in a neutron star can be assumed to be at zero temperature, i.e., the number of quasi-particle excitations is small, they coexist with a normal liquid of electrons in the core and a nuclear lattice in the crusts. Hence, entropy is irreversibly produced due to various dissipative mechanism in the normal fluid even though the superfluid matter is effectively at zero temperature. Formally, the electron liquid and the nuclear lattice in the crusts take over the role of quasi-particle excitations in the superfluid hydrodynamics of liquid He$^4$.

The conservation of the combined mass of the two fluids is given by the continuity equation $$\begin{aligned}
\frac{\partial\rho}{\partial t} + \nabla\cdot J &=&0,
\label{eq:AS2.2}\end{aligned}$$ where the net mass $\rho= \rho_S +\rho_N$ is the sum of the masses of superfluid and normal fluid, $ J = \rho_N {\bf v}_N + \rho_S{\bf v}_S$ is the mass current (hereafter the indexes $S$ and $N$ refer to superfluid and normal fluid, respectively.) The total momentum conservation is $$\begin{aligned}
\frac{\partial J_i}{\partial t} + \frac{\partial
P_{ik}}{\partial x_k} &=& 0, \end{aligned}$$ where $P_{ik}$ is the stress energy tensor. The time evolution of the entropy, $S$, of normal fluid can be written as $$\begin{aligned}
\frac{\partial S}{\partial t}  +  \nabla \cdot S 