final project

Laín Nieto Gómez 2009/2010 Curse

Contents

1.-Introduction…………………………………………………………………2

1.1.-History of wind energy........................................................................21.1.1.-Ancient times and farming use of the wind energy, also

pumping of water...................................................................................................21.1.2.- Modern times and electrical use of the wind energy.............31.1.2.2.-The era of inventions...........................................................31.1.2.3.-Scoping the utility size wind turbines to generate

electricity.....................................................................................................31.1.2.4.-The invention of the three more developed techniques.......4

1.2.-Types of wind turbines.........................................................................61.2.1.-Horizontal axis wind turbines.................................................61.1.2.1.-The number of blades influencing the initial torque and

explanation about the solidity......................................................................61.2.1.2.-Classification of the HAWT turbines in terms of the

direct/indirect contact of the wings with the air .........................................71.2.2.-Vertical axis wind turbines.....................................................81.2.3.-Darrieus rotor..........................................................................91.2.4.-Savonius rotor.........................................................................91.2.5.-Musgrove rotor......................................................................10

2.-Aerodynamic calculus...................................................................................11

2.-Power energy basics and aero dynamical calculus...............................112.1.-NACA airfoils..........................................................................112.2.-Reynolds dimensionless number calculations and variations of

drag-lift coefficients..................................................................................122.3.-NACA report 586, tables, basics and explanations..................142.4.-Relations between the coefficients and the forces, momentum

coefficient..................................................................................................182.5.-Outputpower and its relation with the spinning velocity.........20

2.6.-Step by step calculus of one section´s lift and drag force……23

3.-Mechanical calculus

3.1.-Axis calculation respect to DIN 743..................................................333.2.-Bearings selection..............................................................................34

3.2.1.-6206-RS1 (axis support one) selection justification.............34

Wind Turbine design and modelling 1


3.2.2.-Bearings concerning the vertical axis spinning.....................................................................................................383.3.-Comprobation of tabs according to DIN 6885...................................463.4.-Axis break dimensioning...................................................................48

4.-Electrical calculus……………………………………………………..........58

4.1.-Generator selection............................................................................584.2.-Batteries selection..............................................................................614.3.-Transformation of the electrical output power to utility parameters of

tension voltage and frequency.............................................................................63

5.-Conclusions....................................................................................................64

6.-Bibliography..................................................................................................67

Annexes:

Draftings……………………………………………….................................[-]



1.-Introduction:

1.1.-History of wind energy

Human efforts to harness wind for energy date back to the ancient times, when he used sails to propel ships and boats. Later, wind energy served the mankind by energising his grain grinding mills and water pumps. During its transformation from these crude and heavy devices to today’s efficient and sophisticated machines, the technology went through various phases of development.

1.1.1. - Ancient times and farming use of the wind energy, also pumping of water.

There is disagreement on the origin of the concept of using wind for mechanical power. Some believe that the concept originated in ancient Babylonia. The Babylonian emperor Hammurabi planned to use wind power for his ambitious irrigation project during seventeenth century B.C. Others argue that the birth place of wind mills is India. In Arthasastra, a classic work in Sanskrit written by Kautiliya during 4th century B.C., references are seen on lifting water with contrivances operated by wind. However, there are no records to prove that these concepts got transformed to real hardware.

The earliest documented design of wind mill dates back to 200 B.C. The Persians used wind mills for grinding grains during this period. Those were vertical axis machines having sails made with bundles of reeds or wood. The grinding stone was attached to the vertical shaft. The sails were attached to the central shaft using horizontal struts. The size of the sails was decided by the materials used for its fabrication, usually 5 m long and 9 m tall. By the 13th century, grain grinding mills were popular in most of Europe. The French adopted this technology by 1105 A.D. and the English by 1191 A.D. In contrast with the vertical axis Persian design, European mills had horizontal axis. These post mills were built with beautiful structures. The tower was circular or polygonal in cross-section and constructed in wood or brick. The rotor was manually oriented to the wind by adjusting the tail. The mill was protected against high winds by turning the rotor out of the wind or removing the canvas covering the rotor.

The Dutch, with renowned designer Jan Adriaenszoon, were the pioneers in making these mills. They made many improvements in the design and invented several types of mills. Examples are the tjasker and smock mills. The rotors were made with crude airfoil profile to improve the efficiency. Apart from grain grinding, wind mills were employed to drain marshy lands in Holland. These wind mills reached America by mid-1700, through the Dutch settlers.



This is followed by the water pumping wind mill, which is still considered as one of the most successful application of wind power. The so-called American multi bladed wind turbine appeared in the wind energy history by the mid-1800. Relatively smaller rotors, ranging from one to several meters in diameter, were used for this application. The primary motive was to pump water from a few meters below the surface for agricultural uses. These water pumpers, with its metallic blades and better engineering design, offered good field performance. 1.1.2.- Modern times and electrical use of the wind energy:

1.1.2.1.- The era of inventions:

Over six million of such units were installed in US alone, between 1850 and 1930.The era of wind electric generators began close to 1900’s. The first modern wind turbine, specifically designed for electricity generation, was constructed in Denmark in 1890. It supplied electricity to the rural areas.

During the same period, a large wind electric generator having 17 m ‘picket fence’ rotor was built in Cleveland, Ohio. For the first time, a speed-up gear box was introduced in the design. This system operated for 20 years generating its rated power of 12 kW. More systematic methods were adopted for the engineering design of turbines during this period. With low-solidity rotors and aerodynamically designed blades, these systems could give impressive field performance. By 1910, several hundreds of such machines were supplying electrical power to the villages in Denmark. By about1925, wind electric generators became commercially available in the American market. Similarly, two and three bladed propeller turbines ranging from 0.2 to 3 kW in capacity were available for charging batteries.

1.1.2.3.- Scoping the utility size wind turbines to generate electricity

Turbines with bigger capacity were also developed during this period. The first utility-scale system was installed in Russia in 1931. A 100 kW turbine was installed on the Caspian sea shore, which worked for two years and generated about 20,000 kW.h electricity. Experimental wind plants were subsequently constructed in other countries like United States, Denmark, France, Germany, and Great Britain. A significant development in large-scale systems was the 1250 kW turbine fabricated by Palmer C. Putman. The turbine was put to use in 1941 at the Grandpa’s Knob, near Rutland, Vermont [8]. Its 53 m rotor was mounted on a 34 m tall tower. This machine could achieve a constant rotor speed by changing the blade pitch. The machine operated for 1100 hours during the next five years, i.e., till the blades failed in 1945. The project is considered to be a success as it could demonstrate the technical feasibility of large- scale



wind-electric generation. Some interesting designs of wind turbine were experimented during this period.

1.1.2.4.-The invention of the three more developed techniques

Darrieus G.J.M, a French engineer, put forth the design of Darrieus turbine in 1920,which was patented in United Sates in 1931. In contrast with the popular horizontal axis rotors, Darrieus turbines had narrow curved blades rotating about its vertical axis. During the same period, Julius D. Madaras invented a turbine working on Magnus effect. Magnus effect is basically derived from the force on a spinning cylinder placed in a stream of air. Another significant development at this time was the Savonius rotor in Finland, invented by S.J. Savonius. This rotor was made with two halves of a cylinder split longitudinally and arranged radially on a vertical shaft. The transverse cross-section of the rotor resembled the letter ‘S’ [10]. The rotor was driven by the difference in drag forces acting on its concave and convex halves, facing the wind. Intensive research on the behavior of wind turbines occurred during 1950's.The concept of high tip speed ratio-low solidity turbines got introduced during this period. For example, light-weight constant-speed rotors were developed in Germany in 1968. They had fibre glass blades attached to simple hollow towers supported by guy ropes. The largest of this breed was of 15 m diameter with a rated output of 100 kW.

In the later years, cheaper and more reliable electricity, generated from fossil fuel based plants became available. When the electricity generated from wind cost 12 to 30 cents/kWh in 1940, the same generated from other sources was available at 3 to 6 cents/kWh [7]. Cost of electricity from fossil fuels further declined below 3 cents/kWh by 1970. Fossil fuels were available in plenty at a relatively cheaper rate at that time. Several nuclear power projects were also embarked on, believing that it would be the ultimate source for the future energy needs. Thus, the interest in wind energy declined gradually, especially by 1970.

The oil crisis in 1973, however, forced the scientists, engineers and policy makers to have a second thought on the fossil fuel dependence. They realised that political tampering can restrict the availability and escalate the cost of fossil fuels. Moreover, it was realised that the fossil fuel reserve would be exhausted one day or the other. Nuclear power was unacceptable to many, due to safety reasons. These factors caused the revival of interest in wind energy. Research on resource analysis, hardware development, and cost reduction techniques were intensified. United States entrusted its National Aeronautics and Space Administration (NASA) with the development of6 large wind turbines. As a result, a series of horizontal axis turbines named MOD-0,MOD-1, MOD-2 and MOD-5 were developed . These projects were stopped by mid-1980’s due to various reasons. During the same period, scientists at Sandia Laboratories



focussed their research on the design and development of the Darrieus turbine. They fabricated several models of the Darrieus machine in different sizes during 1980’s.Research and development on wind energy are seen intensified in the later years. A few innovative concepts like the vortex turbine, diffuser augmented design, Musgrove rotor etc. were also proposed during that time. Prototypes of these turbines were constructed and tested. However, only the horizontal axis propeller design could emerge successfully on a commercial scale.

1.2.-Types of wind turbines

1.2.1.-Horizontal axis wind turbines

Horizontal axis wind turbines (HAWT) have their axis of rotation horizontal to the ground and almost parallel to the wind stream (Fig. 2.3). Most of the commercial wind turbines fall under this category. Horizontal axis machines have some distinct advantages such as low cut-in wind speed and easy furling. In general, they show relatively high power coefficient. However, the generator and gearbox of these turbines are to be placed over the tower which makes its design more complex and expensive. Another disadvantage is the need for the tail or yaw drive to orient the turbine towards wind.

1.2.1.1.-The number of blades influencing the initial torque and explanation about the solidity=Sblades / Sswept

Depending on the number of blades, horizontal axis wind turbines are further classified as single bladed, two bladed, three bladed and multi bladed, as shown in Fig. 1.4. Single bladed turbines are cheaper due to savings on blade materials. The drag losses are also minimum for these turbines. However, to balance the blade, a counter weight has to be placed opposite to the hub. Single bladed designs are not very popular due to problems in balancing and visual acceptability. Two bladed rotors also have these drawbacks, but to a lesser extent. Most of the present commercial turbines used for electricity generation have three blades.



Fig.1.4

LeftSingle bladed, two bladed, three bladed and multi bladed turbinesClassification of wind turbines,

RightUpwind and downwind turbines

They are more stable as the aerodynamic loading will be relatively uniform. Machines with more number of blades (6, 8, 12, 18 or even more) are also available. The ratio between the actual blade area to the swept area of a rotor is termed as the solidity . Hence, multi-bladed rotors are also called high solidity rotors. These rotors can start easily as more rotor area interacts with the wind initially. Some low solidity designs may require external starting.

Now consider two rotors, both of the same diameter, but different in number of blades; say one with 3 blades and the other with 12 blades. Which will produce more power at the same wind velocity? As the rotor swept area and velocity are the same, theoretically both the rotors should produce the same power. However aerodynamic losses are more for the rotor with more number of blades. Hence, for the same rotor size and wind velocity, we can expect more power from the three bladed rotor.

Then why do we need turbines with more blades? Some applications like water pumping require high starting torque. For such systems, the torque required for starting goes up to 3-4 times the running torque. Starting torque increases with the solidity. Hence to develop high starting torque, water pumping wind mills are made with multi bladed rotors.

1.2.1.2.-Classification of the HAWT turbines in terms of the direct/indirect contact of the wings with the air



Based on the direction of receiving the wind, HAWT can be classified as upwind and down wind turbines as shown in Fig. 1.5. Upwind turbines have their rotors facing the wind directly. As the wind stream passes the rotor first, they do not have the problem of tower shadow. However, yaw mechanism is essential for such designs to keep the rotor always facing the wind. On the other hand, downwind machines are more flexible and may not require a yaw mechanism. But, as the rotors are placed at the lee side of the tower, there may be uneven loading on the blades as it passes through the shadow of the tower.

1.2.2.- Vertical axis wind turbines

The axis of rotation of vertical axis wind turbine (VAWT) is vertical to the groundand almost perpendicular to the wind direction as seen from Fig.1. 4. The VAWT

Fig.1.5.-Darrieus wind turbine

can receive wind from any direction. Hence complicated yaw devices can be eliminated. The generator and the gearbox of such systems can be housed at the ground level, which makes the tower design simple and more economical. Moreover the maintenance of these turbines can be done at the ground level. For these systems, pitch control is not required when used for synchronous applications. The major disadvantage of some VAWT is that they are usually not self starting. Additional mechanisms may be required to ‘push’ and start the turbine, once it is stopped. As the rotor completes its rotation, the blades have to pass through aerodynamically dead zones which will result in lowering the system efficiency.

There are chances that the blades may run at dangerously high speeds causing the system to fail, if not controlled properly. Further, guy wires are required to



support the tower structure which may pose practical difficulties. Features of some major vertical axis designs are discussed below.

1.2.3.-Darrieus rotor

Darrieus rotor, named after its inventor Georges Jeans Darrieus, works due to the lift force generated from a set of airfoils (Fig. 2.6). In the original design the blades are shaped like egg beaters or troposkein (turning rope) and are under pure tension while in operation. This typical blade configuration helps in minimizing the bending stress experienced by the blades. There are several variations in the Darrieus design of which some are with straight vertical blades, usually called Giromills

Fig.1.7 A low cost Savonius wind turbine with rotors arranged 90º out of phase

Darrieus rotor usually works at high tip speed ratio which makes it attractive for wind electric generators. However, they are not self-starting and require external ‘excitation’ to cut-in. Moreover, the rotor produces peak torque only twice per revolution.

1.2.4.-Savonius rotor

The Savonius wind turbine, invented by S.J. Savonius, is a vertical axis machine consisting of two half cylindrical (or elliptical) blades arranged in ‘S’ shape (Fig.1.7). Convex side of one of the half cylinder and the concave side of the other are facing the wind at a time as shown in Fig. 1.8. The basic driving force of Savonius rotor is drag. The drag coefficient of a concave surface is more than the convex surface. Hence, the half cylinder with concave side facing the wind will experience more drag force than the other cylinder, thus forcing the rotor to



rotate. Sometimes two or more rotors fixed one over the other at 90O offset may be used to smoothen the torque fluctuations during rotation.

Another way to improve the performance is to attach deflector augmenters with the rotor. The augmenter shades the convex half facing the wind and directs the flow to the concave half thus enhancing the performance. Being drag machines, Savonius rotors have relatively lower power coefficient. However, some experimental rotors have shown power coefficient up to 35 percent. These rotors have high solidity and thus high starting torque

Fig. 1.8.-Principle of Savonius rotor

They work at low tip speed ratios, with the maximum of about 1. They are very simple in construction-even can be made from oil barrels cut in two halves lengthwise.Hence they are preferred for high torque-low speed applications like water pumping.

1.2.5.-Musgrove rotor

Musgrove rotor was developed by a research team under Prof. Musgrove at theReading University, UK. It is basically a vertical axis lift machine having ‘H’ shaped blades and a central shaft (Fig. 1.9). At high wind speeds the rotor feathers and turn about a horizontal point due to the centrifugal force.

This eliminates the risk of higher aerodynamic forces on the blades and the structure. Based on driving aerodynamic force, wind turbines are classified as lift and drag machines. Turbines that work predominantly by the lift force are called the lift machines and that by the drag force are called drag turbines. It is always advantageous to utilize the lift force to run the turbine.



Wind turbines are available in various sizes ranging from a fraction of kW to several MW. Based on the size, the turbines may be classified as small (< 25 kW) medium (25-100 kW), large (100-1000 kW) and very large (>1000 kW) machines.

Fig.1.9.-Principle of Musgrove turbine



2.-Power energy basics and aero dynamical calculus.

A wind flow moving with the c velocity carries within it a kinematical energy. This Ec depends on the fluid velocity principally but also is function of the density. The energy which a wind turbine can take from the moving air is:

Ec=C p×12×A×ρ×v3

[2.0]

The coefficient Cp is integrated into the formula to express that it is impossible to get profit of the whole kinematical energy inside the fluid.

2.1.-NACA airfoils:

The NACA airfoils are airfoil shapes for aircraft wings developed by the National Advisory Committee for Aeronautics (NACA). The shape of the NACA airfoils is described using a series of digits following the word "NACA." The parameters in the numerical code can be entered into equations to precisely generate the cross-section of the airfoil and calculate its properties. In our case the airfoil provided by the company has the NACA 4412 profile section. This kind of profiles has a different geometry depending on the number of profile and the Chord (the maximum length of the section). The Klein wind energy company let us choose between the kind of section we want but in a concrete size and in a concrete geometry.

Fig. 2.1: Geometry of an airfoil section, its chord length and other dimensions

The velocity and the angle of attack as the image shows depend on the two components of the velocity. The first one is the air speed, which depends on the height of the wind turbine and the atmosphere, or the surrounding barriers erected. The angle of attack depends on the disposition of the wings and the



relative to wing velocity of the air taking on account the spin of the wings around the axis (namely the relative spin velocity). This compounded velocity is the one we use to search in the abacus of the NACA report taking into account the angle of attack.

2.2.-Reynolds dimensionless number calculations and variations of drag-lift coefficients

Once we know the resultant of the velocities we must look in the NACA report 586 about “Airfoil section characteristics as affected by variations on the Reynolds number”. First of all we must define Reynolds dimensionless number:

Re= c×Lυ

[2.1]

Where (in the International System of Units):

c : The absolute velocity (vector addition of the relative velocity and the air velocity) [m/s]L :The length of the Chord [m]υ : Kinematical viscosity of the fluid in this case the air [m²/s]

The tables provided by the NACA institute need firstly the angle of attack that can be obtained from the geometrical characteristics and angles of the blades that the company provided us and the velocity conditions of the air. Then the Reynolds number where we enter the compound velocity and the Length of that concrete section.

So with the c (compound velocity of the air considering that the wings do not spin so there are the u=ω×r component and the v velocity of the air), L (length of the chord of the wing) and ν the kinematical viscosity of the air (supposing it is constant), we can calculate each of the Reynolds numbers for each sections that will make us to choose between the different curves of the abacus for that chosen NACA blade section type.The type of the airfoil chose for this blade is the following:



Fig.2.2.- NACA 4412 airfoil section appearance

The information provided by the company is the following:

Fig. 2.3: Sizes and parameters of the wings provided by Heyde Windtechnik Deutschland

Last table´s (Fig 2.3) type of blade is the 1,28/8r Wi one. The most important characteristics are the following ones rotor diameter is of 2,55m and at 580 rpm and with a 10m/s wind velocity the nominal power output is of 1400W.

In the 7th page of the 586 NACA report we can find dealing with “Airfoil section characteristics as affected by variations on the Reynolds number”. We can see for the selected section type NACA 4412 the different lift coefficientsCL

, drag coefficientsCd and moment coefficients from the aerodynamic centre Cma . c. for each Reynolds Number.

Now it is going to be explained the use and the information inside the NACA report. The graphics contain the following information. In the first abacus we can see the profile geometry, instead of giving us the polynomial that describes the upper and the down camber they give us graphically the information. Also in this graphic we can find the position of the aerodynamic centre point. That is the point where the resultant force of pressure is applied theoretically. The way they provide us with a.c. point is the Y (vertical distance) and the X (horizontal distance) from a point in the chord that is at a quarter from the attack point.



2.3.-NACA report 586, tables, basics and explanations

Fig. 2.4: Geometry of NACA 4412 airfoil and the placement of the pressure centre.

The numerical data of the first graphic gives us the upper and the lower coordinates of the section in function of the chord coordinate.



Fig. 2.5: Lift coefficients for different Reynolds numbers



Fig. 2.5: Drag and Moment coefficients for different Reynolds numbers



As we can see in the excel page the variation of the Reynolds number for this cases is not so huge and it is in between 164.000 and 331.000, that is the reason for choosing the line with squares as points. With that line we have built up a polynomial that fits the data and that with a mathematical formula provides us with the lift coefficient in function of the angle of attack.

It has been developed taken points from the graphics and a polynomial of 3th degree has been calculated that best fits the data. We can see in the table bellow that for the Reynolds Number range within which we move there are very similar values for the lift coefficient. That is the reason for approximately choosing a first degree polynomial, because are more or less linear and very close one to each other.

1 2 3 4 5 6 7 8 9 10 110

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6f(x) = 0.156510416666667 xR² = 0.968720498626845

Lift coefficient VS Angle of attack

Angle of attack, α

CL

CL = 0,1565*alpha [2.2]

Fig. 2.6: Lift coefficient VS angle of attack

The Pearson coefficient that appears below the deductible by excels polynomial gives a good measure of what has been the polynomial approximation of degree 1. The approximation level is so good that we can use it safely. Once we have the attack angle and the lift coefficient in the second graphic we can get the moment coefficient and the drag coefficients.

For the drag coefficient we have made the same operation, but as the curve’s slope changes like a four degree one the polynomial is different:

Cd = 0,0905α4 - 0,0513α3 - 0,0968α2 + 0,0769α [2.3]



0 0.2 0.4 0.6 0.8 1 1.2 1.40

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

f(x) = 0.09053976502 x⁴ − 0.05131612637 x³ − 0.096847081757 x² + 0.076906412171 xR² = 0.984680390584541

Drag coefficient VS angle of attack

Angle of attack

Cd

Fig 2.7: Drag coefficient VS angle of attack

2.4.-Relations between the coefficients and the forces, momentum coefficient

These coefficients were obtained from the previous abacus used within the following formulas in order to obtain the forces of lift and drag:

FL=CL×ρ2×v2∞×L×1

[2.4]

FD=CD× ρ2×v2∞×L×1

[2.5]

For these formulas we have supposed that the length of the piece of wing which we are calculating measures 1m. Afterwards when we are calculating the integral along the length of the blade we will multiply times a number which is less than the unit.

The supposed density and nominal velocity of the wind are the following:

ρ : Densityv∞: Actual velocity of the airL : It is a variable depending of which is the diameter



From these two equations we can obtain Lift Force and Drag Force. The units are corresponding with Newton in the IS (International System of units). Geometrically the direction of the drag force is parallel to the c velocity vector and goes through the following point obtained from a moment equation at a distance d from the point of attack.

FL×d=CM× ρ2×v2∞×( L×1)×L

[2.6]So finally the distance from the border of attack to the point of application of the drag and the lift forces is “d” that we can express like this:

d=CM × ρ

2×v2∞×L2

FL

[2.7]

In the Fig 2.6 we can see the forces and the distances where in the next image we see a diagram of the lift and drag forces and the angles they form. The point of application of both forces is on the line just at a distance d of the contact point of the air with the profile.

The direction of the drag force vector is the same as the absolute velocity of the air (supposing that the system of reference is spinning with the wing at a velocity equal to ω. The sense of the vector is the same as the projection of the vector velocity. Lift force is perpendicular to the drag force and the sense is directing to the convex part of the profile as showed in the picture.



Fig 2.8: Lift and drag forces representation in a wing

The resulting vector addition of those forces gives us another force that will be split into horizontal and vertical forces. At this point there are two forces; the vertical provides us power for the windmill. The second one must be withstand by the bearings.

For each section must be calculated the torque produced by the vertical component which is proportional to the distance from the section to the rotary axis. With this information for each section and considering the different chord lengths and the different sizes of the profile we can calculate the output torque.

2.5.-Outputpower and its relation with the spinning velocity

We have spoken about the different chord sections and so on and once we have the value of the total torque in the axis, with the velocity of spin we have the output power which is the desirable parameter to maximize. For that we should vary the spin velocity parameter and get the different output powers for different rpm.

This way where obtained the following graphics that show us for a fixed velocity of the air which is the output power varying the spin velocity. This graphic show us that there are some points of maximum output power for each velocity of the air.



These curves represent the different values that our wind turbine can provide us with for each spin velocity. We also have obtained the different third degree polinomic that best suit the points calculated.

200 400 600 800 1000 1200 1400 16000

250500750

1000125015001750200022502500

Output power VS spin velocity

6m/s 6 m/s8 m/s Polynomial (8 m/s)10 m/s Polynomial (10 m/s)

Spin velocity n [rpm]

Output power P [W]

Fig. 2.9: Outputpower [W] VS spinning velocity [rpm], depending on the parameter of the wind velocity

Each of the adjusted curves has an equation as follows:

P(v=14m/s)= 2E-06n3 - 0.0094n2 + 13.002n - 3373.9 R2 = 0.9929 [2.8]

P(v=12m/s)= -2E-07n3 - 0.0021n2 + 3.7719n - 696.61R2 = 0.9853 [2.9]

P(v=10m/s)= -2E-07n3 - 0.0021n2 + 3.7719n - 696.61R2 = 0.9853 [2.10]

P(v=8m/s)= -8E-07n3 - 0.0004n2 + 1.5462n - 165.29R2 = 0.993 [2.11]

P(v=6m/s)= -0.0016n2 + 1.5437n - 173.54R2 = 0.9865 [2.12]

The maximum outputpower points have the following table’s coordinates:

Wind speed 6m/s 8m/s 10m/s 12m/s 14m/sn[rpm] 515 685 825 895 1035

P[W] 191,8864 448,9806 871,97 1376,6285 2231,011



Table 2.1: Maximum outputpower values in both P[W] and n[rpm] coordinates

This all leads us to the conclusion that the power curve needed by the generator must cross the maximum points of the second table. Also there is a polynomial of second degree which suit the data obtained before with a Pearson correlation factor of 0.99, which is quite good.

400 500 600 700 800 900 100011000

500

1000

1500

2000

2500

f(x) = 0.00383393718484898 x² − 1.89857733142903 xR² = 0.991990426353446

Generator Output Power VS n [rpm]

rpm VS Max power output

n [rpm]

Output Power P [W]

Fig. 2.10: Generator outputpower VS n[rpm], for an ideal operation between the wing’s power optimum points and the generator

This graphic indicates different output powers deppending on the revolutions per minute with which the generator must carry out.

About the geometrical specifications of the air turbine wings:

The data provided by the company heyde-windtechnik is the following one. Firstly they provide the geometry of the wings. That we can resume in the following table 2.2:



Geometrical description of the airfoil Width

Radius [m]Section’s

spin θ [deg]Chord length

[m]

Length of the section starting in each

measure[m]1 0,1 36,1 0,1435 0,0252 0,15 28,7 0,1375 0,053 0,2 23,3 0,135 0,0754 0,3 16,1 0,115 0,15 0,4 11,8 0,106 0,16 0,5 9,13 0,098 0,17 0,6 7,2 0,0835 0,18 0,7 5,8 0,0715 0,19 0,8 4,7 0,0595 0,1

10 0,9 3,9 0,0475 0,111 1 3,2 0,0355 0,112 1,1 2,7 0,0235 0,05

Table 2.2: Geometry of the wings given by Heyde Windtechnik

The first column is the number of the section, the second one is the radius of the section. The third one is the section´s spin angle that we have called β. This β angle is the angle in between the chord and the plane of spin of the machine. The other important angle is the one we have named with θ. This angle is the one between the Plane of spin and the absolut velocity so called “c”. This angle is the one we have used to split both drag and lift forces into vertical and horizontal forces. In the following picture all those parameters are drawed.

2.6.-Step by step calculus of one section´s lift force and drag force

We will choose the data provided by “Heyde Windtechnik” company for the blade´s propierties explained before. 1)We fix the density ρ and the kinematic viscosity ν of the air as:

ρ=1.18 kg

m3

[2.13]



υ=0 . 00001m2

s[2.14]

2) The nominal operating point of the Wind Turbine is the following:

n=580 r . p .m[2.15]

ω=2 π60

×580=60 .74rad

s[2.16]

v=10ms

[2.17]

u=ω×r=60 . 74×0,6=36 , 44ms

[2.18]

This was the section number 7 we have choosen this section because the information given by the company was not for this section but we have supposed the section’s length of chord used in the following calculus supposing that this length will vary linearly. We have drawn in a graphic the different chord length and a line was almost crossing them so we will we will use this assumption for for the calculus. So for the given section we have L=0.0835m.

0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.550

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

f(x) = − 0.12 x + 0.1555R² = 0.946123521681998

Chord length VS radius

Chord length [m]

Radius [m]



Fig. 2.11:Chord length decrement with respect to the radius of the section

We establish like in some other books is explained that the approaching speed of wind to the border of attack is two thirds of the v∞ in the sorrounding area.

vreal=23×v∞

[2.19]

Fig. 2.12:

According to the Fig. 2.12 we can calculate the absolute velocity as:

|⃗c|=√u2+v2=37 . 05ms

[2.20]The angle etween the c⃗ and the plane of spin is the following

β=arctan ( uv )=10 .37 °

[2.21]With the module of the vector velocity relative to the airfoil we calculate the Reynolds number for each case, in this section number 7 we have:

Re= c×Lυ

=37 .05×0 .08350 . 00001

=309 . 345[−]

[2.22]

Once calculated for all the other sections, we can note that all the Reynolds numbers for the different sections remain between 1÷4×105 . That is the reason for approximatig the lift and the drag coefficient with the lines of the squares and the triangles. Both of this curves are very close, so we pick up the higher one and approximate it with the exlplained polinomials. This following ones are the polinomials required :

CL=0 .0917 α0+ 0. 4611[2.23]

Wind Turbine design and modelling

v

c

uSpinnig axis

26


CD=0 .2421CL

4 - 0 . 5263CL3+ 0 . 4181CL

2 - 0 .1458CL+ 0 . 0311[2.24]

Where α 0 changes from one section to each other, it is the subtraction of the angle of the c velocity to the angle θ of the position of the section.

The “θ” angle is the spin angle of each section distanced at a radius from the axis is provided by the manufacturer, the second one “β” is dependent on the ω and on the wind velocity “c”. And finally the difference between β and θ is α as follows:

α=β−θ[2.25]

Section's Spin angle θ with the radius

0

5

10

15

20

25

30

35

40

0 0.2 0.4 0.6 0.8 1 1.2

Radius [m]

An

gle

θ [

°]

Reihe1

Fig. 2.13:Section’s spinning angle θ with the radius



Angle between plane of spin "β" and "c" wind absolute velocity

0

10

20

30

40

50

60

0 0.2 0.4 0.6 0.8 1 1.2

Radius [m]

An

gle

β [

°]

Reihe1

Fig. 2.14: Angle between plane of spin β and c wind absolute velocity

Angle of attack α=β-θ [°] VS Radius [m]

0

2

4

6

8

10

12

14

0 0.2 0.4 0.6 0.8 1 1.2

Radius [m]

An

gle

of

atta

ck α

[°]

Reihe1

Fig. 2.15: Angle of attack for each of the sections corresponding with one radius



The lift coefficient varies from 0,7 to 1,25 for the nominal parameters. And for this section is the following one:

CL=0 .0917 α0+ 0. 4611|α0=3, 17=0 .7515[2.26]

With this coefficient, we are able to find out the value of the drag one:

CD=0 .2421CL

4 - 0 . 5263CL3+ 0 . 4181CL

2 - 0 .1458CL+ 0 . 0311|C L= 0 . 75150205=0 .0115

[2.27]

Fig. 2.16: Parts of the turbine and geometrical disposition

Fig. 2.17: Forces and angles used for each section’s forces


FL

Fd

F total

Fu

Fi

V

c Ω r

θ

Ω

29


So the drag and lift forces are descrived as follows for section number 7:

FL=CL×ρ2×v2∞×L×Rl=0 ,7515×0.00001

2×37 . 04722×0 .0835×0 .0835=5 .0814N

FD=CD× ρ2×v2∞×L×Rl=0 .0115×0 .00001

2×37 .04722×0 .0835×0 .0835=0 .0778N

[2.28];[2.29]

With the forces and the angles as drawn in the Fig.2.16 we can obtain the parallel force to the plane of spin and the perpendicular one.We are going to divide both forces in horizontal and vertical directions. First one

is useful so will call it Fu , the other one will be called F i .

Fu=F L×sin (β )−FD×cos( β )=5 .08×sin(10 . 37 °)−0 ,078×cos (10.37 °)=0 .8379N

[2.30]

F i=FL×cos ( β )+F D×sin ( β )=5 . 08×cos (10 .37 )+0 . 078×sin(10 .37 ° )=5 .0124N

[2.31]

The useful force has been so called because for each of the sections there is a momentum created. The integration or sum of the momentums gives us the total momentum. Momentum times spin velocity is equal to power in [W]. And as we have 3 wings the total output power is three times that.

Before calculating the momentum which gives power to the system, we must know for each secction, in this case the number 7 the point of application of the two forces, or the so called a.c. (aerodynamical centre). For that we will use the previously explained formula:

d=CM × ρ

2×v2∞×L2

FL

=0 .1×0 . 00001

2×102×0 .08352

8 .0814=0 . 01111108

[2.32]

This is the distance from the attack point where the force is actuating. It would be useful for terms of torque along the axis of the blade which produce torsion.



But as the wings are not calculated in this work we suppose that for the worst conditions they will support the forces.

Momentum per each section. Considering that the radial length of the wings is the one in between twoo secctions and its centre, as the Fig. 2.16 shows.

M=3×∑i=1

i=12

Fui×ri=756 .4188 [ Nm ]

[2.33]

Fig. 2.16: Momentum sum along the diameter of the turbine

The output power is the momentum times the angular velocity:

P=M ×ω=756 . 4188×60 . 7375=756 . 4187782 [W ][2.34]

3)The variation of the parameters “v∞” and “n”:

In this part we have developed a graphic with the output power obtained with variations in n [r . p . m . ]. Each of the graphics below has different velocity variations.


dM=Fui x rui

31


200 400 600 800 1000 1200 1400 16000

250500750

1000125015001750200022502500

Output power VS spin velocity

6m/s 6 m/s8 m/s Polynomial (8 m/s)10 m/s Polynomial (10 m/s)12 m/s Polynomial (12 m/s)

Spin velocity n [rpm]

Output power P [W]

Fig.2.17 : Pointing of the best outputpower to each of the velocities

The following ones are the correlations more suitable to the different outputpowers for each of the velocities of the wind:

P(n)= 2E-06n3 - 0.0094n2 + 13.002n - 3373.9 R2 = 0.9929

P(n)= -2E-07n3 - 0.0021n2 + 3.7719n - 696.61R2 = 0.9853

P(n)= -2E-07n3 - 0.0021n2 + 3.7719n - 696.61R2 = 0.9853

P(n)= -8E-07n3 - 0.0004n2 + 1.5462n - 165.29R2 = 0.993

P(n)= -0.0016n2 + 1.5437n - 173.54R2 = 0.9865

Signaled points are the maximum output power for each of the velocities of the wind. This points must be the same for the output power of the generator to obtain the maximum power for each wind velocity.



3.-Mechanical calculus

3.1.-Axis calculation respect to DIN 743

There had been calculated two different hipotetical charges, one of them is related to the aerodinamical forces the other one comes from the own weight of the mechanical parts.

The forces coming from the wings are the following ones

Fig. 3.1: Flexional stress on the axis

This charges create the following flexion tensions:

Fig. 3.2: Flexional stress on the axis

It is when the maschine slows down when we can observe a maximum torque along the axis. The value is constant and the distribution goes from the head of


160mm

NA=156 N

VB=20 NVA=80 N

NB=156 N

50 mm

50 mm 160mm

33


the wings till the point where the axis is installed. Due to the torque it can be observed a torsional stress in the axis which distribution along it is the following:

Fig. 3.3: Torque and torsional stress along the axis

The diameter of the axis must be strengthful enough to support both stresses of the torque and the vertical forces. As the DIN 743 asserts we can dimensionate the axis with the following formula:

dmin=3√ 16 T

π τ max

=3√ 16 ∙20000π ∙ 150

=8.79mm → d=30 mm [3.1]

The flexion stress is related to the diameter of the axis as follows:

σ=M f max ymax

I ln

=1.8 ∙0.154 ∙ 10−8 =0.679

Nmm2 [3.2]

This value for the flexional stress is negligible for the resistant stress of the material. By the way, it has been choosen an special iron for the windturbine cosntruction, wich’s resistance is:

σ maxadm=800N

mm2 [3.3]

This is a DIN 1.4401 semisoft steel usually utilized to develop this kind of mechanical elements.

3.2.-Bearings selection

3.2.1.-6206-RS1 (axis support one) selection justification

The turbine plays two different rotational movements. The first one takes part around the horizontal axis. While the axis is rotating and boosting the generator coupling there are some forces coming out that must be abide by our bearings.


T=20 Nm

130mm

34


The opposite reactions of the axis turn out to be the clouts supported by the bearings. Remembering the Fig 3.1 when the resultant forces are isolated then those are the forces wich must be supported by the bearings.

Fig. 3.3: Forces affecting A and B bearings (aerodinamical forces)

There have been calculated both horizontal and vertical forces beeing this ones the results:

Horizontal Load [N] Vertical Load [N]Bearing A 156 80Bearing B 156 20

Table 3.1: Aerodynamical loads in each bearing

For the selection of the bearing there have been choosen the SKF company. For reasons of simplicity both of them will be the same. The web page of SKF provides us with a calculation program wich results for the selected bearing are the following:


A B

35


Fig. 3.4: Equivalent bearing loads and basic rating life.

It has not been taken into account the fact that the wind turbine will be facing gusty winds, which in fact is providing the axis and the whole structure with P (t ) variable charges pending the day.

That is the reason why it must be used the so called Miner's rule or the Palmgren-Miner linear damage hypothesis, which states that where there are k different stress magnitudes in a spectrum, Si (1 ≤ i ≤ k), each contributing ni(Si) cycles, then if Ni(Si) is the number of cycles to failure of a constant stress reversal Si, failure occurs when:

[3.4]

Where C is experimentally found to be between 0, 7 and 2, 2. Usually for design purposes, C is assumed to be 1. This calculus gets us into air wind frequency calculus and estatistical estimations which are not the purpose of this thesis.

On the other hand it is up to the hill to clear the patch between the axis an the bearings. At this point we must look at the following tables taken from spanish



ISO rules which contribute to a hint about how to choose between the different hole adjustment and the shaft adjustment.

Fig. 3.5.a: Guide for radial bearing adjustment, support housing adjustment

Fig. 3.5.b: Guide for radial bearing adjustment, shaft tolerance



If a constant wind velocity facing the turbine is supposed then the loads are constant in value but not in direction. This is because the axis is spinning around while the structure, if the wind direction keeps the same, is fixed. Thus is the second choice in Fig.3.5a H7 for the hole and k5 for the axis.

3.2.2.-Bearings concerning the vertical axis spinning

The wind turbine must be positioned in front of the air orthogonally in order to obtain the most possible quantity of energy from the position where it stands. Therefore it must rotate around a vertical axis.

There has been performed a mechanical construction for letting the machine swivel around this vertical axis but also pass on the forces coming from aerodynamic interactions and the self weight of the construction above.In the following image this construction can be shown.

Fig 3.6: Mechanical construction for vertical turn around, 3D and separated construction

The following lay outs show how is the construction in between the UPN and the rotatory axis:



Fig 3.7: Mechanical construction for vertical turn around, section and model

We assume that the radial bearings do not absorb any axial force and that the whole vertical force is absorbed by the axial bearing. This is the momentum will be absorbed also by the radial bearings. This forces can be obtained from the static planar mechanics:

∑i

M oi=0 ;M +225 V −225 Rv 2−150 Rv 1=0 [3.5]

∑i

V i=0 ;V−Rv2−Rv 1=0 [3.6]

∑i

N i=0 ; N−RN=0 [3.7]

This equations have only three unknown terms, R v2 , Rv 1and RN. M, V and N are calculated as follows.

The “N” force is the resultant of the complete weight of the mechanical construction minus the vertical component of the air impulse.

In the other hand V is caused by the horizontal force coming solely from the aerodynamic forces.

The whole construction weight supposing a homogeneous density of the different iron alloys of 7850 kg /m3is 184,509kg, we must take in account the weight of the wings that more or less are 2kg each wich together can be like 190kg more or less rounding up is about 200kg mass. This in term of force would be like N= 1000N.


MN

V

RN

R v1

R v2

O

39


The V force comes from the aerodynamic drag and lift forces. Therefore the decomposition has been yet made. Resulting for the nominal wind velocity at his optimum point of output power in V=54.04N.The momentum is generated by two forces, one V and the other one the weight. Firstly the gravity center must be traced back.

Fig. 3.8: Catia Screenshot of the centre of gravity and the coorodinates of it according to the assembly coordinate system

Once known the centre of gravity, then the moment can be calculated as follows:


156 N156 N

20 N80 N

1900 N

1

N

VM

40


Fig.3.9: Reduction of the system of forces to the upper point of the pillar bearing junktor

We can calculate the loads in the top part of the metallic piece numbered as 1 that in the following document will be called „pillar bearing junktor“ with the same equations of static equilibrium.

M=2 ∙156 ∙ 228+1900 ∙ 61−80 ∙ 256−20∙100=165 N ∙ m [3.8]

N=1900+100=2000 N [3.9]

V=156+156=312 N [3.10]

These values provide us with the following results for the written equilibrium equations [3.5], [3.6] and [3.7]:

R v1=165+225 ∙ 312−225 ∙Rv 2

150[3.11]

R v1=312−Rv 2 [3.12]

312−Rv 2=165+225∙ 312−225∙ Rv2

150→ Rv 2=75 N

RN=N=2000 N [3.13]

R v1=237 N

R v2=75 N

RN=2000 N

With this data in hand we can check if the three different bearings chosen are the good ones for this shipping. There have been used for three bearing comprobation again the SKF providers of mechanical components program. The results obtained are as follows:

There have been choosen two different ones, for the upper bearing it has been choosen 62307-2RS1, and for the lower one it has been used the RLS 12 (these are the product names that can be selected from the SKF catalog).



Fig 3.10: RLS 12 bearing corresponding to the lower radial one in the costruction

Fig 3.10: 62307-2RS1 bearing corresponding to the upper radial one in the costruction



Fig 3.11: 89413 TN bearing corresponding to the axial one in the costruction

The constuction with those three bearings is explained in the previous Fig.3.7 pag. 23. So the checkout of those three bearings named:

-62307-2RS1

Fig 3.11: 62307-2RS1 bearing life calculus



-RLS 12

Fig 3.12: RLS 12 bearing life calculus



-89413 TN

Fig 3.13: 89413 TN bearing life calculus



3.3.-Comprobation of tabs according to DIN 6885

The mechanical transmission of power between the axis and the generator is worked up at two points in the mechanism by tabs. The first one „1“ transmits wind turbine obtained energy from the wind along the wings into mechanical torque spinning arround the transmission axis.On the other hand there is a second coupling „2“ between the opposite axis part and the generator (See Fig 3.14).

As the Fig 3.14 shows it has been designed with 1 tab per connexion.In order to be able to transmit the not negligible amount of 1.4 kW the following calculated neccesary length of the tabs has been rounded up.

Fig. 3.15: Two mechanical couplings developed with tabs

Firstly we must make sure that for the worst possible consitions of work this mechanical conexion will be accesible to support them. This conditions are fixed by the axis slow down terms. As in the brake mechanical calculus chapter is stated it has been fixed the maximum time expected from the brake to slow down from the maximum velocity of swiveling to cero in 5seconds. More time could have been dangerous for the sorrounding area due to the possibility that the blades get broken by their roots and injury people or cause damages in the nearby goods.

For slowing down as in [3.11] equation is obtained, a braking torque of 165 Nm is required . That is the limiting condition to design the „1“ possition tabs wich will be absorving the torque coming from the air and the kinematic energy stored along the axis.

For the comprobation and dimmensioning of the tabs it has been used the DIN 6885 and the DIN 6892 norm. First of it provides us with the forms of the tabs and the seccond one with the .



There are two different dimensioning systems that involve firstly the shear stress and then the compression strees but as the τ maxinvolved in calculus is much more small than the compression axial stress σ comp .max must be said that before calculating both of them the more restrictive one is τ max.

First of all the force Fu must be calculated. This is the force wich once put at a distance of radius, can perform the torque of transmission. Namely:

Fu=T transmission

d /2= 165

30/2=1333,33 N [3.11]

This whole load must be held by one tab, wich must have the following length „l“:

Firstly the compression tension suffered by the tab is the following one in the upper part of its keyway.

S=Fu

(h−t 1)∙ lt ∙i ∙ k

[3.12]Where

S: Compression stress suffered by the tab/or the resistance of the material when the ltvalue is cleared Fu:Tangential force along the tab lt:length of the tabt 1:distance from the axis cavity of the tag till the end of it once insertedi:Number of tabsk :Coefficient of impairment

The unknown variable is the ltwich must be as follows cleared from the equation. This is going to be the design parameter. Once substituted the values in the equation:

lt=Fu

( h−t 1 )∙ i ∙ k ∙ S= 1333.33

(5−3.1 ) ∙1 ∙1 ∙ 60=12 mm

[3.13]

This length has been calculated the value of the resistance in Mpa of 60N ⁄ mm2. This resistance always must be less than the other two materials. The reason for



this is that when the mechanisim brokes up to broke along the most cheapest part of three of them.

The normalized sizes of the tabs are the following ones for the choosen type:

b x h t 1 Clearance Tighten Needed screw8x5 3.1+02 2+0.1 1.4+0.1 M3x8 (DIN

84)

Table 3.2: Geometrical characteristics of the selecctioned tab according to the norm DIN 6885

The other tab number“2“ is designed as well as the other because it is passing down the same torque and consequently it will have the same size.

3.4.-Axis break dimensioning

It has been installed for security reasons a brake which is able to back away the wind mill. Its design can be showed in the following image Fig. [3.16]. It

Fig.3.16: Selected geometrie and design for the brake, disc brake.

There is a mechanical junktion between the following parts of the wind turbine costrucction: Wings, torque transmission axis, tabs, bearings, and generator rotor. This is why when time to dimensioning the security brake it must be clear the moments of inertia involved in the calculus of the rapid deceleration of the axis.

According with the last paragraph comment there are three different inertia moments wich ares as follows: generator, axis coupligs construction, and the wings.

I TOTAL=I wings+ I axis+coupligs+ I generator rotor [3.14]



Each of them has been calculated in different ways. The most simple calculus is the one of the axis and the couplings one because in CATIA we can obtain it directly. The following Fig [3.17] shows us the value of it:

Fig. 3.17: Catia screenshot involving direct measures of inertia

I_(axis+coupligs)=0.005kg∙m^2 [3.15]

This one is automatically generated by CATIA.

For the other two calculations the method has been less exact done. As the manufacturer does not provide any information about the inertia of the rotor it has been searched in the web one sincronous generator of 1kw wich inertia is as follows:

I generator ,1 KW=0.0075 kg ∙ m2 [3.16]

That was the generator inertia for one generator of less output power, so that we must say that it must have more or less 1.4 times more inertia as our turbine generator will be of this output power. So the final inertia for the generator is:

I generator ,1,4 KW=1.4 ∙ 0.0075=0.0105 kg ∙ m2 [3.17]



The wings are aproximatedly as a beam and it is known that the beams inertia moment in respect to one axis crossing his butt is dependent on the length, the mass as follows.

I wings=3 ∙ I wing=3 ∙13

∙M ∙ L2=M ∙L2=2∙ 1.2752 kg ∙m2 [3.18]

Where:

L: Length of the beam or wing in this case, wich’s value is half of the diameterM: total mass of the wing, wich has been estimated in 2kg

All this three inertias put toghether make a total inertia of 3.25 Kg m^2, wich is the data tied to be calculated. This helped we can calculate helped by the kinematic formulas and the relation between inertia moment and torque the angular acceleration neccesary for stoppig the whole assembly in a short period of time. It has been estimated a time of t=5s, within the wings must have stopped spinning. Calculus are as follow:

First of all we must propose the kinematics formulas for a circular decelerated movemet, with constant deceleration:

θ=θ0+ω∙ t−12

∙ α ∙t 2 [3.19]

ω=ω0−α ∙ t [3.20]

α=constant ? [3.21]

Its also known the relation between the torque and the innertia:

T=I total ∙ α[3.22]

With those equations [3.19]-[3.22] we open up the value of the angular acceleration and finally the value of the necessary torque wich will be provided by the brake M . In last equations the variables have the following values:

Where:

θ :Angle of spin measured in [rad]

θ0: initial angle of spin related to t 0wich in the beginning of swiveling is taken as 0 to make calculus more simple. So θ0=θ (t 0)=θ (t=0 )=0



ω :Angular spinning velocity measured in [rad / s]t :Final value of time, as said before ot has been choosen a final time of t=5s because it is estimated that dynamic problems will not appear in such a little time. Is the time wich the brake needs to slow down from 125 rad/s (the maximum spinning velocity) to 0 rad/s

α : This is the angular deceleration, it is said deceleration because the rpm are decreasing, and that is the reason to it be negative in the equation

Fig. 3.18: Negative angular aceleration of the brake

ω0 : Initial spinning angular velocity wich as it has been stated is 125 rad/s

T :Torque needed from the break to reduce the velocity of the machine. Wich really is the second unknown number of four of the equations. It will be obtained in second term after the obtention of the α .

I total :It is the total junktion momemt of inertia measured aling the turning down axis. It is measured in [kg ∙ m2]

The past equations are solved as:

θ=0+125 rad / s ∙5 s−12

∙α rad /s2 ∙5 s2 [3.23]

0 rad / s=125 rad /s−αrad / s2 ∙5 s [3.24]

α=1255

=50 rad /s2 [3.25]

θ=0 rad [3.26]

T=3.25 kg ∙m2 ∙50 rad /s2=165 N ∙m [3.26]

So this is the essential parameter of the brake the torque neccesary to slow down the velocity from the most possible one 1200 rpm= 125 rad/s to cero.

Once calculated this term it must be calculated the normal force wich the beam must be making against the disc of the brake. This is like that because as we see



in the following Fig3.19 the fuctionament of the brake consists on a long „U“ form beam wich is deformated by an extensor until the swiveling velocity of the axis is too high and can cause damages. After this the extensioner turn around its axis 90º and activates automatically the brake wich as it was said before will expend like 5s to stop with security the axis.

Fig. 3.19 work and functionament of the mechanical security brake.

As the Fig. 3.20 shows there is a circle delimitated by a line wich is the area where the beam is making an estimated costant pressure into the disc. In both sides. In the following screenshot Fig.3.20 it can be seen wich is the size of the area of contact between the beam and the disk. It really simple to measure it without no table or integral just with CATIA.

Fig. 3.20: Screenshot of CATIA drawn semicircle delimitated by a line.


brake slow down torque

brake acctioner force

Wing torque

52


Fig. 3.21: Screenshot of the measured area in CATIA of the Fig 3.20

So, as it is surely known the fricction coefficient is the constant between the normal force and the Fricction force. As the surface of contact between the beam and the disk is symetric, then we can consider both the normal and friccition forces just getting of the centre of gravity from the crossed by lines surface in the Fig. 3.22.

This center of gravity is centered in the axis of the disc at a distance from the centre of gravity of the disk of 57.33 mm as in the Fig 3.21 can be seen.

Once known the distance where both normal and fricction forces actuate, then it is possible to calculate provided the torque the value of the normal needed. It can be calculated with the following equations:



F r ∙ d1=T brake [3.27]

F r=T brake

57.33 mm=165 000 N ∙ mm

57.33 mm=2878 N [3.28]

First of all we must resolve the needed friction force, wich is related to the distance obtained before and to the torque. It is 2878 N as the equation [3.28] states.

The relationship between the normal force and the fricction force is the following one:

F r=μ ∙ N [3.29]Where :

F r : Is the fricction force

μ : Is the coefficient of friction wich really depend on the two materials wich do get friction in between. This coefficient is between two Iron pieces arround 0.15. But the beam will be provided with some speacil materials wich resist the abrassive action and wich give an upper coefficient of friction. This gives us a better (hihger)F r.

N : Normal force actuating against the disk.

N=F r

μ=2878 N

2=1.44 KN [3.29]

This equation needs the value of this coefficient that will be taken as 2 because the material is an specific one with good propierties of heat dissipation, non wear propierties and so on.

Now the unknown parameter to design is the inertia moment and furtherly the secction area of the beam wich produces teh slowing down because of the friction of the axis.

It has been made an aproximation where the beam is considered as in the following Fig 3.22 as a beam with a constant load whose value is the normal



divided times the area of the circular surface part in contact with the rotatory disk.

Fig. 3.22: Table with the deformation necessary in the beam so that it pushes the neccesary load against the disk.

Once it has been calculated the c constant linear load the equation in the Fig.3.22 can be used. It gives us a relationship between the flexion resistance, loads, the positioning of them and the displacement of the most furtherly placed point of the beam.

The following [3.40] equation has as unknown parameters EI, wich is the rigidness of the beam. Young module is E=210 000 Pa. From this value we can get the two dymensions of the beam seccition.



One of them is really the calculated one, the inertia involved on the flexion of the beam wich is drawn in the Fig. 3.23 as I y− y. This value must get out of Equation [3.40]

Fig. 3.23: Inertia axis of the section of the beam involves in the calculus of the folowing equation [3.40]

δ=q ∙ c ∙ a2

3 ∙ E ∙ I [ l ∙(3+( ca )

2)−a ∙(1+( ca )

2

)] [3.40]

Where:

δ : Displacement of the extremal point of the beam wich will be fixed by the designer in order to only have an unknown parameter in the [3.40] equation. It has been fixed to 5mm just not ot be very stiff the beam.

[mm]

q : The linear constant load suffered by the beam. As we only know the normal force we can divide it by the area calculated with CATIA of the contact surface wich is AT=0.03m m2 [mm2]

a : as the diagram in the Fig.2.23 shows, it is the distance from the ligated point of the beam till the centre of the longitudinal linear load.[m]

l : Is the total length of the beam [m]


x-x

y-y

56


c : Is the half of the total length along the load is placed [m]

If the moment of inertia is got from this equation [3.40] it changes to:

E the Young coefficient wich is the value of the slope in the σ−ϵ diagram in the

linear zone of it. It is measured in [ N

mm2 ]

I=q ∙ c ∙ a2

3∙ E ∙ δ [ l ∙(3+( ca )

2)−a ∙(1+( ca )

2

)][3.40]

Substituting the values of the geometrical and force parameters:

I=(20/0.03 ) ∙58 ∙ 732

3∙ 2.1 ∙105 ∙5 [230 ∙(3+( 5873 )

2)−73 ∙(1+( 5873 )

2)]→[3.41]

I=46839m m4

This is the required inertia from the secction to be 5mm less thick as the disk and to provoke on it a torque of 165 Nm wich makes the whole construction to slow dwn in 5s from 125 rad/s to 0rad/s and to stop.

The secction could have been designed in plenty of different ahapes but a rectangle is one of the simplest ines, that is why it has been choosen. The Fig. 3.24 shows the two axis where it works.

There has been fixed the „b“ dymension to 20mm so that „h“ dymension must be obtained from the formula [3.42] wich gives us the relationship between the moment of inertia and the „b“ and „h“ dymensions.

I= 112

b h3 [3.42]

b=20 mm

h=3√( I ∙ 12b )=3√( 46839 mm4 ∙12

20 mm )=42 mm



4.-Electrical calculus

4.1.-Generator selection:

As in the chapter 2 has been explained there is a graphic wich give the output power of the wings against the spinning velocity of the generator. This is the so called graphic:

Output power VS s pin v eloc ity

y = -0.0016x2 + 1.5437x - 173.54R 2 = 0.9865

y = -8E -07x3 - 0.0004x2 + 1.5462x - 165.29R 2 = 0.993 y = -2E -07x3 - 0.0021x2 + 3.7719x - 696.61

R 2 = 0.9853

y = 3E -06x3 - 0.0101x2 + 10.605x - 2255.3R 2 = 0.9986

y = 2E -06x3 - 0.0094x2 + 13.002x - 3373.9R 2 = 0.9929

0

250

500

750

1000

1250

1500

1750

2000

2250

2500

0 200 400 600 800 1000 1200 1400 1600

S pin veloc ity n [rpm ]

Ou

tpu

t p

ow

er P

[W

]

6m/s

8 m/s

10 m/s

12 m/s

14 m/s

Fig: 4.1: Swivelling velocity against the mechanical outputpower and the output generated electricpower supperposed, Ideal generator.

The ideal generator would be that which passes along the point of maximum output power produced. When the producction of this windturbine would be massive then it would be justified to develop an special generator for this characteristics of the machine.

Reality is the Fig 4.2 generator, it has been produced by „Maurer Electromaschinen“ electrical machines constructor this machine has its characteristics in the Fig 4.3 wich the most important ones for us are the nominal power 500W, the tipe of generator trifasic altern current 50=Hz frequency generator with variable intensity generation depending on the rpm of spinning.


IDEAL GENERATOR CURVE

58


Fig. 4.2: Measures and 3D reality appearance of the Generator.

In figure 4.1 it can be seen different points of work to the turbine, wich really depend on the power provided by the wind for each of output powers there is a rotation velocity wich is marked by the generator.

The Fig. 4.2 shows how the generator reacts giving one output power to each of the velocities

It has been choosen the generator from Maurer company it fits perfectly the generator curves needed because really although it is supposed to provide with



1.4 KW with 10m/s wind velocity, this will not be the average velocity of the place where it is placed.

The generator will be moving in terms of outputpower allways beyond the green line of 10m/s, and in average in the 6m/s one. In case that the wind velocity increases dangerously, the generator wont be ablo to catch the whole power, and that is why :

T−T '=Textra=α ∙ I [4.1]

Where

T : the wing’s provided torque measured in Nm

T ' :Torque absorved by the generator wich is then converted into electricity in the rotor of the generator. Measured in Nm

T extra :The difference between them, T and T‘ what makes the rotor gettin accelerated till it would break down or till the infinite if the generator was unbreakable. Measured in Nm

α : Angular acceleration, rotor junktion with all his inertia is accelerated at a constant rate (taking into account that the velocity of the wind keeps like the

starting one) of αrad

s

I :Inertia of the whole construcction measured in kg m2

When the angular velocity achieves 1200 rpm or 125 rad/s, then the security brake starts to decelerate.

The normal voltage generated by the turbine will be at 500 rpm, that as the Fig 4.3 shows provides us with more or less 40V of tension between borns.



Fig 4.3: Rotary speed [rpm] VS Voltage in [V]And current [A] vs rpm

4.2.-Batteries selection

This energy can be collected by four batteries of 12V serial connected. This makes a total of 48V, which is enough for this construcction. The batteries seleccted are those ones of the Fig 4.4 and the Table 4.1 wich shows its Electrical characteristics.



Fig4.4: batterries used to keep the energy roduced by the turbine

Table 4.1: Ah values prizes and voltages of different tipes of bateris, the ones between red lines are the selected ones.

As it has been said in the second pharagraph the generator characteristics show that the generator is a triphasic generator with constant magnest inside 8 pair of poles. The number of poles gives the relation between the spinning velocity and the frecuaency of the generated electricity:

n ∙ p=60 ∙ f [4.2]

As the wind turbine is situated in Europe the frequency is 50Hz, but the generator will provide us with a variable frequency because the spinning velocity will be changins at the same time as the wind velocity does. The Frequency in fact will be:

f =0.13 ∙ n[4.3 ]



Really for each velocity of the air and this concrete turbine there is a direct relation among the wind velocity and the rpm which is spinning at.

As the frequency is variable this does not work for the purpose of the turbine wich really is feeding a house electricity requirements and works at a voltage level of 240V trifasic voltage at a constant frequency.

4.3.-Transformation of the electrical output power to utility parameters of tension voltage and frequency

The use of this windturbine is what gives us the neccesity of a power electronics intermediate exchanger. It is needed a AC/DC transformer, then a set of batteries that will work at 12V and finally an inversor wich provides us with a constant feeding of 50Hz electricity at 240V.

The machine it will not be able to feed all the electrical requirements, and still it must be taken electricity from the facility. But it allows to be less independent from the net, and to save an important amounth of money along the year.

The following Fig4.5 shows us the electrical circuit to feed the house:

Fig 4.5: The electrical construction of AC/DC+batteries+DC/AC



6.-Conclussions:

1)The small wind turbines are as this report shows a good option for house electricity feeding, the invests in this machines can be within 10 years payback amortized.

It is not possible to feed all the energetical requirements of one modern house by means of this technology, but it is possible to get not only a natural benafit from the installation of this turbines but also economical.

2)The wind energy itself appears in a lot of countries where the installation of a net can be very expensive. This utilities also can appear as a dangerous weapon against the animals and the enviroment.

Agreed to this it is very profitable to install this kind of technology in emplacements as farms, etc where installations for nets are very expensive.

3)There is a factor generally aplicated to the productors of energy:

In physics, energy economics and ecological energetics, EROEI (energy returned on energy invested), ERoEI, or EROI (energy return on investment), is the ratio of the amount of usable energy acquired from a particular energy resource to the amount of energy expended to obtain that energy resource. When the EROEI of a resource is equal to or lower than 1, that energy source becomes an "energy sink", and can no longer be used as a primary source of energy.

This is very important to the decission makin, just because this factor when less than 1 clears that is not the well worth to try to invest societies generated energy into it, cause if every source of energy was like that then it would lead to a general lost of energy resources.

In the case of the wind energy Currently (2006) the EROEI in North America and Europe is about 20:1 which has driven its adoption.

By this way theoretically it would be possible to achieve a 100% wind energy production within a long period of time. But this is not real because not all the locations are energetically giving the same energetic payback.


http://en.wikipedia.org/wiki/Primary_energy

http://en.wikipedia.org/wiki/Energy

http://en.wikipedia.org/wiki/Ratio

http://en.wikipedia.org/wiki/Energetics

http://en.wikipedia.org/wiki/Energy_economics

http://en.wikipedia.org/wiki/Physics


In the case of the small energy it is said that this kind of installations can take a 20 years period o life having finished they payback period of time in 10 years. That provides us with a economical and energetically profit.



7.-Bibliography:

Books and PDFS:

[1]Numerische Strömungssimulation von horizontalachsen windturbinen, Florian J. Kronschnabl, TECHNISCHE UNIVERSITÄT MÜNCHENINSTITUT FÜR ENERGIETECHNIK MW7LEHRSTUHL FÜR FLUIDMECHANIK

[2]Ajustes de rodamientos, “Fundación Universidad de Atacama“Especialidad Mecánica Automotriz.

[3[1]DIN 743: Wellen Durchbiegung, Querkraft, Momente Festigkeit, Eigenfrequenzen

[2]DIN 6885 Keyway and keyway details

[3]DIN 1.4401 Stainless steel norms.Resistance of the materials


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6.-Draftings, guide of draftings:

Further contact or questions over: [email protected]


final project

Documents

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wind turbine design

horizontal axis wind

utility size wind turbines

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power energy basics