Respuesta :
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
Explanation:
From the question we are told that
Initial area of frontal area [tex]a_1=18ft^2[/tex]
Final area of frontal area [tex]a_2=14ft^2[/tex]
Distance covered a year [tex]D=12000mile[/tex]
Average speed a year [tex]V_{avg}=55mile/h[/tex]
Density [tex]\rho=50 lbm/ft3[/tex]
Price [tex]P= $3.10/gal[/tex]
Density of air [tex]\rho_{air} 0.075 lbm/ft3[/tex]
Heating value of gasoline [tex]Q=20,000 Btu/lbm[/tex]
Efficiency [tex]\eta=30\%[/tex]
Drag coefficient [tex]CD=0.3[/tex]
[tex]\triangle A=18-14ft^2=4ft^2[/tex]
Generally the equation for drag force [tex]C_D[/tex] is mathematically given as
[tex]F_D=\frac{C_DAPV^2}{2}[/tex]
[tex]F_D=\frac{0.3*18*0.075*(80.685^2)*A}{2}[/tex]
[tex]F_D=73.24Alb[/tex]
where [tex]v=55mil/h*1.467=80.685ft/s[/tex]
Generally the equation for work done W is mathematically given as[tex]W=F_D*L[/tex]
where [tex]L=12000mile*5280[/tex]
[tex]L=63360000[/tex]
[tex]W=73.24A*63360000[/tex]
[tex]W=4.6*10^9A[/tex]
Generally the equation for overall efficiency [tex]\eta[/tex] is mathematically given as
where
[tex]W_{req}=required\ gasoline\ power\ efficiency[/tex]
[tex]\eta=\frac{W}{W_{req}}[/tex]
[tex]W_{req}=\frac{W}{\eta}[/tex]
[tex]W_{req}=\frac{4.6*10^9}{0.3}[/tex]
[tex]W_{req}=1.55*10^{10}A[/tex]
Generally the equation for reduction fee with change in frontal area [tex]\triangle M[/tex] is mathematically given as
[tex]\triangle M =\triangle Vgasoline*cost[/tex]
Where
[tex]\triangle Vgasoline= volume\ reduction\ of\ gasoline[/tex]
[tex]\triangle Vgasoline=\frac{E_{req}}{H*P}[/tex]
[tex]\triangle Vgasoline=\frac{1.55*10^{10}A}{20000*778.169*32.2*50}[/tex]
if
[tex]20000btu/ibm=20000*778.169*32.2(1bm.ft^2/s)[/tex]
[tex]\triangle Vgasoline=\frac{1.55*10^{10}A}{20000*778.169*32.2*50}[/tex]
[tex]\triangle Vgasoline=0.61859A[/tex]
Therefore
[tex]\triangle M =0.61859A*3.10[/tex]
if [tex]1ft^2=7.48gal[/tex]
[tex]\triangle M =0.61859(4)*3.10*7.48[/tex]
[tex]\triangle M = \$ 57.671[/tex]
Generally the equation for reduction of fuel [tex]F_r[/tex]is mathematically given as
[tex]F_r=\frac{\triangle A}{\triangle i}*100[/tex]
where
[tex]\triangle i=18ft^2[/tex]
[tex]F_r=\frac{4}{18}*100[/tex]
Fuel reduction price by reducing front area is 22.22%
Money saved per year is $57
