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 Where 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 straight-line graph. Note that 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