Sun emits EM E at av rate (P)= 3.9 x 1026 J/s
R
• E is radiated in all directions in a spherical shape from the sun.
• Insolation = incoming solar radiation.
• Intensity -power/m2 hitting Earth.
• I = power/Area.
• A = 4r2.
• r orbit ~ 1.5 x 1011m.
• on Earth ~ 1380 W/m2.
Amount of Energy at surface
• Amount hitting surface & absorbed varies:
• Angle of rays (max I when 90o to surface)
• season,
• cloud cover,
• type color of surface,
• reflection
Solar PV 1:30
• http://www.youtube.com/watch?v=Sur89b7afso
Clip 1:30
Solar Themal 1:30Clip 1:30
• http://www.youtube.com/watch?v=64mtITOuXiA
4. A solar panel is installed to heat 1.4 m3 of water from 20 – 50oC. If the average power from the sun hits it with 0.9 kW/m2,
• a. Determine the number of Joules needed to heat the water.
• b. Estimate the area of the solar panel needed to heat the water in 2 hours.
• The density of water is 1000 kg/m3,assume an ideal panel.
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.
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 Av3.– = air density
– A = area of blade sweeps out. Blade = radius.
– v = wind velocity
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 E from the wind is the work done on the blades.
• Pi – Pf
• A = r2.
• P = ½ Av3. Solve for each speed & density.
• (2171 – 171)W = 2000 W.
Power in waves comes from OscillationWave alternate from KE (falling ) to PE rising
• Power per lengthof wavefront
P/L = ½ A2 gv.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.
• = 1000 kg/m3.
• g = 10 m/s2.
• P/l = ½ A2 gv.
• = 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
http://www.youtube.com/watch?v=mcTNkoyvLFs
Pelamis no sound 2 min
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 = Vgh. 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 A2 gρL.• where ρ is the density of the water and g is the acceleration of
free fall.