Viscous fluid under constant force?

Question

For a one-dimensional fluid with viscosity $\eta$ subject to a homogenous acceleration $a$ in periodic boundary conditions, in my understanding the momentum equation is

$$\rho\left(\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x}\right) = \rho a + \eta \frac{\partial^{2} u}{\partial x^{2}}$$

Qualitatively, I expect that starting from zero velocity $u(x, 0)=0$ the fluid should arrive at a terminal velocity given by the equilibrium between the homogenous acceleration and the viscous force. But I can not see this directly by looking at the above equation. In my understanding the effect of the viscosity (or "diffusive") term is to smoothen out and eventually suppress gradients in the velocity profile. The physical reason for this is thermal fluctuations and collisions at the molecular level of the fluid. However, for the exactly same reason I expect the fluid to reach a terminal velocity upon applying a homogenous acceleration. But clearly the viscous term above only affects the velocity in the presence of spatial gradients.

So what am I missing ?

In principle, one could argue that Navier-Stokes equation is not required for this problem and one could just consider a point mass subject to constant force and a drag force given by Stokes formula. This leads to the terminal velocity. However this would not be sufficient for me as I am considering a situation where spatial profiles or gradients of the velocity might eventually enter.

The full problem I am looking at is the following : A compressible fluid within a fixed cubic container is compressed at time $t=0$ by a spatially homogeneous force against one face of the cube. After a long time the mass distribution of the fluid should develop a gradient along the direction of the force. The equilibrium mass distribution should be an exponential, in a similar way to the barometric equation. But directly after turning on the force, and before the fluid has found the equilibrium distribution, intuitively there should be some density oscillations between the two faces of the cube, against which the fluid is compressed.

The density gradient should evolve on a time-scale given by the viscosity $\eta$ and the homogenous force $F$, roughly defined as the ratio of the box size $L$ by the Stokes terminal velocity. The evolution of the density profile can be modelled by the conservation of mass (i.e Fokker-Planck equation for the density) assuming a drift velocity equal to Stokes terminal velocity. This is done for example here J. Phys. 77, 240–243 (2009). But the transient density oscillations are not considered there, and for that the momentum equation should considered. My thought was to add the homogenous force to the momentum equation, and simultaneously solve the momentum and mass conservation equation, with using the velocity field as a convection term in the mass conservation equation.

But adding the homogenous force to the momentum equation does not lead to the terminal velocity.

Answer 1

The solution of the Burger's equation with your initial condition is just: $$ u = gt $$ As you've mentioned yourself, viscosity only acts when your velocity field is inhomogeneous. Even if your initial condition were inhomogeneous, viscosity would act in the transient regime. There is no source of inhomogeneity (PBC, homogeneous force), you therefore don't have a balance of forces. Your heuristic argument is invalid.

For your compressible fluid, if I understand correctly, you are still using Burger's equation and adding the density continuity equation to close your system. I think that it's best to address it in a separate question.

Hope this helps.

Answer to comment

Your starting point is incorrect, there is no natural terminal velocity in your problem. Let's analyse the usual case of an object moving through a fluid. There is a natural constant imposed velocity inhomogeneity due to the no slip boundary conditions. At the surface of the object, the velocity of the fluid is the velocity of the object, while far from it, the fluid is at rest. It is therefore impossible to homogenise the velocity of the fluid, giving you a friction force. This force can be balanced by a force on the object to give a stationary solution.

Since your problem has a homogeneous solution, there is no friction force. The gravity is unbalanced resulting in an increasing velocity. If you want to create some kind of terminal velocity, you could add no slip boundary conditions on the side, and you'll get a Poisefeuille flow.

Your momentum equation is correct, assuming that there is no pressure (this is why it is Burger's equation not Navier-Stokes). If you want to add pressure, you'll need to add a state equation relating pressure $p$ and density $\rho$ to close your system.

Answer 2

I imagine you are assuming that there is slip on the vertical sides of your container, so that there is no viscous drag on the fluid on the sides and the problem is truly 1D spatially.

You have a container holding gas in a zero-g environment, and, at time zero you switch gravity on. You are looking for the time-dependent oscillation until the gas re-equilibrates in response to the imposed gravitational field. Is this correct? If so, I would approach this problem very differently. I would employ a material coordinate system in which the initial vertical coordinate z of a cross section prior to switching on gravity acts as a permanent label of the cross section. Then the amount of mass between material cross sections stays constant at all times. If m is the total mass of gas in the container, then $\frac{m}{h}\Delta z$ is forever the amount of mass between material cross sections z and $z+\Delta z$. The vertical displacement of each material cross section is represented by u(z,t), and the density of the gas at any time is then given by $\frac{m}{Ah}/(1+\frac{\partial u}{\partial z})$.

If you have an interest in pursuing an approach like this (including the handling of the momentum balance in the vertical), I would be glad to flesh it out further.

Answer 3

I've started to work this out for the approach I alluded to in my previous answer. You have a volume of gas in a vertical cylindrical container that is in a zero g environment, and, at time zero you switch-on gravity.

Let z be the coordinate measured vertically from the base of the cylinder before gravity is switched on. This coordinate acts for a permanent label for each cross section of the gas. If m is the total mass of gas in the container, then the initial density of all the gas in the container is $$\rho_i=\frac {m}{Ah}$$and the initial pressure in the container is $$p_i=\rho_0 \frac{RT}{M}=\rho_0 \lambda$$where M is the molecular weight.

For now, I'm only going to focus on the final steady state solution after the gas has equilibrated. In this final state, some of the material cross sections have moved closer together and some of them have moved further apart. This movement of the cross sections can be characterized by the displacement $u=u(z,t)$. At final steady state, the transient movement vanishes, and u= u(z). In terms of the displacement function u, the local density of the gas is $$\rho=\frac{\rho_0}{(1+du/dz)}$$If du/dz is positive, the cross sections have gotten further apart, and the density is less. From the ideal gas law, the pressure at a given cross section is given by $$p=\rho \lambda=\frac{\rho_0 \lambda}{(1+du/dz)}$$The pressure must also satisfy the force balance, which, in terms of our material coordinate system reads: $$\frac{dp}{dz}=-\rho_0 g$$ or $$p=p(0)-\rho_0 g z=\frac{\rho_0 \lambda}{(1+du/dz)}$$So, $$1+\frac{du}{dz}=\frac{\rho_0 \lambda}{p(0)-\rho_0 g z}$$Integrating to get the vertical displacement u yields $$u=-z+\frac {\lambda}{g}\ln{\frac{p(0)}{p(0)-\rho_0 g z}}$$The vertical displacement must be zero at z = h, which gives $$\frac {\lambda}{g}\ln{\frac{p(0)}{p(0)-\rho_0 g h}}=h$$or $$1-\frac{\rho_0 gh}{p(0)}=e^{-\frac{gh}{\lambda}}$$Solving for the pressure at the base $p(0)$ then gives: $$p(0)=\frac{\rho_0 gh}{1-e^{-\frac{gh}{\lambda}}}$$

TRANSIENT ANALYSIS

Now that we've got our feet wet on using the embedded material coordinate approach to solve for the final steady state, let us next set up the PDE for the transient transition.

We focus on the gas contained between embedded material coordinates z and $z+\Delta z$. The mass of gas contained between these material coordinates is $m\frac{\Delta z}{h}$. So the force balance on this slab of gas is $$m\frac{\Delta z}{h}\frac{\partial ^2 u}{\partial t^2}=[\sigma_{zz}A]_{z+\Delta z}-[\sigma_{zz}A]_{z}-m\frac{\Delta z}{h}g$$Where $\sigma_{zz}$ is the component of the tensile stress tensor in the vertical: $$\sigma_{zz}=-p+\frac{4}{3}\left[\frac{\eta}{(1+\partial u/\partial z)}\right] \frac{\partial ^2 u}{\partial t\partial z}$$Combining these two equations, and dividing by $A\Delta z$ gives:$$\rho_o\frac{\partial ^2 u}{\partial t^2}=-\frac{\partial p}{\partial z}-\rho_0 g$$$$+\frac{4}{3}\left[\frac{\eta}{(1+\partial u/\partial z)}\right] \frac{\partial ^3u}{\partial t \partial z^2}-\frac{4}{3}\left[\frac{\eta}{(1+\partial u/\partial z)^2}\right] \frac{\partial ^2u}{\partial t \partial z}\frac{\partial ^2 u}{\partial z^2}\tag{1}$$If, as we have done for the steady state solution, we assume that the temperature is constant during the re-equilibration, the local pressure is related to the deformation by the ideal gas law: $$p=\frac{\rho_0 \lambda}{(1+\frac{\partial u}{\partial z})}$$Substituting this into Eqn. 1 then yields: $$\rho_o\frac{\partial ^2 u}{\partial t^2}=\frac{\rho_0 \lambda}{(1+\frac{\partial u}{\partial z})^2}\frac{\partial ^2u}{\partial z^2}-\rho_0 g$$$$+\frac{4}{3}\left[\frac{\eta}{(1+\partial u/\partial z)}\right] \frac{\partial ^3u}{\partial t \partial z^2}-\frac{4}{3}\left[\frac{\eta}{(1+\partial u/\partial z)^2}\right] \frac{\partial ^2u}{\partial t \partial z}\frac{\partial ^2 u}{\partial z^2}\tag{2}$$

The assumption of constant temperature for this system is a little problematic because there are going to be changes in temperature due to adiabatic compression and expansion of the gas layers, and also due to viscous heating of the gas as a result of the deformations. To properly incorporate these effects, it would be necessary to also include the differential thermal energy balance in the analysis.

Eqn. 2 can be solved numerically, subject to the boundary conditions $u(0,t)=u(h,t)=0$

Eqn. 2 can more conveniently be expressed in terms of two dependent variables, the displacement u and the velocity of materials surfaces v, as:

$$v=\frac{\partial u}{\partial t}$$ and $$\rho_0\frac{\partial v}{\partial t}=\frac{\rho_0 \lambda}{(1+\frac{\partial u}{\partial z})^2}\frac{\partial ^2u}{\partial z^2}-\rho_0 g+\frac{4\eta}{3}\frac{\partial}{\partial z}\left[\left(\frac{1}{(1+\partial u/\partial z)}\right) \frac{\partial v}{ \partial z}\right]$$