I am looking for some clarification on the incompressible, 2D, steady-state, cartesian Navier-Stokes equations for flow through a straight cylinder (flow aligned with the $x$-axis, so $u_y =0$), with negligible gravity:

$$ \rho u_{x}\frac{\partial u_{x}}{\partial x} = - \frac{\partial p}{\partial x} + \mu\left(\frac{\partial^{2}u_{x}}{\partial x^{2}} + \frac{\partial^{2}u_{x}}{\partial y^{2}}\right) \\ \frac{\partial p}{\partial y}=0. $$

Just to be clear, I'm not assuming fully-developed flow, so that $\partial u_x/\partial x\neq 0$.
Also, I am aware that the cylindrical Navier-Stokes equations would be more suitable in this case, but as far as I can tell, using the cartesian Navier-Stokes equations is still a valid formulation.

With that out of the way, upon looking at the cartesian Continuity equation under these same assumptions, the equation simplifies to:

$$ \frac{\partial u_x}{\partial x}=0 $$

Which seems to contradict the non-fully-developed flow assumption from the Navier-Stokes simplification.
I can't wrap my head around how Continuity implies that the flow is fully-developed, yet I never made such an assumption.

My first thoughts are that the Continuity simplification is actually a 1D simplification, not a 2D simplification like my original assumptions. However, since $u_y$ must be zero (because the flow is unidirectional), I can't see how the simplified Continuity equation would be any different from what I have written.

I then thought that this had something to do with the Continuity equation being in differential form so that it's only considering an infinitesimal point. I know that upon considering a control volume (i.e. converting Continuity to integral form), you can arrive at the steady-state conservation of mass equation (indicating that velocity in the $x$-direction does change):

$$ u_1 A_1 = u_2 A_2. $$

Though, this still doesn't really solve my problem, since the cross-sectional area of the cylinder is constant along its length. If anything, this equation just tells me that the flow is developed at every instantaneous point (i.e. it remains fully-developed even under cross-sectional changes).

This is probably the closest I feel to an actual explanation, but the problem still persists that Continuity says that $\partial u_x/\partial x=0$, meaning nothing is stopping me from substituting this into the Navier-Stokes equations, which will ultimately contradict my underlying assumptions.

The only other explanation I can seem to think of is that Continuity explains the quantity changes of the fluid in a system, whereas the Navier-Stokes equations explain the force changes (per unit volume) of the fluid in a system, so they are measuring different qualities of the fluid.

However, I'm not convinced that this has anything to do with the apparent contradiction, as I have seen many cases where a Continuity-derived expression has been directly substituted into the Navier-Stokes equations (e.g. in the derivation of the Reynolds-Averaged Navier-Stokes equations).

I'd really appreciate an explanation of what my misunderstanding is here.

I apologise if this has been asked somewhere else - I really struggled to find another question like this.
Feel free to point me in the right direction if you know where/if this has been asked before.


1 Answer 1


If the flow is not fully-developed, radial component of the velocity is not zero ($u_y\neq0$ and $u_z\neq0$). Intuitively, a piece of liquid near the wall of the tube slows down as $x$ increases, so another piece of liquid that moves behind it at a higher velocity will overtake the first one. To do it, it shifts closer to the center of the tube.

  • $\begingroup$ Yes, that is correct. However, I am not using the assumption of fully-developed flow to infer that $u_{y}=0$, I am using the assumption of 2D flow aligned with the $x$-axis aligned with the cylinder to infer that $u_y=0$. As far as I can tell, this assumption is completely valid (since I am just assuming all of the fluid travels in the same direction and that the flow is laminar). $\endgroup$
    – Sam_Ress
    Commented Mar 29, 2022 at 23:46
  • $\begingroup$ If a fluid travels in the same direction and the flow is stationary, then the velocity is constant. Otherwise it would contradict the conservation of mass: consider for example a small cylinder of radius $R$ with axis parallel to the axis of the tube. Liquid flows only through the bases of the cylinder: the flow through the left base is $u_x(x_1,y,z)\pi R^2$, the flow through the right base is $u_x(x_2,y,z)\pi R^2$. The flows must be equal (or the mass wouldn't be conserved), so $u_x(x_1,y,z)= u_x(x_2,y,z)$ $\endgroup$
    – atarasenko
    Commented Mar 30, 2022 at 11:21
  • $\begingroup$ I see. So what you're saying is that if fully-developed flow is not assumed under 2D conditions, then fundamentally, both $u_y\neq 0$ and $u_x \neq 0$, meaning the flow layers cannot align with a single axis. Upon thinking about this more, I can see why, and understand your initial comment. Thank you for your clarification. Perhaps another query I have is: say I wanted to assume that $u_y$ was small compared to $u_x$. Is it viable to assume that $u_y\approx 0$ under non-fully-developed conditions? $\endgroup$
    – Sam_Ress
    Commented Mar 31, 2022 at 22:28

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