# Question 6.18: A Loan to Pay for a Small Wind Turbine. Suppose that a 900-W......

A Loan to Pay for a Small Wind Turbine. Suppose that a 900-W Whisper H900 wind turbine with 7-ft diameter (2.13 m) blade costs $1600. By the time the system is installed and operational, it costs a total of$2500, which is to be paid for with a 15-yr, 7 percent loan. Assuming O&M costs of $100/yr, estimate the cost per kWhr over the 15-year period if average windspeed at hub height is 15 mph (6.7 m/s). Step-by-Step The 'Blue Check Mark' means that this solution was answered by an expert. Learn more on how do we answer questions. The capital recovery factor for a 7%, 15-yr loan would be $CRF\left(0.07, \ 15 \ yr\right) \ = \ \frac{i\left(1 \ + \ i\right)^{n}}{\left(1 \ + \ i\right)^{n} \ – \ 1} \ = \ \frac{0.07\left(1 \ + \ 0.07\right)^{15}}{\left(1 \ + \ 0.07\right)^{15} \ – \ 1} \ = \ {0.1098}/{yr}$ which agrees with Table 5.5. So, the annual payments on the loan would be $A \ = \ P \ \times \ CRF\left(0.07, \ 15\right) \ = \ \ 2500 \ \times \ {0.1098}/{yr} \ = \ {\ 274.49}/{yr}$ The annual cost, including$100/yr of O&M, is therefore $274.49 +$100 = $374.49. To estimate energy delivered by this machine in 6.7-m/s average wind, let us use the capacity factor approach (6.65): $CF \ = \ 0.087 \bar{V} \ – \ \frac{P_{R}}{D^{2}} \quad \left(\text{Rayleigh winds}\right)$ (6.65) $CF \ = \ 0.087 \bar{V} \ – \ \frac{P_{R}\left(kW\right)}{D^{2}\left(m^{2}\right)} \ = \ 0.087 \ \times \ 6.7 \ – \ \frac{0.90}{2.13^{2}} \ = \ 0.385$ The annual energy delivered (6.59) $\text{Annual energy} \ \left({kWh}/{yr}\right) \ = \ P_{R} \ \left(kW\right) \ \times \ 8760 \ \left({h}/{yr}\right) \ \times \ CF$ (6.59) ${kWh}/{yr} \ = \ 0.90 \ kW \ \times \ 8760 \ {h}/{yr} \ \times \ 0.385 \ = \ 3035 \ {kWh}/{yr}$ The average cost per kWh is therefore $\text{Average cost} \ = \ \frac{\text{Annual cost} \ \left({\}/{yr}\right)}{\text{Annual energy} \ \left({kWh}/{yr}\right)} \ = \ \frac{{\ 374.49}/{yr}}{3035 \ {kWh}/{yr}} \ = \ {\ 0.123}/{kWh}$ That’s a pretty good price of electricity for a small system—cheaper than grid electricity in many areas and certainly cheaper than any other off-grid, home-size generating system.  TABLE 5.5 Capital Recovery Factors as a Function of Interest Rate and Loan Term Years 3% 4% 5% 6% 7% 8% 9% 10% 11% 12% 13% 5 0.2184 0.2246 0.2310 0.2374 0.2439 0.2505 0.2571 0.2638 0.2706 0.2774 0.2843 10 0.1172 0.1233 0.1295 0.1359 0.1424 0.1490 0.1558 0.1627 0.1698 0.1770 0.1843 15 0.0838 0.0899 0.0963 0.1030 0.1098 0.1168 0.1241 0.1315 0.1391 0.1468 0.1547 20 0.0672 0.0736 0.0802 0.0872 0.0944 0.1019 0.1095 0.1175 0.1256 0.1339 0.1424 25 0.0574 0.0640 0.0710 0.0782 0.0858 0.0937 0.1018 0.1102 0.1187 0.1275 0.1364 30 0.0510 0.0578 0.0651 0.0726 0.0806 0.0888 0.0973 0.1061 0.1150 0.1241 0.1334 ## Related Answered Questions Question: 6.15 ## Annual Energy Delivered Using a Spreadsheet. Suppose that a NEG Micon 60-m diameter wind turbine having a rated power of 1000 kW is installed at a site having Rayleigh wind statistics with an average windspeed of 7 m/s at the hub height. a. Find the annual energy generated. b. From the result, find ... ## Verified Answer: a. To find the annual energy delivered, a spreadsh... Question: 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 ...

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