IT202000002731A1 - Liquid gas system for the production of electrical energy from thermal energy at room temperature. - Google Patents

Description

Real description

Reasons for choosing the carrier fluid. Choosing a gas such as air or nitrogen has several advantages. They are gases present in large quantities, and before being inserted into the circuit they are purified of CO2 and other polluting gases, they are easily liquefied, and not asphyxiating (if air is used). Their mean delta T between room temperature and liquefaction temperature? quite large (293 k-97.2 k) and have good capacity? thermal with Cp = 1 kj / kg x? C. The project that will come? described far? with reference to the Nitrogen fluid only, but also for Air the same considerations apply (nitrogen is a little lighter than air and this will be taken into account in the isobar expansions on T1 and T2 turbines). The possible presence of liquid oxygen, in case you want to use liquid air, does not pose particular safety problems to the system, why? during evaporation (see the operation of the VCP valve in the drawing) the liquid air would evaporate without leaving liquid residues. In any case, if necessary,? It is possible to mount a small hydraulic pump, take liquid oxygen from the bottom and push it upwards (in the external sector in Sbl) pulverizing it, or extract it for other purposes (extraction for therapeutic use or for other purposes).

Description of the drawing n? 1

Here the device groups are essentially 7:

The N1 tank in which? high pressure nitrogen present;

The Sb1 heat pipe with the liquid nitrogen deposit at the bottom indicated by liq; The proportional control valve (in which the flow rate always remains the same even if the pressure changes) VCP; The two turbines T1 and T2 with their respective generators G1 and G2; The compressor P1; The Sb2 exchanger incorporated in the Rad radiator which has the task of drawing external thermal energy with the denatured alcohol fluid;

The electronically controlled valves V1-V6 with the pressure gauge M1.

The liquid nitrogen must be inserted (or be present) in the heat pipe Sb1 before each start, otherwise the system cannot? work. Is it evident that the refrigeration circuit? sealed (and insulated) like a normal circuit present in standard refrigerators to avoid the continuous loss of gas which? under pressure, and if not?, the necessary refill of the nitrogen tank N1 is needed.

Simplified description of the circuit: The system, reading the various sensors, sends liquid gas (see drawing) to the external sector in Sb1 with the VCP valve (VCP takes liquid gas at critical pressure and critical temperature at about 101 k) in the 8 Bar storage and sends it to the external sector where the pressure? at 6.8 Bar making it evaporate. This evaporates why? its temperature (101 k)? greater than its critical T (at 6.8 Bar? T crit = 97.2 k) and exchanges energy with the fluid (internal sector) pi? heating in descent (T max input Sb1 = 276.4 k and T min = 101 k). It climbs up in the external sector in Sb1 and enters T2 (produc. En. Mecc. And electrical. On G2) then in the compressor P1 (recompression), enters the exchanger Sb2 to recover heat from the more fluid. hot (Ethyl alcohol. Denatured in the int. sector), which having withdrawn thermal energy from the environment? able to transfer heat to the gas on the way up. On leaving Sb2, it enters T1 (mechanical and electrical production on G1), then exits this to return to Sb1. The plant? insulated and the frictions of the various devices are discharged into the pipes. Compressed gas losses (minimal but unavoidable) are compensated in real time by the high pressure tank N1 (see drawing) with the M1 pressure gauge electronically adjusted to 8 Bar. = 0.6 with minimum air pressures of 6.8 Bar and maximum of 8 Bar, while the flow rate is established for simplicity? at 10 gr / sec.

Description of the components pi? important: The exchanger or rather the Sb1 heat pipe? composed of two sectors (see drawing). The internal one in which the hot fluid enters with T = 276.4 k (this value and the value 97.2 k will be demonstrated later) and the external one in which the vapor-gas rises upwards. The two finned sectors develop very large surfaces and exchange energy with fairly large medium delta T (about 80 k). The internal pressures are equal to 8 bar for the downward fluid (stabilized by M1), while for the upward flow the VCP valve gives it 6.8 (with an average flow rate adjustment always of 10 gr / sec) and a temperature equal to approximately 97.2 k (the compressor P1 draws-intakes from T2, stabilizing the pressure at 6.8 Bar together with the regulating valve V5 and compresses it at 8 Bar, sending it towards the exchanger Sb2).

These values depend on some choices dictated by the gas in question (see gas table Nitrogen, generalized graph, and graph no. 48). At 8 bar its reduced pressure = 8/34 (press crit at 1 Bar) = 0.235 which corresponds to a reduced T = 0.8 (see general compressibility graph.) And therefore critical T = 0.8 x 126, 2 k (critical temp at 1 Bar) = 101 k, while at 6.8 bar it is 97.2 k (at 6.8 Bar? P red = 0.2 which corresponds to a T red = 0.77 then? 0.77 x 126.2 k = 97.2 k).

Now, when referring to a heat pipe, the Sb1 exchanger behaves exactly like a heat pipe. The liquid leaving the VCP re-enters the external sector in Sb1, decreasing from 101 k to 97.2 k if its pressure drops to 6.8 Bar. The liquid in this case is transformed into vapor, trying to recover thermal energy from the other sector (the internal one) that at that moment needs? to liquefy. And so what? even with a delta T of only 3.8 k the fluid in the internal sector liquefies. In fact, by calculating on the generalized gas graph the respective liquefaction energies at critical pressures and critical temperatures, we find that at 8 Bar the enthalpy of liquefaction is 560 joul, while at 6.8 Bar it is 765 joul (in reference to the areas included between the critical temperatures already calculated and the critical T value = 1 considering that nitrogen has liquefied enthalp. = 197.5 kj / kg). It follows that the restart upwards takes the enthalpy of evaporation pi? large (765 j) than that of liquefaction, with a constant delta T = 3.8 k and temperatures also constant of 101 k and 97.2 k up to the value with Z = 0.8 and Z = 0.83.

The liquid-vapor change therefore brings with it a decrease in temperature essentially desired by the first thermodynamic principle. The potential energy in fact, present in all related systems (just like in liquids)? negative with respect to the kinetic energy of the molecules, and the detachment (due to the loss of the liquid state) must necessarily be compensated by the kinetic energy with a decrease in temperature that is - 24.3 k: (97.2 k - 101 k) = - 3.8 k for the pressure decrease and - 20.5 k for the ent. of evap. : (560 j- 765 j) / 10 gr = -20.5 k which added to the -3.8 give - 24.3 k) (see generalized graph).

Description and constructive example of the Sb1 exchanger. In the liquefaction-evaporation zone, even with a delta T = 3.8 k, large quantities are exchanged? of energy. The presence of the heat pipe (with delta T = 3 k) ne? a test, and in the case of using, for example, exchangers with diameters between 200 (copper from 3 nim internal sector) and 280 mm (stainless steel 4 mm external sector) with average distances between liquefaction and evaporation of about 8 cm (distance horizontal average between internal and external sectors), the power exchanged could be around 100,000 w / m2 x? C without considering the surface of the fins and delta T which would greatly increase the exchange. In the practical case in which, for example, the internal surface develops a minimum of 5 m2 (200 mm finned cylinder), the exchange would be: 5 m2 x 100 kw x 3.8 k = 1.9 Mw (theoretical), an exaggerated value and well above the minimum required. Do not ? said for? that under certain conditions it is not possible to choose lower or higher pressures than those

programmed (and in this second case much smaller volumes).

The VCP stabilizes the flow rate for any pressure value, and the liquid with the change of state (with pressure decrease) will carry? in the vapor state on the saturated-liquid line (slightly above this line, (see gas compression graph) with compressibility factor Z equal to about 0.04. Its volume is slightly higher than the liquid volume and until it does not absorb the enthalpy of evaporation (765 j vertically along the dotted line with P rid = 0.2) from the other sector, it cannot increase neither in volume nor in temperature. . int.), will yield the enthalpy of liquefaction to the vapor which is instead at 97.2 k and completely liquefied.

The external steam instead (always on Sbl), stopped at 97.2 k, will remain? cos? (ie 97.2 k) until? will not have? taken all its evaporation enthalpy equal to 765 joul (the enthalpy sampling with press. = cost = 6.8 Bar starts from the value Z = 0.04 and goes up to the value Z = 0.83 after which it will continue? always vertically up to the value T = 252.1 k).

For calculations already? facts (which will be described later), in high entry on Sbl? Fluid T = 276.4 k and bottom liquefaction T = 101 k. In total, in liquefaction the nitrogen releases: (276.4 - 101 k) x 10 grams / sec = 1754 j 560 j (enthalp. Liquef) = 2314 j and the rising vapor takes them all: [(2314 - 765 enthalp. evap) / 10 g = 154.9 k] +97.2 k (vapor liq. temp.) = 252.1 k which corresponds to the temperature of the fluid leaving the high Sbl and entering the turbine T2. (The temperature difference between the input at 276.4 k and the high output at 252.1 k in energy is [276.4 - 252.1] x 10 gr = 243 j? Due to the value of the negative potential energy and the difference between enthalpy of evaporation and enthalpy of liquefaction as lower pressures give larger evaporation enthalpies).

The T2 turbine develops a job with P = so = 6.8 Bar with temperatures between 97.2 k and 252.1 k. The density of the nitrogen at 252.1 k is worth about 1.33 grams / liter and if the flow rate is worth 10 gr / sec (VCP decides with the V5 valve), the volume can? apply: (10 gr: 1.33 = 7.5 lt): 6.8 Bar = 1.1 lt. The work produced at constant pressure has the equation: L = P x delta V = 6.8 x 10 * 5 Pascal x 0.0011 cubic meters (1.1 liters) = 750 theoretical joules. The change in volume? calculated between minimum liquid-vapor volume (which is equivalent to about 10 g: 0.8 (liquid density - vapor) = 12.5 cubic cm which, however, I do not consider) and maximum gas volume (1.1 lt) (The value of Z at 6.8 Bar with T = 252.1 k is worth about 0.99 so the gas can be considered perfect gas in that area, see for this graph 43). The work produced and extracted in T2 generates a decrease in gas temperature with respect to the inlet temperature equal to approximately: 750 x 0.6 (effic.) = 450 j and therefore an outlet T = 252.1 k - (450: 10 gr ) = 207.1 k (decrease of 45 k).

This ? the input value to the compressor P1 (in a work at constant pressure the total energy must be removed from the work produced and extracted while the fluid remains a part of energy represented by an increase in temperature which is: 207.1 k -97.2 k = 109.9 k). Therefore, all the frictions and losses in T2 are added to the gas temperature, and come out together with it with a total value equal to 207.1 k but involve the compressor P1 in a greater job, since the same device has to compress fluid at the inlet to a temperature ap? pi? high.

In fact, P1 must adiabatically re-compress from 6.8 to 8 Bar, therefore: 207.1 k x (8 / 6.8) * 0.286 = 216.9 k with a theoretical delta T of 9.85 k, and practical equal to 9, 85k: 0.6 = 16.4k. Finally, the negative work? = 16.4k x10g = -164 joules with an output T = 223.5k (207.1 - + 16.4 = 223.5k). The gas then enters Sb2 inside which denatured ethyl alcohol circulates. Alcohol remains liquid up to -156? C and has a Cp of 2.84 kj / kg x? C so it can? take energy from the Rad radiator and transfer it to the cryogenic fluid, remaining perfectly liquid.

In this regard, a part of the energy incident on the radiator (indicated by arrows)? an etiostatic plant with solar concentration already? patented by the undersigned (patent n? 0000276838). In this new patent application the Heliostat? inserted in the system to store thermal energy in the water used as an energy vector between the Heliostat and the Radiator. Should the water have an additive (eg denatured alcohol) to prevent it from freezing at 0? C when? in contact with the walls of the radiator in which alcohol flows at about 276 k.

From Sb2 the gas (Nitrogen) with T = 288 k enters T1: density? fluid at 288k = 1, 17 gr / lt and with a flow rate of 10 gr / sec we have: 10 gr / 1, 17 = 8.54 lt / 8 Bar = 1.07 lt while the density? at 223.5 k it is 1.5. Hence the volume? di: 10 gr / 1.5 = 6.6 11/8 Bar = 0.827 lt therefore from 223.5 k to 288 k the variation is: 1.07- 0.827 = 0.243 lt. Isobaric work? finally: 8 x 10 * 5 Pascal x 0.000243 cubic meters = 194 joules x 0.6 (effic) = 116 joules.

The temperature leaving T1 is: 288 k - (116 j: 10 gr) = 276.4 k as programmed at the beginning, (the same thing occurs with the specific R of the nitrogen gas which is 0.297: (288 - 223, 5) x 10 gr x 0.297 x 0.6 = 115 j very close to the 116 j already calculated)

Equation First Principle: The mechanical work produced in the turbines is: (116 j 450 j) = 566 joules which connected internally to the compressor by: 566 j - 164 j = 402 j net mechanics extracted from the system. From the radiator Rad enter 645 j [(288 k - 223.5 k) x 10 gr / sec = 645 j] to which those for enthalpy of evaporation must be removed which how? been already? seen are negative: 645 j - 243 j = 402 j in accordance with the First Principle. The performance, how is it? I was able to ascertain it is worth 1, but in practice, due to the presence of liquid nitrogen (prepared in advance) yes? always below this value (liquid nitrogen does not exist in nature).

For the Second Principle instead? the presence of a drain is required at pi? low temperature represented by the steam at 97.2 k which in order to expand and transform into useful work all the thermal energy that is transferred to it, waits and absorbs all the enthalpy from the internal sector, more? all losses.

Entropy and Kelvin-Plank statement? Have you ever built a machine with unitary thermal efficiency, capable of transforming all the heat received into work ?. Entropy then? a measure of disorder and its law can? be stated as follows:? Will the disorder of an isolated system increase? always or at most? will stay? constant if the system? reversible?. The fluid (starting liquid-vapor) after having crossed Sbl upwards, passes both in the compressor and in the two turbines producing useful work with efficiency = 0.6 and therefore increasing its disorder. In fact at the outlet from? 1 the gas (with respect to the initial steam)? increased both in temperature and in volume. All of this? confirms the law of increasing entropy, but in descent on Sbl (with T input = 276.4 k) its disorder decreases (due to the decrease in temperature) while that of rising steam increases.

This energy exchange for? is realized in? perfect reversibility? as there is neither on the internal nor on the external sector the production of work. To conclude then, the proposed system? a completely reversible system with constant disorder (at least inside our atmosphere, if Sb1 is perfectly isolated, if we do not consider the leaks of fluid to the outside, and not even the wear of the various devices) that operates and develops energy mechanical (and therefore electrical) using electromagnetic energy in transit.

Production and extraction of nitrogen or liquid air: To extract liquid nitrogen from valve V3 (however leaving the liquid quantity in the tank unchanged), it is necessary to intervene automatically on M1 by increasing the pressure beyond the value 8 Bar. The consequence? an increase in the quantity? liquefied in Sb1, why? to an increase in pressure will correspond? a decrease in the enthalpy of liquefaction (the value 560 j decreases as the pressure increases). Leaving the flow rate constant (with the VCP), the prelever? always the same amount? in evaporation (and will always remain a value in enthalpy equal to 765 j) while for the difference it will increase? the quantity? of nitrogen in liquefaction, with the NI tank that will make up for it? the lack of nitrogen in the internal sector. However, liquid nitrogen extraction has an energy cost. In other words, a part of the net electricity produced up to that point would be used to extract Nitrogen. The energy consumed in more? it should absorb the compressor P1 which having sucked up to that moment with a pressure equal to 6.8 Bar would be forced to exceed the 8 Bar output, and increase its work compared to what already? scheduled.

System start-up: The V1-V8 valves are electrically operated n / c valves with the exception of the V5 which? a proportional control butterfly valve. At the start, the valves V2 - V4 - V5 - V6 are partially open so that liquid nitrogen can be loaded into Sb1 under slight overpressure (78 - 80 k and about 1.1 - 1.2 Bar). Filling the cryogenic tank causes the expulsion of a part of air (present before each start) which, passing into V5 and slowly sucked by the compressor P1 (motor speed control P1), comes out along the valve V6. Subsequently the V2 is closed and the V1 is opened to slowly pressurize the circuit with Nitrogen coming from N1 (electr. Control on the M1 pressure gauge) and completely expel the residual air. The N1 tank? isolated and being in high pressure, it feeds regret with gas in decreasing temperature (the expansion of a gas under high pressure causes a strong decrease in temp due to negative potential energy). Does liquefaction at 101-100k go up? the temperature of the liquid deposit from 78-80 k (temperature when it is deposited) up to about 98-100 k, while the pressure on the external circuit in Sb1, regulated by valve V5, will rise? from about 1.1 - 1.2 to 6.8 Bar.

Description of some devices and graphs in the attachment

The Scroll Expander (or Scroll Compressor)? a turbine whose blades have a spiral shape. Tests with different types of gas have on average given yields of around 70% with powers ranging from 1 to 3 kw. The compression ratios vary between 2.5 and 5 and the formation of liquid does not hinder its operation. Wear? almost absent despite being oil-free. Very important ? also the number of revolutions of the impeller which, in reference to the power produced, vary between 1000 and 3600 rpm. For the proportional valve (VCP) instead? the valve body must be in stainless steel (but the other devices must also be stainless steel, or pure copper or pure aluminum). Stainless steel is perfectly suited to liquid cryogenic gases. Also for the butterfly valve (V5) a position transducer must be mounted on the transmission shaft for continuous pressure control (6.8 Bar). The same thing must be done for the PI compressor with a control of the number of revolutions and therefore of the pressure.

The attachment indicated with the n? 48 shows a graph with reference to the Nitrogen fluid. The isobars are drawn vertically while the well-marked line below separates the gas-vapor part from the liquid part. The horizontal line placed at 100-101 kelvins identifies the liquid-vapor critical values at 8 bar, while at the critical value 97.2 k there should be the isobar at about 6.8 bar. The graph instead indicated with the n? 43 shows the values of the compressibility factor? Z for different gases. Applying to each gas its critical pressure and critical temperature they all have the same behavior. At very low pressures they behave like perfect gases regardless of temperature, while at high temperatures they act as perfect gases regardless of pressure. The maximum deviation occurs in the area of the critical point. In essence then, by applying the same system to almost all gases, in theory it would be possible to liquefy them using this system.

Description of the drawing n? 2

To increase the understanding of how the circuit works and therefore complete this patent application, I have inserted drawing n? 2 in which some details are reported, which I think are sufficient to clarify how much already? claimed in the application itself. Before this however, in First Principle equation? it was said that the value of 645 j must be removed the quantity? 243 j for evaporation enthalpy (or potential energy) as this has a negative value with respect to the velocity? of the molecules and the net positive balance is in this case 402 j (645 j - 243 j). Logically? It is also true that when the entire cycle (liquefaction) ends, the total between absorbed and transferred energy (by enthalpy or otherwise) must return to equilibrium: in fact, moving from liquid to vapor with 6.8 bar and T = 101 k the fluid receives: 2314 j from the other sector in liquefaction; - 450 j l? yields in the T2 Turbine; ? 164 j receives them from compression (P1); 645 j receives them from the radiator Rad (amb. External); - 116 j transfers them into the turbine T1; - 2314 j gives them in liquefaction at 8 bar. The total gives a positive balance of 243 j resulting from the recompression from 6.8 to 8 Bar (and also from a slight enthalpy recovery). (These must be extracted from the heat pipe and returned to the Sb2 container with the Reg. Regulator present in drawing n? 2).

The drawing n? 1 then? been completed with the addition of a regulator (Reg) (inserted on the upper part of the heat pipe in the internal section) and with the valves V7 - V8 and? been transformed into drawing n? 2, Does the regulator take a certain amount? of thermal energy in Sb1 (because its fluid is at a lower temp.) and sends it into Sb2 with a value that can? change between a minimum of 223.5 k and a maximum of 247.8 k by acting on the V7 and V8 valves (electronically controlled valves). The valves are required to ensure that the control always falls within the programmed values. Basically, the average liquefaction can? be about 10 gr / sec if the input T always turns around 276.4 k. If, on the other hand, this exceeds them, then the V7 must be closed and the V8 opened (it is necessary that on Sb2 less thermal energy is taken from the Rad radiator as the incoming fluid is hotter than the value 223.5 k). If the internal temperature falls below that value (or rather: if, for example, the ambient temperature entering the Rad radiator falls below 288 k) then the V8 must be partially closed and the V7 partially opened.

The delta T between the Sbl input and the high Sb1 output depends so much on the goodness? of the exchanger. With a difference = 24.3 k you are almost in the worst conditions, and d? probable that in an exchanger with good conductivity, the difference between input and output can be more? small. If, for example, this should be only 6.4 k (therefore T ing T2 = 270 k and not 276.4 k), the evaporation energy withdrawal would exceed the 2314 j which correspond, how can it be? seen, to the total of those transferred in liquefaction then: [(270 k - 97.2 k) x 10 gr = 1728 j 765 j (ent) = 2493 j which subtracted from the 2314 damage: - 179 j J. This situation would lead to an increase in the quantity? liquefied by making the deposit N 1 intervene to equalize the pressure and also the quantity? of molecules present in the internal sector.

To avoid that there? happens, the system must continuously read the difference between the high Sbl inlet and outlet and act on the valves V7 and V8 (temperature and flow control) so that? only the value 64 j, in this case, (6.4 k x 10 gr) enter into pi? actually in Sb2 with respect to the minimum established value which in kelvins is 223.5 k. Together with the Reg regulator in the drawing? The Serbi tank exiting the T2 turbine was also inserted to stabilize the pressure at 6.8 Bar.

A simplification of the circuit? obtainable also working with a single turbine, that is only with the T2, but at the expense of the net power produced.

To conclude then, the system enters: [(402 j (Rad) 164 j (P 1) | = 566 j thermal From the system there are: (450 j 116 j) = 566 j mechanical; Net product and extract 402 j mechanical (566 j - 164 j = 402 j mech).

Unfortunately, many other details have been excluded, (technical tables, accurate description of the exchangers, sensors, algorithms for process control, electrical system, cards for motor speed control, etc. etc.) that I think, however, are not necessary for the acceptance of the application itself.

CO2 splitting and environmental purification: For simplicity? descriptive? The description of the block that purifies and liquefies air has been omitted, from which liquid oxygen and nitrogen are also obtained. Anyway the preparation of liquid air? quite simple but? however necessary to make the entire system autonomous. This involves a decarbonization of the fluid, and more? if you want, even a split of CO2 by taking a part of the energy from the plant itself. Basically, hydrogen and oxygen are first produced by electrolysis, then H2 is used in combination with CO2. The result from graphite, water, heat and oxygen again.

Claims (1)

Claims 1- The plant? characterized by the fact that it takes thermal energy at room temperature, transfers it to the internal cryogenic gas (in this case air or nitrogen, but it is also possible to use another gas useful for the purpose) and transforms it into mechanical and therefore electrical energy. 2- The plant? characterized by the fact that using liquid air or nitrogen it is necessary to purify atmospheric air with the "capture" of CO2, deposit it in a purifier to be then split into Carbon and Oxygen. 3- The plant? characterized by the fact that it can? produce and extract liquid air and nitrogen (and if necessary also other liquid gases) also for purposes other than those described in the plant. 4- The plant? characterized by the fact that a part of the ambient thermal energy incident on the radiator Rad is not the other? that a solar concentration heliostatic plant already? patented by the undersigned (patent n? 0000276838).