264

edits
**Get our free book (in Spanish or English) on rainwater now - To Catch the Rain.**

# Changes

→Hydropower example

To determine the power in such a system, the equation P=Q*H/k is used, where Q is the flow rate, H is the head loss, and k is a constant of 5,310 gal*ft/(min*kW). For a flow of 500 gallons per minute and a head loss of 60 feet, maximum power would be found as: (500 gal/min *60 feet)/5310 gal-ft/min*KW, which comes out to 5.65 kW.

This theoretical maximum is prevented by real world obstacles, such as turbine efficiency, pipe friction, and the conversion from potential to kinetic energy. To determine how much energy can be produced by a plant with a given efficiency, simply multiply the theoretical by the plant's efficiency. Turbine efficiency is generally between 50-80%. For instance, if the same plant had 90% efficiency, 5.65 kW*0.9= 5.085kW.

== Hydropower plants ==