Problem Set 2 (2023)

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Apr 3, 2024

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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

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

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