0
$\begingroup$

We've been given the conservation of mass and conservation of momentum equations for the flow of fluid within a pipe system:

\begin{align}\frac{\partial \rho}{\partial t} + \frac{\partial\rho u}{\partial x} &= 0 \\ \frac{\partial J}{\partial t} + \frac{\partial(Ju + p)}{\partial x} &= \left(\frac{\partial p}{\partial x}\right)_f + \rho g, \end{align} where $\rho(x, t), u(x, t), J(x, t)$ and $p(x, t)$ are the one dimensional density, velocity, momentum and pressure respectively, and $(\partial p)/\partial x)_f$ is a frictional gradient term.

My questions really are:

(i) We're not sure if we're assuming the fluid to be compressible or not. Most literature I find seems to assert that most fluids can safely be assumed to be incompressible. Would it be reasonable to assume this here? It seems strange to see the conservation of mass equation in this form if incompressibility is being assumed.

(ii) More importantly - we need a third (constitutive?) equation to close this system. It doesn't seem that the conservation of energy fits, since temperature is in no way taken into account within this model. We were thinking possibly Boyle's law for ideal (and compressible) gases, but I've never seen this used anywhere before so not sure on this.

For someone that knows this seems like a pretty natural question to ask, but I cannot find anything in the literature that describes an equation to close this system. I really appreciate any advice that can be given.

$\endgroup$
1
$\begingroup$

The way you have written the continuity equation, you are not assuming the flow to be incompressible. If you assume $\rho$ to be constant in time and space, then you will simply get $\frac{\partial u}{\partial x}=0$. In general, assuming constant density is different from assuming incompressibility; assuming constant density implies incompressibility but assuming incompressibility does not imply constant density. If it is reasonable or not to consider the fluid to be incompressible, it really depends on the fluid that you are modelling, if you considering thermal effects and the range of pressures that you are expecting.

Regarding the third constitutive equation, the answer is yes. In the 3D version of the equations, you have

$$ \frac{\partial u}{\partial t} + (u \cdot \nabla) u = \nabla \cdot \tau - \nabla p $$

and, in the simplest case (Newtonian fluid), you assume that $\tau = \mu (\nabla u + \nabla u^T)$ (stress is proportional to strain). This is a trivial constitutive equation in the sense that you have nothing to solve, you just plug it into the momentum equation, that becomes just about velocity and pressure. In more complex fluids this constitutive equation needs to be added to the system, and solved at each time step (this is for instance the case of viscoelastic fluids).

You will also need to consider initial and boundary conditions. One possibility is to start from rest ($u=0$) and ramp it up to a steady state by gradually increasing the pressure gradient or inlet velocity.

$\endgroup$
  • $\begingroup$ Thanks for taking the time to answer @PierreCarre. By plugging that equation into the momentum equation are we not still left with 3 unknowns: density and velocity in the mass equation, and velocity and pressure in the momentum equation and just 2 equations? Or by subbing this in to the momentum equation do we actually get a third equation? $\endgroup$ – Dan Burrows Jun 5 at 21:29
  • $\begingroup$ @DanBurrows The pressure must be linked to the density... Normally you can use some prescribed law to relate them, a simple one can be of power-law type $p(\rho) = a \rho^{\gamma}$ with $a>0$ and $\gamma > 1$. This way you can eliminate pressure from the equations. $\endgroup$ – PierreCarre Jun 5 at 21:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.