Exam M08AAE August 2015_solution_V2.docx

M08AAE/4August 2015Coventry UniversityFaculty of Engineering and ComputingM08AAEAlternative Energy and Smart GridInstructions to candidatesTime allowed: 2 Hours 0 minutesAnswer: Any 3 Questions Total number of questions : 4For this examination you will be supplied with:Answer BooksImportant: You must hand this question paper in at the end of the examination.page 1 of 4M08AEE/4Q1.(a) Critically compare 1st, 2nd and 3rd Generation solar cells considering performance, cost and manufacturability? (2+2+ 2 marks)Solar cell technologies are traditionally divided into three generations. First generation solar cells are mainly based on silicon wafers and typically demonstrate a performance about 15-20 %. These types of solar cells dominate the market and are mainly those seen on rooftops. The benefits of this solar cell technology lie in their good performance, as well as their high stability. However, they are rigid and require a lot of energy in production. (2 mark)The second generation solar cells are based on amorphous silicon, CIGS and CdTe, where the typical performance is 10 - 15%. Since the second generation solar cells avoid use of silicon wafers and have a lower material consumption it has been possible to reduce production costs of these types of solar cells compared to the first generation. The second generation solar cells can also be produced so they are flexible to some degree. However, as the production of second generation solar cells still include vacuum processes and high temperature treatments, there is still a large energy consumption associated with the production of these solar cells. Further, the second generation solar cells are based on scarce elements and this is a limiting factor in the price. ( 2 mark)Third generation solar cells uses organic materials such as small molecules or polymers. Thus, polymer solar cells are a sub category of organic solar cells. The third generation also covers expensive high performance experimental multi-junction solar cells which hold the world record in solar cell performance. This type has only to some extent a commercial application because of the very high production price. A new class of thin film solar cells currently under investigation are perovskite solar cells. Polymer solar cells or plastic solar cells, on the other hand, offer several advantages such as a simple, quick and inexpensive large-scale production and use of materials that are readily available and potentially inexpensive. Polymer solar cells can be fabricated with well-known industrial roll-to-roll (R2R) technologies that can be compared to the printing of newspapers. Although the performance and stability of third generation solar cells is still limited compared to first and second generation solar cells, they have great potential and are already commercialized. (2 marks) b) A photodiode with a band gap energy of Eg = 1.5 eV is exposed to monochromatic radiation (400 THz) with a power density P = 1000 W/m2. The active area of the device is 10 cm X 10 cm. Consider it as an ideal device in series with an internal resistance of 1 m. All measurements were made at 300 K and the open circuit voltage is 0.6 V. i) Estimate the short-circuit current. (3 marks) ii) Calculate the reverse saturation current. (3 marks)c) What are the external parameters of a solar cell and explain how could they be determined? (3 marks)i) The photon flux,=Phf=10006.6*10-34*400*1012= 3.78*1021 photons/m2/s ( 1 marks)The short circuit current, = 6.048 A (2 marks)ii) Power delivered at 100 m load,Voc= KTqln(IshIo+1) k = 1.38 * 10-23T = 3000 kq = 1.6 * 10-19Voc = 0.6 VIo = 5.14 * 10-10 A c) External parameters: short circuit current, open circuit voltage; fill factor, efficiency, eventually series and parallel resistance. (1.5 marks)The external parameters are determined from I-V measurement, when the solar cell is illuminated. The standard illumination conditions are: AM1.5 spectrum, light intensity 1kW/m2, 25C. (1.5 marks)Q2. (a) Discuss why does an alternators voltage drop sharply when it is loaded down with a lagging load? (4 marks)An increase in the load is an increase in the real and reactive power or reactive power drawn from the generator. Such a load increase increases the load current drawn from the generator. Because the field resistor has not been changed the field current is constant therefore flux () is constant. Since the prime mover also keeps a constant speed () and the magnitude of the internal voltage, Ea = k is constant.Assuming the same power factor of the load, change in load will change the magnitude of the armature current IA. However, the angle will be the same (for a Constant PF). Thus, the armature reaction voltage jXsIa will be larger for the increased load. Since the magnitude of the internal generated voltage is constant. If these constraints are plotted on a phasor diagram there is one and only one point at which the armature reaction voltage can be parallel to its original position while increasing in size which is shown in diagram. If the constraints are observed then it is seen that the load increases the terminal voltage decreases rather sharply.b) Explain what is an infinite bus and what constraints does an infinite bus impose on a generator paralleled with it? (2 + 2 marks)An infinite bus is one where injecting power to the grid or drawing power from the grid by connecting machine at the bus. It has infinite capacity as impedance is very less near to zero and the voltage/frequency at this bus is always constant and any downstream disturbances do not affect the voltage/frequency. (2 marks)Often, when a synchronous generator is added to a power system, that system is so large that one additional generator does not cause observable changes to the system. A concept of an infinite bus is used to characterize such power systems. ( 2 marks)b) A three-phase, 450 V, 50 Hz, Y connected 4 pole synchronous generator has a per phase synchronous reactance of 1.2 . Its full load armature current is 60 A at 0.85 PF lagging and the field current has been adjusted so that the terminal voltage is 450 V at no load. What is the terminal voltage of this generator when it is loaded with the rated current at 0.85 power factors (PF) lagging and 0.85 power factor leading ? Draw the phasor diagram for both cases. (7 marks) Marks breakdown: (2 + 2) Marks for voltage calculation & (1.5 + 1.5) marks for phase diagramPer phase generated voltage, Vg= 4503=259.82 VVg=Vt+Ia (Ra+j Xa) Vt is the terminal voltage, Ia is the armature current, Ra is the armature resistance and Xa is armature reactance.Draw the phasor diagram and it can be seen from phasor diagram that two quantities are unknown; phase angle between generated voltage and the terminal voltage & terminal voltage magnitude.If the generator is loaded with 60 A current at 0.85 pf lagging then voltage drop in reactance;jIaXa=j 60-31.790*1.2=7258.210 The phase voltage at the rated load and 0.85 pf lagging is;259.822= (Vt+XaIa sin)2+(XaIacos)267506=Vt+1.2*60sin31.782+(1.2*60* cos(31.78)2Vt = 214.60 V Each phase load or terminal voltage = 214.60 VSo line to line terminal or load voltage = 214.6* 1.732 = 371.67 VIa Xa SinVt = 233.8 VJ Ia XaVg = 265.59 VIa Xa CosIaFor 0.85 pf leading case;IaVgJIaXaVt259.822= (Vt-XaIa sin)2+(XaIacos)2Vt = 290.43The line to line voltage for 0.85 pf leading case would be; 1.732*290.43 = 503 VQ3. a) Briefly explain why some wind turbines have large numbers of blades while others have only two or three. Give examples of the applications of both types of wind turbine. (3 + 1 marks)The main objective of wind turbine design is to maximize the aerodynamic efficiency, or power extracted from the wind. Wind turbines with large number of blades have highly solid swept area and are known as high-solidity wind turbines. Wind turbine with small number of narrow blades is referred to as low-solidity wind turbine. The major factors involved in deciding the number of blades are power coefficient, tip speed ratio and yawing rate to reduce the gyroscopic fatigue. Modern wind turbine are neither built with many rotor blades nor with very wide blades even though turbine with high solidity have the advantage of enabling the rotor to start rotating easily because more rotor area interacts with the wind initially. In theory the more blades a wind turbine rotor has the more efficient it is. However large numbers of blades can interfere with each other so high solidity wind turbines tend to be less efficient overall than low solidity turbines. Of low solidity machines three bladed rotors tend to be the most energy efficient, two bladed rotors are slightly less efficient still. Almost 70% modern wind turbine use 3 blades. ( 3 marks)2 or 3 bladed wind turbines are used for electricity generation because they operate at high tip speed ratios and therefore do not require as high a gear ratio to match the speed of the rotor to that of the generator. Multi-bladed turbines are generally used for water pumping because of their low tip speed ratios and resulting high torque characteristics. ( 1 marks)b) If the wind turbine has blades that are 35 m in length and average wind speed is 10 mph, how much power is produced by the wind turbine? Consider the mechanical to electrical efficiency; the power coefficient and the air density are 80%, 40% & 1.2 kg/m3 respectively. (3 marks)Velocity, V = 10 mph = 10*1.609*1000/(60*60) = 4.47 m/sTurbine blades area, A = pi()*r2 = 3.14*352 = 3848.45 m2Air density, = 1.2 kg/m3Generated power from Wind turbine, P = 0.8*0.4*1.2*3848.45*4.473 = 131.98 kWc) Describe Betzs law and how close are modern wind turbines to this Betz limit? (8 marks) There is a fundamental limit on the efficiency with which power can be extracted from the Wind. This is known as the Betz limit (due to Betz, Lanchester, and Zhukovsky). The total power extracted by the turbine is the difference between the embodied power in and the embodied power out:Pextracted = Pin-Pout = (1)The mass flow rate is given by; Where is the air density, A is the swept area and V is the wind speed.The turbine force,The power extracted by the turbine; (2)Solving equation (1) & (2),The mass flow rate,Now the wind power extracted by turbine;Total wind power,Now the Betz efficiency,The maximum efficiency; when Vout/Vin = 1/3This is the Betz limit on mechanical efficiency for extracting power from the wind. Note that it has nothing to do with thermodynamic efficiency, and is instead a mechanical limit. Well-designed, real turbines will reach efficiencies of 44%. Q4.C) Why do we need to develop the alternative power train such as battery vehicles and hydrogen fuel cell vehicles? (4 marks)Road transport today is dominated by oil-derived fuels and internal combustion engines, which makes it unsustainable both economically and environmentally. In particular, oil price volatility and supply disruptions represent a serious threat to road transport affordability and reliability, the lack of which can affect GDP. Moreover, the use of oil-derived fuels in internal combustion engines is inherently coupled with emissions of greenhouse gases (GHG) as well as of regulated air pollutants such as fine particulate matter (PM10), volatile organic compounds (VOCs) and nitrogen oxides (NOX). Addressing all these issues simultaneously will ultimately require moving away from oil-derived fuels and conventional internal combustion engine powertrains, replacing these with lowcarbon, renewable fuels and alternative low-emission, high-efficiency powertrains. The main technological options that are currently being considered are: improved internal combustion engine vehicles (ICEVs) powered by biofuels, battery electric vehicles (BEVs) and hydrogen fuel cell vehicles (FCVs). Hybrid solutions are also possible, such as battery electric vehicles equipped with range extenders (PHEVs), be they internal combustion engines or fuel cells. All three fuels considered (i.e.: biofuels, electricity and hydrogen) are in principle compatible with the objectives of moving away from oil. They may also significantly reduce GHG emissions from road transport, depending on the primary energy sources and the processes used for their production. Only electricity and hydrogen have zero emissions of air pollutants at point of use. However it should be noted that biofuels that are oxygenates (e.g. ethanol) may also help reduce emissions from combustionb) Why are energy storage techniques of importance when considering renewable energy supplies? Describe three different methods for storing energy and briefly state how they might be used. (2 marks + 4 marks)The future Smart grid will consist of increased quantities of grid connected renewables. This will consist of wind power and solar power. Both forms of energy are intermittent and variable in nature. Therefore storage technologies will be needed to regulate the supply from these sources. Energy storage can also be used to mitigate voltage fluctuations and improve power quality issues such as harmonics. (2 marks)Give 3 examples of the different storage technologies and briefly explain its principle of operation; Batteries; (2 marks)Simply based on two electrodes and an electrolyte. Large utility applications use Lead acid and Sodium Sulphur (NaS).Lead acid have been used for many years in utility applications providing excitation for synchronous generators and as back up auxiliary power supplies. These are cheap but require significant maintenance. Their lifetime is particularly short if discharge too deeply. NaS batteries operate at very high temperatures (300-400 deg C) and have large capacity per unit volume and weight. Li-ion is progressively being used also but primarily for Electric vehicles.NiCd is a robust technology with high energy density. It was traditionally used in power tools and mobile phones etc however it has been displaced by other technologies recently. It has been chosen for a flagship storage project in the USA. Fuel Cells; (1 marks)Using hydrogen and oxygen are another possible storage technology. One of the challenges is the energy re