**PERPETUUM MOBILE OF THE SECOND KIND**

HELLUM, Svein, Olav (Oskleiva 30, 1772 HALDEN, NO)

KRISTOFERSDOTTIR, Lãra Bjørk (Lervikveien 47, 1626 MANSTAD, NO)

EIRIKSSON, Magnus (Östraby 9194, HÖRBY, 242 94, SE)

*;*

**F01K25/04***;*

**F01K25/10**

**F03G7/10**WO2011011983A1 | 2011-02-03 | |||

WO2011012047A1 | 2011-02-03 | |||

WO2005031123A1 | 2005-04-07 |

US20170002691A1 | 2017-01-05 | |||

US20100269503A1 | 2010-10-28 | |||

US20020162330A1 | 2002-11-07 | |||

EP2131105A1 | 2009-12-09 | |||

US20070119175A1 | 2007-05-31 |

CLAIMS 1 A thermodynamic cycle, i.e. a cycle where initial state equals terminal state, characterized by a correlation > 0.8 between the expansion enthalpy difference (DH) and the maximum numerical PV- value appearing in said expansion. 2 A thermodynamic cycle, i.e. a cycle where initial state equals terminal state, characterized in that isochoric heating, with an enthalpy increase (DH) at least 80 % higher than the increase of internal energy (Du), constitute the thermal energy absorption. 3 A thermodynamic cycle, in accordance with claim 1 and 2, characterized in that the drive fluid appears as monophase fluid and as dual phase fluid. 4 A thermodynamic cycle, in accordance with claim 1 and 2, characterized in that the cycle may utilize one temperature reservoir for both cooling and for heating. |

STATE OF THE ART

The possibility of extracting mechanical energy from thermal energy is well known and generally accepted. It requires a drive fluid in a thermodynamic cycle. However, this cycle is useless if it does not produce mechanical energy. Expansion energy must therefore significantly exceed the energy of compression.

Mechanical energy - enthalpy and internal energy internal energy (U) = chemical + thermal energy

enthalpy (H) = internal energy (U) + pressure x volume (PV)

Enthalpy includes, in addition to internal energy, a component for mechanical energy; PV.

PV is the isothermal energy difference between enthalpy H and internal energy U.

Deduced from the definition: Enthalpy is always more than internal energy.

The interconnection is: H = U + PV

DESCRIPTION

The example of a calculation presented below, shows a thermodynamic cycle; a cycle where the fluid state at the starting point equals the fluid state at the finishing point. In the present procedure, the ratio between enthalpy and internal energy is decisive for mechanical energy output, as well as for the absorbed thermal energy.

The present cycle is characterized by the energy output being based on the difference between enthalpy and internal energy. This feature presupposes a considerable PV. A further premise is a high correlation between maximum PV and the enthalpy difference in the same expansion. As a consequence, the principle features are defined and limited by enthalpy and internal energy.

The calculation example demonstrates the function and the impact of the principle features, in practical use. It shows the expansion energy emitted and the thermal energy absorbed, as well as temperature reservoirs. In the calculation example, expansion energy is 3-4 times the amount of energy used for compression. Thus the procedure emits mechanical energy.

Example of calculation thermodynamic cycle with ammonia NH _{3 }

The liquid expands, simultaneously releasing mechanical energy. In the example of calculation, this energy is 44.3 J/g. Liquid expansion is from 678.53 kg/m ^{3 } to 668.6 kg/m ^{3 }. Temperature decrease is from 255 K to 250.45 K.

The subsequent steps lead to increased density, from 668.6 kg/m ^{3 } and back to 678.53 kg/m ^{3 }. Compression energy is significantly less than 44.3 J/g.

During expansion from 668.6 kg/m ^{3 }, temperature falls from 250.45 K to 241 K. The fluid separates into liquid and vapor; quality 0.03284. The liquid entropy decreases proportionally to the temperature. At 241 K, the liquid part, i.e. 96.716 % of the fluid mass, has reached an entropy value of 0.905 J/g* K. Total entropy does not decrease during expansion from 668.6 to 27.5 kg/m ^{3 }. On the contrary, it is presupposed to increase by 1.3 %, from 1.0771 to 1.0912 J/g*K. The vapor entropy must therefore increase to 6.5733 J/g*K. Quality 0.03284 indicates that the vapor part of the fluid constitutes 3.284 % of the mass, at 241 K with a density of 27.5 kg/m ^{3 }.

Reduction of vapor entropy is possible, even with a reservoir temperature of 277 K, and in spite of increasing fluid temperature; from 241 K to 280 K. Entropy level decrease is from 6.5733 to 1.586 J/g*K. A precondition is the densification from 0.94 kg/m ^{3 } to 628 kg/m ^{3 } with T =< 280 K. This compression requires mechanical energy; between 342 and 370 J/g. For 0.03284 gram vapor, it amounts to 11.23 --> 12.15 joule.

The densified vapor, compressed to 628 kg/m ^{3 }, is mixed with the liquid component, density 680.51 kg/m ^{3 }. Total density is then back at 678.53 kg/m ^{3 }, with temperature 242.3 K and entropy 0.929 J/g*K.

Compression energy cost for completing the cycle is 12 joule per gram drive fluid, i.e. less than 1/3 of the emitted 44.3 J/g.

The expected advantage of this procedure is overlapping reservoir temperatures. In the example calculation, the densifving process, from 0.94 to 628 kg/m ^{3 }, and the isochoric heating process, from 242.3 K to 255 K. are both achieved by means of a reservoir temperature between 264 K and 277 K.

As a consequence, entropy decrease and heat energy absorption may be accomplished by using a common temperature reservoir, such as 277 K (+4 °C). In this respect, the present procedure is a contrast to the Carnot-process; a cycle unable to function without a significant difference between the reservoirs TH and TL.

Explanation to components in the device illustration

CP piston in cylinder

connecting rod

crank/crankshaft

At CP, fluid energy is converted into mechanical energy, initially by the isentropic change of pressure, emitting 44.3 J/g. In the calculation sketch, above, this is indicated by the points A -> Bs, optionally to BL _{+ } V (i.e. to 27.5 kg/m ^{3 } and 241 K).

LF liquid filter

This device is located between S _{t + i/ } and Bi/Bv ^{' }n the calculation sketch. Its function is to separate liquid and vapor from the dual phase fluid, with the purpose of processing the vapor fluid.

The liquid 680.51 kg/m ^{3 } (241 K), constituting 96.7 % of the fluid mass, is guided to liquid pump PL 1, state unchanged. In the calculation sketch, this is indicated by BL -> BL.

The vapor part 0.94 kg/m ^{3 }, constituting 3.3 % of the total fluid mass, is guided to C, the compressor. In the calculation sketch, the densification is indicated by Bv -> B _{VD } and fortetting fra 0,94 til 628 kg/m ^{3 }

(= densification from 0.94 to 628 kg/m ^{3 }).

The device liquid filter is known from liquid air production (the Linde process) and from the separation of gas/condensed gas (oil, gas and petroleum processing industry). C compressor

The compressor is driven by mechanical expansion energy; in the present sketch directly by the crankshaft. The compressor performs mechanical work, and the energy demand is subtracted from the expansion energy. In order to keep compression energy at a low level, compression temperature of the fluid is limited to 280 K. This cooling is provided by heat exchanger HEX 1.

The compressor is a static piston pump/compressor or a dynamic radial-/centrifugal compressor, optionally an axial compressor.

HEX 1 heat exchanger 1

Cooling during densification is provided by heat exchanger HEX 1. In the calculation sketch, the compressor C and the heat exchanger HEX 1 are located between Bv and BVD At the point BVD, the density of 628 kg/m ^{3 } is attained, and entropy decreases from 6.5733 J/g*K to 1.586 J/g*K.

In the heat exchanger, a considerable heat transmitting surface is required, as well as sufficient time of exposure, i.e. appropriate volume. This is caused by the substantial amount of heat energy to be extracted from the fluid and returned to the temperature reservoir, in particular the condensing energy of the dual phase step (densification from 4.38 kg/m ^{3 } to 628 kg/m ^{3 }).

The heat exchanger is, by function, the same device as a charging intercooler for turbocharged combustion engines; air-to-air (car) or air-to-water (marine diesel).

PL l and PL 2 liquid pumps

PL 1 is filled with 96.7 % of fluid mass, density 680.51 kg/m ^{3 }, from liquid filter LF.

PL 2 is filled with 3.3 % of fluid mass, density 628 kg/m ^{3 }, from heat exchanger HEX 1.

When PL 1 and PL 2 are being emptied, simultaneously, the valves leading to HEX 2 and to CP are in the open position. The two liquids are then mixed, resulting in 678.53 kg/m ^{3 } and 242.3 K before this fluid enters heat exchanger HEX 2.

As the cylinders of the liquid pumps are being emptied, the working piston in CP moves correspondingly, making the volume increase above the piston top equal to the total volume reduction in PL 1 and PL 2. Unchanged volume means unchanged density; 678.53 kg/m ^{3 }. Thus, the pumps PL 1 and PL 2 do not perform compression work.

The liquid pumps PL 1 and PL 2 are driven by a gear and drive shaft from the crankshaft. The pump pistons are most practically operated by a camshaft.

HEX 2 heat exchanger 2

This heat exchanger corresponds the operation C -> A in the calculation sketch, i.e. isochoric heating from 242.3 K to 255 K. The temperature reservoir for HEX 2 is 264 K in the calculation sketch and 277 K in the device illustration. Both temperatures will work, but heat transmission will be faster using 277 K. Any temperature above 255 K will work, meaning that a 277 K-reservoir may be utilized by both of the heat exchangers.

HEX 2 is, from the inside, exposed to high pressures; in the present version 30 MPa (300 bar). Materials must be selected, reinforced and sized correspondingly.

**Previous Patent:**MULTISPECIFIC BINDING MOLECULES FOR THE PREVENTION, TREATMENT AND DIAGNOSIS OF NEURODEGENERATIVE DIS...

**Next Patent: A ROTARY VANE MACHINE WITH A CAM TRACK AND VANE MECHANISMS**