a. To find the annual energy delivered, a spreadsheet solution is called for. Let’s do a sample calculation for a 6-m/s windspeed to see how it goes, and then present the spreadsheet results.
From Table 6.7, at 6 m/s the NEG Micon 1000/60 generates 150 kW.
From (6.45), the Rayleigh p.d.f. at 6 m/s in a regime with 7-m/s average windspeed is

f\left(v\right) \ = \ \frac{\pi \ v}{2 \bar{v}^{2}} \ \exp \ \left[-\frac{\pi }{4} \ \left(\frac{v}{\bar{v}}\right)^{2} \right] \ = \ \frac{\pi \ \cdot \ 6}{2 \ \cdot \ 7^{2}} \ \exp \ \left[- \frac{\pi}{4} \ \left(\frac{6}{7}\right)^{2}\right] \ = \ 0.10801

In a year with 8760 h, our estimate of the hours the wind blows at 6 m/s is

\text{Hours} \ @ 6 \ {m}/{s} \ = \ 8760 \ {h}/{yr} \ \times \ 0.10801 \ = \ 946 \ {h}/{yr}

So the energy delivered by 6-m/s winds is

\text{Energy} \ \left(@ 6 \ {m}/{s}\right) \ = \ 150 \ kW \ \times \ 946 \ {h}/{yr} \ = \ 141, \ 929 \ {kWh}/{yr}

The rest of the spreadsheet is given below. The total energy produced is 2.85 × 10^{6} kWh/yr.

b. The average efficiency is the fraction of the wind’s energy that is actually converted into electrical energy. Since Rayleigh statistics are assumed, we can use (6.48) to find average power in the wind for a 60-m rotor diameter (assuming the standard value of air density equal to 1.225 kg/m³):

In a year with 8760 h, the energy in the wind is

\text{Energy in wind} \ = \ 8760 \ {h}/{yr} \ \times \ 1134 \ kW \ = \ 9.938 \ \times \ 10^{6} \ kWh

So the average efficiency of this machine in these winds is

\text{Average efficiency} \ = \ \frac{2.85 \ \times \ 10^{6} \ {kWh}/{yr}}{9.938 \ \times \ 10^{6} \ {kWh}/{yr}} \ = \ 0.29 \ = \ 29 \%

c. The productivity (annual energy per swept area) of this machine is

\text{Productivity} \ = \ \frac{2.85 \ \times \ 10^{6} \ {kWh}/{yr}}{\left({\pi}/{4}\right) \ \cdot \ 60^{2} \ m^{2}} \ = \ 1008 \ {kWh}/{m^{2}} \ \cdot \ yr