Question 6.12: Energy Potential for a Wind Farm. Suppose that a wind farm h......
Energy Potential for a Wind Farm. Suppose that a wind farm has 4-rotor-diameter tower spacing along its rows, with 7-diameter spacing between rows (4D × 7D). Assume 30% wind turbine efficiency and an array efficiency of 80%.
a. Find the annual energy production per unit of land area in an area with 400-W/m² winds at hub height (the edge of 50 m, Class 4 winds).
b. Suppose that the owner of the wind turbines leases the land from a rancher for $100 per acre per year (about 10 times what a Texas rancher makes on cattle). What does the lease cost per kWh generated?
a. As the figure suggests, the land area occupied by one wind turbine is 4D × 7D = 28D², where D is the diameter of the rotor. The rotor area is (π/4)D². The energy produced per unit of land area is thus
\begin{matrix} \frac{\text{Energy}}{\text{Land area}} & = \ \frac{1}{28D^{2}}\left(\frac{\text{Wind turbine}}{m^{2} \ \text{land}}\right) \ \cdot \ \frac{\pi }{4} D^{2} \ \left(\frac{m^{2} \ \text{rotor}}{\text{Wind turbine}} \right) \\ & \quad \quad \quad \times \ 400 \ \left(\frac{W}{m^{2} \ \text{rotor}}\right) \ \times \ 0.30 \ \times \ 0.80 \ \times \ 8760 \frac{h}{yr} \\ \frac{\text{Energy}}{\text{Land area}} & = \ 23, \ 588 \left(\frac{W \ \cdot \ h}{m^{2} \ \cdot \ yr}\right) \ = \ 23.588 \left(\frac{kWh}{m^{2} \ \cdot \ yr}\right) \quad \end{matrix}b. At 4047 m² per acre, the annual energy produced per acre is:
\frac{\text{Energy}}{\text{Land area}} \ = \ 23.588 \frac{kWh}{m^{2} \ \cdot \ yr} \ \times \ \frac{4047 \ m^{2}}{\text{acre}} \ = \ 95, \ 461 \frac{kWh}{\text{acre} \ \cdot \ yr}so, leasing the land costs the wind farmer:
\frac{\text{Land cost}}{kWh} \ = \ \frac{\$ 100}{\text{acre} \ \cdot \ yr} \ \times \ \frac{\text{acre} \ \cdot \ yr}{95, \ 461 \ kWh} \ = \ {\$ 0.00105}/{kWh} \ = \ 0.1 \ {\cancel{c}}/{kWh}