text: In order to rewrite Eq. in more familiar form we use the general formula
  Here S is the molar entropy and V the molar volume. With Eq. μ(p,T) can be found by a line integration in the p-T plane. First we integrate from the origin (0,0) to (p,0), so at T =0. Next we integrate from (p,0) to (p,T), so with constant pressure (see figure 6). In the first integral dT=0 and in the second dp=0. With Eq. we obtain
  We are interested only in cases where p is small so that V is practically constant. So
  where V is the molar volume of the liquid at T =0 and p =0. The other term in Eq. is also written as a product of V and a quantity p which has the dimension of pressure
  The pressure p is called the fountain pressure. It can be calculated from the entropy of He which, in turn, can be calculated from the heat capacity. For T =T the fountain pressure is equal to 0.692 bar. With a density of liquid helium of 125 kg/m and g = 9.8 m/s 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. and , Eq. obtains the form
  Substitution of Eq. in  gives
  with ρ = M/V the density of liquid He at zero pressure and temperature.
  Eq. 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. has only mathematical meaning, but in special experimental arrangements p 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 (T=0) and zero pressure (p = 0). 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 cm/s). In the steady state v=0 so Eq. implies
  where the index l (r) applies to the left (right) side of the superleak. In this particular case p = 0, z = z, and p = 0 (since T = 0). Consequently,
  This means that the pressure in the right vessel is equal to the fountain pressure at T.
  In an experiment, arranged as in figure 8, a fountain can be created. The fountain effect is used to drive the circulation of He in dilution refrigerators.