Hello,
We have a new Stirling engine concept.
To understand how it works, see this document:
http://www.monotherme.com/HEAT%20PUMP%2 ... NCIPLE.pdf
If we add a power piston to this incredible heat pump then, we will achieve very high efficiency.
With : Tc = 300k and Th = 600k, helium being the working fluid and dead volume=5%, we could obtain, according to our calculations :
Tc- = 260k (compartment B full) and a Th+ = 670k (compartment A full).
So, about pressure: Pmax/Pmin ≈ 2.3 taking into account all the dead volumes!
Yes: " 2.3 " ... with Tc = 300k and Th = 600k !!!...
For maximum efficiency, the ability of the device to recycle thermal energy (especially the regenerator) is the
paramount constraint (the dead volume of the regenerator is a secondary constraint).
If you want to understand from where this new concept comes, you can read this document :
http://www.monotherme.com/Diathermic%20 ... er%202.pdf
The Fig N°7 show another Stirling engine concept.
I hope you will find this new idea interesting.
What do you think ?
Lucien
A new Stirling engine concept ?
Re: A new Stirling engine concept ?
Lucien,
What says Arthur C. Clarke may be is true, (even), an important demonstration was discovered, once, by a student at a school, when, for fun, a professor asked to him the mathematical proof (considered by all mathematicians as impossible to prove).
However it 's always too easy (and also risky) to speak of "almost zero" in theory.
Build a simple, elementar, prototype, ....and then see if it works. This must be the first step for the Tamil Nadu inventors, and this is especially crucial if are a theorist, and can not really see where the boat is sinking. Before scream at marvel.
I have not looked (also the fault of my bad English) the contents carefully, but I'd be really happy if they pulled a rabbit out of hat. Remember that hell (it's an Italian proverb) is paved with good intentions, and school students are likely to take more mediocre marks, than to prove theorems in mathematics alone.
In short is good have trust; but is better spend time to built a prototype (and give the results), than lose it, quoting Clarke.
Ferraccio
What says Arthur C. Clarke may be is true, (even), an important demonstration was discovered, once, by a student at a school, when, for fun, a professor asked to him the mathematical proof (considered by all mathematicians as impossible to prove).
However it 's always too easy (and also risky) to speak of "almost zero" in theory.
Build a simple, elementar, prototype, ....and then see if it works. This must be the first step for the Tamil Nadu inventors, and this is especially crucial if are a theorist, and can not really see where the boat is sinking. Before scream at marvel.
I have not looked (also the fault of my bad English) the contents carefully, but I'd be really happy if they pulled a rabbit out of hat. Remember that hell (it's an Italian proverb) is paved with good intentions, and school students are likely to take more mediocre marks, than to prove theorems in mathematics alone.
In short is good have trust; but is better spend time to built a prototype (and give the results), than lose it, quoting Clarke.
Ferraccio
Re: A new Stirling engine concept ?
Hello Ferraccio,
We know exactly what is happening in this heat pump. The principle is soooo... simple, soooo... easy to understand. Even a young studient can understand it (with clear explanations).
A small cluster of molecules leaves the cold compartment, raises its temperature by passing through the regenerator (where it absorbs heat) and continues to raise its temperature in the hot compartment (by compression).
If someone wants to dismiss the principle, he will have to demonstrate that the pressure increase do not produce a temperature increase. Good luck! He will have to deeply modify the thermodynamic laws, to write them again.
Yes, building a prototype is a good idea. Many will come (from many places).
We can find some rough calculations here (without dead volumes):
http://www.monotherme.com/Tableau%201bis.xls
Lucien.
We know exactly what is happening in this heat pump. The principle is soooo... simple, soooo... easy to understand. Even a young studient can understand it (with clear explanations).
A small cluster of molecules leaves the cold compartment, raises its temperature by passing through the regenerator (where it absorbs heat) and continues to raise its temperature in the hot compartment (by compression).
If someone wants to dismiss the principle, he will have to demonstrate that the pressure increase do not produce a temperature increase. Good luck! He will have to deeply modify the thermodynamic laws, to write them again.
Yes, building a prototype is a good idea. Many will come (from many places).
We can find some rough calculations here (without dead volumes):
http://www.monotherme.com/Tableau%201bis.xls
Lucien.
Re: A new Stirling engine concept ? Unlikely.
.......Lets see if I can burn you in a paragraphs or two: Reading your document about a heat pump with explainations about how air absorbs and releases heat and then convienently exposing the addition of a power cylinder will result in "a new Stirling concept" will not even get you a date in front of a group of mechanical engineering students. The math may sound great and if so, do you have a Stirling engine or a heat pump? Thermodynamics may lead to interesting mechanical devices but talking much about Thermal principles and a compound regenerator still leaves Stirling without anything new to offer here.
Re: A new Stirling engine concept ?
Dear Longboy,
How do you feel now? I hope you are going well. If you need some help, send me a PM.
Regards,
Lucien01
Hello Ferraccio,
Do you think this is correct or wrong (from the "heat pump" document):
"""a cluster of molecules leaves the hot compartment and its temperature decreases by passing through the regenerator (where it releases heat) then keeps decreasing in the cold compartment (by expansion).""""
""""The gaseous fluid which was already at the same temperature as the one of the cold sink in compartment B will reach, by expansion, a temperature lower than that of the cold sink. This temperature, lower than Tc, is indicated by Tc-. It is not the displacer piston which performs this expansion.""""
?
Lucien01
How do you feel now? I hope you are going well. If you need some help, send me a PM.
Regards,
Lucien01
Hello Ferraccio,
Do you think this is correct or wrong (from the "heat pump" document):
"""a cluster of molecules leaves the hot compartment and its temperature decreases by passing through the regenerator (where it releases heat) then keeps decreasing in the cold compartment (by expansion).""""
""""The gaseous fluid which was already at the same temperature as the one of the cold sink in compartment B will reach, by expansion, a temperature lower than that of the cold sink. This temperature, lower than Tc, is indicated by Tc-. It is not the displacer piston which performs this expansion.""""
?
Lucien01
Re: A new Stirling engine concept ?
Why I would feel........"much better" when you get to the Stirling part of this heat pump Lucien01! Now about that adding a power piston to this pump for more efficiency. Write a paper about that and share it with us sometime! 
Re: A new Stirling engine concept ?
In summary it is good as I thought: on the one hand obvious concepts, and other "pulled by the hair".
The second concept is not defined, whereas a Stirling normally runs from 1000 to 4000 rpm (if mechanical) and very , much faster if acoustic. In fact, in a Stirling does Not exist "expansion" (as after an orifice) but a move (is a displacer, and not a compressor...), and therefore it is impossible that the temperature falls below the cold source, with this RPM is already good if it is not many degrees ABOVE the cold source.
If you rely on abstract concepts (easier in abstruse philosophical!), and were not in strict thermodynamic calculations, (or worse, you think that the basic tools of thermodynamics are the options) will not get anywhere.
It's not me I have to see those calculations.
Here we talk a few about Stirling engines, and believe me, there's already going to miss ..
The second concept is not defined, whereas a Stirling normally runs from 1000 to 4000 rpm (if mechanical) and very , much faster if acoustic. In fact, in a Stirling does Not exist "expansion" (as after an orifice) but a move (is a displacer, and not a compressor...), and therefore it is impossible that the temperature falls below the cold source, with this RPM is already good if it is not many degrees ABOVE the cold source.
If you rely on abstract concepts (easier in abstruse philosophical!), and were not in strict thermodynamic calculations, (or worse, you think that the basic tools of thermodynamics are the options) will not get anywhere.
It's not me I have to see those calculations.
Here we talk a few about Stirling engines, and believe me, there's already going to miss ..
Re: A new Stirling engine concept ?
Nobody seems to understand that, FIRST, we have a Maxwell's demon!
The heat pump have NO power piston !
Please, try to answer THESE questions:
1- Do you think that a Th+ (a higher temperature than the hot source) and a Tc- (a lower temperature than the cold sink) can "appear" ? (without spending mechanical energy !!!!!!!!.....)
yes, no ?
2- Is this really a new heat pump process or principle ? (if it is... then it's very interesting to study, no ?)
yes, no ?
3- What do you think about the efficiency of this heat pump ?
The document to study:
http://www.monotherme.com/HEAT%20PUMP%2 ... NCIPLE.pdf
Regards,
The heat pump have NO power piston !
Please, try to answer THESE questions:
1- Do you think that a Th+ (a higher temperature than the hot source) and a Tc- (a lower temperature than the cold sink) can "appear" ? (without spending mechanical energy !!!!!!!!.....)
yes, no ?
2- Is this really a new heat pump process or principle ? (if it is... then it's very interesting to study, no ?)
yes, no ?
3- What do you think about the efficiency of this heat pump ?
The document to study:
http://www.monotherme.com/HEAT%20PUMP%2 ... NCIPLE.pdf
Regards,
Re: A new Stirling engine concept ?
No! For Th+ I don't think fluid temperature behond its heat input increases thru compression. It can work the other way around without mechanics. A compressed fluid when heated, raises its temperature. The best you can hope for is an increase in fluid temperature thru mechanical compression and the hotter the fluid, greater mechanical energy is required for compression. To increase the temperature thru compression of a pre-heated fluid to behond the input heat source temperature.... I do not know. I would assume your fluid would need to be very hot before compression to energize/ opperate the device its contained in. For Tc.... may be to a point. Can be done statically packing ice around the cold sink to lower ambient temperature of cold sink but how did we get the ice as this is ordinarily a mechanical process of refridgeration? So it still looks like mechanical efforts are definitly needed to do these hot/cold cycles. As for Maxwells demon, regulating individual molocules based on their temperatures thru a gate. And this (gate) is a mechanical means and the second thermodynamic law remains intact. The "apperance" has not been explained without mechanical input. In Stirling, we all wish our models would just self start (appear). We depend very much on mechanics to initiate Stirling principles here.Lucien01 wrote:Nobody seems to understand that, FIRST, we have a Maxwell's demon!
The heat pump have NO power piston !
Please, try to answer THESE questions:
1- Do you think that a Th+ (a higher temperature than the hot source) and a Tc- (a lower temperature than the cold sink) can "appear" ? (without spending mechanical energy !!!!!!!!.....)
yes, no ?
Regards,
-
jesterthought
- Posts: 21
- Joined: Sun May 20, 2007 5:22 am
Re: A new Stirling engine concept ?
I don’t think Auroville’s Remy Mercier is claiming either an engine or a heat pump. What he is describing is a phenomenon that is an incidental by-product of regeneration that is going on in our engines anyway.
Let his displacer piston be balanced by having a push rod at both ends. Then, internal pressure does not push the piston. If the sliding friction is zero and the flow resistance during transfer is also zero, then no external power is required to shuttle working fluid (which I’ll call air for shorthand) through the regenerator from the cold end, B, to the hot end, A, and back.
Assuming there are perfect heat exchangers in the source, at Th, and the sink, at Tc, then the ends of the regenerator will be pegged at Th and Tc as he says, so there will be a constant temperature gradient along it and it will be isothermal. Assume perfect heat exchange as air flows through the regenerator. During displacement from B to A, air will pick up heat in the regenerator and so will be entering A at Th, constant.
What Mercier points out is that subsequent slugs of air passing through the regenerator will be expanded by that temperature rise, thereby compressing the air already in A, which will cause it to be adiabatically heated above Th, to Th+. Note that the air in B will be equally compressed so its temperature will be increased, too.
As the displacement continues, assuming no mixing in A, the earlier air is moved along with the piston by the later air. Thus, the earlier is air being continuously compressed so its temperature is continuously increasing. There is therefore a temperature gradient in A from Th at the inlet to Th+ at the piston. Th+ is increasing and becomes a maximum at the end of the stroke.
As Mercier indicates, no external mechanical power is needed for displacement in this (perfect) machine because there is no back pressure. The air is first heated to Th in passing through the regenerator, taking heat from the regenerator, then its temperature is increased further by adiabatic compression to Th+, the energy coming again from the regenerator.
Note that the pressure is increasing in the regenerator, too because the pressure is equal all round the circuit, so Pa = Pb = Pr. The heat taken from the regenerator is more than that needed to increase the temperature at constant volume because the air is expanding a bit (by, as stated, compressing air in A & B). By expanding, it is doing work (on the rest of the air). That work equals the extra heat taken from the regenerator.
How is the regenerator recharged to maintain its constant temperature? By the return displacement. Assuming no mixing, air exiting A enters the regenerator at Th at first. Air flowing through the regenerator towards the cold end B is cooled, contracts and allows the Air in A (& B) to expand adiabatically and cool a bit. Without doing the sums, I feel intuitively that the process is the reverse of the heating and that the air has cooled to Th by the time it arrives at the exit from A, The hottest air near the piston has more expansion, so it can cool the most to Th. Thus the air heats the regenerator back up to Th at this end. In this perfect machine, it doesn’t need to draw any heat from the source (so the prefect heat exchanger isn’t needed after all). Of course, like all regenerators, the matrix hardly changes temperature at all. That is because the matrix has a huge thermal capacity compared with that of the air shuttling through it.
So Mercier is correct in stating that no mechanical power is required to produce temperatures (in some of the air) above the source temperature, Th, and below the sink temperature, Tc. That is surprising, but I believe it now. What he doesn’t point out is that this will also be happening in all our regenerating engines. Wow! But I’m not sure that the effect is sufficient to be worth any complication to harness it. And I doubt that in any practical engine the effect could ever be measured. But it’s interesting and instructive (to me, anyway).
Let his displacer piston be balanced by having a push rod at both ends. Then, internal pressure does not push the piston. If the sliding friction is zero and the flow resistance during transfer is also zero, then no external power is required to shuttle working fluid (which I’ll call air for shorthand) through the regenerator from the cold end, B, to the hot end, A, and back.
Assuming there are perfect heat exchangers in the source, at Th, and the sink, at Tc, then the ends of the regenerator will be pegged at Th and Tc as he says, so there will be a constant temperature gradient along it and it will be isothermal. Assume perfect heat exchange as air flows through the regenerator. During displacement from B to A, air will pick up heat in the regenerator and so will be entering A at Th, constant.
What Mercier points out is that subsequent slugs of air passing through the regenerator will be expanded by that temperature rise, thereby compressing the air already in A, which will cause it to be adiabatically heated above Th, to Th+. Note that the air in B will be equally compressed so its temperature will be increased, too.
As the displacement continues, assuming no mixing in A, the earlier air is moved along with the piston by the later air. Thus, the earlier is air being continuously compressed so its temperature is continuously increasing. There is therefore a temperature gradient in A from Th at the inlet to Th+ at the piston. Th+ is increasing and becomes a maximum at the end of the stroke.
As Mercier indicates, no external mechanical power is needed for displacement in this (perfect) machine because there is no back pressure. The air is first heated to Th in passing through the regenerator, taking heat from the regenerator, then its temperature is increased further by adiabatic compression to Th+, the energy coming again from the regenerator.
Note that the pressure is increasing in the regenerator, too because the pressure is equal all round the circuit, so Pa = Pb = Pr. The heat taken from the regenerator is more than that needed to increase the temperature at constant volume because the air is expanding a bit (by, as stated, compressing air in A & B). By expanding, it is doing work (on the rest of the air). That work equals the extra heat taken from the regenerator.
How is the regenerator recharged to maintain its constant temperature? By the return displacement. Assuming no mixing, air exiting A enters the regenerator at Th at first. Air flowing through the regenerator towards the cold end B is cooled, contracts and allows the Air in A (& B) to expand adiabatically and cool a bit. Without doing the sums, I feel intuitively that the process is the reverse of the heating and that the air has cooled to Th by the time it arrives at the exit from A, The hottest air near the piston has more expansion, so it can cool the most to Th. Thus the air heats the regenerator back up to Th at this end. In this perfect machine, it doesn’t need to draw any heat from the source (so the prefect heat exchanger isn’t needed after all). Of course, like all regenerators, the matrix hardly changes temperature at all. That is because the matrix has a huge thermal capacity compared with that of the air shuttling through it.
So Mercier is correct in stating that no mechanical power is required to produce temperatures (in some of the air) above the source temperature, Th, and below the sink temperature, Tc. That is surprising, but I believe it now. What he doesn’t point out is that this will also be happening in all our regenerating engines. Wow! But I’m not sure that the effect is sufficient to be worth any complication to harness it. And I doubt that in any practical engine the effect could ever be measured. But it’s interesting and instructive (to me, anyway).
Re: A new Stirling engine concept ?
Yes, exactly. And it is interresting to point out that there is a "dilution effect" minimising the rise in temperature when the hot compartment is filled (because the air enter at Th temperature, not Th+). And there is a "dilution effect" minimising the decrease of the temperature cold side when it is filled (because the air enter at Tc temperature, not Tc-). Of course, when a compartment is drained there is no "dilution effect" for this compartment and the temperature can rise or decrease a lot.jesterthought wrote: What Mercier points out is that subsequent slugs of air passing through the regenerator will be expanded by that temperature rise, thereby compressing the air already in A, which will cause it to be adiabatically heated above Th, to Th+. Note that the air in B will be equally compressed so its temperature will be increased, too.
We have calculated the effect (without dead volumes) and it is not tiny.jesterthought wrote: But I’m not sure that the effect is sufficient to be worth any complication to harness it. And I doubt that in any practical engine the effect could ever be measured.
http://www.monotherme.com/Tableau%201bis.xls
You can modify the temperatures and the gamma (in the yellow cells) and check the results in « RESULTATS ». If we use helium and Th=600k and Tc=300k we have two secondary deltaT :
711-455=256k hot side and 374-244=130K cold side.
Maximum Th+ is 711K and minimum Tc- is 244K.
If we increase the primary deltaT then the effect becomes huge.
And it will be possible (and easy) to use two or three devices together so the secondary deltaT will be two or tree times bigger!... So the dead volume is not problematic (we have calculations with dead volumes).
But, THE great idea is to avoid to use the Th+ and the Tc- to run a Stirling engine or some other kind of thermo-converter device: WE WILL USE THE SECONDARY deltaT (cold side) to run a Stirling engine or some other kind of thermo-converter device!
Why?
Because now we can extract thermic energy from inside the cold side without spending mechanical energy (we produce it!) and without spending the energy of the hot source...
Wow! A Maxwell's demon?
-
jesterthought
- Posts: 21
- Joined: Sun May 20, 2007 5:22 am
Re: A new Stirling engine concept ?
Lucien01, you show that the action has increased the source/sink dT from 300 to 467*k. That effect is much more than I expected, intuitively. It has also reduced the cold side to -29*C! Do you know of any tests to verify this practically?
To save searching through your Sheet, have you assumed the fluid mixes in the cylinder so it is at the same temperature from head to piston? I ask because I have experience of air not mixing when charging long thin bottles. The air already in the bottle is compressed in the far end and at the full adiabatic rise, while the final air to enter is almost at supply temperature. Charging to 20Atm produced a large temperature gradient along the bottle.
If the effect is so large – and it applies to all regeneration – why is it not widely reported in engine analyses? This thread is the first I have heard of it. Since it takes no heat from the source, nor mechanical work input, it cannot affect overall engine performance. Um, or am I missing something?
Further, can your super proposal to use the principle to cool the cold end of a heat engine work, without drawing heat from the source?
Vereeeeeee intriguing.
Jester.
To save searching through your Sheet, have you assumed the fluid mixes in the cylinder so it is at the same temperature from head to piston? I ask because I have experience of air not mixing when charging long thin bottles. The air already in the bottle is compressed in the far end and at the full adiabatic rise, while the final air to enter is almost at supply temperature. Charging to 20Atm produced a large temperature gradient along the bottle.
If the effect is so large – and it applies to all regeneration – why is it not widely reported in engine analyses? This thread is the first I have heard of it. Since it takes no heat from the source, nor mechanical work input, it cannot affect overall engine performance. Um, or am I missing something?
Further, can your super proposal to use the principle to cool the cold end of a heat engine work, without drawing heat from the source?
Vereeeeeee intriguing.
Jester.
Re: A new Stirling engine concept ?
Hello Jester,
Without mass flow, no power. If the useful mass flow is 100% then the secondary deltaT will be reduced from 130 to 65k. If the useful mass flow is 60% then the secondary deltaT (cold side) will be reduced from 130 to 91k.
The cold side is full. The pressure is minimum.
30% of the helium comes out through a heat exchanger HE1. If we don't tap HE1 (and without any losses) then the average temperature of HE1 could be 255k.
40% of the helium comes out through a HE2. The average temperature of the HE2 is 300k. It is "connected" to the surrounding (room temperature = 300k).
30% of the helium comes out through a heat exchanger HE3. If we don't tap HE3 then the temperature of HE3 will be 346k.
The cold side is empty and the pressure is maximum.
Then, HE2 is used to fill the cold compartment at 300k.
So, HE3 is at 346k and can transfer some thermal energy to the surrounding and HE1 is at 255k and can absorb some thermal energy from the surrounding. If we use a big HE3 and if we insulate HE1 then the deltaT will be modified like that: maximum = 310k and minimum = 230k. If we use a big HE1 and if we insulate HE3 then the deltaT will be modified like that: maximum = 390k and minimum = 290k.
This "effect" (the birth of a secondary deltaT cold side and hot side)... the amplitude of this effect is a consequence of the doubling (rough calculations above) of the pressure (without any dead volume and losses in a perfect machine), with and without dilution. Doubling the pressure is a lot. I don't know Stirling engines (tell me if I am right) running with a higher pressure ratio than 1,8.
With a smaller pressure ratio, the effect is smaller.
With a dead volume of 5% we will have to reduce (again) the useful secondary deltaT by 20% about...
With a power piston, we will have to think that about half its capacity is like a huge dead volume killing the remaining "effect"!...
Anyway, there is a few solutions to increase a lot the secondary deltaT.
Lucien01
Without mass flow, no power. If the useful mass flow is 100% then the secondary deltaT will be reduced from 130 to 65k. If the useful mass flow is 60% then the secondary deltaT (cold side) will be reduced from 130 to 91k.
The cold side is full. The pressure is minimum.
30% of the helium comes out through a heat exchanger HE1. If we don't tap HE1 (and without any losses) then the average temperature of HE1 could be 255k.
40% of the helium comes out through a HE2. The average temperature of the HE2 is 300k. It is "connected" to the surrounding (room temperature = 300k).
30% of the helium comes out through a heat exchanger HE3. If we don't tap HE3 then the temperature of HE3 will be 346k.
The cold side is empty and the pressure is maximum.
Then, HE2 is used to fill the cold compartment at 300k.
So, HE3 is at 346k and can transfer some thermal energy to the surrounding and HE1 is at 255k and can absorb some thermal energy from the surrounding. If we use a big HE3 and if we insulate HE1 then the deltaT will be modified like that: maximum = 310k and minimum = 230k. If we use a big HE1 and if we insulate HE3 then the deltaT will be modified like that: maximum = 390k and minimum = 290k.
This "effect" (the birth of a secondary deltaT cold side and hot side)... the amplitude of this effect is a consequence of the doubling (rough calculations above) of the pressure (without any dead volume and losses in a perfect machine), with and without dilution. Doubling the pressure is a lot. I don't know Stirling engines (tell me if I am right) running with a higher pressure ratio than 1,8.
With a smaller pressure ratio, the effect is smaller.
With a dead volume of 5% we will have to reduce (again) the useful secondary deltaT by 20% about...
With a power piston, we will have to think that about half its capacity is like a huge dead volume killing the remaining "effect"!...
Anyway, there is a few solutions to increase a lot the secondary deltaT.
Lucien01
-
jesterthought
- Posts: 21
- Joined: Sun May 20, 2007 5:22 am
Re: A new Stirling engine concept ?
Lucien01, I am not familiar enough with your thinking to learn much from your précis. I presume the figures are from your Sheet.
You show how increasing mass flow reduces the dT. Do you mean you are extracting a mass flow of x% of Mercier’s system flow?
Or do you mean you are circulating some of Mercier’s flow through HE1 for heat exchange?
I fear I can’t perceive enough to understand.
In your earlier post you described what sounded like cascading several Merciers for a large dT.
Any images?
Jester.
You show how increasing mass flow reduces the dT. Do you mean you are extracting a mass flow of x% of Mercier’s system flow?
Or do you mean you are circulating some of Mercier’s flow through HE1 for heat exchange?
I fear I can’t perceive enough to understand.
In your earlier post you described what sounded like cascading several Merciers for a large dT.
Any images?
Jester.
Re: A new Stirling engine concept ?
Yes, two images (of the heat pump). I was speaking "heat pump" (this is a first step, I think the most interesting).
http://www.hostingpics.net/viewer.php?i ... tpump1.jpg
http://www.hostingpics.net/viewer.php?i ... tpump2.jpg
Cold side, we have three heat exchanger E1 (Tc-), E2 (Tc) and E3 (Tc+).
In french "Tcold" is "Tfroide" (Tf)
The "Distributeur" allow to dispatch the air or helium via the pipes C1, C2 or C3.
So... we have a secondary deltaT (and some thermal power) between E1 and E3 without spending mechanical energy.
With a perfect regenerator, I am not able to find a way for the heat (hot source) to go from the hot side to the cold side...
Do you?
"cascading", "larger dT" ?
Yes.
Two devices:
we connect the E1 of a first device to the E3 of the second device.
Now, the E3 of the first device is at a higher temperature and the temperature of the E1 of the second device is at a lower temperature. But we have to disconnect (to the surrounding) the E2 of the first device OR the E2 of the second device OR to disconnect both and connect the remaining E3 or E1 to the surrounding.
http://www.hostingpics.net/viewer.php?i ... tpump1.jpg
http://www.hostingpics.net/viewer.php?i ... tpump2.jpg
Cold side, we have three heat exchanger E1 (Tc-), E2 (Tc) and E3 (Tc+).
In french "Tcold" is "Tfroide" (Tf)
The "Distributeur" allow to dispatch the air or helium via the pipes C1, C2 or C3.
So... we have a secondary deltaT (and some thermal power) between E1 and E3 without spending mechanical energy.
With a perfect regenerator, I am not able to find a way for the heat (hot source) to go from the hot side to the cold side...
Do you?
"cascading", "larger dT" ?
Yes.
Two devices:
we connect the E1 of a first device to the E3 of the second device.
Now, the E3 of the first device is at a higher temperature and the temperature of the E1 of the second device is at a lower temperature. But we have to disconnect (to the surrounding) the E2 of the first device OR the E2 of the second device OR to disconnect both and connect the remaining E3 or E1 to the surrounding.