INJECTORS AND INJECTORS

This invention relates to a process of achieving a total combustion at fuels, with help of at injectors, that would reduce total emissions of pollutants in the atmosphere will be solved, in large part crisis climatic, will stop the greenhouse effect and global warming with which is facing humanity. The invention's field of use is very wide, e.g.; engines at the cars and trucks, engines of propulsion, engines for aircraft, engines of at vessels and of all transport facilities and burning for heat and electricity devices. All types of fuels used until now have been burned without control, through a burning incomplete, in itself, which has led uncontrolled pollution into the atmosphere This has led to global warming, and the greenhouse effect. The total burning of fuels, including wood, can stop global warming and the greenhouse effect. Until now, in all the books of chemistry, there was no explanation about how oxygen or air is endowed to create a total combustion. The books of specialty show a total combustion of methane CH4 with oxygen.

CH ₄ + 20 ₂ -- C0 ₂ + 2H ₂ 0 + energy (formula 1)

As you can see, it isn't explained how oxygen is delivered in order to achiev a total combustion. When the engine with internal combustion was invented by Ohwhen did not work with a total combustion. The first principle of the THERMOCHIJVnSTRY ¹ shows that the engines are produced by manufacturers at a fuel equivalent ratio φ>1 and through combustion, it produces only oxide of carbon CO, and oxide of hydrogen HO, plus fuel. The fuels which not are total combusted, are those that have contributed at the greenhouse effect and at the global warming. Practically the gases which have caused global warming and greenhouse effect, are oxides: CO, HO, NO, SiO, SO, and methane; these are more slight than air. At this moment, all engines which circulate in the world, emit in atmosphere land more than 1000 billion tons of carbon monoxide CO, and oxide of hydrogen HO. There is also known that, without air or oxygen combustion cannot be achieved. This can be seen daily when all types of fuels are burning. Technologies and solutions of combustion known did not provide

URSUT IOSIF CAMPEA AURORA SYDEN ROMULUS enough molecules of air at one molecule of fuel for a total combustion. For example, in the distribution of gas methane to consumers, until now, this was transported by compressed air without being a proportion well established between air and methane.

The disadvantage of this process is that it did not achieve a total combustion of methane and it releases only oxides CO+HO and methane.

The technical problem which my invention proposes to solve is a process which achieves a total combustion of fuels, a process which is capable to endowed, with the necessary amount of air, fuels in order to achieve a total combustion.

My invention solves this problem by using injectors with compressed air into the pipe of air, which has in centre a pipe for fuels. The compressed air will absorb the quantity of fuel used at combustion, achieving a total combustion of fuels through firing. Starting from the formula 1, the total combustion presented in the books of specialists, we reached at the conclusion that you must built an injector which inject into the burning air space molecule by molecule of fuel, because only in this way these a molecule of fuel will be surrounded by the molecules of air needed for achieved a total combustion of fuels. In general, the chemical stoichiometric equation for a total combustion of hydrocarbons with air is.

CxHy + (x+y/4) Air + 3, 76(x+y/4)N ₂ > xC0 ₂ +y/2 H ₂ 0 + 3,76(x+y/4) N ₂

(formula 2)

With this equation, one can calculate the number of molecules of compressed air stoichiometric necessary for a total combustion for a few hydrocarbons. On this base, we created Table 1. For example, for gasoline CsHu, the number of molecules of compressed air for a total combustion are calculated with formula CgHis + (8+15/4) Air +3,76(8+15/4) N ₂ >8C0 ₂ +15/2H ₂ 0+ 3,76(8+15/4N ₂ =

8+7, 50+44, 18=59,68 molecules of air (formula 3)

In Table 1 :

1). Column 1 contains several hydrocarbons and theirs chemical formula.

URSUT IOSIF CA PEAN AURORA SYDEN ROMULUS - 2). Column 2 contains proportion R, which ensure that the total combustion is achieved. proportion R=m _v /mi (formula 4) when. -m _A = air molecules

-niF = fuel molecule

3) . Column 3 contains the number of air compressed molecules stoichiometricaly needed for total combustion at one molecule of hydrocarbons.

4) . Column 4 contains diameter D _A at the pipe for compressed air and diameter DF at the pipe for hydrocarbon, stoichiometricaly, with the proportion R = D _A DF between diameters which ensures that total combustion is achieved. m _A /m _F = D _A DF (formula 5)

In order to achieve a total combustion for different fuels, we use two pipes, a pipe for compressed air having in its center mounted a pipe with fuel dimension selected from Table 1 : Column 1 for gasoline, Column 2 for proportion R = m _A /mF - 59, 68/1 from Column 3. There are 59, 68 molecules of compressed air at one molecule of gasoline, for a total combustion. Column 4 provides the diameter D _A = 59, 68 mm at the pipe for compressed air and DF = 1 mm at the pipe for gasoline. Thus, compressed air passes through the pipe D _A and will absorb gasoline from pipe DF with stoichiometric proportion 59.68/1. On the basis of pressure and diffusion 59.68 molecules of air will surround a molecule of gasoline. This is how total combustion is achieved for the first time in world.

URSUT IOSIF CAMPEAN AURORA SYDEN ROMULUS Table 1

The air compressed at 5 to GO ATM, through the vacuum created, will absorb the fuel from the pipe for hydrocarbon in

URSUT IOSIF CAMPEAN AURORA with the proportion R =m _A /mF , which ensure that is achieved a total combustion will be calculated with formula 6:

F = (P1-P2) nr ⁴ /SLi (formula 6) when: P I = pressure chosen for air compressed from 5 at∞ ATM in the pipe for compressed air.

- P2 = atmospheric pressure in the pipe for hydrocarbons.

- L = 300 mm is the minimum length at the pipe for fuel,

- r - DF/2 from Table 1 , column 4, is radius at pipe for fuel or holes of at nozzle - η = viscosity at fuel or at air.

The hydrocarbon absorbed by the compressed air, will diffusion quickly among the molecules of compressed air, because of its pressure, and will be quickly surrounded by the necessary molecules of air and achieve a rapid homogeneous of mixture for a total combustion. E:g: gasoline from the Column 1 of Table 1 : in Column 2 you find proportion R= 59.68/1, this ensures the total combustion. In Column 3, you can see that there are necessary 59.68 molecules of air for a molecule of gasoline for a total combustion. In Column 4, you can see that we need the diameter D _A - 59.68 mm at pipe for compressed air and the diameter DF = 1 mm at the pipe for gasoline, the report of them being the stoichiometric proportion R = 59.68/1. In the case of an classic internal combustion engine, the pipe for gasoline is concentric mounted in the pipe with compressed air, through a needle with a diameter DF = 1 mm, with conical top which will close at the same time the access of air and gasoline in the cylinder of the engine. In the case of injectors, the pipe for compressed air will be with 90 mm longer than the pipe for fuel. This length of 90mm creates a zone of rapid mixing on basis of the pressure and diffusion. Through the pipe with compressed air, 59, 68 molecules of air at a

The molecule of petrol will be quickly surrounded by 59.68 molecules of air and will achieve a complete mixing. When the mixture will be injected into the atmosphere for combustion, proportion R will ensure the total combustion. The total combustion gives up the following substances; H ₂ 0 +C0 ₂ +N ₂ +energy. Because C0 ₂ is a gas harder than air it will remain at the soil surface and it will be turned into oxygen by plants through photosynthesis. This total combustion is achieved with the help of few different injectors (see Fig, 1, 2, 3, 4, 5, 6, and 7) and they will be to work in an internal combustion engine in two or four-stroke (Fig. 7) or in reactors(Fig. 9.10, 1 1), and namely to supply with heated water and usual water. Some injectors work in an engine with propulsion (Fig. 3), others in welding equipment with acetylene (Fig. 5), others in welding equipment with propane (Fig. 6) and one for distributing gas methane (Fig. 4). This total combustion does not emit monoxide of carbon CO, because the fuel equivalent ratio is φ =1 or smaller. A total combustion must be in accordance with all of the data from Table 1 and it must respect the first principle of TERMOCHEMISTRY ¹ which says: combustion efficiency is η =100% when fuel equivalent ratio φ is φ = 1 or less see (Graphic 1) (Fig, 12/12). A total combustion is possible, when φ = 1, thus the fuel provide maximum energy. Graphic 1 shows the relationship between combustion efficiency when r\ _c =100% and fuel equivalence ratio φ when φ = 1. For φ<1, the products of burning are H ₂ 0+C0 ₂ + N ₂ +air +energy.

For φ=Τ, the products of burning are H ₂ 0+C0 + N ₂ +energy and fuel produces maximum of energy.

For φ>1, the products by burning are only oxides CO + HO + fuel. Engines from today are set by the manufacturer for φ> 1.

In order to achieve a total combustion, it is mandatory to:

1. Comply with all data from Table 1 including Proportion = m _A /mF _.

2. Follow formula 6 for the fuel to be absorbed by compressed air in the mass of air, through the vacuum which is created.

3. To achieve a homogeneous mixture, physical theory shows that pressure and .

URSUT IOSIF CAMPEAiN AURORA SYDEN ROMULUS ^' diffusion performe a rapid mixture homogeneous.

The advantages of this invention used in engines are the following.

- The fuel consumption will be reduced with (50-70) % owing yield 100 % of combustion.

- The performance of the engines will be upgraded with more than (30-40) %, and namely yield at 99.98 %.

- The production cost will be reduced by (30-40) %, because many components from old engines will disappear axis with cams, springs and the valves of intake, carburetor the filters pipes inlet, catalyst, etc.

- The weight of the engines will be reduced through the disappearance of many classic components parts.

- Turbulence is 200 times more than in the case of engines because of the pressure.

- Ignition and explosion is 200 times faster than in the existing engines, on basis of pressure.

- The engine yield grows up to n _m = 99.96 %, on the basis of the total combustion.

- It does not require additional inventions for the engines or any other facilities. This gives liberty of use the existing ones.

Presenting figures.

- Fig. 1 : injector for total combustion with mixture between compressed air, fuel and coal.

- Fig. 2: injector for total combustion with mixture between compressed air and crude oil.

- Fig. 3 : injector for engines with propulsion (used in place of the combustion chamber).

- Fig. 4; injector used for distribution of methane gas.

- Fig. 5: injector for welding with acetylene.

- Fig. 6: injector for welding with propane.

- Fig. 7: injector with compressed air which supplies engines with a mixture of compressed air and fuel.

- Fig. 8: cylinder's geometry with piston.

- Fig. 9: reactor with closed system to produce steam or hot water in a turbine.

URSUT IOSIF CAMPEAN AURORA SYDEN ROMULUS - Fig. 10: reactor with closed system for producing warming and hot water in apartments.

- Fig. 1 1 : reactor with closed system for producing warming and hot water in private houses.

- Fig. 12: Graphic 1 showes the THERMOCHTMISTRY ¹ principle, the relation between Combustion Efficiency, nc = 100% and Fuel Equivalence Ratio Φ, when Φ = 1 or less.

In the following, we present several examples of implementing total combustion with injectors in according with invention.

Example 1. See Fig 1

This example refers to an injector (Fig 1) for a total combustion with a mixture between compressed air, fuel (crude oil) and coal. The injector is made of three different concentric pipes with diameters chosen from Table 1. The pipe 1 for crude oil, is fitted with the tap 2, is mounted parallel with pipe 5 for coal, which is fitted with the tap 6, both pipes are mounted in the center at the pipe 3 for air compressed at 5 to 150 ATM. which is fitted with the tap 4, a flange 11. The mixing of total combustion is achieved in the area 7, which continues with angle 8 for speed, here the mixing is total, the mixture leaving the injector through the angle 9 in the shape of a cone 10 of injection (Fig, 1). When injector begins to work, the taps 2, 4, and 6 are opened of the pipes 1, 3, 5. The air under pressure flowing through the pipe 3 for compressed air, and reaching in the area 7 of mixture, in front at the pipes with crude oil and coal, because of the vacuum which is created, will absorb crude oil and coal with the proportion = m _A /niF in the mass of compressed air. Because of the pressure, diffusion is rapid, and molecules of crude oil and coal will diffuse in the mass of compressed air and will be surrounded by 1 12.16 molecules of compressed air needed for a total combustion, with the proportion R = 107,40/1 + 4, 76/1. This mixture between air, fuel (crude oil) and coal, on the basis of compressed air that passes through angle 8 of 60°, where speed grows and becomes totally homogenized, passing through the angle 9 of 90° is injected in the form of the cone 10, in the reactor or installation for combustion with coal and through ignition, it will realize a total combustion.

URSUT IOSIF The products resulting from the total combustion are H ₂ 0 + C0 ₂ + N ₂ +energy, and, when passing through the water, they will be dissolved and the result is H ₂ C03 +N ₂ . For more safe, taps 2, 4 and 6 for compressed air and fuel will be adjusted into a special facility for fuel equivalent ratio φ = 1 , and will be sealed. The quantity of fuel that will be absorbed in the mass of compressed air, from pipe 1 for crude oil and the pipe 5 for coal is calculated with the following

F = (P1-P2) nr ⁴ /$Li\ (formula 6)

This injector is easily to be manufactured in large serie. Similarly, all installation of combustion for coal, can be equipped with such injector in a year or two. The injector will reduce fuel consumption with 50 %. The diameter of pipe 3 between the angles 8 and 9, is 3 mm and this depends on the size aof the injector. The diameter at the pipe for crude oil is DF^lmm, selected from Table 1, the diameter at the pipe for coal is D _f =lmm (these have same diameters). The pipe with compressed air is with 90 mm longer which form zone 7. The diameters at the pipes mentioned above are calculated in accordance with the Column 4 from Table 1, for fuel selected from the Column 1 from Table 1. The diameter at the pipe with the compressed air is D _A = 1 12, 16 mm which results from the following: diameter D _A = 107, 40 mm selected from Table 1 from Column 4 for crude oil from Column 1 , plus D _A = 4, 76 mm from Column 4 for coal Column 1 in Table 1. For a higher quantity of coal, Column 1 , will consists of 10 times bigger coal pipes, so that the diameter will be DF ⁼ 10 mm, and it is chosen from Table 1 , in Column 4. For a total combustion, the proportion is R = 107, 40/1 + 47, 60/1 = 154, 00/1 =D _A /D _F The quantity of fuel which will be absorbed from the pipe for fuel, by the compressed air through the vacuum created in the pipe of fuel will be calculated with formula 6, namely; F = (P1-P2) -πι· ⁴ / 8ίη L= 300 mm is minimum for a correct absorption and r = DF/2.

URSUT IOSIF Example 2. See Fig 2

This examples refers to an injector for total combustion with mixture between compressed air and fuel (crude oil), intended for all fuels mentioned in Table 1. The injector consists of two concentric pipes, pipe 1 with crude oil is fitted with a tap 2 (Solenoid), the pipe 1 is mounted in center of pipe 3 for air compressed at 10 at 250 ATM, or more, and fitted with solenoid 4, the mixture for a total combustion is achieved in zone 7, which finishes with angle 8 for speed; this is the region where the mixture is totally mixed, the mixture will be injected through angle 9 in the shape of a cone 10, a flange 11 (Fig. 2). A molecule of crude oil needs 107,40 molecules of air for total combustion according to Table 1

From Table 1 ;

- we select fuel from Column 1

- proportion R =m _A /mF =107,40/1 is selected from Column 2

- from Column 3, we see that 107, 40 molecules of compressed air are needed for the total combustion of one molecule of crude oil .

- from Column 4, we select both diameter D _A = 107, 40 mm at the pipe for compressed air and diameter DF = 1 mm at the pipe for fuel (crude oil) (when designing the injector).

The pipe 3 with compressed air is 90 mm, longer than the pipe 1 of crude oil, and creates zone 7. Similarly with the injector from example 1, the resulting products are H ₂ 0 + C0 ₂ + N ₂ + energy

When the injector begins to work, at the same time, solenoids must open the taps 2 and 4 at the pipes 1, 3. The air having the pressure of 10 to 250 ATM or more, will transport 107.40 molecules. When 107, 40 molecules of compressed air reach the zone 7 of homogenized mixture, on basis of the vacuum created in front of the pipe 1 for crude oil, will absorbe, at the same time, o molecule of crude oil from pipe 1. This mixture reaches the angle 8 of 60° for speed, then passes in the angle 9 of 90° and will be injected in reactor in the shape of the cone 10. The quantity of crude oil which will be absorbed by the air for a total combustion calculated with formula 6; F = (PI - Ρ2)·πΓ /8ίη.

URSUT IOSIF CAMPEAN AURORA SYDEN ROM Example 3. See Fig 3

This example refers to an injector for total combustion for engines with propulsion, which will replace the combustion chamber known so far. The injector is composed of a pipe 3 for compressed air, four pipes 1 for the fuel isooctane, arranged at 90° one of the other, fitted with a particular tap 2, the angle 8 of speed and the angle 9 for injection in the shape of a cone 10. The mixture of combustion is realized in zone 7 finished with theangle 8 for speed, the mixture leaving the injector through the angle 9 (Fig 3). The combustion chamber for the engines with propulsion can have the following dimensions: the length, for example, is L = 2149 mm and a diameter D = 1252 mm. We select fuel isooctane from Column 1 in Table 1 and from Column 4 we select D _A = 64 mm and DF = lmm. The proportion R = 64, 00/1 can be found in Column 2, and D _A /D _F: must be equal with D/x diameter D = 1252mm. D _A /D _F = 64, 00/1 and D/x = 1252/x, namely D _F = x = 1252/64, 00 = 21, 04 mm.

Consequently, according to Table 1, Column 4, the diameter at the pipe 3 for air is D _A = 1252 mm, and the diameter of the pipe for isooctane DF = 21 , 04 mm. In Column 2, the proportion R = 1252/21, 04, and Column 3 shows the result: we need of 1252 molecules of air for the total combustion of 21, 04 molecules of isooctane. In order to achieve total combustion we must deduct the cylinder diameter occupied of rotor DR = 400 mm from the diameter D _A = 1252 mm. The result is: D _A i = 852 mm, so that the proportion R = 64, 00/1 is equal with 852/x , 64,00/1 = 852/x, so x = 852/64, 00 = 13,3 1 mm, and so D _F = 13; 31 mm.

The real diameter of the pipe 3 for compressed air will be D _A = 1252 mm.

The diameter D _A i - 852 mm is used to calculate the diameter D _F = 13, 31mm for the pipe 1 for fuel, in order to achieve total combustion.

The engines with propulsion will be designed in accordance with Table 1. The compressor at engine with propulsion must ensure air at a minimum of 5 ATM pressure.

The pipe with compressed air under pressure in the injector becomes the chamber of combustion at an engine with propulsion. In the presented example, L - 2140 minimum mm, D _A =

1 ), than we deduct

example).

URSUT IOSIF This is how the diameter D _A i - 852 mm chosen imaginary from the initial calculation for the diameter of the pipe for fuel. D _A = 1252 mm remains unchanged.

The pipes 1 with fuel isooctane with the length L = 300mm and the diameter DF/4 =13, 3 1/4 mm is selected from Column 4 of the Table 1 . Since the injector for engines with propulsion are provided with four pipes 1 of isooctane, the diameter at each pipe will be DF/4 = 13, 31/4 = 3, 32 mm and these will be fitted up on the circumference of the pipe 3 for compressed air. The pipes of fuel will be fitted up at the distance of 20 mm from the start pipe 3 for compressed air and at the depth of 20 mm, because that is exactly where the flow of compressed air will enter, under pressure. The tap 2 fitted with the pipe 1 for isooctane regulates the flow which is absorbed with a rotation of 360° for the total combustion,.

At the opening taps 2, isooctane must flow easy in the injector. The reservoir with isooctane must be set up above engine in case of an airplane. The length of the zone 7 is of L = 2130 mm for the pipe for fuel up to the turbine in order to achieve a mixture for total combustion The angle 8 is 50° for the mixture prepared in zone 7 and the mixture is totall homogenized. The mixture is than injected through the angle 9 of 30°, than the mixture is ignited to create propulsion of the engine. The quantity of fuel which is absorbed by the compressed air from the pipe for fuel will respect the proportion R = m _A /mF (Column 2 in Table 1 ) and it is calculated with

formula 6. F = (P1-P2) ·πι ^*4 /8Ι_,η. When:

- PI is the pressure (5 at 150 ATM or more) of the compressed air and it is selected.

- P2 is the atmospheric pressure in the pipe for fuel .

- r is radius of the pipe with fuel DF = 3, 32 12 = 1 , 66mm, and because there are four pipes of fuel D _F = 13, 3 1/4 = 3, 32 mm, r = D _F /2 = 3, 32/2 = 1 , 66 mm (Column 4 in Table 1 ).

- L is the minimum length of the pipe for fuel and L = 300 mm.

- η is the viscosity of the fuel or of the air.

Using this injector, the pumps for fuel at the airplanes disappear.

URSUT IOSIF CAM PEA N AURORA SYDEN The consummation of fuel will be 50% lower and the yield engine r| _m will be 98%, because there is total combustion. When the engine of the airplane begins to work, the compressor will send compressed air with high pressure in the pipe 3 of the injector. At the same time, taps 2 on pipes 1 with isooctane will open. The compressed air enters into the pipe 3 of the injector of the engine, with propulsion, and will create vacuum in front of the pipes 1 for fuel and than they will absorbe isooctane in the mass of the air. At the same time, the high pressure will scatter the molecules of fuel among the molecules of compressed air into the zone 7 of mixing. The molecule of isooctane will be surrounded by the molecules of air with the proportion R = 64, 00/1. In the moment of ignition the total combustion of the isooctane is achieved and the propulsion is maximmum.

Example 4. See Fig 4.

This examples refers to the transport of methane gas CH4 with the proportion R = 10, 52/1 at the consumers, where a total combustion is achieved (Fig. 4). In the station for distribution of the methane gas CH4, an injector is fitted up on the pipe of distribution (Fig 4). The pipe 3 for compressed air in the injector has a diameter D _A which is the same with the diameter of pipe for distribution. The compressed air will absorbe (because of the vacuum created) the methane gas CH4 in the mass of the air with the proportion R =10, 52/1. Thus, one molecule of the methane gas needs 10, 52 molecules of air (Table 1, Column 3) for a total combustion. The air is compressed at a pressure between 5 at 150 ATM. The proportion R = 10, 52/1= m _A /mF = D _A /DF for a total combustion is calculated with

fonnula 6:: F = (P1-P2) r ⁴ /SLr\. When:

The pipe 3 with compressed air having the diameter D _A = 10, 52 mm is fitted up in the center of the pipe 1 for methane gas CH ₄ and has the diameter DF = 1 mm. This injector will be fitted up on the pipe for the distribution of methane. In order to deliver a larger quantity of methane, e.g. 100 times more than ussual methane, it is necessary that the distribution pipe has the diameter D _A = 10. 52 xlOO =1052 mm and the pipe 1 of methane has the diameter DF = 100mm. A total combustion and delivery of the methane is achieved with ratio R = 1

air compressed at a pressure of minimum 5 ATM.

URSUT IOSIF CA PEAN AURORA

The products resulting from the total combustion of the methane are: H20 + C02 + N2 + energy.

Example 5. See Fig 5.

This example refers to a total combustion process in the case of welding with acetylene (Fig. 5). The device for welding is provided with an injector of welding with acetylene, and the process will respect the THERMOCHIMISTRY ¹ principle which says: combustion efficiency is η =100% when fuel equivalent ratio φ is φ = 1 or less, and the results of the total combustion are only : H ₂ 0 + C0 ₂ + N ₂ + energy.

For a total combustion, one molecule of acetylene needs 12, 40 molecules of compressed air, between 5 at 100 ATM (Table 1 , Column 3). The air with high pressure (form the vacuum created) absorbs acetylene from the pipe 1 for acetylene in the mass of the air with the proportion R = 12, 40/1 = m _A /niF. The quantity of acetylene absorbed is calculated with

formula 6: F = (P1-P2) ·πι· /8Ι,η. When:

- P I is the pressure at compressed air and it is 5 at 150 ATM or more (selected).

- P2 is the atmospheric pressure in the pipe for fuel.

- r is radius of the pipe for acetylene and r = DF/2 (Table 1, Column 4).

- L is the minimum length of the pipe for acetylene, L = 300 mm.

- η is the viscosity of the fuel (acetylene) or air.

The acetylene which is absorbed by the compressed air in its mass will be diffused rapidly among molecules of air. The molecule of acetylene will be surrounded of 12,. 40 molecules of compressed air with the proportion R = 12, 40/1 and the mixture is total homogenized.

The diameter of the compressed air pipe should be DA = 12, 40 mm and the diameter of the pipe for acetylene DF = 1 mm. The proportion between these two pipes is R = 12, 40/1 = D _A DF. The pipe of compressed air will be with 90 mm longer than the pipe for acetylene. This length creates a zone 7 of quick diffusion and total mixing. When the molecules of compressed air at pressure of minimum 5 ATM reach in front of the pipe for acetylene this is absorbed in the mass with formula 6 and ensure a total combustion.

U RSL IOSI F CAMPEAN AU RORA SYDEN ROMUL The resulting products of the total combustion of acetylene are. H ₂ 0 + C0 ₂ + N ₂ + energy.

The welding is achieved with the help at different tipes of nozzles for welding with dimensions known and available on the market. In the compressed air pipe 3 a tap 4 is fitted with the end in the shape of a beak 12 for speed; the injector also has an angle 8 of 14°, an exit hole 13 of 0, 9 mm. A pipe 3 is concentricaly fitted up in the pipe 1 of acetylene, fitted with a tap 2. The injector also has a zone 7 with the length of 90 mm for creating the mixture between air and acetylene and a shape of a cone 10 for the flame of welding (Fig. 5). The tap 4 fitted on the pipe 3 for compressed air is the first to be opened and the tap 2 on the acetylene pipe 1 is also opened at maximum; the compressed air will absorb acetylene with the proportion R = 12, 40/ 1 which ensures a total combustion.

Example 6. See Fig 6

This example refers to a process of total combustion at the welding with propane (Fig. 6). The device for welding is provided with an injector of welding with propane, which realizes the total combustion of propane according to the THERMOCHTMISTRY ¹ principle which says: combustion efficiency is η = 100% when fuel equivalent ratio φ is φ = 1 or less and the resulting products of a total combustion are only: H ₂ 0 + C0 ₂ + N ₂ + energy.

For a total combustion, one molecule of propane needs 25, 80 molecules of compressed air (between 5 at 100 ATM) (Table 1 in Column 3). The air with high pressure absorbs propane from the propane pipe 1 in the mass of air with the proportion R = 25, 80/1 = m _A /mF. The quantity of propane absorbed is calculated with

formula 6: F = (P1-P2) ·πι· ⁴ /8Ι,η. When:

- P I is the pressure of the compressed air (5 at 1 50 ATM or more) and it is selected.

- P2 is the atmospheric pressure in the pipe for fuel .

- r is radius of the pipe for propane and r = DF/2 (Table 1 , Column 4).

- L is the minimum length of the pipe for propane, L = 300 mm.

- η is the viscosity of fuel propane or air.

URSUT IOSLF CAMPEAN AURORA SYDEN ROMULUS The propane which is absorbed by the compressed air will rapidly diffuse among the molecules of air. One molecule of propane will be surrounded by 25, 80 molecules of compressed air with the proportion R = 25, 80/1 and the mixture is totaly homogenized. The pipe for compressed air should have the diameter D _A = 25, 80 mm and the pipe for propane should have the diameter D _F = 1 mm and the proportion R = 25, 80/1 = D _A /D _F .

The pipe 3 for compressed air will be with 90 mm longer than the pipe 1 for propane. This length of 90 mm creates a zone 7 of diffusion for a total mixing. When molecules of air are compressed at a pressure of minimum 5 ATM in the front of the pipe for propane and these will absorb propane in the mass of air according to formula 6, ensuring a total combustion.

The resulting products of the total combustion of propane are: H ₂ 0 + C0 ₂ + N ₂ + energy. The welding is achieved with the help of some different tipes of nozzles with dimensions known and available on the market, which will use compressed air of at 5 at 150 ATM.

The pipe 3 for compressed air, fitted with a tap 4, has at end the shape of a beak 12 for increasing speed, with an angle 8 of 14° and with the exit hole 13 of 0, 9 mm. The compressed air pipe 3 is concentricaly fitted with a pipe 1 for propane, fitted with a tap 2; the injector also has a zone 7 with the length of 90 mm for realizing the mixture between air and propane and the flame of welding 10 has the shape of a cone (Fig. 6). The tap 4 fitted on the pipe 3 for compressed air is first opened and then tap 2 fitted on the pipe 1 for propane is opened at maximum. Example 7. See Fig 7.

This example refers to a solution for total combustion of fuel, which uses an injector intended to work for delivering the mixture of compressed air and the fuel at engines in two and four-stroke; the solution can use all types of fuels presented in Table 1.

The injector will inject the mixture of compressed air and fuel, of e.g. gasoline, in the cylinder of the engine at the same time, with the proportion

ensures the total combustion at gasoline in engine.

U RSUT IOSI F CAMPEAN AU RORA

The injector is fitted up on the cylinder head of the engine instead of the admission valves. According to Table 1, the tap 2 fitted with the pipe 1 for gasoline, will be regulated for fuel with the equivalence ratio φ when φ = 1.

Each cylinder of the engine will be equipped with such an injector. The injector body 16 is fitted on the cylinder head 14 with the screws 15. The pipe 3 for compressed air is uniting with pipe 27 for compressed air, the pipe 1 for gasoline is fitted with the tap 2, the gasoline flow is regulated with the special tap 17. The injector also has an solenoid 18 which commands the needle (19) of the injector, provided with a channel 20 of 1 mm in the middle, the needle also having four holes 21 at the top.

The solenoid 18 raises and lets down the needle, which closes and opens at the same time the entrance of the mixture between compressed air and gasoline in the cylinder 28 of the engine. There is also a support ring 22, a spiral arc 23, a lid 24, a security ring 25, two tightness rings 26 and a pipe 27.

The mixture of compressed air and gasoline is injected in the cylinder 28 through a spout 29 when the needle opens with the help of the solenoid 18 (Fig. 7). The hole 20 in the middle of the needle has a diameter DF = 1 mm. The pipe 27 for compressed air has a diameter D _A = 59, 68 mm + 10 mm and the diameter of the needle is 10 mm, the diameter of the compressed air pipe 27 is D _A = 59, 68 + 10 = 69, 68 mm.

The four holes 21 create nozzle in the top of the needle 19 and the diameter of the holes 21 is DF 4 = 0, 25 mm; the holes are placed at an angle of 90° one from each other, on the circumference, and they distribute homogeny gasoline in the mass of air with the proportion R = 59, 69/1, ensuring a total combustion.

The needle 19 of the injector has a conical shape at the top. The pipe 27 has a cone 30. The needle closes and opens at the same time the entrance of mixture between compressed air and gasoline in the cylinder of the engine.

The pipe 3 for compressed air has a volume equal with 2* (Vc + VD) (Fig: 8). The diameter D _A of the pipe for compressed air must be selected from Column 4 of the

The tap 17 opens with a rotation of 270° when φ = 1, which corresponds to a maximum rotation, with the help of a device when the accelerator gas pedal of the engine is pressed. When this pedal is released, the mechanical device and the spiral arc will rotate back the tap 17, into the initial position, when φ < 0, 4.

The fuel pipe 1 has a tap 2 fitted on it and the diameter of this pipe is DF = 1mm (Column 4, Table 1 for the coresponding fuel). The tap 2 will be adjusted for φ=1, so to ensure a total combustion.

The solenoid 18 raises and lets down the needle 19 of the injector, when it is electricaly powered. The needle 19 of the injector has a hole 20 in the middle, with the diameter DF = 1 mm. The needle 19 of the injector has a conical shape 30 at the top. The four holes 21 create nozzle at the top at needle, with the diameter DF = ¼ = 0, 25 mm (Column 4, Table 1), and so the fuel is absorbed from the pipe 1 in the mass of compressed air with the proportion R = 59, 68/1, ensuring a total combustion.

The conical angle 30 of the pipe 27 is the same as the conical top of the needle 18 of the injector. The quantity of fuel which will be absorbed by compressed air in the mass of the air, for a total combustion, will be calculated with

formula 6. F = (P1-P2) ·πι· ⁴ /8Ι_η. When:

- PI is the pressure of the compressed air (5 at 150 ATM or more) and it is selected.

- P2 is the atmospheric pressure in the fuel pipe.

- r is the radius of the pipe for gasoline and r = DF/2 (Table 1, Column 4).

- L is the minimum length of the pipe for gasoline, L = 300 mm.

- η is the viscosity of the fuel gasoline or air.

Example 8. See Fig 8

This example refers to a cylinder of an classic engine with piston and the following components: 31 Vc = Clearance volume, 28 Vd = Displacement volume, 33 B = cylinder's diameter, 34 S = stroke, 35 TDC - 'pistons peak dead center" (the upper dead centre, 36 BDC = "pistons bottom dead center" (the bottom dead centre, 37 R = acceding piston stakes length,

crank offset, 40 s = the pistons ' position.

URSUT IOSIF CAMPEAN AURORA When the mixture of compressed air and fuel is injected into the cylinder, it will occupy the entire volume (Vc + V _D ) and will be filled in very quickly andin an effective mod. The air mixed with petrol compressed at least 200 ATM or more, is then injected into the cylinder, and produce a maximum turbulence, providing a perfect homogeneity because of the pressure and diffusion between compressed air and petrol. The mixture is vaporized quickly, and causes the explosion and the total combustion rapidly, obtaining maximum of energy and maximum of rotation at the piston of the engine.

The results of this total combustion are: H ₂ 0 + C0 ₂ + N ₂ + energy. On the basis of compressed air at 200 ATM, the injector starts working when a cycle of intake begins in the engine, the solenoid raises the needle of the injector, then, when compressed air reaches in the front of the holes of at needle, because of the vacuum created. The gasoline is absorbed in the mass of compressed air with the proportion R = 59, 68/1. The pressure of compressed air and the diffusion makes that 59, 68 molecules of compressed air surround a molecule of gasoline when injected through the spout in the cylinder.

At the same time the cycle of compression begins, the solenoid liberates the needle. One spiral arc presses the needle down on the the cone of the pipe 27 which closes both compressed air and gasoline. The spiral arch spiral is supported by a security (safety) ring. Therefore, this high pressure along with diffusion achieves the mixture between compressed air and fuel in the cylinder of the engine. This is why the mixture is 200 times better homogenized than at the known engines. When, this high pressure at compressed air absorbs fuel through the holes of the needle and the mixture is injected through the spout in the cylinder, it creates a high turbulence, which is 200 times higher than at the known engines.

When the piston of the engine reaches the upper dead centre (Fig 8), compression is achieved and the explosion begins. The ignition and explosion are immediately and 200 times faster than at known engines. The combustion is total and will

In order to ensured total combustion when fuel equivalent ratio is φ = 1, the pipe for fuel will be fitted with a tap which will be adjusted for φ = 1 in a special installation (device) sealed to φ = 1. This ensures double guarantee that it is carried out a total combustion, and under a total control, with the provision that the injector is manufactured in accordance to Table 1.

On the body of the injector, in the place where the fuel pipe is connected, a tap is fitted and then linked with a device at the accelerator of the vehicle. The tap is opened for a φ <0, 40 and this coresponds to a minimum rotation. When the accelerator is pressed, the tap rotates with 270° and opens, then rises at φ = 1, which coresponds to the maximum rotation of the engine.

When accelerator is released, the spiral arc and a device will make the tap to rotated at φ < 0, 40, corresponding to the minimum rotation of the engine. Crank angle will be recalculed. The pipe of fuel unites the reservoir of fuel with the injector. Thus, the reservoir of fuel will be mounted above at engine, so that fuel will be flowing freely in the injector. This is why this type of injector can be used at all types of engines that do not need the following components: valves of intake, filter of air, carburator, fuel pump, axis with cams for valves of intake and other elements.

The new engines, which will use my invention do not need a catalyst for that they do not release oxides such as CO, HO, NO; these engines release only H ₂ 0 + C0 ₂ + N ₂ +energy, through a total combustion of all types of hydrocarbons. When using hydrogen as fuel, the result of the combustion is only H ₂ 0 + energy. The water which is obtained can be recover and used in the battery, laboratories and in pharmacies. The injectors will be manufactured using the current technology. The injector can be manufactured .in big series, and in 10 years all new and old engines can be equipped with these injectors.

Example 9, 10 and 11. See Fig 9, 10 and 11

This example refers to a reactor with a closed system without emissions released into the atmosphere, for producing steam or hot water in turbines, respective to produce usually hot water for apartments, villas or private homes and alse^¾y r ^ heating them.

URSUT IOSIF CAMPEAN AURORA

The results of the total combustion (Fig 2) are: ¾0 + C0 ₂ + N ₂ + energy, and these are dissolved in water in the closed system. These reactors are used in closed system and will produce steam or hot water with temperature of 100° C or more, at a pressure of 200 ATM or more. The water is heated in direct contact with the flames of the fire created by the injector. These reactors can be used for total combustion in closed system at all types of hydrocarbons from Table 1, using an injector as presented Fig 2.

The reactors are manufactured with double walls and the water from the closed system is also used for cooling down them.

The reactor presented in Fig 9 is used to produce steam at the pressure of 200 ATM or hot water at pressure of 200 ATM for turbines. The reactor has a radius e.g. of 2 m and the double walls have the length of 7 m. The reactor will utilize an injector as in example 2 (Fig 2) which works at the pressure of 250 ATM in order to produce hot water or steam at pressure of 200 ATM. The reactor presented in Fig 10 also utilizes an injector as in example 2 (Fig 2) which works at the pressure of 15 ATM for heating apartments and institutions and for producing hot usual water.

The reactor presented in Fig 11 also utilizes an injector as in example 2 (Fig 2) which works at pressure of 5 ATM for heating and to producing hot usual water for villas or private homes.

For the total combustion of the hydrocarbons, the injector will inject a flame with the pressure of 250 ATM inside the reactor. Over this flame, water is inject using a compressor with 210 ATM, through a sprayer. The water takes over the entire heating of the flame, and will turn the water into steam or hot water. Inside the reactor temperature and pressure grows because combustion is total and continue. The reactor contains a pipe which leads hot water or steam under pressure into the turbines. The pipe is fitted with a valve which is opened at a pressure of 200 ATM, thus the injector works with 50 ATM , and the compressor will pressed water at 10 ATM. In the turbine, water is conducted for cooling, and will reach the compressor and, so, the system is closed. The spiral pipe for heating the daily consumed water is located in the storage room.

URSUT IOSIF CAMPEAN AURORA The reactors presented in Fig 9, 10 and 1 1 have the form of a cylinder with double walls 41 and 42. At the ends, two oval deadlightsl are mounted through welding. At the upper end there is a pipe for feeding with water, for closed system, through the space between the walls of the reactor and water water is the cooling agent. The pipe has a tap which is closed when the system is filled with water.

The water fills the space between the walls 41 and 42. The pipe comes out from the reactor at the bottom part through a special pulverizer 43 with angle of 150 °, having a tap 44 that opens in case of a damage. The water at pressure of 210, 10 and 5 ATM, through a pulveriser 43 is pulverized over flames of the injector 45. The flames which burn inside the reactors 9, 10 and 11 have different pressures: 250, 15 and 7 ATM. The water pulverized over flames will take all the heat of at flames and will create a pressure in the storage room 46 that has a volume of approximate 25 m ³ .

When the pressure in the reactors 9, 10 and 11 reaches 200, 8 and 2 ATM, the evacuation valve 48 opens; this valve is fitted on the transport pipe 50 toward the turbine, apartments, villas or private homes. The pipe is fitted with a damage tap 49. This pipe, in case of reactor 9 transports water at 200 ATM to the turbine which will be put in function.

In case of reactor 10, the pipe 50 will transport hot water for heating the apartments.

In case of reactor 11, the pipe 50 transport the water for heating the villas or private homes. The water used at turbines, apartments and villas, passing through the return pipe 51, comes back into the double walls 41 and 42 at reactors 9, 10 and 1 l.The tap 56 on the water pipe 55 fills the closed system. Besides the double walls, the water pipe 55 also feeds the compressorclosing the system.

The storage room 46 of the reactors 9, 10 and 11 a spiral pipe 47 with the diameter of 100 mm, 100 mm, and 10 mm for heating usual water. The water enters the compressor which will press the water up to 210, 10 and 5 ATM. The water flows in the spiral pipe 47 located in the feeding pipe 63 having an opened damage tap 64, and then goes to the pump 62 which presses the water through the spiral pipe 47 spiral and leads the water to the distribution pipe 61 toward the apartments, institution and villas; the water comes back through the return pipe into the pump 62.

URSUT IOSIF CAMPEAN AURORA In the injectors of the reactors 9, 10 and 11 the evacuation valve 48 opens at 200, 7, or 3 ATM. The surplus of water will be taken out from the closed system in the reservoir 54 which is coupled at valve 53 mounted on the return pipe 52. The product from the reservoir 54 can be used to irrigated plants which through photosynthesis can produce oxygen 0 ₂ .

The reactor 9 (Fig 9) has a cylindrical form and contains 16 injectors 45 grouped on two or more rows for a total combustion of hydrocarbons. The number of rows with injectors depends on the size of the flames. The evacuation valve 48 opens at the pressure of 200 ATM. The damage tap 49 on the supply pipe 50 opens and lesds the steam to turbine 51: The dimension of the supply pipe 50 is calculated on the basis of neccessary water or steam for the turbine.

The reactor also contains the return pipe 52, a valve 53 connected to the reservoir with a capacity of 25 liters 54 that collects H2C03 +N2, a filling pipe 55 with the tap 56 which is opened. The reactor also contains a shower bath 43 with the angle of 150°, one compressor 57 for 210 ATM. The double walls 41 and 42 are fitted with the water pipe 55 having the tap 56 opened, which will fill the closed system of the reactor with water. The tap 56 will be closed after the system is filled with water. The reactor also consists of a storage room 46 which has 25 m ³ or more. The reactor is hermetically isolated in order to ensure an maximum yield n _m = 99, 98 % and it is equipped with a system of alarm in the case that the maximum temperature and pressure is exceeded.

The reactor from Fig 9 is used to produce steam under pressure of 200 ATM or hot water and must be calculated to resist at a pressure of 260 ATM.

The reactor from Fig 10 is used for heating apartments and institution and to produce usual hot water; it must be calculated to resist at a pressure of 40 ATM. With my invention only one litre of fuel oil is needed for heating a quantity of 120 litres of water at 100° C (in direct contact with the flames). Consuming 720 litres of combustible crude oil in the reactor acoording to my invention (Fig 9, 10) under a total combustion, in 24 hours, the reactor can warm 86400 litres of water at 100°C. In addition, 2400 litres of H2C03+N2 are produced through a total combustion of crude oil.

URSUT IOSIF CAMPEAN AURORA At this moment, the total quantity of water in the closed system is, therefore, 88800 litres, This means that we need only $130 US for heating of 3100 apartments. The price will be very cheap, 1300/3100= 0,40 dollars for an appartment. But 2400 liters must be taken out from closed system for no need to be loaded.

The reactor in Fig 11 is used for daily heat and hot water in villas and private homes and must be calculated to resist at a pressure of 10 ATM. The reactor consumes two litres of crude oil, e.g., in 24 hours and will heat up 240 litres of water in private homes.

The dioxide of carbon C0 ₂ is a gas more difficult than air and when dissolved in water it results: H2CO3 + N ₂ which is a natural fertilizer for plants and forests. The molecular weight of the Air is 28, 966 and of the Carbon Dioxide, C0 ₂ = 44.01.

The gases lighter than air which have created the greenhouse effect and global warming are:

Acetylene = 26.04

Ammoniak = 17

Carbon Monoxide, CO = 28.011

oxide of hydrogen HO = 17,01

Hydroxyl, OH = 17.01

Methane, CH4 = 16,044

Natural Gas = 19.00

Neon, Ne = 20.179

Water Vapor - Steam, ¾0 = 18.02

URSUT IOSIF CAMPEAN AURORA SYDEN ROMULUS