Re: The Reversible engine
Posted: Fri Aug 25, 2023 10:24 am
You have 15 or so paragraphs that to me are a ball of confusion nearly impossible to follow or decipher, I'll read through it again, but in the mean time;
Using simple algebra:
This is my THEORY. I don't necessarily consider it anything more than that. A theory to explain my observations. It could, of course be wrong or require modification, and this is, at any rate a SIMPLIFICATION of a very complex system of interactions. On the other hand I don't want to make it TOO simple, by neglecting important elements, such as potential energy, velocity and momentum, along with displacer(heat input) action.
Along with TDC (Top Dead Center or full compression) and BDC (Bottom Dead Center or full expansion) I'll introduce a new acronym NDC (Neutral Dead Center) to indicate the midway point, which can be thought of as a point of equilibrium (neglecting velocity/momentum)
The forces involved in joules are
J = big J is the given internal energy at the start at NDC (equilibrium).
The working fluid starts out at NDC (Neutral Dead Center). With J joules. (The internal pressure is in balance with external atmospheric pressure so these cancel out)
As a Key:
E internal = J joules (given or starting internal energy)
j1 = joules added or taken in to start the engine (for expansion) including work and/or heat.
jv = joules converted to velocity
js = joules converted to external shaft-work
ja = joules added by atmospheric work input
jd = joules converted to displacement work.
We will assume j1=jv=js=ja=jd for simplicity.
At first the internal energy is J
To start the engine someone pushes the flywheel around by hand to get the engine started while also applying heat. So the combination of applied heat (at a specific time, when the engine revolves enough to lift the displacer near TDC), along with the physical work input results in an increase of internal energy at TDC of j1 joules.
So to start the engine, at startup, E internal = J + j1 joules.
At TDC turning the engine over by hand the j1 joules represents potential energy manifesting as increased pressure at TDC
The engine starts.
At TDC due to the increase in E-internal the pressure is greater than the external atmospheric pressure and the piston moves outward "on its own" (without manual work input to start the engine)
In moving out, some of the internal energy is converted to velocity equivalent to the previous internal energy increase (work goes "out" from the working fluid in joules (jv) as an increase in velocity of the piston) and pressure returns to normal.
E-internal = J + j1 - jv
The piston/engine is now in motion. An object in motion stays in motion (until it meets resistance, Atmospheric pressure at this point is not counted as "resistance" as previously noted the system started out in a state of equilibrium with the surroundings)
At NDC during expansion then:
E-internal = J + (j1 - jv)
If j1 = jv then j1 - jv = 0
So as the engine crosses NDC during expansion:
J + 0 = J therefore
E-internal = J at NDC during expansion. But the engine now has velocity/momentum.
Now at NDC during expansion the displacer covers the hot heat exchanger stopping heat input but the piston continues to move due to momentum and the working fluid expands adiabatically doing displacement work jd while also doing additional shaft-work to drive the engine js.
At BDC the piston slows down, the velocity has converted to shaft-work output.
At BDC
E-internal = J - jd - js
At NDC during expansion the "internal energy" of the working fluid already returned to a balance (but the piston/engine still has velocity/momentum.)
At BDC the gas has expanded from NDC adiabatically (-jd) while also outputting shaft work.(-js).
(jd = displacement work in joules during expansion. js = shaft-work output in joules during expansion)
Now at BDC there is an imbalance and internal pressure is less than atmospheric pressure and the piston stops. The velocity has been converted to "shaft work".(work output external to the "system"). The gas has also expanded doing its own displacement work.
Now due to the low internal energy/pressure the piston begins to move in the opposite direction towards TDC.
At that point (BDC to NDC for about 1/4 cycle) the working fluid is COLD having expanded and lost internal energy to external shaft-work output and so is ready to absorb heat more quickly and easily when heat is made available. (Heat is not yet available, but has been building up in the heat exchanger the hot heat exchanger insulated from the working fluid by the displacer).
Traveling back to NDC atmosphere does "work" on the gas so at NDC the piston has again picked up velocity and:
E-internal = J - jd - js + ja
-jd and +ja cancel leaving -jd so:
E-internal at NDC During "compression" is:
E-internal = J - js
The internal energy is still in deficit at NDC during "compression".
E-internal = J -js joules at NDC but the piston is now in motion (with velocity) moving towards TDC.
From NDC to TDC the velocity of the piston slows down due to increasing pressure and the velocity is converted into heat.
At TDC you again have E internal = J + j1 joules with some heat input from the now exposed hot heat exchanger equivalent to js (previous shaft work output)
But this time the j1 joules are supplied by the momentum of the piston rather than the mechanical input used to start the engine.
And from there the cycle continues.
If the shaft work increases due to the application of an external load this will generally be compensated by an increase in j1 due to the general engine dynamics or the engine will slow down and/or stop is j1 cannot increase to balance the load, but describing the complexities involved in varying the external load and how the engine might respond to that is somewhat beyond this basic description of the engine cycle.
Any engine, if it is running at all is doing some amount of "shaft-work" even if that just amounts to spinning it's own wheels.
To summarize the cycle:
1)NDC (compression): Internal energy = J - js (+ velocity)
(At or rather before startup, of course js would be 0 as the engine hasn't done any work yet and velocity is zero, until the engi e is manually started)
2)TDC: Internal energy = J + j1 (j1 as potential energy/pressure) j1 represents work and heat input to drive 1 expansion.
3)NDC (expansion): Internal energy = J or J (+j1 - jv) pressure drops to normal but now the engine has velocity/momentum.
4)BDC: Internal energy = J - jd - js
During expansion internal energy is lost both in terms of PV displacement work and shaft-work.
I think the problem with the usual "ideal" models, TS diagrams etc. is that no distinction is made between jd ("negligible" displacement work) and js (work output in the form of "external" shaft work)
A typical TS diagram in most textbooks and courses, only accounts for displacement or PV work or erroneously (IMO) assumes displacement work and shaft work to be one and the same which IMO they cannot be.
Calculating shaft work would involve more complex math than represented by a simple PV diagram.
So far, all I've been able to find in relation to calculating shaft work is statements such as, that is beyond the scope of this course. Or that will be covered in (some other) chapter in relation to open systems such as turbines.
There seems to be little recognition that heat engines or a "closed system" like a Stirling engine actually does any external "shaft work" that needs to be accounted for.
Of course, that is understandable as in Caloric theory a water wheel does shaft work by displacement of water from a high level to a low level without any reduction in the volume of the fluid. And heat engine thermodynamics as passed down from Kelvin etc al, is based on Caloric theory.
With heat being energy, however, shaft work does result in a reduction in fluid volume.
When gas expands adiabatically, especially when doing external shaft work, its temperature drops, and so it's pressure drops and so after velocity and momentum are exhausted it's volume will subsequently decrease.
So, for the past 200 years, the unavoidable consequence of heat being energy with an unavoidable subsequent reduction in the volume of the gas or working fluid (in a closed system), with shaft work output has been, apparently, entirely missed or neglected, or gone unrecognized.
As if a gas, upon doing work, does not loose energy or decrease in volume, just like water driving a water wheel passes right through with no apparent loss of energy or volume, but continues on its way down the river.
As if a heat engine only does a negligible amount of displacement work so that the actual shaft work can be ignored.
Thermodynamics started out, entirely steeped in Caloric theory. All the math, diagrams etc and has been, unsuccessfully IMO trying to shore up this caricature and somehow shoehorn new findings and new discoveries into an old worn out obsolete model of heat and heat engine theory, with the unfortunate consequence that it no longer models or represents reality. Not when it comes to Stirling type heat engines anyway.
Where you have mass flow through the system, as in a steam turbine, maybe it has some applicability, but for a Stirling engine, I think some fundamentals, like basic conservation of energy are being overlooked.
How do you add heat as joules, to a closed system, then take out an equivalent amount of energy in joules from the same closed system as work, then imagine it is still necessary to also remove another equivalent amount of heat in joules to return to the initial state?
1 + 1 -1 -1 does not add up to 1
1=1 starting internal energy equals itself
1+1=2 original internal energy plus added heat
1+1-1=1 original internal energy, plus added heat minus work output = original internal energy. (In a closed system at increased volume that would result in "contraction" or a subsequent reduction in volume)
1+1-1-1= -1
You don't have a cycle.
Using simple algebra:
This is my THEORY. I don't necessarily consider it anything more than that. A theory to explain my observations. It could, of course be wrong or require modification, and this is, at any rate a SIMPLIFICATION of a very complex system of interactions. On the other hand I don't want to make it TOO simple, by neglecting important elements, such as potential energy, velocity and momentum, along with displacer(heat input) action.
Along with TDC (Top Dead Center or full compression) and BDC (Bottom Dead Center or full expansion) I'll introduce a new acronym NDC (Neutral Dead Center) to indicate the midway point, which can be thought of as a point of equilibrium (neglecting velocity/momentum)
The forces involved in joules are
J = big J is the given internal energy at the start at NDC (equilibrium).
The working fluid starts out at NDC (Neutral Dead Center). With J joules. (The internal pressure is in balance with external atmospheric pressure so these cancel out)
As a Key:
E internal = J joules (given or starting internal energy)
j1 = joules added or taken in to start the engine (for expansion) including work and/or heat.
jv = joules converted to velocity
js = joules converted to external shaft-work
ja = joules added by atmospheric work input
jd = joules converted to displacement work.
We will assume j1=jv=js=ja=jd for simplicity.
At first the internal energy is J
To start the engine someone pushes the flywheel around by hand to get the engine started while also applying heat. So the combination of applied heat (at a specific time, when the engine revolves enough to lift the displacer near TDC), along with the physical work input results in an increase of internal energy at TDC of j1 joules.
So to start the engine, at startup, E internal = J + j1 joules.
At TDC turning the engine over by hand the j1 joules represents potential energy manifesting as increased pressure at TDC
The engine starts.
At TDC due to the increase in E-internal the pressure is greater than the external atmospheric pressure and the piston moves outward "on its own" (without manual work input to start the engine)
In moving out, some of the internal energy is converted to velocity equivalent to the previous internal energy increase (work goes "out" from the working fluid in joules (jv) as an increase in velocity of the piston) and pressure returns to normal.
E-internal = J + j1 - jv
The piston/engine is now in motion. An object in motion stays in motion (until it meets resistance, Atmospheric pressure at this point is not counted as "resistance" as previously noted the system started out in a state of equilibrium with the surroundings)
At NDC during expansion then:
E-internal = J + (j1 - jv)
If j1 = jv then j1 - jv = 0
So as the engine crosses NDC during expansion:
J + 0 = J therefore
E-internal = J at NDC during expansion. But the engine now has velocity/momentum.
Now at NDC during expansion the displacer covers the hot heat exchanger stopping heat input but the piston continues to move due to momentum and the working fluid expands adiabatically doing displacement work jd while also doing additional shaft-work to drive the engine js.
At BDC the piston slows down, the velocity has converted to shaft-work output.
At BDC
E-internal = J - jd - js
At NDC during expansion the "internal energy" of the working fluid already returned to a balance (but the piston/engine still has velocity/momentum.)
At BDC the gas has expanded from NDC adiabatically (-jd) while also outputting shaft work.(-js).
(jd = displacement work in joules during expansion. js = shaft-work output in joules during expansion)
Now at BDC there is an imbalance and internal pressure is less than atmospheric pressure and the piston stops. The velocity has been converted to "shaft work".(work output external to the "system"). The gas has also expanded doing its own displacement work.
Now due to the low internal energy/pressure the piston begins to move in the opposite direction towards TDC.
At that point (BDC to NDC for about 1/4 cycle) the working fluid is COLD having expanded and lost internal energy to external shaft-work output and so is ready to absorb heat more quickly and easily when heat is made available. (Heat is not yet available, but has been building up in the heat exchanger the hot heat exchanger insulated from the working fluid by the displacer).
Traveling back to NDC atmosphere does "work" on the gas so at NDC the piston has again picked up velocity and:
E-internal = J - jd - js + ja
-jd and +ja cancel leaving -jd so:
E-internal at NDC During "compression" is:
E-internal = J - js
The internal energy is still in deficit at NDC during "compression".
E-internal = J -js joules at NDC but the piston is now in motion (with velocity) moving towards TDC.
From NDC to TDC the velocity of the piston slows down due to increasing pressure and the velocity is converted into heat.
At TDC you again have E internal = J + j1 joules with some heat input from the now exposed hot heat exchanger equivalent to js (previous shaft work output)
But this time the j1 joules are supplied by the momentum of the piston rather than the mechanical input used to start the engine.
And from there the cycle continues.
If the shaft work increases due to the application of an external load this will generally be compensated by an increase in j1 due to the general engine dynamics or the engine will slow down and/or stop is j1 cannot increase to balance the load, but describing the complexities involved in varying the external load and how the engine might respond to that is somewhat beyond this basic description of the engine cycle.
Any engine, if it is running at all is doing some amount of "shaft-work" even if that just amounts to spinning it's own wheels.
To summarize the cycle:
1)NDC (compression): Internal energy = J - js (+ velocity)
(At or rather before startup, of course js would be 0 as the engine hasn't done any work yet and velocity is zero, until the engi e is manually started)
2)TDC: Internal energy = J + j1 (j1 as potential energy/pressure) j1 represents work and heat input to drive 1 expansion.
3)NDC (expansion): Internal energy = J or J (+j1 - jv) pressure drops to normal but now the engine has velocity/momentum.
4)BDC: Internal energy = J - jd - js
During expansion internal energy is lost both in terms of PV displacement work and shaft-work.
I think the problem with the usual "ideal" models, TS diagrams etc. is that no distinction is made between jd ("negligible" displacement work) and js (work output in the form of "external" shaft work)
A typical TS diagram in most textbooks and courses, only accounts for displacement or PV work or erroneously (IMO) assumes displacement work and shaft work to be one and the same which IMO they cannot be.
Calculating shaft work would involve more complex math than represented by a simple PV diagram.
So far, all I've been able to find in relation to calculating shaft work is statements such as, that is beyond the scope of this course. Or that will be covered in (some other) chapter in relation to open systems such as turbines.
There seems to be little recognition that heat engines or a "closed system" like a Stirling engine actually does any external "shaft work" that needs to be accounted for.
Of course, that is understandable as in Caloric theory a water wheel does shaft work by displacement of water from a high level to a low level without any reduction in the volume of the fluid. And heat engine thermodynamics as passed down from Kelvin etc al, is based on Caloric theory.
With heat being energy, however, shaft work does result in a reduction in fluid volume.
When gas expands adiabatically, especially when doing external shaft work, its temperature drops, and so it's pressure drops and so after velocity and momentum are exhausted it's volume will subsequently decrease.
So, for the past 200 years, the unavoidable consequence of heat being energy with an unavoidable subsequent reduction in the volume of the gas or working fluid (in a closed system), with shaft work output has been, apparently, entirely missed or neglected, or gone unrecognized.
As if a gas, upon doing work, does not loose energy or decrease in volume, just like water driving a water wheel passes right through with no apparent loss of energy or volume, but continues on its way down the river.
As if a heat engine only does a negligible amount of displacement work so that the actual shaft work can be ignored.
Thermodynamics started out, entirely steeped in Caloric theory. All the math, diagrams etc and has been, unsuccessfully IMO trying to shore up this caricature and somehow shoehorn new findings and new discoveries into an old worn out obsolete model of heat and heat engine theory, with the unfortunate consequence that it no longer models or represents reality. Not when it comes to Stirling type heat engines anyway.
Where you have mass flow through the system, as in a steam turbine, maybe it has some applicability, but for a Stirling engine, I think some fundamentals, like basic conservation of energy are being overlooked.
How do you add heat as joules, to a closed system, then take out an equivalent amount of energy in joules from the same closed system as work, then imagine it is still necessary to also remove another equivalent amount of heat in joules to return to the initial state?
1 + 1 -1 -1 does not add up to 1
1=1 starting internal energy equals itself
1+1=2 original internal energy plus added heat
1+1-1=1 original internal energy, plus added heat minus work output = original internal energy. (In a closed system at increased volume that would result in "contraction" or a subsequent reduction in volume)
1+1-1-1= -1
You don't have a cycle.