Analysis of Wind Turbine and Design of a Solar Energy System in UK - Assignment Example

Question 1 A 3 blades wind turbine with diameter of 50 m produce 550 kW at a wind speed of 10 m/s. Air density is 1.225 kg/m3. Under these conditions (i) What is the rotational speed of the rotor when it operates at tip speed ratio of 5.0? Blade tip speed =  = 5.0 x 10 = 50m/s (ii) What is the tip speed of the rotor? Blade tip speed =  = 5.0 x 10 = 50m/s (iii) If the generator needs to turn at 1800 rpm, what gear ratio is needed to match the rotor speed to the generator speed? Gear ratio = (iv) What is the overall efficiency of the complete wind turbine (blades, gear box and generator) under these conditions? Theoretical power available = =1202.031kW Efficiency =  Question 2 Draw a power output versus wind speed diagram for a typical industrial wind turbine and label all important parameters on the graph. Explain what are (i) Cut-in speed, (ii) Rated wind speed, (iii) Cut-out (Furling) wind speed. Diagram 1: Power output versus wind speed diagram for a typical industrial wind turbine In order for wind turbines to operate efficiently there are some operating limits that the turbines should operate under depending on the prevailing wind conditions and how the entire system has been designed. The following are the various distinct conditions that have been identified in the graph. Cut - in Wind Speed This is the lowest speed of wind that can allow generation of useful power and when the wind goes below it there will be no useful power that can be produced by the turbine. The typical cut in wind speed is between 3 and 4m/s. Rated Wind Speed This is the wind that is indicated on the nameplate of the wind turbine and is the lowest wind velocity where the turbine is able to generate its full power. The rated wind speed corresponds to the maximum safe electrical generating capacity which can be safely be handled by the electrical generator simply stated it is the electrical rating of the turbine (Sheldahl, R. E. et al.,1980). Usually the rated wind speed has a value of about 15m/, this being about twice the average speed of wind that is expected. So as to keep the turbine operating at wind speeds that is well above the wind speed rating, there may be utilization of control systems in the variation of the pitch of turbine blades, resulting to reduction in rotation speed of rotor and as a result there will be a limit of the mechanical power being subjected to the generator and this will make the electrical output to be relatively constant. As much as turbines will function with wind speeds close to cut-out wind speed, there is automatic reduction in efficiency when the speeds are above the rated speed, such that very little of the available wind energy is captured as a measure of protecting the generator (Worstell, M. H.,1979). As much as it would be possible to install larger generators for extraction of full power from the wind at speeds over the rated wind speed, but doing this would not be economical, because wind speeds above rated wind speed does not occur frequently. Cut - out Wind Speed This is the maximum speed at which the turbine can work without being damaged and beyond this speed the design of the turbine is such that there is a shut down through application of brakes to ensure no damage of the system. The turbine can also be slowed down through furling or stalling in addition to electrical and mechanical braking. Stalling: It is a passive self correcting strategy that is applicable in fixed speed wind turbines. With an increase in wind speed, there is a corresponding increase of wind of attack up to a point where a stalling angle is reached and at that point there is destruction of the lift force turning the blade. However, an increase in the angle of attack is accompanied by an increase in the effective cross section of the blade face-on to the wind, consequently the direct force of the wind and the associated stress being subjected to the blades. A turbine blade which is fully stalled when stopped will be found to have the flat side of the blade directly facing the direction in which the wind is coming from. Furling or Feathering: This is a technique whose origin is associated with sailing where the pitch control of the blades is utilized in the reduction of the angle of attack which in turn results to reduction in the lift that is subjected to the blades in addition to the effective cross section the blades facing in the direction of origin of wind (S. N. L. Staff, 1992). On stopping a fully furled turbine makes the edge of the blade to face into the wind resulting into a reduction of the wind force as well as the stress on the blade. Usually the specification of the cut-out speed is made considerably high in line with the safety requirements and practicality so as to be able to capture most of the available wind energy in the spectrum of the wind speed expected. Question 3 (a) Plot power in the wind versus wind velocity for wind in the range of 1m/s to 15 m/s. Wind speed Power 1 0.6125 2 4.9 3 16.5375 4 39.2 5 76.5625 6 132.3 7 210.0875 8 313.6 9 446.5125 10 612.5 11 815.2375 12 1058.4 13 1345.663 14 1680.7 15 2067.188 (b) Compare the energy at 15C, 1 atm pressure, contained in 1 m2 for wind in the following conditions: (i) 300 hours of 6 m/s winds Power at 6m/s Theoretical power available = =132.3W Energy in 300hrs= 300x60x60x132.3=142.884Mj (ii) 150 hours at 5m/s and 150 hours at 7 m/s (note the average wind is 6 m/s) Power at 5m/s Theoretical power available = =76.56W Energy in 150hrs= 150x60x60x76.56=41.34Mj Power at 7m/s Theoretical power available = =210.09W Energy in 150hrs= 150x60x60x210.09=113.45Mj Total energy for 300hrs = 41.34+ 113.45= 154.79Mj. It is clear that when we consider the two speeds separately and adding the total energy it results to higher energy of 154.79Mj as compared to using a mean velocity of 6m/s in a total time of 300hrs as these results to a lower energy of 142.88Mj. Question 4 a Betz methods Pitch angle formula for Betz method is Where X is the tip speed ration (TSR) , is the angle of attack Chord length formula for Betz method is Where  is the coefficient of lift at the chosen design angle of attack  B is the number of blades Schmitz method Where is angle of relative wind to rotor plane and the other symbols are as defined in the Betz method From the equation for pitch angle of 0 will be equal to angle of attack  We choose the angle of attack that has the largest lift to drag ratio in table 4.1 thus  is 6 = The angle of twist is proportional to the distance of the point away from the hub. Angle of twist (Schmizt) =  X= r/R r Cr Schmitz Betz Schmitz( Betz 0.1 3 0.979639 0.005952 13.58 13.90 0.2 6 0.61071 0.011903 7.01 7.05 0.3 9 0.429806 0.017855 4.70 4.72 0.4 12 0.329011 0.023806 3.54 3.54 0.5 15 0.265788 0.029758 2.83 2.83 0.6 18 0.222685 0.03571 2.36 2.36 0.7 21 0.191498 0.041661 2.02 2.02 0.8 24 0.167919 0.047613 1.77 1.77 0.9 27 0.14948 0.053564 1.57 1.58 1 30 0.134674 0.059516 1.42 1.42 (b) Explain why the blades have to be twisted. Wind travelling on the downwind face of the blade which is rounded, the wind movement should be faster so as to make it possible to reach the end of the blade in time to coincide with the wind that travelled over the upwind face which is flat. With the faster moving air having a tendency of rising in the atmosphere, it will follow that the downwind curved surface will end up with lower pressure above it. There is suction of the low-pressure area by the blade coming from downwind direction, with the effect being referred to as lift. In the side which is upwind of the blade, there is slower movement of wind and this creates an area of higher pressure that is able to push the blade making to slow down. By having turbine blades being twisted, it ensures that an angle is presented that will always take advantage of the ideal lift-to-drag force ratio. Question 5 A yearly histogram of wind speed for a location is shown in Figure 5.1. Find the average wind speed and average power in the wind (W/m²). Speed Time (hrs) Timexspeed 0 24 0 1 276 276 2 527 1054 3 729 2187 4 869 3476 5 941 4705 6 946 5676 7 896 6272 8 805 6440 9 690 6210 10 565 5650 11 444 4884 12 335 4020 13 243 3159 14 170 2380 15 114 1710 16 74 1184 17 46 782 18 28 504 19 16 304 20 9 180 21 5 105 22 3 66 23 1 23 24 1 24 25 0 0 Average wind speed= Power =  References BS EN 61400-12-1(2006). Power performance measurements of electricity producing wind turbines,” Gottschalg R, Infield DG, Kearney MJ (2003) Experimental study of variations of the solar spectrum of relevance to thin film solar cells. Solar Energy Materials & Solar Cells 79: 527-37. Green MA, Emery K, Hishikawa Y, Warta W, Dunlop ED (2013) Solar cell efficiency tables (version 42). Progress in Photovoltaics: Research and Applications 21: 827-37. IEC 61400-12 (1998). Wind turbine power performance testing,”. John B.(2007). Basic Engineering Mathematics published by Elsevier Ltd. How Wind Power Works Sheldahl, R. E. et al. (1980). “Aerodynamic performance of a 5-metre-diameter Darrieus turbine with extruded aluminum NACA-0015 blades,” Sandia National Laboratories ReportSAND80-017, S. N. L. Staff (1992). “Selected Papers on Wind Energy Technology,” SandiaNational Laboratories Report SAND90-161. Worstell, M. H. (1979). “Aerodynamic performance of the 17-metre-diameter Darrieus wind turbine,” Sandia National Laboratories Report SAND78-173. . Photovoltaics In photovoltaic solar cells there is use of semiconductor materials such as silicon where the incoming photons separate positive and negative charge carriers in the cells. Through this there is production of an electrical voltage and consequently an electrical current which is capable of driving the load. With the solar cells being modular, it is possible to have an assembly of units of any desired size. Figure 1 Solar system Solar thermal plants are usually made of parabolic trough and solar power tower plants. The trough collectors make up the solar field of a parabolic trough power plant and encompass large cylindrical parabolic mirrors that are able to concentrate the sunlight on a line of focus. There is installation of several collectors in rows to a length of about 100m with the total solar field being compost of a number of such parallel rows. In a solar thermal plant it is possible for a fossil burner to be used in driving the water steam cycle in the periods when there is no sunlight such as in bad weather or at night. Capacity is guaranteed in solar thermal power plants unlike in the case of the photovoltaic systems. This option makes the solar thermal plants to be more attractive and its quality of planning distribution into the power grid. There is also the possibility of a thermal storage being used as a replacement or complement of fossil burner whereby it will be possible for the power plant to be run with neutral carbon dioxide emissions. In such an arrangement stored heat from storage is used in driving the cycle during the time of sunlight deficit. The advantage of solar voltaic cell over the solar thermal system is that is much simplified since the cells generates electricity directly from sunlight unlike in the later water is heated first so as to be used to drive turbines which would then generate electricity. c) Discuss: how new materials (including nano materials) and design modifications can improve the efficiency of energy harvesting for your system. Use of nanotechnology is important of high level efficiency in solar cells as this technology is used n production of thin film photovoltaic devices. When compared to the commonly used silicon PV panels, the thin film solar panels have advantage when it comes to low-light situations, their flexibility, have low weight and roll-to-roll production capability. The other advantage associated with the nanotechnology photovoltaic devices is of having a higher watt-peak capability for their respective power rating on the basis of net power per day (Gottschalg, 2003). Applying of material science and nanotechnology there has been achievement of efficiency levels of 20.4% in single junction thin film solar cells in CIGS solar cells while in CdTe and dye sensitized solar cells efficiency achievements of 19.6% and 11.9% respectively have been achieved (Gottschalg, 2003) Figure 2Concentrating Solar Power (CSP) mirror mounted on a solar tracker Source: http://www.solarthermalpower.com/ . 2) Select/design an energy storage system. a) Select/design an energy storage system for the energy collector proposed in part 1. Produce a specification for your design. The storage system utilized in the system comprise of batteries which will be used in conversion of the electric energy into chemical energy and later back to electrical energy for use. A deep cycle battery will be the preferred choice used in the solar PV system. This is owing to the fact that deep cycle battery design is such that it can be discharged to a very low energy level and can also be recharged at fast rate or the battery can be cycle charged and discharged on a daily basis for a good number of years. It is important for the battery to be large enough size so that to ensure that enough energy is stored to fulfill the energy requirement of the energy needed by the appliances at night and bad weather days. b) Justification on your design. i. What are the major factors need to be concerned in the design, give details on your assumptions? Some of the issue that need to be considered are the type of appliances that are to consume the power in terms of the rate of consumption of power and number of hours they are used per day. ii. Design calculations For the series charge controller type, the sizing of controller depends on the total PV input current which is delivered to the controller and also depends on PV panel configuration (series or parallel configuration). According to standard practice, the sizing of solar charge controller is to take the short circuit current (Isc) of the PV array, and multiply it by 1.3 Solar charge controller rating = Total short circuit current of PV array x 1.3 Total generated per year = 2500KWh Total energy o be consumed per day =  Total PV panels energy needed =6800x1.3=8840Wh/day Size the PV panel Total Wp of PV panel required = 8840/3.4 = 2600Wp Number of panels required =2600/110 =23.6 modules                                                            The actual required number of modules = 24modules   The system will need to be powered by at least 24modules of 110 Wp PV module. Appliances in used Four 18 Watt fluorescent lamp used 5 hours per day. Two 60 Watt fan used for 2 hours per day. One 75 Watt refrigerator that runs 24 hours per day with compressor run 12 hours and off 12 hours. Inverter sizing Total Watt of all appliances = 4x18 + 60x2 + 75 = 267 W   For safety, the inverter should be considered 25-30% bigger size.   The inverter size should be about 320W or greater. Battery sizing     Total appliances use = (4x18 W x 4 hours) + (2x60 W x 2 hours) + (75 W x 12 hours)     Nominal battery voltage = 12 V     Days of autonomy = 3 days Battery capacity = [(4x18 W x 4 hours) + (2x60 W x 2 hours) + (75 W x 12 hours)] x 3                                                 (0.85 x 0.6 x 12)     Total Ampere-hours required 1141.17 Ah     So the battery should be rated 12 V 1200 Ah for 3 day autonomy. iii. Compare the advantage and disadvantages of different types of storage. The pumped hydroelectric storage and battery storage are some of the methods that can be used to store power of the photovoltaic system (Fthenakis, 2011). The battery system is good since there is little wastage of power. The battery system is also simple compared to the pumped hydroelectric storage which needs to incorporate water and a pump that will pump the water to a higher ground. For electric energy to be obtained the water which is raised is released through a pipe to a lower ground so that it rotates a turbine so that electricity is generated. All this process involves several components and this means high cost at the same time there is high losses of energy. c) Discuss: how new materials (including nano materials) can improve the performance of the storage system? The disadvantages associated with carbon anode lithium ion batteries are twofold: having a modest lithium loading capacity that puts a limitation to the level of energy stored per gram of node material and also the performance is interfered with whenever there is a fall in temperature (Frost, 2009). A nano-infused electrode is able to accommodate higher level loading of lithium and through matching with the required electrolyte the batteries are able to retain performance in a much wider range of temperature variation (Kadirgan, 2006). Test results of electrodes and batteries incorporating nano-silicon anodes have shown superior performance where • There is over 300 percent higher capacity when compared to the already existing carbon anode materials • There is retention of 70 percent of discharge capacity in 300 charge/discharge cycles • 80 percent reduction in anode weight Through this developments there is 25% improvement in energy density in batteries which have been designed as drop-in replacements (Kamat, 2006). 3) Predicate the payback time of your system. a) Assuming: the average price of energy is £0.14p/kWh and the annual increment is 6%; the fixed annual charge rate is 4.0% and the annual operation and maintenance cost is 3% of the initial cost. The government support for Feed-in tariff should not be considered in your calculation. Cost of system Cost of panel The cost of solar panel has a range of 2.40 per watt to 5 per The total Wp for system = 2600 Taking a modest value of cost of solar panel of 4 per watt Total solar panel cost = 2600x4= 10400 For typical application inverter cost =2600 Battery cost = per kWh of storage Total kWh of system = (2 x1420)/1000=2.86 Therefore for the system battery cost = 100x2.86= 286 Installation and maintenance cost = 200 b) Discuss what changes can be made to reduce the payback time. In order for the payback time to be reduced there the cost of the panel need to be reduced to a lower value of say  per watt Total cost = 10400+2600+286+200 = 13486 Table 3 Year 6% cost 4% cost 3% Ment. cost Total Cumul. cot 1 350 14.00 10.50 374.50 374.50 2 371 14.84 10.50 396.34 770.84 3 393.26 15.73 10.50 419.49 1190.33 4 416.86 16.67 10.50 444.03 1634.36 5 441.87 17.67 10.50 470.04 2104.41 6 468.38 18.74 10.50 497.62 2602.02 7 496.48 19.86 10.50 526.84 3128.86 8 526.27 21.05 10.50 557.82 3686.68 9 547.32 21.89 10.50 579.71 4266.40 10 580.16 23.21 10.50 613.87 4880.26 11 614.9696 24.60 10.50 650.07 5530.33 12 651.8678 26.07 10.50 688.44 6218.77 13 690.9798 27.64 10.50 729.12 6947.89 14 732.4386 29.30 10.50 772.24 7720.13 15 776.385 31.06 10.50 817.94 8538.07 16 822.968 32.92 10.50 866.39 9404.46 17 872.3461 34.89 10.50 917.74 10322.20 18 924.6869 36.99 10.50 972.17 11294.37 19 980.1681 39.21 10.50 1029.87 12324.25 20 1038.978 41.56 10.50 1091.04 13415.28 21 1101.317 44.05 10.50 1155.87 14571.15 The payback time is reached at the point when cumulative cost just exceeds the installation cost of13486, from table 3 it can be observed that the payback time is 21 years. References Vasilis Fthenakis (2011). How Long Does it Take for Photovoltaics to Produce the Energy Used? Communities industry Frost & Sullivan (2009 ) . Global solar power markets. San Antonio: s.n.,. Kadirgan, F. (2006). Electrochemical Nano-Coating Processes in Solar Energy Systems. International Journal of Photoenergy. Kamat, Prashant (2006). Carbon Nanomaterials: Building Blocks in Energy Conversion Devices. The Electrochemical Society Interface. Read More