Boundary Layer Theory Questions and Answers

Answer: A

The profile is \(u/U = 2(y/\delta) - (y/\delta)^{2}\).

Apply the three boundary conditions: no slip at the wall, free-stream velocity at the edge, and a smooth join with the free stream (zero slope) at the edge.

\[ \begin{aligned} u(0) = 0 &\Rightarrow a = 0 \\ \left.\tfrac{du}{dy}\right|_{\delta} = 0 &\Rightarrow b + 2c\delta = 0 \\ u(\delta) = U &\Rightarrow b\delta + c\delta^{2} = U \end{aligned} \]

These give \(b = 2U/\delta\) and \(c = -U/\delta^{2}\). The '+' option gives u = 3U at the edge. The 1.5 and 0.5 coefficients belong to a cubic profile and do not satisfy the zero-slope condition for a quadratic.

Answer: B

The displacement thickness is 2 mm.

With \(\eta = y/\delta\) and the profile \(u/U = 2\eta - \eta^{2}\):

\[ \begin{aligned} \delta^{*} &= \delta\int_{0}^{1}\left(1 - \frac{u}{U}\right)d\eta \\ &= \delta\int_{0}^{1}(1 - 2\eta + \eta^{2})\,d\eta \\ &= \delta\left(1 - 1 + \tfrac{1}{3}\right) = \frac{6}{3} = 2\ \text{mm} \end{aligned} \]

Displacement thickness measures the outward shift of the streamlines caused by the slowed fluid, so it must be positive and smaller than \(\delta\); the negative options are impossible. 2.25 mm would come from a different (cubic) profile.

Answer: C

The transition from laminar to turbulent boundary layer on a wing can be delayed by implementing suction through the surface, the key says. Suction removes the slow fluid near the wall, so the boundary layer becomes thinner and more stable. Increasing the angle of attack and adding roughness elements and vortex generators bring on or increase turbulence and so do the opposite.

This is the principle of laminar flow control on experimental wings, where the surface is perforated and a little air is drawn in through it.

Answer: B

The ratio is \(1 - \frac{2}{\pi}\).

The displacement thickness is the integral of the velocity deficit across the boundary layer. With \(\eta = y/\delta\):

\[ \begin{aligned} \frac{\delta^{*}}{\delta} &= \int_{0}^{1}\left[1 - \sin\left(\tfrac{\pi\eta}{2}\right)\right]d\eta \\ &= 1 - \frac{2}{\pi} \approx 0.363 \end{aligned} \]

since \(\int_{0}^{1}\sin(\pi\eta/2)\,d\eta = 2/\pi\). \(2/\pi\) is that integral alone, the flow carried rather than the deficit; \(1 + 2/\pi\) has the sign wrong, and 0 would mean no boundary layer.

Answer: C

The boundary layer becomes 0.5 mm thick.

At a Reynolds number of 1000 the flow is laminar, and the Blasius result for a flat plate is:

\[ \delta = \frac{5x}{\sqrt{Re_{x}}} \propto \frac{1}{\sqrt{U}} \]

At the same location, x is fixed, so increasing the velocity four times divides the thickness by \(\sqrt{4}=2\): 1 mm becomes 0.5 mm.

0.25 mm assumes thickness inversely proportional to velocity, and 2 or 4 mm have the thickness growing with velocity, the wrong direction. A faster stream thins the boundary layer.

Answer: D

The wall shear stress at 30 mm is \(\tau_{w}/\sqrt{2}\).

First check the regime with the local Reynolds number:

\[ \begin{aligned} Re_{x} &= \frac{Ux}{\nu} \\ &= \frac{2 \times 0.015}{1.5 \times 10^{-5}} \\ &= 2000 \\ Re_{x}(30\ \text{mm}) &= 4000 \end{aligned} \]

Both are far below the transition value of about \(5 \times 10^{5}\), so the boundary layer is laminar and the Blasius result applies:

\[ \begin{aligned} \tau_{w} &= \frac{0.332\,\rho U^{2}}{\sqrt{Re_{x}}} \\ &\propto x^{-1/2} \end{aligned} \]

Doubling the distance from 15 mm to 30 mm divides the shear stress by \(\sqrt{2}\). \(\tau_{w}/2\) would need \(\tau_{w} \propto 1/x\). \(\sqrt{2}\,\tau_{w}\) and \(2\,\tau_{w}\) make the stress grow downstream, but it falls, because the boundary layer thickens and the velocity gradient at the wall weakens.

Answer: C

The ratio is \(\frac{1}{6}\).

With \(u/U_{\infty} = \eta = y/\delta\), the momentum thickness is

\[ \begin{aligned} \theta &= \delta\int_{0}^{1} \eta(1 - \eta)\,d\eta \\ &= \delta\left(\frac{1}{2} - \frac{1}{3}\right) = \frac{\delta}{6} \end{aligned} \]

\(1/2\) is the displacement thickness ratio for the same profile, \(\int_{0}^{1}(1 - \eta)\,d\eta\), and \(1/3\) and \(1/4\) do not come from either integral.

Answer: C

The momentum thickness is \(\int_{0}^{\infty} \frac{u}{U_{\infty}}\left(1 - \frac{u}{U_{\infty}}\right) dy\).

Momentum thickness \(\theta\) is the thickness of a layer of free-stream fluid whose momentum flux equals the momentum flux lost because the boundary layer slows the fluid. The mass actually passing through a strip \(dy\) is \(\rho u\,dy\), and each unit of it has lost velocity \(U_{\infty} - u\), so

\[ \begin{aligned} \rho U_{\infty}^{2}\theta \\ = \int_{0}^{\infty} \rho u (U_{\infty} - u)\,dy \end{aligned} \]

which gives the expression above.

\(\int (1 - u/U_{\infty})\,dy\) is the displacement thickness, the mass-flux deficit. \(\int (1 - u^{2}/U_{\infty}^{2})\,dy\) is the energy-type form wrongly built from \(u^{2}\) alone, and \(\int u/U_{\infty}\,dy\) diverges, because \(u \to U_{\infty}\) outside the layer.

Answer: D

The true statement is Bernoulli’s equation cannot be applied in Region I.

Region I is inside the boundary layer. There viscous stresses are significant and the flow is rotational, so mechanical energy is dissipated along a streamline and the inviscid assumption behind Bernoulli's equation fails. It cannot be used there either between arbitrary points or along a streamline.

Region II, outside the boundary layer, is effectively inviscid and, coming from a uniform stream, irrotational. There Bernoulli's equation holds, and because the flow is irrotational it holds between any two points, not only along one streamline. That rules out the statement that it cannot be applied in Region II, and both statements claiming it can be applied in Region I.

Answer: B

The boundary layer thickness grows as \(x^{1/2}\).

For laminar flow over a flat plate, an order-of-magnitude balance of inertia and viscous diffusion in the boundary layer gives \(\frac{U_{\infty}}{x} \sim \frac{\nu}{\delta^{2}}\), so \(\delta \sim \sqrt{\nu x/U_{\infty}}\). The Blasius solution makes this exact:

\[ \begin{aligned} \delta &\approx \frac{5x}{\sqrt{Re_{x}}} = 5\sqrt{\frac{\nu x}{U_{\infty}}} \\ &\propto x^{1/2} \end{aligned} \]

\(x^{-1/2}\) is how the local skin friction coefficient varies, not the thickness, which must grow downstream. \(x^{2}\) grows far too fast, and \(x^{1/4}\) has no basis here; a turbulent layer grows faster, as about \(x^{4/5}\).