Acceleration of a fluid particle

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