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Hydropower

493 bytes added, 09:19, 4 December 2017
Hydropower example
====Hydropower example====
To calculate the theoretical maximum energy that can be produced from a microhydro plant, a few pieces of information are needed. Among these is the Hazen-Williams equation, which allows for the calculation of loss due to friction. The equation is as follows: f=0.2083*(100/c)<sup>1.852</sup>*Q<sup>1.852</sup>/d<sup>14.8528655</sup>, where c is the hazen-williams coefficient and is unique to the piping used, Q is the flow in gallons per minute, and d is the diameter of the pipe in inches. The hazen-williams coefficient c is an indicator of how "smooth" the inner wall of a pipe is; since c is in the denominator, a higher c value correlates to less friction.
For instance, to find the friction loss of a 500 gpm flow through an 8" PVC pipe with a c value of 150, the Hazen-williams equation provides a loss of 0.395 613 feet-loss per 100 feet pipe. Once you have found the feet-loss per 100 feet of pipe, you can multiply this rate by the length of pipe to determine the total feet-loss. You would then subtract the total feet-loss from the total head to find the net head H that you will use in your final power output calculation.
To determine the power in output of such a system, the equation P=Q*H/k is used, where Q is the flow ratein gallons per minute, H is the head lossin feet, 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.
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