Fluid Kinematics Questions and Answers

Answer: A

A free vortex has a core with an irrotational flow, which is the keyed statement. In a free vortex the fluid rotates without any external force, and the velocity varies inversely with the radius, so the flow outside the core is irrotational. A forced vortex is created by an external force like stirring, and it needs a continuous supply of energy and has a constant angular velocity, so the other three statements belong to the forced vortex.

Answer: B

The condition is \(a_{1} + b_{2} = 0\).

For two-dimensional incompressible flow the divergence of the velocity must vanish:

\[ \begin{aligned} \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} &= a_{1} + b_{2} \\ &= 0 \end{aligned} \]

since u = 5 + a₁x + b₁y gives \(\partial u/\partial x = a_{1}\) and v = 4 + a₂x + b₂y gives \(\partial v/\partial y = b_{2}\).

The other combinations pair the wrong derivatives. \(a_{2} - b_{1}\) (not offered) would be the vorticity; \(a_{2} + b_{1} = 0\) is not a condition of continuity at all.

Answer: C

The acceleration magnitude is \(2\sqrt{5}\).

For steady flow the acceleration is purely convective. With \(u = -x^{2} + 3y\) and \(v = 2xy\), at (1, -1): u = -4 and v = -2.

\[ \begin{aligned} a_{x} &= u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} = (-4)(-2) + (-2)(3) = 2 \\ a_{y} &= u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} = (-4)(-2) + (-2)(2) = 4 \end{aligned} \]

So \(|a| = \sqrt{4 + 16} = 2\sqrt{5}\). Zero would be right only if one forgot that steady flow still has convective acceleration; 2 is the x component alone.

Answer: C

The false statement is Its streamlines are given by x = y.

For \(u = Ax\), \(v = -Ay\), the streamlines satisfy

\[ \begin{aligned} \frac{dx}{Ax} &= \frac{dy}{-Ay} \\ \ln x &= -\ln y + c \\ xy &= \text{constant} \end{aligned} \]

so they are rectangular hyperbolas, not the straight line x = y. This is stagnation-point flow.

The other statements are true. \(\partial u/\partial x + \partial v/\partial y = A - A = 0\), so continuity holds. The vorticity \(\partial v/\partial x - \partial u/\partial y = 0\), so the flow is irrotational. And as x tends to 0 the x-velocity vanishes, leaving flow along y only, which is unidirectional.

Answer: B

The flow is steady and two-dimensional.

Differentiate the particle paths to get the velocity, then express it in terms of position:

\[ \begin{aligned} u &= \frac{dx}{dt} = -kx_{o}e^{-kt} = -kx \\ v &= \frac{dy}{dt} = ky_{o}e^{kt} = ky \end{aligned} \]

The Eulerian velocity field (-kx, ky) does not contain t, so the flow is steady, even though each particle's own velocity changes as it moves. It has two non-zero velocity components that depend on two coordinates, so it is two-dimensional (stagnation-point flow).

Calling it unsteady confuses the Lagrangian time dependence of a particle with the field's, and calling it one-dimensional ignores the second velocity component.

Answer: C

The probe records a rate of 18.

A probe moving with the fluid measures the material derivative of temperature:

\[ \begin{aligned} \frac{DT}{Dt} &= \frac{\partial T}{\partial t} \\ &\quad + u\frac{\partial T}{\partial x} + v\frac{\partial T}{\partial y} \end{aligned} \]

Here \(\partial T/\partial t = 4\), \(\partial T/\partial x = 4x + y = 5\), \(\partial T/\partial y = x = 1\), and at (1, 1, 1) the velocity is \(u = 2x = 2\), \(v = y + 3t = 4\):

\[ \frac{DT}{Dt} = 4 + 2(5) + 4(1) = 18 \]

4 is only the local rate \(\partial T/\partial t\), which a fixed probe would record. 14 leaves out the \(y\)-convection term, and 0 has no basis.

Answer: C

The value of \(v(0, -1)\) is 3.

For two-dimensional incompressible flow, continuity requires \(\partial u/\partial x + \partial v/\partial y = 0\). With \(u = 2\sinh x\):

\[ \begin{aligned} \frac{\partial v}{\partial y} &= -2\cosh x \\ v &= -2y\cosh x + g(x) \end{aligned} \]

The condition \(v(x, 0) = \cosh x\) gives \(g(x) = \cosh x\), so \(v = (1 - 2y)\cosh x\). At \(x = 0\), \(y = -1\):

\[ v(0, -1) = (1 + 2)\cosh 0 = 3 \]

1 is the given value at y = 0, which ignores how v changes with y; 2 and 4 do not follow from the continuity equation.

Answer: A

The dividing streamline represents a Rankine's half-body.

A line source sends fluid outward in all directions; a uniform stream sweeps it downstream. Superposing them gives a stagnation point upstream of the source and a dividing streamline that wraps round the source and opens downstream to a constant width \(m/U\), where m is the source strength. Fluid inside never mixes with the outer flow, so the streamline can be replaced by a solid surface: a body with a rounded nose that extends to infinity, the Rankine half-body.

A circular cylinder needs a doublet with the uniform flow, a rotating cylinder adds a vortex to that, and a closed oval or elliptical shape needs a source and a sink of equal strength.

Answer: D

A pathline is shown by injecting a blob of dye and taking a long-exposure picture.

A pathline is the trajectory traced out by one fluid particle over time. Marking a small, single parcel of fluid and recording where it goes over a period, which a long exposure does, captures exactly that track.

Continuous injection with a snapshot shows a streakline, the locus of all particles that passed through the injection point. A continuous injection with long exposure smears that streakline over time. A blob photographed in a single snapshot shows only the blob's position at one instant, not its path. In steady flow these lines coincide, but in unsteady flow they differ.

Answer: D

The total acceleration is \(-1 - 2xt + 2x^{3}\).

The total (material) acceleration in one dimension is the local plus the convective part:

\[ \begin{aligned} a &= \frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} \\ &= -1 + (x^{2} - t)(2x) \\ &= -1 + 2x^{3} - 2xt \end{aligned} \]

The other options get the convective term wrong: they drop the factor 2 from \(\partial u/\partial x = 2x\), or the sign of the xt term, or both. The local term -1 is right in all four, so the convective term is what decides the answer.