Boundary Layer and Drag Questions and Answers

Answer: B

The flow through PS vanishes when \(A_{0} = u_{\infty}(n + 1)\).

The plate is non-porous, so no fluid crosses QR. If nothing crosses PS either, the mass entering through PQ must all leave through RS. Per unit depth:

\[ \begin{aligned} u_{\infty} h &= \int_{0}^{h} A_{0}\left(\frac{y}{h}\right)^{n} dy \\ &= \frac{A_{0} h}{n + 1} \\ A_{0} &= u_{\infty}(n + 1) \end{aligned} \]

\(u_{\infty}/(n + 1)\) is the average speed of a profile with \(A_{0} = u_{\infty}\), the result inverted. The options with n - 1 come from a slip in the power of y when integrating the profile.

Answer: C

The skin friction coefficient varies as \(C_{f} \propto 1/\sqrt{Re_{c}}\).

For a laminar boundary layer on a flat plate the Blasius solution gives the average skin friction coefficient over a plate of length \(c\) as

\[ C_{f} = \frac{1.328}{\sqrt{Re_{c}}} \]

so it falls as the inverse square root of the Reynolds number. A coefficient growing with \(\sqrt{Re_{c}}\) or with \(Re_{c}\) has the trend backwards. \(1/Re_{c}\) falls too fast; that is the behaviour of a creeping, Stokes-type flow, not a boundary layer. (A turbulent boundary layer falls more slowly, roughly as \(Re_{c}^{-1/5}\).)

Answer: A

With Pr = 0.7, the thermal boundary layer is thicker than the velocity boundary layer.

The Prandtl number compares momentum diffusivity with thermal diffusivity. When Pr is below 1, heat diffuses faster than momentum, so the temperature disturbance spreads further from the wall than the velocity disturbance. For laminar flow over a flat plate the ratio is about \(\delta/\delta_{t} \approx Pr^{1/3}\), so \(\delta_{t}/\delta \approx 0.7^{-1/3} \approx 1.13\).

A thinner thermal layer would need Pr above 1, as in oils and water, and equal thicknesses need Pr = 1. The two thicknesses are linked through Pr, so 'no relationship' is wrong.

Answer: C

The shaded area represents the displacement thickness.

The plot is drawn in the non-dimensional velocity \(u/U\) against y. The black region lies between the profile and the line \(u/U = 1\), so its area is

\[ \delta^{*} = \int_{0}^{\infty} (1 - u/U)\,dy \]

which is the definition of displacement thickness: the distance by which the wall would have to be moved out, in an inviscid flow, to carry the same mass-flow deficit that the boundary layer causes.

Boundary layer thickness is a height (where u reaches about 0.99U), not an area. Momentum thickness is \(\int \frac{u}{U}(1 - \frac{u}{U})dy\), which carries an extra factor \(u/U\) and is not the shaded region. The shape factor is the ratio of displacement to momentum thickness, a pure number.

Answer: B

It measures a temperature greater than static but less than total temperature.

An insulated surface reaches the adiabatic wall (recovery) temperature, where the frictional heating in the boundary layer balances the heat conducted away. It is written with a recovery factor r:

\[ T_{aw} = T + r\frac{V^{2}}{2c_{p}} \]

For a laminar boundary layer in air, \(r \approx \sqrt{Pr} \approx 0.85\). Since r is less than 1, the wall recovers most but not all of the kinetic energy, so \(T \lt T_{aw} \lt T_{0}\).

The static temperature would be read only with no recovery at all (r = 0), the total temperature only with full recovery (r = 1), and a temperature above total would need r greater than 1, which laminar flow of air does not give.

Answer: B

The momentum thickness grows as \(\theta \propto x^{1/2}\).

The Blasius solution for the laminar boundary layer on a flat plate gives

\[ \begin{aligned} \theta &= \frac{0.664\,x}{\sqrt{Re_{x}}} \\ Re_{x} &= \frac{U_{\infty}x}{\nu} \end{aligned} \]

so \(\theta = 0.664\sqrt{\nu x / U_{\infty}}\), which is proportional to \(\sqrt{x}\). The displacement thickness and the boundary-layer thickness grow in the same way, as all laminar thicknesses do.

The other exponents do not belong to this solution. \(x^{1/7}\) recalls the one-seventh power velocity profile of a turbulent boundary layer, whose thickness grows roughly as \(x^{4/5}\). A negative exponent would make the layer thin downstream, the opposite of what happens, and \(x^{2/3}\) has no basis here.

Answer: D

The boundary-layer thickness varies as δ \(\propto U^{-1/2}\).

For a laminar boundary layer on a flat plate, the Blasius solution gives

\[ \begin{aligned} \delta &\approx \frac{5.0\,x}{\sqrt{Re_{x}}} \\ &= 5.0\sqrt{\frac{\nu x}{U}} \end{aligned} \]

At a fixed location \(x = L\) and fixed viscosity, \(\delta\) is proportional to \(1/\sqrt{U}\): a faster stream gives a thinner boundary layer, because the momentum carried downstream outpaces the diffusion of viscous effects away from the wall.

The options with positive powers of \(U\) would make the layer thicken as the flow speeds up, the opposite of what happens. \(U^{1/2}\) is the right power with the wrong sign.

Answer: A

\(\delta/x\) is proportional to \(Re_{x}^{-1/2}\).

For a laminar boundary layer on a flat plate, the Blasius solution gives

\[ \frac{\delta}{x} \approx \frac{5.0}{\sqrt{Re_{x}}} \]

The scaling follows from balancing inertia, \(U^{2}/x\), with viscous diffusion, \(\nu U/\delta^{2}\), which gives \(\delta \sim \sqrt{\nu x/U}\). So the boundary layer grows as \(\sqrt{x}\) and \(\delta/x\) falls as \(Re_{x}^{-1/2}\). The exponent −1/5 belongs to the turbulent boundary layer; −1, −3/2 and −2 do not arise for either.

Answer: C

d'Alembert's paradox is the prediction of zero drag by potential flow theory.

In 1752 d'Alembert showed that steady, inviscid, incompressible, irrotational flow past a body gives a pressure distribution symmetric front to back, so the net drag is zero, contrary to every experiment. The paradox is resolved by viscosity: the boundary layer and flow separation create skin friction and pressure drag, which an inviscid theory leaves out.

Departures of drag from \(D \propto v^{2}\) at low speeds (Stokes flow) or high subsonic speeds (drag divergence) are real effects with other names, and shocks in transonic flow are a compressibility effect, not a paradox of potential flow.