Thermodynamic work vs. real work

[VincentG]

This diagram can be used to indicate the operation of an ideal engine operating at the zero point, with the yellow line being 1 bar. In thermo talk, the gas does work on the piston in the area above the yellow line, and the piston does work on the gas in the area below the yellow line.

But in reality, work is also extracted from the area below the yellow line, all the way back to 1 bar. Realistically, there is no back work here and shaft power would be nearly continuous.

Can any interested party add to this diagram to distinguish the difference between the thermodynamic work and the actual work available to be extracted for lifting a weight.

Please save any bickering for another thread.

stirling-cycle-diagram.png (36.15 KiB) Viewed 3384 times

[Fool]

I understand where you are going and the questions you want to answer. Good idea. Thank you.

My lousy attempt:

Forward stroke #1:
The area above the yellow line and below the 'power stroke' line is the work that is produced and given to the piston during the forward stroke. It is the maximum work that accelerates the piston outwards. It is a combination of the total energy produced (power stroke line to absolute zero), minus, the atmospheric PV work (yellow line to zero K).

If adiabatic, no heat is added and a smaller amount of work, than isothermal, is produced. If isothermal more work will be produced, and heat will be added to make up for expansive cooling (Temperature is constant.)

The area below the yellow line, all the way to zero Kelvin, is the atmospheric PV work. It fights the internal gas's work, power stroke line to zero K. The gas work is greater.

Process #2 Dwell:
The piston apparently stops when getting to process #2. Apparently the work is removed from the piston, that work can go into a crank and flywheel, or spring, or other, even used up. Something takes the energy, and stops the piston. Pressure is at ambient, so some external force is needed to stop it. No work is done during the constant volume process. Heat is saved to an apparent regenerator. Temperature goes from Th to Tc.

Return stroke #3:
The area between the return stroke line and the yellow line is the maximum work delivered to the piston during the return stroke. It is also a fight between the gas inside and atmospheric pressure outside. Gas pressure is lower so the work goes, again, into the piston, additive. The crank reversed the direction while storing energy on the flywheel it is additive. The PV work from the atmosphere is the area under the yellow line, the same as the power stroke but additive. You get back from the "atmospheric spring" what was put out into it.

The area under the return stroke line all the way to zero K is the work done to the internal gas, subtractive. The maximum work delivered to the piston is the atmospheric work minus the gas work. This total work is additive for the return stroke.

If adiabatic, zero heat will be rejected, and the work will be lower, and the temperature will rise. If isothermal, the heat of compression will need to be rejected, and work will be greater, temperature will be constant Tc.

Process #4 Dwell:
The piston stops. Heat is added from the regenerator. Zero work. Heat added equals heat saved.

The cycle starts again.

Special note, the area inside the curve, power stroke line, to return stroke line is the thermodynamic work that the cycle has. It is the maximum work output that would happen. Friction will make it less. This is for an ideal cycle. Real cycles cut corners and shrink inward on that cycle making real cycles less powerful and efficient.

It is easier to calculate that area by ignoring the yellow line and adding work from the power stroke line all the way to zero K, minus, the area under the return stroke line all the way to zero K. Same area, work, and heat. That is the way calculus works area to zero. Area under the curve.

It looks like the ideal cycle you've given, uses isothermal prower and return strokes.

Real indicator diagrams can be treated similarly. The integration can be done graphically. And both zero Kelvin, and the yellow line, can be ignored.

If the crankcase is not pressurized and higher engine pressures are used the area moves well above the yellow line, and the yellow line can again be ignored. LOL

I hope this answers some of those questions.

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[VincentG]

If possible, please edit the diagram to visually explain how to represent the thermodynamic work done on the gas vs. the mechanical work done on the piston, based on the actual temperature increase from Tmin to Tmax.

Can we also add ideal values to this ideal diagram? My guess is no.

There is a fundamental problem with the "power stroke line" running down to 0k. If this were the case, the pressure would also be zero, and any cold sink would become a heater like Matt is going on about in his Gamma anomaly thread. The diagram in that case could not be completed, or pressure would have to go into negative numbers, or it would be a different pv diagram all together.

There is a difference between the ideal thermodynamic efficiency that is based on energy from 0k, and the real ideal efficiency that is based on energy added to the ambient energy levels already contained within the gas.

[Fool]

I will try to edit the diagram. It will take a while.

The power stroke line doesn't go all the way to zero. Obvious from the diagram.

I'm describing the area underneath the power stroke line.

In fact I'm describing several ways to calculate the area inside the complete cycle.

To calculate an area four boundary are used in this case. Ceiling, two walls and a floor. They are:

Process #1 the ceiling, power stroke line.
Process #2 the first wall.
Process #3 the floor, return stroke line.
Process #4 the second wall.

Measuring that area involves integration in calculus. Calculus measures an area under a curve. The floor becomes the y axis and two calculations must be used. In thermodynamics that is zero K. The second ceiling is the return stroke line.

So the area under the return stroke and y axis/zero K is calculated. It is subtracted from the area under the power stroke line. Giving the area inside the curve.

The area inside the curve is the thermodynamic work for the complete cycle. It acts reciprocally on the piston.

Heat transfer is determined by how much the lines deviate from a constant Entropy line, adiabatic with work line.

I will try to get the graphic modified to help explain this. It would be easier on a chalkboard in person.

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[matt brown]

Can we also add ideal values to this ideal diagram? My guess is no.

Fool - don't sweat PV for Stirling cycle, Vincent is nearly 100% correct. PV are great for simple stuff with single cylinder (ICE, steam, etc) but their premise is single volume with fixed mass. Once you start combining volumes with different temps and masses, validity falls apart. The typical Stirling PV like Vincent posted is only a representation and not even accurate for an alpha. Anyone can fake it with 4 control points, but the enclosed work area will not be a consistent grid that you can scale work from.

[Fool]

Depends on if you are trying to model it with finite elements, or reading an actual indicator diagram from a real engine.

The real diagram should be way more accurate at showing the pressure on the piston. It will be an average over all the temperature induced pressures from all parts of the engine. Cold, hot, regenerator, dead spaces. But putting in more sensors will just show flow resistances, not big differences.

[Fool]

Total thermodynamic work available for output.

stirling-cycle-diagram (3).png (36.71 KiB) Viewed 3178 times

The area marked in green is the work, Joules, that are developed thermodynamically by the ideal Stirling Cycle shown.

It is the area surrounded by the cycle boundaries. Process #1,2,3,4. Underneath the power stroke, left of the regenerator absorption, above the return stroke, and to the right of the regenerator replacement, lines.

This is the maximum work that can be expected and output through the piston. It is equivalent to the Carnot theorem.

It is a combination of, the forward stroke work, plus, the reverse stroke work, as depicted above and below the yellow line.

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[Fool]

Area of positive work power stroke.

stirling-cycle-diagram (1).png (38.32 KiB) Viewed 3167 times

The area marked in green, is the amount of work done only by the forward stroke as calculated by integration. A function of ln(Vb/Vt). This is the PV work done by the gas to the piston.

The ceiling of this area is the power stroke line. Going from Vt (Volume Top Dead Center, minimum) to Vb (Volume Bottom Dead Center, maximum). It's walls are the processes #2 and #4, regenerator lines. The floor is the Y-axis, also know as zero Kelvin.

This is the energy out, and 100% of the heat in Qh.

Note, the work heat equivalence here is based on the zero K line. As is normal for integration. This is important to remember.

This is the work done by the gas to the piston. Positive work output by the system, one forward stroke. 100% heat to work conversion.

I'm not done.

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[Fool]

Work done to the gas return stroke.

stirling-cycle-diagram (2).png (35.79 KiB) Viewed 3167 times

The area marked in green is the work needed to recompress the gas by the piston to the gas. Negative work because the direction is reversed. Energy is going from some outside source, flywheel crank, through the piston into the gas.

To maintain Tc, the heat of compression must be removed. The heat of compression is equal to the work put in. This is also 100% work to heat conversion. It is negative.

This work is removed from the forward stroke. Subtracted. Note that the subtraction leaves the same area as that enclosed by the cycle as in the first explanation.

This work is also calculated by integration with the floor being the X-axis, zero Kelvin.

If temperature is allowed to rise, Tc will rise towards Th, the area of negative work get larger. Cycle work is reduced and efficiency drops. Less work out per cycle. Same amount of heat in. Without adiabatic lines on the diagram, this is not shown.

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[Fool]

The integration goes to the floor, Y = zero Kelvin, which is the X-Axis, not the Y-axis. I goofed up a few places on that, above. Sorry. Can't edit.

[Fool]

Now. Since people comment that the return stroke can also be a power stroke, the following description needs to be shown. The yellow line.

In integration, the zero point can be moved, let's say to atmospheric pressure, the yellow line. In that case, complicating things, a linear offset is added to the equations. The net effect is to subtract flat, yellow line, PV work from the forward stroke. Then it is added to the reverse stroke. For a net work gain of zero. This also gives the same area enclosed by the complete cycle. That proves the zero work gain.

The next two diagrams show this.

Reduced power of forward stroke.

stirling-cycle-diagram (4).png (34.68 KiB) Viewed 3162 times

Note that the area under curve now has a floor of the yellow line, and is much smaller than going to zero K. The heat to work ratio is no longer 100%. It's less work. Smaller area.

Increased power return stroke.

stirling-cycle-diagram (5).png (34.58 KiB) Viewed 3162 times

Notice that the area is smaller than going to zero K, but it is above the return line making it negative, and since it's a reversed direction, negative again. Two negatives cancel so the return stroke work is positive and added to the forward stroke giving an identical amount of work as in the other two ways of calculating this work.

The efficiency has not changed because of 'buffer pressure '.

The yellow line can be ignored when using calculus to calculate cycle work and efficiency.

The yellow line does show why an atmospheric engine will run with a much smaller flywheel or none at all.

If the piston is allowed to expand further than the yellow line, the atmosphere will provide a great stopping force, PV. The atmosphere becomes a PV spring for just the piston. It allows the engine to run with no crank shaft, or metal springs. Infact it allows the engine to run without a piston as in the laminar flow Stirlings, fluid dynes, Putt Putt boats, Jam Jar and Valveless Pulse Jets. They use the mass of the air/'working fluid' to store positive energy, for the return mechanism, PV Bounce.

It is more effective to have a double acting piston and two displacer chambers. As the Stirling Brothers demonstrated two hundred years ago. And we think we are discovering something new. Oh my!

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[Fool]

If this were the case, the pressure would also be zero, and any cold sink would become a heater

The trouble I see with the cold plate becoming a heater is the gas must become colder than the cold plate. It is well known that there are hysteresis losses in both the the cold plate and hot plate areas when small amounts of gas is left there, called dead space. But it's small amounts. The bulk is in the hot space getting hotter for expansion (expansive cooling) and in the cold space getting colder for compression (compressive heating).

Additionally in a real engine the regenerator isn't 100% effective, so the bulk temperature hitting the cold space will be hotter than Tc, and colder than Th for the hot space.

It is further affected by speed. Less time to come up to Th or down to Tc. Real engine indicator diagrams show this contraction towards the median temperature and pressure as speed increases. Speed increases as load and work output decrease. T-gas is far away from Tc or Th for there respective periods.

In other words, constant T expansion is much lower than Th, and constant T compression is much bigger than Tc. Thus the thermodynamic efficiency is already worse than Carnot before it even starts working. This is why the most successful engines have had large surface areas for the hot cold and regenerator equipment.

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[VincentG]

stirling-cycle-diagram (2).png

The area marked in green is the work needed to recompress the gas by the piston to the gas. Negative work because the direction is reversed. Energy is going from some outside source, flywheel crank, through the piston into the gas.

To maintain Tc, the heat of compression must be removed. The heat of compression is equal to the work put in. This is also 100% work to heat conversion. It is negative.

This work is removed from the forward stroke. Subtracted. Note that the subtraction leaves the same area as that enclosed by the cycle as in the first explanation.

This work is also calculated by integration with the floor being the X-axis, zero Kelvin.

If temperature is allowed to rise, Tc will rise towards Th, the area of negative work get larger. Cycle work is reduced and efficiency drops. Less work out per cycle. Same amount of heat in. Without adiabatic lines on the diagram, this is not shown.

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Unfortunately the image does not copy, but how do you define this area below the pv diagram itself(y axis). All the way to 0 bar?

It is more effective to have a double acting piston and two displacer chambers. As the Stirling Brothers demonstrated two hundred years ago. And we think we are discovering something new. Oh my!

Lol, not much new to discover anymore. Jim Dandy demonstrated that as well, but does it help with thermodynamic efficiency?

[VincentG]

Process #1, in order to be thermodynamically complete would need to continue until internal pressure reached 0 bar. This is effectively impossible, and a paradox emerges in that the gas must also maintain Tmax until 0 bar and can therefore absorb and endless amount of energy from the heater. So where must we draw an end to this essentially endless expansion?

[Fool]

Unfortunately the image does not copy, but how do you define this area below the pv diagram itself(y axis). All the way to 0 bar?

I think you are talking about the area under the return stroke line, process #3, and the Zero Kelvin line, also the X-axis. That is the total work returned back into the gas by the piston. Negative, subtractive.



I ran into that image copying problem myself. It is a royal pain... I've solved it by downloading the image to my phone, and reattaching it.

Area under the PV diagram to zero K.

stirling-cycle-diagram (2).png (35.79 KiB) Viewed 2824 times

Process #1, in order to be thermodynamically complete would need to continue until internal pressure reached 0 bar. This is effectively impossible, and a paradox emerges in that the gas must also maintain Tmax until 0 bar and can therefore absorb and endless amount of energy from the heater. So where must we draw an end to this essentially endless expansion?

I seem to remember a lecture by Edward Bradley, Mathematician TA WSU Pullman Wa., where the inside of the natural log function was finite, despite going to Infinity. He also said a student sarcastically hollered out "Fine! So build one! LOL". Sorry, seemed relevant. Finite area means finite heat, not endless.

I think you just nailed why Carnot has a two part expansion. One at constant Th stopping at atmospheric. And the second adiabatic and stopping at Tc. Perfect stopping points. The process is hard to implement, but easy to understand. First stopping point is where work output goes to zero, and work input starts. The second stopps at Tc, after the expense of work in, because any further would just add hysteresis.

Carnot's idea for the reverse compression stroke is just a mirror image of the two expansions, except at a lower temperature, starting at Tc ending at Th. A beautiful use of solid thermodynamic knowledge, especially for the time period.

He had to understand adiabatic expansion and compression with work, and spontaneous temperature changes. He also understood constant temperature in equates to higher work output. Adiabatic expansion leads to lower work output. Very solid thermodynamics. Thanks Sir Carnot.

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