Question 6 of 6: Laminar boundary layer in an accelerating wind tunnel

Question 6: Laminar boundary layer in an accelerating wind tunnel (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.

Find. The exponent n; δ/x as a function of Re_x; the local skin-friction coefficient C_fx(Re_x); and the average wall shear stress between x = 0 and x = L.

[Figure not reproduced: Figure 5 (redrawn). The contracting upper wall accelerates the outer stream as x 1/3 . The boundary layer thickens at exactly the same rate, which is what makes the wall shear stress constant along the whole tunnel. See the official exam paper.]

Approach. Use the Kármán momentum-integral equation in the form that retains the pressure gradient, substitute the two power laws, and match first the exponents of x (which gives n) and then the coefficients (which gives B). Everything asked for follows from those two results.

1. Write the momentum-integral equation with a pressure gradient. For a two-dimensional incompressible layer, $$\frac{d\theta}{dx}+\left(2\theta+\delta^{*}\right)\frac{1}{U_{0}}\frac{dU_{0}}{dx} =\frac{\tau_{w}}{\rho U_{0}^{2}}=\frac{C_{fx}}{2}.$$ The second term is the one the flat-plate version drops; it is essential here because the outer flow accelerates.
2. Express every ingredient through δ. From the given shape factors, θ = (2/15)δ and δ^* = (1/3)δ. The wall shear follows by differentiating the assumed profile at the wall: $$\tau_{w}=\mu\left.\frac{\partial u}{\partial y}\right|_{y=0} =\mu U_{0}\frac{d}{dy}\!\left[\frac{2y}{\delta}-\frac{y^{2}}{\delta^{2}}\right]_{y=0} =\frac{2\mu U_{0}}{\delta}.$$ And from the given outer flow, $$\frac{1}{U_{0}}\frac{dU_{0}}{dx}=\frac{1}{3x}.$$
3. Substitute the two power laws. Putting δ = Bxⁿ and U₀ = Ax^1/3, the left-hand side becomes $$\frac{2}{15}Bnx^{n-1}+\left[\frac{4}{15}+\frac{1}{3}\right]\frac{Bx^{n}}{3x} =Bx^{n-1}\left[\frac{2}{15}n+\frac{1}{5}\right],$$ where 4/15 + 1/3 = 3/5 has been used, while the right-hand side becomes $$\frac{2\mu U_{0}}{\delta\rho U_{0}^{2}}=\frac{2\nu}{U_{0}\delta} =\frac{2\nu}{AB}\,x^{-1/3-n}.$$
4. Part (a) — match the exponents. The two sides can only agree at every station if the powers of x agree: $$n-1=-\tfrac{1}{3}-n\quad\Longrightarrow\quad 2n=\tfrac{2}{3}$$ $$\boxed{\;n=\tfrac{1}{3}\;}$$ So the layer grows at exactly the rate at which the outer flow accelerates — the two x^1/3 laws are locked together, and that is the design intent behind the curved wall.
5. Match the coefficients to obtain B. With n = 1/3 the bracket becomes 2/45 + 9/45 = 11/45, so $$B\,\frac{11}{45}=\frac{2\nu}{AB}\quad\Longrightarrow\quad B^{2}=\frac{90\nu}{11A}\quad\Longrightarrow\quad B=\sqrt{\frac{90\nu}{11A}} .$$ The condition δ = 0 at x = 0 is automatically satisfied by the power law, so no constant of integration survives.
6. Part (b) — the thickness in Reynolds-number form. The local Reynolds number is $$\mathrm{Re}_{x}=\frac{U_{0}x}{\nu}=\frac{Ax^{4/3}}{\nu}\quad\Longrightarrow\quad x^{-2/3}=\sqrt{\frac{A}{\nu\,\mathrm{Re}_{x}}} .$$ Therefore $$\frac{\delta}{x}=Bx^{n-1}=Bx^{-2/3} =\sqrt{\frac{90\nu}{11A}}\sqrt{\frac{A}{\nu\,\mathrm{Re}_{x}}}$$ $$\boxed{\;\frac{\delta}{x}=\sqrt{\frac{90}{11\,\mathrm{Re}_{x}}}=\frac{2.860}{\sqrt{\mathrm{Re}_{x}}}\;}$$ Both A and ν drop out, leaving a pure function of Re_x. The coefficient is well below the 5.48 that the same parabolic profile gives on a flat plate, which is the expected thinning effect of a favourable pressure gradient.
7. Part (c) — the skin-friction coefficient. Using τ_w = 2μU₀/δ from Step 2, $$C_{fx}=\frac{\tau_{w}}{\tfrac{1}{2}\rho U_{0}^{2}}=\frac{4\nu}{U_{0}\delta} =\frac{4}{\mathrm{Re}_{x}}\cdot\frac{x}{\delta} =\frac{4}{\mathrm{Re}_{x}}\sqrt{\frac{11\,\mathrm{Re}_{x}}{90}}$$ $$\boxed{\;C_{fx}=4\sqrt{\frac{11}{90}}\ \mathrm{Re}_{x}^{-1/2} =\frac{1.398}{\sqrt{\mathrm{Re}_{x}}}\;}$$ Note the tidy structure: the coefficient here is exactly 4 divided by the coefficient in part (b), a consequence of the parabolic profile and nothing else.
8. Part (d) — and the elegant result. Rather than integrating blindly, look at the dimensional wall shear: $$\tau_{w}=\frac{2\mu U_{0}}{\delta}=\frac{2\mu Ax^{1/3}}{Bx^{1/3}}=\frac{2\mu A}{B},$$ in which every power of x cancels. The wall shear stress is the same at every station, so its average over any interval equals its local value: $$\bar\tau_{w}=\frac{1}{L}\int_{0}^{L}\tau_{w}\,dx=\tau_{w}=\frac{2\mu A}{B} =2\mu A\sqrt{\frac{11A}{90\nu}}$$ $$\boxed{\;\bar\tau_{w}=2\sqrt{\frac{11}{90}}\ \rho A^{3/2}\nu^{1/2} =0.6992\,\rho A^{3/2}\nu^{1/2}\;}$$ Equivalently, and more usefully for a test engineer, τ̄_w = ½ρU₀(L)² C_f(Re_L), evaluated with the same 1.398 coefficient.
9. Put numbers on it. Take air at ρ = 1.20 kg/m³, ν = 1.5 × 10⁻⁵ m²/s, with A = 6 SI units and a working length L = 0.40 m. Then B = 4.52 × 10⁻³, U₀(L) = 4.42 m/s and Re_L = 1.18 × 10⁵ — safely laminar, as the question assumes. The wall shear is τ̄_w = 0.0478 Pa at every station, giving a total drag of 0.0191 N per metre of tunnel width over the working length.
10. Why a designer would want this. The general power law U₀ ∝ x^m gives τ_w ∝ x^(3m−1)/2, so m = 1/3 is precisely the exponent that makes the exponent vanish. A tunnel shaped this way presents a uniform wall shear to whatever is mounted on the floor, which is exactly what is wanted for a constant-skin-friction test section or a uniformly loaded surface-shear measurement.