I have the feeling gas cannot have an equivalent of Ohm's law, tying pressure and throughput via some kind of fluid resistance constant depending on the geometry of the obstacle considered. Certainly because gas can be compressible.

However I need a very rough estimate (not a number from experience, a first order model/formula) of the air throughput out of an obstacle of arbitrary geometry of which I know the hollow cross sectional area.

I have done my research but all I can find is Poiseuille's law or pipe flow formulas which apply to very long cylinders (what about if I'm looking at the "resistance" of a complex obstacle?)... And the venturi equation: $$p_i-p_o=1/2(\rho_o v_o²-\rho_i v_i²)$$ With (conservation of mass flow) $$\dot{m}=\rho_i A_i v_i=\rho_o A_o v_o$$ Which gives $$\dot{m}=\sqrt{\frac{2(p_i-p_o)}{\frac{1}{\rho_o A_o^2}-\frac{1}{\rho_i A_i^2}}}$$ Knowing that $$\rho=\frac{m}{V}=\frac{\frac{PVM}{RT}}{V}=\frac{PM}{RT}$$ (M is the molar mass of the gas, R the perfect gas constant)

Is it correct? It's not linear like Ohm's law, but it is a relationship.


I would like this question to be generic, but as an application/illustration, attached is a simplified 3D model of the orifice - the scale is 15mm. I know the area of the side triangles and the front rectangle out of the conduit (top), and I'm wondering what the mass flow is through it.

enter image description here

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    $\begingroup$ Plasmas are fluids & obey Ohm's law. Also, how does this have anything to do with standard-model!? $\endgroup$ – Kyle Kanos Jul 10 '15 at 18:20
  • $\begingroup$ True. I modified my question to reflect this. standard model is a mistake, I wanted to include "models". $\endgroup$ – Mister Mystère Jul 10 '15 at 18:26
  • $\begingroup$ Also: I thought Hagen-Poiseuille's law was valid for incompressible and compressible fluids? (Or at least an alternate form when considering compressibility). I mean, it's really $\Delta P\propto\Delta F$, no? $\endgroup$ – Kyle Kanos Jul 10 '15 at 19:09
  • $\begingroup$ I don't know, CFD is not my domain of expertise... What do you think about the Venturi's equation? I just updated my post with it $\endgroup$ – Mister Mystère Jul 10 '15 at 19:42
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    $\begingroup$ In my opinion this question is much better when leaving out your specific situation. $\endgroup$ – Bernhard Jul 10 '15 at 20:03

An equivalent Ohms law can be applied to gas flow and pressure drop, but only for particular mechanical flow restrictions and limited to a range of flow. But more generally for orifices and tubes the relationship between pressure and flow is quadratic, explained predominantly by the energy equation for flow, also known as Bernoulli's equation.

In the testing of respiratory equipment, companies like Hans Rudolph provide 'linear' flow resistors which approach the ideal linear resistor given by Ohms law. The restrictions in these resistors are accomplished with a screen like diffuser, and their linearity is specified over a restricted range.

So geometry does govern the relationship, but to determine what geometry is required takes CFD software or repeated experimentation.

  • $\begingroup$ That's the hunch I had, thank you for your answer (I'll +1 when I'm on a PC). Regarding my specific application, how would you estimate the mass flow rate? Bernouilli assuming the same mass flow in and out? $\endgroup$ – Mister Mystère Jul 11 '15 at 6:33
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    $\begingroup$ @MisterMystère You are on the right track regarding your initial calculations. But you also have to take thermodynamics into account. For your application can you assume isentropic flow? A good reference is Blaine Andersen's "The Analysis and Design of Pneumatic Systems". Andersen provides a full derivation of the orifice equation. But be careful - he uses $\rho$ to represent specific weight, not density. $\endgroup$ – docscience Jul 11 '15 at 17:34
  • $\begingroup$ Thanks, I'll try to procure it. In the meantime, I will use the venturi equation with, instead of one constant density, 2 different densities = P/rT from the perfect gas law - I understand "on the right track" by "sufficient for rough estimates". $\endgroup$ – Mister Mystère Jul 11 '15 at 21:44
  • $\begingroup$ (I updated my post above, there were also mistakes in the equations) $\endgroup$ – Mister Mystère Jul 11 '15 at 22:28

Application that you have considered is pipe flow. Pipe flow can be assumes as isentropic flow. Isentropic equations are non-linear. (Non-linear i'm using here is algebraic non-linearity. Please note the governing equation of fluid mechanics is also non-linear PDE). So resulting gas equation will be non-linear in nature.

We know ideal gas relation is $p=\rho R_{specific} T$ . This Ideal gas equation can be linearized or make similar to Ohms law if we consider either constant temperature process or constant density process else this is always non linear. I'm sorry to say that, I'm not aware of any easy experiments to visualize that, melting and evaporation are isothermal process. Its tough to do qualitative analysis in those process without equipment.

"Nature is non-linear. Linearity is a sub case of non-linearity".

  • $\begingroup$ You should probably point out that either you're using $R_{specific}=R/M$ or $M=1$, as the eos ought to be $p=\rho RT/M$. $\endgroup$ – Kyle Kanos Jul 13 '15 at 12:50
  • $\begingroup$ Ty. I rarely use universal gas constant. It is $R_{specific}$ and its unit is $JKg^{-1}K$. Till I didn't know there is a standard for $R_{specific}$. Ty for showing that. $\endgroup$ – Arun Govind Neelan A Jul 13 '15 at 13:56

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