Question 6 of 6: Laminar Boundary Layer in a Favourable Pressure Gradient (Momentum Integral)

Question 6: Laminar Boundary Layer in a Favourable Pressure Gradient (Momentum Integral) (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. Parabolic profile with $\delta^{*}/\delta=1/3$, $\theta/\delta=2/15$; external flow $U_0=A x^{1/3}$; $\delta=B x^n$.

Find. $n$; $\delta/x$, $C_{fx}$ and $\bar C_f$ in terms of the Reynolds number.

Approach. Substitute the power laws for $U_0$ and $\delta$ into the Kármán momentum-integral equation with pressure gradient; matching powers of $x$ fixes $n$, and matching coefficients fixes $B$, from which $\delta/x$, $C_{fx}$ and $\bar C_f$ follow.

1. Wall shear and momentum-integral equation. For the parabolic profile $\tau_w=\mu\left.\dfrac{\partial u}{\partial y}\right|_0=\dfrac{2\mu U_0}{\delta}$. With $\theta=\tfrac{2}{15}\delta$ and $\delta^{*}=\tfrac13\delta$, the Kármán equation $$\frac{\tau_w}{\rho}=U_0^2\frac{d\theta}{dx}+(2\theta+\delta^{*})\,U_0\frac{dU_0}{dx}$$ is the governing balance.
2. Exponent $n$, part (a). Substituting $U_0=A x^{1/3}$ and $\delta=B x^n$: the left side $\propto x^{1/3-n}$ while the right side $\propto x^{\,n-1/3}$. Matching powers, $$\tfrac13-n=n-\tfrac13\;\Rightarrow\;\boxed{n=\tfrac13.}$$ (Generally $n=(1-m)/2$ for $U_0\propto x^m$.)
3. Coefficient and $\delta/x$, part (b). Equating coefficients with $n=1/3$ gives $$\frac{2\nu A}{B}=\frac{11}{45}A^2B^2\;\Rightarrow\;B^2=\frac{90\,\nu}{11\,A}.$$ Since $Re_x=U_0 x/\nu=A x^{4/3}/\nu$, one finds $\delta/x=B\,x^{-2/3}=B\sqrt{A/\nu}\;Re_x^{-1/2}$, i.e. $$\boxed{\frac{\delta}{x}=\sqrt{\tfrac{90}{11}}\;Re_x^{-1/2}=2.86\,Re_x^{-1/2}.}$$
4. Local skin friction, part (c). $C_{fx}=\dfrac{2\tau_w}{\rho U_0^2}=\dfrac{4\nu}{U_0\delta}=\dfrac{4}{(\delta/x)\,Re_x}$, so $$\boxed{C_{fx}=\frac{4}{\sqrt{90/11}}\;Re_x^{-1/2}=4\sqrt{\tfrac{11}{90}}\;Re_x^{-1/2}=1.40\,Re_x^{-1/2}.}$$
5. Average coefficient, part (d). Here $\tau_w=\dfrac{2\mu U_0}{\delta}=\dfrac{2\mu A}{B}$ is constant along the plate (both $U_0$ and $\delta$ scale as $x^{1/3}$), so $\bar\tau_w=\tau_w$. Non-dimensionalising with $U_0(L)=A L^{1/3}$, $$\bar C_f=\frac{2\bar\tau_w}{\rho U_0(L)^2}=\frac{2\tau_w}{\rho U_0(L)^2}=C_{fx}(L),$$ hence $$\boxed{\bar C_f=1.40\,Re_L^{-1/2},\qquad Re_L=\frac{U_0(L)\,L}{\nu}.}$$