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EE535: Renewable Energy: Systems, Technology & Economics

EE535: Renewable Energy: Systems, Technology & Economics. Hydroelectric Power. Intro to Hydro. Hydroelectricity is electricity generated by hydropower, i.e., the production of power through use of the gravitational force of falling or flowing water

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EE535: Renewable Energy: Systems, Technology & Economics

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  1. EE535: Renewable Energy: Systems, Technology & Economics Hydroelectric Power

  2. Intro to Hydro • Hydroelectricity is electricity generated by hydropower, i.e., the production of power through use of the gravitational force of falling or flowing water • It is the most widely used form of renewable energy – Particularly in Ireland • Once a hydroelectric complex is constructed, the project produces no direct waste, and has a considerably lower output level of the CO2 than fossil fuel powered energy plants • Worldwide, an installed capacity of 777 GWe supplied 2998 TWh of hydroelectricity in 2006. This was approximately 20% of the world's electricity, and accounted for about 88% of electricity from renewable sources.

  3. Hydro Plant Schematic • Most hydroelectric power comes from the potential energy of dammed water driving a water turbine and generator. • the energy extracted from the water depends on the volume and on the difference in height between the source and the water's outflow • This height difference is called the head. The amount of potential energy in water is proportional to the head. To obtain very high head, water for a hydraulic turbine may be run through a large pipe called a penstock. http://en.wikipedia.org/wiki/Hydroelectricity

  4. Pumped Storage • Pumped storage hydroelectricity produces electricity to supply high peak demands by moving water between reservoirs at different elevations. • At times of low electrical demand, excess generation capacity is used to pump water into the higher reservoir. When there is higher demand, water is released back into the lower reservoir through a turbine. • Pumped storage schemes currently provide the only commercially important means of large-scale grid energy storage and improve the daily load factor of the generation system. • Hydroelectric plants with no reservoir capacity are called run-of-the-river plants, since it is not then possible to store water. • A tidal power plant makes use of the daily rise and fall of water due to tides; such sources are highly predictable, and if conditions permit construction of reservoirs, can also be dispatchable to generate power during high demand periods.

  5. Calculating the amount of available power • A simple formula for approximating electric power production at a hydroelectric plant is: P = ρhrgk, • where P is Power in watts, ρ is the density of water (~1000 kg/m3), h is height in meters, r is flow rate in cubic meters per second, g is acceleration due to gravity of 9.8 m/s2, and k is a coefficient of efficiency ranging from 0 to 1. • Efficiency is often higher (that is, closer to 1) with larger and more modern turbines. • Annual electric energy production depends on the available water supply. In some installations the water flow rate can vary by a factor of 10:1 over the course of a year.

  6. Example • The Dinorwig pumped storage station in Wales has a water capacity of 7x106 m3, which can be released for generating electricity over a 5 hour period. If the effective head is 500m, and the generator efficiency is 90%, calculate the average power output and the total electrical energy produced in 5 hours.

  7. Answer • Generating flow rate r = 7x106 / (5 *60 *60) • r = 388.9 m3/s • P = ρhrgk • P = 1000 x 0.9 x 388.9 x 9.81 x 500 • P = 1.72GW • Energy Output = 5 x 1.72 = 8.6 GWh

  8. Follow – On Question • It takes 6.5 hours to refill the reservoir. Because of frictional drag, the effective head when pumping is 530m. For pumps having an energy efficiency of 90%, calculate the input power to the pumps, the total electrical energy required for pumping, the overall energy efficiency of the pumped storage plant.

  9. Answer • Pumping flow rate = 7x106 / (6.5 *60 *60) • Pumping flow rate = 299.1m3/s • Pumping power required = 1000 x 299.1 x 9.81 x 530 = 1.56GW • Pumps are 90% efficient, so input power to pumps = 1.56/.9 = 1.73GW • Energy Input = 6.5 x 1.73 = 11.2GWh • Overall efficiency = output/input = 8.6 / 11.2 = 77%

  10. Exercise • Check out: • http://www.sei.ie/Renewables/REIO_Library/Case_Studies_and_Other_Reading/kenmarehydropower.pdf • http://www.spiritofireland.org/solution.php • http://en.wikipedia.org/wiki/Hydroelectric_energy_storage

  11. Ardnacrusha http://en.wikipedia.org/wiki/File:Ardnacrusha.jpg

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