Question 4.2: Acceleration of a Fluid Particle through a Nozzle Nadeen is ...

The acceleration following a fluid particle down the center of a nozzle is to be estimated.
Assumptions   1  The flow is steady and incompressible. 2  The x-direction is taken along the centerline of the nozzle. 3  By symmetry, 𝜐 = w = 0 along the centerline, but u increases through the nozzle.
Analysis   The flow is steady, so you may be tempted to say that the acceleration is zero. However, even though the local acceleration ∂\vec{V}/∂t is identically zero for this steady flow field, the advective acceleration (\vec{V}.\vec{\triangledown })\vec{V}  is not zero. We first calculate the average x-component of velocity at the inlet and outlet of the nozzle by dividing volume flow rate by cross-sectional area:

Inlet speed:

u_{inlet} \cong \frac{\dot{V} }{A_{inlet}} = \frac{4\dot{V} }{\pi D_{inlet}^2} = \frac{(5.30 \times 10^{-5}  m^3/s)}{\pi (0.0170  m)^2} = 0.589  m/s

Similarly, the average outlet speed is u_{\text{outlet}} = 3.19 m/s. We now calculate the acceleration in two ways, with equivalent results. First, a simple average value of acceleration in the x-direction is calculated based on the change in speed divided by an estimate of the residence time of a fluid particle in the nozzle, \Delta t = \Delta x/u_{avg} (Fig. 4–10). By the fundamental definition of acceleration as the rate of change of velocity,

Method A:      a_x = \cong \frac{\Delta u}{\Delta t} = \frac{u_{outlet} – u_{intel}}{\Delta x/u_{avg}} = \frac{u_{outlet} – u_{inlet}}{2 \Delta x/(u_{outlet} + u_{inlet})} = \frac{u_{outlet}^2 – u_{inlet}^2} {2 \Delta x}

The second method uses the equation for acceleration field components in Cartesian coordinates, Eq. 4–11,

a_x = \frac{∂u}{∂t} + u\frac{∂u}{∂x} + \upsilon \frac{∂u}{∂y} + w\frac{∂u}{∂z}
Cartesian coordinates: a_y = \frac{∂\upsilon}{∂t} + u\frac{∂\upsilon}{∂x} + \upsilon \frac{∂u}{∂y} + w\frac{∂\upsilon}{∂z}      (4–11)
a_z = \frac{∂w}{∂t} + u\frac{∂w}{∂x} + \upsilon \frac{∂w}{∂y} + w\frac{∂w}{∂z}

Method B:            a_x=\underbrace{\cancel{\frac{\partial u}{\partial t}}}_{\text {Steady }}  +  u \frac{\partial u}{\partial x}  +  \underset{v=0 \text { along centerline }}{v \underbrace{\cancel{\frac{\partial u}{\partial y}}}}+\underset{w = 0 \text { along centerline }}{w \cancel{\frac{\partial u}{0 z}}} \cong u_{\text {avg }} \frac{\Delta u}{\Delta x}

Here we see that only one advective term is nonzero. We approximate the average speed through the nozzle as the average of the inlet and outlet speeds, and we use a first-order finite difference approximation (Fig. 4–11) for the average value of derivative ∂u/∂x through the centerline of the nozzle:

a_x \cong \frac{u_{outlet}  +  u_{inlet}}{2}\frac{u_{outlet}  –  u_{inlet}}{\Delta x} = \frac{u_{outlet}^2  –  u_{inlet}^2}{2  \Delta x}

The result of method B is identical to that of method A. Substitution of the given values yields

a_x \cong \frac{u_{outlet}^2  –  u_{inlet}^2}{2 \Delta x} = \frac{(3.19  m/s)^2  –  (0.589  m/s)^2}{2(0.0991  m)} = 49.6  m/s^2

Discussion   Fluid particles are accelerated through the nozzle at nearly five times the acceleration of gravity (almost five g’s)! This simple example clearly illustrates that the acceleration of a fluid particle can be nonzero, even in steady flow. Note that the acceleration is actually a point function, whereas we have estimated a simple average acceleration through the entire nozzle.