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

freezing

deposition

condensation

explanation;

melting also called fusion is the process where solid becomes liquid, sublimation is the process where solid becomes gaseous without first changing to liquid, deposition involves change from gaseous to solid state, freezing involves change from liquid to solid, boiling or evaporation involves change of liquid to gaseous state, while condensation is the change from gaseous state to liquid state.

endothermic reactions involves absorption of heat from the surroundings or the environment while exothermic reactions involves loss of heat energy to the surroundings.

in this case, sublimation, evaporation and melting are endothermic reactions (gain energy from the environment) while, freezing, deposition and condensation are exothermic (loss of energy to the surroundings) reactions.

Answer from: Quest

e/c, where e is its total energy

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