Question 6 of 6: Boundary Layer in an Accelerating External Flow

Question 6: Boundary Layer in an Accelerating External Flow (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. External flow $U_0=A\,x^{1/3}$; parabolic profile $u/U_0=2(y/\delta)-(y/\delta)^2$ with $\delta^\ast=\delta/3$, $\theta=2\delta/15$, wall slope $\partial u/\partial y|_0=2U_0/\delta$; growth law $\delta=B\,x^n$.

Find. $n$; $\delta/x(Re_x)$; $C_{fx}(Re_x)$; $\overline{C_f}(Re_L)$.

Approach. Apply the von Kármán momentum-integral equation with a pressure gradient, insert the profile shape factors and $U_0(x)$, match powers of $x$ for $n$, then match coefficients for $B$, and finally form the shear coefficients.

1. Momentum-integral equation. With an external pressure gradient, $$\frac{\tau_w}{\rho U_0^2}=\frac{d\theta}{dx}+\frac{2\theta+\delta^\ast}{U_0}\frac{dU_0}{dx}.$$ For the parabolic profile $\tau_w=\mu\,\partial u/\partial y|_0=2\mu U_0/\delta$, so the left side is $2\nu/(U_0\delta)$.
2. Insert shape factors and $U_0=A\,x^{1/3}$. Using $\theta=\tfrac{2}{15}\delta$, $\delta^\ast=\tfrac13\delta$, and $dU_0/dx=U_0/(3x)$, $$\frac{2\nu}{U_0\delta}=\frac{2}{15}\frac{d\delta}{dx}+\left(\frac{2}{15}\cdot2+\frac13\right)\delta\cdot\frac{1}{3x}=\frac{2}{15}\frac{d\delta}{dx}+\frac{1}{5}\frac{\delta}{x}.$$
3. (a) Match powers of $x$. With $\delta=B x^n$ the left side scales as $x^{-1/3-n}$ and the right side as $x^{\,n-1}$. Equality for all $x$ requires $$-\tfrac13-n=n-1\ \Rightarrow\ 2n=\tfrac23\ \Rightarrow\ \boxed{n=\tfrac13.}$$
4. (b) Coefficient $B$ and $\delta/x$. Matching coefficients with $n=\tfrac13$ (so $2n+3=\tfrac{11}{3}$), $$\frac{2\nu}{AB}=B\,\frac{2n+3}{15}=B\,\frac{11}{45}\ \Rightarrow\ B^2=\frac{90\nu}{11A},\quad B=\sqrt{\tfrac{90}{11}}\,\sqrt{\tfrac{\nu}{A}}.$$ Since $\delta/x=B\,x^{-2/3}$ and $Re_x=U_0x/\nu=A x^{4/3}/\nu$ gives $x^{2/3}=(Re_x\,\nu/A)^{1/2}$, $$\boxed{\frac{\delta}{x}=\sqrt{\tfrac{90}{11}}\;Re_x^{-1/2}=\frac{2.86}{\sqrt{Re_x}}.}$$
5. (c) Local shear coefficient. $C_{fx}=\dfrac{\tau_w}{\tfrac12\rho U_0^2}=\dfrac{4\nu}{U_0\delta}$. Substituting $\delta=2.86\,x\,Re_x^{-1/2}$ and $U_0=Re_x\nu/x$, $$\boxed{C_{fx}=\frac{4}{\sqrt{90/11}}\,Re_x^{-1/2}=\frac{1.40}{\sqrt{Re_x}}.}$$
6. (d) Average coefficient. For $U_0=A x^{1/3}$ the wall shear $\tau_w=\tfrac12\rho U_0^2 C_{fx}$ works out independent of $x$ (the $x$-dependences cancel), so $\overline{\tau_w}=\tau_w$. Dividing by $\tfrac12\rho U_0^2$ evaluated at $U_0=A L^{1/3}$, $$\boxed{\overline{C_f}=\frac{1.40}{\sqrt{Re_L}},\qquad Re_L=\frac{A L^{4/3}}{\nu}.}$$ The average coefficient has the same constant as the local one precisely because the wall shear is uniform in this accelerating flow.