Accodring to this answer to an older question of mine, dissipation in a fluid in W/kg is proportional to U³/L (U characteristic speed, L Length). I was wondering why this is independent of viscosity, and tried a different approach. I assumed dissipation to be the energy lost to friction.
If you have a Volume that is enclosed by to Areas A at distance d and a flow in the plane of one of those A with speed u, the friction force should be $F = A u \eta d^{-1}$, with viscostiy $\eta$.
Multiply by speed and we should arrive at a power. Divide by Volume time density and we have a power/mass. If we say that $A \sim d^2$ and the Volume $V \sim d^3$ we arrive at
$$ P_{dissipation} \sim \frac{u^2}{d} * \frac{\eta}{\rho}$$
Units check out.
Now let'S do a thought experiment: We have stirred tank. We know that once the tank contents are in motion and we have a stable flow, the dissipation over the complete volume should equal the power of the mixer - conversation of energy. Now we look at two cases: We operate the mixer at 1 kW and at 8 kW. If we compare the two cases with the relation for dissipation as $\sim \frac{u3}{d}$, we expect the characteristik speed to double with the eight-fold increase in power.
If we use 'my' relation, we expect the speed to increase $\sqrt[3]{8} = 2.828$-fold. This does not match, obviously. What is wrong with my approach?
