## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## WriteWise AI Model by Holooly Genius

Your Ultimate AI Essay Writer & Assistant.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

Products

## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## WriteWise AI Model by Holooly Genius

Your Ultimate AI Essay Writer & Assistant.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

## 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.

## Verified Solution

The physical arrangement is shown in Figure 14.7. A laminar boundary layer is expected to transition at Rex ≤ 5 × 105. Thus we can solve for the distance xC to the transition point by using Eq. 14.12a to write: xC = (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 = xC we use Eq. 14.12b to calculate the thickness as

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

δ(xC) = 5.0 xC(Rex)−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 FD = 1.328 ρU2wL/$\sqrt{Re_L}$. In this case, we must also be careful to insert L = xC, 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 ReL = 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 lbf

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 FD = 1.328/$\sqrt{Re_L}$ = 0.0019.

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