Question 6 of 6: Drag on a plate standing in a turbulent boundary layer

Question 6: Drag on a plate standing in a turbulent boundary layer (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A thin plate of streamwise length $L$ and height $b$, wetted on both faces, standing normal to a wall and aligned with the flow. The approaching wall boundary layer is fully turbulent with thickness $\delta$ and profile $u(y)=U_\infty (y/\delta)^{1/7}$, and the plate is taken to lie wholly inside it, $b\le\delta$. Fluid properties $\rho$ and $\nu$ are uniform.

Find. (a) a closed-form drag coefficient for the plate, and (b) the ratio of that drag to the drag the same plate would feel in a uniform stream $U_\infty$.

Approach. Treat the plate as a stack of independent strips (the strip, or blade-element, approximation), assign each strip the standard turbulent flat-plate friction coefficient evaluated at its own local Reynolds number, integrate over the height, and normalise.

1. Part (a) — state the flat-plate friction law used for each strip. For a fully turbulent plate the local skin-friction coefficient follows the one-seventh-power correlation $c_{fx}=0.0266\,Re_x^{-1/7}$, whose integral over a plate of length $L$ gives the plate-average value $$\begin{aligned} C_F&=\frac{1}{L}\int_0^{L}c_{fx}\,\mathrm{d}x=\frac{7}{6}(0.0266)\,Re_L^{-1/7}=0.031\,Re_L^{-1/7}\cr Re_L&=\frac{U L}{\nu} \end{aligned}$$ This is the standard result quoted in White and is valid for $5\times10^{5}<Re_L<10^{7}$.
2. Write the force on one strip. A strip at height $y$, of height $\mathrm{d}y$, sees the local speed $u(y)$ and is wetted on both faces, so its drag is $$\mathrm{d}F=2\,C_F\!\left(\frac{u(y)L}{\nu}\right)\tfrac{1}{2}\rho\,u(y)^{2}\,L\,\mathrm{d}y =0.031\,\rho L\left(\frac{L}{\nu}\right)^{-1/7}u(y)^{2-1/7}\,\mathrm{d}y .$$ The exponent collects to $u^{13/7}$; the strips are treated as independent because the spanwise pressure gradient along the plate is negligible for a thin aligned plate.
3. Insert the power-law profile and integrate over the height. With $u=U_\infty(y/\delta)^{1/7}$ the integrand becomes $y^{13/49}$, and $$\int_0^{b}u^{13/7}\mathrm{d}y =U_\infty^{13/7}\,\delta^{-13/49}\!\int_0^{b}\! y^{13/49}\mathrm{d}y =\frac{49}{62}\,U_\infty^{13/7}\,b\left(\frac{b}{\delta}\right)^{13/49},$$ since $\tfrac{1}{1+13/49}=\tfrac{49}{62}$. Hence the total drag is $$F=0.031\,\rho\,L^{6/7}\nu^{1/7}U_\infty^{13/7}\left(\frac{49}{62}\right) b\left(\frac{b}{\delta}\right)^{13/49}.$$
4. Normalise to a drag coefficient. Referring the force to the total wetted area $2bL$ and the free-stream speed $U_\infty$, $$C_D\equiv\frac{F}{\tfrac{1}{2}\rho U_\infty^{2}(2bL)} =0.031\left(\frac{49}{62}\right)Re_L^{-1/7}\left(\frac{b}{\delta}\right)^{13/49}$$ $$\boxed{\;C_D=0.0245\,Re_L^{-1/7}\left(\frac{b}{\delta}\right)^{13/49}\;}$$ in which the Reynolds number is formed on the free-stream speed, $Re_L=U_\infty L/\nu$. If the planform area $bL$ is preferred as the reference, the numerical constant doubles to 0.049 and the functional form is unchanged.
5. Part (b) — form the reference drag in a uniform stream. Immersed in a uniform stream $U_\infty$ every strip sees the same speed, so the same correlation gives $$\begin{aligned} F_0&=0.031\,Re_L^{-1/7}\cdot\tfrac{1}{2}\rho U_\infty^{2}(2bL)\cr C_{D,0}&=0.031\,Re_L^{-1/7} \end{aligned}$$
6. Compare the two. Dividing, everything except the profile factor cancels: $$\boxed{\;\frac{F}{F_0}=\frac{C_D}{C_{D,0}} =\frac{49}{62}\left(\frac{b}{\delta}\right)^{13/49}\;}$$ The ratio is always less than unity, so the plate inside the boundary layer always feels less drag than the same plate in a uniform stream, simply because most of it is bathed in fluid moving slower than $U_\infty$. The reduction is modest, because the one-seventh profile is very full: even a plate reaching exactly to the edge of the layer, $b=\delta$, feels only $49/62=0.790$ of the uniform-stream drag, a 21 per cent saving. A shorter plate saves more — at $b=0.75\delta$ the ratio is 0.732, and at $b=0.10\delta$ it is 0.427.
7. Check the algebra on a concrete case. For air at $U_\infty=30$ m/s, $\nu=1.5\times10^{-5}\ \text{m}^{2}/\text{s}$, $\rho=1.2\ \text{kg/m}^{3}$, $L=2$ m, $\delta=80$ mm and $b=60$ mm, the closed form gives $F=0.335$ N against $F_0=0.458$ N, a ratio of 0.732 exactly as predicted, and a strip-by-strip numerical integration reproduces $F$ to five figures.