Physics, 13.04.2021 01:00 TSZRobloxian8290

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

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Answer from: FlayMaster101

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.

22.22%

$57

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

$F_D=\frac{C_DAPV^2}{2}$

$F_D=\frac{0.3*18*0.075*(80.685^2)*A}{2}$

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

$W_{req}=required\ gasoline\ power\ efficiency$

$\eta=\frac{W}{W_{req}}$

$W_{req}=\frac{W}{\eta}$

$W_{req}=\frac{4.6*10^9}{0.3}$

$W_{req}=1.55*10^{10}A$

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

$\triangle M =\triangle Vgasoline*cost$

Where

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

$\triangle Vgasoline=\frac{E_{req}}{H*P}$

$\triangle Vgasoline=\frac{1.55*10^{10}A}{20000*778.169*32.2*50}$

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

$\triangle Vgasoline=\frac{1.55*10^{10}A}{20000*778.169*32.2*50}$

$\triangle Vgasoline=0.61859A$

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

$F_r=\frac{\triangle A}{\triangle i}*100$

where

$\triangle i=18ft^2$

$F_r=\frac{4}{18}*100$

Fuel reduction price by reducing front area is 22.22%

Money saved per year is $57

Answer from: Quest

Rain is pulled by gravity which is a non contact force

Answer from: Quest

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

so, refractive index of medium = 300000/200000 = 1.5

1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

∞ refractive index is usually for opaque mediums. so third medium is opaque medium.

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