Fluid Kinematics Questions and Answers

Answer: C

Only statements (i) and (iii) are true, so the answer is (i) and (iii) only.

(i) is true: for constant density the continuity equation \(\partial\rho/\partial t + \nabla\cdot(\rho\vec{V}) = 0\) reduces to \(\nabla\cdot\vec{V} = 0\), whether or not the flow is steady. (iii) is true: non-Newtonian fluids such as shear-thinning and shear-thickening fluids have stress varying nonlinearly with strain rate.

(ii) is false: circulation is the line integral of velocity round a closed curve, \(\Gamma = \oint \vec{V}\cdot d\vec{l}\), equal by Stokes' theorem to the surface integral of vorticity. (iv) reverses the derivation: integrating Euler's equation along a streamline in steady flow gives Bernoulli's equation, not the other way round. Every option containing (ii) or (iv) is therefore wrong.

Answer: C

The value is \(p = \frac{2}{5}\).

A streamline is everywhere tangent to the velocity, so \(\frac{dx}{u} = \frac{dy}{v}\):

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

Through (2, 1) the constant is 2, so at \(x = 5\), \(5p = 2\) and \(p = 2/5\).

5/2 is what results from writing \(y/x\) = constant, which ignores the minus sign in \(u = -x\). 2 and 5 would require the streamline to be horizontal or to pass through the points on a line through the origin.

Answer: C

The slope is \(-\frac{u}{v}\).

Along a potential (equipotential) line \(\phi\) is constant, so

\[ \begin{aligned} d\phi &= \frac{\partial\phi}{\partial x}dx + \frac{\partial\phi}{\partial y}dy \\ &= u\,dx + v\,dy = 0 \end{aligned} \]

giving \(\frac{dy}{dx} = -\frac{u}{v}\).

\(\frac{v}{u}\) is the slope of a streamline, which is tangent to the velocity. The two slopes multiply to \(-1\), the familiar result that streamlines and equipotentials cross at right angles. \(\frac{u}{v}\) and \(-\frac{v}{u}\) have the right pieces in the wrong arrangement.

Answer: B

The locus is a streakline.

A streakline joins, at one instant, all the particles that have passed through a given fixed point, such as the line of dye or smoke released continuously from a nozzle.

A streamline is everywhere tangent to the velocity field at an instant. A pathline is the trajectory of a single particle over time. A timeline is a line of particles that were marked at the same instant. In steady flow streamlines, pathlines and streaklines coincide, but in unsteady flow only the streakline matches this definition.

Answer: C

The vorticity in the y-z plane is \(\frac{\partial w}{\partial y} - \frac{\partial v}{\partial z}\).

Rotation in the y-z plane is rotation about the x axis, so it is the x-component of \(\nabla \times \vec{V}\):

\(\omega_x = \dfrac{\partial w}{\partial y} - \dfrac{\partial v}{\partial z}\)

\(\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}\) is \(\omega_z\), rotation in the x-y plane, and \(\frac{\partial u}{\partial z} - \frac{\partial w}{\partial x}\) is \(\omega_y\), rotation in the x-z plane. \(\frac{\partial v}{\partial y} - \frac{\partial w}{\partial z}\) is a difference of normal strain rates, not a rotation.

Answer: A

The velocity is a two-dimensional function of \(r\) and \(z\).

In the converging section the flow area shrinks along the axis, so the axial velocity grows with \(z\) to keep the flow rate constant, and the walls turn the stream inward, giving a radial component that also varies with \(r\). Because the section is axisymmetric, nothing changes with the angle \(\theta\).

A function of \(r\) only describes fully developed flow in the straight part of the pipe, not the converging end. A function of \(z\) only would ignore the no-slip variation across the section. A dependence on \(\theta\) contradicts the axial symmetry of the geometry.

Answer: B

Both [1] and [2] are NOT correct.

The two statements swap the definitions. The Lagrangian description follows individual fluid particles and records their position and properties as functions of time. The Eulerian description fixes attention on points in space and records what happens there as different particles pass through.

Statement [1] gives the Lagrangian definition under the Eulerian name, and Statement [2] gives the Eulerian definition under the Lagrangian name, so neither is correct, and the options claiming one or both are correct fail.

Answer: A

The constant is \(P = 5\).

For incompressible flow, conservation of mass requires zero divergence of the velocity field:

\[ \begin{aligned} \nabla \cdot \vec{V} &= \frac{\partial(5x)}{\partial x} + \frac{\partial(-Py)}{\partial y} + \frac{\partial(3)}{\partial z} \\ &= 5 - P + 0 = 0 \end{aligned} \]

so \(P = 5\). \(P = -5\) would make the divergence 10, and 8 or 2 come from wrongly including the constant \(z\) component 3 (5 + 3 or 5 − 3), whose derivative is zero.

Answer: B

The stream function is defined by \(V_{r} = \frac{1}{r}\frac{\partial \psi}{\partial z}\) and \(V_{z} = -\frac{1}{r}\frac{\partial \psi}{\partial r}\).

A stream function must satisfy the continuity equation identically. Substitute these definitions:

\[ \begin{aligned} \frac{1}{r}\frac{\partial}{\partial r}\left(\frac{\partial \psi}{\partial z}\right) + \frac{\partial}{\partial z}\left(-\frac{1}{r}\frac{\partial \psi}{\partial r}\right) \\ = \frac{1}{r}\psi_{rz} - \frac{1}{r}\psi_{zr} = 0 \end{aligned} \]

The mixed derivatives cancel, as required. Without the \(1/r\) in \(V_{r}\) (option A) the first term becomes \(\frac{1}{r}(\psi_{z} + r\psi_{rz})\), which does not cancel. With a plus sign in \(V_{z}\) the terms add instead of cancelling, and without the \(1/r\) in \(V_{z}\) the two terms differ by a factor \(r\).

Answer: A

The product remains always constant.

A doublet is the limit of a source and a sink of strength \(m\) a distance \(a\) apart as \(a \to 0\). If \(m\) stayed fixed, the two would simply cancel. To keep a finite flow pattern, the strength is increased as the distance shrinks so that the product \(m a\), the doublet strength, is held constant throughout the approach.

A decreasing product would make the pattern vanish, an increasing one would make it blow up, and 'first increases and then decreases' does not describe any limiting process used to define the doublet.