Stirling cycle

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Stirling cycle in the pV and TS diagram

Scheme of the Stirling engine comparison process

The Stirling cycle consists of two isothermal changes of state and two isochoric changes of state and is usually represented with the pV and TS diagram . The Stirling cycle can be implemented using a machine with two pistons and a regenerator. The diagram opposite shows a possible arrangement. The piston positions marked with (1,2,3,4) are the corner points in the pV and TS diagrams.

The Stirling engine approximately implements this cycle.

Stirling cycle

The ideal Stirling cycle can be described by four changes in state, which are shown in the adjacent pV diagram. The red line corresponds to a hot volume, the blue line to a cold volume. It is a cycle of two isotherms and two isochors in which the working medium is periodically expanded and compressed . The work done by the machine corresponds in the pV diagram to the area enclosed by the lines.

The individual process steps Ito are IVexplained below.

pV diagram

Discontinuous ideal Stirling engine

Isothermal expansion

Process step I- line 1 → 2: The isothermal expansion from volume V ₁ to V ₂ takes place at a constant temperature T _H , with the heat Q _{12 being} absorbed and work W _{12 being} given off. The gas volume increases, the pressure decreases, but the temperature is kept constant by the heater. The gas does work by moving the working piston.

{\ displaystyle Q_ {12} = Q _ {\ text {zu}} = - W_ {12} = n \ cdot R \ cdot T _ {\ text {H}} \ cdot \ ln {\ frac {V_ {2}} {V_ {1}}} = {\ frac {m} {M}} \ cdot R \ cdot T _ {\ text {H}} \ cdot \ ln {\ frac {V_ {2}} {V_ {1}} }}

n = amount of substance of the gas in mol

m = mass of the gas in g

M = = molar mass of the gas in g / mol ${\ displaystyle {\ frac {m} {n}}}$

C _v = molar heat capacity at v = const. in J mol ⁻¹ K ⁻¹

R = universal gas constant in J mol ⁻¹ K ⁻¹

Isochoric cooling

Process step II- line 2 → 3: The isochoric cooling takes place at a constant volume ( V ₂ = V ₃ ) at which the heat Q _{23 is given off} by the gas to the regenerator. If the volume remains the same, the temperature and pressure of the gas change and the regenerator stores the heat. Moving the pistons does not require any work, as the same pressure acts on both.

{\ displaystyle Q_ {23} = \ Delta E _ {\ text {Reg}} = n \ cdot C _ {\ text {V}} \ cdot (T _ {\ text {H}} - T _ {\ text {K}} )}

Δ E _Reg = change in the thermal energy of the regenerator

Isothermal compression

Process step III- line 3 → 4: The isothermal compression from volume V ₃ to V ₄ takes place at a constant temperature T _K , the heat Q _{34 being} given off and the work W _{34 being} supplied. The gas volume decreases, the pressure increases, but the temperature is kept constant by the cooling. Moving the piston requires work.

{\ displaystyle Q_ {34} = Q _ {\ text {ab}} = W_ {34} = n \ cdot R \ cdot T _ {\ text {K}} \ cdot \ ln {\ frac {V_ {3}} { V_ {4}}}}

Isochoric warming

Process step IV- line 4 → 1: The isochoric heating takes place at a constant volume ( V ₁ = V ₄ ), at which the heat Q _{41 is} absorbed by the gas and given off by the regenerator. If the volume remains the same, the temperature and pressure of the gas change and the regenerator emits the IIheat stored in step . Moving the pistons does not require any work, as the same pressure acts on both.

{\ displaystyle Q_ {41} = - \ Delta E _ {\ text {Reg}} = n \ cdot C _ {\ text {V}} \ cdot (T _ {\ text {K}} - T _ {\ text {H} })}

The entire process description applies to the uncharged Stirling engine. The pressure behind the working piston is therefore always lower than in the cylinder.

Summary

Why can the Stirling engine give up work? You need a heat gradient. In the expansion phase, the gas must be prevented from cooling down, and in the compression phase, the heating of the gas must be suppressed. The efficiency corresponds to the thermal efficiency.

{\ displaystyle \ eta = {T _ {\ text {H}} - T _ {\ text {K}} \ over T _ {\ text {H}}}}

T _H = hot temperature
T _K = cold temperature

In the process step I, the isothermal expansion at the high temperature T _H , the gas in the closed cylinder absorbs heat and converts it completely into work. The pressure p of the gas generates a force F (= p * A ) on the area A of the working piston . If this piston now moves upwards by the distance Δ s , the work done is:

{\ displaystyle | W | = F \ cdot \ Delta s = p \ cdot A \ cdot \ Delta s = p \ cdot \ Delta V}

In the PV diagram of the ideal Stirling process can clearly recognize the work done as the area under the line 1 → 2, of the isotherms at the temperature T _H , again.

During the process step III, the isothermal compression at low temperature T _K , less work has to be done, the area under the line 3 → 4, the isotherm at temperature T _K , is smaller. During one revolution of the motor, therefore, the area 1234 enclosed by the circuit is precisely the work W _ab that is output as a whole.

{\ displaystyle W _ {\ text {ab}} = W_ {12} -W_ {34}}

The larger the area 1234 shown, the more work the motor can produce with one revolution. The greater the ratio of W ₁₂ to W _{34, the} better the efficiency.

Real Stirling process

The ideal Stirling process, like all other ideal circular processes, cannot be precisely implemented. The following pv diagram shows with the area ( yellow ) the real power that remains for use compared to the ideal process diagram above.

pV diagram

The following list of the reasons for this is also an introduction to the problem of the Stirling engine .

Reasons for loss of efficiency

Some reasons why the real process deviates from the ideal:

mechanical friction
A discontinuous piston control is only limited realizable

In order to improve the efficiency (the process is better extended in the corners) and to keep the dead space as small as possible, a discontinuous piston control makes sense. The disadvantage is higher wear due to mechanical stress and the development of noise.

pV diagram Effect of the type of movement on the useful work (hatched area)

Due to the high gas velocity , isothermal changes in state are poorly realized
Regenerator efficiency is limited
Dead space effects

Ideally, the entire working medium (gas) is in the expansion and compression space. For engines built up to 1999, the dead space is around 30 to 50% of the total volume. The heat exchanger units such as heaters, regenerators, coolers are usually located in these dead spaces (also called dead spaces). The resulting changed volume ratios also bring about changed pressure ratios, which have a very unfavorable effect on the overall efficiency.

Loss of heat through the material

This heat loss is caused by the heat flow along the cylinder outwards in the direction of the temperature gradient.

Dissipation due to loss of working gas and pressure

This loss occurs more intensely in Stirling engines with a nominal speed of more than 200 rpm. The compression and expansion take place so quickly that the heat flow, which would be necessary for isothermal energy, can no longer keep pace. The result is the pressure rise during compression or a steep pressure drop during expansion.

Web links

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