# Do non-linear passive pneumatic throttles exist?

On of the more simple throttles is a channel, which is described by the Hagen-Poiseuille equation: $$\Delta P = \frac{8\mu L Q}{\pi r^4}$$ This equation, when rearranged, tells us that the Flowrate $Q$ is linearly dependent on $\Delta P$. Do passive non-linear pneumatic devices exist, which are not linearly dependent on $\Delta P$?

For example, is there a pneumatic equivalent to the Thermistor - which has a non-constant $R(U)$-dependency, and as such a nonlinear Flow vs Voltage curve?

• Well, above the Reynolds number, you'll go nonlinear, but probably not in a useful way. More to the point: can you post the problem you're trying to solve, rather than asking how to design the solution you've posited? Jun 17, 2016 at 12:20
• @CarlWitthoft In the application of interest, I want an approximate constant flowrate for a certain pressure range (e.g. 1-5 bar) through a device. I know there are devices that can accomplish that, but there are other constraints which wont allow me to use them. I think my question is a little too broad, I will rephrase it to be more specific.
– JHK
Jun 18, 2016 at 9:17
• When you rephrase the question, also include the solutions that won't work for you
– mart
Jun 21, 2016 at 7:04
• Give this link a look, might answer your question. If not, rephrase the question to look something like in this link: (engineering.stackexchange.com/questions/10733/…) Feb 16, 2017 at 14:21
• Well, a typical pressure regulator is entirely mechanical and produces constant output pressure for allowed range of input pressures; with constant output gas consumption you obtain constant flow.
– SF.
May 18, 2017 at 14:28

In the 'pass' direction it's very much linear per Hagen-Poiseuille equation. In the reverse direction for tiny $$\Delta P$$ it's linear, but as the pressure increases, the flow increase is vanishingly low. I'm not sure what the exact relation is in the inverse direction, but it's way far from linear - and the device is entirely passive, no moving parts whatsoever.