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Q1:1. (25 pts) A solar panel is mounted facing due south in a location at 40° N latitude, with a tilt angle of 38° (south facing). If the sun shines on the panel for the entire day on the spring equinox with clear skies, calculate the total energy incident per square meter on the panel during the course of the day in kWh/m². (Use the direct normal values shown in the table below). Hour 1 2 3 4 5 6 7 8 9 10 11 12 13 15 16 17 18 19 20 21 22 23 24 Normal Irradiance [w/m2] 0 0 0 0 0 130.00 480.00 756.00 875.00 892.00 908.00 915.00 908.00 892.00 875.00 756.00 480.00 130.00 0 0 0 0 0 0 2. (10 pts) A house at 48° N latitude has a roof that faces due south, and is elevated to an angle of 28°. A solar panel is mounted to the roof. What is the angle of incidence between the sun and the array. 1 p.m. on June 1?See Answer
Q2:Problem 1 (10 pts) Assume that coal has a sulfur content of 3% by weight. If all the sulfur is converted into SO2 during the combustion process, (a) how much SO₂ is produced per tonne of coal? (b) How much is produced per megajoule of energy produced? The molecular mass of sulfur is 32 gram mole. Sulfur content in 1 tonne of coal is given (or may be found from what's given); you need to find out the content of SO2 when I tonne of coal is burned. So, find out the ratio (of molar mass) of S in SO2; from here you can determine the SO₂ content./nProblem 2 (10 pts) A typical gasoline-powered vehicle requires about 3.7 MJ of primary energy (i.e., energy content of the gasoline) to travel 1 km. How many kilograms of CO2 are produced annually by a vehicle that is driven 40,000 km per year? From the energy content of gasoline (octane), you may find out the amount (kg) of gasoline required to drive the vehicle annually. The chemical reaction for burning octane is given by: 2C8H18 + 250216CO2 + 18H20 From here, you can find out how many grams of octane yields what amount of CO2, which may then be used to find the total CO2 produced.See Answer
Q3:Course work: Design a heat pump to heat the James Watt Building South This course work assignment is part of the final assessment of this course, and it accounts for 25 marks (25% of the total assessment). Objective: The peak heating demand of the James Watt Building South is assumed to be 400 kW. Its heating network system delivers heat by pumping hot water at 80 *C through radiators. The measured river water temperature for most of the heating season is in the range 6-10 °C. After extracting heat, the cold water should be returned to the river with a temperature no less than 4 °C. You are expected to apply the learned thermodynamics knowledge and skills to conduct a conceptual design of a two-stage water source heat pump to extract heat from the water of river Kelvin to heat the James Watt Building South.See Answer
Q4:1 The temperatures of the hot and cold reservoirs are: = 1050°F = 565°C = 839 K T=554°F = 295°C = 563 K This gives an ideal Carnot efficiency of 100 ×(1-563K/839K) = 33 % Thus the total solar insolation must be 1 GW/0.33 = 3 GW. The installation in Figure 9.4 is in California and from Figure 8.2 the average insolation can be estimated to be 225 W/m. The area needed to produce the necessary power is 1.3 x 10 m. For a circular array this corresponds to an area about 4 km in diameter. This is too large of an area. Why?See Answer
Q5:2. Problem 10.1 (10 pts) The power output in watts is given as P = (0.602 kg/m³)4 nv³ The velocity as a function of height isSee Answer
Q6:3. Problem 10.2 (10 pts) Area per turbine is (3.14) x (5 m) = 78.5 m² per turbine. The power per turbine will be P = 4.08 × 10³ W The average spacing of the wind turbines downwind will be 10 times the rotor diameter or 100 m and the crosswind spacing will be 3 times the rotor diameter or 30 m. Thus each turbine will occupy a land area of (30 m) x (100 m) = 3000 m². Thus, 1 km² square of land area will accommodate (10 m/km)/(3000 m²) = 333 turbines with a total output of 1.36 MW (b) From equation (9.3), P = 30.2 MW. This is more than 20 times the wind power output.See Answer
Q7:4. Problem 10.3 (10 pts) P = (0.602 kg/m )nAv. Solving for n givesSee Answer
Q8:Scenario, Assignment Brief and Guidance: Scenario: You are an engineer working on a wind turbine "green energy" project for BCP City Council. Their wind turbine blades are not lasting as long as predicted, and in-service problems do not match predicted life from laboratory test results. You have been tasked with reviewing and evaluating a number of possible materials for the improved manufacture of wind turbine blades, so that they last longer and are more reliable. Blades have failed in the past because of a build-up of electromagnetic charge, excessive heat, corrosion, mechanical strain and failure. Assignment Brief:/nAssignment Brief: Activity 1: Some wind turbine blades have cracked, failed or twisted (reduced 'angle of attack' and hence performance) in-service in much shorter times than predicted. Using the data from your laboratory tests, prepare and present a report that compares and contrasts the following: • torsion test results of four materials (two metallic & two non-metallic) against theoretical results (you may use the theoretical data from APPENDIX 1: Material Shear Modulus Data) • beam deflection test results of four materials (two metallic & two non-metallic) against theoretical results (you may use the theoretical data from APPENDIX 2: Modulus of Elasticity Material Data) Evaluate any differences and suggest why real-world & laboratory test results for wind turbine blades could differ from theoretical data. NOTE: Include evidence, photos and results of your laboratory tests in your report. So that your audience can understand the context of your report, you will need to describe how the structures of the different materials relate to their properties. You will also need to refer to the types of degradation that are most common in the four materials, and how relevant that is to the wind turbine blade application. Use citation/references using the Harvard referencing system where relevant. Show and explain any formulae used and calculations.See Answer
Q9:Instructions: Need to write a discussion in 200 words on any 1 option. Support your discussion with facts, graphs, pictures and follow all the instructions in the file. Need to write 2 peer responses in 150 words each. Posts and opinions will be provided later.See Answer
Q10:Question 5 You are preparing a one-line electrical drawing for a grid-tied PV system with battery backup, which of the following is the right components order (starting from PV modules and ending at the utility Grid): O Inverter-disconnect-meter-PV array-battery O PV array-disconnect-inverter-meter-battery O PV array-battery-inverter-disconnect-meter O PV array-battery-meter-inverter-disconnect Question 6 The Installer is responsible for: 1 pts O Inspect PV system for code compliance 1 pts Designing the PV system according to the local and national standards Installing PV system according to the local and national standards O Designing the PV system according to the local and national standards AND Installing PV system according to the local and national standardsSee Answer
Q11:Question 11 Based on the data presented in both of the PV 2014 IEA roadmap and the STE 2014 IEA roadmap, which solar technology has greater solar market share? O STE is leading the global solar market and PV comes as second place O PV has greater global solar market share than STE O CPV is leading the global solar market The market of the solar technologies is equally split Question 12 Based on Question 11, what is the main reason that one technology outnumbered the other? 1 pts O STE is much simpler technology that uses mirrors and with larger plants it makes economical sense to be leading the solar market with cheaper initial cost O PV prices are going down and the development of new CSP still cannot compete with the low PV prices O CPV generates more per unit area power and it occupies less land space which makes it more economical choice to outnumber PV technology O Since the market is equally shared, that indicates that the prices are all the same for STE, CPV, and PV 1 ptsSee Answer
Q12:Question 9 A Commercial PV systems can fall into which of the following PV market segments? O Residential O Utility scale O Non-residential O All of above Question 10 Commercial PV systems are cheaper to install when compared to: O Residential O Utility scale O Non-residential O All of the above 1 pts 1 ptsSee Answer
Q13:Incorrect Question 1 You had a conversation with friends about PV technology, before taking the lesson. Some people used the term PV modules and other used PV panel. Which of the following is correct? A "Module" is a combination of PV cells They are the building block for PV array "Modules" are also known as PV "Panel" and they can be used interchangeably 0/1 pts O All of the aboveSee Answer
Q14:Question 4 On the current-voltage characteristics of PV module, the highest current occurs: At maximum power point When Voltage equals Zero At open circuit condition None of the above ptsSee Answer
Q15:ct Question 5 On the power-voltage characteristics of PV module, the highest power occurs: At maximum power point condition At short circuit condition At open circuit condition None of the above 27 0/1 ptsSee Answer
Q16:Incorrect Question 7 On the current-voltage characteristics of PV module, the Voltage changes with: ⒸHigher irradiance and temperature Lower irradiance and temperature The operating point corresponding to the load requirements All of the above 0/1 ptsSee Answer
Q17:Question 8 You have the choice to design PV system in one of the following cities: Denver, CO (High irradiance with Low temperature) Phoenix, AZ (High irradiance with high temperature) Seattle, WA (Low irradiance with average temperature) Which city will offer ideal operating condition (better energy yield): Seattle, WA Phoenix, AZ Denver, CO They are all the same 27 0/1 ptsSee Answer
Q18:Question 11 The bypass diode is installed on PV module so that: Current always flows through it to prevent cell damage Current flows through it once a day when voltage across the modules equals Zero Cells can be shaded without any damage so this diode is almost never used 0/1 pts Provides alternative path for current to flow when cells are shaded to prevent cell damage when module is shadedSee Answer
Q19:ct Question 12 Blocking diode is used to: Blocks modules from operating at short circuit Blocks over voltage when modules are shaded Protect PV string from acting like a load for other parallel strings when modules are shaded in that string All of the above Need final answers for all the answers 0/1 pts 4See Answer
Q20:Question 13 You are purchasing solar PV modules, you noticed STC and NOCT specification are listed differently. To make things more interesting, you found out PTC test standards are different as well!! What is the reason? (choose all that apply): PTC simulates how close to the real field operating conditions the modules are NOCT and STC specify the operating temperature differently ✔PTC, NOCT and STC can be used interchangeably so numbers don't count STC is the only standard we should use 0/1 pts Need final answers for all the answersSee Answer

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