Question 6 of 6: Integral Boundary Layer over a Moving Belt
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — December 2019, 16-Mec-A6 Advanced Fluid Mechanics. Open book, 3 hours. Six questions of equal value (20 marks); any five constitute a complete paper. All six are solved here as a study resource.
Given. Zero-pressure-gradient laminar boundary layer over a belt that moves with the flow at $U_b=0.5\,U_0$; a quadratic velocity profile is assumed, with $\eta=y/\delta$.
Given data
Free-stream speed
$U_0$ (constant, $dp/dx=0$)
Belt speed
$U_b=0.5\,U_0$
Trial profile
$u/U_0=a+b\,\eta+c\,\eta^2$, $\eta=y/\delta$
Boundary condition at LE
$\delta=0$ at $x=0$
Find. $a,b,c$; $\delta(x)$; $C_{fx}(x)$; and $\overline{\tau_w}$ over $[0,L]$.
Figure 5. The belt moves with the stream at $0.5\,U_0$, so the profile rises from $0.5\,U_0$ at the wall to $U_0$ at the edge with zero slope at $y=\delta$.
Check: the paper is inconsistent ($U_b=0$ in one line versus $U_b=0.5\,U_0$ in the figure and heading). The figure value $U_b=0.5\,U_0$ is used throughout — it is consistent with the trial profile’s wall condition ($a=0.5$).
Approach. Fix $a,b,c$ from the wall and edge conditions, evaluate the momentum thickness ratio $\theta/\delta$, and close the Kármán momentum-integral equation $\tau_w=\rho U_0^2\,d\theta/dx$ (since $dp/dx=0$) to get $\delta(x)$; the wall-shear definitions then give $C_{fx}$ and $\overline{\tau_w}$.
(a) Profile coefficients. Three conditions: at the wall $u(0)=U_b=0.5U_0\Rightarrow a=0.5$; at the edge $u(\delta)=U_0\Rightarrow a+b+c=1$; smooth match $\left.\partial u/\partial y\right|_{\delta}=0\Rightarrow b+2c=0$. Solving,
$$\boxed{a=\tfrac12,\quad b=1,\quad c=-\tfrac12,\qquad \frac{u}{U_0}=\tfrac12+\eta-\tfrac12\eta^2 .}$$
Momentum thickness. $\dfrac{\theta}{\delta}=\displaystyle\int_0^1\frac{u}{U_0}\Big(1-\frac{u}{U_0}\Big)\,d\eta=\frac{7}{60}$, and the wall shear is $\tau_w=\mu\left.\dfrac{\partial u}{\partial y}\right|_{0}=\dfrac{\mu U_0}{\delta}\,b=\dfrac{\mu U_0}{\delta}$.
(b) Boundary-layer thickness. The momentum integral $\tau_w=\rho U_0^2\,d\theta/dx$ becomes $\dfrac{\mu U_0}{\delta}=\rho U_0^2\dfrac{7}{60}\dfrac{d\delta}{dx}$, i.e. $\delta\,d\delta=\dfrac{60}{7}\dfrac{\nu}{U_0}\,dx$. Integrating from $\delta(0)=0$,
$$\delta^2=\frac{120}{7}\frac{\nu x}{U_0}\ \Rightarrow\ \boxed{\delta=4.14\sqrt{\frac{\nu x}{U_0}}=\frac{4.14\,x}{\sqrt{Re_x}}.}$$
(c) Local skin-friction coefficient. $C_{fx}=\dfrac{\tau_w}{\tfrac12\rho U_0^2}=\dfrac{2\mu}{\rho U_0\delta}=\dfrac{2\nu}{U_0\delta}$; substituting $\delta$,
$$\boxed{C_{fx}=\frac{2}{4.14}\frac{1}{\sqrt{Re_x}}=\frac{0.483}{\sqrt{Re_x}}.}$$
(d) Average wall shear stress. Since $\tau_w=\dfrac{\mu U_0}{\delta}\propto x^{-1/2}$,
$$\overline{\tau_w}=\frac{1}{L}\int_0^{L}\tau_w\,dx=2\,\tau_w(L)=\boxed{\frac{0.483\,\rho U_0^2}{\sqrt{Re_L}}\ }\quad\Big(\overline{C_f}=\frac{0.966}{\sqrt{Re_L}}\Big).$$
