Discussion
Doubling the free-stream velocity over a flat plate in fully laminar flow changes the total friction drag force by a factor of
Answer: C
The drag goes as \( C_F \times V^2 \), and \( C_F \) varies as \( Re^{-1/2} \), that is as \( V^{-1/2} \). The product therefore scales as \( V^{1.5} \), so doubling the speed multiplies the drag by \( 2^{1.5} = 2.83 \) rather than by four.
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