What does it mean for the grand potential $\Phi$ to be minimised for a process at constant $T,V,\mu$, when $\Phi$ is a function of $T,V,\mu$? I've read from a few places like Kjellander, R. (2019). Statistical Mechanics of Liquids and Solutions: Intermolecular Forces, Structure and Surface Interactions Volume I. p.83. and this Physics Stack Exchange post that (quote from Kjellander):

The grand potential of a system always decreases for a spontaneous
  process at constant temperature, volume and chemical potential.

My confusion with this statement is that; since the grand potential $\Phi$, is a function of $T$, $V$, and $\mu$, if $T$, $V$, and $\mu$ are fixed, then it seems like $\Phi(T, V, \mu)$ should also be fixed, so what is there to minimize?
 A: The spontaneous process that the statement is referring to is meant to bring the system from some unstable state to a stable equilibrium state. One function $\Phi$ describes only one of these states, usually the stable equilibrium state with constant chemical potential throughout. The other state may not even have the same chemical potential throughout, it can consist of several subsystems where chemical potential has different values.
For example, if in the system there are two phases in a an unstable ratio (e.g. too much liquid and too little vapor, for the given $T,V$), then chemical potential can be different in the liquid and in the vapor phase. Let the grand potential per unit volume be given by function $\phi_{liq}(T,\mu_{liq})$ and for vapor $\phi_{vap}(T,\mu_{vap})$.
There may be a state where we have non-equilibrium volumes and chemical potentials of liquid and vapor, e.g. too much liquid and too little vapor. The grand potential in this non-equilibrium state has value
$$
\Phi_1 = V_{liq,1}\phi_{liq}(T,\mu_{liq,1}) + (V-V_{liq,1})\phi_{vap}(T,\mu_{vap,1}).
$$
The system then may evolve towards a stable state where liquid volume goes from $V_{liq,1}$ to $V_{liq,2}$ (some liquid will evaporate), both phases have the same temperature $T$, and the same chemical potential $\mu_2$ . The grand potential of the system in the equilibrium state will be
$$
\Phi_2 = V_{liq,2}\phi_{liq}(T,V,\mu_2) + (V-V_{liq,2})\phi_{vap}(T,V,\mu_2).
$$
The decrease of grand potential means that grand potential of the later equilibrium state is lower or same than the grand potential of the original non-equilibrium state:
$$
\Phi_2 \leq \Phi_1.
$$
A: Imagine a bottom of water，although $p$, $T$, $\mu$ are constant for both liquid and gas contents，however，some liquid may convert to gas and the total grand potential changes.
