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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 $$ u(x+dx) = u(x)+\frac{\partial u}{\partial x}dx + \mathcal{O}(dx^2) $$

answer: Applying the conservation of mass by evaluating the mass flux

$$\dot{m} = \rho uA \tag{1}\label{1}$$

at the borders and the change of mass inside the control volume gives $$ \rho u(x)A(x) = \rho\frac{dA}{dt}dx + \rho u(x + dx)A(x + dx) \tag{2}\label{2} $$

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 \tag{3}\label{3}\\ \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}}\tag{4}\label{4}\\ 0&= \frac{\partial A}{\partial t} + \frac{\partial (Au)}{\partial x}\label{mass_flux_border} \end{align}

Question:

  • How can we go from (1) to (2) step by step? Where does the $t$ variable in (2) comes from
  • Going from (3) to (4) implies that $A(x)u(x) - u(x+dx)A(x+dx) = \frac{\partial A}{\partial t}$. How can we mathematically / physically make that assumption?
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Your problem deals with both position $x$ and time $t$. Going from equation $1$ to $2$ can be summed up as below from the conservation of mass.

mass flow rate in at $x$ - mass flow rate out at $x + dx$ = rate of change in mass of element $dx$ w.r.t. time

The left-hand side arises immediately from the mass flow definition. The right-hand side comes from the change in mass based on differences in element volume caused by the cross-sectional area $A$ over time. The mass $m$ of the differential element $dx$ is density times the differential volume or $\rho A dx$. The starting equation can then be taken as below.

$$\rho u(x,t) A(x,t) - \rho u(x + dx,t + dt) A(x + dx,t + dt) = \frac{m(t + dt) - m(t)}{t + dt - t}\\= \frac{\rho A(x^*,t + dt)dx - \rho A(x^*,t)dx}{dt}$$

I've explicitly indicated the $x$ and $t$ dependences of $A$ and $u$. The $t + dt$ doesn't change anything about what you've already done for the expansions in $x$ as $dt$ just approaches zero in the end. The focus of the first equation is the position.

For the last equation, I've indicated $x^*$ as a position between $x$ and $x + dx$ which provides the differential volume of an element $dx$. This is a consequence of the mean value theorem for integrals. If you apply the Taylor series expansion for $t$ for the last equation, you should arrive at equation $2$. As $A$ is a function of $x$ and $t$, that total derivative should be a partial derivative in your equation $2$.

Going from equation $3$ to $4$ is answered by your equation $2$ by dividing out the constant density. So that implication is purely from mass conservation.

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  • $\begingroup$ Many thanks for your answer @JeqHar. I think that what is not clear to me yet is why we state the starting point of the problem. I rephrased the question accordingly $\endgroup$ – ecjb Dec 24 '19 at 9:58
  • $\begingroup$ I've edited my response to account for this. $\endgroup$ – Harjeq Dec 24 '19 at 10:34
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First question (pass from Eq. 1 to Eq. 2)

Let me first state that Eq. 1 is perhaps confusingly informal. What this equation expresses is that the flux of fluid through all the boundaries must be equal to the increase of mass in the interior volume. This is the mass conservation principle. Since the only boundaries with a flux are the the cross sections at coordinates $x$ and $x + dx$, the total mass increase rate between them is $$\dot{m} = A(x)u(x)-A(x+dx)u(x+dx) \tag{*}\label{*},$$ since the velocity is entering the volume through $A(x)$ and exiting through $A(x+dx)$ if $u > 0$.

On the other hand, at any given moment, the total mass bounded between the two cross sections can be calculated as $$m = \int_{A(x)\cup A(x+dx)}{\rho} \, \mathrm{d}\Omega = \int_{x}^{x+dx} \rho A(s, t)\, \mathrm{d}s$$ So $$\dot{m} = \int_{x}^{x+dx} \rho \frac{\partial A(s, t)}{\partial t}\, \mathrm{d}s \tag{**}\label{**}$$ Then, applying the Taylor expansion approximation and neglecting order-2 terms one gets (by the Fundamental Theorem of Calculus) $$\dot{m} = \dot{m}|_{x} + \frac{\partial \dot{m}}{\partial x}|_{x}dx + O(dx^2) = 0 + \rho \left ( \frac{\partial A}{\partial t}|_{x} \right ) + O(dx^2) = \rho\frac{\partial A(x)}{\partial t} \\ $$ where the last equality follows from neglecting order-2 terms. Note that in order to obtain the second equality, only the top integral limit in \eqref{**} must be assumed to change (the bottom one is fixed at s=x). Combining this equation with \eqref{*} you will obtain your Eq. 2.

Second question (mass conservation assumption) We have covered this in the previous answer: It is just an expression of the mass conservation principle (the rate of change of the mass equals the flux entering the domain) for this case, where

  1. The control domain is the volume contained between the cross sections $A(x)$ and $A(x+dx)$.
  2. The velocity of the fluid is assumed to be one-dimensional and aligned with the $x$-coordinate
  3. The tube walls cannot be traversed
  4. Density is constant
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  • $\begingroup$ Oh, I see that my answer is similar to that of @JeqHar. Well, since I was editing it before he posted it, I will leave it, since the explanation is not exactly equal. $\endgroup$ – Guillermo BCN Dec 24 '19 at 12:06

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