Question 14.2: A box-shaped truck body 2.5 m wide, 3 m high, and 7 m long (...

We will first use Eq. 14.16a to calculate the maximum Reynolds number at the down-stream edge x = L of each panel. With viscosity data from Appendix A, we obtain

Re_L=\frac{UL}{\nu}=\frac{(20 \ m/s)(7\ m )}{1.51×10^{−5}\ m^2/s}=9.3\times 10^6

This is just within the applicable range of Eqs. 14.16a–14.16h. The drag force on each panel is calculated by using Eq. 14.16g. For the top we obtain

Re_x=\frac{Ux}{\nu}                           (14.16a)

\frac{\delta (x)}{x}=0.370(Re_x)^{-1/5}                            (14.16b)

\frac{\delta ^*(x)}{x}=0.0463(Re_x)^{-1/5}                            (14.16c)

\frac{\Theta (x)}{x}=0.0360(Re_x)^{-1/5}                            (14.16d)

\tau_W (x) = 0.0288ρU²(Re_x)^−1/5                               (14.16e)

C_f(x) = 0.0577(Re_x)^−1/5                                            (14.16f)

C_D = 0.072(Re_L)^−1/5                                           (14.16h)

F_{D_{top}} = 0.036ρU²wL(Re_L)^−1/5

=\frac{0.036(1.204\ kg/m^3)(20\ m/s)^2(2.5\ m)(7\ m)}{(9.3×10^6)^{1/5}}
= 12.3 N
The drag on each side panel is found using the same formula with w = 3 m instead of w = 2.5 m. Thus we can write F_{D_{side}} = (3/2.5)F_{D_{top}} = 14.8 N. The drag of all three panels is now calculated as F_{D} = 2F_{D_{side}} + F_{D_{top}} = 2(14.8 N ) + 12.3 N = 41.9 N. To find the boundary layer thickness and wall shear stress on the top panel, we use Eqs. 14.16b and 14.16e, respectively. Writing these explicitly in terms of x we have

τ_W(x) =0.0288ρU^2\left(\frac{Ux}{\nu} \right)^{-1/5}

= 0.0288(1.204 kg/m³)(20 m/s)² \left(\frac{20\ m/s}{1.51\times 10^{-5}\ m^2/s} \right)^{-1/5} x^{-1/5}

= 0.83 x^−1/5 (N/m²)(m^1/5)

Note that these results for δ(x) and \tau_W (x) also apply to the side panels. The maximum value of the shear stress and boundary layer thickness on each panel will occur at x = L. Inserting the data we find:

δ(L) = 0.022L^4/5m^1/5 = 0.022(7 m )^4/5m^1/5 = 0.104 m = 10.4 cm

\tau_W (L) = 0.83L^−1/5(N/m²)(m^1/5) = (0.83)(7 m )^−1/5(N/m²)(m^1/5) = 0.56 N/m²