# Potential energy of Hydroelectric plant before the dam was built

So the energy potential of a Hydroelectric dam is the difference between the head and base. Before the dam was built where does all that energy go? I can understand that for a long valley that is dammed up the the energy was released from the headwaters to where the dam was built over a large area. However in a possibly simpler case of a waterfall, you have the "instantaneous" power of difference between the bottom and the head. Where is all that energy going? Is it just dissipated though vibration and sound?

The energy is converted to heat. The friction of the water with the river bed, and with itself, converts the energy to heat.

• So I see this answer. Can you add some figures for some prominent falls or temperature differences before and after to quantify? Oct 27, 2014 at 17:38
• @QueueHammer an example is worked out here: physics.stackexchange.com/questions/107266/… Oct 27, 2014 at 18:07
• Could the energy extracted by the plant have actually been reaching the river's drain before, and getting dumped into whatever sea/lake/ocean it terminates in? It would be nice to see a reference confirming that the properties of the water flow at the drain (volume/speed/temperature) are not changed significantly by building a plant 100km upstream. Apr 16, 2015 at 8:45
• But, for that matter, it also dissipates into sound, as the author of the question states. Feb 24, 2019 at 12:21

Two energy sinks:

• Heat, as stated in the other answer.
• mechanical work on solids: rocks get ground to gravel, gravel to smaller grave etc.,
• mechanical work on the fluid, it will mst likely flow faster downstream, unless all that energy is dissipated as mechanical work as described above

Let's take a closer look at the heat. We look at a reasonable upper limit for thermal energy imparted on the fluid, so no energy is lost to the surroundings, no mechanical work is done on solids and the flow velocity is the same before and after. Then the power of the river is $\dot{m}*g*\Delta H$, mass flow rate times local gravity time head. The thermal power expanded is $\dot{m}* c * \Delta T$, mass flow rate times thermal capacity times temp. diff. With thermal capacity for water we get somewhere near 2,34 mK per m head difference. Not that this value could in theory be larger after a rough patch in the stream with lots of dissipation, but it's very likely that heat echange with the surrounding will drown out this effect.