Fluid flow: Force acting on the fluid and the Navier-Stokes equation

Question

Consider a one dimensional fluid flow in a rectangular tube. Typical streams are the poiseuille streams. Consider the case in wich we apply a force on the fluid. The Navier-Stokes equation (for incompressible fluids) is formally: $$ \rho_f \frac{d \vec{v}}{dt}=-\nabla p+\rho_f \vec{f}+\eta \nabla^2 \vec{v}$$ The flow is $1D$ so: $\frac{\partial \vec{v}}{\partial t}=\frac{d \vec{v}}{dt}$. Consider inviscid flow: $\eta=0$. $$ \rho_f \frac{\partial \vec{v}}{\partial t}=-\nabla p+\rho_f \vec{f}$$
The lengths along the tube is denoted by $s$. Let`s apply the force: $$\vec{f}=q\sin s.\hat{s}$$ Where $q$ is just a constant to match the appropriate units of force per kg and $\hat{s}$ the unit vector in the positive $s$ direction. We don't have a pressure difference so the equation of motion reduces to: $$ \rho_f \frac{\partial \vec{v}}{\partial t}=\rho_f q\sin s.\hat{s}$$ taking the dot product with $\hat{s}$: $$ \frac{\partial v}{\partial t}= q\sin s$$ Where $\vec{v} \cdot\hat{s}=v $ So: $$v(t,s) = qt \sin s$$ The velocity in the other directions is $0$. So we have an inconsistency with the continuity equation: $$\nabla \cdot \vec{v} = \frac{\partial v}{ds}=qt \cos s \neq 0$$ How is this possible? Is the assumtion of incompressibility incorrect? Maybe there is a pressure due to the force?

To go a bit further:
Consider the case when the tube is closed like a torus. there are viscous effects and there is a non-conservative force. furthermore the fluid is incompressible. What equation describes the motion of this problem? The above Navier-Stokes equation gives a contradiction.

Thanks.

Answer 1

I agree with user3823992 that it was incorrect to neglect the pressure differential. With the steady sinusoidal body force that's given, it's basically a hydrostatics problem with the pressure differential balancing the body force. Consider the Navier-Stokes momentum equation:

$$ \frac{\partial \mathbf{v}}{\partial t}+(\mathbf{v} \cdot \nabla)\mathbf{v}=-\frac{\nabla p}{\rho}+\mathbf{f} +\nu \nabla^2 \mathbf{v} $$

If we assume the velocity $\mathbf{v}$ is zero then it reduces to the hydrostatic case, where:

$$ \frac{\nabla p}{\rho}=\mathbf{f} $$

$\mathbf{f}$ only has a component in the s-direction, therefore so will $\nabla p$:

$$ \begin{eqnarray*} \frac{dp}{ds} \cdot \hat s&=&\rho q \sin(s) \cdot \hat s\\ \int_{p_0}^pdp&=&\rho q \int_0^s \sin(s) ds\\ p-p_0&=&\rho q (1- \cos (s)) \end{eqnarray*} $$

($p_0$ is just an arbitrary reference pressure that may have been present before the force was applied).

$\mathbf{v}=0$ obviously satisfies the continuity equation, although I think any other solution with a constant $\mathbf{v}$ would also satisfy the equation. This would just be bulk fluid motion that was present before the force was applied, and would tend to zero in steady-state if viscous drag on the walls is included.

In the case where the tube is closed like a torus, the flow is still governed by the Navier-Stokes equations. The momentum equation in polar coordinates (r, $\theta$, z) can be reduced to:

$$ \begin{eqnarray*} \theta: f_\theta&=&-\nu (\frac{1}{r} \frac{\partial}{\partial r}(r \frac{\partial V_\theta}{\partial r})+\frac{\partial^2 V_\theta}{\partial z^2}-\frac{V_\theta}{r^2})\\ r: \frac{\partial p}{\partial r}&=&\rho \frac{V^2_\theta}{r} \end{eqnarray*} $$

The only component of velocity is in the $\theta$ (circumferential) direction. The first line is the body force balanced by the wall friction and the $\frac{\partial p}{\partial r}$ in the second line is necessary to provide the centripetal force for the curved streamlines. However, this is now a 2-dimensional PDE and I think it's pretty unlikely that you'd be able to integrate or find a simple function to satisfy it - to find the velocity profile you would probably have to resort to CFD at this point.

The Poiseuille equation has a nice, straightforward solution because it is axisymmetric and effectively 1-dimensional.

Answer 2

Hello @abcdef this is a very interesting problem

As of today no (general) analytical solution for the NS-Equations has been found, in fact we would all hear about it, since it is worth one million dollars (see Millennium Prize).

And, even though it has not been proven, there is a lot of evidence that the NS-Equations do describe fluid motion as long as the free path length between the fluid molecules is sufficiently small enough so that in can be treated as a continuum.

In special cases, that is problem sets which permit simplifications (as inviscid or incompressible flow) solutions to the NS-Equations can be found. It has to pointed out here, that those simplifications introduce an error into the result of the calculation. However, this error is usually small enough to be neglectable (e.g. incompressibility for $M\ll0.3$) or the error can be modelled and its influence therewith reduced (e.g. drag/pressure-loss can be modelled assuming $\Delta p_{t} = f\left(p_{dyn}\right)$)

The question contains some simplifications

1. 1D-Flow
2. incompressible flow
3. inviscid flow
4. no pressure difference
5. one-atomic gas

Based on this simplifications no physical solution can be found. Which basically means one or more of the simplifications are not justifiable. It does not mean that the NS-Equations do not or cannot describe this flow.

As already pointed out in other answers [1] neglecting or preventing a pressure difference might be the major problem.

Assuming very small amplitudes ($\,q\,$) of the force the pressure difference might play a minor role. However, by neglecting possible viscosity-effects makes the NS-Equations unbalanced. Since the basic idea of the NS-Equations is, that the momentum is balanced.

It seems as if there exists no simple solution to this problem and an iterative approach needs to be tried out in order to balance pressure, body-forces, and velocities. Depending on the actual application of such flows also the assumption of one-atomic gases might need some reconsideration.

[1]: Comments by @user3823992