Question 1. The table below gives the average number of hours each year of mean wind speeds at 10m height at an open country site near Auckland where the roughness length z0= 0.02 m. The data is based upon statistics from over 20 years or records. The number of hours of each level includes those ±0.5 m/s of that level. For example the hours for a 9 m/s mean wind includes all those where 8.5£V<9.5 m/s.
| V(m/s) |
Hrs |
| 0 |
1084 |
| 1 |
280 |
| 2 |
518 |
| 3 |
1207 |
| 4 |
1158 |
| 5 |
1054 |
| 6 |
971 |
| 7 |
722 |
| 8 |
690 |
| 9 |
431 |
| 10 |
323 |
| 11 |
135 |
| 12 |
78 |
| 13 |
65.5 |
| 14 |
22.5 |
| 15 |
13.5 |
| 16 |
4 |
| 17 |
2 |
| 18 |
1 |
| 19 |
0.5 |
| Total |
8760 |
The general distribution of the data may be fitted by a Weibull distribution, which takes the form Q (> V) is the annual probability that a mean wind speed V is exceeded. The constant A is the fraction of the year when there is at least some wind (in this case A = 1 - 1084/(365*24) = 0.876) and the constants C and k are obtained by fitting the data. One method for doing this is to calculate the proportion of the year when each wind speed is exceeded (Q(>V)). For example Q(>7.5 m/s) can be determined by adding the hours for wind speeds 8, 9, 10 ...19 and dividing by the total hours in a year (8760 hrs) . Once data for > 0.5, > 1.5, > 2.5 m/s etc has been calculated plot ln(ln(A)ln(Q(>V))) against ln(V). It may be noted that taking double logarithms of the Weibull equation gives which means that we expect a straightline graph.
The gradient of the linear fit is k and the y axis intercept -kln(C). Fit a Weibull distribution to the data in the table and hence determine
k and C. Include your graph of ln(ln(A)ln(Q(>V))) against ln(V). In addition plot a graph which compares the data supplied with the equivalent data obtained from the fitted Weibull.
Note that the number of hours of wind speeds around 12 m/s can be calculated from
Question 2 Wind turbine calculations
A 90m diameter horizontal axis wind turbine is being considered for the site mentioned in Question 1. The turbine has a cut-in mean wind speed of 4.5 m/s and a rated wind speed of 11 m/s. Assume that between the cut in wind speed and the rated wind speed the overall power coefficient is constant at 0.35. Above the rated wind speed the power output is maintained at the rated level. The cut-out mean wind speed is 25 m/s. Take air density as r = 1.2 kg/m3.
a) Determine the rated power output of this turbine.
b) Two tower heights are under consideration, one will place the hub at a height of 80 m, the other at a height of 100 m. Calculate the capacity factors and annual energy output (in kWh) with these two options. Your answer must show clearly how you have calculated these.
c) Placing the hub at the higher level will obviously incur additional cost. What percentage increase in overall cost do you think can be justified based on the increased power generation? Explain your reasoning.
Notes:
1. You may assume a logarithmic velocity profile with z0= 0.02 m applies at all relevant heights.
2. For the same site, a Weibull distribution determined from measurements at one height may be modified to give that at a different height by using the same values for A and k, but modifying the value of C in proportion to the mean velocity at the two heights.