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Pyotr Leonidovich Kapitza's 1938 test compared helium flow through optically flat glass disks. Suspension set the tube-to-bath level difference; a cathetometer read levels. The gap was estimated as $0.5\mu \mathrm{m}$. Levels barely moved over minutes above the transition, but equalized in seconds below. A laminar flow viscosity calculation gave only an upper bound.^[ref]

The essential control is to compare the same restriction and level readout at different temperatures. A changed pressure head is not itself evidence for a changed material response: the driving head must be included when interpreting the flow. For an idealized comparison at equal vapor pressure, negligible surface tension corrections and nearly constant mass density $\rho$, hydrostatic equilibrium givesHere $\Delta h$ is the difference in free-surface heights and $g$ is gravitational acceleration. This explains why following a liquid level provides both a driving pressure and a transported volume; it does not turn a finite-resolution observation into proof of zero viscosity.

A superleak is a narrow or porous connection that strongly suppresses the viscous normal component of a superfluid while allowing appreciable superfluid flow. Treating ${\mathbf{v}}_n\simeq 0$ is a regime assumption, not a perfect material filter for particular atoms. The thermomechanical effect uses that distinction to maintain a temperature difference during chemical equilibration. The idealization fails when normal leakage or dissipative quantum vortex motion becomes appreciable.

The thermomechanical effect couples a temperature difference to a pressure difference across a superleak. For the closed-chamber model in the apparatus figure, consider neighboring equilibrium states with small velocities, negligible height difference and pure helium-4. Let ${\mu }_m$ be chemical potential per unit mass, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. The Gibbs free energy differential per unit mass givesThe superleak allows the superfluid to redistribute until the two sides have equal ${\mu }_m$. Setting $d{\mu }_m=0$ therefore givesThe sign matters: heating one side at initially equal pressure lowers its ${\mu }_m$; inflow raises its pressure until chemical balance is restored. The hotter side has the higher pressure. The integral follows the constant-${\mu }_m$ path; it need not be evaluated at fixed pressure when material properties vary appreciably.

For an original local example, stipulate constant $\rho =150\mathrm{kg}{\mathrm{m}}^{-3}$, $s=100\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $T_h-T_c=0.020\mathrm{K}$. The predicted pressure difference is $300\mathrm{Pa}$. These are teaching assumptions, not helium measurements. This static thermodynamics relation predicts neither the required heater power nor the filling time: those require heat leakage, channel conductance and chamber heat capacity. An outlet would introduce kinetic energy and a flowing balance, so the static $300\mathrm{Pa}$ is not a predicted fountain height.