Fontes Renováveis Não-Convencionais. Parte II
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1 Fontes Renováveis Não-Convencionais Parte II Prof. Antonio Simões Costa Prof. Tom Overbye, U. of Illinois
2 Power in the Wind Consider the kinetic energy of a packet of air with mass m moving at velocity v 1 KE mv Divide by time and get power 1 m passing though A Power through area A v t The mass flow rate is (r is air density) m passing though A m= = ra v t 1
3 Power in the Wind Combining previous equations, 1 Power through area A ra vv 1 3 PW ra v (6.4) Power in the wind P W (Watts) = power in the wind ρ (kg/m 3 )= air density (1.5kg/m 3 at 15 C and 1 atm) A (m )= the cross-sectional area that wind passes through v (m/s)= wind speed normal to A (1 m/s =.37 mph)
4 Power in the Wind (for reference solar is about 600 w/m^ in summer) Power increases with the cube of wind speed Doubling the wind speed increases the power by eight Energy in 1 hour of 0 mph winds is the same as energy in 8 hours of 10 mph winds Nonlinear, so we cannot use average wind speed 3
5 Power in the Wind 1 3 PW ra v Power in the wind is also proportional to A For a conventional HAWT, A = (π/4)d, so wind power is proportional to the blade diameter squared Cost is somewhat proportional to blade diameter This explains why larger wind turbines are more cost effective (plus, as we shall see, because they are higher, the winds are stronger) 4
6 Example: Energy in 1 m of Wind 1 3 Energy ra v t 100 hours of 6 m/s winds 1 3 Energy 1.5 kg/m (1m ) 6 m/s h=13,30 Wh 50 hours of 3 m/s winds and 50 hours of 9 m/s winds - *the average wind speed is 6 m/s 1 3 Energy (3 m/s) 1.5 kg/m (1m ) 3 m/s 3 50 h=87 Wh 1 3 Energy (9 m/s) 1.5 kg/m (1m ) 9 m/s 3 50 h=,36 Wh Don t use average wind total = 3,15 Wh speed! 5
7 Air Density for Different Temperatures and Pressures r P M.W. 10 RT 3 P = absolute pressure (atm) M.W. = molecular weight of air (g/mol) = 8.97 g/mol T = absolute temperature (K) R = ideal gas constant = m3 atm K-1 mol-1 Air density is greater at lower temperatures For example, in comparing 90º F (305 K) to 10º F 65.3 K), ratio is about
8 Air Density Altitude Correction dp P( z dz) P( z) grdz dp dz rg We get a differential equation in terms of pressure: dp P dz H P1 atm e where H is in meters 7
9 Air Density Temperature and Altitude Impacts Variation in density with respect to temperature and altitude is given by r 353.1exp( z/ T) T kg/m 3 where T is in kelvins (K) and z is in meters above sea level With z=0, T= then ρ = 1.5kg/m 3 With z=00, T= , then ρ = 1.5kg/m (about 91% of sea level, 15 degree C value) 8
10 Impact of Elevation and Earth s Roughness on Windspeed Since power increases with the cube of wind speed, we can expect a significant economic impact from even a moderate increase in wind speed There is a lot of friction in the first few hundred meters above ground smooth surfaces (like water) are better Wind speeds are greater at higher elevations tall towers are better Forests and buildings slow the wind down a lot 9
11 Characterization of Elevation and Earth s Roughness on Wind Speed v v H H 0 0 α = friction coefficient given in Table 6.3 v = wind speed at height H v 0 = wind speed at height H 0 (H 0 is usually 10 m) Typical value of α in open terrain is 1/7 For a large city, α = 0.4; for small town, α = 0.3, for high crops,, α = 0., for calm water or hard ground, α =
12 Friction coefficient 0
13 Impact of Elevation and Earth s Roughness on Wind speed Alternative formulation (used in Europe) v ln( / ) H l v ln( H / l) 0 0 l is the roughness length given in Table 7. Note that both equations are just approximations of the variation in wind speed due to elevation and roughness the best thing is to have actual measurements 11
14 Roughness length 0
15 Impact of Elevation and Earth s Roughness on Power in the Wind Combining earlier equations we get 3 3 P v H = P0 v0 H0 The other constants in the power in the wind equation are the same, so they just cancel: P P 0 1 r 1 r Av Av
16 Impact of Elevation and Earth s Roughness on Windspeed Figure 7.16 For a small town, windspeed at 100 m is twice that at 10 m Areas with smoother surfaces have less variation with height 13
17 Example Rotor Stress Wind turbine with hub at 50-m and a 30-m diameter rotor, α = 0. Find the ratio of power in the wind at highest point to lowest point 50 m 65 m P P = m Power in the wind at the top of the blades is 45% higher! Picture may not be to scale 14
18 Maximum Rotor Efficiency Two extreme cases, and neither makes sense- Downwind velocity is zero turbine extracted all of the power Downwind velocity is the same as the upwind velocity turbine extracted no power Albert Betz There must be some ideal slowing of the wind so that the turbine extracts the maximum power 15
19 Maximum Rotor Efficiency Constraint on the ability of a wind turbine to convert kinetic energy in the wind into mechanical power Think about wind passing though a turbine- it slows down and the pressure is reduced so it expands Figure
20 Power Extracted by The Blades 1 P b m v v d (7.1) ṁ = mass flow rate of air within stream tube v = upwind undisturbed wind speed v d = downwind wind speed From the difference in kinetic energy between upwind and downwind air flows 17
21 Determining Mass Flow Rate Easiest to determine at the plane of the rotor because we know the cross sectional area A Then, the mass flow rate is m r Av b (7.) Assume the velocity through the rotor v b is the average of upwind velocity v and downwind velocity v d v b v vd = m v v r A d 18
22 Power Extracted by the Blades Then 1 v vd Pb r A v v Define d (7.3) v d v, will be less than 1.0 (7.4) Then substituting for v d to get the power extracted 1 v v P b ra v v (7.5) 19
23 Power Extracted by the Blades 1 v v P b ra v v Pb v v = v - v + v v v - v 3 v = v = r Av P W = Power in the wind C P = Rotor efficiency 0
24 Maximum Rotor Efficiency Find the wind speed ratio λ that maximizes the rotor efficiency, C P From the previous slide C P 1 1 = Set the derivative of rotor efficiency to zero and solve for λ: C P C P C P = =3 1 0 = maximizes rotor efficiency 1
25 Maximum Rotor Efficiency Plug the optimal value for λ back into C P to find the maximum rotor efficiency: C P = 59.3% (7.9) The maximum efficiency of 59.3% occurs when air is slowed to 1/3 of its upstream rate Called the Betz efficiency or Betz law
26 Maximum Rotor Efficiency Rotor efficiency C P vs. wind speed ratio λ 3
27 Tip-Speed Ratio (TSR) Efficiency is a function of how fast the rotor turns Tip-Speed Ratio (TSR) is the speed of the outer tip of the blade divided by wind speed Rotor tip speed rpm D Tip-Speed-Ratio (TSR) = (7.30) Wind speed 60v D = rotor diameter (m) v = upwind undisturbed wind speed (m/s) rpm = rotor speed, (revolutions/min) One meter per second =.4 miles per hour 4
28 Tip-Speed Ratio (TSR) TSR for various rotor types If blade turns too slow then wind passes through without hitting blade; too fast results in turbulence Rotors with fewer blades reach their maximum efficiency at higher tip-speed ratios Figure 7.18 A higher TSR is needed when there are fewer blades 1
29 Example 40-m wind turbine, three-blades, 600 kw, wind speed is 14 m/s, air density is 1.5 kg/m 3 a. Find the rpm of the rotor if it operates at a TSR of 4.0 b. Find the tip speed of the rotor c. What gear ratio is needed to match the rotor speed to the generator speed if the generator must turn at 1800 rpm? d. What is the efficiency of the wind turbine under these conditions?
30 Example a. Find the rpm of the rotor if it operates at a TSR of 4.0 Rewriting (7.30), Tip-Speed-Ratio (TSR) 60v rpm D 4.060sec/min 14m/s rpm = 6.7 rev/min 40m/rev We can also express this as seconds per revolution: 6.7 rev/min rpm = rev/sec or.4 sec/rev 60 sec/min 3
31 Example b. Tip speed From (7.30): rpm D Rotor tip speed= 60 sec/min Rotor tip speed = (rev/sec) D Rotor tip speed = rev/sec 40 m/rev = 55.9 m/s c. Gear Ratio Generator rpm 1800 Gear Ratio = = = 67.4 Rotor rpm 6.7 4
32 Example d. Efficiency of the complete wind turbine (blades, gear box, generator) under these conditions From (7.7): PW A v = kw 4 Overall efficiency: 600 kw 8.4% 11 kw 5
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