Drag of a Plate Immersed in a Turbulent Boundary Layer: 22-Mec-A6 Fluid Machinery, May 2017, Q6

Question 6: Drag of a Plate Immersed 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 two-sided plate of streamwise length $L$ and height $\delta$ standing in a turbulent boundary layer whose approach velocity varies with height as $u(y)=U_\infty(y/\delta)^{1/7}$; the plate height equals the boundary-layer thickness $\delta$.

Find. (a) the plate drag coefficient; (b) the ratio of this drag to the uniform-stream ($U_\infty$) drag.

Figure 6: Splitter plate ($L\times\delta$) aligned with the flow, immersed in a $1/7$-power-law turbulent boundary layer; each height $y$ sees local free-stream $u(y)$.

Approach. Treat each horizontal strip $dy$ at height $y$ as an independent turbulent flat-plate boundary layer developing over length $L$ at the local free-stream speed $u(y)$, use the standard turbulent friction law $C_{D}=0.074\,Re_L^{-1/5}$ per side, and integrate over $0\le y\le\delta$.

Strip drag. For a strip of height $dy$ at local speed $U(y)$, both sides wetted, the turbulent flat-plate law gives $$dF=2\left[0.074\left(\frac{U(y)L}{\nu}\right)^{-1/5}\right]\tfrac12\rho\,U(y)^2\,L\,dy=0.074\,\rho L\left(\frac{L}{\nu}\right)^{-1/5}U(y)^{9/5}dy.$$
Integrate over the profile. With $U(y)=U_\infty(y/\delta)^{1/7}$, $U(y)^{9/5}=U_\infty^{9/5}(y/\delta)^{9/35}$, and $$\int_0^{\delta}\!\Big(\tfrac{y}{\delta}\Big)^{9/35}dy=\delta\int_0^1\xi^{9/35}d\xi=\frac{35}{44}\,\delta.$$ Hence the total (two-sided) drag is $$F=0.074\left(\frac{35}{44}\right)\rho\,\delta L\,U_\infty^{2}\,Re_L^{-1/5},\qquad Re_L=\frac{U_\infty L}{\nu}.$$
Drag coefficient. Referencing the one-side planform area $A=L\delta$ and dynamic pressure $\tfrac12\rho U_\infty^2$ (with both sides included in $F$), $$C_D=\frac{F}{\tfrac12\rho U_\infty^2\,L\delta}=2(0.074)\frac{35}{44}Re_L^{-1/5},$$ $$\boxed{C_D=0.118\,Re_L^{-1/5}.}$$
(b) Compare with a uniform stream. In a uniform stream $U_\infty$ the same plate (two sides) has $F_{\text{unif}}=0.074\,\rho\,\delta L\,U_\infty^2\,Re_L^{-1/5}$. The ratio is exactly the profile shape factor, $$\boxed{\frac{F_{\text{shear}}}{F_{\text{unif}}}=\frac{35}{44}\approx0.795.}$$ The sheared boundary-layer flow produces about 20% less drag, because the slow near-wall fluid carries less momentum than a uniform $U_\infty$ would.

Question 6 — results
Quantity	Result
Total drag	$F=0.074\,\tfrac{35}{44}\,\rho\,\delta L\,U_\infty^2\,Re_L^{-1/5}$
Drag coefficient (both sides, ref. $L\delta$)	$C_D=0.118\,Re_L^{-1/5}$
Shear-flow / uniform-flow drag ratio	$35/44\approx0.795$

Topic: Turbulent flat-plate skin friction and the strip (local-similarity) method for a non-uniform approach flow.

Key relations: turbulent one-side drag law $C_D=0.074\,Re_L^{-1/5}$ ($5\times10^5\lesssim Re_L\lesssim10^7$); $1/7$-power velocity profile; strip integration $F=\int dF(y)$.

Why this works: when the plate height equals the boundary-layer thickness, the local free-stream seen by each horizontal strip is the boundary-layer velocity at that height. Because turbulent friction scales as $U^{9/5}$, integrating the $1/7$-power profile reduces the drag by the pure number $\int_0^1\xi^{9/35}d\xi=35/44$ relative to a uniform stream — a clean measure of how much momentum the retarded near-wall fluid removes from the loading. The result is independent of $\delta$ and $\nu$ except through $Re_L$.

Common pitfall: using the exponent on $U$ carelessly — the drag integrand goes as $U^{9/5}$ (the $-1/5$ from $Re_L^{-1/5}$ combines with $U^2$), not as $U^2$; and forgetting the factor of two for a two-sided plate. Mixing the laminar Blasius law into a stated turbulent profile is also a frequent error.

Source: White, Fluid Mechanics, 8th ed., §7.4 (turbulent flat-plate drag); Fox, Pritchard & McDonald, 10th ed., Ch. 9.

Question 6 of 6: Drag of a Plate Immersed in a Turbulent Boundary Layer

Question 6: Drag of a Plate Immersed in a Turbulent Boundary Layer (20 marks)