293510 4 2 3483\u03a9 27 10 8 \u03a9 14210 6 293510 4 2 1306\u03a9 These resistors are

# 293510 4 2 3483ω 27 10 8 ω 14210 6 293510 4 2

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2.935×10 −4 ? 2 = 3483Ω ? ???? = 𝜌 ???? × ? 𝐴 = 2.7 × 10 −8 Ω ⋅ ? × 1.42×10 6 ? 2.935×10 −4 ? 2 = 130.6Ω These resistors are arranged in parallel: ? ????? = ( 1 3484Ω + 1 130.6Ω ) −1 = 125.9Ω There are 8 cables in parallel: ? ????? = ( 1 ? ???𝑙𝑒 × 8) −1 = ( 1 125.9Ω × 8) −1 = 15.7Ω ≈ 16 Ω Alternative Method accepted: Calculate the cross-area of each metal for all 8 cables, and then calculate the total resistance of the transmission line, treating the two metals as two resistors in parallel. b. How much current flows through the transmission lines? What is the power loss in the line? What is the percentage of power loss relative to the power generated (1890 MW)? [10 points] ANS-v1: The current 𝐼 = ????? ?????𝑔? = 1.89×10 9 ? 5.33×10 5 ?−(−5.33×10 5 ?) = 1773𝐴 ≈ 1770 A The power loss ? ???? = 𝐼 2 ? = (1773𝐴) 2 × 63.0Ω = 198?𝑊 ≈ 2.0x10 2 MW (or 200 MW) The percentage of power loss 198?? 1890?? = 10.4% ≈ 10.% [10.% to 11.% are fine due to numerical calculations] ANS-v2: The current 𝐼 = ????? ?????𝑔? = 1.89×10 9 ? 5.33×10 5 ? = 3546𝐴 ≈ 3550 A The power loss ? ???? = 𝐼 2 ? = (3546𝐴) 2 × 15.7Ω = 197?𝑊 ≈ 2.0x10 2 MW (or 200 MW) The percentage of power loss 197?? 1890?? = 10.4% ≈ 10.% c. What would the losses be if the power were transmitted using 333 kV AC transmission lines, which are popular for shorter distance transmission in Southern Africa (assume the same cables are used). What is the percentage of power loss relative to the power generated (1890 MW)? [10 points] ANS: Assuming the 8 cables are in parallel. You should get: Resistance = 15.7Ω Current = 𝐼 = ????? ?????𝑔? = 1.89×10 9 ? 3.33×10 5 ? = 5676𝐴 The power loss ? ???? = 𝐼 2 ? = (5676𝐴) 2 × 15.7Ω = 506?𝑊 ≈ 510 MW Percentage loss = 506?? 1890?? = 26.8% ≈ 27% (lower transmission voltage results in larger power loss during transmission)
3. Innovation of solar photovoltaic technology: the learning curve & PURPA [40 points for ER100/PP184 and 50 points for ER200/PP284] The cost of solar photovoltaic technology has declined dramatically over the past 40 years, for the silicon wafer based modular solar panels (PV modules) (the blue dots in the figure). Over the past decade, First Solar, a top U.S. solar manufacturer, championed the deployment of cheaper thin film PV modules (the red triangles in the figure). The learning curve model can be used to describe the cost reduction in relation with the increasing cumulative installation: the more we install PV modules, the better and cheaper we can make them. According to the learning curve for the Si wafer PV modules, the cumulative installation by the end of 1985 is 107 MW, and the price (in US\$2010) is \$12.526/W. The model suggests that the price falls to \$1.455/W by the end of 2011, with 56.872 GW of cumulative installation. According to the learning curve for the First Solar’s thin film PV modules, the cumulative installation by the end of 2006 is 110. MW, and the manufacturing cost (in US\$2010) is \$1.586/W. The model suggests the price falls to \$0.614/W by the end of 2013, with 8.823 GW of cumulative installation. Note: The data for thin film PV modules is presented in cost, while that for Si wafer is presented in price. The learning curve model for Si wafer PV modules uses prices, whereas the learning curve model for thin film PV modules uses cost.

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• Summer '19
• Photovoltaics, pUC, pv modules, Si wafer PV