}$.

The volume of nickel in US quarter = $\frac{1}{3}$.

Volume fraction of copper $f_{Cu}=0.667$

Volume fraction of nickel $f_{Ni}=0.334$

Calculation of volume of quarter,

  $V_{quarter}=\frac{\pi }{4}\times d^2\times t$

Conversion of d from inch to cm, by multiplying it with 2.54 cm,

  $\begin{array}{l}\text{d}=\frac{15}{16}\text{inch}\times 2.54\text{cm}\\ \text{d}=2.381\text{cm}\end{array}$

Conversion of thickness to cm, by multiplying it with 2.54 cm,

  $\begin{array}{l}\text{t}\text{=}\frac{1}{16}\text{inch}\times 2.54\text{cm}\\ \text{t}\text{=}0.1587\text{cm}\\ V_{quarter}=\frac{\pi }{4}\times {\left(2.381\right)}^2\times 0.1587\\ V_{quarter}=0.7066{\text{cm}}^{\text{3}}\end{array}$

Thus, the volume of quarter = $V_{quarter}=0.7066{\text{cm}}^{\text{3}}$.

Calculation of volume of copper using rule of mixing concept,

  $f_{Cu}=\frac{V_{Cu}}{V_{quarter}}$

Volume fraction is defined as the ratio of volume of copper to the volume of quarter. On rearranging the equation,

  $V_{Cu}=f_{Cu}\times V_{quarter}$

Volume fraction of copper $f_{Cu}=0.667$

  $V_{quarter}=0.7066{\text{cm}}^{\text{3}}$

Putting the values,

  $\begin{array}{l}{V}_{Cu}=0.667\times 0.7066\\ V_{Cu}=0.471{\text{cm}}^3\end{array}$

Therefore, the volume of copper $V_{Cu}=0.471{\text{cm}}^3$

Calculation of volume of nickel using volume fraction relation,

  $\begin{array}{l}{f}_{Ni}=\frac{V_{Ni}}{V_{quarter}}\\ V_{Ni}=f_{Ni}\times V_{quarter}\end{array}$

  $V_{quarter}=0.7066{\text{cm}}^{\text{3}}$

Volume fraction of nickel $f_{Ni}=0.334$

  $\begin{array}{l}{V}_{Ni}=0.334\times 0.7066\\ V_{Ni}=0.2355{\text{cm}}^3\end{array}$

Thus, the volume of nickel $V_{Ni}=0.2355{\text{cm}}^3$.

Calculation of weight of copper using the relationship of weight with density given as,

  $W_{Cu}=V_{Cu}\times {\rho }_{Cu}$

The values of density and volume are,

  $\begin{array}{l}{\rho }_{Cu}=8.93g/c{m}^3\\ V_{Cu}=0.471{\text{cm}}^3\end{array}$

Putting the values in the formula of weight,

  $\begin{array}{l}{W}_{Cu}=0.471\times 8.93\\ W_{Cu}=4.206\text{gram}\end{array}$

Calculation of weight of nickel using the relationship of weight with density given as,

  $W_{Ni}=V_{Ni}\times {\rho }_{Ni}$

The values of density and volume are,

  $\begin{array}{l}{V}_{Ni}=0.2355{\text{cm}}^3\\ {\rho }_{Ni}=8.902\text{g}{\text{/cm}}^{\text{3}}\end{array}$

Putting the values,

  $\begin{array}{l}{W}_{Ni}=0.2355\times 8.902\\ W_{Ni}=2.096\text{gram}\end{array}$

Calculation of cost of copper and nickel using the relation,

  ${\left(\text{Cost}\right)}_{Cu}=\left(\text{Cost}\text{of}\text{copper}\text{per}\text{kg}\right)\times \left({\text{W}}_{\text{Cu}}\right)$

Conversion of per pound to per kg by multiplying it with 2.205

  ${\left(\text{Cost}\right)}_{Cu}=\left(\text{cost}\text{of}\text{copper}\text{per}\text{kg}\right)\times \left({\text{W}}_{\text{Cu}}\right)$

  $W_{Cu}=4.206\text{gram}$

Cost of copper per kg = $\$1.10\times 2.205$ per kg

  $\begin{array}{l}{\left(\text{Cost}\right)}_{Cu}=\left(\$1.10\times 2.205\right)\times \left(4.206g\times \frac{1\text{kg}}{1000\text{gram}}\right)\\ {\left(\text{Cost}\right)}_{Cu}=\$0.0109\end{array}$

Calculation of cost of nickel,

  ${\left(\text{Cost}\right)}_{Ni}=\left(\text{cost}\text{of}\text{nickel}\text{per}\text{kg}\right)\times \left({\text{W}}_{\text{Ni}}\right)$

Conversion of per pound to per kg by multiplying it with 2.205,

  ${\left(\text{Cost}\right)}_{Ni}=\left(\text{cost}\text{of}\text{nickel}\text{per}\text{kg}\right)\times \left({\text{W}}_{\text{Ni}}\right)$

  $W_{Ni}=2.096\text{gram}$

Cost of nickel per kg = $\left(\$4.10\times 2.205\right)$

Putting the values,

  $\begin{array}{l}{\left(\text{Cost}\right)}_{Ni}=\left(\$4.10\times 2.205\right)\times \left(2.096\text{gram}\times \frac{1\text{kg}}{1000\text{gram}}\right)\\ {\left(\text{Cost}\right)}_{Ni}=\$0.01894\end{array}$

The total cost of quarter is the sum of cost of nickel and copper,

  $\begin{array}{l}{\left(\text{Cost}\right)}_{quarter}={\left(\text{Cost}\right)}_{Cu}+{\left(\text{Cost}\right)}_{Ni}\\ {\left(\text{Cost}\right)}_{quarter}=\$0.0109+\$0.01894\\ {\left(\text{Cost}\right)}_{quarter}=\$0.02984\end{array}$

Thus, the total cost of quarter ${\left(\text{Cost}\right)}_{quarter}=\$0.02984$.

Weight of nickel based on the volume of quarter is calculated as,

  $\begin{array}{l}{W}_{nickel}=V_{quarter}\times {\rho }_{nickel}\\ W_{nickel}=0.7066{\text{cm}}^3\times 8.902\text{g}{\text{/cm}}^{\text{3}}\\ W_{nickel}=6.290\text{gram}\end{array}$

Calculation of cost of quarter, if it is made of nickel,

  $\begin{array}{l}{\left(\text{Cost}\right)}_{Ni}=\left(\text{cost}\text{of}\text{nickel}\text{per}\text{kg}\right)\times \left({\text{W}}_{\text{Ni}}\right)\\ {\left(\text{Cost}\right)}_{Ni}=\left(9.0405\right)\times \left(6.290\text{gram}\times \frac{1\text{kg}}{1000\text{gram}}\right)\\ {\left(\text{Cost}\right)}_{Ni}=\$0.0568\end{array}$

Thus, the cost of quarter made up of nickel ${\left(\text{Cost}\right)}_{Ni}=\$0.0568$

The comparison of total cost of quarter made up of copper and nickel with cost of quarter made up of only nickel is given by,

  $\frac{{\left(\mathrm{Cos}t\right)}_{quarter}}{{\left(\text{Cost}\right)}_{Ni}}=\frac{\$0.02984}{\$0.0568}=0.525$

Thus, the cost of quarter made up of copper and nickel is half of the cost of quarter made up of nickel.

Conclusion

The cost of quarter made up of copper and nickel is ${\left(\text{Cost}\right)}_{quarter}=\$0.02984$.

The cost of quarter made up of nickel is ${\left(\text{Cost}\right)}_{Ni}=\$0.0568$

Thus, the cost of quarter made up of copper and nickel is half of the cost of quarter made up of nickel.