How long do the phases of sunset take? How long does it take from the moment the Sun star 'touches' the horizon, to the time it totally disappear? (I'm not talking about the radiation (rays), but just about the appearance of the star by itself. I don't care of the rays). If I were near a beach or plane I'd check it, but I'm in crowded urban region so it's difficult to check it I believe).
@Calc-You-Later's answer is correct in places where the path of the Sun is perpendicular to the horizon (at the equator during the equinoxes, somewhere between the tropics the rest of the time). There, the sunset lasts 2 minutes. However, at other latitudes, the sunset may last a bit longer. 

Let $\alpha$ be the angle the path of the Sun makes with the horizon. This angle depends on your latitude and the date. 



The speed of the Sun along its path is $\vec{v_s} = $ 0.25°/min. This speed can be projected along the vertical and horizontal axes. The vertical speed is now $\vec{v_s} . \sin (\alpha)$. The sunset length is still the apparent angle of the Sun divided by the vertical speed of the Sun. 

So at higher latitudes, if the Sun sets, for instance, at a 45° angle, then the sunset will last 2 min / sin(45°) = 2.85 minutes = 2 minutes 51 seconds. If the Sun sets with a 60° angle, the sunset will last 4 minutes. 



To get the value of $\alpha$, you can add your latitude and the solar declination:


Determine how many days have passed since December 21st (i.e since the winter solstice). Let's call this number $x$
Multiply this number by the amount of turning the Earth does in one day
$ y = x * 360/365.25$
Get the cosine of this number and multiply it by the tilt of the Earth $z = -23.5 * cos(y)$. This is the solar declination
Add the latitude $\theta$ of your location $\alpha = z + \theta $




This does not account for atmospheric effects. The atmosphere will bend the rays of light, and can make it possible to see the Sun even though it is under the horizon. 

The atmospheric effects, however, depend on the atmospheric conditions (temperature gradient,...), and are more difficult to quantify.