Question 6 of 6: Boundary layer in an accelerating suction wind tunnel

Question 6: Boundary layer in an accelerating suction 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 that makes the assumed growth law satisfy the momentum-integral equation, the resulting δ/x and local skin-friction coefficient in terms of Re_x, and the average skin-friction coefficient over 0 ≤ x ≤ L referred to the exit velocity.

Approach. Substitute the two power laws and the profile constants into the Kármán momentum-integral equation, then match the powers of x to get n and match the coefficients to get B; everything else follows algebraically.

1. Write the momentum-integral equation for a flow with a pressure gradient. For a two-dimensional incompressible boundary layer beneath an external velocity U₀(x), $$\frac{\mathrm{d}\theta}{\mathrm{d}x} +\left(2\theta+\delta^{*}\right)\frac{1}{U_0}\frac{\mathrm{d}U_0}{\mathrm{d}x} =\frac{\tau_w}{\rho U_0^{2}} .$$ The bracket is what carries the effect of the acceleration; it disappears only for a flat plate in a uniform stream.
2. Evaluate the wall shear from the assumed profile. Differentiating $u/U_0=2(y/\delta)-(y/\delta)^{2}$ at the wall, $$\tau_w=\mu\left.\frac{\partial u}{\partial y}\right|_{y=0}=\frac{2\mu U_0}{\delta} \quad\Longrightarrow\quad \frac{\tau_w}{\rho U_0^{2}}=\frac{2\nu}{U_0\delta}.$$
3. Part (a) — substitute the two power laws. With $\theta=\tfrac{2}{15}\delta$, $\delta^{*}=\tfrac{1}{3}\delta$, $\delta=Bx^{n}$ and $U_0=Ax^{m}$ with m = 1/3, the left-hand side becomes $$\frac{2}{15}Bn\,x^{n-1}+\left(\frac{4}{15}+\frac{1}{3}\right)Bx^{n}\frac{m}{x} =Bx^{n-1}\left[\frac{2}{15}n+\frac{3}{5}m\right],$$ while the right-hand side is $$\frac{2\nu}{U_0\delta}=\frac{2\nu}{AB}x^{-m-n}.$$
4. Match the exponents of x. The two sides can agree at every station only if $$n-1=-m-n \quad\Longrightarrow\quad n=\frac{1-m}{2}=\frac{1-\tfrac{1}{3}}{2},$$ $$\boxed{n=\tfrac{1}{3}}$$ so the layer grows as x^1/3. This is slower than the x^1/2 of a flat plate, which is exactly what a favourable pressure gradient should do: the accelerating core keeps thinning the layer relative to its own distance from the leading edge.
5. Match the coefficients to obtain B. Substituting n = m = 1/3 into the bracket, $$\frac{2}{15}\left(\frac{1}{3}\right)+\frac{3}{5}\left(\frac{1}{3}\right) =\frac{2}{45}+\frac{9}{45}=\frac{11}{45},$$ so $B^{2}\dfrac{11}{45}=\dfrac{2\nu}{A}$ and $$\boxed{B=\sqrt{\frac{90\,\nu}{11\,A}}}$$
6. Part (b) — convert to a Reynolds-number form. Since $\mathrm{Re}_x=U_0x/\nu=Ax^{4/3}/\nu$, we have $x^{-2/3}=\sqrt{A/(\nu\,\mathrm{Re}_x)}$, and 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 have cancelled, so the result is universal for this profile and this acceleration. For comparison the same parabolic profile on a flat plate gives 5.48/√Re_x, so the favourable gradient here thins the layer by very nearly a factor of two.
7. Part (c) — form the local skin-friction coefficient. From step 2, $$C_{fx}=\frac{\tau_w}{\tfrac{1}{2}\rho U_0^{2}}=\frac{4\nu}{U_0\delta} =\frac{4}{\mathrm{Re}_x}\left(\frac{x}{\delta}\right) =\frac{4}{\mathrm{Re}_x}\cdot\frac{\sqrt{\mathrm{Re}_x}}{2.860},$$ $$\boxed{C_{fx}=4\sqrt{\frac{11}{90}}\,\mathrm{Re}_x^{-1/2}=\frac{1.398}{\sqrt{\mathrm{Re}_x}}}$$ The coefficient of δ/x and the coefficient of C_fx are locked to one another through the profile constant, since C_fx = 4 ÷ (coefficient of δ/x).
8. Part (d) — discover that the wall shear is constant along the tunnel. Writing the dimensional shear out, $$\tau_w=C_{fx}\frac{\rho U_0^{2}}{2} =1.398\left(\frac{\nu}{Ax^{4/3}}\right)^{1/2}\frac{\rho A^{2}x^{2/3}}{2} =\frac{1.398}{2}\rho A^{3/2}\nu^{1/2}\,x^{0},$$ so τ_w is independent of x. In general $\tau_w\propto x^{(3m-1)/2}$, and m = 1/3 is precisely the exponent that makes it uniform. That is the whole reason a suction tunnel is contoured this way: a test section with constant wall shear has a boundary layer whose growth is predictable and whose wall heat-transfer and skin-friction instrumentation reads the same at every station.
9. Complete the averaging. Because the integrand is constant, $$\bar\tau_w=\frac{1}{L}\int_0^{L}\tau_w\,\mathrm{d}x=\tau_w ,$$ and dividing by the dynamic pressure formed on the exit velocity U₀ = AL^1/3 gives $$\boxed{\overline{C_f}=\frac{1.398}{\sqrt{\mathrm{Re}_L}},\qquad \mathrm{Re}_L=\frac{AL^{4/3}}{\nu}}$$ i.e. the average coefficient is numerically equal to the local coefficient evaluated at x = L. On a flat plate the average is 2 times the trailing-edge local value; here the ratio is exactly 1, which is a compact way of stating the same uniform-shear result.
10. Put numbers on it. For air (ρ = 1.2 kg/m³, ν = 1.5 × 10⁻⁵ m²/s) with A = 5 SI units and a 1.0 m test length, the exit velocity is 5.0 m/s, Re_L = 3.33 × 10⁵, δ(L) = 4.95 mm, C_fx(L) = 2.42 × 10⁻³ and τ_w = 0.0363 Pa. Recomputing the shear directly from the profile, 2μU₀/δ = 2(1.8 × 10⁻⁵)(5.0)/0.00495, returns the same 0.0363 Pa, and evaluating it at x = 0.2 m and x = 2.5 m returns 0.0363 Pa there too — the uniformity claimed in step 8.