WO2016178037A1 - Combined impeller-pressure type wind power plant - Google Patents

Abstract

A combined impeller— pressure wind power plant uses wind speed from 5km/h to 250km/h. A wind speed of 250km/h produces 15,625 times more power, than today's wind power units when using maximum allowed wind speed. If (R_MAX) is the biggest distance of rotating points from shaft rotation axis, then wind aggregate's inertia moment of resistance to rotation rises with (R_MAX ³), and the total moment of the forces rises with (R_MAX ²). Top and bottom amplifiers increase the density (ρ) air streamlines. Decreased (R_MAX) increases angular velocity (ω) but decreases the total moment, so that increased air mass and energy efficiency coefficient multiplied by 3.rd potention of maximum wind speed can produce enormous energy. Greater output height than input height of air streamlines creates underpressure. Increased angular velocity and underpressure throw burnt out molecules with exhausted kinetic energy out from the power unit. Control microprocessors bring out maximum possible power by turning heating water and electric net on and off, but they also control centrifugal acceleration and opening and closing of roof door. A small generator with permanent magnets provides excitation for bigger generators.

Description

COMBINED IMPELLER-PRESSURE TYPE

WIND POWER PLANT

DESCRIPTION OF INVENTION :

Invention-related area :

Wind turbines and pressure windmills are patented classification as F03d.

EXSICITION wind turbines able to be compressive, or designate as propeller's, because it faster rotate then to-day's wind turbines from a more reasons.

The dimensions of wind pressure aggregate dictate dimensions of the rest components parts of power plant.

Without next fisic's lawfulness, we there is no knowing the substance of this patent.

Wind energy is a kinetic energy→ work of air masses in motion, and the power is the work performed in 1 second : mvj²

m is air mass, and Vj is wind speed.

2

If vertically through flat surface S the parallel's air streamlines pass with the speed of

Vj , then the mass m of these air streamlines which pass in 1 second is equal to the product of the volume S Vj and specific air density p. Accordingly, the mass of the air streamsline of one second in our case equals Illisek = p S Vj

If we calculate the kinetic energy in one second, then we calculate the power:

C p ( p S Vj ) Vj² C p p S Vj³ C p is the energy efficiency coefficient

E IN_ISEK = = = Cs Vj³ if : Cs = 0,5 Cp p S

2 2

Nobody in the world has ever so far thoroughly enough used the fact to maximally use the wind power and the fact that the wind power depends about speed wind on 3.rd potention.

We know that Perpetum mobile does not exist. If the same mass of the air streams hits the same surfaces ASi— Si, then the same force Fi = F₂ and the same power Pi = P₂ impacts on both surfaces. If surface ASi is at a distance Γι of the axis rotation shaft, and if surface AS₂ is at a distance Yi of the axis rotation shaft, and that the power P equals the product of the moment M and rotational velocity (0, and we have that is :

Ρι= Μιωι = Ρ₂= Μ₂ω₂

The moments of forces depend against a length the force arm rotation about the shaft, and we have that is : Mi = riFi , and M2 = r₂F₂, and for two the same surfaces, where is Fi = F₂ = F we have that is : nFoi = r₂Fo₂ za ASi = AS₂ ωι = (r₂ : ri)a>₂ = ηω₂

IF n= r₂ : i^*i THEN ω₂= η ¹ωι

IF the duck with surface S2 has n times greater height and equal width as surface Si, THEN surface S₂ is n times greater and also the force and power wind is n times greater : r₂ = nr 1 F₂ = nFi and M₂ = r₂ F2 = nri nFi = n²Mi

P₂ = Μ₂ω₂ = n² Micoi : n = n Pi

If moment M of force continually work on the body, the body will rotate with the turning moment. The fundamental law of dinamic rotation the momentary time→ curent is M = α I, where a is the angular acceleration of rotation, and I is the inertia moment define with formula : I = ΙΪΗΓΐ² + ΓΙ12Γ2² + ΪΙ13Γ3² + . +^' . + ΠΙΝΓΝ²

Analysis Inertia moment of the resistance at rotation is :

If rotation axis go through the gravity centre of the rectangle T [ 0, y, h 2], formula for Inertia moment of the resistance a rectangle is : IY RECT = m ( h²/12 + dpL²/ 12)

If rotation axis go through the bottom of a rectangle, formula for I γ RECT with Steiner's rule is h is height of the duck.

(IDUCK is portliness.

I Y_ RECT = 111 RECT [h²/ 3 + dpL²/ 12] m RECT = γ h adoucK

For dDucK « h the approximate result is :

If only height of 2. rectangle's duck is enlargged for proportion n = r₂ : ri , and width in direction Y axis

and his portliness in direction X axis is equal, then is this case the fizic's formulas bound for the distance r of points from the symmetrical line of the rotation axis the shaft are next :

Ii : 11 = b.2³ : hi³ = n³ : n³ =n ³

a2 = M ₂ :l2 ai = M i:Ii a₂ : ai = (M2li) : (M1I2) = (M2 :M i)* (Ii :I2) The proportion of the angular acceleration is : an ai =( Γ2² : ri² ) :(Γι ³ : n³)] = (ri : Γ2) =n^_1,

In the beginning without the wind, the angular velocity (0 = 0, and in the short dim time on the beginning when the wind begin to work, the angular velocity 1. momentary curent time is

1. momentary curent time become prior time. CO PRIOR = ( CUR_I

COCURJ = (OPRIOR + a Δί, and we can again to write : CO2 = n^_1 (Oi

P2 α 2 = η-¹αι θ 2 = η^_1ωι

ac_2 = ω2² Γ2 ac_i = θι² Γι = η²ίθ2² Γι = η²ω2²Γ2 ΐΊ Υι = ^η½2² Γ2 :n=nac_2 When the most distant point on compressive wind 1. aggregate rotate on the radius R max_i =0.4m, and the most distant point on compressive wind 2. aggregate rotate on the radius Rmax_2 = lni, we have : n = 2,5. On 2,5 times smaller radius, moment Mi is 6,25 times smaller, what it make possible the using power amplifiers because of increased mass air streams for the compressive rotorblades, when the wind speed Vj > 200km/h. The angular velocity (Oi and the centrifugal acceleration are 2,5 times greater, but the centrifugal forces are equal because we have :

Fc_i = mi * a c_i = ni2 : 2,5 * a c_2* 2,5 = m 2* a c_2 = Fc_2

The inertia moment to rotation of 1 windmill has 15,25 times smaller if much greater if the C!DUC portliness is equal, and that is enable 1 windmill has even number 18 rotorblades, and reduced depth entry air streams inside rotorblades, how the centrifugal acceleration faster throw the molecules of the air streams out the wind aggregate, if this molecules give path of their kinetic energy for the rotation of 18 rotorblades. On this way the new air streams run blockade, and this new molecules can to give path of their kinetic energy for the rotation of rotorblades. For the roller is a different situation. Mass of the roller is : m = ρΚ²π p is specific density in (kg/m3).

Formula for the inertia moment of the rotation for the roller is :

Now the inertia moment of the rotation is 39,0625 times smaller, because X dimension and Z dimension are equal, and we must remember :

IF di : d₂ =2,5 THEN the roller 39,0625 times more slowly to rotate. ( lakse zarotira )

Top of to-day's a rotorblade of windmill go in the sharp, and his flat surface more resemble on the isosceles triangle, then on the rectangle.

In the different book write how the rotorblade round with bigger speed of wind speed. Till now nowhere I don't read, that only the top's part of rotor blade has bigger speed but the bottom's part of rotor blade has smaller speed. The significance is ??? The top's part of rotorblade break out of the top's air streams. The top's air streams check the rotation and reduce the power. IF the top's surface is bigger the brake is bigger, and because of top of the rotorblade is narrow. Because of the rotorblade more resemble on the triangle, then on the rectangle, but I wish to know something about the inertia moment of the resistance o rotation of the rotor blade. Because of I drew small deform the isosceles triangle on FIG.7, where I increase dimensions in direction X axis and Y axis because beter looking the way execution formulas.

At the vector algebra when + X axis rotates towards + Y axis, the direction of the progression of the right screw shows + Z axis above the X Y plane, the formula of which is Z = 0.

The segment's form equation of the left leg triangle with FIG.7 in X Z plane is :

X Z a Z

+ -=1 or 1. limit is Xi=--(1 )

-½a h 2 h

The segment's form equation of the right leg triangle with FIG.7 in X Z plane is :

X Z a Z

— + -=1 or 2. limit is X₂ = -(l--)

½a h 2 h

Now follow the calculation the inertia moment for FIG.7 by formula :

m y₂ z₂ x₂

Ix = - {i[i(z² + y²)[dx]dz}dy

m y₂ h z

Ix = -{i (i(z²+y²)a*(l )dz]dy}

V y, 0 h

V y, 0 h h V

y₂ h³ h⁴ h²y² d/2 h³ hy²

Ix = ya{i (-+hy²- -)dy} = ya{ f (-+ )dy}

d/2 h³y hy³

Ix = ya I + J

-d/2 12 6 h³d hd³

Ix = ya ( 1-— )

12 24

Triangle's surface on FIG. 7 has volume V = 0.5 a* v * d, and mass this the body is product m = γ * 0.5 a* h * d , where a is base and h is height of the isosceles triangle, but d is portliness of the rotor blade and p is specific density (kg/m3) of the rotorblade's substance. The denominators are not equal therefore because the position of the starting the coordinate Z axis there is on the bottom of the height h, and for the coordinate X axis there is on the middle of the portliness d.

h² h³

If d « h for the triangle we have : I_x ~ ml— ~ γ * a* d *

6 12

Rectangle's surface has 2 times bigger mass and we have : I_x~ γ *a*d * h³: 3

Mass of the rectangle is 2 times bigger of the triangle, and the inertia moment to rotation is 4 times bigger, but IF masSRECT = IHSSSTRIA , the inertia moment to rotation in this case is 2 times bigger, because the bigger mass has smaller radius rotation of the triangle and smaller mass has bigger radius rotation of the triangle. Beacuse of 4 times smaller the inertia moment to rotation the triangle has bigger sensitive, and the rotation of the rotor blades begin with the smaller speed wind, then if the the rotor blades resemble to the rectangle with a more biger height h of the width a. The top part of the rotor blades rotate with the bigger speed of wind speed, and this part must subdue the resistance of the air where the smaller surface lose smaller energy. Now we know why the rotor blades sensitive to the isosceles triangle. The inertia moment to rotation is proportional with height h³ of the isosceles triangle, where h is the distance between top of the triangle and the axis of the shaft.

If r₂ = 5ri SENSOR, then we have : Ρ₂ = 5Ρ₁ Μ₂ = 25Μι I₂ = 125li ωι = 5 ω₂ The propeller of the airplane rotate faster of to day wind turbine but SENSOR wind turbine has the amplifier power and nickname propeller, if generally Γ2 » 5ri_sENSOR .

The formula for the power of the wind without a losses is : Pvj = 0,5 pSvj³

The bounds of a pressure WINDMILL run away of the wind, and formula for the power on the wind aggregate is :

Only tangential speed wind Vj tang [ m/s ] with the arm r produces the moment of the rotation, and has equal direction with Vector of angular velocity of ω.

Avj = Vj_tang— oR The points on blades with or speed is escaping from Vj tang speed of the air streamlines. R variable show the distance between any point on the blade and axis of the shaft→ the centre of the rotation.

p is density of the air [ kg/ m³ ] p = 1,293 kg/m³

C p is coefficient of the kinetic efficiency energy air mass, which is kvocient between

The air streams pass through S [m²] surface on rotorblades. Dimension of S depend upon the form all amplifiers power and depend of dimension RMAX . The most distant horizontal line dY long for RMAX of axis rotation shaft.

In our case S = 2,07 RMAX * dY, and we have : Pvj = 0,5 /T>2,07RMAX* dY(Avj)³

For any speed wind Vj depend upon vista blade and all amplifiers power work maximum power on the wind aggregate, only IF generator produces exactly determined value of contra moment in Nm. Only when contra moment of generator is equal to a moment of rotation on the wind aggregate, the rotation stabilize own angular velocity ω exactly on determined value value. I carry into in my programs the determined value for contra moment of generator for getting the maximum power. I play the work micro proccesor for the supervisory program of the supervision arrangement, and my programs gives results which are equal to the results, which obtain with till now bring formulas. Now I shall get the results of my different programs with carry into programs the determined value for contra moment of generator for getting the maximum power.

For R MAX = 0,32m and dY= 4m

For Vj = 30m/s 0,5

0,6465* 5,2992 [m²]*30³ = 92500,1856 W C p = 27425,4998 : 92500,1856 = 0,29649 = 29,649%

For vj = 40m/s 0,5 p2,07R_MAx*dYVj³ = 3,4259328* 40³= 219259,6992 W

C_P = 65008,4447 : 219259,6992 = 0,29649 = 29,649%

For Vj = 50m/s 0,5 p2,07 RMAX* dYVj³=3,4259328[m²]*50³= 428241,6 W

C p = 126970,674 : 428241,6 = 0,296493 = 29,649%

For R MAX = 0,5m and dY= 5m

For Vj = 40m/s 0,5 p2,07R_MAx *dYVj³ = 6,691275* 40³= 428241,6 W

C p = 247926,433 : 836409,375 = 0,296417 = 29,6417%

For R MAX = lm and dY= 5m

For vj = 40m/s 0,5 p2,07R_MAx*dYVj³ = 13,38255* 40³= 856483,2 W

C p = 253921,079 : 856483,2 = 0,29649 = 29,64694%

For Vj = 50m/s 0,5 p2,07 RMAX * dY Vj³ = 13,38255* 50³ = 1672818,75 W

C p = 495939,619 : 1672818,75 = 0,296469 = 29,6469%

For R MAX = 1,5m and dY= 5m

For vj = 40m/s 0,5 p2,07 RMAX *dYVj³ = 20,073825* 40³= 1284724,8 W

C p = 380879,593 : 1284724,8 = 0,2964678 = 29,64678%

My programs show that if values are R MAX < lm, L_DucK=dY=4m OR dY=5m without a regard on wind speed for the chosen configuration power amplifiers and a horizontal's air streams, the coefficient efficiency kinetic's energy C_P is always on 4 decimal's digit that is 29,64%.

This patent is possible realize only, if we make the comparison different dimensions and the different fisic's variables, and theirs interacting dependence.

The average speed wind on the mountain of Croate's sea-side is Vj = 40m/s = 144 km/h.

SENZOR pressure WINDMILL with R MAX = 0,32m and dY= 4m has Inertia moment of resistance to rotation only 1=24,56 kgm². The speed wind Vj = 40m/s produce 65008 W.

SENZOR pressure WINDMILL with R_MAX = 0,32m and dY= 5m has Inertia moment of resistance to rotation only 1=30,6 kgm². The speed wind Vj = 40m/s produce 81261 W.

R MAX= 0,4m and dY= 4m has Inertia moment of resistance to rotation only 1=46,596 kgm². The speed wind Vj = 40m/s produce 81643 W.

If R MAX= 0,5m and dY= 5m has the Inertia moment of resistance to rotation I = 113,276 kgm² is not SENZOR. vj = 40m/s produce 126964W. Consequently 1,556 times bigger power must conquer 3,7 times bigger Inertia moment of resistance to rotation then what SENZOR pressure windmill must conquer.

Small blades with R MAX = lm and with dY= 5m has the Inertia moment of resistance to rotation I = 841,645 kgm². Vj = 40m/s produce 253921 W. Consequently 3,125 times bigger power must conquer 27,5 times bigger Inertia moment of resistance to rotation then what SENZOR pressure windmill must conquer.

R MAX= 1,5m and dY= 5m has the Inertia moment of resistance to rotation I = 2793.8

Vj - 40m/s produce 380880 W. Consequently 4,687 times bigger power must conquer 91,3 times bigger Inertia moment of resistance to rotation then what SENZOR pressure windmill must conquer.

R MAX= 1,6m and dY= 5m has the Inertia moment of resistance to rotation I = 3376.75 kgm². Vj = 40m/s produce 404330 W. Consequently 4,9757 times bigger power must conquer 110,35 times bigger Inertia moment of resistance to rotation then what SENZOR pressure windmill must conquer.

Consequently the top comparison was for the wind speed Vj = 40m/s = 144 km/h.

Fisic's laws of the UNDERPRESSURE must to study with help the

equation of condition gas

pV=nRT

The quantity substance n is equal proportion between the whole mass misec of the gas that is air streams which run crash in 1 second on the small bound of the aggregate and minor's mass of the gas M. If parallel air streamlines with the speed of Vj vertically press surface SINP width dY and height hiNP = 2,07m for R MAX =lm, then the mass HI i sec air streamlines, which run for 1 secunde on the surface SINP is equal product belong volume SINP* Vj and specific weight air p or IT1 isec = p hmp dY Vj , and we have that is :

poViNP = pohiNP dYVjRT : M

If input mass of molecules air enter inside the duck of small blades without lose, then the pressure PPRIOR of air streamlines change still before their kick in the ducks of bound, and after that the molecules lose the path own kinetic energy. Now the density of the air po≠ 1,293 kg/m³, because the density of the air on equal widths dY of duck compress with input height hiNP = 2,07m on the height

and also the density of the air increase on :

/JPRIOR = 1,293 * 2,07 : 0,54 kg/m³ = 4,9565 kg/m³.

The volume decreased about 3,83 time, because the height decreased about 3,83 time, and the pressure PPRIOR increased about 3,83 time.

Toward Boyle Mariot's law is :

0,54p_PRioR = 2,07p₀ OR p_PR,oR = 3,83 p₀

If houT>3,45m on width dY~ 5m, then we have ρουτ = 2,07 po : 3,45 ~ 0,6 po with the difference pressure Δρ=(2,69ρο— 0,6po)/5m ~ 0,418po/lm

The question is set : How much the percent of molekules air streamlines not pas with the pressure po on the pressure 3,83 po inside the wind aggregate, although for a

underpressure we have ρουτ ~ 0,605 po, and inertia moment of the resistance to rotation of the pressure WINDMILL is small, than the molekules will run over the roof out.

Because of prof. dr.Virag suggested the roof as on FIG. 3, where the roof climb of the point T₂(©474,Y,107) to the point -Si6(0497,Y,©653).

Now we have houT> 6,75m on length of about 5m widths dY on the duck, and then we have : ρουτ = 2,07 po : 6,75 ft 0,307 po, and Δρ = (2,69 po - 0,307 po ) / 5m ft 0,477 po / lm

I don't believe then the underpressure ρουτ ft 0,307 po can to pull the air streamlines in small bound of the wind aggregate on the pressure PPRIOR = 3,83 po , IF I > 2000 kgm², but if the beginning of lower powers amplifier decrease the beginning height for 4cm, and the end we omit 2 lowest part height powers amplifier, then we have :

(0,54 + 4)p_PRioR = (2,07- 07)po OR PPRIOR = 2,362 po

a = arc sin(0,42) = 24,83458759° β = a + 20° RMIN = 42 : sin β = 59,57cm

We know that molecules air which will enter inside the duck of small blades it will climb of the angle a to the angle β much more of 4mm, so that the depth of a duck of 40cm is completely enough, that a molecules of air which lose the part own kinetic energy because of the transformer in energy rotation, don't fall on the bottom of the duck. The bottoms of the duck are on figures sign with ai and with a₂ with lengths calculate in cm. The flood— gate of the horizontal's air streamlines of 42cm contains equal a mass air streamlines as that the flood-gate don't exist. The horizontal's air streamlines transformed in climbing air streamlines lower amplifier with much smaller radial component force, but with much bigger tangential component force, which produce the rotation. If we wish decrease the pressure PPRIOR , we must decrease dimensions the lower amplifier for RMAX> 40cm, in the relation on the looking the lower amplifier for RMAX < 40cm with extremely small inertia moment of the resistance. See there tabulated cm dimensions list the lower amplifier for RMAX= 32cm and I = 30,6 kgm² and for

1,6m and I =3376,75 kgm².

© = RMAX : 100cm— is the multiplicator for X or Z coordinate, if this sign there are registered before X or Z coordinate.

S₃(©107,Y,©42) T₃(©90,Y,©46) «→ T₄(©158,Y, ©30) <→ T₅(©392,Y,0)

T₅(©392,Y,0) ^ Te(©732,Y, - ©30) <→ T₇(©1228,Y,- ©62) <→ T₈(©2010,Y, - ©100) ©90 =0,32*90 =28.8 for R_MAx= 32cm ©90 = 1,6*90 = 144 for

l,6m

1. path is of (28.8 ,Y, 14.72) to (50.56,Y,9.6) of (144,Y,73.6) to (252.8,Y,48)

2. path is of (50.56,Y,9.6) to (125.44,Y,0) of (252.8,Y,48) to (627.2,Y,0)

3. path is of (125.44,Y,0) to (234.24,Y,-9.6) of (627.2,Y,0) to (1171.2,Y,-48)

4. path is of (234.24,Y,-9.6) to (392.96,Y,- 19.84) of (1171.2,Y,-48) to (1964.8,Y,- 99.2)

5. path is of (392.96,Y,-19.84) to (643.2,Y,-32) of (1964.8,Y,- 99.2) to (3216,Y,- 160)

With regard to the multiplicator © for X and Z coordinate, if something we reckon for

this results are important for equal path. If for we change 1. path so that the flood-gate of the horizontal's air streamlines is on height of 30cm, and R MAX = 1,6m has not 4. path and 5. path, then we have :

100-30 =70 0,7ppRioR = (2,07- 07)po ili PPRIOR = 1,957 po

Normally because of I =3376,75 kgm², we must lose something of power. The best visibility of this is on FIG. 28.

RMAX= 1,6m and dY= 5m has I = 3376,75 kgm², and because of huge resistance to rotation I think that is impossible that Vj = 50m/s produce 789183,5W or that Vj = 60m/s produce 1363985,8W.

When we change 1. path and throw all 4. path and 5. path, that Vj = 50m/s produce 529283W and Vj = 60m/s produce 914600W.

I consider that Cp coefficient increase if on every roof of the pressure wind

aggregate, over every press's bound we place propeller wind power plant, because of increasing the resistance running over roof out to air streamlines. Numerous molecules will abandon to execute the rotation of 6 small rotor blades. Perhaps we can to have 12 smaller rotor blades on the smaller height of the roof of the pressure windmill. Between 12 rotor blades existed greatly small empty increasing C p the coefficient of efficiency energy but the resistance running molecules over roof out will be rise.

The horizontal's air stream on FIG. 1 keep striking on the height's difference between point T2 and point Tio transform in the air streamlines of lower power amplifier SENZOR propeller wind aggregate and rises their power proportional height of the roof.

If R MAx=lm and length of the rotor blades is 5.6m, let the inclination is smaller of 38°. We wish rise the power of a impeller wind power plant for 40% increasing p→ specific air density, what is impossible attain if the impeller wind power plant mount on the tower height 1 km.

If for RMAX>0, 5m on the FIG. 1 molecules run over roof out, dr. Virag pro-dean engine university in Zagreb propose the roof of wind aggregate as on FIG. 3 because of the increasing the resistance running of air streamlines over roof out. The output height for molecules will increase for 2 times will cause increasing under pressure for pressure windmill. The pressure senzor windmill with

will be more economical then today wind turbine, and none the risk not exist and we can produce 1. PROTOTYPE as on FIG. 1 and 2. PROTOTYPE as on FIG. 3, because I = 30,6 kgm² is equal and extremely little for both case.

After the comparison result for FIG. 1 and for FIG. 3, we can more easily choose cheaper economical solution.

When I penetrated through north (bura) wind as boy, I don't feel none the pressure on my legs, but I feel the big pressure on my breast and on my head. Because of the beggining lower amplifier must begin on height 2m on horizontal's plain, but bottom level wind aggregate air streamlines must to have the passage with bigger the resistance of the resistance passage inside small bound in wind aggregate. The measuring will show under which angle will be pulled plane for the drainage the lowest air streamlines for FIG. 1 and for FIG. 3. The roof for FIG. 1 begins of the point T₂(©474,Y, ©107) to the point Tio(0,Y, ©380). The roof for FIG. 3 begins of the point T₂(©474,Y, ©107) to the point -Si6(-©474,Y,©653), and senzor impeller wind power plant is placed over Sio(0,Y,©380).

The beginning height of the lower amplifier for impeller wind power plant if the pressure windmill has RMAX =lm is for FIG. 1 and for FIG. 3 is equal h = 273cm.

For RMAX=0,32m, h=273cm* 0,32=87,36cm.

The small height of the roof can'ot increase the power for 40%. For

and for dY= 5m the inertia moment of the resistance is only 30,6 kgm², and impeller wind power plant is not needing increase the esistance air streamlines. If is this the thrue, then impeller wind power plant can to look like FIG. 5.

For lm and for dY= 5m the inertia moment of the resistance is 841,6466 kgm², and this is good that propeller wind turbine is placed on the roof for FIG. 1 or for FIG. 3, that increase the resistance running to air streamlines over roof out. Because of the inertia moment of the resistance to rotation I for R MAX < 0,5m, I increase the lower amplifier about 3 times in the relation on earlier solution and because of PPRIOR = 3,83 po for entry in wind aggregate. For RMAX > 0,5m, we need do the measuring for how much we must to decrease the length of lower amplifiers.

This analysis show, that if on the beginning when the wind begin puff, then perhaps the pressure senzor windmill with RMAX< 0,4m with I < 50 kgm², and if on the beginning electrical's power charge not exist, perhaps the pressure windmill can to oneself on output air streamlines lift up the roofs door.

If this is not possible, impeller wind power plant can lift up the roofs door to everyone the pressure windmill. After that the press windmill are more sensitive of propeller wind aggregate. The pressure windmill has the least generator with permanent magnets which give curent for activate magnets of the excitation's generator, and the excitation's generator give curent for activate excitation magnets of the bigest main

generator with electro— magnets.

The pressure windmills with RMAX= 0,5m and with dY= 5m are not SENZOR, because it has 1= 113,276 kgm², and the roofs door must be for output air streamlines very much lift up, that the output pressure will be what smaller. On the start speed wind is smaller and I> 113,276 kgm², and the bigest main generator with electro— magnets has not the none load, how contra moment of generator does not exist.

The pressure windmills with RMAX > lm and with dY= 5m have big I the inertia moment to rotation. The angle's acceleration is small, and we need a long time that angular velocity G) will be enough big that G) produces enough a bigger centrifugal acceleration of the gravitation acceleration for getting the exhaust air. Too much a long time produce air's cork. The engine of any SENZOR ELECTRIC POWERS rotate it's the shaft to enough angular velocity o, and after that same the engine will lift up the roofs door on the output and open the entry air streamlines. On the start its generators with electro— magnets has not the load of the resistance. If one part of duck of the small bounds rotate with equal speed as wind speed, then this part don't produce energy without the resistance to rotation. If one part of duck rotate with smaller speed of the wind speed, then this part will produce energy with the resistance to rotation. If one part of duck rotate with biger speed of the wind speed, then this part will expend the energy and draw in oneself air streamlines with the preventionth produce air's cork on the start rotation with slowly speed wind. If this incident is occur the during working the pressure windmill, then the accumulation energy of the rotation shaft will begin to dissipate. Respectively when wind speed decrease the accumulation energy of the rotation shaft will dissipate and decrease the angle's speed (0 the rotational shaft where there is generator.

Naturaly every wind turbine must have CONTROL UNIT, that with the heaters of a water attain such the contra moment of generator, which will produce the maximal bigest power. Momentarily amplified wind speed, not follow always increased the load on the power electric net, and the angle's speed ω of wind aggregate will increase, what will decrease value Avj, toward formula :

Avj = Vjjang- OOR.

The same such at night electric consumer on the power grid is smaller, and the contra moment of generator is smaller and biger the angle's speed a>, and the power because of Avj will be smaller of possible the bigest power.

Only tangencial speed wind Vjjang [m/s] produce moment of the rotation and has same direction as COR speed with which the points on the duck of the small blade run away speed Vjjang, and formula for the power is : PEF = 0,5 pS (AVj)³.

In the both top's case, CONTROL UNIT will give in addition thermal electric resistance for warming water in a basins or in a system central heating, that every speed of wind give what biger energy. Besides at if any wind turbine has not CONTROL UNIT, anglular velocity ω can rises so big, that big ω destroy the duck on small blades.

Coriolis's force

If any body move inside the rotational room, then besides centrifugal force, Coriolis's force performs on this body inside the rotational space.

2 m BUR A CO ROT X VBURA OR 3 COR— 2 CO ROT X VBURA gdje je CO— 27Γ I

When north eastern wind→ BURA blow with north, vector speed_BuRA = VBURA go of +X axis toward - X axis, OR

VBURA =— V i This is equation of the horizontal's air stream.

Angle's speed ω is always vertical on momentary position a duck on small blade. If the duck momentary there is on position angle φ, then mathematical^ formula for the vector angle's speed is next :

CO = ω (— s ttp i + cos(p k) in I. quadrant, or ω = ω (— sincp i— cos(p k) in II.

quadrant. Because of simple writing a formulas, we do not put→ over vectors and over digits i, j and k, and the formulas will take without numerical's values.

OR a_CoR = 2G)(- sin(pi + cosipk) X(-vi) = -2 (OVcos(|>j for I. quadrant

OR for comparison a CENTR

On sadness Coriolis's force will keep declining molecules of air in direction— Y axis, and centrifugal acceleration in direction angle φ. This declination in direction— Y axis at small angle φ is enough problematic and because of the width duck on small blade must be biger of the height duck, if the pressure wind aggregate will function good, because exist possibility correction of force under pressing.

OR a _CoR = 2ro(- sin(pi - cos(pk) X(-vi) = +2 (OVcos<pj for II. quadrant

Coriolis's force is equal vector product, only radijal's components speed air

streamlines in direction angle φ of small blade realise Coriolis's force.

VBURA = V(— cosPi + sinPk) This is lift up air streamlines with the lower power amplifier.

OR a _CoR = 2ro(- sin9i + cos9k) X [V(- cos i + sinPk)] for I. quadrant

OR a COR = 2ων (- cosi cosp j + sin( sinPj )

On sadness if the angle β is small the declination molecule of air in direction— Y axis is insignificantly decrease, but if angle φ is the sufficiently big, arrive to annul dictate of Coriolis's force.

VBURA = V(— cosPi— sinPk) This is falling air streamlines with the upper power amplifier. OR a_CoR = 2ro(- sin(pi + cosipk) X [ V(- cosPi - sinpk)] for i, quadrant

OR a_CoR = 2ων (- cos<pcosPj - shKpsinPj )

On sadness the angle β is small, but the angle φ must be the sufficiently big, that the declination molecule of air in direction— Y axis will reduced.

For II. quadrant we must to define angle δ = φ— 90°, and we have that is :

OR a_CoR = 2(Q(- cos6i - sin6k) X [V(- cosPi - sinPk)] for II. quadrant OR a coR = 2ων (sindcosPj— cosdsinPj )

IF cosp > cos8 THEN sinP < sin6 This is good, that is angle p < δ.

We know that molecules of the air always wish to go where more easily, because the molecules avoid to rotate small blades, and molecules drag through small hole because of the air can press on the smaller volume. If the blade has 18 small blades, then the interspace between small blades is Δφ = 360° : 18 = 20°. The most adversely case is one small blade is on position 80°, and next small blade is on position 100°. The interspace between the highest horizontal with aces Z= lm and the highest horizontal line of small blade is about 1,52cm, and the loss is more 2 times smaller, then then when the blade has 12 small blades. Besides this between air streamlines exist the friction, and the air streamlines of the upper amplifiers press horizontal's streams towards down and reduce its flight with the most effectively for case about 1,52cm, the horizontal's streams will decrease angle P for the air streamlines of the upper amplifiers. The small blades will have bigger angle φ and with new angle p will decrease depth to streams of the upper amplifiers inside duck and increase the EXHAUST, and decrease Coriolis's acceleration toward formula

a COR = 2 ων (— cos(pcosPj— sin( sinPj ).

Besides decreasing depth of entry to streams is decreasing area of Coriolis's acceleration and decreasing shift in the direction Y axis, or in the direction the length of the shaft. Because of the reason I increase height of a obstruction on 46cm, and the lowest horizontal's stream kick in the duck of small blade at the angle

φ = arc sin 0,46 + 20° = 27,3871075° +20° =47,3871075°

RMINJS = 46cm : sin 47,3871075° = 62,50469 cm

The depth of entry the lowest horizontal's stream is : 100cm— 62,50469cm =

37,49531cm, and now for

the duck can go in the depth of entry only 40cm.

The smallest effect toward formula a CO_R = 2ων (— cos(pcosPj + shK sinpj), Coriolis's acceleration has at air streamlines of the lower amplifier power.

Every air streamlines lower amplifier has smaller radial's component at angle (p, then horizontal's air streamlines at equal angle (p.

Because of formula a _CoR = 2ων (— cos(pcosPj + suHpsinPj), the action Coriolis's acceleration for air streamlines lower amplifier is more heavily programs, then the action centrifugal or gravitation's acceleration with formula : a _REZ ⁼ CO²R— 9,81 sin(p .

Because of before mentioned I increase the length of activity lower amplifier about 2 time, and decrease the length upper amplifier and angle of the climbing.

The upper amplifier for RMAX =lm go of coordinates (80,Y,100) until coordinates (474,Y,107) [cm].

The upper amplifier climbs only for 107 - 100 = 7cm with angle a =1,017838237°. I suppose that air streamlines refuse of the amplifier with the angle : β =2,035676474°.

The lowest air streamlines has formula :

Z- 107 = tgp (X9,5cm 474) OR Z = 0,0355442 X+ 90,15204

The orbit every point on the top duck is circle with equation : Z² + X² = 100².

If equation Z = 0,0355442 X+ 90,15204 of the lowest streamline insert in equation of circle we have :

1,0012634X²+ 6,408767 X - 1872,609 = 0

The solution is : Xi = 40,16426 cm a = arc cos (Xi : 100) = 66,3190944°

Now we must calculate where air streamlines Z = 0,0355442 X+ 90,15204 kick small bound which equation Z =Xtg(20°+66,3190944°) in X Z plane.

The solution is : X = 5,812994 cm RMIN = X : COS 86,3190944° = 90,5479cm

The lowest air streamlines of upper amplifier go entry in bound only 100cm— 90,54cm ~ 9,5cm, and because between air streamlines exist the friction, air streamlines of upper amplifier will press the horizontal's air streamlines toward down and dicrease flight of air streamlines above the roof of wind aggregate. Toward the law action and reaction the horizontal's air streamlines decrease angle β, and if(p is enough big, toward the formula for Coriolis acceleration we have :

a COR = 2ων (— cos(pcosPj— sinq>sinPj)

The depth of entry air streamlines decreased, and make easy throwing molecules execute the working and lose kinetic energy out, and decreased efect of Coriolis acceleration. Now is possible for

use triangle static for the fraim with entry's duck of 40cm. My program for different RMAX and for different the depth of entry, calculate dimensions of the fraim and for the duck. STATE OF THE ART

Generally at all WIND TURBINES, the bigger wind speed produce faster the rotation blades. The power grows with the wind speed on 3.rd potential, and the centrifugal acceleration grows with the angle's speed on 2.nd potential. When to big wind speed arrive, then rotor blades of the wind aggregate change the angle, so that the wind don't keep striking vertically on the surface of the rotor blades already with onether angle because of decrease the force. The force decrease moment of the rotation. The decreasing moment decrease the angle's speed and the centrifugal acceleration because of the decreasing on the top of the rotors blade. Consequently all known wind turbine begin to limit the power at wind speed about 80km/h, and it forced on this, because theirs rotor blades are longer of 10m. This is clearly that if wind aggregate has only 3 rotor blades, the wind streams run out between rotor blades more if rotor blades are longer. Consequently much more the wind streams run out, of the wind streams force into work, but at the press windmill we effort that little the wind streams run out, because the wind streams must to rotate a shaft. If molecules of the air decrease speed and lose own kinetic energy for a rotations rotorblades, the pressure windmill need throw out this molecules with decrease kinetic energy, that don't disturb the entry new molecules with biger speed and biger kinetic energy, and we good enough explain in prior text.

With R AX we designated the most distance lines points on the duck of the axis rotation shaft. I the inertia moment of the resistance to the rotation grow with RMAX on 3.rd potential. Because of we use the small radius RMAX of the rotation. Moment M is product the diferent tangential's wind speed and rotation's speed points CoR and with the arm of a radius the rotation. The angle's acceleration a_AKc is equal quotient momenta M and I the inertia moment of the resistance to the rotation. IF I is smaller, the acceleration ¾AKC is bigger and faster increase angular velocity ω up to this till contra moment of the generator is equal as curent moment M of wind, when suma of all moment is zero.

The power of the wind grow with wind speed on 3.rd potential, but at extreme small the resistance to the rotation, the acceleration a_AKc terible faster increase the angular velocity ω, that the generator with a electro magnets must immediately connect resistive loads if the thunder eliminate electric net, that CONTROL UNIT save the wind aggregate of destruction because of too quick a rotation. Because of in generator's room on equal shaft there is the smallest generator with permanents magnets, which give the current to excitation generator, and which give the current to the excitation of a electro magnets of the greatest generator wind turbines.

As soon wind aggregate rotates the system of 3 generators always produce curent for the loads and for own CONTROL UNIT, which must immediately conect resistive loads on the greatest generator with a electro magnets. All 3 generators are inside Faraday's cage, and the thunder cann't disconnected the working pressure windmills.

At night the consumption of energy on the electric net is smaller, and because of contra moment of generator is smaller, and the production of power is not maximal. Then control unit will connect resistive loads for warming water in the basin.

The engineers well know next : Induce the electrical's force of generator ( voltage ) is equal product of the number of a coil with the change magnetic's flux upon time.

n_NAvojA*A<I> IINAVOJA* C_GEN (O_GEN We supposed : ΔΦ = CGEN OOGEN

At At When wind speed is so small, that the voltage on the generator of the wind turbine smaller of the voltage on the electric net, that the nergy can use for warming water. Besides that at the surprise very strong kick of wind, the electric net can'ot to insure big resistive loads. Because of the electronics must immidiately connected the resistive loads in basen with water because the inertia moment of the resistance to rotation of pressure windmill is very small. The working area for variable wind speed is about 2,5 times bigger of to days wind turbine with upper limit for wind speed about lOOkm/h.

Generator with a electro magnets of pressure windmill can to produce the alternating current with frequency in area of 10Hz to 100 Hz, and energetics electronic dissimulator first the alternating current change in direct current and after that produce the curent of 50 Hz. If we need the curent only for warming water in basin, that energetics electronic dissimulator is not necessary.

Tehnical problem

When let molecule wind stream kick on the rotorblade, its speed decrease on the speed of the point which rotate round axis of the rotation. Must exist the exhaust for molecules air with decreasing speed or kinetic energy that go out of rotorblades, but must exist enough big under pressure. This under pressure will draw out from wind aggregate, because we must prevent to produce the cork for molecules with biger speed or with biger kinetic energy that kick on the small blades and so produce energie rotation with decreasing speed for Avj. If η is the energy efficiency coefficient, then we have formula for the power :

The middle of wind aggregate is a shaft with small blades. At the side from where wind arrive, the door of the roof are close and touch the upper amplifiers, then the air streamlines skate down of the upper amplifier and up lower amplifiers on the duck of the small blades. The air streamlines on the bigger heigh of 107 cm for RMAX =lm reject of the roof and go above roofs top of 3,8m out. On the opposite side the roof of wind aggregate is equal, but the door of the roof are open and the line of the top on the door has height 3,2m. Much more the difference between output's heigh of 3,2m and input's height of 1,07m exist enough big under pressure, that molecule's air which give own part energy for the rotation small blades and the shaft go out of the wind aggregate. For the case when the rain begin to fall the top on the door decrease height on the 2.35m, because the rain don't allowed break in entry of the wind aggregate.

Normally when rain falling down, then wind jugo ( south's wind ) has opposite situation with the door of wind bura ( north's wind ) .

Centrifugal force at even number rotorblades has opposite drift of the force produce the efforts on tension on the shaft and on the radial axial thrust bearing without produce pair of the force.

Control unit of generator with permanents magnets with FIG. 22 must on the beginning to fortify with own SENZORS from where wind blow and on this input's side close roofs door, but on the output of exhaust air streamlines must to lift the door, because this is fundamental condition that press wind aggregate can to work.

Because of we need one strong senzor wind power plant, which has generator with small permanents magnets, that it can to arrange positions of roofs door on the input and on the output. Analysis throwing out air streamlines is relative simple, because the analysis execute always with physical and mathematically formulas only for one small blades on one position, and this analysis is complete equal if blade has 12 small blades or 18 small blades. The method throwing depend of the depth penetration of air streamlines in small bound, and at the angle φ, because the drift of vector speed CORMAX is tangent on the small blade, and at the angular velocity (0, because of formula for resultant acceleration is 3CE ⁼ C0²R— 9,81* sin φ is component of the gravitation's acceleration. How any molecul move and increase R, centrifugal acceleration o²R continualy increase, but the gravitation's acceleration increase because of increasing the angle φ at the rotation in I. quadrant, and this is not uniform hurriedly moving. Siii(p is Mac Laurin's order with factorials and because of not exist none exact formula. Because of PASCAL program continually calculate variables with changed different variables on every part of lcm. Program found in loop the value acE — ,81* sin(p, and after that multiplikate average speed ( V_SR) and time At, and then we have that path is lcm.

0,01 m

The overturned path must be precise lcm, and value of R is a distance between the middle of lcm and symmetry of the rotation shaft.

Program calculate prior speed on the beginnig of this cm and curent speed on the end of this path of lcm, and half-addition give the average speed.

VcUR-VpRIOR + VsR*At

On next lcm of path, VPRIOR is equal the speed VCUR of prior lcm path. During the travelling of the molecule at a duck of 1 cm, the molecule rotate and change radial component gravitation with βίηφ and the drift of vector speed (DRMAX. Program calculate the variables of one happening and make next the tabulate list and after that program will draw diagrams of this variables.

In this table the beginning action of horizontal air streamlines is on the angle φ= 19,268°, when air streamlines no come in the depth inside small bounds. The angle's speed is 9 rad/sec.

On position φ= 19,268°, the beginning the depth inside small bound is only 3cm.

On lcm path of h=97cm to h=98cm has :

0,0229961419sek* 0,43855552 m/sek = 0,010085m ~ 0,01m = lcm.

On 2. cm of h=98cm to h=99cm is : 0,00885362055s*l,12948142m/s=0,0099999999m~lcm. On path of h =99cm to h=lm has : 0,00627362582s* l,59397456m/s = 0,00999999m -lcm. The small bound on the beginning there is on the position h=97cm and φ=19,268°, and after that on the position h=98cm and φ=31,734°.

The position h = 99cm has the angle φ = 36,2997°.

The position h=lm has the angle <p=39,53478° and then output current speed was l,79869737m/s. Because of continually increasing the middle speed between h=99cm and h= lm was the bigest and for orbit of 3,235° needed only 0,00627362582s, but for path between h=97cm and h= lm needed 0,0381233882s. Vector speed of the rotation had constant value

but the angle changed contionaly. At the end the resultant speed between output speed of molecules and vector speed of the rotation is (9² + 1.8²)¹' ² = 9,178 m/s.

The subject matter of invention

Formula for the power WIND TURBINE is : P = EKINJSEK = 0,5 C P p S Vj³

CP is the energy efficiency coefficient or turbine power coefficient,

p is specific density of the air or air density.

S is circular surface, which describe the top of rotor blade of wind turbine.

Vj is wind speed.

Mass of triangle's blade depth d is : m = γ* 0,5a* h * d, where a is base of triangle with height h, and γ specific density of material of triangle's blade. The axis of the rotation is on bottom of height h, but on the middle depth d, and the inertia moment of the resistance to rotation is :

Ix =ya [( h ³d : 12) + (h ³d :24)] ~ya( h ³d :12).

Formula for power the press wind turbine is : PEF = 0,5 pS (Avj)³ = 0,5 pS ( Vjjang - G)R ) ³.

Only wind speed Vj in drift tangent produce moment of the rotation. The points on the duck of small blades has 0)R speed in drift tangent as and speed Vjjang , which come up this points with the speed AVj = Vjjang— G)R.

S is vertically level surface on the drift movement horizontal wind stream on the entry inside amplifier power of the wind aggregate. To-days wind turbine work normally until speed wind about 80km/h after that the rotation slow down, and when wind speed increase over lOOkm/h, the rotation stop. This signify wind speed 200km/h produce biger power for (200 : 80 )³ = 15,625 time of the wind speed 80km/h. 200km/h produce for lh equal energy as 80km/h for 15h and 37,5min.

I Y_RECTANGLE = IIIRECT [ h²/ 3 + dpi / 12] IHRECT = γ hadpL

For dpL « h we can tell that is : I Y RECT ~ γ h³adpL : 3

If RMAX and RMAX_I are height of 2 rectangles, and if the proportion their heights is :

: RMAXJ , and fisic's formulas proportion in connection with a distance points of the axis rotation shaft are next :

IF n=r2 : ri

Toward prior exposition primary meaningfulness of invention is that pressure windmill and senzor propeller agregate are using energy wind when maximal wind speed is biger of 200km/h for their location.

When today's wind turbine work until Vj = 80km/h, then wind produce 6,25 times smaller moment, than when speed wind of Vj = 200km/h produce 6,25 times biger moment, because moment air streamlines rise with wind speed on 2.nd potention. However moment air streamlines increase with R MAX on 2.nd potention. R MAX is the bigest distance between the axis rotation shaft and the uppermost horizontal line. Because of we must to use small radius rotation RMAX and for pressure windmill and for senzor propeller wind power plant. But bigger width small blades catch bigger mass air streamlines at pressure windmill, and we use the width of 5m or of 4m, because the widths are much bigger of R MAX. With regard on expense material wind agregate with R MAX =

0.32. produce cheaper lkWh energy of the wind agregate with RMAX = 1,6m.

Secondary meaningfulness is that wind turbines used lower amplifiers and upper amplifier with using small radius rotation multiply with biger width small blade give bigger surface SBIGER = SINP from where arrive horizontal's air streamlines, because the bigger surface produce biger energy. If air streamlines compress with biger surface SINP on smaller surface, then the pressure on entry in the small bounds of wind aggregate will increase multiplication with kvocjent SINP : S. Now we have a question : will the air streamlines avoid to go on positions, where the pressure is 3 times bigger, if one can run away over roof out. Consequently must exist something what will draw air streamlines that one entry in wind aggregate. At for all that must be filled even 5 conditions.

1. condition is that air streamlines come out through surface SOUT » SINP , that the pressure on output will be smaller, because the under pressure decreased with multiplication kvocjent SOUT : SINP . The under pressure must draw the molecules out of the wind aggregate, one kick in duck of small bound and path of their kinetic energy transform in the rotation.

2. condition is that the centrifugal acceleration is enough bigger of the gravitation acceleration everywhere inside small bounds, because the exhaust air must throw molecules out of small blades wind aggregate, one lose path of their kinetic energy because of the rotation shaft.

3. condition follow from fact, that bigger wind speed linear increase angle's speed, and this fact dictate method working of CONTROL UNIT. 2 times bigger wind speed produce 2 times bigger angle's speed and 4 times bigger angle's acceleration. Control unit will must sometimes abstain of getting maximum power.

Wind speed Vj =15m/s produce for 7,1 sec angular velocity ω > 6 rad/s.

For Vj < 15m/s need too much time for getting necessary angular velocity ω and we can expect the creation air's cork. Decreasig load increase ω. So Vj = 20m/s produce maximal power when load moment of generator is 3830Nm and start need time t=l,9 seconds that come angle's speed ω > 6 rad/s. But if load moment of generator is only 800 Nm, then need t < 0,8 sec that come angular velocity ω >7 rad/s.

Wind speed Vj = 40m/s need for the start time 0,4 sec that come angular velocity ω >7 rad/s. Wind speed Vj = 50m/s need for the start time 0,2 sec that come angular velocity ω >7 rad/s.

For t < 0,8 sek do not come at start working of wind aggregate the creation air's cork.

4. condition is that propeller senzor wind power plant with theirs engines will rotate shaft of a pressure wind turbine with start's angle's speed ω > 6 rad/s, and beginning load for shaft will be only the excitation of the least generator with permanent magnets which give curent for activate magnets of the excitation's generator, and after that will place a height of input and the height of output for the air streamlines toward FIG. 1 for toward FIG. 3.

5. condition is fact that only tangential component air streamlines produce the rotation and do not create Coriolis's aceleration, but the radial component of the air streamlines produce radial pressure and Coriolis's force on the radial axial rotor bearings. The climbing air streamlines of the lower amplifiers on position equal angle entry less inside small blades than horizontal's air streamlines and one has bigger a tangential's component air streamlines and one has smaller radial's component air streamlines than horizontal's air streamlines. The falling air streamlines of the upper amplifiers on position equal angle come in deeper inside small blades than horizontal's air streamlines and one has smaller a tangential's component air streamlines and one has bigger radial's component air streamlines than horizontal's air streamlines. I decrease the upper amplifiers, but because of the air can pressure on smaller volume, the lower amplifiers power increased more of 2 times in relation on my earlier decision.

This is a reason when R MAX > 0,5m, we must measure how much must reduce in size the lower amplifiers. The main news in relation on my earlier decision that when R MAX < 0.4m, we have much more increasing of the lower amplifiers, because Inertia moment of the resistance a rotation is only I < 47 kgm², and this decision will be economic the most payable.

Nobody living does not know, what experimental measuring with thus much heterogeneous combinations exactly will show, and specially for R MAX =l,5m with I > 2700 kgm², and specially for RMAX =l,6m with I > 3300 kgm².

The third meaningfulness is that senzor propeller wind power plant produced as soon as more energy with using the lower amplifiers power and upper amplifier power with the prevent that air streamlines run out over top roto blades. The upper amplifier power will be protection of the tunder. Toward senzor propeller wind power plant is on the roof of pressure windmill and produce the resistance for run out and because of this part molecules come in the pressure windmill. Molecules which run over roof will increase power of the senzor propeller wind power plant.

The fourth meaningfulness is possibility that excitation propeller wind power plant with small RMAX because of extremely inertia moment of the resistance to rotation has big angular velocity, what enable using lOpoles generator without using 2 multiplier.

The fifth meaningfulness is that every press windmill has the smallest generator with permanents magnets, which give the current to excitation generator. The excitation

generator will give a curent for excitation electro magnets of main the bigest generator. All 3 generator of press windmill there are in middle room little house, which during rain is

Faradey cage because of the protection of the tunder.

All a wind turbines have the smallest generator with permanent magnets because of own excitation and own control unit.

The sixth meaningfulness is that control unit enable energy with warm resistance for warming water, if consume energy on electric net is insufficient.

The seventh meaningfulness is that main control unit contionally supervise and manage work whole system different a type WIND TURBINES because of different reasen. LIST OF FIGURES

FIG. 1 show a view of a roof 1. variant of a pressure windmill. This side of a roof from where arrive wind has closed roof 's door of the roof to last position upper amplifier. The roof on the opposite side has open roof 's door because of decreasing pressure (under pressure). Centrifugal acceleration press a molecules out of a small blades which one spended a part own energy, but under pressure draw this molecules out of the wind aggregate. On floor lower amplifier are the rotational flooring door for completely closing a passage air streamlines in the wind aggregate. The small generator with permanents magnets give a current to excitation generator. The excitation generator give a current to servo engines, which dictate the positions roof 's doors and the rotational flooring doors. 2 roofs over the pressure windmill are lower amplifiers for the rotor blades aggregate. We know on the floor ground the wind does not blow, and if a point T8 lifted over ground, slantig surface drain air streamlines under whole construction, which a carriers carry on this way, that air streamlines can between the carriers run out.

The roof under generator's room has not a door and expanded until level ground is prolonged a lower amplifier power for the rotor blades aggregate.

FIG. 2 show a way action a horizontal's air streamlines, the air streamlines shortened the lower amplifier power, if this will be necessary and air streamlines upper ampifier power on the bound with 18 small bounds of the pressure windmill.

FIG.3 show 2. variant of pressure windmill with one propeller aggregate under every one bound. Now we have not roofs door, but because of sheer roof must rotate as children's swing round own middle if the wind change drift. This is very good that all 2. variant is lifted under ground with similar a carriers as on FIG. 1.

Although now the roof increase under pressure, I think that the lower door need, because may to need one propeller wind aggregate with own engines rotate 2 shaft of the pressure windmill with enough big angular velocity that will be CO²RMIN » 9,81m/s².

FIG. 4 present Y Z plan the left full shaft of I. bound with small bound 4 and 13, and plan a hollow shaft is upon a left full shaft with 2 multiplier and with grounded three- phase generator. The hollow shaft there is upon a symmetrical right full shaft of II. bound with small bound 4 and 13, because only so we can prevent undesirable vibration.

Normally that for RMAX= 0,32m or for RMAX = 0,4m, we can Use 16 poles generator instead 4 poles generator with 2 multiplier.

FIG. 5 show independent work a propeller wind power plant with own lower power amplifier and with additional upper power amplifier increase density of the air. The upper power amplifier beside a help underpressure prevent that air streamlines run under the propeller wind power plant.

FIG. 6 show decrease vertical loss molecules of the air under blade with 18 small blades in comparison with loss under blade with 12 small blades.

FIG. 7 has plan triangular surface volume V= 0,5 a* V*d, and mass of this volume is : m = p* 0,5a*v*d, where a is isosceles triangle with height v, and with fatness d, and p is density of material from small bounds.

For d « v in drift X axis formula for the inertia moment of the resistance is :

v² v³

Iv ~ m*— ¾ p *a*d*—

6 12 In FIG. 7 there is also slanting projection roller where in drift X axis d = R, and

R² pR^ formula for the inertia moment of the resistance for roller is : I_Y = m *— =

2 2

FIG. 8 show a dimensions a metal frame and the duck for RMAX = 0,32m. Drawn is the deepest entry of 12cm on a duck of horizontal's air streamlines, when the duck is on position angle φ = 47,3871°. The duck mount on the frame with enough quality exhaust if (0 > 11 rad/s.

FIG. 9 show that for RMAX = 0,32m and L = 4m have that is :

For wind speed of Vj = 5m/s, if generator of the pressure windmill is very little loaded, then need 19,1 sec, that angular velocity is ω > ll r/s, and that is ll² * 0,19- 9,81* sin 47,3871° =15,77m/s², and for time start phase working pressure windmill, then control unit will not open the door, but if prior wind speed was considerably biger, then control unit will disconnect electric net with the generator, and connect reduced resistive loads for warming a water with generator.

If ω= 13, 5r/s the danger of creation air's cork will not exist.

For wind speed Vj = 8m/s, if generator of the pressure windmill is much loaded, then need 10,1 sec that angle's speed ω > ll r/s, and control unit similarly react as at Vj = 5m/s.

For wind speed Vj =18m s, if generator of the pressure windmill is much loaded, then need only 1,1 sec that angle's speed ω > 11 r/s, and control unit can connect a generator with electric net.

FIG. 10 show that for RMAX = 0,32m and L = 4m at wind speed Vj > 30m/s, a generator can not be without loading because of extreme little the inertia moment of the resistance to rotation and angle's speed ω> 117 rad/s and centrifugal acceleration ac=0)²R rapidly increase and destroy wind aggregate. Because of generator with electro magnets must have generator with permanents magnets with own control unit for momentary connect resistive loads for warming a water with generator.

FIG. 11 show on top of a page for RMAX = 0,32m and L = 4m, how centrifugal's acceleration ac=0D²R has the action, if ω=50 rad/s. If Vj = 50m/s and anti moment of generator is 1960Nm produce maximal power about 127 kW and angle's speed 00= 64,78 rad/s =618,6turn/min.

If Vj=60m/s with anti moment of generator's load of 2830Nm produce maximal power about 219,4 kW and angular velocity ω=77,5284^/8^740ί^η/ιηΗΐ~61,6Ηζ for lOpoles generator, and we do not need multipliers. On the bottom of page is tabulated list for Vj = 60m/s, which show summary results for any active small bounds with list of variables of moments, tangentials and radials forces for any small part lower power amplifier and for upper power amplifier and for horizontal's air streamlines. We need observe that is amount all radial's force about 2568 N, because this need for dimensions of radial's axial's bearings, and special need observe that the inertia moment of the resistance to rotation is extremely small about 24,43 kgm².

FIG. 12 show a dimensions a metal's frame and the duck for RMAX = 0,4m. The deepest brunt of the horizontal's air streamlines is 15cm inside the duck mount on the frame with enough quality exhaust if ω = 10 rad/s.

FIG. 13 show that for RMAX = 0,4m and L= 4m have that is : For wind speed Vj = 5m/s, if generator of the press wind turbine is little loading, we need 22,5 sec. for getting angular velocity ω > lOr/s and 10² * 0,24 - 9,81 * sin 47,3871° = 16,78m/s², and in start's phase of wind aggregate, control unit does not open the door, but if prior wind speed was considerably biger, then control unit will disconnect electric net and generator, and connect reduced resistive loads for warming a water with generator. If ω= 12,l r/s~115turn/min the danger of creation air's cork will not exist.

For wind speed Vj = 8m/s and for Vj = lOm/s, if generator of the pressure windmill has little loading, nevertheless need much time for getting angular velocity ω > lO /s, and control unit will react similarly as at Vj = 5m/s.

For wind speed Vj =20m s, generator of the pressure windmill can be normally loading, because we need 1 sec. for getting angular velocity ω > lOr/s.

FIG. 14 show for R MAX = 0,4m and L = 4m at wind speed Vj = 40m/s for 0,2 sek angular velocity increase on ω> 10 rad/s and can increase more of 40 rad/s, and at wind speed Vj = 30m/s angular velocity increase on ω>31 rad/s.

If Vj = 50m/s with anti moment of generator's with 3070Nm produce maximal power about 158,7kW and angle's speed W= ~51,7rad/s=493,7turn min.

If Vj = 60m/s with anti moment of generator's with 4420Nm produce maximal power about 274 kW and angular velocity ω= 62 rad/s ~592turn/min~49Hz for 10 poles generator, and multipliers are not necessary. On the bottom of page is tabulated list for Vj = 60m/s with a summary results for any active small bounds with list about variables of moments, tangentials and radials forces for any small part lower power amplifier and for upper power amplifier and for horizontal's air streamlines. We need observe that is amount all radial's force about 3208 N, because this need for dimensions of radial's axial's bearings.

The Inertia moment of the resistance to rotation is about 46,6 kgm².

FIG. 15 show a dimensions a metal's frame okvira and the duck for RMAX = 0,5m. The deepest brunt of the horizontal's air streamlines is 19cm inside the duck mount on the frame with enough quality exhaust if G> = 9 rad/s.

FIG. 16 show that for RMAX = 0,5m and L = 5m have that is :

For wind speed Vj =10m/s, if generator of the pressure windmill has very little loading, we need 4sec. for getting angular velocity ω > 9r/s and 9² * 0,31 - 9,81 * sin 47,3871° = 17,89m/s², and in start's phase of wind aggregate, control unit does not open the door. If prior wind speed was considerably biger, then control unit will disconnect electric net and generator, and connect reduced resistive loads for warming a water with generator, because ω= 14,789 r/s> 9 r/s.

For wind speed Vj =20m/s, generator of the pressure windmill can be normally loading, because we need lsec. for getting angular velocity ω > 10 r/s. If generator remain without load, for 4sec angular velocity increase on ω > 100 r/s.

FIG. 17 show that for RMAX = 0,5m and L = 5m at wind speed Vj = 30m s at normally loading generator for 0,4sec angular velocity increase on ω>9 rad/s and after that increase more of 24,5 rad/s. At wind speed Vj = 40m/s angular velocity increase on ω>33 rad/s.

If Vj = 60m/s with anti moment of generator's with 6000Nm produce maximal power about 428,5 kW and angular velocity ω= 49,8 r/s ~ 475,555 turn/min~39,6Hz za lOpoles generator ili79Hz for 10 poles generator, and multipliers are not necessary. On the bottom of page is tabulated list for Vj = 60m/s with a summary results for any active small bounds with list about variables of moments, tangentials and radials forces for any small part lower power amplifier and for upper amplifier and for horizontal's air streamlines. We need observe that is amount all radial's force about 5000 N, because this need for dimensions of radial's axial's bearings. The inertia moment of the resistance to rotation is about 110,6 kgm².

FIG. 18 show a dimensions a metal's frame and the duck for RMAX = lm. The deepest brunt of the horizontal's air streamlines is 37,5cm inside the duck mount on the frame with enough quality exhaust if ω = 6rad/s.

FIG. 19 show that for RMAX = 0,5m and L = 5m we have that is :

For wind speed Vj =10m/s, if generator of the pressure windmill is very little loading, we need 5,3 sec. for getting angle's speed ω > 6r/s and 6² * 0,62 - 9,81 * sin 47,3871° = 15,lm/s². If generator is normally loading for wind speed Vj = 15m/s, we need 7,1 sec. for getting angular velocity ω > 6r/s and after 60 sec the angular velocity will be only ω = 6,228 r/s.

Normally loading generator for wind speed Vj = 20m/s need 1,9s. for getting angular velocity ω > 6 r/s and for angular velocity Vj = 25m s after 60 sec the angle's speed will be ω = 10,3597 r/s.

FIG. 20 show that for RMAX =lm and L= 5m generator normally loading for wind speed Vj=30m/s need 0,8 sec. for getting ω >7r/s and 7² * 0,62 - 9,81 *sin47,3871° =

23,16m/s², and after 60 sec the angular velocity will be ω = 12,414 r/s. If generator is without loading the angular velocity increase on ω ~ 50 r/s.

FIG. 21 show that for

and L= 5m and Vj = 50m/s depth entry of 38cm, and we have 38 vectors, where 1. small circle is time moment when molecule abandon wind aggregate, and 2. small circle is the end of resultant vector of radial's vector centrifugal acceleration decrease for radial's component gravitation acceleration and tangential vector of rotation duck, when angular velocity is ω = 20 r/s.

Vj = 50m/s with R MAX = 1 ni with anti moment of generator with 23970Nm has 2 time bigger maximal power 496 kW, and 2 time smaller angular velocity ω= 20,69 r/s

~197,5turn/min of RMAX = 0,5m.

Vj = 60m/s with

and 4 times bigger a anti moment with 34500Nm of RMAX = 0,5m produce maximal power about 857 kW and angular velocity ω= 24,84 r/s ~

237turn/min, and we need multipliers. On the bottom of page is tabulated list for Vj = 60m/s, where we have summary results for any active small bounds with list of variables for moments, tangentials and radials forces for any small part lower power amplifier and for upper power amplifier and for horizontal's air streamlines. We need observe that is amount all radial's force about 10000 N. The Inertia moment of the resistance to rotation is about 841,6 kgm².

FIG. 22 show control unit generator with permanents magnets, which one control correct working as correct exhaust, and correctly connect generator with electric net and guide care that generator with electro magnets have continually a curent for excitation electro magnets, and stabilization realized on electronic's way with binary tree resistive loads for warming a water with generator.

FIG. 23 show that if for RMAX =lm and Inertia moment of the resistance of 843kgm², we must decrease pressure PPRIOR on entry in small blades wind aggregate abandoning underlying the depth lower amplifier of point T₇(©1228,Y,-©62) to point T₈(©2010,Y,-©100), that press wind aggregate begin to work, then we have :

0,58p_PR,oR =(2,07-0,38)p„ OR p_PR,oR = 2,91 p„. Throwing out 18,357% air streamlines, or 38 cm height of totally height of 207 cm, and when we compare the results of FIG. 21 and of FIG.23 we have the loss power of 18,023%.

FIG. 24 show that for RMAX =l,5m, L = 5m and for Vj = 20m/s we must have 4 times smaller load in compare with a load for getting maximal power, we need 1,6 sec for getting angle's speed ω > 6r/s. With Vj = 30m/s for getting maximal power about 160 kW, we need 1,8 sek for getting angle's speed ω > 7r/s. With Vj = 40m/s we can get maximal power about 380 kW.

FIG. 25 show that for RMAX =l,5m, L = 5m and for diametar shaft d = 19cm, a Inertia moment of the resistance to rotation I = 2796 kgm², and for d = 18cm, I = 2793,8 kgm². Because of so huge a Inertia moment of the resistance, we will decrease length lower emplifier power, because we must decrease the pressure of air streamlines on entry in small bounds of the wind aggregate and increase under pressure and exhaust of the wind aggregate. How much we need decreased the amplifiers depend upon results of experiments.

If we do not need to decrease length of lower amplifiers, then wind speed Vj = 50m/s produce maximal power of 740 kW, if anti moment of generator's is 53820 Nm, and wind speed Vj=60m/s produce maximal power of 1,279 MW, if anti moment of generator is 77200 Nm.

When on the top in the beginning fall 4 cm of height and on the bottom omit 70 cm height, then wind speed Vj = 50m/s produce maximal power of 496 kW, if anti moment of generator's is 35950 Nm, and the wind speed Vj = 60m/s produce maximal power of 857 kW, if anti moment of generator is 51764 Nm. For the wind speed Vj = 60m/s, a amount all radial' s force is 15000 N.

FIG. 26 show on the top a dimensions a frame for RMAX = 1,6m and for diametar of shaft d=19cm. For length duck L=4m, inertia moment of the resistance to rotate 1=2710 kgm². The wind speed Vj = 30m/s produce maximal power of 136kW, if anti moment of generator is 17600 Nm.

Vj =40m/s produce maximal power of 323 kW, if anti moment of generator is 31300 Nm.

For length of the duck L = 5m, inertia moment of the resistance to rotate is I = 3377 kgm². The wind speed Vj = 30m/s produce maximal power of 170kW, if anti moment of generator is 22000Nm.

Vj=40m/s produce maximal power of 404kW, if anti moment of generator is

39020Nm.

FIG. 27 show for RMAX =l,6m, L = 5m and for diametar of shaft d = 18cm, the Inertia moment of the resistance to rotate is I = 3381,8 kgm², and for L = 4m, I = 2714,8 kgm².

For length of the duck L = 4m, wind speed Vj = 50m/s produce maximal power of 420 kW, if anti moment of generator is 32784 Nm. The wind speed Vj = 60m/s produce the maximal power of 726kW, if anti moment of generator is 47200 Nm.

For length of the duck L = 5m, wind speed Vj = 50m/s produce a maximal power of 525kW, if anti moment of generator is 40980 Nm. The wind speed Vj = 60m/s produce a maximal power of 907 kW, if anti moment of generator is 59000 Nm. For Vj = 60m/s, the amount all radial's force is 9196N.

FIG. 28 show for RMAX =l,6m, L = 5m and for diametar of the shaft d = 19cm, the inertia moment of the resistance to rotate is I = 3384,168 kgm², and for d = 18cm, the inertia moment is I = 3381,8 kgm². When on the top in the beginning fall 4 cm of height and on the bottom omit 70 cm height, then wind speed Vj = 50m/s produce maximal power of 529 kW, if anti moment of generator is

40900 Nm. The wind speed Vj = 60m/s produce maximal power of 915kW, if anti moment of generator is 58900 Nm. If in the beginning fall 4 cm of height of lower amplifier, the amount all mass air's molecules is unchanged.

When we do not decrease length of lower amplifiers, then wind speed Vj = 50m/s produce maximal power of 789kW, if anti moment of generator is 61000Nm.

Wind speed Vj = 60m/s produce maximal power of 1,364 MW, if anti moment of generator is 77200 Nm.

FIG. 29 show Synchronous generator mount on the hollow shaft. The smallest generator with permanents magnets give the current to excitation generator. The excitation generator will give a curent for excitation electro magnets of main the bigest generator. The bigest generator produce AC alternating current, which one change in DC direct current, and on the end the direct current change in alternating current with frequency of 50Hz. At small power the electronics will decrease alternating current on this way, that the output voltage with frequency of 50Hz of the bigest generator is biger of the phase's voltage of electric net.

Control unit is equal as on FIG. 22 .

D E T A I L E D D E S C R I P T I O N O F S E V E R A L W A Y S O F A C H I E V I N G I N V E N T I O N

The pressure wind aggregate will produce with the lower and upper amplifiers power. The pressure wind aggregat has a small radius rotate RMAX, but with the big extended length 1 duck. Surface S of air streamlines is proportional with ( RMAX + AIILOWER + Δ-IUPPER ) and with length 1 duck. Length Ah LOWER and AIIUPPER are dependent of greatness RMAX. Arm of the force depend of greatness RMAX. The small radius rotate RMAX produce smaller moment of the force but Inertia moment of the resistance to rotate is proportional with R MAX³ and one is very little. Power is proportional with Vj³, and with wind speed Vj > 50m/s, we can produce big power.

Control unit with electronics is show on FIG.22. Control unit measure current and frequency on the current's transformer on output of generator and frequency on electric network if generator gives the current to electric net. If the current on the electric net turn off FM frequency's instrument will show zero, and switch 6 will open contacs toward electric net on all 3 phase. If the current exist, switch 6 has closed contacs on all 3 phase. Control unit on wind turbine has generator with permanents magnets and excitation generator and generator with electro magnets if need first will include switchs of SKI to SK5 for all 3 phase resistances which belong a wind turbine.

Every phase has equal combinations resistances, where numbering with 1 and 2 are completely equal.

Numbering resistances with 3 have 2 time smaller power of resistance with number 2. Numbering resistances with 4 have 2 time smaller power of resistance with number 3. Numbering resistances with 5 have 2 time smaller power of resistance with number 4. The regulation the power and frequency execute with a method the binary tree. If micro processor of control unit fortify that at night very little loading, and all possible power of WIND AGGREGATE do not use, adequate combination warming resistances will be parallel to switch with electric net on all 3 phase.

If wind speed is so little, that the loading with for generatora with warming resistances produce too small angle's wind and centrifugal acceleration, and because of can'ot blow out air's molecules, which give a path kinetic energy for the rotation shaft, then switchs of SKI to SK5 make combination with very little loading or generator do not have loading. If anti moment of generator do not exist, angle's speed will enough increase, that centrifugal acceleration blow out molecules of the air, and the cork with falling behind air's molecules don't exist.

This is a method calculation resistances.

For example, the biggest posible wind speed of Vj = 60m/s=216km/h appear on any location.

Wind aggregate with RMAX = 0,32m and with dy = 4m will rotate the shaft with angle's speed (0= 77,5284478 r/s= 740,34 turn/min. This is enough big angle's speed and we don't need the multipliers. If we can use 16poles generator, we have that is 3000 : 8 = 375 or for getting frequency of 50 Hz we need 375 turn/min. Toward this the wind speed Vj = 60m/s will produce frequency of f₆o = 98,7 Hz, because we have that is 50 *740,34 : 375 = 98,7 Hz.

If we decide that phase's voltage is UR=US =UT~240 V, if wind speed Vj = 10m/s=36 km/h, and the shaft will rotate with angle's speed ω= 14,382 r/s= 137,34 turn/min, and produced a frequency of fio = 18,3Hz, because is 50* 137,34 : 375 = 18,3. If with 18,3 Hz has 240 V, then will has with a frequency 98,7 Hz, a voltage of 240* 98,7 : 18,3 = 1294 V.

A power for Vj = 60m/s is 6³ = 216 times biger then power for Vj =10m/s. Program calculated that :

IF Vj = 60m/s and RMAX = 0,32m and dy = 4m

THEN Peo_MAX = 219405 W.

Mathematical's fisic's result of POWER for Vj = 10m s is 219405 W : 216 = 1015,7 W, but PASCAL program calculated 1006,8W, and the difference is less than 1%. Toward this CONTROL UNIT on FIG. 22 on one phase must have 6 resistances in series, beause one is 1294 : 6 = 216V.

Numbering resistances with 1 and 2 are completely equal.

Numbering resistances with 3 have 2 time smaller of resistance with number 2.

Numbering resistances with 4 have 2 time smaller of resistance with number 3.

Numbering resistances with 5 have 2 time smaller of resistance with number 4.

This is finely regulation the resistances with a method the binary tree.

Whole power of 219405W distribute on 3 phase, and one phase wll have 73135 W. The power 73135 W distribute on numbering resistances with 1 and 2, and we have, when appear wind speed Vj = 60m/s through any the curent has 73135 W : 2 : 1294 V = 28,3 A.

Consequently for RMAX ^:

For Vj = 10m/s ω = 14,382 r/s n= fio = 18,3Hz

PIO_MAX =1007W

For Vj =20m/s G>= 25,797 r/s n^: f₂₀=32,8Hz U₂o=431V P₂O_MAX =8126W

For Vj

For Vj =40m/s ω= 51,553 r/s n= f₄₀ = 65,64 Hz U₄₀=861V P₄O_MAX=

65008W

For Vj =50m/s ω= 64,78 r/s n=

For Vj =60m s ω = 77,528 r/s n=

Consequently for RMAX = lm and for dy = 5m we have : For Vj = 60m/s ω = 24,84 r/s n = 237,32 okr/min and for 4 poles generator need = 3000 turn/min need m triplication of 12,6473 if we wish to have f₆o = 100Hz.

For Vj =20m/s (0=8,287 r/s n =49,722 turn/min f₂₀=33,36Hz U₂₀=333,6V P₂O_MAX =31740W

For Vj =30m s G)=12,414 r/s n= 118,5 turn/min f₃₀=49,975Hz U₃₀=500V P₃O_MAX =

107123W

For Vj =40m/s 0=16,55r/s n=158turn/min f₄„=66,626Hz U₄₀=666V P₄O_ AX =253921 W For vj =50m s ω= 20,86 r/s n= 199 turn/min f₅₀=84Hz U₅₀=840V PSO_MAX =495940 W

For vj =60m/s ω= 24,84 r/s n= 246,34 turn/min f₆₀ = 100Hz U₆₀ = 1000V P₆O_MAX =

856983W

1000V : 200V=5 One series will have 8 resistance. Any phase will load with 285661 W. Now numbering resistances with 1, 2, 3 and 4 are completely equal, and numbering resistances with 5, 6, 7 and 8 will use for finely regulation the resistances with a method the binary tree.

Consequently 4 series of 5 resistances will contain power of 285661 W or one series will be loading with the power of 71415W, and the phase's current will have 71,4 A only if wind speed will be Vj = 60m/s = 216km/h. But I think that for R MAX = lm and for dy = 5m and for Vj = 60m/s, that because of the inertia moment of the resistance to the rotation is 841,6kgm² and because of the lower amplifier will be shortened with produce a reduced power.

The production GENERATOR and electronic's energy's transformer is expensive for maximal voltage 1294 V, than for maximal voltage lkV. However for RMAX =lm and for dy = 5m, PASCAL programs show for wind speeds Vj > 20m/s 4 poles generator can produce the current for electric network with change load because than danger stop of creation air's cork.

If speed of rotation is bigger GENERATOR faster cool own coils, and one is easier and cheaper and for electronic's energy's transformer is all the same do 100Hz or 50Hz transform in direct current. Perhaps the shaft of 4 poles generator at maximal wind speed thanks to 2 multipliers can rotate about 3000 turn/min for produce current of 100 Hz, electronic's energy's transformer without problems transform in current with a frequency of 50 Hz.

The stabilization angular velocity for any wind speed attain only then if produced moment of forcess on the wind aggregate is exactly equal with produced electrical anti moment of generator.

Every pressure wind power plant must have generator with permanents magnets and control unit, how in everyone time moment can produce the stream for the excitation electro magnets, because the excitation do not allowed about electical net with 50 Hz.

Generator with permanents magnets for equal power is much expensive of the generator with electro magnets. Because of the smallest generator with permanents magnets give the stream to the excitation for the excitation's generator. The excitation's generator give the stream to the excitation for the generator with electro magnets. The generator with electro magnets has internal resistance only about 10 ηΐΩ, and one have not problem with a cooling, and this is main reason for small expensive. If the thunder disconnect the electrical network, because of a safety control unit of the pressure windmill will connect a resistances for warming water in basin or of the central warming on the generator with electro magnets.

When angle's speed become too big, we must decrease angle's speed with connection parallel combination a resistances, because the power real loads and anti moment of generator increase, what decrease angle's speed. When we connect the real loads, of the fundamental importance is that the loads will connect directly on generator, without intervention protection fuse, because we wish protection function of the loads. Because of is very important that a chosen resistances has very big reliability. If the chosen resistances definite power made for 220V, for action maximal wind speed, because smaller wind speed will produce smaler voltage of 220V.

Notwithstanding a prior action, the frequency of generator increase over upper limit may be 100Hz, we need have the option with closed door because of the prevent the arrival air streamlines.

When the door close, we need to think about a way running air streamlines. Because of I anticipate that a closed door with horizontal's line always make the angle enough smaller of 45°, and air streamlines will turn aside over the roof of the wind aggregate. If the closed door with horizontal's line make the angle bigger of 45°, the wind will destroy the door. The wind with bigger speed don't allow opening the door, and we can instal barrier to air Streamlines before the door. After that he door will can without problem again lift up.

When pressure windmill work only as warming station, control unit will connect a resistances for warming water in basin with water toward fisic's law of electrotehnic.

AH manufacturers know that 4 poles generator better and more quality of 2 poles generator, and 3 phase, as and three-phase generators because of rotation magnetic field give 50 % more energie then one-phase generators. 4 poles generator are easier and cheaper of three-phase generators with bigger number of 4 poles.

The possibility exist, that with a roof of the pressure windmill flow a water in a cistern for the water on a top mountain. Toward this The pressure windmill is simple and chieper, because one is not high. Toward this only the pressure windmill can supply the users with electrical energy and with the water. Thermal electrical station can save a life to many rock-climbings or preserve their health of the consequence ice-cold temperature. In thermal station with the pressure windmill many people will can cook, sleep, and the best warm if wind strongly blow, because strongly wind produce huge electrical energy, and many rock-climbings can wait that the wind reduce and that bad weather pass.

The people can use the water of a cistern for washing and for cooking and for drink, but if on the mountain burst a fire, this a water of a cistern can on beginning extinguish a fire.

CALCULATE DIMENSIONS OF THE WIND AGGREGATE

Let us presume that the shaft is made of steel with next tehnical characteristics :

Solidness of material Rm TOL = 500 Mpa Specific pressure P_ TOL = 100 Mpa

Tension (TTENS_ TOL = 120 Mpa Pressure (TTL_ TOL ⁼ 120 Mpa

Bending < BEND_ TOL = 120 Mpa Sheer stress T STRESS_TOL = 96 Mpa

Twisting TTWIST TOL = 90 Mpa Specific density p = 7900 kg/m³

CALCULATION the diameter of the full shaft for the windmill

If the shaft is made of steel, and module of the steel sliding is Gki = 80* 10⁹Pa.

With the old_timer calculation I concluded that the greatest difficulty satisfy the condition that the allowed swivel angle 1) must be within the limits from 0,25° to 1 m. The specific deformation of the swivel angle is calculated following the formula :

= ( Mkriio * 180 ) : (Gki Ipo π )

The polar moment of the full shaft is I_po = d⁴ 71 : 32

For different value RMAX and for different value wind moment which will stabilize a rotational one wing at stable uragan the wind speed Vj = 60m/s = 216km/h, we will calculate the polar moment and after that is the diameter of the full shaft if the swivel angle in this time moment is 0.25°.

The rotational of two wing transmit on the middle of the hollow shaft, which direct rotate generator or indirect with the multiplier.

Ip=(Mi_wi„g* 180):(GO7T)

d⁴ π : 32 = (Miwing * 180) : (G υ π)

Then is d⁴ =(Mi_win_g* 5760):(GO π²) = Miwing* 2,91805*10^"8

For the hollow shaft formula is : d_v ⁴ - du⁴ = 2 Miwing * 2,91805* 10 ^{^ 8}

dv is the outside diameter and du is the internal diameter of the hollow shaft.

Letis : du=d+lmm k=d_v:du k⁴*du⁴=du⁴ + 2Miwin_g*2,91805*10-⁸ k⁴= l + 2Mkriia* 2,91805* 10 ^~8 : du⁴ at the latter end : d_v =k* du

d = 0,08016m ~ 8cm r = 4cm du = 0,081m k= 1,307 d_v ~ 10,6cm

For RMAX=0,4m and for dY= 4m and for Mi wiNG=2210Nm, IOTP_ROT= 46,596 kgm² d⁴ = 6,44889* 10-⁵m

d = 0,0896m ~ 9cm r = 4,5cm du = 0,091m k = 1,3028 d_v~12cm

For R MA = 0,5m and for dY= 5m and for Mi WING =4300 Nm, IOTP_ROT= 110,63 kgm² d⁴ = 1,25476* 10 ^" m

d = 0,1058m ~ 11cm r = 5,5 cm du = 0,111m k= 1,266 d_v~ 14,5cm

For RMAX=lm and for dY= 5m and for Mi WING =17250 m, IoTP_ROT=841,8466kgm² d⁴ = 5,0336364* 10-⁴m

d = 0,14978m ~ 15cm r= 7,5cm du = 0,151m k= 1,309 d_v~20cm

For RMAx=1.6m and for dY= 5m and for Mi WING =43925 Nm, IOTP_ROT= 3341 kgm² d = 0,1892m « 19cm r = 9,5cm du = 0,191m k = 1,3079 d_v~25cm

For

Moment of one wing is 17250 Nm.

Twisting TTWISTTOLERANCE = 90 Mpa

TTWIST = Mi WING : WPO = 17250 Nm : 6,6268 * 10^ m³ = 26,03 Mpa « TTWIST_TOL

Mass of the shaft ( vratilo ) is mvR = ΰ²*π:4 * 6 * 7900 = 0,15² * π * 11850 = 837,6 kg

Weight of the shaft is GVR = 837,6kg* 9,81- 8217N

Now we must calculate the dimensions frame, because after that we will can calculate centrifugal force.

The pressure windmill has 18 rotorblades.

IF 1st blade is on position—10° = 350° and 2nd blade is on position 10°

THEN 3rd blade is on position 30°

Radius of the shaft RsH=40cm, and the topmost horizontal edge of the rotorblade rotate on radius RMAX= lm, and duck go in the depth d = 40 cm.

IF R_SH=40cm and RMAX = lm and d=40cm

THEN BEGIN

100sinl0°-7,5sin20° 50-7,5sin20°

a = Arc tg = 9,194345° β = Arc tg = 30,80565505°

100cosl0°-7,5cos20° 100cos30°-7,5cos20° T,o<→T₃o={[(100cosl0⁰- 0cosa)-(100cos30°- 40cosP)]²+ [(50- 40sinP)-(100sinlO°-

40sina)]²}^½

Tio ^ Taso = 2* (100sinl0°- 40sina) = 21,947 cm

The height frame = [(100cosl0°- 7,5 )²+( 100 sinl0°)²]^½ = 92,623 cm

END ;

mDUCK_i8_BLADE = 0,01* 5* (0,4 + 0,1973+ 0,4 + 0, 21947)* 9* 1500 = 821,32 kg

For RMAX= lm height is 0,4 m.

Consequently a duck width 5m coarse 1cm, and long for 18 rotorblades about 10,06m with density 1500kg/m³ have a mass about 825 kg , and weight about 8093N.

The wind speed Vj = 60m/s = 216km/h produce anle's speed ω = 24,84 r/s, and now we can calculate a centrifugal force on the duck.

m_DUCK_1BLADE~ 45,63 kg

1

The force of Ϊ rotorblade is Fc = mo² Jdr = 45,63* 24,84² ( 1- 0,6) = 11262N

0,6

Centrifugal force for 18 rotorblades is FCUK_IS = 202716 N

Resultant forces weight the duck and the shaft together with resultant forces of radial's forces if wind speed Vj = 60m/s is : GUK = 8217 + 8093+ 10018 = 26328 N is about 7.7 times less of a total centrifugal force of the duck. This is main reason why I insist that we have even number rotorblades.

Centrifugal force with opposite centrifugal force produce tension efforts without arm for a centrifugal's moment, and efforts on tension is main variable for dimension radial axial bearing. The efforts on tension is not dangerous for the shaft, because the shaft of the steel is long 6m.

For Vj = 60m s and for RMAX= 0,32 m and for dy = 4m, angle's speed ω = 77,528 rad/s, and with regard on linear's proportionality dimension in drift X and Z axis.

ninucKj = 45,63* 0,8* 0,32 = 11,68128 kg

IIIDUCKJS = 210,263 kg GDUCK is = 2062,68 N

0,32

Force of 1 rotorblade is Fc = ma>² Jdr = 11,681* 77,528²(0,32- 0,19) = 9127N

0,19

Centrifugal force for 18 rotorblades is FDUCK IS = 164290 N

Mass of the shaft is !HSHAFT = d² * 7T :4 * 5 * 7900 = 0,08² * π * 9875 = 198,5 kg

GSHAFT = 198,5 * 9,81- 1948 N

Weight of the duck and for the shaft is GUK = 2063 + 1948 = 4011 N is about 41 times less of a total centrifugal force.

MTOT FRAME = (0,03+0,02)* 0,008m * 0,25m * 2700kg/m³ = 0,27 kg

Weight 5 vertically frame is about 14 N.

Centrifugal force of duck and of 2 vertically frame for vj = 55 m/s is :

Fc = 0.21875 *( 16,875+0,27 ) * ( 29,5584rad/s)² * l² ~ 3276N.

m_HOR_ FRAME = ( 0,03 +0,02)* 0,0075 m² * 5m *2700kg/m³ = 5,0625 kg* 5,0625kg ~ 3273 N. ∑ Fc = 8661,212 N + 3276 N + 4423,1 N + 3273,1 N ~ 19633,4 N

If Vj = 60 m/s resultant force all Weight together with resultant forces radials force is : GUK = 4011 + 114 + 2568 = 6693 N is about 3 times less of a total centrifugal force.

2. rotorblade is load with the bigest moment of about 720 Nm.

Consequently because of correctness result for surface any carrier, we must calculate quotient 720Nm : 5 =144Nm and we have : r³ = ( 144 Nm : 50000000Pa )* 2 : π = 0,00000183464944

Radius is : r = 0,012239m ~ 1,25cm,

And surface is Γ²π ~ 4,91cm².

Mass any carrier for RMAX= 0,32m is m = 0,3m* 0,000491m² * 7900kg /m³ = 1,16367kg. 5 carrier together have about 5,82 kg, or weight is about 57 N.

METHOD OF APPLYING THE INVENTION

We know very well, that we can with using force compress the air and almost all the gas on much smaller volume, because air is compressible. Bernull's equations are not correct for the air, because the air's fluids are compressible. The water is not compressible, and

Bernull's equations for the water are correct.

Of universal a gas's law pV= nRT prove Boyle Mariott's law pV= const and prove for our case equation : 0,54 PPRIOR = 2,07 po or PPRIOR = 3,83 p_Q po is input pressure.

First path of lower amplifier for RiviAX = lm has coordinates :

T₃(©90,Y,©46) T₄(©158,Y, ©30)

For R MAX = 0,32m all ©X and ©Z coordinate multiplicate with 0.32. For

all ©X and ©Z coordinate multiplicate with 0,4. Consequently if shorten first path of lower amplifier, then its beginning in the point S₃(©107,Y,©42) instead the point T₃, we prove the equation :

0,58 PPRIOR = 2,07 p₀ or PPRIOR = 3,569 p₀. PPRIOR is the pressure before the duck. I = the inertia moment of the resistance to rotation.

If R MAX = 0,32m I has only 24,4 kgm², but R MAX = 0,4m I has 46,4 kgm²

The under pressure on output has ρ₀υτ~ 0,6 p₀ can drag molecules, which one lose 29% kinetic energy for the rotation rotorblades on wind aggregate, and because of the pressure after the duck is PAFTER ~ 0,71p_PRi_OR ~ 2,855 p₀

If the under pressure is not enough, then wind aggregate will look toward idea of dr. Virag→ sub-dean mechanical's university in Zagreb as on FIG. 3, but on the roof up rotor blades will mount the pressure windmill, which one increase resistance a flight to air streamlines over the roof out and compel air streamlines that one enter in the pressure wind aggregate. FIG. 3 produce under pressure on output ρουτ~ 0,307 p₀.

For RMAx=lm the Inertia moment of the resistance to rotation of 843kgm² is enough big, and if the pressure wind aggregate don't begin to work, then we must decrease pressure PPRIOR on the entry in duck of rotor blades wind aggregate with decreasing a depth lower amplifier power. If throw out the lowest part of the lower amplifier :

Of the point T?(©1228,Y,-©62) to point T₈(©2010,Y,-©100),

Then we have : 0,58p_PRioR =(2,07-0,38)p₀ or PPRIOR = 2,91 p₀.

Throwing out 38 cm height air streamlines of total height of 207 cm produce the loss of 18,375% air streamlines, and if compare a results of FIG. 21 with FIG. 23 we have the loss power of (856983,7W-702528,924W) : 856983/7W = 18,023%

For FIG. 23 we have : d⁴ = Mkriio* 2,91805* 10 ^{" 8} = 14250* 2,91805* 10 - ⁸

Consequently the diametar of the full shaft decrease for 6mm. d = 0,1428m ~ 144mm

If throwing out 38 cm is not enough, then we can decrease PPRIOR, if throwing out the path of lower amplifier of the point T₆(©732,Y,-©30) to T₇(©1228,Y,-©62), and then we have :

0,58 PPRIOR = ( 2,07 - 0,7 ) p₀ or ρ_Ρβ,_οκ = 2,362 p₀. _:

R MAX is the bigest distance between the axis rotation shaft and the uppermost horizontal line.

I believe that then under pressure for FIG. 3 of ρ₀υτ~ 0,307 p₀ will be eonough that begin to work wind aggregate with R.MAx=lm, although one have the Inertia moment of the resistance to rotation of 843kgm². We must to know that decrease depth lower amplifier is also decrease the power of wind aggregate.

If coefficient reduction is a proportion between RMAX : RMAX_BIGGER =n

Force's moment on a rotorblades decrease with coefficient reduction on 2.nd potention, and we can use energy wind speed bigger of 200 km/h. Although amplifiers increase PPRIOR, wind aggregate can perfectly work. The Inertia moment of the resistance to rotation decrease with coefficient reduction on 3.rd potention on the duck with increase action of the under pressure. Because of the pressure windmill mast have small R MAX with bigger width of the duck dy because of getting bigger power.

The air streamlines like to avoid the working and also the rotation rotor blades, and if we mount the propeller wind power plant on the roof, rather number of air streamlines will abstain of flight over the roof out, and one will compel to rotate rotorblades of the pressure windmill. However the roof at the same time affect as lower emplifier power for PROPELLER wind power plant. All the same the upper amplifier above the PROPELLER wind power plant will prevent running air streamlines above PROPELLER rotorblades.

Extremely little the Inertia moment of the resistance to rotation demand quickly a reaction of control unit program wizh using the electronic and micro processor.

List of symbols used in the text

0 Χ,Υ,Ζ are coordinates a point in the midlle drawing little circle.

R→ are signs after that there are equation surface.

The pressure windmill has 3 room, and in the middle room is generator, and in the left room is windmill with 18 duck's rotorblades, and in the right room is windmill with 18 duck's rotorblades.

1 is sign for 1. rotorblade. 2 is sign for 2. rotorblade. 3 is sign for 3. rotorblade.

4 is sign for 4. rotorblade. 5 is sign for 5. rotorblade. 6 is sign for 6. rotorblade.

7 is sign for 7. rotorblade. 8 is sign for 8. rotorblade. 9 is sign for 9. rotorblade. 10 is sign for 10. rotorblade. 11 is sign for 11. rotorblade. 12 is sign for 12. rotorblade. 13 is sign for 13. rotorblade. 14 is sign for 14. rotorblade. 15 is sign for 15. rotorblade. 16 is sign for 16. rotorblade. 17 is sign for 17. rotorblade. 18 is sign for 18. rotorblade. Vj and v are vectors speed air streamlines.

Δ is vector difference in drift movement air streamlines,

φ is angle rotorblade toward horizontal axis simmetry the rotation shaft,

δ is angle between different vectors speed Vj for climbing or for falling or for horizontal's air streamline and vector G)R vertical on rotoblade. exsicition

TV is the pressure windmill.

STV is the senzor pressure windmill.

SPV is the senzor propeller wind power plant.

a is angle between any plane amplifier and horizontal plane with equation Z = constant, β is angle between air streamlines and horizontal plane with equation Z = constant.

K is the roof.

1 is sign on FIGURES wind aggregate for the left room of pressure windmill.

KI is the roof over the left room of windmill. 11 is sign on FIGURES wind aggregate for the right room of pressure windmill. KII is the roof over the right room of windmill.

KG is the roof over the room of generator.

Ml is sign of force moment for climbing or for falling or for horizontal's air streamlines which work on the duck of 1. rotorblade on the position of 27,3871° to 47,3871°.

Fit is sign of tangential force for climbing or for falling or for horizontal's air streamlines which work on the duck of 1. rotorblade.

Fin is sign of radial force for climbing or for falling or for horizontal's air streamlines which work on the duck of 1. rotorblade.

M3 is sign of force moment for climbing or for falling or for horizontal's air streamlines which work on the duck of 3. rotorblade on the position of 67,3871° to 87,3871°.

F3t is sign of tangential force for climbing or for falling or for horizontal's air streamlines which work on the duck of 3. rotorblade.

F3n is sign of radial force for climbing or for falling or for horizontal's air streamlines which work on the duck of 3. rotorblade.

M4 is sign of force moment for climbing or for falling or for horizontal's air streamlines which work on the duck of 3. rotorblade on the position of 87,3871° do 107,3871° and etc.

L is length of the shaft.

RMAX is the bigest distance between the axis rotation shaft and a uppermost horizontal line on duck.

Tn— n is sign for ordinal number the point and author will give to manufacturer precise X Y and Z coordinate.

© = RMAX : 100cm is factor of multiplication with which need multiplicate X or Z coordinate, if this sign registered before X or Z coordinate.

Ti - the point Ti has coordinate ©80,Y, ©100 and the abbreviation is Ti(©80,Y, ©100) -Ti - is the abbreviation for -Ti(-©80,Y, ©100 )

T₂(©474,Y, ©107) T₃(©90,Y,©46) S₃(©107,Y,©42) T₄(©158,Y, ©30) Ts(©392,Y,0) T₆(©732,Y, - ©30) T₇(©1228,Y,- ©62) T₈(©2010,Y, - ©100) T₉(©243,Y,©240) Tio(0,Y,©380 ) Sio(0,Y,©380 ) So(0,Y,©900 ) Tn(©506.5,Y,©200)

Ti2(©100,Y,©150) Ti₃(©30,Y,©230) Ti₄(0,Y,©240) -Ti₅(-©108,Y,©276)

Ti6(©474,Y,©320) -Si6(-Si6(-©474,Y,©653) Tn(©105,Y,0) Tis(©105,0, - ©120Z) Ti9(©35,15,Z) T₂₀(©120,30,Z) T₂i(©35,45,Z) Z is coordinate of bottom,

-Ti9(-©35,15,Z) -T2o(-©120,30,Z) -T₂i(-©35,45,Z) but coordinate X and Y are T₂9(©35,200,Z) T₃i(©35,230,Z) T₃o(©120,215,Z) worth for whole height carrier. T₃2(©35,400,Z) T₃₄(©35,430,Z) T₃₃(©120,415,Z) and so on

T₃₉(©435,15,Z) T4o(©520,30,Z) T₄i(©435,45,Z)

T42(©435,400,Z) T₄4(©435,430,Z) T₄3(©520,415,Z) and so on

T₄9(©835,15,Z) T₅o(©920,30,Z) T₅i(©835,45,Z)

Ts2(©835,400,Z) T₅₄(©835,430,Z) Ts3(©920,415,Z) and so on

Y coordinates don't have before the sign ©, because Y coordinates have constant value for every

RMAX was a value of the pressure aggregate.

KU is sign for control unit.

IF RMAX is the bigest distance between the axis rotation shaft and the top rotor blades aggregate

THEN © = R AX : 740cm is factor of multiplication with which need multiplicate X or Y or

Z coordinate, if this sign registered before X or Y or Z coordinate.

Pi(-©560,-©750,0) P₂(©560,-©750,0) P₃(0,-©750,©420)

Poi(0,0,©420) Po2( 0,0, ©1170) for Rmax = 740cm P₀₃(0,0,-©1920)

P₄(-©560, ©750,0) Ps( ©560, ©750,0) Ρ₆(0,©750,©420)

Pz are points on the same botom plane wih Z coordinate under wind aggregate. EX is sign for the smallest generator with permanent magnets and with a excitation generator.

SG is a Synchronous Generator.

ST is a current's transformer.

Ii and I₂ are current of the current's transformer.

SKO, SKI, SK2, SK3, SK4, SK5 and SK6 are electrical switchs.

G is generator.

Gp is generator with permanent's magnets.

Gex is excitation generator.

Gel is generator with electro magnets.

SG is synchrony generator.

FG is frequency of generator.

FM is frequency of electrical net.

R, S and T are phase of three-phase electrical net.

Nu or 0 are sign for zero electric wire.

N is North pole of magnet.

Nw is North's wind.

S is South pole of magnet.

Sw is South's wind.

AC is alternating current.

DC is direct current.

Mp is motor drive.

dy is width of the roof.

Pr Vr are rectangular door.

Tr_Vr are trapezoidal door.

* is multiplier

P MAX is sign that diagrams on figures correspond using maximal power.

I is sign for the inertia moment of the resistance to rotation on diagrams of FIGURES.

Claims

Patent claims

1. COMBINED impeller-presser type wind POWER PLANT characterized by the fact, where a word exsicition implicitly wind aggregate with extremely little the inertia moment of the resistance to rotation, because the resistance decrease with (RMAX) on 3.rd potention, where (RMAX) is a mark for the bigest distance between the axis rotation shaft and the uppermost horizontal line, and of course propeller rotor blade or blade of pressure windmill with very little wind speed produce energy, and a word type for a pressure windmill denote that the bigest height at l.st type is on the middle roof increase with RMAX on 1.St potention, or at 2.nd type has a rotational roof with the lowest height of roof on its brink line and with the biggest height of roof on its brink line with opposite side and so we have 2 times bigger total height, how we can have about 2 times lower output underpressure.

2. POWER PLANT in accordance with claim 1 characterized by the fact, what the house of the press windmill has 3 a room, where l.st and 3.rd the room have only the lateral walls and wind aggregate with 18 rotorblades, but the middle 2.nd room is with walls completely closed and together with the roof constitute Faraday's cage because of the protection for generator of the thunder.

3. POWER PLANT in accordance with claim 1 characterized by the fact, where for

1.st type air streamlines enter under clossed a roofs door and come out on opposite side with a openly highly lifted door and for 2.nd type air streamlines enter under the lowest height of a roof on its brink line and air streamlines come out on opposite side with the biggest height of the roof and so output side on the roof of 2.nd type is 2 times bigger of 1.st type, what increase UNDER PRESSURE.

4. POWER PLANT in accordance with claim 1 characterized by the fact, that at the pressure windmill total moment decrease with (RMAX) on 2.nd potention and the inertia's (I) resistance to rotation decrease with (RMAX) on 3.rd potention, and because of for (RMAX) < 0.4m the resistance is extremely little what give a possibility for the biggest emplifier and economy, but the lower emplifiers and upper emplifier increase the pressure before the duck on the rotorblades, and IF I resistance increase THEN we must need experiments for decreasing the lower emplifiers because of getting effective UNDER

PRESSURE.

5. POWER PLANT in accordance with claim 1 characterized by the fact, that at propeller wind aggregate* total wind's moment on rotorblades decrease with (RMAX) on 2.nd potention, and one have enough long the lower amplifier power, and well earthed the upper metal amplifier power use as a protection for generator of a blow the thunder, but the upper amplifier prevent retreat air streamlines under the uppermost on the top of the rotorblades.

6. POWER PLANT in accordance with claim 1 and 3 characterized by the fact, that the ROOF is lower power amplifier and together with propeller wind power plant increase a resistance to retreat air streamlines of the rotation duck's blades at the pressure windmill.

7. POWER PLANT in accordance with claim 1 characterized by the fact, that a generators propeller wind power plant with its motors rotate a holow shaft in the middle generator's room of the pressure windmill, and when this rotation produce the rotation of full shaft in 1.st and in 3.rd room and angular velocity with centrifugal acceleration enough bigger of gravitation acceleration on any point duck for the quality EXHAUST, a motors will open the door for entry and for exit in the pressure wind aggregate to air streamlines, but control unit must switch a warming resistance in basin with water at the start generator with electro magnets if wind speed is little.

8. POWER PLANT in accordance with claim 2 characterized by the fact, that the vibrations are much weakly, and because of the rotation of 2 full shafts must produce the rotation on the middle holow shaft, but if we use 2 multipliers for bigger (RMAX ) , we must mont first multiplier on left and second multiplier on right side of generator's shaft because of increasing angular velocity ((QGEN) without the vibrations.

9. POWER PLANT in accordance with claim 2 and 8 characterized by the fact, how the vibrations will be much weakly, 2 carriers frame for the duck of 2 rotorblades are connected together in the distance of 40°, but the tops of neighbouring carriers for 18 neighbouring rotorblades have always the distance of 20° create so a triangle statically immovable figures, what prevent to change interspace neighbouring carriers because of the pressure of the air streamlines.

10. POWER PLANT in accordance with claim 9 characterized by the fact, that tangential's type carrier mount on definite depth of 2 neighbouring radial's carriers inside one pair radial's carriers and between two pair radial's carriers, how both pair's

combinations have a triangle statically immovable figures, because we must mont the duck on both pair's combinations because the prevention theentance for air streamlines in depth under the duck toward the shaft.

11. POWER PLANT in accordance with claim 1 characterized by the fact, that the pressure wind aggregate ban the strikes of the air streamlines into the 3^rd and the 4^th quadrant of the XZ plane, and that the horizontal's air streamlines of small angles are transformed into climbing slanting air streamlines with increasing tangential component force because of producing the rotation with decreasing sterile radial's force for produce the rotation, with deacreasing Coriolis's force and pressing radial aksial bearing and shaft.

12. POWER PLANT in accordance with claim 1 characterized by the fact, that any the pressure wind aggregate has the smallest generator with permanents magnets because of giving a current to excitation of excitation generator, and one is enough big for giving current to excitation electro magnets of the bigest main generator.

13. POWER PLANT in accordance with claim 1 and 12 characterized by the fact, that any the press wind aggregate with (RMAX) < 0,4m and with extremely little the inertia moment of the resistance to rotation have 10 poles generator with electro magnets without using 2 multipliers and has maximal long lower emplifier power, and then during the winter on location with wind speed bigger of 150 km/h, the average price for lkWh electrical energy will be the smallest between all today known source energy.

14. POWER PLANT in accordance with claim 1 and 12 characterized by the fact, that own control unit lead care maximal using energy a pressure windmill, but if the pressure windmill need only for warming a water, then energetics electronic dissimulator for getting curent of 50Hz is not necessary and control unit realized the stabilization on electronic's way with binary tree resistive loads for warming a water.