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I'm trying to figure out the unit of impedance in the hydraulic analogy of electronic networks. Assuming


with $P$ as pressure difference, $Q$ as volumetric flow rate and $Z$ as impedance, the unit would be

$$\frac{Pa}{m^3/s} = \frac{N/m^2}{m^3/s} = \frac{kg/ms^2}{m^3/s} = \frac{kg}{m^4s}$$

Intuitively, this unit doesn't make sense. Did I get some part of the unit conversion wrong? If not, could someone help me understand the unit?

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@ColinK The answer "there is no deeper meaning" is intuitively unappealing to me :) – Andreas Mar 5 '12 at 17:39
@andreas Well, units are not meant for intuition. They're just a notation to simplify/standardise stuff. – Manishearth Mar 5 '12 at 18:03
up vote 1 down vote accepted

If you use a basic forms loss (like the free jet condition) you have:

$$\Delta P = \frac{k}{2} \rho v^2$$

The difference between $\Delta P$ and $P$ doesn't bother me, because I think you might just as well switch that in your question. But I need to get things in terms of $Q$.

$$Q = \frac{m^3}{s} = v \times A$$

$$\Delta P = \frac{k \rho}{2 A^2} Q^2$$

Now this is close to what you want but unsatisfactory because of the exponent. Well, let's switch from this to consideration of specifically laminar flow. In particular, let's use the Darcy friction factor.

$$k = f \frac{L}{D}$$

$$f = \frac{64}{Re}$$

$$Re = \frac{G D}{\mu} = \frac{\rho Q D}{A \mu}$$

$$k = \frac{L}{D} 64 \frac{A \mu}{\rho Q D} = \frac{64 A \mu L}{\rho Q D^2} $$

Back to the head loss equation.

$$\Delta P = \frac{64 A \mu L}{\rho Q D^2} \frac{\rho}{2 A^2} Q^2 = \frac{32 \mu L}{A D^2} Q $$

Ok there, for a specific case of laminar flow through a pipe you can have your resistor analogy with some physical basis. Now, your question was about units. Here are the units of the above equation:

$$Pa = \frac{kg}{m s^2} = \left( \frac{kg}{m^4 s} \right) \left( \frac{m^3}{s} \right)$$

So yes, right off the bat there doesn't seem to be anything wrong with this. But I'll take the 2nd approach.

$$Z = \frac{32 \mu L}{A D^2} \rightarrow \left( \frac{kg}{m s} \right) (m) \left( \frac{1}{m^2} \right) \left( \frac{1}{m^2} \right) = \frac{kg}{m^4 s} $$

Yep, we're good.

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Thanks for the multiple verifications! Can I ask, where's your last formula from? $Z=\frac{32\mu L}{AD^2}$ – Andreas Mar 5 '12 at 17:44
Two equations above has the form $\Delta P = Z Q$. That defines $Z$, the expression is copy and pasted. – Alan Rominger Mar 5 '12 at 17:50

Its fine. Never try to extract any sense from higher units. Does the unit of energy $\mathrm{kgm^2/s^2}$ make sense? Nope, not until you relate it with a formula and break it into bits. Writing your unit as $\mathrm{\frac{Pa}{m^3/s}}$ is the most intuitive thing to do. By no means should you look at this as 'rate of change of mass in hyperspace'. Multiple different quantities can have the same unit. Torque is by no means the same as energy, for example.

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Thanks, I think I didn't realise that there's different kinds of metres involved. – Andreas Mar 5 '12 at 17:44
@andreas no no no.. I meant that different, unrelated quantities, like torque and energy (clearly different) can have the same unit(joule). But yes, the meters have different sources, so aggregating them destroys their intuitive meaning. – Manishearth Mar 5 '12 at 18:00

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