Physic: Calculated Waterflow based on Power generation of turbine

Question

I have a turbine which produces 50MW, the water falls down from three meters height.

How much water does flow per second?

Please do not answer this question, but rather give an explanation how I can calculate it myself.

Answer 1

This is physics, not math.

If a mass $m$ of water falls through a height $h$ then the potential energy change is $mgh$.

If $\dot{m}$ of water passes through a height $h$ then the potential energy change per unit time is $\dot{m}gh$.

If water has an average density of $\rho$, and the volume flow rate is $\dot{V}$ then the mass flow rate is $\dot{m} = \rho \dot{V}$.

If all the potential energy is converted to usable work $P$ in the turbine then we have $P = \rho \dot{V} gh$.

You have $P,h,g,\rho$. Compute $\dot{V}$.

Addendum:

$\rho$ is the density of water, we usually take $\rho = 1000\ kg/m^3$.

$g$ is the acceleration due to gravity, we usually take $g = 9.81 \ m/s^2$.

In the above, $h=3 \ m$, $P= 50 \times 10^6 \ W$.

The dot as in $\dot{m}$ usually means the rate of change with respect to time. So, $\dot{V}$ is the change of volume per unit time, or flow rate.