How does calculating Hubble time work? For calculating we take $t=\frac dV$, where $V=Hd$ . But velocity isn't uniform for an object as distance also increases with time. And if velocity isn't constant the equation $t=\frac H d$ also isn't applicable. So how is the Hubble time the age of the universe intuitively?
 A: You're right, $V$ is not necessarily constant, and there are so many different ways to model the universe.
The fact that $H$ is related to the "Hubble time" comes from when we look at these models, we find that estimates of the age of the universe are usually (for simple models at least)
$$t_0\propto\frac{1}{H_0}$$
where $t_0$ is the amount of cosmic time that has elapsed between the beginning of the universe and now, and $H_0$ is the Hubble parameter evaluated today.  Note that in general models, $H$ itself is not constant, but a function of time, but the relationship
$$V=Hd$$
is still true (just now it's not a simple linear relationship, since $H$ is not constant).
In a very simple universe which is flat and has no cosmological constant, (called an Einstein-de-Sitter universe), one can show that the age of the universe is
$$t_0=\frac{2}{3H_0}$$
Note also that the value of the Hubble parameter is $H_0\sim70\text{ km s}^{-1}\text{ Mpc}^{-1}$ or $\sim2\times10^{-18}\text{ s}^{-1}$ in sensible units - i.e. inverse time.  So it makes sense that $1/H_0$ is a time-scale for the universe, but it just so happens that using $t_0=H_0^{-1}$ is just a simple estimation.
To properly answer this question, you need to look at the details of cosmology and the cosmological field equations (Friedmann equations).
A: The unit of hubble parameter is $1/s$ so simply we can define Hubble time as $\frac {1} {H}$, which the unit becomes second. You dont have to think about the distances or etc. But if you want to you can write
$$t=\frac {r} {V}=\frac {r} {Hr}=\frac{1} {H}$$ In the derivation distances cancel out.
For current hubble parameter, $H_0=70kms\pm 7^{-1}Mpc^{-1}$ Hubble time is, $H_0^{-1}=14.0\pm 1.4Gyr$
