An automobile with projected area 2-6 m2 is running on a road with a speed of 120 km per hour- The mass density and the kinematic viscosity of air are 1-2 kg-m3 and 1-5- 10-5 m2-s- respectively- The drag coefficient is 0-30-The metric horse power required to overcome the drag force is

The drag force on the automobile may be given as nbsp; nbsp; nbsp;F_D=C_DA×ρ V^22 Here, nbsp;C_D=0.30, A=2.6 m^2, ρ =1.2 kg/m^3, V=120 kmph ∴ F_D=0 ...

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Fluid Mechanics

Drag on a Sphere

An automobile...

Question

An automobile with projected area $2.6 m^{2}$ is running on a road with a speed of 120 km per hour. The mass density and the kinematic viscosity of air are $1.2 k g / m^{3}$ and $1.5 \times 10^{- 5} m^{2} / s$ , respectively. The drag coefficient is 0.30.
The metric horse power required to overcome the drag force is

33.23

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23.23

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20.23

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Solution

The correct option is C 23.23
The drag force on the automobile may be given as
$F_{D} = C_{D} A \times \frac{ρ V^{2}}{2}$
Here, $C_{D} = 0.30, A = 2.6 m^{2}, ρ = 1.2 k g / m^{3}, V = 120 k m p h$
$∴ F_{D} = 0.30 \times 2.6 \times \frac{1.2 \times (120 \times 1000)^{2}}{2 \times (60 \times 60)^{2}}$
$= 520 N$

Power required to overcome the drag = Drag force × velocity
$= 520 \times \frac{120 \times 1000}{60 \times 60} = 17333.33 W a t t$
(1 mhp = 735.5 watt)
$= \frac{17333.33}{735.5} m e t r i c h o r s e p o w e r$
$= 23.567 m e t r i c h o r s e p o w e r$

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An automobile with projected area 2.6 m2 is running on a road with a speed of 120 km per hour. The mass density and the kinematic viscosity of air are 1.2 kg/m3 and 1.5×10−5 m2/s, respectively. The drag coefficient is 0.30. The metric horse power required to overcome the drag force is