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?