Editing Superfluid helium-4 (section)

=== Fountain pressure ===
In order to rewrite Eq.{{EquationNote|1|(1)}} in more familiar form we use the general formula

{{NumBlk2|:|<math display="block">\mathrm{d} \mu = V_m\mathrm{d}p - S_m\mathrm{d}T.</math>|2}}

Here <math display="inline">S_m</math> is the molar entropy and <math display="inline">V_m</math> the molar volume. With Eq.{{EquationNote|2|(2)}} <math display="inline">\mu(p,T)</math> can be found by a line integration in the <math display="inline">p</math>–<math display="inline">T</math> plane. First we integrate from the origin <math display="inline">(0,0)</math> to <math display="inline">(p,0)</math>, so at <math display="inline">T=0</math>. Next we integrate from <math display="inline">(p,0)</math> to <math display="inline">(p,T)</math>, so with constant pressure (see figure 6). In the first integral <math display="inline">\mathrm{d}T=0</math> and in the second <math display="inline">\mathrm{d}p=0</math>. With Eq.{{EquationNote|2|(2)}} we obtain

{{NumBlk2|:|<math display="block">\mu (p,T)=\mu (0,0)+\int_{0}^{p} V_{m}(p^\prime,0)\mathrm{d}p^\prime
-\int_{0}^T S_{m}(p,T^\prime)\mathrm{d}T^\prime.</math>|3}}

We are interested only in cases where <math display="inline">p</math> is small so that <math display="inline">V_m</math> is practically constant. So

{{NumBlk2|:|<math display="block">\int_{0}^{p} V_{m}(p^\prime,0)\mathrm{d}p^\prime = V_{m0}p</math>|4}}

where <math display="inline">V_{m0}</math> is the molar volume of the liquid at <math display="inline">T=0</math> and <math display="inline">p=0</math>. The other term in Eq.{{EquationNote|3|(3)}} is also written as a product of <math display="inline">V_{m0}</math> and a quantity <math>p_f</math> which has the dimension of pressure

{{NumBlk2|:|<math display="block">\int_{0}^T S_{m}(p,T^\prime)\mathrm{d}T^\prime=V_{m0}p_{f}.</math>|5}}

The pressure <math display="inline">p_f</math> is called the fountain pressure. It can be calculated from the entropy of <sup>4</sup>He which, in turn, can be calculated from the heat capacity. For <math display="inline">T=T_\lambda</math> the fountain pressure is equal to 0.692 bar. With a density of liquid helium of 125&nbsp;kg/m<sup>3</sup> and {{mvar|g}} = 9.8&nbsp;m/s<sup>2</sup> this corresponds with a liquid-helium column of 56 meter height. So, in many experiments, the fountain pressure has a bigger effect on the motion of the superfluid helium than gravity.

With Eqs.{{EquationNote|4|(4)}} and {{EquationNote|5|(5)}}, Eq.{{EquationNote|3|(3)}} obtains the form

{{NumBlk2|:|<math display="block">\mu(p,T) = \mu_0 + V_{m0}(p-p_{f}).</math>|6}}

Substitution of Eq.{{EquationNote|6|(6)}} in {{EquationNote|1|(1)}} gives

{{NumBlk2|:|<math display="block">\rho_0 \frac{\mathrm{d} \vec v_s}{\mathrm{d}t} = - \vec\nabla (p + \rho_0gz-p_{f}).</math>|7}}

with <math display="inline">\rho_0 = M_4/V_{m0}</math> the density of liquid <sup>4</sup>He at zero pressure and temperature.

Eq.{{EquationNote|7|(7)}} shows that the superfluid component is accelerated by gradients in the pressure and in the gravitational field, as usual, but also by a gradient in the fountain pressure.

So far Eq.{{EquationNote|5|(5)}} has only mathematical meaning, but in special experimental arrangements <math display="inline">p_f</math> can show up as a real pressure. Figure 7 shows two vessels both containing He-II. The left vessel is supposed to be at zero kelvins (<math display="inline">T_l=0</math>) and zero pressure (<math display="inline">p_l = 0</math>). The vessels are connected by a so-called superleak. This is a tube, filled with a very fine powder, so the flow of the normal component is blocked. However, the superfluid component can flow through this superleak without any problem (below a critical velocity of about 20&nbsp;cm/s). In the steady state <math display="inline">v_s=0</math> so Eq.{{EquationNote|7|(7)}} implies

{{NumBlk2|:|<math>p_{l}+\rho_0gz_{l}-p_{fl}=p_{r}+ \rho_0gz_{r}-p_{fr}</math>|8}}

where the indexes <math display="inline">l</math> and <math display="inline">r</math> apply to the left and right side of the superleak respectively. In this particular case <math display="inline">p_l = 0</math>, <math display="inline">z_l = z_r</math>, and <math display="inline">p_{fl} = 0</math> (since <math display="inline">T_l = 0</math>). Consequently,

<math display="block">0=p_{r}-p_{fr}.</math>

This means that the pressure in the right vessel is equal to the fountain pressure at <math display="inline">T_r</math>.

In an experiment, arranged as in figure 8, a fountain can be created. The fountain effect is used to drive the circulation of <sup>3</sup>He in dilution refrigerators.<ref>{{cite journal|doi=10.1016/0375-9601(75)90087-0|title=A dilution refrigerator with superfluid injection|year=1975|last1=Staas|first1=F. A.|last2=Severijns|first2=A. P.|last3=Van Der Waerden|first3=H. C.bM.|journal=Physics Letters A|volume=53|issue=4|page=327|bibcode = 1975PhLA...53..327S }}</ref><ref>{{cite journal|title=<sup>3</sup>He flow in dilute <sup>3</sup>He-<sup>4</sup>He mixtures at temperatures between 10 and 150 mK|doi=10.1103/PhysRevB.32.2870|pmid=9937394|year=1985|last1=Castelijns|first1=C.|last2=Kuerten|first2=J.|last3=De Waele|first3=A.|last4=Gijsman|first4=H.|journal=Physical Review B|volume=32|issue=5|pages=2870–2886|bibcode = 1985PhRvB..32.2870C |url=https://research.tue.nl/nl/publications/3he-flow-in-dilute-3he4he-mixtures-at-temperatures-between-10-and-150-mk(d7aefa27-5cff-4379-9b28-e0623ec7de38).html}}</ref>
[[File:Counterflow heat exchange 01.jpg|thumb|Fig. 9. Transport of heat by a counterflow of the normal and superfluid components of He-II]]