Answer:
Amount reduction is 29.06 gal/year
Cost of reduction is $90.09 / year
Explanation:
Assume The effect of reduction of the frontal area on the drag coefficient is
negligible
Given that:
The drag coefficient ([tex]C_D[/tex]) = 0.4 for a passenger car.
A = frontal area of car = 18 ft²
Velocity (V) = 55 mile per hour = (55 × 1.4667) ft/s = 80.6685 ft/s, density of air (ρ) = 0.075 lbm/ft³
The frontal drag force can be calculated by:
[tex]F_D=C_DA\frac{\rho V^2}{2} =0.3 *18*\frac{0.075*80.6685^2}{2} =1317.75\\F_D=1317.75lbm.ft/s^2=(1317,75*\frac{1}{32.2})lbf=40.9lbf[/tex]
The amount of work done to overcome this drag force is calculated by:
d = 12000 miles per year
[tex]W_{DRAG}=F_D*d=(40.9lbf)*(12000miles/year)*\frac{5280ft}{1mile}*\frac{1lbf}{778.169 lbf ft} =3.330* 10^6 Btu/year[/tex]
and the required energy input is:
[tex]E_{in}=\frac{W_{DRAG}}{\eta_{car}} =\frac{3.330*10^6Btu/year}{0.30} =11.1*10^7Btu/year[/tex]
Heating value (HV) = 20000 Btu/lbm
[tex]m_{fuel}=E_{in}/HV=1.11*10^7/20000\\[/tex]
Amount of fuel = [tex]\frac{m_{fuel}}{\rho{fuel}} = \frac{1.11*10^7/20000}{50}=11.1ft^3/year *\frac{7.4804gal}{1ft^3}=83.03 gal/year[/tex]
Cost of fuel = Amount of fuel × Price of fuel = 11.1 ft³/year × ($3.1/gal) × (7.4804 gal / 1 ft³) = $257.4 per year.
The percent reduction in the fuel consumption due to reducing frontal area (reduction ratio) is given by:
Reduction ratio = [tex]\frac{A-A{new}}{A}=\frac{20-13}{20} =0.35[/tex]
Amount reduction = Reduction ratio × Amount of fuel = 0.35 × 83.03 gal/year =29.06 gal/year
Cost of reduction = Reduction ratio × cost of fuel = 0.35 × $257.4 = $90.09 / year
