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Chapter 14

Q. 14.2

A box-shaped truck body 2.5 m wide, 3 m high, and 7 m long (Figure 14.9) is traveling at 20 m/s in 20°C air. Calculate the contributions to the total drag of the truck from the sides and top of the truck body. Assume that a sheet metal seam near the leading edge of each panel causes the boundary layer to be turbulent for the full length of the panel. Also find the wall shear stress and boundary layer thickness along the top panel, and the maximum value of the wall shear stress and boundary layer thickness on this panel.

14.9

Step-by-Step

Verified Solution

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(Rex)−1/5                               (14.16e)

Cf(x) = 0.0577(Rex)−1/5                                            (14.16f)

CD = 0.072(ReL)−1/5                                           (14.16h)

F_{D_{top}} = 0.036ρU2wL(ReL)−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

\delta (x)=0.370x\left(\frac{Ux}{\nu} \right)^{-1/5}=0.370\left(\frac{20\ m/s}{1.51\times 10^{-5}\ m^2/s} \right)^{-1/5} x^{4/5}=0.022x^{4/5}\ m^{1/5}

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)(m1/5)

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) = 0.022L4/5m1/5 = 0.022(7 m )4/5m1/5 = 0.104 m = 10.4 cm

\tau_W (L) = 0.83L−1/5(N/m2)(m1/5) = (0.83)(7 m )−1/5(N/m2)(m1/5) = 0.56 N/m2