22-Mec-A6 Fluid Machinery · December 2018

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 2018, 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.

Reference texts. J.D. Anderson, Modern Compressible Flow, 3rd ed. (Q1); F.M. White, Fluid Mechanics, 8th ed. and Kundu, Cohen & Dowling, Fluid Mechanics, 6th ed. (Q2, Q5); F.M. White, Viscous Fluid Flow, 3rd ed. (Q3, Q4, Q6); Schlichting & Gersten, Boundary-Layer Theory, 8th ed. (Q6); Fox & McDonald, Introduction to Fluid Mechanics, 10th ed. (general).

Question 6: Integral Boundary Layer over a Moving Belt (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. A zero-pressure-gradient laminar boundary layer whose wall (the belt) moves at $U_b=0.5U_0$ in the flow direction, with an assumed quadratic velocity profile in $\eta=y/\delta$.

Given data
Free-stream speed	$U_0$ (constant, $dU_0/dx=0$)
Belt (wall) speed	$U_b = 0.5\,U_0$
Trial profile	$u/U_0=a+b\eta+c\eta^2,\ \ \eta=y/\delta$
Conditions	$\delta(0)=0$; laminar; $W/\delta\gg10$ (2-D)

Find. (a) $a,b,c$; (b) $\delta(x)$; (c) $C_{fx}$; (d) $\bar\tau_w$ over $[0,L]$.

Figure 6. Laminar boundary layer over a belt that translates at $U_b=0.5U_0$; the profile runs from $U_b$ at the wall to $U_0$ at the edge.

Approach. Fix the profile coefficients from the wall and edge conditions, evaluate the momentum thickness and wall shear for that profile, then close $\delta(x)$ with the von Kármán momentum-integral equation and differentiate/average for the shear coefficients.

(a) Coefficients. Impose $u(0)=U_b=0.5U_0$, $u(\delta)=U_0$, and a smooth edge $\partial u/\partial y|_\delta=0$: $$a=0.5,\qquad a+b+c=1,\qquad b+2c=0 \;\Rightarrow\; \boxed{a=0.5,\ b=1,\ c=-0.5.}$$ So $u/U_0 = 0.5+\eta-0.5\eta^{2}$.
Momentum thickness and wall shear. With $f=u/U_0$, $$\frac{\theta}{\delta}=\int_0^1 f(1-f)\,d\eta=\frac{5}{6}-\frac{43}{60}=\frac{7}{60},\qquad \tau_w=\mu\frac{\partial u}{\partial y}\Big|_0=\frac{\mu U_0}{\delta}\,b=\frac{\mu U_0}{\delta}.$$
(b) Boundary-layer thickness. The zero-pressure-gradient momentum integral $\tau_w=\rho U_0^2\,d\theta/dx$ gives $$\frac{\mu U_0}{\delta}=\rho U_0^{2}\frac{7}{60}\frac{d\delta}{dx}\;\Rightarrow\; \delta\,d\delta=\frac{60}{7}\frac{\nu}{U_0}dx .$$ Integrating from $\delta(0)=0$: $$\boxed{\delta(x)=\sqrt{\frac{120}{7}}\sqrt{\frac{\nu x}{U_0}}=4.14\sqrt{\frac{\nu x}{U_0}}\;,\qquad \frac{\delta}{x}=\frac{4.14}{\sqrt{Re_x}}.}$$
(c) Local shear coefficient. With $\tau_w=\mu U_0/\delta$, $$C_{fx}=\frac{\tau_w}{\tfrac12\rho U_0^{2}}=\frac{2\nu}{U_0\,\delta}=\frac{2}{\sqrt{120/7}}\frac{1}{\sqrt{Re_x}}\;\Rightarrow\;\boxed{C_{fx}=\frac{0.483}{\sqrt{Re_x}}.}$$
(d) Average wall shear. Since $\tau_w=\mu U_0/\delta\propto x^{-1/2}$, its average over $[0,L]$ is exactly twice the local value at $L$: $$\bar\tau_w=\frac1L\int_0^L\tau_w\,dx = 2\,\tau_w(L)\;\Rightarrow\; \bar C_f=\frac{0.966}{\sqrt{Re_L}},\qquad \boxed{\bar\tau_w=\frac{0.483\,\rho U_0^{2}}{\sqrt{Re_L}}\;,\quad Re_L=\frac{U_0L}{\nu}.}$$

Question 6 — results
Quantity	Result
Profile coefficients	$a=0.5,\ b=1,\ c=-0.5$
Momentum thickness	$\theta=\tfrac{7}{60}\delta$
BL thickness	$\delta/x = 4.14/\sqrt{Re_x}$
Local shear coeff.	$C_{fx}=0.483/\sqrt{Re_x}$
Average shear	$\bar\tau_w=0.483\,\rho U_0^2/\sqrt{Re_L}$ (i.e. $\bar C_f=0.966/\sqrt{Re_L}$)

Topic. The Kármán–Pohlhausen momentum-integral method for a laminar boundary layer, adapted to a moving (belt) wall.

Key relations. Assumed profile with $u(0)=U_b$; momentum integral $\tau_w=\rho U_0^2\,d\theta/dx$; $\theta=\int_0^\delta \tfrac{u}{U_0}(1-\tfrac{u}{U_0})dy$; $\tau_w=\mu\,\partial u/\partial y|_0$; $C_{fx}=\tau_w/(\tfrac12\rho U_0^2)$.

Why this works. The integral method needs only a plausible profile satisfying the wall and edge conditions; substituting it into the momentum balance yields a first-order ODE for $\delta(x)$ that integrates in closed form. The moving wall enters solely through the boundary condition $u(0)=U_b=0.5U_0$, which halves the wall velocity gradient (here $b=1$ instead of $2$) and hence lowers the friction relative to a stationary plate. Because $\tau_w\propto x^{-1/2}$, the length-average is simply twice the trailing-edge local value.

Common pitfall. Using the free-stream speed at the wall (i.e. $a=0$). The belt is not at rest: $u(0)=U_b$, so $a=0.5$; forgetting this gives the ordinary stationary-plate coefficients ($\delta/x=5.48/\sqrt{Re_x}$) and overstates the drag. Also, the exit velocity slope must vanish at the edge for a smooth match, which supplies the third condition.

Source. Schlichting & Gersten, Boundary-Layer Theory, 8th ed., Ch. 8 (integral methods); White, Viscous Fluid Flow, 3rd ed., Ch. 4.

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