Shear stress, displacement thickness, drag on plate

Find the Reynolds number first to decide whether the boundary layer is laminar or turbulent, then use the matching formulas.

Assumptions

Air at about 20 °C: $\rho = 1.2\ \text{kg/m}^3$, $\nu = 1.5\times10^{-5}\ \text{m}^2/\text{s}$. $U = 6.5$ m/s, $L = 2$ m, $B = 1.5$ m, $x = 1.8$ m.

Reynolds numbers

$$Re_x = \frac{Ux}{\nu} = \frac{6.5(1.8)}{1.5\times10^{-5}} = 7.8\times10^5,\qquad Re_L = \frac{6.5(2)}{1.5\times10^{-5}} = 8.67\times10^5$$

Both exceed the critical value $5\times10^5$, so the boundary layer is taken as turbulent (1/7 power law, valid for $Re < 10^7$).

Boundary layer thickness and displacement thickness at x = 1.8 m

$$\begin{aligned} \delta &= \frac{0.37\,x}{Re_x^{0.2}} = \frac{0.37(1.8)}{(7.8\times10^5)^{0.2}} = 0.0442\ \text{m} \\ \delta^* &= \frac{\delta}{8} = 0.00552\ \text{m} = 5.52\ \text{mm} \end{aligned}$$

Shear stress at x = 1.8 m

$$\begin{aligned} C_f &= \frac{0.0576}{Re_x^{0.2}} = 0.003819 \\ \tau_0 &= C_f\cdot\tfrac12\rho U^2 = 0.003819(0.5)(1.2)(6.5)^2 = 0.0968\ \text{N/m}^2 \end{aligned}$$

Drag force on one face

$$\begin{aligned} C_D &= \frac{0.074}{Re_L^{0.2}} = 0.004805 \\ F_D &= C_D\cdot\tfrac12\rho U^2\,(L B) = 0.004805(0.5)(1.2)(6.5)^2(2\times1.5) = 0.365\ \text{N} \end{aligned}$$

If the laminar length at the leading edge is allowed for, $C_D = 0.074/Re_L^{0.2} - 1700/Re_L = 0.002843$ and $F_D = 0.216$ N.

Answer: $\tau_0 \approx 0.097$ N/m², $\delta^* \approx 5.5$ mm, drag on one face $\approx 0.37$ N (about 0.22 N if the initial laminar part is included).