Fluid Kinematics Questions and Answers

Answer: B

\(\nabla^{2}\psi = 0\) holds when the flow is incompressible and irrotational.

A stream function exists for two-dimensional incompressible flow, because defining \(u = \partial\psi/\partial y\) and \(v = -\partial\psi/\partial x\) satisfies continuity automatically. Substituting these into the vorticity gives

\[ \begin{aligned} \omega_{z} &= \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \\ &= -\nabla^{2}\psi \end{aligned} \]

so \(\nabla^{2}\psi = 0\) is exactly the condition of zero vorticity. Both conditions are needed: incompressibility for \(\psi\) to exist in this form, and irrotationality for it to satisfy Laplace's equation. Neither alone is enough, and compressible flow does not have this stream function.

Answer: A

The flow that is not incompressible is \(u = x^{3} + xy^{2},\ v = y^{3} + yx^{2}\).

A two-dimensional flow is incompressible when its divergence vanishes, \(\partial u/\partial x + \partial v/\partial y = 0\). For this flow

\[ \begin{aligned} \nabla \cdot \vec{V} &= (3x^{2} + y^{2}) + (3y^{2} + x^{2}) \\ &= 4(x^{2} + y^{2}) \end{aligned} \]

which is not zero except at the origin.

The others pass: 10t - 10t = 0 for the second; in the third u depends only on y and v = 0, so both derivatives vanish; and in the fourth u = 2y, v = 2x give 0 + 0.

Answer: A

The product is -1.

Along a line of constant \(\phi\), \(d\phi = u\,dx + v\,dy = 0\), so its slope is \(dy/dx = -u/v\). Along a streamline, \(d\psi = -v\,dx + u\,dy = 0\), so its slope is \(dy/dx = v/u\). Their product is

\[ \left(-\frac{u}{v}\right)\left(\frac{v}{u}\right) = -1 \]

which means the equipotential lines and streamlines cross at right angles everywhere (the flow net). The product would be 0 or infinite only where one velocity component vanished, which the question excludes, and +1 would mean the two families were parallel or mirror images, not orthogonal.

Answer: D

The flow is built from a uniform flow, a doublet and a vortex.

A uniform flow plus a doublet gives the potential flow past a stationary circular cylinder: the doublet's streamline of zero value is a closed circle. Adding a free vortex at the centre superposes a circulation around the cylinder without changing the circle as a streamline, which models the rotation and produces the Magnus lift (Kutta-Joukowski lift per unit length is ρUΓ).

Uniform flow and a doublet alone describe a non-rotating cylinder. A source or a sink with a uniform flow gives an open half-body, not a closed cylinder, so adding a vortex to them does not represent a rotating cylinder either.

Answer: A

The wall condition is \(\vec{V} \cdot \vec{n} = 0\).

The Euler equations describe inviscid flow. Without viscosity the fluid can slip along a solid wall, so the only physical requirement is that it does not pass through it: the velocity component normal to the wall must vanish, \(\vec{V} \cdot \vec{n} = 0\) (the no-penetration or slip condition).

\(\vec{V} = 0\) is the no-slip condition, which belongs to the viscous Navier-Stokes equations and cannot be imposed on the lower-order Euler equations. \(\nabla \cdot \vec{V} = 0\) is the incompressibility condition throughout the field, not a boundary condition. \(\vec{V} \times \vec{n} = 0\) would force the flow to be perpendicular to the wall, the opposite of what an inviscid wall allows.

Answer: B

The velocity field is \(\vec{V} = 2(x^{2} - 3y^{2})\hat{i} - 4xy\,\hat{j}\).

With \(\psi = 2x^{2}y - 2y^{3}\) and the usual convention \(u = \partial\psi/\partial y\), \(v = -\partial\psi/\partial x\):

\[ \begin{aligned} u &= 2x^{2} - 6y^{2} = 2(x^{2} - 3y^{2}) \\ v &= -4xy \end{aligned} \]

Check: \(\partial u/\partial x + \partial v/\partial y = 4x - 4x = 0\), so the flow is incompressible, as it must be. The option with \(+4xy\) has the wrong sign on v and fails this continuity check. The two options with \(2x^{2}y\) for u are just the stream function's first term, not its derivative.

Answer: C

The acceleration is 24.1 m/s².

With \(u = 2xy^{2}\) and \(v = 3xyt\), the material acceleration is the local rate of change \(\partial \vec{V}/\partial t\) plus the convective terms \(u\,\partial\vec{V}/\partial x\) and \(v\,\partial\vec{V}/\partial y\). At \((1, 1, 1)\), \(u = 2\) and \(v = 3\):

\[ \begin{aligned} a_{x} &= 0 + u(2y^{2}) + v(4xy) = 4 + 12 = 16 \\ a_{y} &= 3xy + u(3yt) + v(3xt) = 3 + 6 + 9 = 18 \\ |a| &= \sqrt{16^{2} + 18^{2}} = \sqrt{580} = 24.1 \end{aligned} \]

16.0 and 18.1 are close to the two components on their own, and 34.1 adds them as if they were in the same direction.

Answer: C

The x-velocity is \(u = 5x + 2y\).

Incompressibility requires \(\partial u/\partial x + \partial v/\partial y = 0\). With \(v = 2x - 5y\), \(\partial v/\partial y = -5\), so \(\partial u/\partial x = 5\) and \(u = 5x + f(y)\).

Irrotationality requires \(\partial v/\partial x - \partial u/\partial y = 0\). Here \(\partial v/\partial x = 2\), so \(f'(y) = 2\) and \(f(y) = 2y\) (up to a constant). Hence \(u = 5x + 2y\).

\(u = 5x - 2y\) satisfies continuity but gives vorticity \(2 - (-2) = 4\). \(u = 2x \pm 5y\) gives \(\partial u/\partial x = 2\), which violates continuity.

Answer: D

The distance is \(aa' = \frac{m}{2U}\).

The matching stream function is \(\psi = Uy + \frac{m}{2\pi}\theta\). The stagnation point lies on the negative x-axis (\(\theta = \pi\)), so the stagnation streamline is \(\psi = m/2\). Where it crosses the y-axis, \(\theta = \pi/2\):

\[ \begin{aligned} Uy + \frac{m}{2\pi}\cdot\frac{\pi}{2} &= \frac{m}{2} \\ Uy &= \frac{m}{4},\quad y = \frac{m}{4U} \end{aligned} \]

By symmetry \(a'\) is at \(-m/(4U)\), so \(aa' = m/(2U)\). \(m/U\) is the full width of the half-body far downstream, where the streamline becomes parallel to the x-axis, and \(2m/U\) and \(8m/U\) come from mixing up the factors of \(2\pi\).

Answer: A

The key gives \(\phi = \frac{K\theta}{2\pi} - U_{\infty} r \cos\theta\).

This form uses the sign convention \(\vec{V} = -\nabla\phi\), found in several textbooks. With it, a uniform flow \(U_{\infty}\) in the \(+x\) direction has \(\phi = -U_{\infty}x = -U_{\infty} r\cos\theta\), and a clockwise vortex, whose tangential velocity is \(v_{\theta} = -\frac{K}{2\pi r}\), has

\[ \begin{aligned} v_{\theta} &= -\frac{1}{r}\frac{\partial\phi}{\partial\theta} \\ \Rightarrow \phi &= \frac{K\theta}{2\pi} \end{aligned} \]

Potentials add for combined flows, giving the keyed expression. (With the other convention, \(\vec{V} = \nabla\phi\), every sign flips, \(\phi = U_{\infty}r\cos\theta - \frac{K\theta}{2\pi}\), which describes the same flow.)

The other options cannot be right in either convention. \(U_{\infty} r\sin\theta\) is the stream function of a uniform flow, not its potential, and \(K\ln r\) is the vortex's stream function, while the potential of a vortex depends on \(\theta\).