Kinematics of Fluid Flow

(a) Classification of flow

(b) Streamline, pathline, streakline

• Streamline: an imaginary curve drawn so that the tangent at every point gives the direction of the velocity vector at that instant. No flow crosses a streamline.
• Pathline: the actual path traced by a single fluid particle over time (Lagrangian concept).
• Streakline: the locus of all particles that have passed through a fixed point earlier, e.g. a line of dye injected steadily at a point.

Equation of streamline

Let $d\vec s = dx\,\hat i + dy\,\hat j + dz\,\hat k$ be an element along the streamline. It is parallel to $\vec V = u\hat i + v\hat j + w\hat k$, so $d\vec s \times \vec V = 0$. Hence

$$\frac{dx}{u} = \frac{dy}{v} = \frac{dz}{w}$$

For 2D flow, $dy/dx = v/u$, which is integrated to get the streamline family.

When they coincide

In steady flow the velocity field does not change with time, so streamlines, pathlines and streaklines are identical. In unsteady flow they differ.

Definitions

Circulation $\Gamma$ is the line integral of the tangential velocity component around a closed curve:

$$\Gamma = \oint \vec V \cdot d\vec s$$

Vorticity $\vec\zeta$ is twice the angular velocity of a fluid particle, $\vec\zeta = 2\vec\omega = \nabla \times \vec V$. For 2D flow in the $xy$-plane:

$$\zeta_z = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}$$

Circulation round a rectangle

Take an element $dx \times dy$ with velocity components $u$, $v$ at the lower left corner, and go anticlockwise.

• Bottom side (to the right): $u\,dx$
• Right side (up): $\left(v + \dfrac{\partial v}{\partial x}dx\right)dy$
• Top side (to the left): $-\left(u + \dfrac{\partial u}{\partial y}dy\right)dx$
• Left side (down): $-v\,dy$

Adding:

$$d\Gamma = \left(\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}\right) dx\,dy = \zeta_z\, dA$$

So circulation per unit area equals vorticity. This is the basis of Stokes' theorem, $\Gamma = \iint \zeta\, dA$.

Rotational and irrotational flow

For irrotational flow a velocity potential $\phi$ exists, with $u = \partial\phi/\partial x$ and $v = \partial\phi/\partial y$.

Velocity components

Using $u = \dfrac{\partial\psi}{\partial y}$ and $v = -\dfrac{\partial\psi}{\partial x}$:

$$\begin{aligned} u &= 3x^2 - 3y^2 \\ v &= -6xy \end{aligned}$$

(a) Continuity

$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 6x - 6x = 0$$

Continuity is satisfied (as it always is when a stream function exists).

(b) Irrotationality

$$\zeta_z = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = -6y - (-6y) = 0$$

The vorticity is zero, so the flow is irrotational.

(c) Velocity potential

Since $u = \partial\phi/\partial x$:

$$\phi = \int (3x^2 - 3y^2)\,dx = x^3 - 3xy^2 + f(y)$$

Then $\partial\phi/\partial y = -6xy + f'(y)$. This must equal $v = -6xy$, so $f'(y) = 0$. Taking the constant as zero,

$$\phi = x^3 - 3xy^2$$

(d) At the point (1, 2)

$$\begin{aligned} u &= 3(1) - 3(4) = -9\ \text{m/s} \\ v &= -6(1)(2) = -12\ \text{m/s} \\ V &= \sqrt{9^2 + 12^2} = 15\ \text{m/s} \end{aligned}$$

Discharge between (1, 1) and (2, 1)

$$\begin{aligned} \psi_1 &= 3(1)(1) - 1 = 2 \\ \psi_2 &= 3(4)(1) - 1 = 11 \\ q &= \psi_2 - \psi_1 = 9\ \text{m}^2/\text{s per unit depth} \end{aligned}$$

Answer: $u = 3x^2 - 3y^2$, $v = -6xy$; irrotational; $\phi = x^3 - 3xy^2$; at (1, 2) $V = 15\ \text{m/s}$; $q = 9\ \text{m}^2/\text{s}$

Local and convective acceleration

Using the Eulerian description, the acceleration of a particle is the material derivative:

$$\vec a = \frac{D\vec V}{Dt} = \frac{\partial \vec V}{\partial t} + (\vec V\cdot\nabla)\vec V$$

• Local (temporal) acceleration $\partial \vec V/\partial t$: change of velocity with time at a fixed point. It is zero in steady flow.
• Convective acceleration $(\vec V\cdot\nabla)\vec V$: change of velocity due to the particle moving to a different position. It is zero in uniform flow.

In 2D:

$$\begin{aligned} a_x &= \frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} \\ a_y &= \frac{\partial v}{\partial t} + u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} \end{aligned}$$

Velocity at the point

At $(1, 2)$ and $t = 1$:

$$\begin{aligned} u &= 2(1) + 1 = 3\ \text{m/s} \\ v &= -2(2) + 2(1) = -2\ \text{m/s} \end{aligned}$$

Derivatives

$\dfrac{\partial u}{\partial t} = 1$, $\dfrac{\partial u}{\partial x} = 2$, $\dfrac{\partial u}{\partial y} = 0$, $\dfrac{\partial v}{\partial t} = 2$, $\dfrac{\partial v}{\partial x} = 0$, $\dfrac{\partial v}{\partial y} = -2$.

Acceleration components

$$\begin{aligned} a_x &= 1 + (3)(2) + (-2)(0) = 7\ \text{m/s}^2 \\ a_y &= 2 + (3)(0) + (-2)(-2) = 6\ \text{m/s}^2 \\ |a| &= \sqrt{7^2 + 6^2} = 9.22\ \text{m/s}^2 \end{aligned}$$

Steadiness

Since $u$ and $v$ depend on $t$ ($\partial u/\partial t \neq 0$), the flow is unsteady.

Answer: $a_x = 7\ \text{m/s}^2$, $a_y = 6\ \text{m/s}^2$, $|a| = 9.22\ \text{m/s}^2$; flow is unsteady