Fluid Kinematics Questions and Answers

Answer: C

The answer is option C: \(\phi = \tfrac{x^{3}}{3} - xy^{2}\), \(\psi = x^{2}y - \tfrac{y^{3}}{3}\).

Use the convention of most Indian fluid dynamics texts (the one the paper's own Q76 key follows): \(u = -\partial\phi/\partial x\), \(v = -\partial\phi/\partial y\), and \(u = -\partial\psi/\partial y\), \(v = \partial\psi/\partial x\). For option C:

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

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

Both pairs reproduce \(u\) and \(v\), and the flow is irrotational and incompressible, so both functions exist. With the opposite sign convention (\(u = \phi_{x}\), \(u = \psi_{y}\)) the functions are \(\phi = xy^{2} - \tfrac{x^{3}}{3}\) and \(\psi = \tfrac{y^{3}}{3} - x^{2}y\), which is option B with its two labels interchanged, so B is wrong under either convention. Options A and D have the wrong signs on the cubic terms.

Answer: B

The image is a doublet of strength \(\mu\) at \(z = -\bar{z}_{0}\) with inclination \(\pi - \alpha\).

The options place the image at \(-\bar{z}_{0}\) or \(-z_{0}\); \(-\bar{z}_{0}\) is the mirror image of \(z_{0}\) in the imaginary axis, so the straight line here is the \(y\)-axis. The image of a doublet in a line is the reflected doublet: same strength, at the mirror point, with its axis reflected.

Reflecting a direction at angle \(\alpha\) to the \(x\)-axis in the \(y\)-axis turns it into the direction at angle \(\pi - \alpha\). Formally, for \(f(z) = \mu e^{i\alpha}/(z - z_{0})\) the image system in the \(y\)-axis is \(\overline{f(-\bar{z})}\):

\[ \begin{aligned} \overline{f(-\bar{z})} &= \frac{-\mu e^{-i\alpha}}{z + \bar{z}_{0}} \\ &= \frac{\mu e^{i(\pi - \alpha)}}{z + \bar{z}_{0}} \end{aligned} \]

a doublet of strength \(\mu\) at \(-\bar{z}_{0}\) inclined at \(\pi - \alpha\).

\(-z_{0}\) is the reflection in the origin, not in a line, and keeping the inclination \(\alpha\) at the mirror point would not make the line a streamline.

Answer: D

The answer is stream lines, path lines and streak lines are coincident.

If the velocity field does not depend on time (steady flow), the streamline pattern never changes. A fluid particle then moves along the streamline through its starting point, so path lines coincide with streamlines. All particles passing through a fixed point follow the same path, so the streak line through that point is that same curve.

All three families coincide, not just two of them. They differ only in unsteady flow.

Answer: B

The answer is flow is inviscid, incompressible and irrotational.

A velocity potential \(\phi\) with \(\mathbf{q} = -\nabla\phi\) (or \(+\nabla\phi\)) exists exactly when the flow is irrotational, \(\nabla \times \mathbf{q} = 0\), on a simply connected region. In the classical theory of ideal fluids, potential flow is studied for inviscid, incompressible, irrotational motion, where \(\phi\) also satisfies Laplace's equation.

Incompressibility or inviscidity alone does not give a potential, and a rotational flow by definition has no velocity potential. So option B is the only one that includes irrotationality.

Answer: A

The stream lines are \((x - t)^{2} - (y - t)^{2} = \) constant.

With \(\mathbf{q} = -\nabla\phi\), \(u = -(y - t)\) and \(v = -(x - t)\). Stream lines at a fixed time satisfy \(dx/u = dy/v\):

\[ \begin{aligned} \frac{dx}{y - t} &= \frac{dy}{x - t} \\ (x - t)\,dx &= (y - t)\,dy \\ (x - t)^{2} - (y - t)^{2} &= c \end{aligned} \]

(The sign convention for \(\phi\) does not change the curves.) These are rectangular hyperbolas centred at \((t, t)\). The sum of squares would describe circles, which is what a vortex, not this potential, gives, and the options in \(x^{2}\) and \(y^{2}\) have lost the shift by \(t\).

Answer: D

The keyed stagnation point is \(z = a + m/U\).

The key follows the convention common in Indian textbooks, where the complex potential is \(w = \phi + i\psi\) with velocity \(\mathbf{q} = -\nabla\phi\), so \(dw/dz = -u + iv\). Then \(w = Uz\) is a uniform stream in the negative \(x\)-direction, and a source of strength \(m\) at \(a\) is \(-m\log(z - a)\). The total potential and its derivative are

\[ \begin{aligned} w &= Uz - m\log(z - a) \\ \frac{dw}{dz} &= U - \frac{m}{z - a} = 0 \\ z &= a + \frac{m}{U} \end{aligned} \]

Physically, the stagnation point lies upstream of the source, here on the side of larger \(x\), at distance \(m/U\). (Under the opposite convention, with the stream along the positive \(x\)-axis, it is at \(a - m/U\), which is not offered.) The other options have the ratio inverted or no dependence on \(a\).

Answer: C

The field satisfying conservation of mass is \(u = 2xy - x^{2}\), \(v = 2xy - y^{2}\).

For incompressible plane flow the continuity equation is \(u_{x} + v_{y} = 0\). Checking each option:

• A: \(6x + (-8xy) \neq 0\).
• B: \(2x - 2y \neq 0\) in general.
• C: \((2y - 2x) + (2x - 2y) = 0\) everywhere.
• D: \(2xy + 2xy = 4xy \neq 0\).

Only option C has zero divergence at every point, so only it describes a possible incompressible flow.

Answer: B

The motion is possible only if \(a + d = 0\).

Incompressibility requires the divergence of the velocity to vanish:

\[ \begin{aligned} \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} &= a + d \\ &= 0 \end{aligned} \]

The constants \(b\) and \(c\) do not enter the divergence; they decide instead whether the flow is irrotational (\(c - b = 0\), i.e. \(b = c\)). So \(b = c\) and \(b + c = 0\) concern vorticity, not incompressibility, and \(a = d\) would give divergence \(2a \neq 0\).

Answer: A

The velocity components are \(u = x^{2}y - 2x - \dfrac{y^{3}}{3}\), \(v = 2y - xy^{2} + \dfrac{x^{3}}{3}\).

With the convention \(\mathbf{q} = -\nabla\phi\), used in many fluid dynamics texts and by this key:

\[ \begin{aligned} u &= -\frac{\partial\phi}{\partial x} = -2x - \frac{y^{3}}{3} + x^{2}y \\ v &= -\frac{\partial\phi}{\partial y} = 2y - xy^{2} + \frac{x^{3}}{3} \end{aligned} \]

which is option A. Option D drops the cubic terms, and options B and C mix up which derivative goes with which component.

Answer: B

The image is a source of strength \(m\) at \(z = a^{2}/f\) and a sink of strength \(m\) at the origin.

By the circle theorem, a source at \(z = f\) outside the cylinder \(|z| = a\) has complex potential \(-m\log(z - f)\). Its image is obtained from \(-m\log(a^{2}/z - f)\), which equals \(-m\log(z - a^{2}/f) + m\log z\) plus a constant. That is a source of strength \(m\) at the inverse point \(a^{2}/f\) and a sink of strength \(m\) at the centre.

The sink at the origin keeps the net flux out of the cylinder zero, so no fluid is created inside it. The other options get the signs of the two image singularities wrong.