A 1/17th scale model of a new hybrid car is tested in a wind tunnel at the same Reynolds number as that of the full-scale prototype. The drag coefficient of the model is the same as that of the prototype. Assuming the model and prototype are both tested in air, find the ratio of the drag on the scale model (Fm) to the drag on the prototype (Fp), i.e., Fm/Fp. Since the size of the model is 1/17th of the size of the prototype, the projected area of the model is (1/17)2 of the projected area of the prototype. Round your answer to the nearest tenth.

Answer:

The ratio of the drag coefficients [tex]\dfrac{F_m}{F_p}[/tex] is approximately 0.0002

Step-by-step explanation:

The given Reynolds number of the model = The Reynolds number of the prototype

The drag coefficient of the model, [tex]c_{m}[/tex] = The drag coefficient of the prototype, [tex]c_{p}[/tex]

The medium of the test for the model, [tex]\rho_m[/tex] = The medium of the test for the prototype, [tex]\rho_p[/tex]

The drag force is given as follows;

[tex]F_D = C_D \times A \times \dfrac{\rho \cdot V^2}{2}[/tex]

We have;

[tex]L_p = \dfrac{\rho _p}{\rho _m} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_m} \right)^2 \times L_m[/tex]

Therefore;

[tex]\dfrac{L_p}{L_m} = \dfrac{\rho _p}{\rho _m} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_m} \right)^2[/tex]

[tex]\dfrac{L_p}{L_m} =\dfrac{17}{1}[/tex]

[tex]\therefore \dfrac{L_p}{L_m} = \dfrac{17}{1} =\dfrac{\rho _p}{\rho _p} \times \left(\dfrac{V_p}{V_m} \right)^2 \times \left(\dfrac{c_p}{c_p} \right)^2 = \left(\dfrac{V_p}{V_m} \right)^2[/tex]

[tex]\dfrac{17}{1} = \left(\dfrac{V_p}{V_m} \right)^2[/tex]

[tex]\dfrac{F_p}{F_m} = \dfrac{c_p \times A_p \times \dfrac{\rho_p \cdot V_p^2}{2}}{c_m \times A_m \times \dfrac{\rho_m \cdot V_m^2}{2}} = \dfrac{A_p}{A_m} \times \dfrac{V_p^2}{V_m^2}[/tex]

[tex]\dfrac{A_m}{A_p} = \left( \dfrac{1}{17} \right)^2[/tex]

[tex]\dfrac{F_p}{F_m} = \dfrac{A_p}{A_m} \times \dfrac{V_p^2}{V_m^2}= \left (\dfrac{17}{1} \right)^2 \times \left( \left\dfrac{17}{1} \right) = 17^3[/tex]

[tex]\dfrac{F_m}{F_p} = \left( \left\dfrac{1}{17} \right)^3[/tex]= (1/17)^3 ≈ 0.0002

The ratio of the drag coefficients [tex]\dfrac{F_m}{F_p}[/tex] ≈ 0.0002.

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