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ECE 333 Homework 4 Solutions :: Spring 2010 1 A horizontal-axis wind turbine with rotor 20 meters in diameter is 30-percent efficient in 10 m/s winds at 1 atmosphere of pressure and 15o C temperature. a. How much power would it produce in those winds? # 1 0.3"1 3 2 3 !"# $ A v ! !"# P wind !"# 1.225 20 10 ! P 57.73 kW P !"# 2 2 4 b. Estimate the air density on a 2500-meter 2500 -meter mountaintop at 10o C. From Table Table 6.1 the density at 10o C is 1.247 kg/m3, yielding a K T ratio of 1.247/1.225 = 1.018. Using (6.13), the density ratio, K A, can be found: 0.9109 "4 ! 0.743 P P 0 exp !"1.185x10 #2500 $% 0.9109 & K A! 1.225 The air density at 2500-meters can now be found using (6.14): 3 0.927 kg m "!1.225 K T K A!1.225"1.018"0.743 # Note that (6.7) could also be used to find the same solution. c. Estimate the power the turbine would produce on that mountain with the same windspeed assuming its efficiency is not affected by air density den sity.. Turbine power is proportional to air density therefore: 0.927 43.7 kW " P P 57.73 kW ! 1.225 !
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2 An anemometer mounted 10 m above a surface with crops, hedges and shrubs, shows a windspeed of 5 m/s. Assuming 15o C and 1 atm pressure. Determine the following for a wind turbine with hub height 60 m and a rotor diameter of 60 m: a. Estimate the windspeed and the specific power in the wind (W/m2) at the highest point that a rotor blade reaches. The windspeed (6.15) and specific power at 90 m: 0.20 1 90 2 3 3 P 90! " v !0.5"1.225"7.759 # P 90 286 W m 7.759 m s v 5 !v 90 90 2 10 b. Repeat (a) at the lowest point that a rotor blade falls to. The windspeed and specific power at 30 m: 0.20 1 30 2 3 3 ! " v !0.5"1.225"6.229 # P 30 148 W m P 6.229 m s v 5 !v 30 30 30 2 10 c. Compare the ratio of wind power at the two elevations using the results from (a) and (b) and compare that with the ratio obtained using (6.17). Ratio of wind power from (a) and (b): 2 286 W ! m 1.93 2 148 W ! m Ratio of wind power using (6.17):
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90 1.93 !" 30 P 0 H 0 The ratios using (a) and (b) and (6.17) agree. P
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3 Consider the probability density function shown in Figure P6.3 for wind speed. a. What is an appropriate value of k for this to be a legitimate probability density function? In order for the plot to be a probability density function, the area under the curve must equal 1. Since the plot is a simple triangle, it is easy to find the area: 1 1 A bh 10 k 1 ∴ k=0.2 2 2 !
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So f ! V !" 0.02 V ! 0 "V !10 m ! s ! b. What is the average power in these winds (W/m2) under standard (15o C, 1 atm) conditions? 1 3 3 3 P AVG ! " A ! V ! AVG note that V AVG!" V ! AVG 2 10 # 5 10 3 3 3 ! V ! AVG !" f ! V !"V dV !" 0.02V#V dV ! 0.02 " 400 5 0 0 1 2 P AVG ! "1.225"400 # P AVG 245 W m 2 4 Suppose the wind probability density function is just a constant over the 5 to 20 m/s range of windspeeds as shown in Figure P6.4. The power curve for a small 1 kW wind turbine is also shown in the figure. a. What is the probability that the wind is blowing between 5 and 15 m/s? This can be found by taking the fraction of the total area between 5 and 15 m/s or 2/3. Note you could also find the same solution by integrating the pdf from 5 to 15 m/s. b. What is the annual energy that the wind turbine would generate? Since the wind only blows between 5 and 20 m/s and the wind turbine only responds to winds below 15 m/s, we only need to consider the probability that the wind blows between 5 and 15 m/s, found in (a) to be 2/3. 2 10 8760 hr ! yr !1 kW 5840 kWh yr 2.1 x 10 J ! yr E 3 c. What is the average power in the wind? 1 3 P AVG ! " A ! V ! AVG Similar to 3a, the area under the curve must be 1 ∴ k = 1/15. 2 20 # 1 3 1 3 3 4 4 ! V ! AVG !" f ! V !"V dV !" V dV ! " 20 # 5 $! 2656 m ! s 15 60 0 5 1 2 P AVG ! "1.225"2656 # P AVG 1626 W m 2 !
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All tables, figures and equations are taken from chapter 6 of [1]. References [1] G. M. Masters, Renewable and Efficient Electric Power Systems. Hoboken, NJ, John Wiley & Sons, Inc., 2004, pp. 307-378.
