Consider the Navier-Stokes equation in an infinitely long, translation invariant channel of rectangular cross-section for an incompressible fluid in steady-state. Suppose the Reynold's number is low (laminar flow regime). Further suppose that the vollumetric flow rate entering the channel is held constant (flow-rate-driven flow).

Question: What is the velocity field of the flow in such a channel?

Note that the aforementioned problem is similar to Poiseuille-flow problem, where the flow is driven by pressure, but in the above problem, instead of fixing the pressure difference between the inlet and outlet of the channel, we fix the volumetric flow rate.

My first approach was to solve the Poiseuille-flow problem in such rectangular channels, which results in

$$ \begin{aligned} v_{x}(y, z)=\frac{4 h^{2} \Delta p}{\pi^{3} \eta L} \sum_{n, \text { odd }}^{\infty} \frac{1}{n^{3}}\left[1-\frac{\cosh \left(n \pi \frac{y}{h}\right)}{\cosh \left(n \pi \frac{w}{2 h}\right)}\right] \sin \left(n \pi \frac{z}{h}\right) \\ \\ \qquad z\in [0,h], \quad y\in[-w/2, w/2], \end{aligned} $$

then obtain the pressure-difference in terms of the flow rate, then use that expression to eliminate the pressure-difference in the equations. In the end, I get

$$ \begin{aligned} v_{x}(y, z) \sim \frac{48Q}{h \pi^{3}} \frac{1}{\left[w\left[1-0.630 \frac{h}{w}\right] \right]} \sum_{n, \text { odd }}^{\infty} \frac{1}{n^{3}}\left[1-\frac{\cosh \left(n \pi \frac{y}{h}\right)}{\cosh \left(n \pi \frac{(w)}{2 h}\right)}\right] \sin \left(n \pi \frac{z}{h}\right) \end{aligned} $$ where $w,h$ are the width and the height of the channel, and $Q$ is the flow rate.

However, since the original equation was derived by keeping the pressure constant, I can't provide much argument for why the flow-rate driven flow profile should have the above form.


1 Answer 1


For an infinitely long channel, the pressure drop between inlet and outlet would be infinite. But the pressure drop over any $1$ meter section would be the same as any other $1$ m section. Given a $1$ m section with constant flow, you would expect constant pressure.

  • $\begingroup$ so it would mean the flow-rate driven flow would be the same as pressure-driven flow? $\endgroup$
    – Our
    Jun 8, 2022 at 18:13
  • $\begingroup$ I was assuming that the flow must be driven by a pressure gradient. The pressure would be adjusted to keep the flow constant as opposed to keeping the pressure constant. Though I expect it would come out to the same thing. $\endgroup$
    – mmesser314
    Jun 8, 2022 at 20:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.