Wind Energy Systems MASE 5705 Spring 2017, Feb. 1,6: L5 + L6

1. Key points of L3 +L4 2a. WT output estimation for ideal conditions and for Weibull- distribution 2b. HW2 3. Bin Width, and Calculation of bin frequency or probability and the corresponding pdf 4. Estimation of AEO* of a proposed site 5. AEO Estimate of a Specific WT in a proposed site 6. Review of Ch. 2 through applications to a realistic database (Mod-2 2.5 MW WT) 7. Comments on HW 1 involving graphical solution of Weibull Parameters 8. HW. 4 *AEO ≡ Annual Energy Output (kWh/y)

1. Key points of L3 + L4

Weibull Distribution p. 60 (2.60-61)

Finding Weibull Parameters k and c Least squares fit for the observed values of bin frequency fi (e.g. subroutine: lsqcurvefit), (not in the text.) Analytical–Empirical Approach (pp. 60-61) Graphical Approach (Lect. 3-4, pp. 81-84) (Lect. 5-6, pp. 98-103)

p. 60 (2.66) Eq. (2.66) page 60, 2nd edition

Energy Pattern Factor (2.69) p. 61 Eq. (2.69), p. 61, 2nd Edition

Eq. (2.69), p. 61, 2nd Edition

p. 61 Rayleigh p. 61 in 2nd Edition

Variation of parameters with Weibull k shape factor Table 2.4 p. 61 k Ke 1.2 0.837 3.99 2 0.523 1.91 3 0.363 1.40 5 0.229 1.15

Why bother about σU/Ū and Ke? σU/Ū = Turbulence intensity ( L7 + L8, 2-7-12) = a measure of turbulence impact on 1) extracted power from a proposed site, 2) random loads of WT Ke = Energy pattern factor = very useful in calculating annual energy Output (AEO) of a proposed site (an example follows)

Rayleigh distribution A special case of Weibull Distribution with shape factor k = 2 # 2.58-2.59 (p. 59)

Eqs. (2.59), (2.66), pp. 59-60, 2nd edition

Rayleigh Distribution (2.58-2.59) p. 59 Eqs. (2.58-2.59) p. 59, 2nd Edition

2a. WT Output Estimation for Ideal Conditions and for Weibull Distribution

PROJECT #1 Spring 2017 (Due Feb. 20) A PROJECT #1 Spring 2017 (Due Feb. 20)   A. For each of the two data bases for Dodge City and for Kansas, compute the following from the data only: 1.) Mean wind velocity, U-bar 2.) Standard deviation of wind velocity, U 3.) Mean of the wind velocity cubed (NOT the cube of the mean!). 4.) k and c for Wiebull distribution using the approximate analytical method. 5.) The Wiebull estimation of the mean of velocity cubed. B. Assume that the first bin (1 m/sec) has a range of 0 to 1.5 m/sec. Assume that bins 2-27 have a range of V-mid +/- 0.5 m/sec, and assume that bin 28 has a range of 27.5 m/sec to infinity. With that assumption, for each of the two data sets: 1.) Compare the fi(U) of each bin with the prediction from Wiebull. 2.) With your U-bar, U, k, and c for the Wiebull distribution––and assuming a COP of 40%, a cut-in speed of 2.5 m/sec, and a cut-out speed of 22.5 m/sec––compute the estimate of power that you will generate in one year at each site.

Dodge City Eastern Kansas (m/sec, m) 1 47 2 5 3 82 4 65 5 140 6 219 7 266 8 276 9 198 10 314 11 155 12 177 13 141 14 142 15 133 16 96 17 102 18 101 19 48 20 78 21 28 22 21 23 23 24 12 25 19 26 8 27 2 28 9 1 300 2 161 3 127 4 261 5 188 6 294 7 151 8 347 9 125 10 376 11 67 12 207 13 67 14 91 15 29 16 51 17 19 18 39 19 1 20 7 21 0 22 2 23 0 24 1 25 1 26 0 27 0 28 0

1. Ideal Conditions: 1a. Maximum WT efficiency (Betz factor) mechanical . electrical =  = 1 (ideal transmission) (physics of Betz factor Cp,max ; derivation in ch.3) 1b. Carlin’s one-two-three equation (2.82), p. 64 1c. Wind speed that gives maximum power for Weibull distribution.

  p. 63 (2.75) 2nd Edition, (2.75), page 63 1/2 𝜌𝐴(𝑈 ̅ )^3

U2= U3 U4 U1=Ū (1) (2) (3) (4) U1=Ū U2= U3 U4 = far wake velocity (4)   U2= U3 U4 U1=Ū (1) (2) (3) (4) U1=Ū U2= U3   U4 = far wake velocity   Force on disk Force on fluid (4)

In Chapter 3, we will show that for optimum conditions      

    = 2A2

Protor = Pextracted by the rotor = Pinput – P“output” lost in far wake Protor = P1 – P4  

Ideal Power

Power =

An Ideal Machine η (2.72) P. 63

Carlin’s 1-2-3 Equation (2.69, p. 61)  

Carlin’s One-Two-Three Equation (2.82 p. 64) (2.5.10) p. 61

B.2.8 Annual Power Estimation Problem-Betz Type Machine (p. 618) Estimate the annual production of a 12 m diameter horizontal axial wind turbine which is operating at the standard atmospheric conditions ( = 1.225 kg/m3 ) in a 8 m/s average wind speed regime. You are to assume that the site wind–speed probability density is given by the Rayleigh density distribution.

Pr. 2.8 , p. 618 (2.82), p. 64   (2.5.10) p. 61

= 40.14 x 24 x 365 = 40.14 x 8760 = 352,000 kWh/y =352 MW/year year * kWh year = 40.14 x 24 x 365 = 40.14 x 8760 = 352,000 kWh/y =352 MW/year

2b. Homework 2

2b. Homework (not 2017) Find: Use:  

Hints (not really!) and solve: 1/k

k should be 1/k (show that it is true)

In other words:

3. Bin Width, and Calculation of Bin probability and the corresponding pdf

The probability density function f(U) has the following fundamental properties

4. Estimation of Average Energy Output (AEO) of a proposed site

Estimation of AEO (kWh/y) of a proposed site with wind speed Bin Data or Wind Speed Histogram Bin width (e.g. 0.5 m/s) For each bin, Umax, Umin and number of hours in that bin are measured.

Wind Data Distribution for Hypothetical Site 0.9998 We used this database earlier, in estimating Ū for the site. Now, we use the same database for site AEO estimation, neglecting bin 16 ( = 1.15 kg/m3). (From: Energy Systems Engineering, F.M. Vanek and L.D. Albright, McGraw Hill, 2008, with permission)

Method of Bins for site AEO estimation We will compute available power (η = 1).

Remarks The preceding method of bins is the most reliable and it is always applicable. This method, however, requires a comprehensive database involving hourly measurements for at least one year for each bin: Umax, Umin and number of hours. For several sites only Ū is available.

When only Ū is available, the Rayleigh approximation of the number of hours in each bin is found to be satisfactory for several sites. In this case, we create our own bins and estimate the number of hours for each bin from Rayleigh.

2nd Edition page 61

Let us apply these two approximate methods for the same earlier-treated hypothetical site (Ū = 5.57 m/s) .

The analyst-created Bin Data for Hypothetical site when only Ū = 5 The analyst-created Bin Data for Hypothetical site when only Ū = 5.57 m/s is known,  = 1.15 kg/m3.

AEO Estimation for Hypothetical Site ,

Explanation Consider bin 6: Umin = 4 m/s , Umax = 5 m/s, Ū6 = 4.5 m/s Calculate the probability that the wind-speed measurements belong to this bin, and then the number of hours in that bin.

* More accurate calculations follow Comparison of Annual Output from Observed Bin Data and from Rayleigh-Estimated Bin Hours Observed (kWh/m 2 ) Estimated (kWh/m 1 0.0 3 1.0 1.2 4 7.2 8.5 5 29.9 27.9 6 65.7* 62.4 7 119.2 108.2 8 162.2 155.9 9 172.0 194.2 10 193.9 214.1 11 218.4 212.2 12 218.3 191.2 13 193.3 158.0 14 139.3 120.5 15 84.9 85.1 Total 1605.3 1539.4 * 65.7 = (52.4 x 1254)/1000 (Bin 16 is not included) Annual Output BIN * More accurate calculations follow

Repeat : BIN # 1/2 Ū (hr/year) 1 80 2 0.07 204 1.94 496 4 8.98 806 5 Observed Dynamic Power Observed Duration 1/2 Ū 3 (hr/year) 1 80 2 0.07 204 1.94 496 4 8.98 806 5 24.65 1211 6 52.4 1254 7 95.67 1246 8 157.91 1027 9 242.58 709 10 353.12 549 11 492.99 443 12 665.63 328 13 874.5 221 14 1123.05 124 15 1414.72 60 16 8984.38 8760

Observed and Estimated AEO of Hypothetical Site

AEO (kWh/m2) Bin Number

Site AEO (kWh/m2): Observed = 1605 Rayleigh-Estimated = 1539 (4.1%) EPF-based = 1662 (3.6%) For this hypothetical site, these two approximations provide excellent results. After all, yearly wind-speed averages show ±10% variations.

5. AEO estimate of a specific WT in a proposed site

5b. For a proposed site for which only Ū is 5b. For a proposed site for which only Ū is measured or known and the number of hours for each bin is estimated from Rayleigh distribution. (Homework 3)

A specific WT ≡ the power curve is known (the most important officially issued document for that WT)

Measurements of site Bin Data and WT power output data for Mod-2 2.5 MW WT (300 ft diameter) * * Reasons for such a wide bin width are not known. Lower values are more likely than higher values. Take Ū18= 17.25 m/s

(Measured) Power Curve for Mod-2 2. 5 MW HAWT (Uci = 6 (Measured) Power Curve for Mod-2 2.5 MW HAWT (Uci = 6.25 m/s , Uco = 22.4 m/s)

(1980) The 2.5 MW Mod-2 HAWT, the second generation turbine in the large-scale segment of the Federal Wind Energy Program.

What is Efficiency at 2.5 MW? U = 14m/sec, ρ = 1.225kg/m3, D = 91.4m (1/2)(1.225)(π/4)(91.4)2(14)3 = 11,000,000 watts = 11,000 KW = 11.0 MW COP = 2.5/11.0 = 23% η = (27/16)(.22) = 0.38

Power Curve of Wind Turbine Measured power output versus wind speed is referred to as power curve. “The power curve is a wind turbine’s official certificate of performance, which has to be guaranteed by manufacturer. This is why the accurate description and confirmation of the power curve is of special significance…”

“Wind turbines having a power curve based only on theoretical calculations are sold only in exceptional cases. As a rule, the power curve of all commercial wind turbines is measured and certified by independent institutions ...” (from: Wind Energy, Erich Hau, Springer 2006.)

Ur Uci Uco

“ The development of German 3-MW Growian 100 m dia HAWT in 1982 represented the greatest technology leap of the times, as well as the highest technological risks. … It never operated satisfactorily and was dismantled after only limited testing time.”

AEO Estimate of a Specific WT Method of Bins in a proposed site Rayleigh Distribution (only Ū is known)

AEO of a Specific WT in a Proposed Site AEO ≡ Annual Energy Output Specific WT ≡ WT with a specific power curve, and Uci and Uco are known Proposed Site ≡ Wind-speed histogram or bin data (Umax, Umin and number of hours in each bin) is known

Power Curve for a hypothetical case (for illustration) Power Curve for a 1.5 MW WT Uci = 3m/s, Uco=21 m/s, Dia = D= 69 m  (hub height) = 1.15 kg/m3 Power Output (kW)

AEO of a specific 1.5 MW WT in a proposed site (Ū = 8.4 m/s) * See next figure

Power Curve for 1.5-MW turbine for wind speeds from 0 to 25 m/s

Method of Bins

Energy-Extraction Efficiency of the 1.5 MW Wind Turbine (Depends on density.)  

Left curve, windspeed histogram Left curve, windspeed histogram. Center curve, power curve for Carter 25* . Right, energy in wind and hatched part is the energy captured by the wind turbine. * 9.8 m–dia 25 kW machine (From : Wind Characteristics , J. Rothatgi and V. Nelson, Alternative Energy Institute , West Texas A and M University, Canyon, TX, 1994)

AEO Estimate of a Specific WT in a Site for which only Ū is known.

Create your own bin data, and estimate the number of hours in each bin by using the Rayleigh distribution. (Similar exercises have been earlier treated for estimating the site potential when only Ū is known.)

Homework (not 2017) For the preceding 1.5 MW WT with the given power curve in slide 82. (Include a cut-in speed of 3 m/sec and cut-out sped of 21 m/sec. Calculate the AEO for this turbine with Ū = 8.4 m/s, and a Rayleigh wind-speed distribution. Use the bin widths from slide 81.

6. Review of Chapter 2 through applications to a realistic database (Mod-2 2.5 MW WT )

Measurements of site Bin Data and WT power output data for Mod-2 2.5 MW WT * * Reasons for such a wide bin width are not known. Lower values are more likely than higher values. Take Ū18= 17.25 m/s

8760 bin Ūi Duration Observed 1 3.13 2147 2 6.5 416 3 7 440 4 7.5 458 468 6 8.5 470 9 466 9.5 453 10 435 10.5 410 11 381 12 11.5 349 13 314 14 12.5 278 15 242 16 13.5 208 17 175 18 17.25 648 19 30 8760

7. Comments on HW1 involving graphical solution of Weibull parameters “Graphical” = Electronically graphical (Using Excel)

Text, p. 59 Graphical : log-log plot “Using this method, a straight line is drawn through a plot of wind speed, U, on the x-axis and F(U) on the y-axis of log-log paper. The slope of the straight line gives k. Then, the intersection of a horizontal line with F(U) = 0.632 gives an estimate of c on the x-axis.”

The first sentence (“Using this method …) is ambiguous. Details of F(U) = 0.632

Graphical Method (Hints) y = a x + b Finally,  k = a and c = exp (-b/k)

y = m x + b For U = c, F(U) = F(c) = 1 – 1/e = 0.632

Our approach

For the Hypothetical Site (Ū = 5.57 m/s , σU = 2.86 m/s) Weibull Parameter Least squares empirical graphical k 2.1170 2.0663 2.04 6.2925 6.291 6.125

Mod-2 2.5MW WT Database (Ū = 8.6655 m/s) Weibull Parameter Least squares empirical graphical k 2.6603 Not done 2.6763 9.7491 10.0065*  

Homework (not 2017) We have designed a HAWT and make the following claim: “Our turbine with a blade length of 40 meters achieves its rated output of 1.5 MW in winds of 20 MPH.” ( = 1.15 kg/m3) What is the flaw in this statement? Use appropriate calculations to show the flaw. (Based on energy Systems Engineering, ibid.)