Question

Assuming 100% efficient energy conversion, how much water stored behind a 50

centimeter high hydroelectric dam would be required to charge the battery?

Answer 1

Let us calculate the hydropower of the dam (see https://en.wikipedia.org/wiki/Hydropower#Mechanical_power)

$P = \eta \dot{m}g h$ , where $\eta$ is the efficiency, $\dot{m}$ is the mass flow rate, $g$ is the acceleration due to gravity and $h$ is the height.

If we calculate the energy provided by dam, we should multiply the formula by time, so we'll get

$E=\eta mgh.$

There are many different types of batteries. AAA batteries have typical capacity of ~ 1 Wh = $3.6\cdot10^3\,$J.

Therefore, $m = \dfrac{E}{\eta gh} = \dfrac{3.6\cdot10^3\,\mathrm{J}}{1.00\cdot9.81\,\mathrm{N/kg} \cdot0.50\,\mathrm{m}} = 734\,\mathrm{kg}.$

If we consider D batteries, we'll read they have capacities of ~ 10 Wh (see https://en.wikipedia.org/wiki/D_battery). Therefore,

$m = \dfrac{E}{\eta gh} = \dfrac{3.6\cdot10^4\,\mathrm{J}}{1.00\cdot9.81\,\mathrm{N/kg} \cdot0.50\,\mathrm{m}} \approx 7340\,\mathrm{kg}.$