Solar and Hydroelectric Power By Abby Counihan, Ella Counihan, Deqa Mahammud, Paige Carlson, Fiona Kelliher
Photovoltaic Cells solar cell produce electricity by using photoelectric effects convert sunlight to electricity part of the PV system size can vary ( roughly 1-3 inches across in width) connect to form modules much bigger ( several feet wide and across) the modules are then connected to form photovoltaic arrays
Photovoltaic Cells light can be reflected, absorbed or pass right through the cell only the light absorbed though can be made into electricity made from semiconductor material (one positive and one negative) the freed electrons naturally migrate to the positive layer, which creates a voltage differential like a battery made from silicon
Photovoltaic Cells The Photovoltaic cells only use up the light from the sun and not the heat coming off the sun. This is opposite from what is used for Solar Heating Panels. Since this uses the light and not the heat this process is not 100% efficient.
Photovoltaic Cells Sankey Diagram
Solar Heating Panel Sankey Diagram:
Solar Heating Panel The silicon panel collects the sun's heat then transforms it into electrical energy that is used to heat up the water. This process uses the sun's heat but the light given off from the sun isn't used. The solar heating panel is not able to use up all of the heat given off by the sun because it is only about 70% efficient. expensive
Solar panels light gets absorded electrons get excited, and get moved up to higher energy levels so able to more freely, so can move around freely which creates a current radiation energy after the engery has been converted into electrical energy it goes though a converter to change from DC to AC
Solar Panels silicon panels surrounded by a metal frame been around for over 50 years photovoltaic efficiency of 18% new technology: thin glass film panels less bulky cheaper only 8% photovoltaic efficency
Solar panels vs Photovoltaic cells Photovoltaic Cells Efficiency: 70% Energy Cost: $6,000 per system 2 panels/80 sq ft. Solar Photovoltaic Efficiency: 12-15% Energy Cost: $48,000 per system requires 32 panels/456 sq ft.
Solving Photovoltaic Cell Problems 1. For the school bus to work, it needs 1.1kW. The bus company would like to make it more eco-friendly, and decides to use photovoltaic cells to power the bus. a) What is the area of the panels needed? Assume efficiency is 15% and light intensity is 650 W/m^2. b) If the bus roof is 2.5m by 4.5m, is it possible to power the bus?
Solutions a) area = input power/light intensity input power = output power/efficiency = 1.1 kw/.15 = 7.3333 kw area = 7.3333 kw/650 W/m^2 = 11.3 m^2 needed b) 2.5m*4.5m = 11.25 m^2. No, there's just barely not enough room :(
Solar Panel Heating Problem 2. How big is the area of a pool if it uses solar panels and the water energy gain per day is 22.48 kj and the rate of solar energy received is 562 kj/m^2?
Solution Water energy = (Rate of solar energy received) * (Area) Area = (Water energy)/(rate of solar energy received) (22.48*10^6 J)/(562 kj/m^2) = 40 m^2
Types Hydroelectric Schemes Run-of-River project Diversion hydropower project Pumped Storage Microhydropower projects Impoundment hydropower project
Run-of-River Scheme The run of the river scheme relies on the natural flow of the river. Either with a heavier flow rate and lower head or higher head but lower flow rate. Head is the height of the waterfall
Diversion Scheme A diversion scheme involves taking a supply of water from a dammed river to a remote powerhouse with a turbine and generator. The water is transported through a canal or low pressure tunnel and then back to the river.
Pumped Storage Scheme A Pumped storage scheme incorporates two reservoirs that are used during times of low demand and cheap electricity (night time). The electricity is bought to pump water from a lower basin to the upper basin. The water is released to create power when electricity is in high demand.
Microhydropower Schemes Microhydropower schemes produce 100 kilowatts of power or less and are seen in farms, ranches family homes, and third world countries.
Impoundment Schemes Impoundment Schemes are the common type. They store water from a dam in a reservoir. When released the water flows through a turbine which activates a generator to produce electricity.
Energy Transformations Part 1 1. Intake gates open, water from reservoir flows in. As it goes through penstock, it builds up pressure, because it's forced through a small space. 2. Water hits turbine, begins to turn blades (Francis Turbine, curved blades, 172 tons and 90 rpm). 3. As turbine turns, so do magnets inside generator. The magnets rub against copper coils, moving electrons and producing an AC current. 4. Transformer inside powerhouse takes the current and turns it into higher-voltage power. 5. Energy is sent through power lines and out into the world. Used water goes out through outflow pipe and back into the river, moving downstream.
Energy Transformations Part 2 Stored energy Kinetic Energy Kinetic/Potential Electrical Energy
energy transformation from turbine to generator Kinetic energy of water -> kinetic enegy of blades kinetic energy of blades -> kinetic energy of magnets kinetic energy of magnets -> electrical energy
Solving Problems w/hydroelectric Schemes 1a) How much electrical power can a scheme produce if the net head is 90m high, the time it takes for the water to fall is 1 sec, and the mass flow rate of the water is 3.8 tons/sec? b) What is the flow rate of the hydroelectric scheme using the flow rate found in the previous problem? (Assume efficiency of the scheme is.8)
The Equations P=mgh/t P is power in kw, m is mass flow rate in tons/sec, g is acceleration due to gravity, and h is head (or height of water drop), t is time in seconds P=phrgk P is power in watts, p is density of water, h is head (or height of water drop), r is flow rate in m^3/sec, k is coefficient of efficiency from 0-1 with 1 being most efficient
Solutions 1. P=mgh/t P=(3.81 tons/sec*9.81 m/sec^2*90m)/1 sec P=3,364 kw 2. P=rphgk r = P/phgk r = 3,364,000/(1000kg/m^3*90m*9.81*.8) r = 48 m^3/second
What is the Solar Power Incident? The solar power incident is the way that the tilt of the earth affects the amount of heat energy from the sun on different parts of the earth during different times of the year.
How does it work? Conceptual Level Earth turns on a 24 degree angle in relation to the sun. This causes uneven heating of the earths surface. During the summer earth's Northern Hemisphere points towards the sun, this creates warmer and longer days. During the winter earth's Northern Hemisphere points away from the sun which causes less direct sunlight and shorter days.
How does it work? Equation The measure of intensity of the solar radiation that strikes earth's surface is called irradiance. It can be measured by the equation I0 is extraterrestrial irradiance on a plane perpendicular to the Sun s rays and N is the day of the year such that for January 1st, n = 1. However not every point on earth is perpendicular to the rays from the sun. Thus, the cosine effect must be taken into account.
How does it work? Cosine Effect When the sun's rays hit a point perpendicular from the sun to the earth's surface, the power of the ray is at full irradiance, however not every point on earth is perpendicular to the sun. The cosine effect works so that the solar irradiance falling on any part of the earths surface will be reduced by the cosine of the angle Θz between the perpendicular surface and a central ray from the sun.
Power Incident per unit area of Earth's Surface The use of the equation allows you to determine the amount of solar radiation at any point on the earth on any given day of the year. I0= irradiance outside the atmosphere on a plane perpendicular to the sun's rays
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