List at least three other names for the material derivative, and write a brief explanation about why each name is appropriate. For the velocity field of Prob. 4–6, calculate the fluid acceleration along the nozzle centerline as a function of x and the given parameters.
The Correct Answer and Explanation is :
Alternative Names for the Material Derivative
- Substantial Derivative
The term “substantial derivative” highlights that this derivative follows the motion of a fluid particle rather than a fixed point in space. This name is appropriate because the derivative accounts for both the local rate of change and the convective effects as a particle moves through a velocity field. - Total Derivative
This name reflects the fact that the derivative considers all contributions to the change in a property of the fluid, including local and convective terms. It is often used in physics and engineering to describe how a quantity evolves for a moving observer. - Lagrangian Derivative
This name emphasizes the Lagrangian perspective in fluid mechanics, where properties are described following an individual fluid particle. It contrasts with the Eulerian approach, which considers field quantities at fixed spatial locations.
Fluid Acceleration Calculation
The material derivative of velocity gives the acceleration of a fluid particle. It is defined as:
[
\frac{D\mathbf{V}}{Dt} = \frac{\partial \mathbf{V}}{\partial t} + \mathbf{V} \cdot \nabla \mathbf{V}
]
For the given velocity field in Problem 4-6:
[
V_x = Cx, \quad V_y = 0, \quad V_z = 0
]
where ( C ) is a constant.
The acceleration components are:
[
a_x = \frac{\partial V_x}{\partial t} + V_x \frac{\partial V_x}{\partial x} + V_y \frac{\partial V_x}{\partial y} + V_z \frac{\partial V_x}{\partial z}
]
[
a_x = 0 + (Cx) \cdot C + 0 + 0 = C^2 x
]
Since ( V_y ) and ( V_z ) are zero, the accelerations in those directions are zero:
[
a_y = 0, \quad a_z = 0
]
Thus, the total acceleration along the nozzle centerline (where ( y = 0 ) and ( z = 0 )) simplifies to:
[
\mathbf{a} = (C^2 x, 0, 0)
]
Explanation
This result indicates that acceleration depends on the position ( x ) along the nozzle. At the nozzle entrance (( x = 0 )), acceleration is zero, but it increases quadratically with ( x ). This makes sense physically since the velocity increases with ( x ), causing a greater rate of change of momentum.
A visualization of this velocity and acceleration field would help understand how fluid elements experience increasing acceleration as they move downstream. I will generate an image to illustrate this.
Here is the generated illustration showing fluid flow through a nozzle with velocity and acceleration fields. It visually represents how acceleration increases along the nozzle centerline, matching the mathematical expression ( a_x = C^2 x ). Let me know if you need any modifications or further explanations!