## Chapter 14

## Q. 14.1

A thin ﬂat plate, 10 ft tall and 1 ft wide, forms the leading edge of a banner towed by an aircraft at 100 mph on a 70°F day. How far from the leading edge of the plate does the laminar portion of boundary layer extend? What is the boundary layer thickness at the downstream end of the laminar boundary layer? Find the drag force on the plate contributed by the laminar boundary layer, and the corresponding drag coefficient.

## Step-by-Step

## Verified Solution

The physical arrangement is shown in Figure 14.7. A laminar boundary layer is expected to transition at Re_{x }≤ 5 × 10^{5}. Thus we can solve for the distance x_{C} to the transition point by using Eq. 14.12a to write: x_{C} = (5 × 10^{5} \nu)/U. Using U = 100 mph = 146.7 ft/s and, from Appendix A, \nu = 1.64 × 10^{−4} ft^{2}/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 × 10^{5})^{−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 ρU^{2}wL/\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/ft^{3} from Appendix A, inserting the other data, and noting that Re_{L} = 5 × 10^{5}, we obtain

F_D=\frac{1.328ρU^2wL }{\sqrt{Re_L}}

=\frac{1.328}{\sqrt{5\times 10^5}} (2.329 × 10^{−3} slug/ft^{3})(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)