Question

o reduce the drag coefficient and thus to improve the fuel efficiency, the frontal area of a car is to be reduced. Determine the amount of fuel and money saved per year as a result of reducing the frontal area from 18 to 14 ft2. Assume the car is driven 12,000 mi a year at an average speed of 55 mi/h. Take the density and price of gasoline to be 50 lbm/ft3 and $3.10/gal, respectively; the density of air to be 0.075 lbm/ft3, the heating value of gasoline to be 20,000 Btu/lbm; and the overall efficiency of the engine to be 30 percent. Take the drag coefficient as CD

Answer 1

Complete question

To reduce the drag coefficient and thus to improve the fuel efficiency, the frontal area of a car is to be reduced. Determine the amount of fuel and money saved per year as a result of reducing the frontal area from 18 to 14 ft2. Assume the car is driven 12,000 mi a year at an average speed of 55 mi/h. Take the density and price of gasoline to be 50 lbm/ft3 and $3.10/gal, respectively; the density of air to be 0.075 lbm/ft3, the heating value of gasoline to be 20,000 Btu/lbm; and the overall efficiency of the engine to be 30 percent. Take the drag coefficient as CD=0.3 for a passenger car.

Answer:

22.22%

$57

Step-by-step explanation:

From the question we are told that

Initial area of frontal area
`a_1=18ft^2`

Final area of frontal area
`a_2=14ft^2`

Distance covered a year
`D=12000mile`

Average speed a year
`V_(avg)=55mile/h`

Density
`\rho=50 lbm/ft3`

Price
`P= $3.10/gal`

Density of air
`\rho_(air) 0.075 lbm/ft3`

Heating value of gasoline
`Q=20,000 Btu/lbm`

Efficiency
`\eta=30\%`

Drag coefficient
`CD=0.3`

`\triangle A=18-14ft^2=4ft^2`

Generally the equation for drag force
`C_D` is mathematically given as

`F_D=(C_DAPV^2)/(2)`

`F_D=(0.3*18*0.075*(80.685^2)*A)/(2)`

`F_D=73.24Alb`

where
`v=55mil/h*1.467=80.685ft/s`

Generally the equation for work done W is mathematically given as
`W=F_D*L`

where
`L=12000mile*5280`

`L=63360000`

`W=73.24A*63360000`

`W=4.6*10^9A`

Generally the equation for overall efficiency
`\eta` is mathematically given as

where

`W_(req)=required\ gasoline\ power\ efficiency`

`\eta=(W)/(W_(req))`

`W_(req)=(W)/(\eta)`

`W_(req)=(4.6*10^9)/(0.3)`

`W_(req)=1.55*10^(10)A`

Generally the equation for reduction fee with change in frontal area
`\triangle M` is mathematically given as

`\triangle M =\triangle Vgasoline*cost`

Where

`\triangle Vgasoline= volume\ reduction\ of\ gasoline`

`\triangle Vgasoline=(E_(req))/(H*P)`

`\triangle Vgasoline=(1.55*10^(10)A)/(20000*778.169*32.2*50)`

if

`20000btu/ibm=20000*778.169*32.2(1bm.ft^2/s)`

`\triangle Vgasoline=(1.55*10^(10)A)/(20000*778.169*32.2*50)`

`\triangle Vgasoline=0.61859A`

Therefore

`\triangle M =0.61859A*3.10`

if
`1ft^2=7.48gal`

`\triangle M =0.61859(4)*3.10*7.48`

`\triangle M = \$ 57.671`

Generally the equation for reduction of fuel
`F_r`is mathematically given as

`F_r=(\triangle A)/(\triangle i)*100`

where

`\triangle i=18ft^2`

`F_r=(4)/(18)*100`

Fuel reduction price by reducing front area is 22.22%

Money saved per year is $57