Chapter 15, Problem 22P

To determine

The amount of fuel and money saved per year.

Explanation of Solution

Given:

Frontal area of the car is $18{\text{ ft}}^2$.

New frontal area is $15{\text{ ft}}^2$.

Distance travelled per year is $12000\text{ mi}$.

Average speed is $55\text{mi}/\text{h}$.

Density of gasoline is $50\text{lbm}/{\text{ft}}^3$.

Price of gasoline is $\$3.10/\text{gal}$.

Density of air is $0.075\text{lbm}/{\text{ft}}^3$.

Heating value of gasoline is $20000\text{Btu}/\text{lbm}$.

Overall efficiency of the engine is 30%.

Calculation:

From Table 15-2, the drag coefficient of a passenger car is $C_D=0.3$.

The drag force is,

  $\begin{array}{c}{F}_D=C_{D}A\frac{\rho V^2}{2}\\ =0.3\left(18{\text{ ft}}^2\right)\frac{\left(0.075\text{lbm}/{\text{ft}}^3\right){\left(55\text{mi}/\text{h}\right)}^2}{2}{\left(\frac{1.4667\text{ft}/\text{s}}{1\text{mi}/\text{h}}\right)}^2\left(\frac{1\text{ lbf}}{32.2\text{lbm}\cdot \text{ft}/{\text{s}}^2}\right)\\ =40.924\text{ lbf}\end{array}$

Work done to overcome the drag is,

  $\begin{array}{c}{W}_{drag}=F_{D}L\\ =\left(40.924\text{ lbf}\right)\left(12000\text{mi}/\text{yr}\times \frac{5280\text{ft}}{1\text{ mi}}\right)\left(\frac{1\text{ Btu}}{778.169\text{ lbf}\cdot \text{ft}}\right)\\ =3.33\times {10}^6\text{Btu}/\text{yr}\end{array}$

Energy required to overcome the drag is,

  $\begin{array}{c}{E}_{in}=\frac{W_{drag}}{\eta }\\ =\frac{3.33\times {10}^6\text{Btu}/\text{yr}}{0.30}\\ =1.111\times {10}^7\text{Btu}/\text{yr}\end{array}$

The amount of fuel required to supply the energy is,

  $\begin{array}{c}\text{Amount of fuel}=\frac{m_{fuel}}{{\rho }_{fuel}}\\ =\frac{{\text{E}}_{in}/HV}{{\rho }_{fuel}}\\ =\frac{1.111\times {10}^7\text{Btu}/\text{yr}}{\left(20000\text{Btu}/\text{lbm}\right)\left(50\text{lbm}/{\text{ft}}^3\right)}\\ =11.11{\text{ft}}^3/\text{yr}\end{array}$

The cost of fuel supplied per year is,

  $\begin{array}{c}\text{Cost}=\left(\text{Amount of fuel}\right)\left(\text{Unit cost}\right)\\ =\left(11.11{\text{ft}}^3/\text{yr}\times \frac{7.4804\text{gal}}{1{\text{ ft}}^3}\right)\left(\$3.10/\text{gal}\right)\\ =\$257.6/\text{yr}\end{array}$

Calculate the reduction ratio.

  $\begin{array}{c}\text{Reduction ratio}=\frac{A-A_{new}}{A}\\ =\frac{18-15}{18}\\ =0.1667\end{array}$

Calculate the reduction in fuel amount.

  $\begin{array}{c}\text{Amount reduction}=\left(\text{Reduction ratio}\right)\left(\text{Amount of fuel}\right)\\ =0.1667\left(11.11{\text{ft}}^3/\text{yr}\times \frac{7.4804\text{gal}}{1{\text{ ft}}^3}\right)\\ =13.854\text{gal}/\text{yr}\end{array}$

Calculate the reduction in the cost.

  $\begin{array}{c}\text{Cost reduction}=\left(\text{Reduction ratio}\right)\left(\text{Cost of fuel}\right)\\ =0.1667\left(\$257.6/\text{yr}\right)\\ =\$42.942/\text{yr}\end{array}$

Thus, the amount of fuel and the money saved per year is $13.854\text{gal}/\text{yr}$ and $\$42.942/\text{yr}$.