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Renewable Energy. Solar Hydroelectric Wind Wave. Sun emits EM E at av rate (P)= 3.9 x 10 26 J/s E is spread in all directions in a spherical shape from the sun. The power/m 2 hitting Earth is the Intensity. Insolation = incoming solar radiation. I = power/Area A = 4 p r 2 .
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Renewable Energy • Solar • Hydroelectric • Wind • Wave
Sun emits EM E at av rate (P)= 3.9 x 1026 J/sE is spread in all directions in a spherical shape from the sun.The power/m2 hitting Earth is the Intensity. Insolation = incoming solar radiation. I = power/Area A = 4pr2. r orbit ~ 1.5 x 1011m.on Earth ~ 1380 W/m2. R
Energy at surface is either reflected (scattered) or absorbed • Amount hitting surface & absorbed varies: Angle of rays (max I when 90o to surface) season, cloud cover, type color of surface…
Renewable Energy Sources • Solar – thermal & photovoltaic (PV) • Hydroelectric • Wind • Waves
Solar • Solar Panels Photovoltaic PV Cells • heating only. Produce current
Clip 1:30 Solar PV 1:30 • http://www.youtube.com/watch?v=Sur89b7afso Solar Themal 1:30 Clip 1:30 • http://www.youtube.com/watch?v=64mtITOuXiA
Hydroelectric - Dams • Falling water spins turbine PE >>KE >> Electric • Can calculate energy • mgh is E of water Mass is also density x volume. • At home look at example Hamper pg 192 – 193.
Clip 2:15 Hydro elec 3 min • http://www.youtube.com/watch?v=wvxUZF4lvGw
Pumper Storage German Version 2.5 min • http://www.youtube.com/watch?v=GJ7ltJlMY9E
1. Hydro Sample Prb • The flow rate of water over a dam of height 40-m is 500 L/s. • Find the power generated. • density of water is 1 kg/L or 1000 kg/ m3.
Power = rate E transformed. • GPE = mgh • P = E/t (each second 500 L of water fall = V/t) • P = mgh/t • Need mass water • D = mass/vol dV = mass sub dV for mass. • P = (dV) gh • t • (1 kg/L) (500 L/s) (10 m/s2) (40 m) • P = 200,000 W
Wind Turbine • Cylindrical volume of air with velocity v, spins blades. KE air > KE blades> elec E. • Power P, = ½ Arv3. • r = air density • A = area of blade sweeps out. Blade = radius. • v = wind velocity
Can calculate Area circle. KE/s = ½ r (pr2) v3. Power P, = ½ Arv3. Mass = (r) vol
Wind 1.21 min • http://www.youtube.com/watch?v=0Kx3qj_oRCchttp://www.youtube.com/watch?v=sLXZkn2W-lk&feature=player_detailpage
2. Wind • Wind with density of 1.2 kg/m3.goes through a windmill with 1.5m blades with vi = 8m/s. The wind slows to 3 m/s, and density changes to 1.8 kg/m3 after leaving the blades. • Find the max power of the windmill.
The DE from the wind is the work done on the blades. • Pi – Pf • A = pr2. • P = ½ Arv3. Solve for each speed & density. • (2171 – 171)W = 2000 W.
Wave Poweruses up & down motion of waves (KE>PE) • Pelamis Oscillating • Hot dog Water column
Power in waves comes from OscillationWave alternate from KE (falling ) to PE rising • Power per lengthofwavefront P/L = ½ A2rgv. A = amplitude (half height) g ~ 10 m/s2. v = wave velocity
3. Waves. A wave with v = 4.8 m/s and height = 10 m is 2 m long. Find the power. • A = ampl = 5 m. • r = 1000 kg/m3. • g = 10 m/s2. • P/l = ½ A2rgv. • = ½ (5 m)2(1000)(10)(4.8 m/s) = 6 x 105 W/m • For 2 m wave = 1.2 MW
Oscillating Water Column2:45 • http://www.youtube.com/watch?v=gcStpg3i5V8 Pelamis no sound 2 min http://www.youtube.com/watch?v=mcTNkoyvLFs
Hints. • Often use density to get a mass. • Must approximate shapes for volume calculation. • Ep = mgh for both hydroelectric & derivation of wave energy. • Rates L/s or kg/s often can be used somehow to get at power which is a rate. • Ex P = mgh = rVgh. V is a flow rate. t t t
Derivation of Equation • A wave on the surface of water is assumed to be a square-wave of height 2A, as shown. • The wave has wavelength λ, speed v and has a wavefront of length L. For this wave, • (i)show that the gravitational potential energy EP stored in one wavelength of the wave is given by • EP = ½ A2lgρL. • where ρ is the density of the water and g is the acceleration of free fall.
Mass water = rV = r (1/2 AlL). • Height fall = A • mgh = r (1/2 AlL) A = 1/2A2lrg