Conical nozzle

Homework Help
1

for steady flow through a conical nozzle, the axial velocity is approximately u_z=U\left(1-x/L\right)^{-2}, where U is the entrance vleoicty and L is the distance to the apparent vertex of the cone. Derive a general expression for the axial acceleration du/dt and the acceleration at the entrance if U=5m/s and L=2m

2

This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook

PK Bot’s answer:

  1. ENGINEERING CONTEXT:
    Although most fluid-mechanical analysis is done in an Eulerian framework, Lagrangian analysis is useful for some analyses. The Lagrangian or convective acceleration that a fluid element feels is illustrative especially when we use the nozzle to accelerate something other than the fluid element, as is the case in a sandblaster. Regions of very high fluid acceleration in a sandblaster will lead to a speed lag between the fluid and the particles.

  2. PHYSICS:
    There really isn’t any physics in this problem. It is an exercise in understanding what the total derivative is and understanding the difference between partial and total derivatives. The flow is steady, so \partial u/\partial t=0 everywhere, but du/dt\neq 0.

  3. ASSUMPTIONS:
    None are needed. The problem specifies all that is needed–this is a calculus problem.

  4. SOLUTION TECHNIQUE:

Because the flow is steady, the Eulerian acceleration \partial u/\partial t=0.

Given the axial velocity profile

u(x)=U\left(1-\frac{x}{L}\right)^{-2}\,,

its spatial derivative is

\frac{du}{dx}=\frac{2U}{L}\left(1-\frac{x}{L}\right)^{-3}\,.

Thus the material (or Lagrangian) acceleration is

a=\frac{du}{dt}=\frac{\partial u}{\partial t}+ \frac{\partial u}{\partial x}\frac{\partial x}{\partial t}= u\frac{\partial u}{\partial x}
a=\frac{du}{dt}=\;U\left(1-\frac{x}{L}\right)^{-2}\cdot\frac{2U}{L}\left(1-\frac{x}{L}\right)^{-3}=\frac{2U^2}{L}\left(1-\frac{x}{L}\right)^{-5}\,.

At the entrance, where x=0, the acceleration becomes

a(0)=\frac{2U^2}{L}\,.

Substituting U=5\,m/s and L=2\,m yields

a(0)=\frac{2(5^2)}{2}=\frac{2\cdot25}{2}=25\,m/s^2\,.

or 2.5 Gs of acceleration.

  1. REFLECTION:
    This exercise is a reminder that fluid particles in a steady flow are accelerating.

To formulate this response, I reviewed the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2). Below, I have organized the sections based on their relevance. If they are marked as relevant, I encourage you to explore them:

Textbook sections:

  1. Chapter 1 Engineering Models: Quasi-1D Compressible Flow (HIGHLY RELEVANT)
  2. Chapter 1.5: Isentropic quasi-1D flow relations Part 1 (HIGHLY RELEVANT)
  3. Chapter 1 Fundamentals: The area–velocity relation is derived by combining conservation of mass, co… (HIGHLY RELEVANT)
  4. Chapter 5.4: Euler Equation: Streamwise Direction i.e. Bernoulli Equation Part 2 (HIGHLY RELEVANT)
  5. Chapter 5.4: Euler Equation: Streamwise Direction i.e. Bernoulli Equation Part 5 Part 1 (HIGHLY RELEVANT)

Disclaimer:

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3

We talk, especially in an intro fluid mechanics class in UG engineering, mostly about steady flows. But by definition fluid is moving around, even when the flow is steady, and it is usually accelerating in some way, and the distinction between Eulerian acceleration (partial derivative \frac{\partial \vec{u}}{\partial t}) and Lagrangian acceleration (total derivative \frac{d\vec{u}}{dt}) is important.