NO20180312A1 - Method for extracting mechanical energy from thermal energy - Google Patents

Description

Method for extracting mechanical energy from thermal energy

STATE OF THE ART

It is known that mechanical energy can be extracted from thermal energy. This requires a drive fluid in a thermodynamic cycle. Such a cycle has no practical value if it does not emit net mechanical energy. The expansion energy must be greater than the compression work.

Mechanical energy – enthalpy and internal energy

internal energy (symbol U) = chemical energy thermal energy

enthalpy (symbol H) = chemical energy thermal energy pressure x volume (PV)

Enthalpy contains a component for mechanical energy, PV, which is the isothermal energy difference between enthalpy H and internal energy U. It follows from this that enthalpy is always greater than internal energy. The relationship can be expressed as follows: H = U PV

PATENT DESCRIPTION

The calculation example shows a thermodynamic cycle. That is, a cycle where the opening state is identical to the closing state. In the described method, the size ratio between enthalpy and internal energy is decisive for mechanical energy yield and for absorption of thermal energy. It is characteristic of the cycle that the energy yield is based on the difference between enthalpy and internal energy, and it is therefore assumed that PV is significant. A high correlation is also assumed between the expansion's maximum PV and the same expansion's enthalpy difference. Characteristic and principled features must therefore be defined and delimited by reference to enthalpy and internal energy.

The calculation example shows the function and significance of the characteristic features in a practical implementation. It shows expansion energy and thermal energy absorption, with temperature reservoirs. In the calculation example, the expansion energy is 3-4 times greater than the compression energy.

The procedure thus provides a net mechanical energy yield.

Calculation example thermodynamic cycle with NH3

Mechanical energy is released when liquid – here NH3 – expands. In the calculation example, the energy is 44.3 J/g. The expansion is from 678.53 to 668.6 kg/m<3>, and the temperature decreases from 255 K to 250.45 K. Through the described cycle, the density increases – from 668.6 kg/m<3> and back to 678 .53 kg/m<3>. The compression energy must be significantly lower than 44.3 J/g.

When expanding from 668.6 to 27.5 kg/m<3>, the temperature drops from 250.45 K to 241 K, and the fluid separates into two phases. The condition ends at steam quality 0.03284. The entropy in the liquid part decreases with temperature. This way you get – at 241 K – a mass where 96.716% has the entropy value 0.905 J/g*K.

Total entropy does not decrease through the expansion (2nd main theorem) from 668.6 to 27.5 kg/m<3>. On the other hand, it is expected to increase by 1.3% – from 1.0771 to 1.0912 J/g*K. Gas entropy necessarily rises. It becomes 6.5733 J/g*K. Steam quality 0.03284 means that the gas proportion is 3.284% of the mass at 241 K and 27.5 kg/m<3>.

The entropy in the gas part can be lowered - also with a reservoir of 277 K. This can happen despite increasing fluid temperature - from 241 K to 280 K. The entropy can be lowered from 6.5733 to 1.5860 J/g*K. The prerequisite is densification from 0.94 kg/m<3>to 628 kg/m<3>with T =< 280 K. Densification requires mechanical work, 342 – 370 J/g. For 0.03284 grams of gas, this becomes 11.23 --- > 12.15 joules.

The condensed gas, condensed to 628 kg/m<3>– is mixed with the liquid part, which holds 680.51 kg/m<3>. Thus, the density is back at 678.53 kg/m<3>– with temperature 242.3 K and entropy 0.929 J/g*K. The operation costs 12 joules in compression energy, i.e. less than 1/3 of the delivered 44.3 J/g. The net mechanical energy yield is thus 32 J/g.

The supposed advantage of this method is the overlapping temperature reservoir. In the calculation example, both the process densification from 0.94 to 628 kg/m<3> and the process isochor heating from 242.3 to 255 K can take place at a reservoir temperature between 264 K and 277 K.

This means that entropy reduction and energy absorption can take place using one and the same temperature reservoir, for example 277 K (+4<0>C). This creates a clear contrast to the Carnot process, which does not work without a clear difference between the reservoirs TH and TL.

Principle sketch with process data

Calculation example – detailed – from point to point

A to BSisentropic 1.0771 J/g*K

A 678.53 kg/m<3>255.00 K 30 MPa H = 283.82 J/g

BS668.60 kg/m<3>250.45 K 0.16836 MPa H = 239.52 J/g Isentropic process/pressure change in liquid gives energy yield 44.3 J/g

BStil BL V

BS668.60 kg/m<3>250.45 K 0.16836 MPa H = 239.52 J/g

BL V27.50 kg/m<3>241.00 K 0.10745 MPa H = 242.02 J/g Entropy increase 1.3% from 1.0771 to 1.0912 J/g*K

NIST isochoric properties 27.5 kg/m<3>

The expansion from 668.6 to 27.5 kg/m<3>occurs with pressure between 0.16836 MPa and 0.10745 MPa. Released mechanical energy is thus negligible.

Total condition values for BL V

K MPa kg/m<3>U J/g H J/g S J/g*K phase 241 0.10745 27.5/0.03284 238.11 242.02 1.0912 liquid/vapor Vapor quality is 0.03284. This means that 3.284% of the mass is in the gas phase. 96.716% is liquid.

Condition BL

K MPa kg/m<3>U J/g H J/g S J/g*K phase 241 0.10745 680.51 197.0 197.16 0.905 liquid State BV

K MPa kg/m<3>U J/g H J/g S J/g*K phase 241 0.10745 0.93976 1448.9 1563.2 6.5733 vapor

From point BL V to point C

Point C must have the same mass and density as point A, i.e. 1 gram and 678.53 kg/m<3>. The temperature should be as close as possible to 241 K. Procedure:

The mass in state BL (liquid 0.96716 grams) does not change. The mass with the state BV (vapor 0.03284 grams) is condensed from 0.94 to 628 kg/m<3> (to BVD on the sketch).

K MPa kg/m<3>U J/g H J/g S J/g*K phase 241 0.10745 0.93976 1448.9 1563.2 6.5733 vapor 280 0.55092 628 374.12 375.00 1.5860 liquid/vapor

-4-

The compression energy is calculated at 342 J/g. For 0.03284 grams it becomes 11.23 joules.

The masses BLog the condensed BV – which has become the liquid BVD – are mixed adiabatically and proportionally with respect to density, entropy and energy:

temperature mass proportion S J/g*K H J/g density kg/m<3>

BL241 K 96.716 % 0.905 197.16 680.51

BVD280 K 3.284% 1.586 375 628

According to NIST, the new state will be:

vapor & liquid fluid data – combined and mixed 678.53 kg/m<3>

temp K pressure MPa quality internal U enthalpy H entropy S phase 242.30 0.11457 7.8697e-07 202.79 202.96 0.92901 liquid/vapour This is the state that will constitute point C.

The density in C is 678.53 kg/m<3>, as in point A. The entropy is lowered from 1.0912 to 0.92901 J/g*K.

With a temperature of 242.3 K, the fluid can absorb heat from the reservoir of 264 K, so that the fluid temperature rises to 255 K.

From point C to point A

The density is kept at 678.53 kg/m<3>, and there is an isochoric temperature increase. The gas portion is negligible (quality 7.8697e-07) at 242.3 K, and saturation occurs at 242.6 K. At 243 K, there is a visible difference between U-increase and H-increase.

The temperature in the reservoir, 264 K, is at any point significantly higher than in the fluid. The increase of 80.86 J/g (H) must be transferred in its entirety from the reservoir 264 K.

K MPa kg/m<3>U J/g H J/g S J/g*K phase

242.3 0.11457 678.53 202.79 202.96 0.92901 liquid/vapor 255 30 678.53 239.61 283.82 1.0771 liquid

Because the fluid is in pure liquid phase almost the entire temperature interval, the enthalpy increase�H is more than twice as high as the increase in internal energy ΔU:

ΔU ΔH

increase from 242.3 K to 255 K 36.82 J/g 80.86 J/g

After isochoric heating with energy from reservoir 264 K, the state is back in A.

Claims (4)

PATENT CLAIMS 1. Thermodynamic cycle, i.e. initial state equal to final state, characterized by a correlation > 0.8 between the expansion's enthalpy difference (ΔH) and the maximum PV value that occurs in said expansion. 2. Thermodynamic cycle, i.e. initial state equal to final state, characterized by thermal energy absorption being an isochoric heating with an enthalpy increase (ΔH) at least 80% higher than the increase in internal energy (ΔU). 3. Thermodynamic cycle, in accordance with claims 1 and 2, characterized in that the drive fluid of the cycle acts as a monophase fluid and a 2-phase fluid. 4. Thermodynamic cycle, in accordance with requirements 1 and 2, characterized in that the method's thermodynamic cycle can use the same temperature reservoir as heat drain (TL) and as heat source (TH). SUMMARY A method of extracting mechanical energy from thermal energy. The method deviates in principle from known technology in that the energy yield is based on the difference between enthalpy and internal energy. The procedure is consistent with a thermodynamic cycle. In the description, the temperature reservoir for cooling is 277 K (+4<0>C). The heat reservoir can have any temperature above 255 K (-18<0>C). It can thus be 277 K. The procedure is calculated with fluid data from the National Institute of Standards and Technology (USA). The thermodynamic cycle and energy calculations have been analyzed and checked by experts, through Tel-Tek/SINTEF, using the process simulation program Aspen HYSYS. For EntroMission AS Olav Hellum Olav Hellum