Question

For those who think electric cars are sissy, Keio University in Japan has tested a 22-ft long prototype whose eight electric motors generate a total of 590 horsepower. The "Kaz" cruises at 175 mi/h. If the drag coefficient is 0.35 and the frontal area is 23.5 ft2, what percent of this power is expended against sea-level air drag? Round the final answer to three decimal places.

Answer 1

Answer:

The percent of power extended against sea-level air drag is 50.15%

Explanation:

Length L is 22 ft, Power P is 590 hp,

Speed of air V = 175 mi/h. To convert to ft, we will multiply by 1.46

V = 175 × 1.46 = 255.5 ft

Frontal Area A is given as 23.5 ft

Coefficient of drag, Cd is 0.35

From text, Density of Air at sea level ρ = 0.00237 slug/ft³

The formula for calculating Drag Force F is

F = Cd × $\frac{1}{2}$ × ρAV²

Substitute the values above into the formula

F = 0.35 × $\frac{1}{2}$ × 0.00237 × 23.5 × 255.5²

= 0.35 × $\frac{1}{2}$ × 3635.78

= 636.26 lbf

But Power P = FV

Therefore we substitute for F and V to get P

P = 636.26 × 255.5

= 162564.43 ft.lbf/s

To convert to horsepower, we multiply by 0.00182 hp

= 162564.43 × 0.00182

= 295.87 hp

The percent of power extended against sea-level air drag is

= (295.87/590) × 100

= 50.15%

Answer 2

Answer:

51.693%

Explanation:

Answer is divided into two parts first we will calculate drag force that car experiences and then how much work does the force required to overcome drag do in one second, as you will see that this work would be power required to overcome drag force.

(a) Drag force

In Aerodynamics formula for drag force is.

                            $F_{d} =\frac{1}{2} pv^2C_{p} A$

All the variables in this formula are self explanatory, their values are.

    $p = 1.225kg/m^2$  Air density at sea level.

    $v = 78.2232m/s$  Velocity if car.

    $C_{P} =0.35$    Drag Coefficient.

     $A = 2.183221m^2$    Frontal Area.

I have converted all units to SI system of units.

Plugging all this in Formula for Drag force will give, as follow.

   $F_{d} = \frac{1}{2} *1.225kg/m^3*(78.232m/s)^2*0.35*2.183221m^2 = 2864.4464N$

That is Drag force, as it is logical, force required to our come drag force is also 2864.4464N.

(b) Work that force required to overcome drag force does.

  Formula for work is  $W = F.d$ .

here d is distance covered in one second which would be.

$d = vt = 78.2232\frac{m}{s} *s =78.2232m$.

With this in work formula we get.

$w = 2864.446N*78.232m = 224320.3319J$

that is work that electric car is doing in one second to overcome drag force

it is 51.963% of work that car is capable of generating in one second.

Note! Work done per unit time is power.

This car is wasting more than half of it's power in overcoming drag force due to air, think about it.!