is it possible to calculate the flow of water out of a pipe where gravity is the only force acting upon the water and therefore generating the flow. The pipe diameter is X and there is minimal fall (say 6 inches?).
The real life scenario here is that I am trying to calculate the flow of water out of a washing machine waste when the filter plug is removed at the bottom?
The Bernoulli equation will give you a pretty good estimate. It says that: $$ P_{1} + \frac{1}{2}\rho v_{1}^{2} + \rho g h_{1} = P_{2} + \frac{1}{2}\rho v_{2}^{2} + \rho g h_{2} $$ You pick 2 points in a flow (1 and 2). In this case, you can ignore the pressure terms, $P$, and focus on velocity, $v$, height, $h$, and gravitational acceleration, $g$. In this case, you can use the free surface of the water in the machine as point 1, meaning that $P_{1} = v _{1} = 0$, $h_{1}= 0.15 m$, then point 2 is at the outlet of your pipe, and $P_{2} = h_{2} = 0$. Then: $$ v_{2} = \sqrt{2gh_{1}} $$ gravitational acceleration is $9.81m/s^2$. If you want to add a (probably small) degree of accuracy to your calculations , the so-called head-loss calculations are useful. You can use this calculator: https://apps.engineeringtoolbox.com/head-loss-water-pipe-a_15.html
Which will give you a pressure loss in kPa, and velocity loss in m/s. You can adjust your prediction as: $$ v_{adjusted} = v_{2} - v_{loss} $$ so your average flow velocity will You'll notice, though, that you need to know the flow rate:$$Q_{2} = v_{2}\pi r_{pipe}^{2}$$ before you can calculate the head loss.
