Question

Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehicle. Consider a car with 1.85 m width and 1.75 m height, with a drag coefficient of 0.30. Determine the amount of fuel and money saved per year as a result of reducing the car height to 1.50 m while keeping its width the same. Assume the car is driven 25,000 km (15,000 miles) a year at an average speed of 100 km/h. Take the density and price of gasoline to be 0.74 kg/L and $1.04/L. Also assume the density of air to be 1.20 kg/m3, the heating value of gasoline to be 44,000 kJ/kg, and the overall efficiency of the car’s drive train to be 30%.

Answer 1

`\Delta V = 209.151\,L`,
`\Delta C = 217.517\,USD`

Step-by-step explanation:

The drag force is equal to:

`F_(D) = C_(D)\cdot (1)/(2)\cdot \rho_(air)\cdot v^(2)\cdot A`

Where
`C_(D)` is the drag coefficient and
`A` is the frontal area, respectively. The work loss due to drag forces is:

`W = F_(D)\cdot \Delta s`

The reduction on amount of fuel is associated with the reduction in work loss:

`\Delta W = (F_(D,1) - F_(D,2))\cdot \Delta s`

Where
`F_(D,1)` and
`F_(D,2)` are the original and the reduced frontal areas, respectively.

`\Delta W = C_(D)\cdot (1)/(2)\cdot \rho_(air)\cdot v^(2)\cdot (A_(1)-A_(2))\cdot \Delta s`

The change is work loss in a year is:

`\Delta W = (0.3)\cdot \left((1)/(2)\right)\cdot (1.20\,(kg)/(m^(3)))\cdot (27.778\,(m)/(s))^(2)\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25* 10^(6)\,m)`

`\Delta W = 2.043* 10^(9)\,J`

`\Delta W = 2.043* 10^(6)\,kJ`

The change in chemical energy from gasoline is:

`\Delta E = (\Delta W)/(\eta)`

`\Delta E = (2.043* 10^(6)\,kJ)/(0.3)`

`\Delta E = 6.81* 10^(6)\,kJ`

The changes in gasoline consumption is:

`\Delta m = (\Delta E)/(L_(c))`

`\Delta m = (6.81* 10^(6)\,kJ)/(44000\,(kJ)/(kg) )`

`\Delta m = 154.772\,kg`

`\Delta V = (154.772\,kg)/(0.74\,(kg)/(L) )`

`\Delta V = 209.151\,L`

Lastly, the money saved is:

`\Delta C = \left((154.772\,kg)/(0.74\,(kg)/(L) )\right)\cdot (1.04\,(USD)/(L) )`

`\Delta C = 217.517\,USD`