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For the fundamental passive $RLC$ electrical circuit, analogies are often used to describe the physics of fluid flow where $L$ represents the inertance analogous to inductance, $C$ is compliance analogous to capacitance, $R$ is the fluidic resistance analogous to electrical resistance , $Q$ is flow analogous to electrical current, and $P$ is pressure analogous to voltage.

Between these analogies one can determine the stored energy in the $L$ and $C$ components as

$$E_L=\frac{1}{2}Li^2 \text{ or analogously } E_L=\frac{1}{2}LQ^2$$

and

$$E_c=\frac{1}{2}Cv^2 \text{ or analogously } E_c=\frac{1}{2}CP^2$$ respectively.

And at least for the electrical resistance we have the energy dissipated as

$$E_R=R\int{i^2}dt$$

But what of the power dissipated by a fluidic resistance?

We can write

$$E_R=R\int{Q^2}dt$$

and the units indeed units of energy. But for electrical resistance we know the energy is dissipated by heat whereas its not so clear at all to me how energy is dissipated in the fluidic resistance. Does the analogy hold? If so how is energy being dissipated?

Text books will say "loss of head" but that's just pressure drop. And you have that occurring also in the electrical circuit as voltage drop. So where is the power, energy going in the fluidic circuit?

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  • $\begingroup$ Perhaps viscous heating would take a form similar to ohmic losses $\endgroup$ – Maxim Umansky Apr 16 '18 at 14:59
  • $\begingroup$ @MaximUmansky That makes sense, but is it true physically? In an electrical circuit the energy is dissipated externally as radiative or convective heat from the surface of the resistor. What of a fluidic circuit? Perhaps energy both radiated externally and carried by the fluid itself - unlike electrical current. I know analogies indeed limited, but in many cases the quantitative calculations indeed carry over. So I guess the main point of my question - is that last integral equation correct? Is there indeed physical energy losses that can be quantitatively figured by this equation? $\endgroup$ – docscience Apr 16 '18 at 15:57
  • $\begingroup$ I think that the analogy sort of breaks down because in the case of fluids if you consider a fluid element there is energy associated with not just the pressure on the fluid element (analogous to the V on an electron), but also there is energy associated with the kinetic energy of the fluid element and for fluids energy can be exchanged between these two forms as a fluid element travels along its circuit. In fact, according to the Bernoulli equation, energy of a fluid element is conserved among the components of pressure energy, kinetic energy, and gravitational potential energy. $\endgroup$ – Samuel Weir Apr 16 '18 at 17:23
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    $\begingroup$ @doscience In the Zeldovich & Raizer book "Physics of shock waves etc." Eq. 1.94 contains the derivation of the viscous heating term $\mu (du/dx)^2$ which is I believe what you are looking for $\endgroup$ – Maxim Umansky Apr 16 '18 at 18:20
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    $\begingroup$ In fact, a full solution for viscous heat dissipation rate worked out for Poiseuille flow in a channel is available, e.g., on mech.kth.se/~luca/Smak/rec8.pdf. Their expression for the dissipation rate const $\times \mu U^2$ where $U$ is the flow bulk velocity is a direct analogy to ohmic heat $I^2 R$. $\endgroup$ – Maxim Umansky Apr 16 '18 at 22:41
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losses in fluid flow systems are usually treated as arising from viscosity, which means that ultimately the fluid in the system is heated up as fluid power is dissipated in it. and yes, the constituitive equations for fluid flow have near-perfect electrical analogues (just as you have written out) at least to first order, when the fluid flow is subsonic and incompressible.

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    $\begingroup$ Agreed. My process engineering experience from years back clearly indicated that lost power turns into heat in the case of pumped fluids. $\endgroup$ – David White Apr 17 '18 at 1:47
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    $\begingroup$ @DavidWhite, yep- which is why hydraulic systems have heat exchangers in them! $\endgroup$ – niels nielsen Apr 17 '18 at 2:44

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