Chapter 15, Problem 98RQ

To determine

The money wasted per year.

Explanation of Solution

Given:

The diameter of the ball is, $D=2.62\text{ in}$.

The equivalent roughness factor is, $\epsilon /\text{D}=1.5\times {10}^{-3}$.

Distance travelled per year is, $L=16000\text{ miles}$.

Average speed of the car is, $V=55\text{ mph}$.

The overall efficiency is, $\eta =0.308$.

The density of the fuel is, ${\rho }_{fuel}=50.2\text{lbm}/{\text{ft}}^3$.

The heating value of the fuel is, $HV=1.47\times {10}^7\text{ft}\cdot \text{lbf}/\text{lbm}$.

Cost of the fuel is $\$4.00/\text{gal}$.

Calculation:

Obtain the following properties of air:

  $\begin{array}{l}\text{Density,}{\rho }_{air}=0.07518\text{lbm}/{\text{ft}}^3\\ \text{Viscosity,}{\mu }_{air}=1.227\times {10}^{-5}\text{lbm}/\text{ft}\cdot \text{s}\end{array}$

The Reynolds number is,

  $\begin{array}{c}\mathrm{Re}=\frac{{\rho }_{air}VD}{{\mu }_{air}}\\ =\frac{\left(0.07518\text{lbm}/{\text{ft}}^3\right)\left(55\times 1.466\text{ft}/\text{s}\right)\left(2.62/12\text{ ft}\right)}{1.227\times {10}^{-5}\text{lbm}/\text{ft}\cdot \text{s}}\\ =107900\end{array}$

The drag coefficient for a sphere for the above Reynolds number is $C_D=0.505$.

The drag force is,

  $\begin{array}{c}{F}_D=C_{D}A\frac{{\rho }_{air}{V}^2}{2}\\ =C_D\left(\frac{\pi }{4}{D}^2\right)\frac{{\rho }_{air}{V}^2}{2}\end{array}$

The power required to overcome the resistances is,

  $\begin{array}{c}{\dot{W}}_{drag}=F_{D}L\\ =C_D\left(\frac{\pi }{4}{D}^2\right)\frac{{\rho }_{air}{V}^{2}L}{2}\end{array}$

The energy provided by the fuel is,

  $\begin{array}{c}{E}_{req}=\frac{{\dot{W}}_{drag}}{\eta }\\ =\frac{C_D\left(\frac{\pi }{4}{D}^2\right)\frac{{\rho }_{air}{V}^{2}L}{2}}{\eta }\end{array}$

The quantity of fuel required is,

  $\begin{array}{c}{\text{V}}_{fuel}=\frac{E_{req}/HV}{{\rho }_{fuel}}\\ =\frac{C_D\left(\frac{\pi }{4}{D}^2\right)\frac{{\rho }_{air}{V}^{2}L}{2}}{\eta \left(HV\right){\rho }_{fuel}}\\ =\frac{\left(0.505\right)\left[\frac{\pi }{4}{\left(2.62/12\text{ ft}\right)}^2\right]\left[\frac{\begin{array}{l}\left(0.07518\text{lbm}/{\text{ft}}^3\right)\times {\left(55\times 1.466\text{ft}/\text{s}\right)}^2\\ \times \left(16000\times 5280\text{ft}\right)\end{array}}{2}\right]}{0.308\left(1.47\times {10}^7\text{ft}\cdot \text{lbf}/\text{lbm}\right)\left(50.2\text{lbm}/{\text{ft}}^3\right)}\\ =0.05343{\text{ ft}}^3\times \frac{7.481\text{ gal}}{1{\text{ ft}}^3}\\ =0.3997\text{ gal}\end{array}$

The extra cost of fuel is,

  $\begin{array}{c}\text{Additional Cost}={\text{V}}_{fuel}\times \text{Cost}\\ =\left(0.3997\text{ gal}\right)\left(\$4.00/\text{gal}\right)\\ =\$1.6\end{array}$

Thus, the money wasted per year is $\$1.6$.

Since the additional cost is minimal, Jamie need not remove the tennis ball.