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

Question 6: Drag on a plate standing inside 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 exactly $\delta$, standing edge-on to a wall so that both of its faces are wetted, and sitting entirely inside a fully turbulent boundary layer of thickness $\delta$ whose profile is $u(y) = U_\infty (y/\delta)^{1/7}$. The fluid has density $\rho$ and kinematic viscosity $\nu$. The plate's top edge coincides with the edge of the boundary layer, so it sees the full range of the profile from $u = 0$ at the wall to $u = U_\infty$ at $y = \delta$.

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

[Figure not reproduced: Figure 6 (redrawn). Each horizontal strip $dy$ of the plate is treated as its own two-sided flat plate of length $L$ immersed in a uniform stream of the local speed $u(y)$, and the strips are integrated over $0 \le y \le \delta$. See the official exam paper.]

Approach. Treat each horizontal strip of the plate as an independent two-sided flat plate at its own local free-stream speed $u(y)$ and hence its own local Reynolds number, apply the standard turbulent flat-plate friction law to each strip, and integrate over the plate height. The whole derivation collapses to one power integral.

1. Write the friction law for a single strip. The average friction coefficient for one side of a smooth flat plate in a fully turbulent layer, obtained by integrating the local law $C_{f,x} = 0.0266\,Re_x^{-1/7}$ from the leading edge, is $$C_D = \frac{7}{6}(0.0266)\,Re_L^{-1/7} = 0.031\,Re_L^{-1/7}$$ A strip of height $dy$ at elevation $y$ has wetted area $2L\,dy$ (two faces) and sees the local speed $u(y)$, so its own Reynolds number is $Re_L(y) = u(y)L/\nu$ and its drag is $$dF = 2\left[0.031\left(\frac{u L}{\nu}\right)^{-1/7}\right]\frac{\rho u^{2}}{2}\,L\,dy = 0.031\,\rho\,L\left(\frac{\nu}{L}\right)^{1/7} u^{13/7}\,dy$$ The exponent is the key structural result: the $u^{2}$ of the dynamic pressure is reduced by the $u^{-1/7}$ of the friction law, leaving $2-\tfrac17 = \tfrac{13}{7}$.
2. Substitute the profile and integrate over the height. With $u(y) = U_\infty (y/\delta)^{1/7}$ the velocity raised to the thirteen-sevenths power is $u^{13/7} = U_\infty^{13/7}(y/\delta)^{13/49}$, and the plate spans the full layer, $0 \le y \le \delta$: $$\int_{0}^{\delta}\left(\frac{y}{\delta}\right)^{13/49} dy = \delta \cdot \frac{1}{1+\tfrac{13}{49}} = \frac{49}{62}\,\delta = 0.7903\,\delta$$ so the total drag is $$F_D = 0.031\,\rho\,L^{6/7}\,\nu^{1/7}\,U_\infty^{13/7}\;\frac{49}{62}\,\delta$$ Note that this is an ordinary power integral, not a boundary-layer integration — the momentum-integral machinery is not needed and never enters.
3. Non-dimensionalise to get the drag coefficient (part a). Referring the drag to the total wetted area $A = 2L\delta$ and the free-stream dynamic pressure, $$C_D \equiv \frac{F_D}{\tfrac12 \rho U_\infty^{2}\,(2L\delta)} = 0.031 \cdot \frac{49}{62}\left(\frac{\nu}{U_\infty L}\right)^{1/7}$$ $$\boxed{C_D = \frac{49}{62}\,(0.031)\,Re_L^{-1/7} = 0.0245\,Re_L^{-1/7}, \qquad Re_L = \frac{U_\infty L}{\nu}}$$ The Reynolds number is built on the free-stream speed $U_\infty$ and the plate length $L$, so the formula is directly comparable with the textbook flat-plate result. If the plate height were some general $b$ rather than exactly $\delta$, the same integration gives the more general $C_D = 0.0245\,Re_L^{-1/7}(b/\delta)^{13/49}$, of which the boxed result is the case $b = \delta$.
4. Form the uniform-stream reference (part b). The same plate held in a uniform stream $U_\infty$ — every strip now seeing the full free-stream speed — has $$F_{unif} = 0.031\,Re_L^{-1/7}\;\tfrac12\rho U_\infty^{2}\,(2L\delta)$$ which is the ordinary two-sided flat-plate drag with no height integral at all.
5. Take the ratio and interpret it. Every factor cancels except the height integral: $$\frac{F_D}{F_{unif}} = \boxed{\frac{49}{62} = 0.790}$$ so the plate sitting inside the boundary layer feels about 79 % of the uniform-stream drag — only 21 % less. The instructive point is how weak that sheltering is. Intuition suggests a large reduction, because the lower half of the plate sits in fluid moving far slower than $U_\infty$. Two effects defeat that intuition. First, a one-seventh power law is extremely blunt: at only 1 % of the layer thickness the local speed is already $0.01^{1/7} = 52\;\%$ of $U_\infty$, so most of the plate is in fast-moving fluid. Second, the drag scales as $u^{13/7}$ rather than $u^{2}$, so the friction law itself partially compensates for the lower local speeds. A designer should therefore not expect much protection from burying a fin inside a turbulent boundary layer.