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Question

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.

Application

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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  • $\begingroup$ 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
    Commented Jul 10, 2015 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
    Commented Jul 10, 2015 at 19:42
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    $\begingroup$ In my opinion this question is much better when leaving out your specific situation. $\endgroup$
    – Bernhard
    Commented Jul 10, 2015 at 20:03
  • $\begingroup$ I learned this in college, but have forgotten. I'm sure you could find it in the CRC Handbook of Chemistry and Physics. Also, the equation changes when the flow is transonic. $\endgroup$
    – Warren Dew
    Commented Jul 10, 2015 at 20:33
  • $\begingroup$ Electrical resistance (in a resistor) is just a fluid (free electrons) flowing through a lattice under pressure difference (voltage gradient) and bouncing off atoms, making them hot. Seems to me it's practically the same thing as flow in pipes at low speed. At higher speeds you get pressure relative to speed squared, so that's different. The latter is Bernoulli's venturi equation, so you probably want to stick to pipe flow. $\endgroup$
    – Mike Dunlavey
    Commented Jul 10, 2015 at 20:41

3 Answers 3

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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.

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  • $\begingroup$ 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
    Commented Jul 11, 2015 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
    Commented Jul 11, 2015 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
    Commented Jul 11, 2015 at 21:44
  • $\begingroup$ (I updated my post above, there were also mistakes in the equations) $\endgroup$
    – Mister Mystère
    Commented Jul 11, 2015 at 22:28
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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".

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  • $\begingroup$ 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
    Commented Jul 13, 2015 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. $\endgroup$
    – AGN
    Commented Jul 13, 2015 at 13:56
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This it not a propper answer, but something to think about. Let's talk DC first, if one applies a DC voltage, to a speaker, other than heating it up, its voice coil would move either backwards or frontwards, according to polarity, right? What if I hermetically seal two earphones to syringe like tube, at each end, with a piston that somehow has a 0 position (this 0 position may be obatained with springs, or, magnetically if it's a metallic piston, you name it) and apply DC to one of the phones? I think, if it pulls the voice coil on one side, which will pull the piston, it will pull the voice coil on the other side and vice versa, which I guess, intuitively, it would produce reversed polarity voltage on the passive earphone's terminals. Also, if an external force is applied to the piston, against the force inducted onto the piston, it would attenuate or even null the current trhough the voice coil to which the voltage was applied. If my assumption is correct, I suppose the same would happen with AC voltages. I guess such hypothesis, if right, would be an anology of ohm's law.

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