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Module 1: Solar and Wind Energy

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Wind Energy Systems

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Hi friends, now we will discuss on the topic wind energy. As you know, the wind is nothingbut the flow of air or the motion of air. All we have seen that in case of fan, we getair motion and we have to apply electricity, then electricity helps the movement of thefan blade and that is electrical to mechanical energy conversion and then movement of thefan blade gives some motion to the air, so that is the air is getting some motion, andit is getting some energy inside it that is the kinetic energy of the air when it is inmotion.So that way, electrical energy is converted to the kinetic energy of the air in case offan. Now if you can develop a certain device which can work in the reverse principle ofthe fan, then we can be able to convert the kinetic energy stored in the air into electricalenergy. So, the principle of wind energy is this one and we will discuss more about thiswind energy in two classes.In this class, our content is the wind as a source of energy, then wind energy system.We will also discuss types of wind machines and then energy production from wind and windenergy computation and the nature of wind and selection of site.Now, we will see wind as a source of energy. As you know that wind is nothing but the motionof air, so air gets some velocity and this is because of the uneven heating of the earth’ssurface. Earth’s surface we have land area, we have water. So land and water both arenot equally heated. So we know that the land gets heated very quickly with respect to waterand again gets cooled very quickly with respect to water. So during daytime, the land areagets heated quickly and becomes warmer than the water.So air at the land being warm up goes upward and the cold air on the water moves towardsthe land area. So we get one movement of air or the motion of air that is called wind,so that wind is generated in the daytime in that way, and the nighttime the reverse directionswe can get. So, this is the reason why we can get the wind and this wind gives somekinetic energy to the air and that kinetic energy we can convert into the electricalenergy.Once we want to get the electrical energy or mechanical energy from the wind, then wehave to get good velocity of the wind and we know that the more wind speed more is thevelocity of atmospheric air and more is the kinetic energy.So now we will see the wind energy is the kinetic energy associated with the movementof atmospheric air. It has been used for hundreds of years for sailing, grinding grains andfor irrigation. So these are the ancient applications of the wind energy and recent applicationis the electricity production. So wind energy systems convert this kinetic energy into moreuseful forms of the power.Then in 2017, report says that in the world we have 539,581 megawatts wind energy productioncapacity and the total production of this 539,581 megawatt electricity is produced inthe world, out of this around 85% that is equal to 456,572 megawatt is produced in 10top countries and this 15% is produced in the rest of the world. So if we think aboutthe top 10 countries around the world producing wind energy, the China ranks the first position,it is 188,232 megawatt capacity and its share is 35%.So in China around 35% energy is produced from the wind. For US, the second position,89,077 megawatt that is 17% share, for Germany the third position 56,132 megawatt and theshare is 10%, for India we have 32,848 megawatt with a share of 6% energy. So out of the totalenergy produced in the country, we are having 6% percent of the wind energy. Spain, we have23,170 megawatt and 4% that is their share, then United Kingdom 18,872, it has 3% share.Then France 13,759 megawatt with 3% percent share, then Brazil 12,763 megawatt with 2%share, Canada 12,239 megawatt with 2% percent share, and Italy 9,479 megawatt with 2% share.So this is the statistics of the wind energy productions around the globe as per 2017 data.Now if we think about the wind energy system, then we can get 3 major terms; one is windmachine, then windmill, and wind turbines. So wind machines machine convert the windenergy into useful form of energy that is a very generic common name wind machine. Thenwindmill works on the principle of converting kinetic energy of the wind to mechanical energy,so windmill really when the kinetic energy of wind is converted to mechanical energyspecifically that is windmill, and wind turbines that transform the energy in the wind intomechanical power which is further transferred into electrical energy.So the wind turbine is the latest development and it helps to generate electricity fromthe wind. Now how the kinetic energy of air is transferred into electrical energy thatis the point we have to understand. So when the wind is available when the air is in amotion and we have some windmill, say windmill has some blades, some rotor, and then windflows it is blocked by the blades, and then downwind of the blades, there is some lowpressure creation and that low pressure gives a lift to the blades, that is one force theblade is getting.Another is a drag force when the wind is blowing and then there is some resistance the bladeis giving some resistance there were some a drag force on it, and then the drag forceas well as the lift both helps to rotate the rotor. This is the mechanism, the rotor rotatesand then the blades also rotate. So if the rotor is fixed with a shaft, then shaft willalso rotate, and one side of that shaft can be connected to turbine and then turbine togenerator, so then electricity can be produced.So the force of the lift is actually much stronger than the winds force against thefront side of the blade which is called drag, and the combination of lift and drag is whatcauses the rotor to spin. This rotating action then turns a generator which creates electricity.So, this is the mechanism through which electricity is produced. Now the wind power which is availablethat is proportional to the cubic root of the velocity of the wind.So if the wind speed doubles, then the power potential will increase 8 times. So wind energytechnologies use the energy in wind for practical purposes such as generating electricity, chargingbatteries, pumping water and grinding grains, already we have discussed.Now we will see the wind energy systems. So here, this figure shows us 2 systems, oneis your old windmill and another is a new wind turbine. So old windmill we have bricktower where in both the cases we see the blades or the rotor is placed at certain elevation,so not at the immediate surface of the earth, it is put at certain height. So the heightis required because as you go up the wind velocity increases.At the surface, the resistance is more so wind velocity is less, if we go up, then thevelocity will be higher, and more the velocity, obviously more will be the energy potential.So around 30 meters height, the windmill is installed here, so that is the old model andthis in case of new model we are getting that is new wind turbine the blades are different,the fiberglass blades we are getting are here we are having wooden blades. So this is thebasic difference between 2 types of windmills and wind turbine.For getting optimum efficiency or considerable performance, the wind speed should be 16 to20 and the height may be at 50 meter. So these are the typical values which give us optimumresult. Now we see the wind turbine, the size is 2 megawatt, from handle to 2 megawatt capacitywind turbine is available and typical is 250 kilowatt and then rotor diameter 25 to 30meter and tower height is 30 to 40 mete. So this is the descriptions of wind system.Then we will see the types of wind machines. So wind machines that helps to convert electricalenergy to any useable form of energy, so that wind machines may be of 2 types, one is yourhorizontal axis machines and vertical axis machines. So there are 2 types of machines.So here we see this is your horizontal axis machine, in this machine, we are seeing herethe axis of rotation is parallel to the direction of the wind stream. So this is our axis ofrotation, so we need also flowing in these directions and rotor is also there.So we see here rotor axis is horizontal and can be adjusted so that it is parallel tothe direction of the wind. So this rotor which we are having here this is axis is parallelto the direction of the wind, and it is not necessary that every time the wind is flowingin the same direction, so the direction will change, when the wind direction will change,the rotor movement the direction can also be changed in this design.So horizontal axis windmill can further be classified as single blade, double blade,and multi blades. So these blade, we have 3 blades we are seeing that may be single,double, or multi blade system we may have.Then we will see vertical axis machines, so this is an example of vertical axis. So thisis our axis vertical and these are our blades. So in this case, this is the simplest of themodern types of windmill which works like a cup anemometer and rotor axis is verticaland fixed and then the machine has become popular since it requires relatively low airspeed or air velocity or wind velocity. So low velocity of winds can even generate electricityby this type of machines with respect to horizontal axis windmill.Now we will see the comparison between these 2, horizontal axis and vertical axis machines.So in case of horizontal axis, a controller is required for proper orientations. So whenthe wind speed is changing, then the directions of the rotor will also change by use of somecontroller, this is very important for horizontal axis machines, but in case of vertical axisnot necessary, no controller is required because this is fixed.Then variable speed power transmission for better control, then simpler power transmissionsystem for the rotor to the axis, variable blade pitch can tolerate wider fluctuationsin wind speed in case of horizontal axis machines. For vertical the machine performance evenat low wind velocity running 8 kilometer per hour to 16 kilometer per hour. The horizontalaxis is lightweight and vertical axis it is bulky having much material.Horizontal axis high tower ensures greater wind speed and low height installation leadsto lower available wind speed in case of vertical axis. So it is very clear to us that verticalaxis height is lesser than that of the horizontal axis machines. Now we will see how to calculatethe power productions or power efficiency or power potential of wind machines or wind.So power production from wind. Now we will see the kinetic energy of a mass of m movingat a speed of v is 1/2mv2. We know that kinetic energy is equal to 1/2mv2. So what will bethe power potential, the power potential will be 1/2 x mass flow rate, so this m will bemass flow rate, x v2, that means that this much of power I can get for unit time, thisis the power potential.Now the mass of air passing in unit time through an area A, so this is our area A, cross sectionalarea, wind is flowing in this direction, so what will happen in this case, what will themass of it, mass is volume x density, so volume of air x density of air. So volume of airmeans area x the linear velocity. So that is A is the area x v where v is the wind speed,so this is our volume of air which is passing through this cross sectional area in unittime, so that is the v.So if we multiply it with the rho of air, so this is the mass flow rate is equal torho x A x v. So in this expression, what we can get 1/2 mv square, 1/2 m is equal to rhointo A into v and v square, so we are getting 1/2 into rho into A into v cube where A isa constant for this particular A, but this is our power potential, 1/2 into rho intoA into v cube.Now the power potential which is available in the wind that wind will not be convertedor will not be available for its transfer because after getting resistance at the blade,the wind will reduce its velocity to 0, that is why the energy which is available withthe wind the complete conversion will not be possible and it has been proved by differentresearchers particularly the Albert Betz in 1929, he has recommended and proved that around59.3% of energy associated with the wind may be transferred in the turbine.So the power potential we can express or the turbine power we can express as 1/2 into rhointo A into v cube which we had into some constant that is Cp, so Cp is called the powercoefficient, and this Cp is the power coefficient, so what we are getting now P is equal to 1/2into rho into A into v cube into Cp. Now A, Cp, rho and 1/2 or A may change, rho is theproperty of the wind, and Cp is also a power coefficient, so these are constant and 1/2into this equal to constant, we are considering is equal to K, so this is equal to K.So K into A into v cube, this is our power potential and this 1/2 into rho into Cp thatis almost around to 0.37, it has been proved by many experimentation and approximatelythis can be 0.37 for air. So if this is, then what will be our total power potential interms of electricity, so that will be this into 0.37 into A into v cube into some efficiencyfactors, this is for wind energy which will be available, then wind to mechanical conversionand mechanical to electrical conversions and there will be transmission.So another 3 types of efficiency we have to multiply, efficiency mechanical, efficiencyelectrical, and efficiency transmission so that we have to consider.So overall power will be getting 0.37 into efficiency of mechanical conversion, kineticto mechanical, then mechanical to electrical, and thereafter transmission, so then the machineryis used for the transmission so that there will be always having some efficiency term.So this and we have v/10 cube because this power efficiency we are considering thesepower potential it is considered in terms of kilowatt, so here we are having what unit,this unit if we want to convert into kilowatt that is divided by 1000 and that is writtenv/10 to the power cube.So v cube we had already, now v cube/10, 1000 is there, so v/10 to the power cube that hasbeen made. So this is the expression which is used to calculate the power potential inkilowatt unit when the v is meter per second, the wind speed is in meter per second. Sothis way we can calculate the power potential of the windmill.Now as you know that the power potential is proportional to the v cube that is the windvelocity. Now we have one blade, so if we have a rotor, then we have some blades, sothese blades will be moving and then we will get energy. So here, this is the blade tip,so it may be the end of it or any position it may be, so this wind linear velocity isgiving the rotational movement to the rotor and to the blade also and this movement theRPM will be varying, if we have 2 different machines with different diameter of the rotor,the RPM will be varying.So to compare these variations in the RPM in spite of using RPM, another term can beused, that is tip-speed-ratio. So tip speed ratio is defined as the blade tip speed meterper second by wind speed, what is the wind speed and what is the blade tip speed we aregetting, so that way we can get one number and that is easy to compare 2 different machines.So this TSR range is 1 to 15 and it is defined as this blade tip speed/wind speed. Now whatis the blade tip speed in meter per second.If we know the RPM, then 2pi r into N, N is the revolution per minute and per 1 revolutionit 2pi r it moves, r is the radius of the blade. So 2 pi r into N, N is the RPM, sothat will be the linear distance covered by the tip. So this is the blade tip speed 2pir N and then this is wind speed so per minute revolution, so time is our 1 minute, so 60second, so 60 second into v, so v is the wind speed.So now we are getting tip speed ratio is equal to 2 pi r N/60 into v, where r is the radiusat which SR is calculated, that it speed ratio is calculated and v is the wind speed in meterper second and N is the RPM. Now we can calculate the speed at any position also, it may ber, many many values, so that way that is called speed ratio. So here we have a speed, herewe have a speed because this point has to travel more distance, this point has to travelless distance. So SR that is speed ratio will be different for these 2 cases where windspeed is fixed.So that is the concept has been developed to determine the tip speed ratio to comparedifferent turbine blades and machines.Next we will see the wind shear. So if we go up, we will see that the wind speed isincreased. So once the wind speed is increased, then obviously the power potential will alsobe increased. So there are some mathematical relationships the people try to correlatethese increasing the speed with the height and two types of equations are available,that is logarithmic equation and power equation. So this is our logarithmic equations, it isclear that velocity unknown is equal to velocity known into ln height desired by beta/ln heightknown by beta.So if we have height 1, here is one machine, here one machines we like to put, say thisis your h1 from the base, this is our h2 from the base, so I know there is a wind speedhere, so it is v1 and this is v2, I do not know. So we will be using here that is yourheight, so velocity unknown, say v2, it is not known, so here v2 is equal to v1 intoyour ln height of v1 that is known and height of v2 that is h2, height h2/beta.So this beta is called a roughness constant and this beta value will depend upon the roughnessof the surface and it will vary from place to place, different types of surface, differenttypes of beta value you will get. Those beta value normally ranges from 0.0001 to 3.0 m.So it has wide variations in the beta depending upon the nature of the surface.The other expression which is used to predict the wind speed at any height if we know thewind speed at any other height that is equal to that is a power equation, so velocity atunknown is equal to velocity of known into height of desired and then height of known,so height desired means that is for unknown and this is for known velocity and heightand then to the power alpha, that is a power expression.So this alpha is called a wind shear exponent and one thumb rule is that this alpha valueis around 1/7th, so a generally recognized rule of thumb is that wind speed increasesas the 1/7th power of the height above ground and here this table give some example of differentterrain and different alpha values, so it is somewhere it is 0.10, 0.14, 0.16, 0.2,0.2 to 0.24 and these are the difference conditions.Now we will see some wind energy computation. So how the wind energy can be calculated withsome numerical problems, we will discuss now. Say a horizontal axis wind turbine with 30m rotor diameter produces 1 megawatt electricity at a wind speed of 6 kilometer per hour. Thencalculate the following, blade tip speed for a tip speed ratio of 4.25 and then overallpercent conversion efficiency of the wind turbine from wind energy to electricity, airdensity is 1.225 kg per meter cube. So this is our problem statement.We have to calculate the blade tip speed. That means if we have a blade here, so thistip speed we have to calculate and it is given as that our TSR is equal to 4.25. TSR is equalto we have come to know that tip speed/wind speed. So tip speed is equal to wind speedinto TSR. So here TSR is equal to 4.25 and wind speed is equal to 60 kilometer per hour,so 60 into 1000 meter/3600 per second. So we are getting now 70.833 meter per second.So if we convert it into kilometer per hour, then it is 254.9988 kilometer per hour.So this is our blade tip speed here, linear velocity of the blade tip we are getting.Second overall percent conversion efficiency we have to calculate. So now we will see whatamount of electricity we are producing here and what was the maximum potential we havehere. So to calculate potential, we need to get the value of wind speed and here we havev equal to 60 kilometer per hour that if you have to convert it into meter per second andso 60 into 1000 meter/3600 that is meter per second, so that is 16.6666 meter per second,so this is our v.Then what will be the theoretical energy available, what is the theoretical energy available 1/2into what is this we need v cube and we need A and we need rho. So P is equal to 1/2 intorho into A into v cube. So 1/2 into rho is equal to 1.225 kg per meter cube and thenA is equal to the area, this is the area, the blade which is having the diameter ofhow much 30 meter rotor diameter. So pi r square, we are having pi r square/4, pi dsquare/4 by four. So we are getting A is equal to pi d square/4 or equal to pi r square.So this is the area we are having, so that area we will get here and put this value here,rho value is given here, and v value I have calculated here this one. So by putting theseexpressions, we are getting the available energy in the wind is equal to 1/2 into rhointo A into meter cube, so this is equal to this megawatt. This will be 2.003361 into10 to the power of 6 watt, that is equal to megawatt we are getting.So this is the megawatt energy available in the wind and as electricity we are getting1 megawatt. So the conversion efficiency we can get1/2.003361 into 100 that is equal to49.91% or approximately 50% efficiency we are getting.Now we will see another problem statement. So this is a figure, it shows that one horizontalaxis windmill is there and what we see the distance is H and this height and the velocityis related with this formula V is equal to KH square. Then figure 1 shows the variationof the height from the surface of the earth to wind speed, so this is our velocity andthis is our surface of the this is the trend at which the velocity changes with height.Then it is given as V is equal to KH square, K is equal to 1 per meter per second.Calculate the minimum height at which a horizontal axis windmill should be adapted to generatea wind power density of 10 kilowatt per meter square normal to the direction of the wind.So this is our 10 kilowatt per meter square energy potential or energy density that wehave to calculate the height at which the mill has to be put. So now the other informationis given the electrical efficiency and mechanical efficiency of the system are both 0.9 andwhereas the transmission efficiency is 0.52.So mechanical equal to efficiency electrical equal to 0.9 and efficiency of transmissionequal to 0.52 it is given, so what is our case. Our P is equal to you know 0.37 inton into efficiency of electricity, efficiency of transmission and then we have V/10 to thepower cube, this is the formula we have. So this formula we have, now in this formulaour P is given 10 kilowatt per meter square, so this is kilowatt per meter square, so kilometerunit, and then we are having so A term is not there per meter, so A is not here, A hascome to this place.So then we are having this n is equal to 0.9, eta e is equal to 0.9, eta t is equal to 0.52and V, V is equal to how much that I know that V/10 to the power cube, this is the expression.So from these expressions, we can calculate the value of V, so value of V is equal tothis much 64 to the power 1/3 that is 4 into 10 meter per second, so 4 into 10 meter persecond, then 40 meter per second we are getting that V and that if we get the value of V,K is equal to 1, so H square equal to we are getting 40, so H is equal to 1/2 of 40, sothat is equal to 6.32 meter.So at 6.32 meter if we put this turbine, then we will get this 10 kilo watt per meter squarepower density.Next we will see the nature of wind and selection of site. Nature of wind means whether it willbe having sufficient speed or not, that will depend upon many factors, that is your latitudeof the place, altitude of the place, topography of the place, and scale of the hour, month,and year at what time, in which month and which year; so all those things will influencethe wind speed and the type of wind that we mean. Then suitable sites for the wind energyinstallation that is the best sites at offshore and on the sea coast.It is as we know that day time and night time, there will be wind, in both the cases therewill be direction of wind, so at offshore and on sea coast, this is a best place forthe installation of the wind turbines. The second best sites are in the mountains wherethe height is high, so air velocity will be more, so that is also a good location forthe installation of the wind turbines. The lowest level of wind energy is found in planes.So planes it is not so important or so economic or so feasible to install the wind turbines.So then for wind turbines, it also requires lot of land, so we need lot of area availablefor the installation of the wind turbines because if I put one wind turbine here, soI have to keep some space between 2 wind turbines for the installation. For example, you seethis is our arrangement of wind turbines, so this is one, another one, another one,another one. So these and these are rows and these columns also we are having turbine here,turbine here, turbine here.So this one turbine to turbine distance and this turbine to turbine distance it has tobe maintained. People try to understand and then to optimize the distance required andbasically after turbine the downstream of that the wind speed becomes less and thenwe need to give some more time to recover this wind velocity and then before reachingto another turbine. So this gap is necessary. So once the gap is necessary, what will bethat land requirement, that will depend upon the number of turbines I need to install orthe capacity of the plant.The overall capacity of the plant I want to get from the windmills that should be knownto us, so that we can calculate the land requirement. So thumb rule says that spacing of turbinesbetween 5 and 10 rotor diameter apart. If prevailing winds are generally from the samedirection, turbines may be installed 3 or 4 rotor diameters apart in the direction perpendicularto the prevailing winds and under multi directional wind conditions spacing of between 5 and 7rotor diameter is recommended.Here it is mentioned that a simple rule of thumb the turbines follow the spacing ruleof 4 rotor diameters apart horizontally and 7 rotor diameters apart vertically in thewind’s direction. So we have made some discussions on wind energy productions. Thank you foryour patience.