= Thermomechanical effect
{title2=$dP=\rho s\,dT$}
{wiki=Fountain_effect}
= Fountain effect
{synonym}
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 $\mathrm{J\,kg^{-1}}$. The <Gibbs free energy> differential per unit <mass> gives
$$
d\mu_m=-s\,dT+\frac{1}{\rho}\,dP.
$$
The <superleak> allows the <superfluid> to redistribute until the two sides have equal $\mu_m$. Setting $d\mu_m=0$ therefore gives
$$
\left.\frac{dP}{dT}\right|_{\mu_m}=\rho s>0,\qquad
P_h-P_c=\int_{T_c}^{T_h}\rho(T,P(T))s(T,P(T))\,dT.
$$
The 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 $\rho=150\,\mathrm{kg\,m^{-3}}$, $s=100\,\mathrm{J\,kg^{-1}\,K^{-1}}$ 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.
Back to article page
