Energy Technology Exam Practice PDF

Exam practice for energy technology and resources *Question 1:* What are the advantages and disadvantages of renewable energy resources? *Answer:* Advantages - More energy diversity will lead to better energy supply security - Abundant availability - Creates new job opportunities - B...

Exam practice for energy technology and resources *Question 1:* What are the advantages and disadvantages of renewable energy resources? *Answer:* Advantages - More energy diversity will lead to better energy supply security - Abundant availability - Creates new job opportunities - Better for the environment compared to conventional fuels. - Long economic lifetime and lower levelized cost of energy (LCOE) Disadvantages - Most is use-it or lose-it energy due to variable availability - Lower energy density - Requires energy storage for balancing the demand - Large use of land - Marine and land ecosystem disruption. *Question 2:* As part of your exam preparation for FE407, you had a very good breakfast equivalent to 1728 kJ. If your metabolism during the exam (the rate at which your body burns food) is 120 W, how long will it take for your body to completely burn off the breakfast? *Answer:* Converting kJ to joules: 1728 kJ = 1728 \* 1000 = 1,728,000 J Since 1 Watt = 1 joule the conversion here is pretty simple **Formula for power: Power = Energy / Time** Since we are wondering about time, we need to rearrange the formula to solve It **Time = Energy / Power** Time = 1,728,00 J / 120 J/S = 14,400 seconds Then we need to convert it to hours to make it more readable 14,400 seconds = 14,400 / 60 = 240 minutes = 240 / 60 = 4 hours It will take 4 hours for the body to completely burn of the 1728 kJ breakfast with a metabolic of 120 W (: *Question 3* Define capacity factor, availability and system efficiency *Answer:* **Capacity factor** The actual energy output divided by the maximum possible energy output over a certain time period. **Availability** **The percentage of time or number of hours that the technology is available to produce power. Normally it would be around 98 -- 99%.** **Efficiency** **Useful energy output divided by the input energy of power plant or heating technologies. It refers to mechanical, electrical or thermal efficiencies of the technology.** *Question 4* What are the main differences in the working principles between reaction turbines and impulse turbines? Provide examples of reaction turbines and impulse turbines. *Answer:* 1. Impulse turbines only use kinetic energy of water while reaction turbine uses both kinetic energy and pressure energy. 2. Impulse turbines are mostly used for large water heads but for reaction turbines it's for low water heads. 3. The blades for impulse turbines are symmetrical and for reaction turbines they are asymmetrical. 4. The turbine size of impulse turbines is smaller but has the same power output as a reaction turbine. *Question 5* Imagine that the pipe length of a hydropower plant is 300 meters, and the diameter is 1.5 meters. The pipes are made of galvanized iron. The water flow in terms of Reynolds number (Re) inside the pipe is 6×1056 \\times 10\^56×105. If the water velocity inside the pipe is 2 m/s, how large will the head loss be in the pipe? (10 points) The formula for head loss is given as: Et bilde som inneholder Font, håndskrift, hvit, kalligrafi Automatisk generert beskrivelse - f is the friction factor, - L is the length of the pipe (m), - D is the diameter of the pipe (m), - v is the water velocity (m/s), - g is the acceleration due to gravity (9.81 m/s²). *Answer:* The procedure is the following. To calculate absolute roughness, you have to use 0.15mm or 0.00015 / the diameter which is 1.5. That equals to 0.0001 the same as 10\^-4 The friction factor here by reading the diagram like this  Will be around 0.013 f = 0.013 L = 300M D = 1.5m v = 2m/s g = 9.81 m/s\^2 galvanized iron = 0.15 e(mm) Now we can calculate the formula Simply putting in the numbers and we get 0.53m is the head loss in the pipe. *Question 6* We have a hydropower plant with a head (H) of 100 meters and a water flow rate (Q) that varies throughout the day, as shown in Table 1. The type of turbine used is a Francis turbine, and its efficiency (ηt​) at different flow rates is also shown in Table 1. The density of water is 1000 kg/m³, and gravity is 9.81 m/s². How much is the total daily electrical energy production? Show your calculations and complete the table for the power and energy production for each hour. The formula for power is: P(W)= η⋅ρ⋅Q⋅g⋅H Where: - η is the efficiency, - H is the net head (in meters), - Q is the water flow rate (in m³/s), - ρ is the water density (1000 kg/m³), - g is the acceleration due to gravity (9.81 m/s²). *Answer:* For P(W), hour 1 it would be: 92 \* 1000kg/m\^3 \* 1.48 m\^3/s \* 9.81m/s\^2 \* 100 = 1334 MW Etc... to 24 To calculate the total electricity production that day it would be sum of all the MW = 26.57 MW *Question 7* The characteristic curve for current (I) versus voltage (V) for a given solar module at different solar irradiance levels is shown in Figure 1. Answer the following questions based on Figure 1 (10 points): a\. What are the approximate **maximum power point current** and **maximum power point voltage** of the solar module at 1000 W/m² irradiance? b\. What is the approximate **maximum power** of the solar module at 1000 W/m² irradiance? c\. If the efficiency of the solar module at 600 W/m² irradiance is 16.5%, what is the **area** of the solar module? The efficiency formula is: η(%)=Pmax / I \* A Where: - η is the efficiency of the solar module (%), - Pmax​ is the maximum power of the solar module (W), - I is the solar irradiance (W/m²), - A is the area of the solar module (m²). *Answer:* A. Et bilde som inneholder tekst, line, diagram, Plottdiagram Automatisk generert beskrivelse - The maximum current point for 1000W/m\^3 will be **9 A** - The maximum power point for voltage is **24V** B. Here they are asking for the irradiance(innstråling) which is taking the 9 A \* 24V = **216 W** C. **For C you have to use the formula to find out A for areal.** N = 16,5% = 0.165 Pmax = 216 W I = 600 W/m\^2 **We will have to rearrange the formula** **A = Pmax / N \* I** **That makes it: 216 / 0.165 \* 600W/m\^2** **216 / 99 = 2.18 m\^2** *Question 8* List the components of a complete solar power system without grid connection that delivers both DC and AC power. *Answer:* 1. Is PV array or arrays that converts sunlight to DC electricity 2. Wiring and mounting to install the PV arrays and connect it all. 3. Then you need a charge controller, this will help you with not overcharging the battery 4. Then you need somewhere to store it, like a battery bank 5. To use the energy, you will need to convert it, so an inverter that converts DC electricity to AC 6. Power meter or system monitor, this is good to measure how much you are using and sending to the grid. *Question 9* You have used SAM for techno-economic evaluation of solar and wind energy projects. Write down the general steps you must follow when performing a techno-economic evaluation of grid-connected solar systems in SAM. *Answer:* *Question 10* Assume that you want to install a wind turbine in your yard. You have calculated the wind turbine\'s maximum power based on the average wind speed. Later, you found out that the average wind speed was off by 5%. How much does this affect the maximum power? *Answer:* We need to substitute the wind speed change. If the wind speed is 5% higher: Let's say Poriginal represents 1 or 100% to express changes U = wind speed Take the formula Pnew = Poriginal \* (unew / u) \^3 If the wind speed is 5% higher: Pnew = Poriginal \* (1.05u / u) \^3 = Poriginal \* 1.05\^3 1.05\^3 = 1.157625 = 115% but -- 1 bcs 1 represents the original value or 100% of the power before any changes. So, 115% - 1 = 15.76% when increased by 5% power If the wind speed is 5% lower: Pnew = Poriginal \* (0.95u / u) \^ 3 = Poriginal \*0. 95\^3 0.95 \^3 = 0.857375 which is 85% but -- 1 which is equal to -- 14.26%, meaning it will decrease with 14.26% So, a 5% change in measuring the wind speed can result in pretty significant changes from 14 -16% in power output. *Question 11* The wind speed characteristics at a given location can be represented using the Weibull probability distribution by utilizing the shape parameter (k) and the scale parameter (c). What does a higher k-value and lower c-value tell us? What does a lower k-value and higher c-value tell us? *Answer:* **High K value** **In the Weibull distribution the K value becomes tall and narrower, which means smaller the spread in the wind speeds. (Like a tall mountain).** **Low K value** For low K values the graph will look a lot flatter and more "spread out". Like for gusty locations which normally leads to a lower k value because of wind changes quicky and frequently. **High C (large scale) value** A high C value indicates that wind speeds are generally higher, meaning that it's peaking at a higher wind speed. It also means that the failures are more spread out over a longer time period. **Low C (small scale) value** Winds speeds are generally lower, and failures will occur in a short time period. Question 12: ------------ Det årlege vindrosediagrammet for øya Fedje er gitt som vedlegg. Svar på følgjande spørsmål basert på vindrosediagrammet. Answer: ------- Question 13 ----------- Define the three main types of bioenergy resources with examples and outline the biofuel policy for transport in Norway. Answer: ------- 1. Solid Biomass: This is organic materials such as wood, agricultural residue, and forestry waste that can be burned by energy. 2. Liquid Biofuels: This is fuels that can be used directly in engines from biomass or as additives. Example: Ethanol and biodiesel are used in transportation. 3. Biogas: This is fuel that is produced when organic matter, such as food or animal waste is broken down. Methane is produced from food waste which can be used to fuel for example a bus. Question 15 ----------- What are the advantages and limitations of hydrogen in the energy system? Answer: ------- Advantages: - Clean energy source: Hydrogen only produces water vapor as a byproduct when used in fuel cells or burned. - Energy storage: Hydrogen can be stored a long time which is useful for balancing supply and demand. - High energy density: Hydrogen has a high energy density by weight, meaning it can be storing more energy per kilogram compared to other fuels. Limitation: - Production challenges and cost: Green hydrogen is really expensive and energy intensive, most of the hydrogen today is grey which emits Co2 from natural gas. - Also very costly to store and transport, requires a high-pressure storage and cryogenic cooling. - Safety concerns: Hydrogen is highly flammable and need careful handling.