Question 14.1: A thin flat plate, 10 ft tall and 1 ft wide, forms the leadin...

The physical arrangement is shown in Figure 14.7. A laminar boundary layer is expected to transition at Re_x ≤ 5 × 10⁵. Thus we can solve for the distance x_C to the transition point by using Eq. 14.12a to write: x_C = (5 × 10⁵ \nu)/U. Using U = 100 mph = 146.7 ft/s and, from Appendix A, \nu = 1.64 × 10⁻⁴ ft²/s for air, we find

Re_{x}=\frac{Ux}{\nu }       (14.12a)

x_C=\frac{5×10^5[1.64×10^{−4} (\mathrm{ft}^2/s)]}{146.7\ \mathrm{ft}/s}=0.56\ \mathrm{ft}

Transition begins to occur just beyond the midpoint of the plate, and the rear portion of the plate has a turbulent boundary layer. To find the laminar boundary layer thickness at x = x_C we use Eq. 14.12b to calculate the thickness as

\frac{\delta(x)}{x}=5.0(Re_x)^{-1/2}        (14.12b)

δ(x_C) = 5.0 x_C(Re_x)^−1/2 = 5.0(0.56 ft)(5 × 10⁵)^−1/2 = 0.00396 ft = 0.048 in.

Thus the laminar boundary layer is only 0.048 in. thick when transition occurs. To find the drag force on both sides of the plate due to the laminar boundary layer, we must multiply Eq. 14.12g,

F_D=\frac{0.664\rho U^2wL}{\sqrt{Re_L} }            (14.12g)

which gives the drag on one side, by 2 and obtain F_D = 1.328 ρU²wL/\sqrt{Re_L}. In this case, we must also be careful to insert L = x_C, since this defines the portion of the plate covered by the laminar boundary layer. Using ρ = 2.329 × 10⁻³ slug/ft³ from Appendix A, inserting the other data, and noting that Re_L = 5 × 10⁵, we obtain

F_D=\frac{1.328ρU^2wL }{\sqrt{Re_L}}

=\frac{1.328}{\sqrt{5\times 10^5}} (2.329 × 10⁻³ slug/ft³)(146.7 ft/s )²(10 ft)(0.56 ft) = 0.53 lb_f

The total drag on the plate is actually much larger than this since we have not accounted for the drag of the turbulent boundary layer. The drag coefficient for the laminar portion of the boundary layer, which refers to one side of the plate only, is given by Eq. 14.12h as F_D = 1.328/\sqrt{Re_L} = 0.0019.

C_D=\frac{1.328}{\sqrt{Re_L} }           (14.12h)