Solar and Hydroelectric Power. Abbie Thill Becca Mattson Grace Nordquist Keira Jacobs Miyabi Goedert

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1 Solar and Hydroelectric Power Abbie Thill Becca Mattson Grace Nordquist Keira Jacobs Miyabi Goedert

2 Photovoltaic Cell vs Solar Heating Panel Photovoltaic cells power things such as calculators and satellites. Photo means light and voltaic means electricity. Photovoltaic cells convert sunlight into electricity, hence their name. Many times they are called PV cells. PV cells are made of semiconductors, the most common being silicon. When light hits the semiconductor, some of the energy is absorbed by the semiconductor. Then the energy knocks loose the electrons so they can flow freely. One or more electric fields inside the PV cell force the electrons to flow in one direction. The flow creates a current that can be extracted for external use. The current and the voltage crated by the electric field make the photovoltaic cell's power.

When light hits the semiconductor, some of the energy is absorbed by the semiconductor. Then the energy knocks loose the electrons so they can flow freely.

3 Photovoltaic Cell vs Solar Heating Panel Solar Heating panels work by heating water. The heat that is gathered by the solar collector is then transferred to the water supply. The water is then stored until it is used. A conventional water heater would take hot water from the solar storage tank instead of using household energy to heat water in the boiler. This technology is a good source of renewable energy. In the US, the first commercial solar water heaters were sold in California in the 1890 s, and today this method is most commonly used to heat pools.

A conventional water heater would take hot water from the solar storage tank instead of using household energy to heat water in the boiler.

4 Solar Heating Panel Photovoltaic Cells

5 Solar Heating Panel vs. photovoltaic cells Solar Thermal Efficiency: 70% Energy Production Cost: ~$6,000 per system Daily Output: 22.3 kwh requires 2 panels/80 sq ft. Size: King-sized mattress Solar Photovoltaic Efficiency: 12-15% Energy Production Cost: ~$48,000 per system Daily Output: 22.3 kwh requires 32 panels/456 sq ft. Size: covers an entire roof

system Daily Output: 22.3 kwh requires 2 panels/80 sq ft.

6 Photovoltaic Cell vs Solar Heating Panel Even though solar heating panels are much more efficient than photovoltaic panels, solar heating panels are used for heating water, mostly in pools and household water heaters. To use the sun's energy to power other devices photovoltaic cells are used.

are used for heating water, mostly in pools and household water heaters.

7 Sankey Diagram For Photovoltaic Cell The photovoltaic cell only uses the light from the sun and not the heat so it isn't 100% efficient. The Semi-conductor transforms only about 20% of the sun's energy into current for outside use.

8 Sankey Diagram For Solar Heating Panel The Solar Panel collects the sun's heat and turns it into energy to heat the water. It uses the sun's heat but not the light, and the panel cannot absorb all of the heat either. The panel is about 70% efficient.

9 Applications of Photovoltaic cells You want to install photovoltaic cells on the roof of your cottage. The cottage has roof area of 139 m 2. What is the available electrical power considering the total cell efficiency is 14% and the constant light intensity is 980Wm -2? Solution: Power input = (light intensity)*(roof area) = (980 Wm -2 )*(139m 2 ) = 136 x 10 3 W = 136 kw Since Efficiency = (Power output)/(power input), Power output = Efficiency*(Power input) = 0.14*(136 kw) = 19 kw of electricity is available

What is the available electrical power considering the total cell efficiency is 14% and the constant light intensity is 980Wm -2?

10 Back to the Future Application In Back to the Future, the DeLorean needs 1.21 GW in order to operate. If Doc decides to go green and use photovoltaic cells to power his trip, (a) what is the area of panels needed? Assume that the efficiency is 16% and the constant light intensity of 965 Wm -2. (b) How many 1.30 m 2 tiles are needed? (c) Is this realistic? Solution: a) Efficiency = (output power)/(input power) Input power = 1.21*10 9 W/0.16 = 7.56 GW Area = Power Input/light intensity Area = 7.56*10 9 /965 Wm -2 = 7.84*10 6 m 2 b) Number of tiles = 7.84*10 6 m 2 /1.30 m 2 = 6.03 *10 6 tiles c) There are 6050 m 2 in a football field 7.84*10 6 m 2 /6050 m 2 = over 1,295 football fields This is definitely not realistic!

Assume that the efficiency is 16% and the constant light intensity of 965 Wm -2. (b) How many 1.30 m 2 tiles are needed? (c) Is this realistic?

11 Application of Solar Heating Panels A solar-powered water heater absorbs solar energy at the rate of 10.2 MJm -2 per day. (a) Calculate the total water energy gain per day if the collector area is 6.1 m 2, the collector efficiency is 65% and the water volume is 430 dm 3. (b) Calculate the daily change in the tank's water temperature. Solution: a) Water energy gain per day = (rate of solar energy received)* (collector area) Water energy gain per day = (10.2*10 6 MJm -2 )*(6.1m 2 ) = 6.22*10 7 J b) Q = mc T, where the density of water is 1.0 kg dm -3, m = (density of water)*(volume) = (1.0 kg dm -3 )*(430 dm 3 ) = 430 kg T = Q/(mc) = 6.22*10 7 J/(430 kg*4186 J o C -1 kg -1 ) =35 o C. The water temperature increases 35 0 C daily.

(b) Calculate the daily change in the tank's water temperature.

12 Seasonal variations in Earth's solar power incident -Solar radiation differs with the seasons because of a change in the way the sun's radiation is spread. -In the summer, the earth's north pole is tilted towards the sun, so the sun's radiation is more concentrated in the North and more spread out in the South.

-In the summer, the earth's north pole is tilted towards the sun, so the

Regional variations in Earth's solar power incident. - The sun's radiation is most intense when the rays strike perpendicular to the surface of the earth.

13 Regional variations in Earth's solar power incident. - The sun's radiation is most intense when the rays strike perpendicular to the surface of the earth. Therefore, the suns radiation is strongest at the equator, and much less at higher and lower latitudes. - The reason for this is at latitudes farther away from the equator, the same amount of radiation is spread over a larger area.

14 Main Components of Hydroelectric Power Plants Reservoir: A place where a natural body of water is stored, at a level higher than the turbine. Dam: Obstructs the water flow from the reservoir. Stopping the water helps harness the energy. The bottom of the dam has gates that can be lifted to let the water through. Penstock: Connects the reservoir to the turbine propeller. When the gates are lifted the force of gravity makes the water flow through the penstock, the potential energy stored in the dam is converted into kinetic energy. Turbine: The kinetic energy created by the penstock turns the blades in the turbine. The turbine has a shaft connected to the generator Generator: When the blades in the turbine rotate, the shaft turns a motor which produces an electric current in the generator. Power Lines: Transports the power produced by the generator to various stations through the power lines.

When the gates are lifted the force of gravity makes the water flow through the penstock, the potential energy stored in the dam is converted into kinetic energy.

15 Main Components of Hydroelectric Power Plants

16 Hydroelectric Schemes 1. Impoundment hydropower projects 2. Run-of-river projects 3. Microhydropower projects 4. Diversion hydropower projects 5. Pumped storage projects

17 Impoundment Hydropower The water is dammed and may be released to meet either changing electricity demands or to maintain a constant water level.

18 Run-of-River Utilizes the natural flow of the river and can be used with either large flow rates and low heads or small flow rates and high heads. The head of a hydropower project is the height of the diversion structure.

19 Microhydropower A hydroelectric plant which produces 100 kilowatts or less of power. Used by isolated home owners and third world countries. Microhydropower plants can use either low or high heads.

20 Diversion Hydropower Channels a portion of the river through a canal or penstock and may or may not require a dam. A penstock is the channel that carries the water to the turbines.

21 Pumped Storage Pumps water from low levels to high levels during times when demand for electricity is low, and releases it to fall back down when demand is high.

22 Energy transformations in hydroelectric schemes The water stored in the reservoir behind the dam has gravitational potential energy because it is stored at a higher elevation. When the water falls through the penstock toward the turbines it gains translational kinetic energy. This kinetic energy turns the turbines and gives rotational kinetic energy to the generator. Electromagnets in the generator create electrical current from the rotational energy gained from the turbine, and then a transformer turns this AC current into a higher voltage current which is then sent through power lines to homes and businesses.

23 Sankey Diagram For Hydroelectric Not all of the potential energy from the falling water is turned into kinetic energy. Not all of the kinetic energy in the turbine gets turned into current in the generator. Some of it is turned into heat in the process that's why it is not a 100% efficient process.

24 Calculating Power and Flow Rate in Hydroelectric Schemes P=(mgh)/t P= power in kw m= mass flow rate in tons/second g= gravity h= head height t= time Q= 0.83((b*d)/144)(100/t) Q= flow rate in cubic feet per second b= average stream width d= average stream depth t= time it takes for stream to drift 100 feet

25 Solving Problems with Hydroelectric Schemes How much electrical power per second can a hydroelectric scheme produce when the net head of the water is 120m high and the mass flow rate of the water is 5.5 tons/second? P=(mgh)/t P=(5.5*9.8*120)/1 P=6500 kw Calculate the flow rate of a hydroelectric scheme if the river is 240 inches wide with an average depth of 72 inches and it takes 23 seconds for the river to drift 100 feet. Q=0.83(bd/144)(100/t) Q=0.83((240*72)/144)(100/23) Q=430 cubic feet per second

Lab 10. Solar and Wind Power

1 Name Lab 10. Solar and Wind Power INTRODUCTION Sunlight can be used to create heat or generate electrical power. This is referred to as solar energy. It is a clean form of energy production, which doesn't