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Consider the case of a one-dimensional incompressible, non-viscous fluid flowing down a vertical pipe under the influence of gravity. Since we assume the flow is constant along the cross section of the pipe from the one dimensional assumption, let us denote the velocity of the fluid down the pipe to be $v$, with all other velocity components zero. Similarly, let us orient our axis such that the positive x axis points down the pipe. The Navier-Stokes equations in this case consist of three separate conservation equations — conservation of mass, momentum, and energy. Under the assumption of incompressibility, conservation of mass simply gives: \begin{equation} \tag{1} \frac{dv}{dx} = 0 . \end{equation} The momentum equation gives us the following equation: \begin{equation}\tag{2} \rho v \frac{dv}{dx} = -\frac{dp}{dx} + \rho g, \end{equation} where $\rho$ is the fluid density, $p$ the fluid pressure, and $g$ the gravitational acceleration. Finally, the energy equation relates the internal energy per unit mass, kinetic energy per unit mass, and the work done against pressure, and by the gravitational body force. We have the following relationship: \begin{equation}\tag{3} \rho v \frac{de}{dx} = -p \frac{dv}{dx}, \end{equation} where $e$ is the internal energy per unit mass of the fluid. This equation was obtained from the application of equation $(2.55)$, from this article to the present case assuming no viscosity, no thermal heat transfer, and no heat source.

Using the equation of state for an ideal gas, we have the following relationship between the internal energy of a gas and its temperature: $$ U = \frac{3}{2} NkT,$$ where we assumed the fluid to be a monoatomic gas, with number of molecules $N$, and temperature $T$. However, we also have the following relationship for an ideal gas: $$ p = \left(\frac{N}{V}\right)kT = nkT,$$ where $n$ is the number density of the fluid.

Now, the internal energy density per mass of the fluid $e$ can be written as: $$ e = \frac{U}{\rho V} = \frac{3}{2 \rho}nkT = \frac{3}{2} \frac{p}{\rho}.$$ Finally, under the assumption of incompressibility and equation $(1)$, equation $(2)$ simply becomes: \begin{equation}\tag{4} \frac{dp}{dx} = \rho g. \end{equation} Similarly, plugging in $e$ into equation $(3)$ and setting the right hand side to zero yields the following: \begin{equation}\tag{5} \frac{dp}{dx} = 0. \end{equation} It seems like these two equations for the pressure of the fluid are inconsistent. Yet, I can't find any assumption that is incorrect, which leads me to believe that the assumption of incompressibility is incorrect and it is not possible to have an incompressible fluid under a constant cross section with a body force acting on it. The fluid necessarily needs to be compressed.

However, I am unsure and I might have made a mistake in my assumption or have a flaw in my understanding. Would the system have been consistent if the fluid were compressible and why?

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    $\begingroup$ Does the contradiction still arise if you assume an incompressible liquid instead of a gas? Because the isothermal compressibility $-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T$ of the constitutive model you’re using is $\frac{1}{P}$, which of course is not zero. This may or may not be relevant. $\endgroup$ Jul 26, 2023 at 0:34

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There is nothing inconsistent, a priori, about an absolutely incompressible fluid. The problem with your analysis of the Euler equations (they are not really the Navier-Stokes equations if the viscosity is zero, since the viscous term is a singular perturbation) is that you have assumed two different equations of state—one incompressible, one compressible.

An incompressible fluid does not obey the ideal gas law, or anything close to it. The ideal gas law, written in the form $p=nkT$, has the pressure increase smoothly as the density $n=\rho/m$ does. This is not incompressibility. Incompressibility demands that the density $\rho$ not change, always taking the value $\rho_{0}$. So $p=0$ at $\rho=\rho_{0}$, but it increases infinitely quickly for $\rho<\rho_{0}$; whatever external pressure is applied to the fluid, it can resist it without the volume decreasing. There is not a nice functional form for this equation of state, because of this singular behavior. Instead, you should treat $\rho=\rho_{0}$ as a constant, and $p$ should be allowed to take whatever value is needed in order to satisfy the Euler equations.

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