Problem Set 2 (2023)

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Student Name: Lauren Eusebio ENGN-0020 Problem Set 2 “Cape Wind” INSTRUCTIONS FOR SUBMITTING YOUR WORK • Note that each problem starts on a new page. Please leave that structure alone. • Type your answers directly into this document, leaving your answers in red text . • If you have handwritten material Sometimes it is difficult to type mathematical expressions, so we invite you to hand write those answers. However, put the handwritten work directly into this .docx document before creating the .pdf of it. The best way to do this. • Take a picture of your handwritten work with a cellphone camera, transfer the picture to your computer, and insert the picture directly into this document. • Convert this document to a .pdf format. (Different operating systems and software have different methods for accomplishing this. It is an important skill to learn if you don’t already know!) • Construct a single .pdf document with all elements of the assignment. This may require you to create a .pdf of this completed document and merge it with .pdf of other components you create. • Submit your work electronically into Canvas using Gradescope . Version 2; 2/20/2019

1) You are advising a small town that wants to install a single wind turbine to power homes in the community. The results of a wind survey are shown in the table below. The wind turbine manufacturer has 3 models at different price points, also shown in the table. 1.1) Complete the table of numbers. For each turbine, compute the maximum power it can generate (in megawatts) and the installation cost per megawatt . The maximum power that a wind turbine can generate is given by Power maximum = 0.59 x ½ ρ A V 3 Assume: ρ = 1.2 kg/m 3 (air density) Recall that the 0.59 is the “Betz Factor” and accounts for the fact that it is not physically possible to extract all of the kinetic energy from moving air. Wind Survey Results Wind Turbine Manufacturer Information Computed Performance Hub Height (m) Average Wind Speed (m/s) Rotor Diameter (m) Installed Cost Max. Power Generated (MW) Installation Cost ($’s per MW) 75 3.0 60 $1,000,000 0.009008 $9,008 100 6.0 85 $4,000,000 0.072316 $289,264 140 8.0 110 $18,000,000 0.215307 $3,875,526 Height: 75 m ● Power(max) = 0.59 * [1/2ρAv 3 ] ● Power(max) = 0.59 * [½ * 1.2 * π(30 2 ) * (3.0) 2 ] = 9008 W = 0.009008 MW ● Installation cost: 0.009008 * 1,000,000 = $9,008 Height: 100 m ● Power(max) = 0.59 * [1/2ρAv 3 ] ● Power(max) = 0.59 * [½ * 1.2 * π(42.5 2 ) * (6.0) 2 ] = 72316 W = 0.072316 MW ● Installation cost: 0.072316 * 4,000,000 = $289,264 Height: 140 m ● Power(max) = 0.59 * [1/2ρAv 3 ] ● Power(max) = 0.59 * [½ * 1.2 * π(55 2 ) * (8.0) 2 ] = 215307 W = 0.215307 MW ● Installation cost: 0.215307 * 18,000,000 = $3,875,526 Page 2 of 11

1.2) If the primary concern of the town is to maximize the power generated per dollar of their investment , which turbine do you recommend they install? 1- 0.009008/1,000,000 = 9.008 x 10 -9 MW/$ 2- 0.072316/4,000,000 = 1.807 x 10 -8 MW/$ 3- 0.215307/18,000,000 = 1.19615 x 10 -8 MW/$ Based on this, I recommend they install turbine #2 in order to maximize the amount of power generated per dollar they spend on installation. 1.3) If the typical home in the community uses 50 kWh of energy each day, how many homes could be powered by the largest of these turbines? Largest turbine- turbine #3 1 kWh = amount of energy it takes to run a 1,000 watt object for 1 hour Turbine #3 produces 215,307 watts of energy→ 1 kWh = 1,000 watts 215,307/1,000 = 215.307 kWh 215.307/50 = 4.31 In total, 4 homes could be powered by the largest of these turbines. 1.4) OPTIONAL. ANSWER WILL BE PROVIDED, BUT UNGRADED. In 1.1 above, you calculated the maximum power generated by the biggest wind turbine. Let’s assume that, due to capacity factor, the average power over the course of a year is just half that. Further assume that we sell every kWh of electricity we produce. If the electricity is sold to at $0.25 / kWh, how many years will it take to earn back our installed cost of $18 million? Page 3 of 11

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2) For your personal edification, read some history behind the power unit “horsepower.” And then proceed to this idea of using Tesla batteries for community power! 2.1) What is the peak power output of the Tesla Model X Plaid (in horsepower). 1.020 horsepower. 2.2) How many watts of power is this equivalent to? (You can use a conversion factor you find, rather than dimensional analysis.) 1 horsepower = 745.7 watts 1,020 * 745.7 = 760,614 watts 2.3) A community wants to set up a way for owners of Tesla Model X Plaid vehicles to use their cars’ battery packs as a kind of neighborhood emergency back-up power source. Perhaps the community has a renewable source (solar or wind) that could be used to charge up these “volunteer batteries.” For a single Tesla Model X Plaid vehicle, how many homes could we supply electricity to? Assume that every home requires 2,000 watts of electrical power to function adequately. • Electric vehicles are essentially giant batteries on wheels, so this isn’t a terrible idea. We just need volunteer Tesla owners in the neighborhood who are OK with supplying electricity to the neighbors in a pinch! • In the next problem, we’ll get to how long the battery can supply power. For this problem, we just want the number of 2,000 watt loads we could supply by pulling peak power out of a single Tesla battery. 760,614 watts/2,000 watts = 380.307 We could supply electricity to 380 homes with one Tesla Model X Plaid. 2.4) The battery pack in a single Tesla Model X Plaid can store 120 kWh of electrical energy. If we powered all of the homes that you calculated in 2.3 (above) with 2,000 watts, how long would the power stay on before we need to switch back to the grid? • Start with “Energy = Power x Time” or “Time = Energy / Power.” • Assume that the single Tesla battery can keep putting out its full peak power until it’s completely drained. (Probably not a great assumption, but we’ll go with it anyway!) • Assume that the Tesla owner is OK with waking up to find their car completely dead because the neighborhood needed power overnight during a snowstorm. (Also probably not a great assumption.) ● Power required = 760,614 watts Page 4 of 11

● 1 kWh = 3.6 * 10 6 J ○ 120 kWh = 432,000,000 J Time = Energy/Power Time = 432,000,000/760,614 Time = 567.96 seconds/60 = 9.47 The power would stay on for about 9 ½ minutes before we would need to switch back to the grid. Page 5 of 11

3) Capacity factor and wind farms In the equation for the practical amount of electricity that can be generated by a wind turbine, there is a “capacity factor” (CF) term. Power practical = CF x 0.59 x ½ ρ A V 3 3.1) Explain briefly what “capacity factor” refers to, and whether low or high CF is preferred at a prospective wind farm site. (2–3 sentences) Capacity factor refers to the fraction of time that wind blows at optimal speed, which is essentially a measure of how frequently the turbine is also running at its optimum power. Because of this, a high CF is preferred at a prospective wind farm site. 3.2) The graph below shows two types of data for Ontario, Canada: 1) Month-by-month fluctuation of wind capacity factor over several years (2006–2012). These are the different color curves, ranging from approximately 10%–45% (left side y-axis). 2) Typical month-by-month fluctuation of electricity demand (black curve), ranging from approximately 15,000–18,000 MW (right side y-axis). What do you conclude from this graph about whether relying fully on wind energy would be good energy policy for Ontario? Explain how the graph supports your conclusion. (2–3 sentences) I don’t believe that relying fully on wind energy would be a good energy policy for Ontario. As the graph indicates, capacity factor is highest during the first 6 Page 6 of 11

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and last 4 months of the year, but we see a sharp increase from around June to September every year. When this happens, the demand for electricity skyrockets at such a sharp increase that I would imagine it would be difficult to supply all of this extra demanded electricity on a whim. Due to natural wind patterns out of our control, wind energy does not always make for the most reliable source of energy. Page 7 of 11

4) Residential wind power installation An environmentally conscious family mounted 3 medium-sized wind turbines along the roofline of their north-facing house (see scale drawing). Each is rated at 2 kW peak power output. On really windy days (6 m/s) when the wind direction is from the north or south, the combined output of their system indeed reaches the expected 6 kW (3 turbines x 2 kW/turbine). But when those same strong winds (6 m/s) are blowing from the east or west, the total power never gets much above 3 kW. They are disappointed by this unanticipated poor performance of their system when the wind isn’t from the north or south. 4.1) What is the most likely reason for the drop in performance of the system when the wind is blowing from the east or west? Explain briefly. (2–3 sentences) The most likely reason for the drop in performance when the wind is blowing from east or west is the positioning of the wind turbines on the roof. When wind blows from the east or west, it likely blows at a position that makes it difficult to adequately turn the rotor and blades. Consequently, the generator is not able to operate at its optimum ability if the rotor and blades aren’t turning properly. 4.2) Apart from the occasionally disappointing performance, the homeowners installed the system in hopes that it would eventually pay for itself by selling excess electricity generation back to the local utility. (This is sometimes called “net metering.”) Assumptions Total system cost (all 3 turbines @ $10,000 each) = $30,000 Average continuous power generated by system = 4 kW Average continuous power used in the owner’s home = 2 kW Sale price of excess power to utility = $0.20 / kWh How many years will it take for the system to pay for itself? That is, how many years will it take to earn back $30,000 by selling excess electricity to the local utility? 30,000/0.20 = 150,000 kWh 150,000 hours/24 hours = 6,250 days/365 days = 17.12 It will take a bit over 17 years to earn back the $30,000 by selling excess electricity. Page 8 of 11

5) Offshore wind farm construction cost See the required reading document “Offshore Wind Market Report: 2022 Edition.” Dominion Energy, a company undertaking a large project off the coast of Virginia, has disclosed their estimates for construction costs. These are summarized in Table 21 on page 82. 5.1) Construction costs are those labeled “$/kW”. What is the total construction cost, given the stated project size of 2,587 MW? 2,587 MW → 2,587,000 kW 2,587,000 kW * $5,874/kW = $ 15,196,038,000 5.2) Since the wind farm operates for 30 years, what will be the total “operating and maintenance (O&M) costs?” • Use the stated project size of 2,587 MW for the calculation. 2,587 MW → 2,587,000 kW 2,587,000 kW/30 years = 86,233.33 kW per year 86,233.33 kW * $50/kW-yr = $4,311,667 5.3) How much total revenue will be generated over those 30 years if the wind farm sells electricity for $0.20 / kWhr? • Since the project size is 2,587 MW and the capacity factor is 43.3%, assume that the average power output will be 1,120 MW. • Assume that there is no down time at all during those 30 years, meaning that the wind farm generates 1,120 MW continuously (not a great assumption, but let’s use it anyway). Avg power output = 1,120 MW → 1,120,000 kW 30 years * 8760 hours/year = 262,800 hours 1,120,000 kW * 262,800 hours = 294,336,000,000 kWh 294,336,000,000 kWh * $0.20/kWh = $58,867,200,000 5.4) Based on this quite high-level and simplistic analysis of costs (5.1 and 5.2) and revenue (5.3) over the 30 years of operation, is this wind farm a profitable business? The construction costs ($ 15,196,038,000) combined with the operation and maintenance costs ($4,311,667) add up to a grand total of $15,200,349,667. The revenue generated from this project is $58,867,200,000, so this would be Page 9 of 11

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a profitable business because $58,867,200,000 - $15,200,349,667 = $ 43,666,850,333. Page 10 of 11

6) Business considerations All for-profit businesses need eventually to show a positive number at the bottom line. (The “bottom line” refers to the last line of the profit-and-loss financial statement. When it is positive, the business is profitable; when it is negative, the business is losing money and may be headed toward bankruptcy.) Compared to a traditional natural gas electricity generating company , over the span of a 30-year lifetime of large wind turbines, describe two general factors that would contribute to favorable profitability for a grid-scale wind farm. 6.1) A factor contributing to favorable profitability. One factor contributing to favorable profitability for a grid-scale wind farm is the fact that the large amount of electricity produced by wind turbines can be sold to power utility companies (such as in 5.3) and sent to a larger power grid. Thus, a wind turbine can pay for itself after a certain number of years and eventually make profit. 6.2) An additional factor contributing to favorable profitability. Another factor contributing to favorable profitability for a grid-scale wind farm are tax incentives for renewable energy sources like turbines. For example, the Renewable Electricity Production Tax Credit (PTC) lets owners of renewable energy sources like turbines get tax credits for each kWh of electricity generated during a 10-year period. This incentivizes high electricity production in return for profit from the tax credit. Page 11 of 11