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The drag force Fd, imposed by the surrounding air on a vehicle moving with velocity V is given by:
Fd =C dA 2​ pV2
where Cd is a constant called the drag coefficient, A is the projected frontal area of the vehicle, and \rhorho is the air density.
Determine the power, in hp, required to overcome aerodynamic drag for an automobile moving at
(a) 25 miles per hour,
(b) 70 miles per hour.
Assume Cd=0.28,
A= 25ft2
and p=0.075Ib/ft2

Sagot :

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Answer:

Explanation:

a)

Given that:

V = 25 mi/hr

To ft/sec, we have:

[tex]V = 25 \times \dfrac{5280}{3600} ft/s[/tex]

[tex]V = \dfrac{110}{3} ft/s[/tex]

[tex]\rho = 0.075 \ lb/ft^3[/tex]

[tex]\rho = 0.075 \times \dfrac{1 \ lbf s^2/ft}{32.174 \ lbm}[/tex]

[tex]\rho = \dfrac{0.075}{32.174 } lbf.s^2/ft^4[/tex]

[tex]C_d = 0.28[/tex]

A = 25ft²

Recall that:

The drag force [tex]F_d =\dfrac{C_dA \rho V^2}{2}[/tex]

[tex]F_d =\dfrac{1}{2}\times 0.28 \times 25\times \dfrac{0.075}{32.174}\times (\dfrac{110}{3})^2[/tex]

[tex]F_d =10.967 \ lbf[/tex]

[tex]P = F_dV \\ \\ P = 10.97 \times (\dfrac{110}{3}} \\ \\ P = 402.3 \ hp[/tex]

For 70 miles per hour, we have:

[tex]V = 70 \times \dfrac{5280}{3600} ft/s[/tex]

[tex]V = \dfrac{308}{3} ft/s[/tex]

The drag force [tex]F_d =\dfrac{C_dA \rho V^2}{2}[/tex]

[tex]F_d =\dfrac{1}{2}\times 0.28 \times 25\times \dfrac{0.075}{32.174}\times (\dfrac{308}{3})^2[/tex]

[tex]F_d =85.99 \ lbf[/tex]

[tex]P = F_dV \\ \\ P = 85.99 \times (\dfrac{308}{3}}) \\ \\ P = 8828.2 \ hp[/tex]

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