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In deriving the acoustic wave equation, the momentum equation is used.

$$\frac{\partial \mathbf{u}}{\partial t}+ (\mathbf{u}\nabla)\mathbf{u}=-\frac{1}{\rho} \nabla p$$

Intuitively, the convection term $(\mathbf{u}\nabla)\mathbf{u} $ represents a component of acceleration, but how is this acceleration originate?

P.S. What is the difference between $(\mathbf{u}\nabla)\mathbf{u} $ and $(\mathbf{u}\cdot\nabla)\mathbf{u} $?

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    $\begingroup$ They presumably mean the same thing, but the former is extremely confusing while the latter is perfectly clear. $\endgroup$ – G. Smith Aug 1 at 5:47
  • $\begingroup$ Acoustic wave is linear, you should not need this term at all (unless there is some background flow velocity u$_0$). $\endgroup$ – Maxim Umansky Aug 1 at 6:25
  • $\begingroup$ @G.Smith Is $(\mathbf{u}\cdot\nabla)\mathbf{u}=(\frac{\partial u_x}{\partial x}+\frac{\partial u_y}{\partial y}+\frac{\partial u_z}{\partial z})\mathbf{u}? $\endgroup$ – ecook Aug 1 at 12:20
  • $\begingroup$ You left out the final $ so the MathJax didn’t display as math. $\endgroup$ – G. Smith Aug 1 at 17:03
  • $\begingroup$ $$(\mathbf{u}\cdot\nabla)\mathbf{u}=\left(u_x\frac{\partial}{\partial x}+u_y\frac{\partial}{\partial y}+u_z\frac{\partial}{\partial z}\right)\mathbf{u}$$ This is very different from what you had. The $\mathbf{u}\cdot\nabla$ is a scalar differential operator, and is not the divergence $\nabla\cdot\mathbf{u}$. $\endgroup$ – G. Smith Aug 1 at 17:11
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The momentum equation written above is written for an incompressible fluid, otherwise the density $\rho$ would have to be written inside the partial derivative: $$ \frac{\partial\rho\mathbf{u}}{\partial t} $$

Hence, the momentum in a Control Volume can change by either:

  • a change in velocity of the fluid
  • convection (transport) of momentum through the CV boundaries. This is the meaning of the term $(\mathbf{u}\nabla)\mathbf{u}$.
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  • $\begingroup$ Thank you for the answer $\endgroup$ – ecook Aug 5 at 7:17

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