This is answering the second part of your question to get a mathematical relationship for the parameters that are involved.
We write down the continuity equation in integral form.
$$\int_{V}\dfrac{\partial \rho}{\partial t}dV+\oint_S\rho \mathbf{u}\cdot\mathbf{n}dS=0$$
The Greek letter $\rho$ is the density of your fluid. $\mathbf{u}\cdot\mathbf{n}$ is the dot product of the velocity at the surface $\mathbf{u}$ and the outward normal unit vector $\mathbf{n}$.
For stationary flow, the volume integral vanishes. At the inlet side, we assume a uniform velocity $u_{\text{i}}$, which is parallel with the unit vector of the surface, and a surface area of $A_{\text{i}}$. Same procedure at the outlet with $u_{\text{o}}$ and $A_{\text{o}}$. Evaluation of the surface integral is only necessary at the inlet and outlet. We get:
$$-\rho u_{\text{i}}A_{\text{i}}+\rho u_{\text{o}}A_{\text{o}}=0.$$
Note, that $u_{\text{i}}A_{\text{i}}$ is the volumetric flowrate at the inlet and $\rho u_{\text{i}}A_{\text{i}}$ is representing the massflowrate at the inlet. Assuming circular inlet and outlet geometries we obtain:
$$u_{\text{i}}\frac{\pi}{4} d^2_{\text{i}}=u_{\text{o}}\frac{\pi}{4} d^2_{\text{o}} \implies u_{\text{i}} d^2_{\text{i}}=u_{\text{o}}d^2_{\text{o}}.$$
So if you know the flow rate at the inlet $Q_{\text{i}}=u_{\text{i}}A_{\text{i}}$ (can be determined by measuring the volume of output in a certain time). You can easily infer the velocity at the outlet for a given outlet $d_{\text{o}}$ and vice versa.
In order to get more information about the system behavior it is possible to apply Bernoulli's equation with empirical pressure loss $\Delta p_{\text{loss}}$ due to dissipative effects (can be looked up in tables for different pipings):
$$p_{\text{i}}+\frac 12\rho u^2_{\text{i}}+\rho g h_{\text{i}}=p_{\text{o}}+\frac 12 \rho u^2_{\text{o}}+\rho g h_{\text{o}}+\Delta p_{\text{loss}}.$$
We now assume that the pressure at the outlet and the inlet is the same, which is a good assumption for slowly moving air as surrounding "fluid". Hence, we can drop the pressure terms. Subtract $\frac 12 \rho u^2_{\text{i}}$ and use the continuity equation to replace $u_{\text{i}}=\frac {A_{\text{o}}}{A_{\text{i}}}u_{\text{o}}$. Additionally, we subtract $\rho g h_{\text{o}}$ and $\Delta p_{\text{loss}}$. We get the following equation:
$$\rho g(h_{\text{i}}-h_{\text{o}}) -\Delta p_{\text{loss}}=\frac 12 \rho u^2_{\text{o}}\left(1-\frac{A_{\text{o}}}{A_{\text{i}}} \right).$$
Solving this equation for $u_{\text{o}}$:
$$u_{\text{o}}=\sqrt{\frac{2g(h_{\text{i}}-h_{\text{o}}) -2\frac{\Delta p_{\text{loss}}}{\rho}}{1-\frac{A_{\text{o}}}{A_{\text{i}}}}}.$$
If you additionally assume the following simplifications
- Assume that there is no pressure loss: $\Delta p_{\text{loss}}=0$
- Assume that the inlet area is much larger than the outlet area: $\frac{A_{\text{o}}}{A_{\text{i}}} \ll 1$
you will obtain Torricelli's law as a crude estimate:
$$u_{\text{o}} \approx \sqrt{2g(h_{\text{i}}-h_{\text{o}})}.$$
