The average drag coefficient for a laminar boundary layer over a flat plate was obtained as 0.018. If all other factors remain unchanged, and length of the plate is increased by 4 times its original value, this average drag coefficient would change to

0.0036

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0.0056

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0.008

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0.009

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Solution

The correct option is D 0.009
$C_{D} = \frac{1.328}{\sqrt{R e_{L}}} = \frac{1.328}{\sqrt{\frac{ρ v L}{μ}}} ∴ C_{D} α \frac{1}{\sqrt{L}} ∴ \frac{C D_{1}}{C D_{2}} = \frac{\sqrt{L_{2}}}{\sqrt{L_{1}}} \Rightarrow \frac{0.018}{C_{D_{2}}} = \frac{\sqrt{4 L_{1}}}{\sqrt{L_{1}}} \Rightarrow C_{D_{2}} = 0.009$

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Similar questions

δ_{L}

δ_{L}

(R e_{x})^{- 1 / 2}

(ρ = 1.2 k g / m^{3}

v = 2 \times 10^{- 5} m^{2} / s)

C_{f} = \frac{1.328}{{\sqrt{R e}}_{x}}

List - I	List - II
A. Boundary layer thickness	1. Decrease in the flow direction
B. Shear stress at the plate	2. Increase in the flow direction
C. Pressure gradient along the plate	3. Remains unchanged

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