In a centrifugal separator how does the heavy or light phase move upwards? I've been studying about the working principles of a centrifugal separator. Even though i understand how centrifugal force separates the liquid into 2 phases(heavy and light) i cant make sense of how these 2 phases move upward. Is it because there is nowhere else for the fluids to go? If the separator wasn't full whould the 2 phases still move upwards? Someone please help  
 A: The two phases move upwards for essentially the same reason that water flows up out of an artesian well, except centrifugal force replaces gravity. As long as the outlets lie "below" the hydraulic head - the equipotential top surface of a connected body of liquid - the phases can flow uphill to the outlets. If the separator is not full, the flow will stop when the centrifugal liquid surface no longer reaches the outlets.
Your diagram appears to be a dairy paring disc separator. Whole milk (dark blue) enters with some pressure $P_W$, and skim milk (light blue) and cream (yellow) come out with pressures $P_S$ and $P_C$.  The inlet for the whole milk is on the axis-of-rotation, but the outlets for the skim milk and cream are at radii $r_S$ and $r_C$ from this axis. The separator rotates at some angular velocity $\omega$.
The question is about why the phases flow upwards, and the answer would be the same even if the two phases had equal density $\rho$, so let's assume that. (Cream is only a few percent lighter than skim milk, so this isn't far off.)  As long as the inlets and outlets are all at the same vertical height, we can ignore any gravitational pressure differences, and more generally we can neglect gravity because the separator spins so fast, e.g. 6000-9000 rpm, that the centrifugal acceleration ($\omega^2 r$) is much greater than the local gravitational acceleration $g$, except for the central few millimetres. (At 9000 rpm,  $\omega^2 r >g$ for $r> 0.5$ mm.)
If the separator is full (with no air pockets), then the outlet pressures are the sum of the input pressure and the centrifugal pressure:
$$P_S=\frac{1}{2}\rho \omega^2 r_S^2 + P_W$$
$$P_C=\frac{1}{2}\rho \omega^2 r_C^2 + P_W$$
As long as $P_S$ and $P_C$ are greater than zero, the phases will flow up and out of the separator. If $P_W>0$, they will flow even if the separator is not spinning. If $P_W\approx 0$, they will flow up and out as long as $\omega^2>0$ and there is continuous input.
If the separator is not full, the surface of the liquid will lie along an equipotential surface.
Under gravity, the top surface of a static liquid follows a locally flat horizontal gravitational equipotential.  When the separator is spinning at high speed, however, the the liquid surface will lie on a centrifugal equipotential that is a cylinder around the axis of rotation.
The liquid will continue to flow up and out until its cylindrical surface falls outside the outlets, i.e. is piled up against the out wall.
The details change for the curved surface of the liquid at low rotational velocities where $\omega^2 r \sim g$, but I believe the qualitative points are the same.
