Chapter 6

Q. 6.19

Price of Electricity from a Wind Farm. A wind farm project has 40 1500-kW turbines with 64-m blades. Capital costs are $60 million and the levelized O&M cost is $1.8 million/yr. The project will be financed with a $45 million, 20-yr loan at 7% plus an equity investment of $15 million that needs a 15% return. Turbines are exposed to Rayleigh winds averaging 8.5 m/s. What levelized price would the electricity have to sell for to make the project viable?

Step-by-Step

Verified Solution

We can estimate the annual energy that will be delivered by starting with the capacity factor, (6.65):

CF \ = \ 0.087 \bar{V} \ – \ \frac{P_{R}}{D^{2}} \quad \left(\text{Rayleigh winds}\right) (6.65)

CF \ = \ 0.087 \bar{V} \ \left({m}/{s}\right) \ – \ \frac{P_{R}\left(kW\right)}{\left[D\left(m\right)\right]^{2}} \ = \ 0.087 \ \times \ 8.5 \ – \ \frac{1500}{64^{2}} \ = \ 0.373

For 40 such turbines, the annual electrical production will be

\begin{matrix} \text{Annual energy} & = \ 40 \ \text{turbines} \ \times \ 1500 \ kW \ \times \ 8760 \ {h}/{yr} \ \times \ 0.373 \\ & = \ 196 \ \times \ 10^{6} \ {kWh}/{yr} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \end{matrix}

The debt payments will be

\begin{matrix} A \ = \ P \ \cdot \ \left[\frac{i\left(1 \ + \ i\right)^{n}}{\left(1 \ + \ i\right)^{n} \ – \ 1}\right] & = \ \$ 45,000,000 \ \cdot \ \left[\frac{0.07\left(1 \ + \ 0.07\right)^{20}}{\left(1 \ + \ 0.07\right)^{20} \ – \ 1}\right] \\ & = \ \$ 4.24 \ \times \ {10^{6}}/{yr} \quad \quad \quad \quad \quad \ \ \ \end{matrix}

The annual return on equity needs to be

\text{Equity} \ = \ {0.15}/{yr} \ \times \ \$ 15,000,000 \ = \ \$ 2.25 \ \times \ {10^{6}}/{yr}

The levelized O&M cost is $1.8 million, so the total for O&M, debt, and equity is

\text{Annual cost} \ = \ \$\left(4.24 \ + \ 2.25 \ + \ 1.8\right) \ \times \ 10^{6} \ = \ \$ 8.29 \ \times \ {10^{6}}/{yr}

The levelized price at which electricity needs to be sold is therefore

\text{Selling price} \ = \ \frac{\$ 8.29 \ \times \ {10^{6}}/{yr}}{196 \ \times \ 10^{6} \ {kWh}/{yr}} \ = \ \$ 0.0423 \ = \ {4.23\cancel{c}}/{kWh}