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Wave Propagation in elastic Tubes

When fluid flows in a region with elastic boundaries, such as blood in blood vessels, the dynamics of the flow are influenced by the moving boundaries and vice versa (fluid-structure interaction). If the flow is unsteady (e.g. pulsatile flow), one particular consequence of this interaction is the propagation of pressure pulses. To get a first impression of this phenomena, we consider fluid in an elastic tube and a disturbance wave propagating along the tube. The following assumptions hold:

  • The flow is incompressible ($\rho$ = const).
  • The fluid is inviscid and initially at rest.
  • The amplitude of the disturbance is smaller and its wave length much bigger than the radius of the vessel.
  • Pressure is independent of the cross-stream coordinate, hence $p = p(x, t)$, where $x$ is the streamwise coordinate.
  • The tube is horizontal (gravity does not have to be considered)

In figure below:

enter image description here

a material volume is depicted at the location of the pressure pulse. Apply conservation of mass by evaluating the mass flux $\dot{m} = \rho uA$ at the borders and the change of mass inside the material volume. Hint: Estimate the state at the right border of a given quantity by a Taylor expansion, e.g. for velocity $$ u(x+dx) = u(x)+\frac{\partial u}{\partial x}dx + \mathcal{O}(dx^2) $$

\textbf{answer}: Applying the conservation of mass by evaluating the mass flux $\dot{m} = \rho uA$ at the borders and the change of mass inside the control volume gives $$ \rho u(x)A(x) = \rho\frac{dA}{dt}dx + \rho u(x + dx)A(x + dx) \quad (4) $$

At the right border the velocity and the cross-sectional area can be expressed by a Taylor expansion:

\begin{align*} \require{cancel} u(x + dx) &= u(x)+ \frac{\partial u}{\partial x}dx + \cancel{\mathcal{O}(dx^2)} \\ A(x + dx) &= A(x)+ \frac{\partial A}{\partial x}dx + \cancel{\mathcal{O}(dx^2)} \\ \Rightarrow u(x+dx)A(x+dx) &= A(x)u(x) + A(x)\frac{\partial u}{\partial x}dx + u(x)\frac{\partial A}{\partial x}dx + \cancel{\frac{\partial A}{\partial x}\frac{\partial u}{\partial x}dx^2}\\ 0 &= A(x)u(x) - u(x+dx)A(x+dx) + A(x)\frac{\partial u}{\partial x}dx + u(x)\frac{\partial A}{\partial x}dx \\ \end{align*}

Remark: the terms of order $dx^2$ ($\mathcal{O}(dx^2)$ and $\frac{\partial A}{\partial x}\frac{\partial u}{\partial x}dx^2$) are neglected.

We can then rearrange terms as follows:

\begin{align} 0&= \frac{\partial A}{\partial t} +\underbrace{ A(x)\frac{\partial u}{\partial x}+ u(x)\frac{\partial A}{\partial x}}_{\frac{\partial (Au)}{\partial x}}\label{eq_cons_mass_assign3_ex1a}\\ 0&= \frac{\partial A}{\partial t} + \frac{\partial (Au)}{\partial x}\label{mass_flux_border} \end{align}

Question: Can we say that $A(x)u(x) - u(x+dx)A(x+dx) = \frac{\partial A}{\partial t}$?

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