Innovative Low-Carbon Energy Conversion Systems PDF

Innovative Low-Carbon Energy Conversion Systems CHEG 403 – Fall 2024 5 - Renewable energy and power Faisal Abdulla AlMarzooqi FRIDAY, OCTOBER 18, 2024 2 ku.ac.ae Solar Energy Solar energy can be harvested throu...

Innovative Low-Carbon Energy Conversion Systems CHEG 403 – Fall 2024 5 - Renewable energy and power Faisal Abdulla AlMarzooqi FRIDAY, OCTOBER 18, 2024 2 ku.ac.ae Solar Energy Solar energy can be harvested through two main routes: 1. Solar Photovoltaic (PV) Technologies (solar energy to electricity) 2. Solar Thermal Technologies (solar energy to heat) 3 ku.ac.ae Solar Energy Solar Photovoltaic Technologies Mohammed bin Rashid Al Maktoum Solar Park in Dubai holds till date the World record for the lowest solar energy cost at 0.06 AED per kWh 4 ku.ac.ae Solar Energy Solar Photovoltaic Technologies 5 ku.ac.ae Solar Energy Solar Photovoltaic Technologies - Fundamentals In order to encourage the conversion of photons to electrons using the photoelectric effect, the two layers of silicon that constitute a silicon-based PV cell are modified so that they will be more likely to produce either: 1. Loose electrons 2. Holes in the molecular structure where electrons can reattach In one PV cell design, the upper or n-type layer is doped with phosphorus, with 5 valence electrons, while the lower or p-type layer is doped with boron, which has 3 valence electrons. This type of molecular structure is known as a p-n junction. 6 ku.ac.ae Solar Energy Solar Photovoltaic Technologies - Fundamentals An arriving photon possessing an energy value equal to or greater than the bandgap energy EG is able to break loose an electron from the structure of the PV cell. ℎ𝑐 𝐸= 𝜆 E = energy of a wave [J] ; h = Planck’s constant = 6.626x10-34 J.s c = speed of light = 3x108 m/s ; 𝜆 = wavelength [m] ℎ𝑐 𝐸! = 𝜆! 7 ku.ac.ae Solar Energy 1.37 kW/m2 Losses in PV Cells and Gross Current Generated by Incoming Light Losses within the cell: 1 kW/m2 - Quantum losses: photons with E < EG are unable to produce the photoelectric effect, so the energy they contain is not available to the Earth system, constituting one type of loss. Also, each photon can only produce 1 electron, regardless of the energy of the photon, so the “excess” energy is also not available for producing current. - Reflection losses: photons arriving at the surface of the PV cell are subject to a fractional reflection loss 𝜌(E), where 0 < 𝜌 < 1; the amount of Atmosphere reflection varies with the energy of the photon. PV cells are typically coated with an antireflective coating to reduce 𝜌(E). - Transmission losses: because PV cells possess finite width (i.e., thickness of the cell in the direction of travel of the photon), it is possible for a photon to pass through the cell without colliding with an atom in the structure. In this circumstance, the photon has been “transmitted” through the cell, leading to transmission losses. The amount of these losses are a function of both the thickness of the cell (W) and the energy of the photon (E), and are written t(E,W), where 0 < t < 1. 8 ku.ac.ae Solar Energy Losses in PV Cells and Gross Current Generated by Incoming Light Losses within the cell: (continued) - Collection losses: in a real-world PV cell, some fraction of the electrons released will permanently reabsorb into the structure before they have a chance to leave the cell, so that the collection efficiency, denoted 𝜂"#$ (E), will be less than 100%. For a range of photons with energy E such that EG < E < E*, where E* is some approximate upper bound that is specific to the design of cell in question, 𝜂"#$ (E) is close to unity for many types of cells. However, especially at very high photon energy values, empirical observation shows that electrons released are very likely to reabsorb, and 𝜂"#$ (E) falls to nearly 0. Thus, collection losses constitute another type of significant loss in the cell. The gross current generated by the PV cell, also known as the light current IL [A/m2]: ' 𝐼% = 𝑒 * 𝜂"#$ 𝐸 𝑆 𝐸 𝛼 𝐸, 𝑊 𝑑𝐸 &! where e = unit charge per electron = 1.6022x10-19 C = 1.6022x10-19 A0s = 1.6022x10-19 J/V ; S(E) = frequency of arriving photons as a function of photon energy ; W = thickness of the cell 9 ku.ac.ae Solar Energy Losses in PV Cells and Gross Current Generated by Incoming Light ' 𝐼% = 𝑒 * 𝜂"#$ 𝐸 𝑆 𝐸 𝛼 𝐸, 𝑊 𝑑𝐸 &! To simplify notation, define the absorption coefficient 𝛼(E,W) to take into account reflection 𝜌 𝐸 and transmission 𝜏 𝐸, 𝑊 losses as follows: 𝛼 𝐸, 𝑊 = 1 − 𝜌 𝐸 − 𝜏 𝐸, 𝑊 In practice, the curve for wavelength of arriving photons is observed to have a shape approximated by an appropriate statistical function with a single peak and zero probability of negative values, such as a Weibull distribution. 10 ku.ac.ae Solar Energy Example 1 A solar cell has a constant collection efficiency [𝜂"#$ (E)] of 95% between energy values (E) of 1.89×10−19 and 5.68×10−19 J, and 0% outside of this range. The absorption coefficient [𝛼(E,W)] is 80% for all energy values at the given thickness of the solar cell. Photons arrivals are distributed as a function of energy such that S(E)=(2.5×1021 particles/s·m2)(f(E)), where f(E) is a Weibull(𝛼 = 3, 𝛽 = 3) distribution with E measured in units of 10−19 J. Calculate IL. 11 ku.ac.ae Solar Energy The Dark Current, ID: An additional energy loss occurs in the PV cell due to the potential difference created from the presence of extra electrons in the n-type layer and extra holes in the p-type layer. This potential difference is common to any semiconductor, and any current resulting from its presence is called the dark current, or ID, since it can occur regardless of whether the PV cell is in sunlight or not. The dark current has the following functional form: 𝑒𝑉 𝐼( = 𝐼) exp −1 𝑚𝑘𝑇 where I0 = saturation current ; V = voltage ; m = a parameter whose value depends on the conditions of operation of the device ; k = Boltzmann constant = 1.38x10-23 J/K ; T = operating temperature in [K] The value of m depends on the voltage at which the device is operating, with the minimum value of m = 1 occurring at ideal device voltage under full or nearly full sun, and 1 < m < 2 for partial voltages. For simplicity, we will focus on device operation under ideal conditions with voltage near the maximum value possible, and use m = 1. 12 ku.ac.ae Solar Energy The Dark Current, ID: (continued) Because ID flows in the opposite direction from IL , the net current from the PV cell is the difference between the two, that is: *+ I = IL – ID ; 𝐼 = 𝐼% − 𝐼) exp −1 ,-. At V = 0, the amount of current produced is the short- circuit current ISC, which is equal to the light current (IL) since exp(0) = 1 and ID makes no contribution, that is, ISC = IL. Due to the exponential nature of ID, its effect on I grows rapidly above a certain value, so that at the open- circuit voltage VOC, the device ceases to produce a positive current. 13 ku.ac.ae Solar Energy The Dark Current, ID: (continued) It is now possible to solve for VOC by setting I = 0 by definition, there is no current flow in an open circuit: 𝑒𝑉/0 0 = 𝐼% − 𝐼) exp −1 𝑚𝑘𝑇 𝑚𝑘𝑇 𝐼% 𝑉/0 = ln +1 𝑒 𝐼) We are now able to characterise the function of the cell, since we know the range of voltages over which it operates, 0 < V < VOC, and we can calculate the current at any value V given fixed device parameter I0, value of m at the operating condition, and temperature T. 14 ku.ac.ae Solar Energy Calculation of Maximum Power Output The question we ask here: how can we calculate the maximum current, IM and voltage VM Recall: Power, P = IV We can take the derivative of this equation and set it to zero to find the maximum current and voltage: 12 13 43" = 𝐼 𝑑𝑉 + 𝑉 𝑑𝐼 = 0 à = 1+ 1+ +" Additionally, the values of IM and VM must satisfy: 𝑒𝑉5 𝐼5 = 𝐼% − 𝐼) exp −1 𝑚𝑘𝑇 Now have two equations for two unknowns, IM and VM. The solution to this system of equations does not exist in analytical form; however, an approximate solution that gives satisfactory results is available. 15 ku.ac.ae Solar Energy Calculation of Maximum Power Output (continued) Define parameters a and b such that a = 1 + ln(IL/I0) and b = a/(a+1). IM and VM can be written in terms of a and b as follows: 78 9 𝐼5 = 𝐼% 1 − 𝑎 46 ; 𝑉5 = 𝑉/0 1 − 9 it is clear that a device with: IM = ISC and VM = VOC would achieve the maximum possible power output from the device. However it is impossible to have a device operating at ISC and VOC at the same time. Therefore we introduce the Fill Factor (FF) which is used to evaluate the actual maximum value of P relative to this upper bound. 16 ku.ac.ae Solar Energy Calculation of Maximum Power Output (continued) 3" + " 78 9 𝐹𝐹 = = 1 − 𝑎 46 1− 3# +$% 9 In actual devices, fill factor values of FF > 0.7 are considered to be an acceptable result. Example 2 Consider a PV cell with light current (IL) of 0.025 A/cm2, saturation current (I0) of 1.0×10−11 A/cm2, open circuit voltage (VOC) of 0.547 V, and operating temperature of 20°C. For simplicity, assume a value of m = 1 and constant saturation current across its operating range. What is the combination of voltage and current that maximizes output from this device, and what is its fill factor? 17 ku.ac.ae Wind Energy Use of wind energy for human purposes entails the conversion of the kinetic energy that is present intermittently in the wind into mechanical energy, usually in the form of rotation of a shaft. From there, the energy can be applied to mechanical work or further converted to electricity using a generator. Wind speed in the UAE on average 7 to 12 mph 18 ku.ac.ae Wind Energy Wind speeds can be categorised by ’Bin,’ such that each Bin takes a wind speed range. We can calculate an average speed, Uaverage: 𝑈average = ∑=:;< 𝑝: 𝑈:,JKLMJNL where pi = percentage of year that wind speed is in i, Ui,average = average speed for the bin i The calculation based on the data in the table gives, Uaverage = 5.57 m/s Note that the 2 h for bin 16 are not included in the average speed calculation since the bin average speed is unknown. 19 ku.ac.ae Wind Energy For a given wind speed U, the power P in watts available in the wind per square meter of cross-sectional area is calculated as follows: 𝑃 = 0.5𝜌𝑈 > where 𝜌 = density of air [kg/m3] Using Rayleigh’s function, the probability distribution function and the cumulative distribution function we can define the probability that the wind speed will be at or below a given value U, given a known value of Uaverage: ? −𝜋 𝑈 𝑝 windspeed ≤ 𝑈 = 1 − exp 4 𝑈average Example 3 Calculate the probability that the wind is in bin 6, if Uaverage = 5.57 m/s 20 ku.ac.ae Wind Energy Example 3 Calculate the probability that the wind is in bin 6, if Uaverage = 5.57 m/s 21 ku.ac.ae Wind Energy Effect of height: The effect of height above the ground can be approximated by the following equation: 𝑧 A 𝑈 𝑧 = 𝑈 𝑧@ 𝑧@ where z = height above the ground ; zr = some reference height above the ground for which wind speed [U(zr)] is known ; U(z) = the wind speed as a function of height ; 𝛼 =s the wind shear coefficient. A typical value of the wind shear in flat locations might be a = 0.2. 22 ku.ac.ae Wind Energy Example 4 Suppose that the wind data gathered at 30 m elevation revealed an average wind speed of 5.57 m/s. A wind analyst evaluates the data and sumrises that if the turbine height is increased, the site may be viable for a commercial wind venture. The analyst believes that an average wind speed of 7.5 m/s will make the site commercially viable. At what height will this wind speed be obtained, if the wind shear value of a = 0.2 is used? 23 ku.ac.ae Hydropower Hydropower is a renewable energy resource resulting from the stored energy in water that flows from a higher to a lower elevation under the influence of the earth’s gravitational field. Three types: 1. Impoundment hydropower: a dam that holds water and then that water is used to generate hydropower (picture on the right) 24 ku.ac.ae Hydropower Three types: (continued) 2. Diversion or run-of-river hydropower: extracting a portion of the total energy contained in flowing water itself (picture on the right) 25 ku.ac.ae Hydropower Three types: (continued) 3. Pumped storage: During periods when electricity demand is low, electricity is used to pump water from a lower reservoir up to a higher elevation, where it is stored. When electricity demand increases, the flow is reversed, and electricity is produced as water passes from the higher storage reservoir back to the lower one. (animation on the right) 26 ku.ac.ae Hydropower DEWA Hydropower project in Hatta Pumped storage type 1,500 MWh storage capacity 27 ku.ac.ae Hydropower The hydropower output can be represented by a simple formula: Power = (total hydraulic head) x (volumetric flowrate) x (efficiency) 1 𝑃𝑜𝑤𝑒𝑟 = 𝜌𝑔𝑍 + ∆ 𝑣 ? ×𝑄×𝜀 2 𝜌 = density of water [kg/m3] ; g = gravitational constant [9.81 m/s2] ; Z = height of water head [m] ; Q = volumetric flowrate [m3/s] ; ∆ 𝑣 ? = the difference in the square of the inlet and exiting fl uid velocity across the energy converter 𝜌𝑔𝑍 = static head contribution 28 ku.ac.ae Hydropower 29 ku.ac.ae Nuclear Energy The general concept Nuclear Radiation 30 ku.ac.ae Nuclear Energy The general concept Nuclear Radiation 31 ku.ac.ae Nuclear Energy The general concept Nuclear Radiation 32 ku.ac.ae Nuclear Energy Fuel Cycle 33 ku.ac.ae Nuclear Energy UAE Nuclear Power Barakah 14 GigaWatts by 2030 34 ku.ac.ae Nuclear Energy UAE Nuclear Power Barakah 14 GigaWatts by 2030 35 ku.ac.ae Nuclear Energy General Nuclear Power Plant layout 36 ku.ac.ae Nuclear Energy The Nuclear Reactor 37 ku.ac.ae Nuclear Energy The Nuclear Reactor 38 ku.ac.ae Nuclear Energy 39 ku.ac.ae Nuclear Energy Nuclear Science Revision: Z = atomic number = number of protons in the nucleus = number of electrons around the nucleus for a neutral atom A = atomic mass = the sum of the number of protons and neutrons in the nucleus of that atom 1 proton has a mass = 1 atomic mass unit (AMU) [actually, the mass of a proton is 1.673×10−27 kg] 1 neutron has a mass = 1 atomic mass unit (AMU) [actually, the mass of a proton is 1.675×10−27 kg] Nuclei with the same number of protons but different numbers of neutrons are isotopes of that element For example, Uranium 233, or U-233, is an isotope of Uranium with 92 protons and 141 neutrons, U-238 is an isotope with 92 protons and 146 neutrons, and so on. 40 ku.ac.ae Nuclear Energy The mass of a nucleus: When protons and neutrons bind together to make a nucleus, energy is released. This energy is called the binding energy or the mass defect since energy and mass are interchangeable based on Einstein: E = mc2 🧐 we can calculate the mass of a nucleus with neutrons and protons (N, Z) and binding energy B(N, Z) using: 𝐵 𝑁, 𝑍 𝑀 𝑁, 𝑍 = 𝑁𝑚= + 𝑍𝑚B − 𝑐? mn = mass of neutron [kg] ; mp = mass of proton [kg] ? ).DDE 𝑍 𝑍−1 𝐴 − 2𝑍 𝐵 𝐴, 𝑍 = 𝐶< 𝐴 + 𝐶? 𝐴 + 𝐶> + 𝐶F 𝐴).>>> 𝐴 C1 = 2.5×10−12 J ; C2= −2.6×10−12 J ; C3= −0.114×10−12 J ; C4= −3.8×10−12 J 41 ku.ac.ae Nuclear Energy The general concept Nuclear Radiation 42 ku.ac.ae Nuclear Energy Reactions Associated with Nuclear Energy The most common reaction used to release nuclear energy for electricity production today is the fissioning of uranium atoms, and in particular U-235 atoms, since fissioning of U-235 creates by-products that are ideal for depositing energy into a working fluid and at the same time perpetuating a chain reaction in the reactor. The chain reaction for splitting U-235 uses a “thermal” neutron, which, after being released by the fissioning of a heavy nucleus, is slowed by collisions with other matter around the reaction site, reducing its energy. The most probable energy Ep of the thermal neutron is a function of temperature T in degrees kelvin, and can be approximated as: Ep = 0.5kT where k = Boltzmann’s constant = 1.38x10-23 J/K = 8.617×10−5 eV/K ; T = temperature around the reactions site [K] For example, if the temperature around the reaction site is 573 K (300 C), Ep = 0.0247 eV = 3.957x10-21 J 43 ku.ac.ae Nuclear Energy Reactions Associated with Nuclear Energy When a U-235 nucleus is struck by a thermal neutron, the neutron is absorbed and then one of two reactions follows: where n = neutron ; Kr = Krypton ; Ba = Barium ; 𝛾 = energetic photon or high energy electromagnetic particle 44 ku.ac.ae Nuclear Energy Uranium 235 enrichment Technique Separation factor 235UF - 238UF 6 6 Distillation (a) 1.00002 Chemical exchange(K) 1.0016 Separation factors Gaseous diffusion & !"" # 1.00429 $ ! $ !" ! % ! " Gas centrifuge & $$ #$% ( $%! ' $%& )#&! " # !! 1.162 - 1.3 % ! !" " Aerodynamic 1.03 - 1.15? Dr. Ian W. Cumming – Imperial College London - 2005 45 ku.ac.ae Nuclear Energy 2.5 Uranium 235 enrichment Separation factor, ⍺ Gas Centrifuge 2 𝑀𝑊? − 𝑀𝑊< 𝑉9? 1.5 𝛼 = exp 2𝑅𝑇 1 6 235UF /238UF It is necessary to have uranium 6 with 2 to 3% 235U 0.5 concentration for Nuclear power plants 0 0 100 200 300 400 500 600 700 800 900 1000 1100 1200 Axial speed of the gas flow within the centrifuge, Va [m/s] Dr. Ian W. Cumming – Imperial College London - 2005 46 ku.ac.ae Nuclear Energy Separative Work Units (SWU): SWU = WV(xw) + PV(xp) - FV(xf) SWU used to describe the capacity of an enrichment plant 1−𝑥 𝑉 𝑥 = 1 − 2𝑥 ln 𝑥 F = Feed(kg), P = Product(kg), W = Waste(kg) xf = Feed composition, xp = Product composition, xw = Waste composition Note that units of SWU are kg SWU price ~ 368 AED/SWU Dr. Ian W. Cumming – Imperial College London – 2005 Mueller, H. and Laucht, J., 2005. Status report on the cost and availability of enriched uranium for research reactors (No. INIS-XA-C--117). 47 ku.ac.ae Nuclear Energy Example 5 Calculate the cost of Uranium 235 enrichment for the Barakah Nuclear project. Consider that the plant has four APR1400 nuclear reactors each consuming 150,000.00 kg of Uranium 235 per year. Uranium ore contains 0.7% by mass 235U and the nuclear plant feed needs to be 3% by mass 235U. The waste from a Uranium enrichment plant is usually around 0.2% by mass. The electricity Tariff in the UAE is approximately 0.23 AED per kWh. The total operating cost of running one APR1400 nuclear plant is 457.50 million AED per year. - Calculate the Separative Work Units cost per year - Calculate the total profit per year ~ 30 Billion AED per year (DEWA 2023 ~ 9 Billion AED per year) Dr. Ian W. Cumming – Imperial College London – 2005 Mueller, H. and Laucht, J., 2005. Status report on the cost and availability of enriched uranium for research reactors (No. INIS-XA-C--117). 48 ku.ac.ae Nuclear Energy Energy consumption for Uranium enrichment Process kW.h electrical/SWU Electromagnetic ~ 25,000 Aerodynamic ~ 3000 Modern gas diffusion 2,400 - 2,500 Centrifuge 50 - 60 Dr. Ian W. Cumming – Imperial College London – 2005 Mueller, H. and Laucht, J., 2005. Status report on the cost and availability of enriched uranium for research reactors (No. INIS-XA-C--117). Thank You ku.ac.ae