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.
Best Answer
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.