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 f

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

in progress 0
Thông Đạt 3 weeks 2021-08-30T02:06:07+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-08-30T02:07:20+00:00

    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 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 asW=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}

    if

    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_ris 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

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )