IT202100028205A1 - SYSTEM AND PROCEDURE FOR WORK GENERATION FROM SURFACE FREE ENERGY - Google Patents

Description

SYSTEM AND PROCEDURE FOR GENERATION OF WORK FROM ENERGY

FREE SURFACE

Field of invention

The present invention concerns the alternative energy sector and in particular has as its object a plant and process for generating work from surface free energy, according to a system that can be indicated with the acronym LES (Work from Superficial Energy).

State of art

To date the field of alternative energies, that is? systems for extracting mechanical, electrical and thermal energy from alternative sources, ? generally dominated by renewable natural sources such as solar radiation, wind, wave or tidal motion, geothermal phenomena, etc. These renewable sources currently used provide energy as long as present, but they can be discontinuous and decidedly variable even when present, what? obviously constituting a problem that on the one hand we are trying to resolve or at least minimize, on the other it pushes us to search for further sources and solutions that can ensure this continuity, as well as generally providing supplementary resources to global energy needs.

Summary of the Invention

The fundamental purpose of the present invention? that of providing a system (system and method) that effectively exploits a source other than traditional natural ones, and in particular the latent surface free energy, present at the interface between different elements, which manifests itself through the action of natural forces such as those of adhesion and cohesion.

The target ? that of using these forces to carry out work (raising a certain volume of liquid) and then exploiting, through a special system, the energy variation that this work can? generate. To work, the system must do that the energy obtained (through the transformation from superficial to potential) is supplied continuously? and therefore allows the forces of adhesion and cohesion to cyclically raise the liquid to carry out work and recreate the energetic conditions of a natural process (capillarity?) in itself? irreversible and made reversible (except for entropic losses) thanks to the external energy supplied, both in terms of work expended and heat absorbed; otherwise, without contextual variations, the work of the adhesion and cohesion forces would result in the one-off rise of the liquid. The invention therefore aims to ensure cyclical nature of the system. that creates the conditions for having energy that is continuous over time and substantially stable in value.

These aims and purposes, together with other accessories, are achieved by the system and method of generating work from free surface energy according to the present invention, the essential characteristics of which are the subject of the attached independent claims.

Brief description of the figures

The characteristics and advantages of the system and method according to the present invention will be more clearly from the following description of its embodiments, made by way of example and not by way of limitation, with reference to the attached drawings in which:

? figure 1? a top view of a unit? basis of the system according to the invention, in a phase 0 and in a phase 1 as will be detailed below;

? figure 2? a top view of a unit? functional of the system in figure 1; ? figure 3? a top view of a unit? main of the system according to the invention;

? figure 4? again a top view similar to that of figure 1, with a group of slides in the process of moving from a zone Z1 to a zone Z2 of the unit;

? figure 5? a longitudinal section view of the unit? based on the plane indicated by the arrows A-A of figure 1, in a first phase (or phase 1) of operation;

? figure 6? a view similar to that of figure 5, in a second phase (or phase 2) of operation after the passage of a volume of water ?V;

? figure 7? a representative diagram of the energy balance of the system according to the invention;

? figure 8 represents a thermodynamic cycle according to which the system according to the invention operates.

Detailed description of the invention

Will the methods be explained below? of creating a system suitable for obtaining work from the adhesion and cohesion forces due to the variation in surface energy when different solid, liquid and gaseous elements are brought into contact.

Does it show that? Is it possible to obtain energy from properties? of matter thanks to the forces that are exerted between molecules due to essentially electromagnetic interaction: these forces are called adhesion (between different molecules) and cohesion (between identical molecules) and the useful work obtained from the combination of these forces can? be, for example, transformed into electrical energy both directly and by accumulation of potential energy.

To obtain useful work from the aforementioned forces, we exploit in particular the effect of the surface tension that is created in the liquid, at the interface with gas and solid, as the resultant force of the interactions between the molecules of the different physical elements.

Surface tension creates a pressure difference, which can be determined with Laplace's law (higher pressure value in the liquid compared to the gas) and this difference gives rise to the phenomenon of capillarity, with greater differences in level. or less accentuated, if the liquid is in contact with solids of suitable materials, dimensions and distances. To exploit the capillary phenomenon to obtain energy, the fundamental idea behind the present invention is? that of varying the level of the liquid contained in two different volumes of a container through the presence/absence of the capillary phenomenon, that is? by cyclically raising and lowering the level through appropriate movements of mobile elements of a system.

These mobile elements move in a field of forces (adhesion and cohesion) due to surface energy; this field causes a difference in the level of the liquid between the elements, i.e. it determines the variation (increasing) of the contact surfaces of the materials and therefore the variation (decreasing) of their surface energy: this variation? the energy source that the system uses by transforming it into potential mechanical energy (difference in level of the liquid between the two volumes of the container).

? it is clear that, so that? the system produces useful work, the potential energy obtained, which, as will be seen? more? after you, ? demonstrably lower than the variation in surface energy, it must be higher than the work spent to move the elements of the system. Will we see? precisely that the work spent to move the mobile elements can? be lower than that obtained only if the system respects certain construction conditions, as the capillarity/viscosity parameters come into play of the elements used. This means that useful energy depends on the modes? construction of the system, modality? which must tend to optimize the quantity? of energy obtained by maximizing the activity? of the forces of adhesion and cohesion while the work expended must be the minimum necessary to guarantee that this quantity? of energy is produced continuously? from the forces. The quantity? of work produced by the forces of adhesion and cohesion? independent of the movement of the mobile elements, that is? from the mechanical work spent to move these elements, while it depends on the system: as we will see from the calculations, once the elements have been chosen, the liquid raised? constant in the height-distance surface of the capillary component and therefore independent of the work expended in displacement. In particular, we will see that on the ratio ?energy obtained / work expended? significantly affects the dimensional ratio of width/height of the mobile elements: this particular situation? also highlighted by the number of capillaries? which indicates how much the capillary forces prevail over the resistant viscosity ones.

For the aforementioned demonstrations and related calculations, the use of a system defined in terms of materials and related dimensions is assumed according to the hypotheses described below. In particular, the following elements will be taken into consideration at room temperature (around 20?-25?): as a solid element, glass; as a liquid element, water; as a gaseous element, air. The choice of these elements? motivated by the fact that they exploit adhesion and cohesion forces well, are common, easy to use and non-polluting. The choice of glass, in addition to its cost compared to other solid materials, also lies in the fact that it has a high surface tension value compared to water (about 16 times) and this allows the glass to be easily ?wettable? (good adhesion forces) as this possibility? ? all the more? true how much more? the surface free energies of the solid are high and the surface free energies of the liquid are low.

In order to obtain better results and/or reductions in the size of the system, can other highly wettable materials be used? that technology can? offer and at the same time another liquid or aqueous mixture can? be chosen to increase the surface tension (without at the same time significantly increasing the contact angle), while respecting the costs/benefits, sustainability? environmental and possible deterioration of the elements used.

With reference to the figures mentioned above, the operation of the system according to the invention is based on the use of the difference in water level that the adhesion and cohesion forces create in two separate volumes of a container thanks to the raising of the water by capillary action. among the mobile elements of the system; in practice the work done by the forces? similar to that of a lifting pump which moves the water from one part of the same volume to another, thus creating? a difference in height. This difference in height d? gives rise to a flow between the two volumes of the container which alternately increase or decrease the quantity? of water depending on whether or not there are glass elements inside them. In this way the flow of water can for example, passing through an impeller that transforms the energy received into potential energy to power an electric turbine.

Does the system include one or more? unit? main PUs (figure 3), each of which in turn includes one or more? unit? base UB (figure 1), an impeller G and pipes T which, as explained further? next, they connect the impeller G with the units? basic UB. The unit? basic UB? the unit? where does the activity take place? which is the basis of the invention and? logically subdivided into a certain number of units? functional UF (figure 2) as clarified more? after you.

A unit? basic UB? formed by a container C having two volumes of water separated by a dividing element or diaphragm or fixed support D which divides longitudinally in half? the container and by a series of movable glasses V which all move together orthogonally to the fixed support in sliding openings specifically defined by it. The container? then divided in width into two equal areas indicated with Z1 and Z2; in each area there?? a water surface S1 always free from moving glass while the remaining water surface is occupied (surface S3) or free (surface S2) by moving glass. The height of the water measured with respect to the base of the container can be indicated for brevity? also as water level.

During the cyclical operation of the system, the glasses, along their entire length, move orthogonally to the two zones, both on the outward and return journey, passing from one zone to the other through special openings in the fixed support. The fixed support? made in such a way as not to let the water pass from one area to another, therefore, in addition to resisting the low pressure of the water (which, taking into account the heights involved, is of the order of a thousandth of an atmosphere), it presents also openings that perfectly coincide with the shapes of the moving glass, with the exception of the upper part, which is open to allow the connection between the glasses of the series. The glass must slide adhering to the support, that is? watertight, to be ?clean? from the drops of water which will thus fall into the same area from where the glass is coming out, therefore the layer of water rises due to capillary action. on the glass (for brevity also referred to as capillary water) returns by falling into the relevant area thanks to the precise adhesion of the glass to the support: in practice there must be no exchange of water between the two areas of the container C. The aforementioned adhesion it must however be such that the glass support friction tends to zero; in this regard, can the support, for example, be made of plastic material such as Teflon? (PTFE), at least along the openings where the glass slides, as Teflon? it has very low values of static and dynamic friction coefficients (around 0.04). In this embodiment the series of glasses are considered to be moving and the container C is fixed but in practice it will be possible to the effectiveness of the conceptually inverse solution be verified, that is? which involves moving the container while keeping the series of glasses fixed.

The glass panes are preferably rectangular in shape (the simulations generally indicate that the square shape should be discarded in compliance with the aforementioned width/height dimensional ratio): the best results in terms of extractable energy are obtained for tall and narrow glasses with a thickness minimum possible; with the increase in the width of the glass the useful work produced is reduced until we end up with a loss-making system where the work spent becomes greater than that obtained. The glass must have perfectly smooth and clean surfaces in order not to reduce wettability; the height of the glass corresponds to the maximum height at which it can? arrive the water when the phenomenon of capillarity acts? and this height? approximately also the height of the unit? basic UB, of the unit? functional UF and of the unit? main UP.

The mobile series of glasses can be connected to each other using rigid physical supports, such as thin steel wires (or other material) such as to rigidly hold the glasses together and aligned, which can be moved, for example, with pulleys placed on the upper part, or the movement can? take place via a single motor arm that moves the lattice to which the glasses are attached in their upper part; differently, as already? mentioned, the glass can be kept fixed and the container moved. However the sliding is achieved, the creation of the system involves the use of technologies capable of eliminating friction or in any case reducing it to the minimum possible. The friction losses of some moving parts such as impellers, pulleys or glass-support sliding will be neglected in the simulation calculations of the system under consideration or in any case evaluated indirectly by maximizing other losses.

The unit? functional UF, with particular reference again to figure 2, ? logically composed of a movable glass V and a body of water S between two movable glasses: ? It is in this part of the system that the capillary phenomenon occurs. Its length corresponds to the distance between two adjacent sets of glass (length of the water space plus the thickness of the glass). The dimensions of the unit? functional UF are in the order of a millimeter (which is the order of magnitude below which the capillary phenomenon has the greatest effect) so part of the calculations will refer to this unit? of measurement.

To describe more? in detail the functioning of the system, consider at this point the following list of quantities/variables involved (with related symbols), their relationships based on known physical laws, definitions and assumptions.

Quantities and general physical relationships

m = meter; kg = kilogram-mass; sec = second

N = Newton = kg*m/sec<2>; J = Joule = N*m; W = Watts = J/sec

g = acceleration of gravity? = 9.81 m/sec<2>

? = density? water (water mass / volume) = 1,000 kg/m<3 >?*g = 9810 N/m<3 >?v = density? glass (glass mass / volume) = 2,500 kg/m<3 >?v*g = 24,500 N/m<3 >? = viscosity coefficient? of water = 10<-3 >N*sec/m<2 >(? can be seen as the momentum lost per unit of surface in the flow of water between the slides, ? = m*v/ S = ?*v*l with m=mass of water, v=flow speed, S and l respectively surface area and length of movement involved)

? = surface tension of water = 73*10<-3 >N/m (surface tension of glass approximately 1.2 N/m)

As mentioned, to improve performance and/or reduce the size of the system, ? to check the possibility? to use liquids or liquid mixtures with ? *what? more? high of the water (where is the contact angle) and at the same time with a ? not too high compared to the water; furthermore, if the surface tension of the chosen liquid approaches or exceeds that of glass, it will go? another material has been identified to replace glass to maintain good wettability.

Laplace's law: ?p = ? * (1/R1 1/R2); ?p ? the pressure difference at the water-air interface and R1 and R2 are the radii of curvature, orthogonal to each other, at the water-air interface between two slides

Dimensional parameters

- Slides (they have orthogonal displacement with respect to the direction of elongation, or length, of the functional unit UF and of the basic unit UB):

lv = length (in the direction of the width of the basic unit UB)

sp = thickness (in the direction of the length of the basic unit UB)

hv = height

Unit? functional UF:

- d sp = length (with d = distance between two adjacent slides)

- lv = width

- hv = height

nUF = number of units? functional units UF (subsequently, in the sizing of the system, this number is combined with a one meter long basic unit UB) Unit? basic UB:

- nUF * (d sp) luS1 = length (with luS1 = length of the part of the UB basic unit always free from glass slides; length of the S1 surface)

- 2*lv = width

- hv = height

nUB = number of units? basic unit UB (subsequently, in the sizing of the system, this number is combined with a main unit UP one meter wide) zones Z1 and Z2: they have length and height equal to the unit? basic UB but they are half as wide? of the unit? basic UB (width of each zone = lv)

Unit? main UP:

- length = unit length? basic UB is the space in length for pipes and impeller

- width = nUB * unit width? basic UB = nUB*(2*lv)

- hv = height

The units? main UP are units? flat which can also be superimposed on different horizontal planes; the vertical distance between the planes must be greater than hv, (in addition to the height of the slide, the thickness of the base of the container of the UB base unit must be taken into account, as well as a minimum detachment of the slide from this base for the passage of the water attracted by capillarity from the surface volume S1 (arrow F1 in figure 5) and of a minimum detachment of the slide from the upper base of another UB base unit also to accommodate the slide movement mechanism).

water surfaces of each area of the unit? basic UB:

- S1 = luS1 * lv (surface of the water area always free from glass)

- S2 = nuf * (d sp ) * lv (surface of the water area in which the slides move and temporarily free from slides)

- S3 = nuf * d * lv (surface of the water area in which the slides move and temporarily occupied by slides; S3 is the sum of the water surfaces between the slides)

And therefore:

- the total surface area of an area? given by S1+S2

- the total surface area of an area minus the surface area occupied by the slides? given by S1+S3

- the surface occupied by the slides? given by S2 ? S3 = nuf * sp * lv

To simplify the subsequent formulas, the following surfaces can then be defined: SA = S1 S2

SB = S1 S3

Are the following levels still defined in the areas of the unit? basic UB:

- h0 = reference level of the two zones with respect to the base of the UB

- hc = height of the water raised by capillarity? by the adhesion and cohesion forces between the slides

- hb = S1 level under the effect of the capillary phenomenon, that is? when slides are present in the area (hb < h0)

- hs = level of the zone in the absence of the capillary phenomenon, that is? when there are no slides in the area (hs > h0)

Finally, the following volumes of water in the unit are defined: basic UB:

- Vc = volume of capillary water between the slides

- V0 = volume of water in each zone as volume relative to the reference level h0 during the system operating cycle

- ?V = volume of water exchanged between zones Z1 and Z2 during the cycle phases for which:

V0 = volume of water plus? low content in an area during the cycle V0 ?V = volume of water more? high content in an area during the cycle To provide a simplified, but still representative, description of the functioning of the system and its efficiency, through the calculations that follow, it is assumed that:

I1. Losses due to dissipative effects such as friction are zero;

I2. for the calculation of the pressure difference ?p at the water-air interface, in Laplace's law only the radius pi? small, what will be indicated? as R1, which takes into account the curvature of the water orthogonal to the two slides; the other ray R2 ? practically infinite as the slides are rectangular in shape, but at least as regards the dimensions of the system the pseudo circumference is considered having as its length the perimeter of the water-air interface between two slides (longitudinal or parallel curvature to the two slides); therefore, for the size of the system it will be considered? R2 negligible (R2 >> R1) in the condition in which the length of the slides lv >> d (at least lv >= 15*d); for the work spent will be considered? however 1/R2 is negligible, unless it is evaluated separately in Appendix C at the end of this description with the related considerations. Furthermore, in order not to consider the capillary effect in the area temporarily free from slides and in the area always free from slides, it is also required that the length of each zone is at least 15 times the d value, i.e. nuf * (d sp) >= 15 *d and therefore for safety nuf >= 15 and luS1 >= 15*d.

I3. in Laplace's law we set R1 = d; from tests carried out on distances between the slides of interest for the present invention, R1 ~= d; the tests were carried out with the following d values expressed in millimetres: 1.2; 1.0; 0.8; 0.6; 0.4; 0.2. For each value of d indicated above ? the value of h at which ? the water arrived and rose by capillary action; given that R1 = ? /(?*g*h), for each value of h obtained by varying the distance d between the slides, ? It was possible to determine R1 and compare it with d. The average values in millimeters obtained were the following: (d=1.2; h=6.5 ?> R1=1.14); (d=1.0; h=8.0 ?> R1=0.93); (d=0.8 ?> h=9.0 ?> R1=0.82); (d=0.6 ?> h=12.0 ?> R1=0.62); (d=0.4 ?> h=19.0 ?> R1=0.39); (d=0.2 ?> h=38.0 ?> R1=0.195).

I4. FR = constant force applied to the single slide for its horizontal movement; this force will be? applied at constant acceleration

I5. To optimize (reduce) the dimensions of the system, the following conditions derived from the subsequent calculations are set:

- height of the slides hv = h0 hc (i.e. equal to the maximum rise of water compared to the base of the UB basic unit);

- hb = 0 (or in any case close to 0); in this case, a very low level water collection must be provided under the slides, for example by slightly lowering the UB base in the surface occupied by the slides (figure 5), to ensure that during the capillary phenomenon the water passes from the surface S1 towards the slides (arrow F2 of the same figure 5) and that the base of the same remains slightly immersed in water since with hb = 0 the surface S1 is completely emptied of water I6. Wall thicknesses of the unit container base UB and fixed support negligible in the calculations compared to the other dimensions involved.

OPERATING CYCLE

The operation of the system according to the invention is cyclical and can be described by analyzing the operation of the unit? basic UB. The operating cycles of the unit? basic UB can be logically divided into more? phases.

Phases of the UB operating cycle:

Phase 0 (or initial cycle)? assuming we start the cycles with the slides in the Z1 zone, the volume of water V0 is introduced into this zone and the volume of water V0 ?V into the Z2 zone.

Phase 1 ? the operating cycle begins. On the surface of the Z1 zone occupied by the slides the capillary phenomenon occurs, that is? the entire volume of water V0 (with the hypothesis I5 hb = 0) passes from the surface S1 to the surface S3; in this situation we have that in Z1/S1 the level ? hb = 0, in Z1/S3 ? hb + hc = hc (figure 5) while throughout the Z2 zone the level ? hs (hs > h0)

Phase 2? with zones Z1 and Z2 in communication via pipes and impeller, the water passes from zone Z2 (initial level hs) to Z1/S1 (initial level hb=0) producing work on the impeller until ? the entire volume ?V has passed, bringing the Z2 and Z1/S1 zones to the same level h0. To have the maximum height difference hs - hb (i.e. hs) of the water between zone Z2 and Z1/S1? it is necessary to provide a mechanism for closing/opening the connection with the impeller (opening at height difference hs and closing at zero difference in level). Between the slides, in Z1/S3, the water is rise to the maximum level h0 + hc which for the hypothesis I5 ? also the height hv of the slide (figure 6). At this point the slides are moved from zone Z1 to zone Z2 (figure 4).

Phase 3? this phase coincides with Phase 1 with inverted zones, that is, what is? occurred in zone Z1 now occurs in zone Z2.

Phase 4? with the reverse procedure compared to Phase 2, the same quantity? of water ?V passes from zone Z1 to zone Z2, putting the impeller in motion again; in this case the system must provide for the reverse rotation of the impeller or the opening/closing of valves to reverse the path of the water, or alternatively another impeller. The slides are moved in the opposite direction and from the Z2 area they return to the Z1 area: with this movement the cycle ends. The system finds itself in Phase 1 and a new operating cycle begins.

For each cycle we will have useful work for the impellers from which the work spent for moving the moving series of slides in one direction and the other must be deducted. From the calculations below it is demonstrated that in certain conditions the work expended? inferior to the work produced by the forces of adhesion and cohesion and thanks to these entities? of the forces at play? It is possible to obtain useful work, even if in practice you will have to? take into account any losses overlooked here.

The appendices below contain calculation simulations relating to:

Appendix A - useful work produced in a cycle by one?unit? basic UB Appendix B- work spent in one cycle for the operation of a UB Appendix C - energy balance of the system

Appendix D - example system parameters spreadsheet.

Net work and power of the unit? basic UB

Ln (cycle UB) is defined as the net work of a unit? basic UB for each cycle of the system, we will have from [15, Appendix A] and [R22, Appendix B]:

Ln(UB cycle) = Lu ? Ls (J) [LU1]

Cycle time? 2*tl therefore the power obtained from a UB is:

P = Ln(UB cycle) / (2*tl) (W) [LU2]

Limits of the time tl of met? cycle and the length lv of the slide.

So that? the system produces energy must be:

Lu > Ls and therefore Lu / Ls > 1

from which:

place ? =

you have:

tl(minimum) = [LU3]

and therefore it must be tl > tl(minimum)

<l>v<(maximum)> [LU4]

and therefore it must be lv < lv(maximum).

Unit sizing? main UP (1m<2>) and unit? basic UB (1m)

With UP (1m<2>) we define a?unit? main UP which occupies a flat surface of 1 m<2>. For the calculation of the parameters, the spaces occupied by the pipes, the impellers and the slide handling system are not taken into account. Considering that we are talking about elements with dimensions in the order of magnitude of the millimetre, we calculate how many units? functional UF are in a?unit? basic UB one meter long which will be indicated? as UB(1m). With the quantities expressed in millimeters we have:

nuf * (d sp) luS1 = 1000

from which: nuf = int[(1000 - luS1) / (d sp)] [D1] with nuf >= 15 Let's evaluate how many UBs in width fit into 1 meter:

nub *(2*lv ) = 1000

from which: nub = int(500 / lv) [D2] with lv(maximum)> lv >= 15*d Therefore, a UP(1m<2>) contains a number of UB(1m) equal to nub while the UB (1m) contains a number of units? functional UF equal to nuf.

System parameters summary

- nuf >= 15

- lv(maximum)> lv >= 15*d

- hv = h0 hc = h0 ? /(?*g*d) [P1]

- hb = 0 from which for [7, Appendix A]: V0 = Vc and developing the calculations we obtain: h0 = Vc / (S1 S2) = [ ? /(?*g)] * nuf / [luS1 nuf*(d+sp )] [P2]

and replacing [P2] in [P1]:

hv = h0 hc = [ ? /(?*g)] * [nuf / (luS1 nuf*(d+sp )) 1/d] [P3]

The system parameters can be divided as follows:

- independent

sp (slide thickness)

d (distance between adjacent sets of slides)

luS1 (surface length S1) with luS1 >= 15*d ---- > alternatively you can? decide to take hv as an independent parameter

lv (slide length) where 15*d =< lv < lv(maximum) from the condition [LU4] tl (half cycle time; lv movement time of the slide to pass from one area to another) in compliance with the condition [LU3]

so tl > tl(minimum)

- employees

nuf (number of UF functional units that are in a UB(1m))

hv (slide height)

hv = h0 hc = [ ? /(?*g)] * [nuf / (luS1 nuf*(d+sp )) 1/d]

nub = int(500 / lv) [nub = unit number? basic UB in a UP(1m<2>)]

S1 = luS1 * lv

S2 = nuf * (d sp ) * lv

S3 = nuf * d * lv

SA = S1 S2

SB = S1 S3

V0 = Vc = nuf * lv * ? /(?*g)

?V= Vc * (S1 S3 ) / (S1 S2 ) = Vc * SB / SA

hc = ? /(?*g*d)

h0 = Vc / (S1 S2) = Vc / SA

hs = (Vc + ?V) / (S1 S2) = Vc * (SA + SB) / (SA)<2>

Work and power of the system:

Lu = (J)

or also:

Lu <= > (J)

mmax = [4*?*tl/d ?v *sp ] *hv * lv (kg)

Ls = 4*nuf*(lv /tl)<2>*mmax = 4*nuf*(1/tl)<2>*[4*?*tl/d ?v *sp ] *hv *(lv)< 3 >(J)

from which you more? note that the work expended increases with the cube of the length of the slide so, in proportion, the system will have to have tall and short slides, as well as very thin in thickness. In other words, as the length of the slide increases, a situation arises in which if lv >= lv(maximum) the work expended equals or exceeds the useful work since the useful work increases with the increase in the length of the slide slide, but only in proportion [Lu ? lv ; Ls ? (lv)<3 >].

To highlight the proportionality? of Lu with lv let L = (luS1/nuf) d, from which:

Lu <= > (J)

Ln(UB cycle) = Lu ? Ls (J)

P(UB) = Ln(UB cycle) / (2*tl) (W)

Indicating with ?% the percentage yield of the system we will have:

?% = 100*Ln(UB cycle) / Lu = 100*(Lu ? Ls)/Lu = 100*(1 ? Ls/Lu)

? = 73*10<-3 >N/m

? = 10<-3 >N*sec/m<2>

? = density? water (water mass / volume) = 1,000 kg/m<3 >?*g = 9810 N/m<3 >?v = density? glass (glass mass / volume) = 2,500 kg/m<3 >?v*g = 24,500 N/m<3>

System dimensions in relation to the power supplied

The calculated dimensions are the theoretical ones; in reality? they will be a little larger to take into account the spaces for:

- the pipes, the impellers and the slide handling system

- management and maintenance of the system

- the system for transforming the potential energy obtained into electrical energy (water collection tank, turbine(s), discharge tank).

Having defined UC (Cubic Unit) as the space of 1 m<3>occupied by the PU(1m<2>), we evaluate how many PU(1m<2>) are in 1 m<3>, that is? how many units? main UPs of 1 m<2>can be overlapped in a height of one meter.

Taking into account what has already been done? said about the height of a unit? basic UB, i.e. that in addition to the height of the slide other spaces must be taken into account between a slide and the next superimposed one, it is assumed that the necessary space is at least 5 mm, therefore, indicating with nup the number of UP( 1m<2>), you will have:

nup*(hv + 5) = 1000 from which:

nup = int(1000 / (hv 5))

Defined as ?Palace? a 1000 m <3 > building (10 meters x 10 meters, 10 meters high), the number of units? cubic UC contained in the building will be:

n_UC(Building) = 10<3>

Defined ?1Km<2>? one square kilometer of space occupied by the Palaces (1000 meters x 1000 meters, height 10 meters), the number of Palaces contained in 1Km<2 >will be: n_Palazzi(1Km<2 >) = 10<4>

If P ? the power of a UB(1m), as determined by [LU2], the power obtained by varying the size of the system can be calculated as:

P = P[UB(1m)]

P[UP(1m<2>)] = nub *P[UB(1m)] = nub * P

P(UC) = nup * P[UP(1m<2>)] = nup * nub * P

P(Building) = 10<3 >* P(UC) = 10<3 >* nup * nub * P

P(1Km<2 >) = 10<4 >* P(Building) = 10<7 >* nup * nub * P

Energy requirements.

In 2018 the Italian energy requirement? state of 321 TWh with an average power of approximately 35 Gw obtained from:

- 51% fossil fuel thermoelectric power plants (mainly gas and coal) 163 Twh

- 35% renewable sources 114 TWh

- 14% import from abroad 44 TWh

the share from renewable sources (114 TWh)? was obtained from:

43% hydroelectric; 20% solar; 17% bioenergy; 15% wind; 5% geothermal.

Using a spreadsheet such as the one shown in Appendix D? It is possible to size the system according to the invention based on the energy requirement required (in particular, an example is reproduced to give an idea of the ratio of energy obtained / volumes occupied by the system).

? susceptible to further experimental verification how much work spent to move the slides is transformed into heat and consequently lowers the value of surface tension; the same reasoning applies to the increase in the ambient temperature in which? entered the system: this is not the case? was taken into account in the theoretical calculations since the reduction of surface tension? proportionally limited for the temperature variations involved. For example, for water, the surface tension is 73 mN/m at 20/25?, approximately 70 mN/m at 40? and about 66 mN/m at 100? while it increases around 75 mN/m at 0?; therefore, if the system works in a temperature range between 15 and 35 degrees, the variation in energy produced can fluctuate by approximately ?6/7% compared to that calculated at 20/25? with ? at 73 mN/m.

Although the system according to the invention occupies a certain volume in relation to the power supplied, the following advantageous features are evident:

- energy production? constant over time (apart from the small oscillations mentioned above due to temperature variations);

- energy production? continuous over time (other renewable energy sources such as the sun and wind may be absent and not constant while the intermolecular forces of adhesion and cohesion are always present at the solid-liquid-gas interface in the environmental conditions in which this system works);

- the plant, once fully operational, has practically zero raw material costs (it does not require a continuous supply of water or any aqueous solutions but only supplies for leaks or for any replacements in the long term);

- there are no polluting emissions;

- in case of need? disposal of the components over time, the materials are not polluting and can be easily recycled, unless the elements used are verified in the case of aqueous solutions;

- low maintenance costs: in case of faults or breakages you can? think about a system of substitutions for Unit? Cubic with unit extraction? malfunctioning and insertion of the new one.

In summary, the system according to the invention envisages making the system work continuously. natural forces (adhesion and cohesion) to obtain alternative and renewable energy to be used through possible transformation (for example into electricity). The system has, according to a constructional aspect, the following characteristics:

a) unit? basic UB:

- container in light plastic material

- lifting means with capillary effect obtained with a solid element, preferably glass;

- means of dividing the container, for example such as a diaphragm made of Teflon? and rising from the bottom of the container;

- the cyclic displacement movement between the container and the capillary system can? be carried out in two ways: with movement of the capillary system or with movement of the container;

b) unit component? main UP:

- composed of one or more? unit? basic UB

- impeller and connecting pipes

c) arms and motor for the cyclic movement of the capillary system (slides) inside the units? basic UB.

The system can be composed of more? unit? main UPs and the impeller(s) can: - be connected directly to a turbine for the production of electricity - lift a volume of water to obtain potential mechanical energy to be transformed.

Functionally, the method according to the invention requires the container to be divided into two zones with volumes of liquid at different levels as a condition for having an energy (potential difference in level) that the system must make continuously renewable. over time and the most possible constant in value.

For the use/conversion of energy, the areas of the container are connected via pipes and impellers with possible energy grouping in a liquid collection tank and related transformation turbine or direct connection of the impellers to the turbine.

Which components are necessary to obtain and cyclically renew potential energy, given that the use of this energy by the conversion means cancels the difference in level between the two containers for which the system must continuously recreate? this difference in height can be defined as:

a) non-capillary part of each zone of the container: ? the part where the capillary effect due to the surface tension of the adhesion and cohesion forces is? negligible or practically nil;

b) capillary part of each zone of the container: active part of the system where the adhesion and cohesion forces occur which carry out the work of raising the liquid; these forces can be assimilated to a lifting pump that extracts liquid from the non-capillary sector referred to in point a); this part of each area of the container ? alternately capillary and non-capillary;

c) moving part: series of flat laminar elements (slides) where the capillary phenomenon occurs which, with the simultaneous movement of the entire series between one area and another, determines whether the part of each area of the container referred to in point b )? or not ? capillary, realizing the cyclicity? of the system necessary to recreate the difference in height between the two areas of the container and therefore to renew the potential energy. For the cyclic movement of this part of the system, work must be provided (work spent Ls) to overcome the inertia forces and any friction; ? to keep in mind, as highlighted by the calculations in Appendix B, that the entity? of this work is, in particular, strongly dependent on the length dimensions of the capillary elements (slides) and this dimension must therefore be optimized.

This invention? has been described so far with reference to preferred embodiments. ? it is to be understood that other embodiments may exist that relate to the same inventive core, as defined by the scope of protection of the claims reported below. In fact, this project adopts any type of system capable of functioning in a conceptually equal or similar way to what has been described (obtain useful energy with continuity from the adhesion and cohesion forces) and therefore adopts and includes any project that uses surface tension through systems also made with materials, variants, configurations and sizes different from the particular system described and sized here.

APPENDIX A - Calculation of the useful work produced in a cycle by a UB Laplace's law

For hypothesis I2 Laplace's law becomes:

?p = ? * (1/R1 1/R2) = ? * (1/R1)

therefore for hypothesis I3: [L0].

In Phase 1 and Phase 3, between two adjacent slides the water rises due to capillary action. at a height hc compared to the level that the water would have without the capillarity phenomenon; this quantity? of water has such a weight as to create a downward pressure equal to ?*g*hc, so the water rises by capillary action? until the pressure exerted upwards by the surface tension and the downward pressure due to the weight of the water column have equal intensity; therefore at equilibrium we will have:

?p = ?*g* hc and therefore from [L0]:

? /d = (?*g)* hc from which you can? obtain hc :

[L1]

Do you notice the values of ? And ? of the water and the dimensions involved in the order of the millimetre, in the specific case in question the [L1] can? be written in the following terms:

hc(mm.) * d(mm.) = ? /(?*g) ~= 7.44 mm.<2 >[L2]

From [L2] we note that, once the materials have been chosen (in this case glass and water), the surface hc*d of the liquid rising by capillary action? remains constant as the distance between the interfaced solids varies (i.e. as d varies).

Operation phases of the UB

Starting cycle with slides in Z1.

In Phase 0:

- the volume of water in Z1 is: V0 = h0 * (S1 + S2) = h0 * SA [1] - the volume of water in Z2 is: V0 ?V [2] with ?V = h0 * (S1 + S3 ) = h0 * SB [3] - Vc = nuf * lv * hc * d [4]

and for [L1]:

Vc = nuf * lv * ? /(?*g) [5]

the level throughout Z1? h0 while throughout Z2 ? hs.

At the end of Phase 1 we have the following situation:

- in Z2 the level? hs

- between the slides, that is? in Z1/S3, the level ? hb hc = hc for hb = 0

- in Z1/S1 the level ?:

hb = (V0 ? Vc) / (S1 S3) [6]

from which, for hb = 0 we have:

V0 = Vc [7]

In this case (hb = 0) all the volume of water V0 is found between the slides sucked in by capillary action.

At the end of Phase 2 we have the following situation:

- the volume of water ?V ? passed from Z2 to Z1 spreading onto the surface S1 S3 of Z1 free from glass slides

- the level in Z2? dropped from hs to h0

- the level in Z1/S1 from hb = 0? returned to h0 and in Z1/S3 (between the slides) ? rose from hc to h0 + hc

- the volume of water ?V passed from Z2 to Z1 from [3], [1] and [7] can? be written as follows:

?V= h0 * (S1 S3) = Vc * (S1 S3 ) / (S1 S2 ) = Vc * SB / SA [8]

from which, for hb = 0 we can obtain hs:

hs = (V0 ?V) / (S1 S2 ) [9]

and for [7] and [8]:

hs <=> [10]

from [1] and [7] we have:

h0 = Vc / (S1 S2) = Vc / SA [11]

The useful work of the impeller obtained when the volume of water ?V passes through Z1 is: Lu1 = (?*g) * ?V * hs / 2 [12]

and replacing from [5], [8] and [10]:

Lu1 [13]

Lu1 ? calculated taking into account that during Phase 2 the head x of the water goes from hs to 0; why is it possible? write:

from [10] placing hs variable in x, it results: Vc = x * (SA)<2>/(SA+SB) from which:

At the end of Phase 3 the situation is the same as Phase 1 but the roles of Z1 and Z2 are reversed; therefore the useful work of the impeller obtained when the volume of water ?V passes through Z2? same as that of Phase 1 with reversed flow:

Lu2 = Lu1 [14]

and at the end of the cycle you will have? a total useful work Lu as the sum of the work obtained in Phase 1 and Phase 3:

Lu = Lu1 + Lu2

hence, the useful work produced in one cycle by a UB will be:

[15]

How can you? note, the useful work depends on the square of the surface tension so having a liquid with a higher surface tension? high water increases the possibility? to obtain energy with more systems? reduced in size.

For example, mineral oils can have a surface tension approximately six times higher than that of water, but solutions of water and salts can also have a surface tension greater than water alone (a 50% solution of water and potassium carbonate has a tension of 107*10<-3 >N/m).

Water-soluble solutions are made up of polar substances such as sugar, table salt, acids, bases. However, non-polar (non-polar) substances are not soluble in water, i.e. organic ones such as oil and naphthalene which are soluble in each other. Polar substances such as water have molecules with a dipole moment other than zero while non-polar ones have molecules with a dipole moment equal to zero, that is, the center of gravity of the positive charges coincides with the center of gravity of the negative charges (they are spatially neutral molecules).

How already? said ? However, the convenience of using solutions other than water, both in economic and environmental terms, must be verified. Furthermore ? to evaluate the use of this liquid taking into account its viscosity coefficient? ? which could make it less convenient than water due to the excessive increase in labor expended compared to the increase in useful labor.

Performance of the system with respect to the work done by the adhesion and cohesion forces (surface tension).

?f% indicates the percentage efficiency given by the ratio between the useful work produced by the system and that carried out by the adhesion and cohesion forces. This efficiency answers the question: how much work done by surface tension can be obtained with this system?

The volume of water raised between two slides, that is? raised in a unit? functional UF?:

hc*d*lv and therefore the raised weight?:

(?*g)*hc*d*lv but we know that hc*d = ? /(?*g) so the raised weight is: ? *lv The adhesion and cohesion forces, through the surface tension that occurs between two adjacent glass slides, are able to move the volume of water determined above, raising its center of gravity by hc/2 so that the work done is ? given by the weight for half? of the capillary height, or, indicating this work with Lts:

Lts = ? *lv*hc/2 and replacing hc from [L1], we get:

Lts = [( ? )<2>/(2*?*g)] * lv /d [N1]

Since Lu? calculated for a cycle (Lts is double in a cycle) and for a number of water lifts, or slides, equal to nuf, the percentage yield will be? given by: ?f% = 100*Lu/(2*nuf*Lts) [N2] and replacing Lu from [15] and Lts from [N1] we obtain: [N3] APPENDIX B - Calculation of the work spent in a cycle for the operation of a unit? basic UB

The total resistance force that opposes the movement of a slide in one direction and the other during an operating cycle depends on:

- by the inertia of the mass of the slide

- by the inertia of the mass of displaced water

- from the resistance due to the effect of viscosity? for water-slide flow; do the slides slide parallel to each other at the same speed? and at the same time (they are closely connected to each other), therefore, for the purposes of calculating the viscosity? resistant to displacement? as if the slides were stationary and, in the opposite direction to the movement, the water flowed between them (as in the real case in which the series of slides are held still and the container is moved)

- for hypothesis I1, any losses due to rubbing of the slide as it passes through the opening of the support are neglected

The resisting force varies as the displacement of the slide varies but with hypothesis I4 we only take into consideration the maximum value which is indicated by Fr at constant acceleration; therefore Fr will be? the constant value of the force that will be applied? to the slide for displacement, although a smaller variable force could be applied for all other points of displacement of the slide. With this hypothesis it is believed to compensate, at least in part, the neglected losses, as well as to simplify the calculations.

The movement of the slide is divided into two phases of the cycle:

a) slide coming out of Z1 and entering Z2

b) slide coming out of Z2 and entering Z1

The calculation of the resisting forces can? be half done? cycle since the resisting forces of the second half? of the cycle have the same value and direction, they differ only in the opposite direction of movement.

Let Fr(x) indicate the total resistant force (variable with the displacement x of the slide) and let us take into consideration the phase a above. Given that the displacement x of the slide goes from 0 to lv, can we? obtain Fr(x) as follows:

Fr(x) = Fv(x) Fiv(x) Fia(x) [R1]

Where:

- Fv(x) ? the force due to viscosity:

Fv(x) = Fvz1(x) Fvz2(x) [R2]

with Fvz1(x) and Fvz2(x) resisting forces due to viscosity, respectively in the Z1 zone and in the Z2 zone

- Fiv(x) ? the inertia force due to the mass of the glass: ? a force independent of the displacement x since the mass of the glass and the acceleration are constant, so we can indicate Fiv(x) with Fiv

- Fia(x) ? the inertia force of the mass of water moved by the slide in Z2 (the slide in Z1 does not occupy a volume of water but on the contrary frees it as it is coming out of it)

Therefore:

Fr(x) = Fvz1(x) Fvz2(x) Fiv Fia(x) [R3]

Calculation of Fvz1(x) and Fvz2(x).

Fvz1(x) = ? * v(x) * Sv1(x) / d [R4]

Where:

v(x) = velocity? with which the slide is moved (i.e. all the moving series of slides) Sv1(x) = surfaces of the slide present in Z1 during the movement:

Sv1(x) = 2 * hv * x (during the movement of the slide the height of the water between the slides remains at the maximum height under the capillary effect: hv = h0 + hc; in reality it would tend to rise to hs hc if the slide had this height; the factor 2 takes into account the two faces of the slide that see the water flow)

d, how already? said, ? the distance between two slides, that is? ? the thickness of the water that two glass slides, assumed to be stationary, see moving

Is the speed defined? set to the slide as follows:

v(x) = a*t(x) with: a = constant acceleration; t(x) = slide movement time,

arises:

- t(0)=0 (start of slide movement from Z1 towards Z2)

- t(lv)=tl (end of slide movement: from completely inserted in Z1 to completely inserted in Z2 after a movement x=lv)

and for the above you can? write:

t(x) = (tl/lv)*x

therefore:

v(x) = a * (tl/lv) * x

and therefore the speed? of the slide goes from 0 to the maximum value (a*tl) when the slide ? completely out of Z1 and then completely inserted into Z2; at the end of the slide? blocked, it reverses the direction of movement starting again from v(x) = 0 to exit Z2 and enter Z1 completing the cycle.

For the above, [R4] becomes:

Fvz1(x) = 2*?*a*(1/d)*(tl/lv )*hv*x<2 >[R5] for x --- > from 0 to lv t(x) --- > from 0 to tl v(x) --- > from 0 to (a*tl)

Fvz2(x) = ? * v(x) * Sv2(x) / d [R6]

Where:

Sv2(x) = surfaces of the slide present in Z2; during the movement the height of the water between the slides changes from hv to hc as the water level is lowered in Z2; therefore:

Sv2(x) = 2*[hv ? (hv ? hc)*(x/(2*lv))]*x for x that goes from 0 (slide entering Z2) to lv (slide completely inserted in Z2)

and [R6] becomes:

Fvz2(x) = 2*?*a*(1/d)*(tl/lv )*[hv ? (hv ? hc)*(x/(2*lv))]*x<2 >[R7] for x --- > from 0 to lv t(x) --- > from 0 to tl v(x) --- > from 0 to (a*tl)

Calculation of inertia forces:

indicating with mv the mass of the slide and with ma(x) the mass of the water displaced when the slide leaves the support, we have:

Fiv = mv *a = (?v *hv *lv*sp ) * a [R8]

The volume ma(x) increases proportionally as the slide enters further and further. in Z2 (and therefore increases proportionally as x increases) but also depends on the height of the displaced water; the height of the displaced water passes from h0 to 0 (hypothesis hb = 0). Therefore:

Fia(x) = ma(x)*a = ?*sp *x*[h0*[1 - (x/lv)]]* a [R9] Taking into account [R5], [R7], R8] and [R9] can you? write [R3] as follows:

Fr(x) = Fvz1(x) Fvz2(x) Fiv Fia(x) = mtot (x) * a [R10] with:

mtot (x) = 2*?*a*(1/d)*(tl/lv )*[2*hv ? (hv ? hc)*(x/(2*lv))]*x<2 >+ (?v *sp *hv *lv ) ?*sp *h0*[1 - (x/lv)]*x [ R11]

Given the above, we are in a position to calculate the maximum resistance force Fr, therefore we will take the maximum mass mtot (x) which we will call mmax, so we will have:

Fr = mmax*a [R12]

and therefore for [R12] we will have, as mentioned, a = constant in displacement time. We calculate Fr starting from the following condition:

<[R13]>

but t = (tl/lv)*x and therefore dt = (tl/lv)*dx so t*dt = (tl/lv)<2>*x*dx

from which, replacing in [R13]:

<[R14]>

Taking [R12] into account, [R14] becomes:

<[R15]>

from which and therefore:

Fr = 2 * lv * mmax / tl<2 >[R16]

To calculate mmax we ask:

mtot (x) = k1*x<3>+ k2*x<2 >+ k3*x k4 [R17]

with:

k1 = - ?*tl*(hv ? hc)/[d*(lv)<2>] (dimension: kg/m<3>) k2 = 4*?*tl*hv /(d*lv) - ? *sp *h0 /lv (dimension: kg/m<2>) k3 = ?*sp *h0 (dimension: kg/m) k4 = ?v *sp *hv *lv (dimension: kg) if the term is neglected k1*x<3 >negative (k1<0; term that tends to reduce the mass) and small compared to the others since these are millimetric displacements (cubic), can it? write [R17] as follows:

mtot (x) ~= k2*x<2 >+ k3*x k4 [R18]

[R18] describes a concave parabola upwards for which the mass has an increasing trend with the displacement and therefore it is possible? believe that the mass? maximum for x = lv

and substituting in [R18]:

mmax = mtot (lv) = k2*(lv)<2 >+ k3*lv k4 = [4*?*tl/d ?v *sp ] *hv * lv [R20]

The maximum work spent to move the slide in one cycle (movement of lv in one direction for phase a and of lv in the opposite direction in phase b) will be:

Lsv = 2*Fr*lv = 4*(lv /tl)<2>*mmax = 4*[4*?*tl/d ?v *sp ] *hv * (lv)<3>/(tl)< 2 >(J) [R21]

The work spent on the entire UB will be:

Ls = nuf * Lsv (J) [R22]

Indicating with ca the number of capillaries? (dimensionless), we have:

ca = ?*v/ ? with v = speed? of displacement between glass and water

if we consider the speed? system average vm = lv/tl we obtain:

ca = ?*vm/ ? = (?*lv)/( ? *tl) [R23]

with the values involved relating to water and considering that the dimensions of the slide are of the order of millimeters and the time? lower or of the order of the second, you can? evaluate ca as follows: ca = (10<(-3)>/73)*[lv(mm.)/tl(sec.)] ~ 13.7*10<(-6)>*[lv(mm. )/tl(sec.)].

Therefore ca? of the order of 10<(-5) >for which it is possible? state that during the movement between water and glass the capillary forces prevail over the resistant viscosity ones.

APPENDIX C - System energy balance.

Energy used by the capillary phenomenon.

In Phase 1 the rise by capillarity? any water between the slides? due to the surface tension that is the energy per unit? of surface at the water-air-glass interface. Briefly, without going into too much detail, can it? simply say that water and glass are two polar elements; thanks to this peculiarity? the water molecules, with the positively polarized hydrogen, are attracted by the glass molecules composed mainly of silicon and oxygen: in particular, the positively polarized hydrogen of the water ? attracted by the oxygen of the negatively polarized glass. This attraction occurs between glass and water, with wettability? of the glass by the water, why? the surface energies of glass are greater than those of water, i.e. the cohesion forces between the water molecules are lower than the cohesion forces between the glass molecules and therefore at the interface the greater forces of the glass attract the molecules by adhesion of the water giving rise to spreadability? of the liquid onto the solid.

At equilibrium the surface free energies per unit? of surface, at the interfaces between water, glass and air result from the Young equation:

Ts = Tls Tl*cos?

where Ts = surface tension of the solid (glass)

Tl = surface tension of the liquid (? of water)

Tls = liquid-solid interfacial tension

? = liquid-solid contact angle

For ? = 0 (cos? = 1) we have complete wettability, that is? the liquid spreads completely on the solid as practically happens with water-glass (perfectly smooth and clean glass); in this case we have:

Ts = Tls Tl [B1]

that is? Ts > Tl which confirms that to have wettability? the surface tension of the solid must be greater than the surface tension of the liquid. The work per unit? of surface spent by molecules in water-glass adhesion is obtained from the Dupr? equation:

Wa = Ts Tl ? Tls [B2]

this work corresponds to the variation (decrease) in surface energy of the elements before and after the rise of the water by capillary action? between the slides; in practice, the water-glass system (see the UF functional unit), at the moment of solid-liquid contact by capillary action, loses the energies existing on the free surfaces of the solid and liquid and at the same time acquires the contact energy solid-liquid. In the present case, for each side of the slides we have that:

a) Ts? the energy lost by the glass interface molecules

b) Tl = ? ? the energy lost by the water interface molecules

c) Tls? the energy acquired by the water and glass molecules in contact with each other. Obtained Ts from [B1] and replaced in [B2], we obtain:

Wa = Tls Tl Tl - Tls = 2* ? [B3]

(in reality the value of Ts in [B1] is related to the value of the surface tension of the glass at the interface with the air while the value of Ts in [B2] is related to the value of the surface tension of the glass at the interface with the vacuum but this difference can be neglected since Tl in the air is approximately Tl in the vacuum, as specified below in NOTE1) Wa ? the free energy is G (variation of the Gibbs free energy) per unit? of surface that the system according to the invention can? theoretically use at every half cycle of operation: energy lost when the water rises due to capillarity, energy regained when the water falls following the Ls work applied. Per unit? of the surface, the energy variations at the end of the rise of the capillary water are the following: a) the glass has lost the surface free energy Ts

b) at the water-glass interface yes? an energy Tls lower than Ts has been established

c) energy? = Ts ? Tls? the surface energy lost by the glass in contact with water and Tl = ? ? the surface energy lost by the water

In the present system the water is raised by capillary action. between two slides; surface tension? can? be imagined as an elastic membrane stretched between the two slides and considering that the surface of each slide, wetted by capillarity, is S = hc*lv, can it be imagined? obtain the free surface energy Es lost/gained during the half cycle:

Ex = Wa*(2*S) = 4 * ? * hc * lv [B4]

In reality, to this energy Es we should subtract/add the surface energy gained/lost by the water on the two sides in contact with the air equal to 2*? *hc *d but in the case of the present invention, for the hypothesis I2, it results d << lv so this energy value is ? negligible compared to that lost/bought.

How already? calculated in Appendix A, the capillary work carried out at the water-glass contact is obtained. The volume of water raised between two slides is: hc*d*lv

and therefore the raised weight is: (?*g)*hc*d*lv

but we know that hc*d = ? /(?*g) so the raised weight is: ? *lv

The forces of adhesion and cohesion, through the surface tension that occurs between two adjacent slides, are able to move the volume of water determined above, raising its center of gravity by hc/2 so that the work done is ? given by the weight for half? of the capillary height, or, indicating this work with Lts:

Lts = ? *lv*hc/2 [B5]

from which, comparing [B5] with [B4] we have:

Lts = Es / 8 [B6]

It is therefore noted that only 1/8 of the energy Es (?G) lost by the system, on the water-glass interface surfaces that come into contact by capillarity, is? used to carry out capillary work; the other 7/8 can be considered lost due to thermodynamic dissipation (unusable latent heat) and subsequently recovered during the detachment of the water during the movement of the slides. In the cyclic functioning of the system there is a dissipation of energy for all those inevitably irreversible phases inherent in the thermodynamic processes between the system and the external environment with an increase in entropy as per the second law of thermodynamics; an example ? the hysteresis of the contact angle which implies an irreversible process with energy dissipation: the liquid-solid contact angle ? generally greater when the liquid spreads than when it shrinks (in the invention the shrinkage is forced by the work provided by moving the slides); another example of dissipation? given by the possible irregularities? and impurities in the glass and from friction during cyclic movement. ? to verify how and in how much time the aforementioned dissipation can bring about substantial changes to the parameters of the contact angle and surface energies with aging of the elements and related need? replacement, as well as? to an increase in system temperature.

This means that, for each cycle, there will be a constant tendency to have less useful work and more heat exchanged with the external environment.

For [B3] it is known that, for every half cycle of operation of the system, the surface energy varies by 2* ? per unit? of surface, that is:

- decreases if the wetted surface increases (free surface lost); in the system according to the invention, 1/8 is transformed into potential energy due to capillarity? (and transformed into useful work) and 7/8 into heat released to the environment as thermodynamic work due to water-glass adhesion, or less freedom? of movement of molecules. The first law of thermodynamics can? be expressed as follows: the system loses energy to the external environment, 1/8 as expansion work (capillary rise of the water) and 7/8 in the form of heat due to movement lost by the glass-water molecules in favor of of air molecules in the external environment - increases if the wetted surface decreases (acquired free surface); in the system according to the invention, a part of the energy is provided by the work spent to remove the capillary water from the slides and a part? heat taken from the environment as thermodynamic work due to the greater movement of the water and glass molecules, free to move in cohesion and no longer? subject to the membership link. The supply of external work makes the natural process of capillarity reversible, without prejudice to the essential entropic processes. The first law of thermodynamics can? be expressed as follows: the system receives energy from the external environment, partly as induced work of compression of the system (breaking of the adhesion bonds and therefore of the surface tension with descent of the capillary water due to gravity?) and mostly under form of heat as movement provided by the air molecules of the external environment to the water-glass molecules that have regained their ?freedom?? superficial.

In reality? losses due to irreversible processes that occur during operating cycles with an increase in entropy between system and environment must also be taken into account.

Consider the energy exchanged in half a cycle (water rising between the slides and water falling for moving and cleaning the slides); for the reversible process only, the energy that exits the system towards the outside is equal to the energy that enters the system from the outside, but heat also escapes from the system towards the outside due to irreversible dissipation processes (increase in entropy) due to the friction of the moving parts (slides, impellers, etc.). The free energy that comes out of the system? composed of useful work (capillary energy) and heat lost from the surfaces in capillary contact, part of which is lost due to irreversible dissipation processes (increase in entropy) due to the deterioration of the elements used (as mentioned: hysteresis, irregularities and impurities of the glass, etc.). The energy that enters the system from outside? composed of work expended to move the slides and heat absorbed by the free contact surfaces.

? to underline the fact that, for each cycle, the system will be? able to receive quantities? increasingly less energy from the external environment as the increase in entropy due to the aforementioned dissipations (and therefore the deterioration of the elements) will reduce? each time the quantity? of free energy of the system for which it will be reduced? the quantity? of heat received from the external environment with the aforementioned consequences for the water and glass elements and with reduction of the useful work produced by the system. How already? mentioned, ? to verify how much time, compared to the minimum performance required of the system, will be? it is necessary to replace the elements damaged due to the aforementioned dissipation.

Place, per unit? of surface:

- Ec = capillary energy

- Lus = useful work (obtained from Ec)

- Lss = work spent (to ?clean? the slides from capillary water)

- Ltc = thermodynamic work (heat) transferred to the environment

- Ltp = thermodynamic work (heat) taken from the environment

- ?C = Ltp ? Ltc = work (heat) taken from the environment in half? operating cycle of the system

we'll have:

Ec = Wa/8 = (1/4)* ?

Ltc = Wa ? Ec = 2* ? ? (1/4)* ? = (7/4)* ?

Ltp = Wa ? Lss = 2* ? - Lss

in the reversible cycle: Ltp Lss = Ltc Ec

therefore: ?C = Ltp ? Ltc = Ec ? Lss ~= Lus ? Lss [B7] if losses in the transformation from Ec to Lus are excluded.

[B7] highlights that the useful work obtained by the system according to the invention comes from the heat absorbed from the surrounding environment, that is: the surrounding environment provides the thermodynamic work in the form of capillary energy which the system according to the invention transforms into useful work (and to a small extent in losses; from calculations the efficiency of the system in the transformation of capillary energy into useful work exceeds 95%).

Since the work spent? lower than the capillary energy (Lss < Ec) it turns out that the heat that the system absorbs from the surrounding environment is greater than the heat transferred (Ltp > Ltc); for this reason, how already? said, the thermodynamic work absorbed by the system and transformed into capillary work, due to the inevitable losses due to friction and molecular interactions, can lead over time to an increase in system temperature and to possible changes to the parameters of the elements used.

Energy synthesis.

The energy exchange between the system according to the invention and the external environment per unit? of surface in half a cycle can? be verified as follows.

?G = ?H ? T*?S = 2* ? with:

?G = change in usable free energy

?H = change in the enthalpy of the system, i.e. the energy content that the system can? exchange with the outside

T*?S = caloric variation due to losses (irreversible part of the process with increase in entropy; T=absolute temperature in Kelvin degrees; ?S=change in entropy) In the reversible process relating to half a cycle of the system we have that, in the first part of the half cycle, ?G exits the system towards the external environment in the form of work Ec (capillary expansion) and heat lost from the free surfaces; in the second part of the half cycle (following the "cleaning" of the slides with recovery of the aqua-glass surfaces obtained by moving the glass through the Ls work) ?G passes from the external environment towards the system in the form of Ls work (breaking of the capillarity? and reduction of the volume of water) and heat supplied to the freed surfaces.

In reality, how already? said, there are further irreversible losses in the system with an increase in entropy and aging of the elements (hysteresis, roughness and glass impurities, etc.); to take into account these losses, which increase every half cycle, can we? define ?GD as the change in free energy available at each half cycle: ?GD = ?G ? ?i T*?Yes [B8]

where ?Yes ? the entropy variation (due to hysteresis, etc.) at the ith half cycle of operation of the system and ?i indicates the ith summation of the entropic heat loss.

In the first part of the half cycle we have: ?GD = Ec Qs ~= Lu Qs [B9] with Lu = useful work obtained from Ec; Qs = heat that passes from the system to the environment.

In the second part of the half cycle we have: ?GD = Ls Qa [B10] with Ls = work spent on recovering capillary surfaces?; Qa = heat that passes from the environment to the system.

But from [B8] it appears that ?GD decreases every half cycle and therefore from [B9] it appears that both Qs and Lu decrease since Lu ? approximately 1/8 of the available energy, or approximately 1/8 of ?GD.

From [B10] instead we have that only Qa decreases since the work spent Ls ? applied consistently; it follows:

a) as long as Lu > Ls the system provides net work at the expense of ambient heat what difference Qa? Qs

b) with the constant decrease in Lu the system would reach the condition of Lu = Ls (absence of net work) and then of Lu < Ls with a loss-making system.

NOTE1

Young's equation: Tsa = Tls Tla*cos?

where Tsa = surface tension of the solid (glass) in contact with the air

Tla = surface tension of the liquid, i.e. ? of water in contact with air Tls = liquid-solid interfacial tension

Doupr equation? (per unit area): Wa = Ts Tl ? Tls

where Ts = surface tension of the solid (glass) in contact with vacuum

Tl = surface tension of the liquid, i.e. ? of water in contact with vacuum Tls = liquid-solid interfacial tension

Place Tla ~= Tl and ? = 0 (cos? = 1), the Young equation becomes: Tsa = Tls Tl But Tsa < Ts since the surface molecules of the solid are more free to move in a vacuum with respect to the air and therefore from Tsa we must subtract the difference Ts -Tsa from which: Tsa = Tls Tl ? (Ts ? Tsa) and therefore:

Ts = Tl Tls which replaced in the Dupr equation? leads to the conclusion that the energy lost by the system per unit? of surface in the water-glass capillary coupling results precisely: Wa = 2*Tl = 2* ? as per [B3]

Reversible and irreversible transformations of the system? operating cycle.

As stated previously, for the purposes of thermodynamic exchanges we schematize the system according to the invention in its fundamental part of operation: rise and fall of the capillary water between a pair of slides (half cycle) taking into account the reversible and irreversible parts of the system .

In Fig.7 ? Was an isolated system (universe) divided between the external environment (which we will call the environment) and the system of units indicated? functional UF of the system (capillary water between two slides which we will call system); the system ? been divided between the irreversible part and the reversible part, indicating with:

Ls = total work spent (for moving slides or container)

Ls1 = work spent minus friction losses of moving parts (heat qattr) Ls2 = work spent only for viscosity? of the water (excluding the work spent to move the inertial masses of the glasses and the basic water); Ls2 ? the actual work spent to bring the capillary water back to the base level, canceling the variation ?V of the capillary volume; ? the work of viscosity? which breaks the water-glass adhesion bonds with breaking the surface tension; at the beginning of the movement the system is at a lower temperature than the ambient temperature and the heat qe passes from the environment to the system

S2 = following viscosity work? the capillary water and the glasses regain free surfaces, that is, they regain the lost internal energy (?U>0 and therefore increase in temperature up to T0 ?T) and increase in pressure by the system (?p>0; see law of Laplace with p = p ambient ?p; the water has dropped to the base level with ?V<0 up to ?V=0; after which, due to surface tension, the water will rise again capillary between the glasses to compensate for its greater pressure compared to the air in the external environment)

Es = total energy exchanged between system and environment in half a cycle; ? the energy received by the system is the sum of the viscosity work? Ls2 and heat qe and ? also the energy exiting the system as the sum of the capillary energy Ec and the heat qu S3 = in this phase the capillary water? rise between the glasses due to the surface tension and the internal pressure of the water has equaled that of the external environment (?p<0 until reaching ?p=0 and therefore p = p environment); ?V>0 with expansion of the capillary water, loss of internal energy (?U<0 and therefore decrease in temperature up to T0 - ?T)

Lu = useful work obtained from the capillary energy of rising water (in reality we should speak of capillary energy Ec but we can consider Lu~=Ec)

Lu1 = effective useful work less than Lu due to entropic losses (hysteresis, etc.)

qu = quantity? of heat that passes from the system to the environment (from the glass, to the water and then into the air) due to water-glass adhesion with reduction of free surfaces and therefore loss of surface free energies

In Fig.8 ? represented the thermodynamic cycle of the system.

State values at point A:

- PA pressure = Patm? /d with Patm = atmospheric pressure

- capillary volume Vc(A) = 0

- temperature TA = T0 + ?T

State values at point B:

- pressure PB = Patm

- capillary volume Vc(B) = ?V = hc*lv*d

- temperature TB = T0 - ?T

In this reversible cycle there are two transformations: from B to A and from A to B. From the first law of thermodynamics (?U = Q ? L) the following convention is adopted: Q and L are positive if they enter the system, negative if they leave .

Transformation from B to A.

In this transformation the work Ls2 is performed on the system by moving the slides; this movement allows the system to free the glass from the capillary water which by gravity goes down to basic level; the water-glass surfaces are no longer in adhesion and the system regains the lost energy from the environment (Es = ?U = Q+L = qe Ls2 > 0) Transformation from A to B.

In this transformation the system carries out work Lu (~Ec) on the environment through the surface tension which raises the water; this rising water determines the adhesion of the water-glass surfaces and the system loses the surface free energy to the environment (Es = ?U = -Q-L = -qu - Lu < 0)

System energies in half? cycle, or energies in a thermodynamic cycle of the system.

Ex = 2* ? *(2*S) = 4* ? *S with S=capillary surface of the single slide and therefore Es = 4* ? *hc*lv

Es = qu Lu = qe Ls2

Lu = Ex/8

Ls2 = k*(Es/8) where k = Ls2/Lu with 0<k<1

qu = Es ? Lu = 7*(Ex/8)

qe = Es ? Ls2 = (1 ? k/8)*Ex

Taking into account the energy Es exchanged between the system and the environment, the system performance is:

? = (Lu ? Ls)/Ex

so the maximum efficiency is for Ls which tends to 0 and therefore:

?max = Mo/H = 1/8

?% < 12.5%

Approximate calculation of ?T.

The energy Es heats/cools the system during the thermodynamic cycle, therefore we can set:

Es = CT * ?T with CT capacity? temperature of the water-glass system

CT = ca*ma + cv*mv = ca*ma + 0.2*ca*mv taking into account that cv*~= 0.2*ca with ca = specific heat of the water, ma = mass of the water, cv= heat specific to glass, mv = glass mass

ma = ?a*Va mv = ?v*Vv

with ?a = density? of the water at m<3>, ?v = density? of the glass at m<3>, Va = hc*lv*d = volume of the capillary water, Vv = volume of the glasses involved in the heat exchange. In the hypothesis Vv ~= Va the result is: CT = (?a + 0.2*? v)* ca*Va but ?v ~= 2.5*?a therefore: CT = 1.5*ca*?a*Va

therefore:

?T = Es / CT = (4* ? *hc*lv) / (1.5*ca*?a*Va) = (4* ? ) / (1.5*ca*?a*d)

but ?a = 1000 kg/m<3>, ca = 4186 J/(kg*K)

therefore: ?T ~= 46.5*10<-6 >/ d ?K (?C) with d in millimeters (having chosen the elements such as water-glass we see that the temperature variation depends only on the inverse of the distance between the slides)

for example: for d = 0.3 mm there is a temperature variation of the cycle 2*?T ~= 310 ??K and if the room temperature ? of 300 ?K the ratio 2*?T / T0 ~= 10<-6 >Change in system-environment entropy.

The variation of the reversible entropy ?Ss of the system at each half cycle is: ?Ss = qe / (T0 ? ?T) - qu / (T0 + ?T) = (1 ? k/8)*Es / (T0 ? ?T) ? 7*(Es/8) / (T0 + ?T) neglecting ?T with respect to T0 results in: ?Ss ~= [(1 ? k)/8]*(Es/T0) > 0 being k<1 The cycle ? reversible so the reversible entropy variation ?Sa of the environment will be:

?Sa = -?Ss since, in a reversible system the entropy can? be considered null by idealizing the transformations of the cycle as the sum of infinitesimal transformations (from the second law of thermodynamics: dS = dqrev./T).

Therefore, the activity? of the plant according to the invention will produce? in general in the isolated system ?universe? the following variation ?STOT of total entropy:

?STOT = ?Srev ?Sirrev having indicated with ?Srev = ?Sa ?Ss the reversible entropy variation and with ?sirrev the irreversible entropy variation; but ?Srev = 0 so:

?STOT = ?Sirrev = ?Sattriti + ?Shysteresis > 0

where: ?Sattriti = positive change in entropy due to friction; ?Systeresis = positive change in entropy due to hysteresis, etc.

Example of heat removed from the environment.

Specific heat references:

- water 1 Kcal/(kg*C) 4,186 KJ/(kg*C)

- air 0.241 Kcal/(kg*C) approximately 1 KJ/(kg*C)

(1 kg of water that cools/heats by 1 degree is equivalent to approximately 4 kg of air that heats/cools)

density? air: 1.225 kg/m?

Take for example a plant measuring 1 km<2>that produces 400 MW; in this case we have that in one second the withdrawal of heat from the environment is equivalent to:

- approximately 95,500 kg of water decreased by 1 degree (95.5 m?)

- or: approximately 400 tons of air decreased by 1 degree (approximately 326,000 m? of air decreased by 1 degree)

With this example, net of dissipations, i.e. net of irreversible processes, in about 30 years the system lowers the temperature of the air having the surface of Italy as its volume for a height of 1000 meters by 1 degree. From the calculations: air = 300*10<12 >m<3 >x 1.225 kg/m<3 >= 367.5*10<12 >kg * 1 KJ/(kg*C) = 367.5*10<12 >KJ/C; necessity? calories from the environment: 400*10<3 >KJ/sec * 31,536*10<6 >sec/year = 12,614*10<12 >KJ/year; hence 367.5/12.614 ~= 30 years/C.

Verification of the surface energy used in a cycle by a UB.

Is this energy calculated as half? cycle (the one relating to the entire cycle is worth double). For energy calculation? It is necessary to determine the surfaces affected by the "wetting" variation.

In Z1 the wetted surface between the slides ? increasing due to capillarity; placing Sa this surface we have:

Sa = nuf*hc*2*(lv+d) (1)

in Z1/S1 the wetted surface passes from the height h0 to hb=0 and then returns to h0 with the passage of the volume of water ?V so there is no change in the wetted surface after the two phases 1 and 2 (there will be It is however a thermodynamic variation of energy with zero summation at the end of the two phases).

In Z2 the wetted surface decreases due to the passage of the volume of water ?V into Z1; setting Sd this surface we have:

Sd = 2*[lv + nuf(d+sp ) luS1]*(hs-h0) (2)

? I note that:

Vc = nuf*lv*hc*d, hs-h0 = Vc*(S1+S3)/(S1+S2)<2>, lv >> d

so (1) and (2) can be rewritten as follows:

Sa = 2*Vc/d (1.1)

Sd = 2*(Vc/lv) * [(S1+S3)/(S1+S2)<2>] * [(lv)<2 >+ S1+S2] (2.1)

but (lv)<2 ><< S1+S2 so (2.1) can? be simplified as follows:

Sd = 2*(Vc/lv) * (S1+S3)/(S1+S2) (2.2)

setting k = (S1+S3)/(S1+S2) with k < 1 (2.2) becomes:

Sd = 2*k*(Vc/lv) (2.3)

Place Sv = Sa ? Is the surface area increased in wettability? in half? cycle, from (1.1) and (2.3) we have:

Sv = 2*[Vc/(d*lv)] * (lv ? k*d) (3)

but lv >> k*d so (3) becomes:

Sv = 2*Vc/d (3.1) or Sd negligible compared to Sa

substituting Vc we have:

Sv = 2*nuf*lv*hc (3.2)

so the surface energy lost in a cycle is:

Ep = 2*Wa*Sv (3.3)

Ep = 8* ? *nuf*lv*hc (3.4) and remembering that hc = ? /(?*g*d) we have:

Ep = 8*[( ? )<2>/(?*g)]*nuf*(lv/d) (3.5)

(3.4) corresponds to [B4] Wa*(2*S) = 4 * ? * hc * lv compared to all slides for a cycle, i.e. Ep = 2*nuf*Wa*(2*S)

Work spent due to opposition of surface tension.

In Appendix B? Was the work spent to move the slides calculated taking into account the resistance due to the inertia of the glass and water masses and the viscosity? of the water during movement, neglecting, in addition to friction, the resistance exerted by surface tension.

Applying Laplace to the rectangular shape assumed by the water between two glass slides, we have two mutually orthogonal components of the surface tension: the one proportional to ? /R1 perpendicular to the slides and that proportional to ? /R2 parallel to the slides.

But for the rectangular shape of the R2 slides --- > ? why is it possible? say that there is no component of tension that opposes the motion because:

- the component? /R1 ? nothing as it is orthogonal to the motion

- the component? /R2 ? parallel to the motion but has intensity? practically nothing.

What if you don't want to neglect the component anyway? /R2, considering that the water-air interface is not parallel to the glass? linear but forms two right angles on the two sides of the glass which descend vertically (at most a small curvature for each of the two corners of the glass), in the worst case scenario we can consider R2 the radius of a pseudo circumference of length equal to the perimeter of the glass, that is:

2*?*R2 = 2*(hc+lv) from which: R2 = (hc+lv)/?

so the adhesion force F?p parallel to the direction of movement of the slides, due to the surface tension which opposes the movement of the surface capillary water S=hc*d orthogonal to the movement, for a pair of slides is equal to:

F?p = ( ? /R2)*S = (?* ? )* hc*d/(hc+lv)

and the work spent L?p = is:

L?p = F?p * lv = [?*( ? )<2>/(?*g)]*lv/(hc+lv)

and then in a UB cycle:

L?p (cycle) = 2*nuf *L?p = 2*nuf *[?*( ? )<2>/(?*g)]*lv/(hc+lv)

and with the chosen materials it is approximately: L?p (cycle) ~= 3.4*nuf *lv/(hc+lv) ?J In conclusion, the fact that it is the component of the surface tension proportional to ? /R1 and orthogonal to the slides and to bring water and glass back to the initial conditions, with their surface energies free from water-glass interfacial contacts, is a displacement which is opposed, among other forces, by the tension proportional to ? /R2 << ? /R1 and parallel to the slides, allows useful work to be obtained from the system.

APPENDIX D - example of system parameters spreadsheet

Claims (10)

1. Work generation system comprising at least one unit? base (UB) in turn comprising: a container (C) capable of containing a liquid mass; means (D) for internally dividing, hydraulically, said container (C) into at least two zones (Z1, Z2); movable means (V) for capillary lifting of the liquid, movable alternatively with respect to said container along a direction of motion parallel to a bottom of said container, supported or guided in a sealed manner by said subdivision means (D), so as to be suitable for be positioned predominantly in one or the other of said zones (Z1, Z2), said mobile means being configured to promote a lifting of the liquid by capillarity? and consequently the creation of a difference in liquid level between the at least two zones depending on their prevailing positioning in one or the other zone; means for alternative movement of said capillary lifting means or of said container; means (T) for hydraulic connection between the at least two zones, external to the container, comprising energy conversion means (G) suitable for transforming the hydraulic kinetic energy of the hydraulic flow generated in said connection means by said difference in height into mechanical energy and/or or electric. 2. System according to claim 1, in which said capillary lifting means (V) comprise a plurality of of laminar bodies placed side by side, arranged parallel to said direction of motion and having a prevailing development along a direction of elevation from said bottom of the container. 3. System according to claim 2, in which said laminar bodies are made of glass. 4. System according to claim 2 or 3, in which said laminar bodies have a rectangular shape. 5. Plant according to any of the previous claims, wherein said liquid mass is? made up of water. 6. System according to any one of the previous claims, wherein said energy conversion means comprise at least one impeller (G). 7. System according to any of the preceding claims, wherein said subdivision means comprise a diaphragm (D) made of Teflon? with passage windows for said mobile capillary lifting means (V). 8. System according to any of the previous claims, in which said zones (Z1, Z2) of said container have the same volume between them. 9. System according to any of the previous claims, in which one or more? of these units? (UB) are combined into one?unit? main (UP), said energy conversion means (G) being shared between two or more? of these units? basic (UB). 10. Method for generating work which involves alternatively moving mobile capillary lifting means (V) between two zones (Z1, Z2) of a container (C) containing a liquid mass, so as to position said mobile means (V) predominantly in one or the other of said zones (Z1, Z2), hydraulically isolated from each other inside the container, and cyclically cause, by capillarity, a difference in level of liquid between the two zones, and to transform the ?potential energy effect of said difference in height in hydraulic kinetic energy of a hydraulic flow generated in connecting means (T) between the areas external to the container, and in turn said kinetic energy in mechanical and/or electrical energy.