The theory is there but the real world application seems to be unrealistic. The hot air would produce upward force, but to create a downward pull it would seem that the air inside the cylinder would have to be colder than the air outside, otherwise, it becomes static.
What is really going to spin your head, is when you begin to study sympathetically connecting symbiotic systems with a combined COP greater than 1.0. Say like a heat pump system COP 11.0, with a TEM system COP(.1). Or a heat pump system COP 5.0 with a stirling generator system COP (.25).... okay i'll stop here before i get branded as heretic.... Hint: The combined COP is figured by directly multiplying the COP's together!
This exercise uses only expansion and contraction of a gas in gaseous state: we are bound to just under 2.0 COP. When we take off the kid gloves and include the ability to expand and contract through evaporation and condensation, COP goes hyper. Carnot in his limited scope, puts the theoretical max at COP 200. There is a whole realm of detail, data, math, and observation to explore and record. New formulas must and will be developed to account for this uncharted area of study and focus.
If your perfect generator is able to harness "1x" energy upon the upward stroke, that would imply that there is exactly zero energy available in order to lift the piston. In other words, your generator can only harness a maximum of "1x - Eg" where Eg is the amount of energy necessary to overcome the gravitational force upon the piston. The force used to overcome the gravitational force converts kinetic energy due to gas expansion into gravitational potential energy. COP = 1.0
You are technically correct, but only on one aspect, However overall, you are missing the point. In this IDEAL exercise 2.0 COP is actually a limit and not actually attainable. Brake horsepower at 2.0 COP would stall the piston. Anything approaching but not including 2.0 COP would be valid (I.E. 1.999 COP). As long as the piston actually moves! The part your missing, and the point of the vid.... is the portion from 1.0 COP upwards towards 2.0 COP.
@PulseFuelNerd Are you saying the following?: If the first expansion stroke is adiabatic (no heat out of the system) then the 1x input heat energy is converted into 1x work due to gas expansion. At top dead center, the cylinder stops and the second process begins in which, ideally, 100% of remaining heat energy is bled off to the environment, thus converting this "vacuum stroke" into an additional 1x work.
I just want to get on the same page for the sake of discussion.
@PulseFuelNerd Alright, well assuming we are on the same page (and if I'm off, please direct me as necessary), this still seems problematic. When the piston moves up due to expansion, in an ideal case, 100% of that heat energy is converted into mechanical energy. Let, for example, the only limitation on the movement of the piston be the external pressure. This would imply that the pressure within the cylinder has equalized with that of the environment (but not necessarily the temperature).
@PinellasPatriot As such, if the temperature of the environment happens to still be less than that of the gas within the cylinder, then some heat transfer to the environment could still occur, thus allowing a return stroke and C.O.P. greater than 1.0. Where I take issue, however, is that this is just a transient condition. My point is as follows: after the return stroke (in ideal case of C.O.P. = 2.0) the temperature within the cylinder will have equalized with the environment.
@PinellasPatriot In your second cycle (just like the first), when all of the internal energy is converted to mechanical in the second expansion stroke, the temperature after the fact will return to its base state before heat addition: that of the environment. In this case, there is no mechanism to power the return stroke. This gets me to my point (finally): You could achieve COP = 2.0 (under ideal circumstances) in a transient condition but not at steady state.
Addendum: Q1 - Q2 is not LESS THAN the work done through the piston...that's what I meant to say. In other words, the mechanical work you get out is never more than the heat energy you put in, that is not rejected. You put heat in, only a portion of it turns to mechanical, the rest is rejected.
If you put a mechanical load on it that requires a greater amount of energy to complete than the heat has that it can do work with, you can't expect it to complete the task.
@randommagnum Please think in terms of Ideal theory, set aside applied physics. System is sealed and no loses to environment, until such time is choosen. state1 heat is put into the space between cylinder and piston, the load is progressive just as the pressure applied to the piston is, at end of piston movement, energy leaves system and exteral pressure pushes piston back to rest state energy can again be harnessed from the environment, heat pumps are a type of heat system, and cop>4 in many.
@d3adp001 "heat pumps are a type of heat system, and cop>4 in many."
That is a non-applicable argument because a head pump has work put into it. COP doesn't work the same way with a heat pump as it does with a heat engine.
You & PFN fail to see 3 things: the gas, when expanding/doing work would NOT be isothermal; if there is waste heat (heat leaves system) then work done on piston is LESS than input energy; return stroke energy is stored energy that came from power stroke.
@randommagnum First off you neglect the first premise, IDEAL, second off heat pumps are in fact a heat system in which the exchanged must travel through the engine, which is the same as what is occuring in the carnot, yes the difference is one is turned and one is being turned, but you neglect to observe and explain the environmental effect. the work done after heat leaves is not stored energy its a return to base state
@d3adp001 IF no heat leaves the system then the piston would move due to the expansion of gas. IF the heat never left the system the piston would stay in this excited state. The work performed would be equal to the heat put in less losses which there are none so 1 unit in 1 unit out. Now allow the heat to leave the system. The piston will revert back to original state. the energy comes into the system from the environment. which is what pushes the piston back to rest position.
@d3adp001 So either the max energy that can be extracted is 1/2 for the expansion cycle and 1/2 for the condensation cycle, or its 1 for expansion cycle and potential for 1 on the condensation cycle, Now if you want to say that only the difference between environmental state and excited state can be extracted on the expansion and the environmental can be extracted on the condensation then thats fine. But note one thing, you didnt say it I did, because I am already aware of the possiblities.
Your not really participating in a discussion and not contributing, your simply spitting out what was put into your mouth be someone else. If you have an understanding, then explain it, but to be antagonistic is a waste of your time. Contribute productively or shut up.
Go to Wikipedia and type in "Carnot Cycle". Even in that idealized example, assuming no losses (isentropic) the heat you put in, or Q1 (what you're calling X), minus the amount of heat rejected into the environment, or Q2, is not going to be a greater amount of energy than the work done through the piston. The work the piston does coming down was part of the work the gas did pushing it up (gravity storage).
Whatever energy gets put into living the cup will not be there to escape into the environment.
Let's say you apply heat to the air, but the cup is LOCKED so it can't expand; as the heat goes into the environment, MORE heat will do so because it stayed compressed. If you allow it to push the cup up, it expands and looses temperature, just like refrigerant gas (or an air duster) getting cold when it expands.
The energy into the cup + energy bled into the environment = 1X (1 COP).
You are completely wrong. You do not understand IT at all. Energy is only trasfered. If you transfer energy in a isobaric system, it will cause increase in volume thanks to more kinetic energy of the particles. If you want to get some energy from the piston, you must let it apply some force to generator. THIS FORCE IS TRANSFORMED FORCE FROM VIBRATING PARTICLES, YOU GET ENERGY FROM THEM. BY LETTING THE PISTON TO DO SOME WORK (PRODUCING ENERGY) YOU ARE CAUSING ENERGY DROP IN THE AIR.
Then show the equation defining the action step. show me the equation of energy into the pressure and then transfered out and into kinetic. I am sorry but some stand alone statement that means nothing. if x amount of heat is applied to the gas, the gas applies pressure on the piston, the piston can the move some mass, how high will it lift it? and what is the energy of position vs the input heat energy. answers please.
If I wanted energy from the piston I would burn it
If you can not get into the other half of 2.0COP from simple gas expansion and contraction due to temperature changes... Then you will really be at a loss when we discuss the crazy COP from PHASE CHANGE of liquid state to gaseous state.
Hey folks, There is a new world of discovery to explore and conquer!
New World or Square World?
Red Pill Or Blue Pill?
Please try and apply yourself, until your mind can grasp and see the free other half of temperature change.
the second thing is that say I have a volume of gas. I exert heat on that volume. I can predict the change in volume of gas based of very specific formula. If that volume expands, and then cools at the height of the piston. the same gas has to contract from cooling to let the piston down. Without additional heat that volume of gas will not push the piston up. Just look at hot air balloons. they need additional heat to expand the gas to rise again.
well sir then I ask you to do the math. Here are the conditions STP, 1 G, vol start 200cc, gas hydrogen, no friction, piston = 0 kg. during the heating cycle, the system is a perfect insultator, cooling cycle the system is a perfect conductor.
heat input energy 1kj.
imagine that the piston pushes a 1 kg wieght up during heating, and imagine the piston is connected to rachet that again lifts the 1 kg wiegh as the piston moves down.
What is the energy stored in the 1kg wieght at the end.
First thing. There will be resistance in the sides of the piston going up. Let alone the resistance of the bearings, or lever or whatever pushes the generator that creates the energy to feed back.
Is it possible to do the same experiment with a perfect electromagnet with a perfect coil at the bottom from the cylinder, and a perfect magnet as a piston? Instead of contracting air you would have a collapsing magnetic field and gravity that makes the piston come down....the same physics....am I correct?
I am trying to understand, I'm with you guys, friendly people here.
I am not completely understanding what your getting at tuna, perfect coil/mag etc I understand, but what the piston is doing is eluding me. Are you saying push the magnet up with coil energy and then collect when it comes back down?
But yes, that is exactly what I mean. Pushing up a magnet with a coil, and collect the energy from the upwards AND the downwards movement with a collecting coil. With electronics you can even forget the piston idea and just use a rotor with magnets instead, or a even a motionless energizer with coils only and no magnets at all. This is what Floyd Sweet was doing. Motionless Energy Generator,or MEG. I can't help thinking that it's all the same physics, am I wrong?
with the piston tuna this is what will happen. You shoot it upward, moving freely it will move x distance. with an ideal generator it will move 1/2 x because the generator will slow the piston. The difference is that the speed of the piston indicates how far it moves and friction o it slows it, but the pressure on the piston will keep on pushing, the friction slows it, but cant stop it(well enogh would)
I doubt that the output energy from the combined movement from the expanding and contracting air inside the most perfect cylinder is greater then the input energy that is needed to heat up the air in order to get movement to begin with. Unless you use a solar array or geothermal energy as a source. But I have seen such a system before, it was something with expanding gas and solar heat, and cooling it back down again in order to generate electricity. Some students project as I can remember....
I will try to look it up for you, but it has been some time ago that I have seen it here on youtube. If I can't find it here I will search for it on google. I will let you know if I have it.
I see your point, and I can imagine that it's efficient.
I am not sure if I understand what casting doubt means, but I have no doubts about your integrity. I could not find the video I mentioned before, I'm sorry for that. but I did find another interesting page.
Boyn, I have studied stirlings and other heat engines extensively. can you point to how the mental exercise put forth is different than a stirling engine. Please review those before you come back.
this is an ideal system mental exercise, therefore conversion and losses dont exist. If you can not do that, then your particpation in this excerise will be greatly inhibited.
The heat in a heat pump moves out of the heated and cooled coils by the gradiant created.
the removal of the excited state is for free, the differencial created allows it to happen.
What good is the exercise? Nothing is ideal. "Removal of the excited state" is NOT free. We still have conversion rates with passive or mechanical. Yes you gain higher efficiency from the environment and see higher cop because of that but even at .000001% loss is still a loss.
Love to be with you on this D3 but I need better convincing.
I am not trying to convince you of anything. mental excerises and mathmatics can be ideal f=ma e=.5mv^2 and so on are all ideal.
What happens when 1000joules is placed into a volume of 250ml of ideal gas. Tell me about pressures. and if that ideal gas is contained in a cylinder, with a piston and the pressure before in the 250ml is 1atm, and the pressure on the backside of the piston is 1 atm. Assume perfect insulator for cylinder.
Rufeo, your comment is completely moot, and void until you can give the mechanical reason why, preferably with some form of mathmatics to go along with it.
Why?For instance the piston moves with a force of say 10. You capture some of this with a generator say 5 (not even getting into losses) the rest of the 5 is used to actually push the atmosphere backwards and out the cylinder. The atmosphere then pushes back with 5 which your generator can capture 2.5. So total input energy is 10 and output is 7. This is better than the normal 5 that would be captured normally from the initial expansion. But it can never be twice the more than the original input
The theory is there but the real world application seems to be unrealistic. The hot air would produce upward force, but to create a downward pull it would seem that the air inside the cylinder would have to be colder than the air outside, otherwise, it becomes static.
Fuzzy3676 5 months ago
What is really going to spin your head, is when you begin to study sympathetically connecting symbiotic systems with a combined COP greater than 1.0. Say like a heat pump system COP 11.0, with a TEM system COP(.1). Or a heat pump system COP 5.0 with a stirling generator system COP (.25).... okay i'll stop here before i get branded as heretic.... Hint: The combined COP is figured by directly multiplying the COP's together!
PulseFuelNerd 7 months ago
This exercise uses only expansion and contraction of a gas in gaseous state: we are bound to just under 2.0 COP. When we take off the kid gloves and include the ability to expand and contract through evaporation and condensation, COP goes hyper. Carnot in his limited scope, puts the theoretical max at COP 200. There is a whole realm of detail, data, math, and observation to explore and record. New formulas must and will be developed to account for this uncharted area of study and focus.
PulseFuelNerd 7 months ago
If your perfect generator is able to harness "1x" energy upon the upward stroke, that would imply that there is exactly zero energy available in order to lift the piston. In other words, your generator can only harness a maximum of "1x - Eg" where Eg is the amount of energy necessary to overcome the gravitational force upon the piston. The force used to overcome the gravitational force converts kinetic energy due to gas expansion into gravitational potential energy. COP = 1.0
PinellasPatriot 7 months ago
@PinellasPatriot
You are technically correct, but only on one aspect, However overall, you are missing the point. In this IDEAL exercise 2.0 COP is actually a limit and not actually attainable. Brake horsepower at 2.0 COP would stall the piston. Anything approaching but not including 2.0 COP would be valid (I.E. 1.999 COP). As long as the piston actually moves! The part your missing, and the point of the vid.... is the portion from 1.0 COP upwards towards 2.0 COP.
PulseFuelNerd 7 months ago
@PulseFuelNerd Are you saying the following?: If the first expansion stroke is adiabatic (no heat out of the system) then the 1x input heat energy is converted into 1x work due to gas expansion. At top dead center, the cylinder stops and the second process begins in which, ideally, 100% of remaining heat energy is bled off to the environment, thus converting this "vacuum stroke" into an additional 1x work.
I just want to get on the same page for the sake of discussion.
PinellasPatriot 7 months ago
@PinellasPatriot
Ya, close enough for now, what's up?
PulseFuelNerd 7 months ago
@PulseFuelNerd Alright, well assuming we are on the same page (and if I'm off, please direct me as necessary), this still seems problematic. When the piston moves up due to expansion, in an ideal case, 100% of that heat energy is converted into mechanical energy. Let, for example, the only limitation on the movement of the piston be the external pressure. This would imply that the pressure within the cylinder has equalized with that of the environment (but not necessarily the temperature).
PinellasPatriot 7 months ago
@PinellasPatriot As such, if the temperature of the environment happens to still be less than that of the gas within the cylinder, then some heat transfer to the environment could still occur, thus allowing a return stroke and C.O.P. greater than 1.0. Where I take issue, however, is that this is just a transient condition. My point is as follows: after the return stroke (in ideal case of C.O.P. = 2.0) the temperature within the cylinder will have equalized with the environment.
PinellasPatriot 7 months ago
@PinellasPatriot In your second cycle (just like the first), when all of the internal energy is converted to mechanical in the second expansion stroke, the temperature after the fact will return to its base state before heat addition: that of the environment. In this case, there is no mechanism to power the return stroke. This gets me to my point (finally): You could achieve COP = 2.0 (under ideal circumstances) in a transient condition but not at steady state.
PinellasPatriot 7 months ago
Addendum: Q1 - Q2 is not LESS THAN the work done through the piston...that's what I meant to say. In other words, the mechanical work you get out is never more than the heat energy you put in, that is not rejected. You put heat in, only a portion of it turns to mechanical, the rest is rejected.
If you put a mechanical load on it that requires a greater amount of energy to complete than the heat has that it can do work with, you can't expect it to complete the task.
randommagnum 8 months ago
@randommagnum Please think in terms of Ideal theory, set aside applied physics. System is sealed and no loses to environment, until such time is choosen. state1 heat is put into the space between cylinder and piston, the load is progressive just as the pressure applied to the piston is, at end of piston movement, energy leaves system and exteral pressure pushes piston back to rest state energy can again be harnessed from the environment, heat pumps are a type of heat system, and cop>4 in many.
d3adp001 7 months ago
@d3adp001
"Please think in terms of Ideal theory"
We ARE assuming no losses.
"set aside applied physics"
We cannot set aside physics/thermodynamics altogether.
"at end of piston movement, energy leaves system"
..which is less energy than was put into it, again, even assuming no losses, because energy was expended to move the piston..
"exteral pressure pushes piston back to rest"
which is only the second wave of energy that originated from the initial stroke.
randommagnum 7 months ago
@d3adp001 "heat pumps are a type of heat system, and cop>4 in many."
That is a non-applicable argument because a head pump has work put into it. COP doesn't work the same way with a heat pump as it does with a heat engine.
You & PFN fail to see 3 things: the gas, when expanding/doing work would NOT be isothermal; if there is waste heat (heat leaves system) then work done on piston is LESS than input energy; return stroke energy is stored energy that came from power stroke.
Total junk science.
randommagnum 7 months ago
@randommagnum First off you neglect the first premise, IDEAL, second off heat pumps are in fact a heat system in which the exchanged must travel through the engine, which is the same as what is occuring in the carnot, yes the difference is one is turned and one is being turned, but you neglect to observe and explain the environmental effect. the work done after heat leaves is not stored energy its a return to base state
d3adp001 7 months ago
@d3adp001 IF no heat leaves the system then the piston would move due to the expansion of gas. IF the heat never left the system the piston would stay in this excited state. The work performed would be equal to the heat put in less losses which there are none so 1 unit in 1 unit out. Now allow the heat to leave the system. The piston will revert back to original state. the energy comes into the system from the environment. which is what pushes the piston back to rest position.
d3adp001 7 months ago
@d3adp001 So either the max energy that can be extracted is 1/2 for the expansion cycle and 1/2 for the condensation cycle, or its 1 for expansion cycle and potential for 1 on the condensation cycle, Now if you want to say that only the difference between environmental state and excited state can be extracted on the expansion and the environmental can be extracted on the condensation then thats fine. But note one thing, you didnt say it I did, because I am already aware of the possiblities.
d3adp001 7 months ago
Your not really participating in a discussion and not contributing, your simply spitting out what was put into your mouth be someone else. If you have an understanding, then explain it, but to be antagonistic is a waste of your time. Contribute productively or shut up.
d3adp001 7 months ago
@d3adp001 Wow...this is really turning into a mind-numbing discussion.
I'm going to see about making a video to respond to what you are saying here...maybe that way you'll be more willing to hear me out.
Stand by...
randommagnum 7 months ago
Let me explain this another way...
Go to Wikipedia and type in "Carnot Cycle". Even in that idealized example, assuming no losses (isentropic) the heat you put in, or Q1 (what you're calling X), minus the amount of heat rejected into the environment, or Q2, is not going to be a greater amount of energy than the work done through the piston. The work the piston does coming down was part of the work the gas did pushing it up (gravity storage).
Do you understand?
randommagnum 8 months ago
Whatever energy gets put into living the cup will not be there to escape into the environment.
Let's say you apply heat to the air, but the cup is LOCKED so it can't expand; as the heat goes into the environment, MORE heat will do so because it stayed compressed. If you allow it to push the cup up, it expands and looses temperature, just like refrigerant gas (or an air duster) getting cold when it expands.
The energy into the cup + energy bled into the environment = 1X (1 COP).
:/
randommagnum 10 months ago
You are completely wrong. You do not understand IT at all. Energy is only trasfered. If you transfer energy in a isobaric system, it will cause increase in volume thanks to more kinetic energy of the particles. If you want to get some energy from the piston, you must let it apply some force to generator. THIS FORCE IS TRANSFORMED FORCE FROM VIBRATING PARTICLES, YOU GET ENERGY FROM THEM. BY LETTING THE PISTON TO DO SOME WORK (PRODUCING ENERGY) YOU ARE CAUSING ENERGY DROP IN THE AIR.
Stepanfo1 2 years ago
Then show the equation defining the action step. show me the equation of energy into the pressure and then transfered out and into kinetic. I am sorry but some stand alone statement that means nothing. if x amount of heat is applied to the gas, the gas applies pressure on the piston, the piston can the move some mass, how high will it lift it? and what is the energy of position vs the input heat energy. answers please.
If I wanted energy from the piston I would burn it
d3adp001 2 years ago
If you can not get into the other half of 2.0COP from simple gas expansion and contraction due to temperature changes... Then you will really be at a loss when we discuss the crazy COP from PHASE CHANGE of liquid state to gaseous state.
Hey folks, There is a new world of discovery to explore and conquer!
New World or Square World?
Red Pill Or Blue Pill?
Please try and apply yourself, until your mind can grasp and see the free other half of temperature change.
PulseFuelNerd 2 years ago
Hello Rhombic Sterling engine
rascalwind 2 years ago
the second thing is that say I have a volume of gas. I exert heat on that volume. I can predict the change in volume of gas based of very specific formula. If that volume expands, and then cools at the height of the piston. the same gas has to contract from cooling to let the piston down. Without additional heat that volume of gas will not push the piston up. Just look at hot air balloons. they need additional heat to expand the gas to rise again.
thanks for reading.
Greg
wanjockey 2 years ago
well sir then I ask you to do the math. Here are the conditions STP, 1 G, vol start 200cc, gas hydrogen, no friction, piston = 0 kg. during the heating cycle, the system is a perfect insultator, cooling cycle the system is a perfect conductor.
heat input energy 1kj.
imagine that the piston pushes a 1 kg wieght up during heating, and imagine the piston is connected to rachet that again lifts the 1 kg wiegh as the piston moves down.
What is the energy stored in the 1kg wieght at the end.
d3adp001 2 years ago
First thing. There will be resistance in the sides of the piston going up. Let alone the resistance of the bearings, or lever or whatever pushes the generator that creates the energy to feed back.
wanjockey 2 years ago
Does the word ideal mean anything to you?
what your saying is the equal of if you let go of a rock it will fall.
So if I asked how fast would it fall in a vaccuum would you try and tell me about wind resistance?
d3adp001 2 years ago
STERLING ENGINGE
BUT YOUR DONT WORK
MRJOSEPHKING1971 2 years ago
What?
Common people, full sentences!!!
PulseFuelNerd 2 years ago
STIRLING ENGINE with an I not with an E unless its made from silver.
PFN apparently a "thought excercise" is wasted on this guy, he is out of shape up stairs.
d3adp001 2 years ago
Is it possible to do the same experiment with a perfect electromagnet with a perfect coil at the bottom from the cylinder, and a perfect magnet as a piston? Instead of contracting air you would have a collapsing magnetic field and gravity that makes the piston come down....the same physics....am I correct?
I am trying to understand, I'm with you guys, friendly people here.
insAneTunA 2 years ago
I am not completely understanding what your getting at tuna, perfect coil/mag etc I understand, but what the piston is doing is eluding me. Are you saying push the magnet up with coil energy and then collect when it comes back down?
Need input stephanie, iiinnnnppuuuttt
d3adp001 2 years ago
I remember that movie :-))
But yes, that is exactly what I mean. Pushing up a magnet with a coil, and collect the energy from the upwards AND the downwards movement with a collecting coil. With electronics you can even forget the piston idea and just use a rotor with magnets instead, or a even a motionless energizer with coils only and no magnets at all. This is what Floyd Sweet was doing. Motionless Energy Generator,or MEG. I can't help thinking that it's all the same physics, am I wrong?
insAneTunA 2 years ago
ok so to make it easy for you.
part 2
wanjockey 2 years ago
with the piston tuna this is what will happen. You shoot it upward, moving freely it will move x distance. with an ideal generator it will move 1/2 x because the generator will slow the piston. The difference is that the speed of the piston indicates how far it moves and friction o it slows it, but the pressure on the piston will keep on pushing, the friction slows it, but cant stop it(well enogh would)
d3adp001 2 years ago
all of those other things have may variables to consider, to many for this method of communication.
d3adp001 2 years ago
I doubt that the output energy from the combined movement from the expanding and contracting air inside the most perfect cylinder is greater then the input energy that is needed to heat up the air in order to get movement to begin with. Unless you use a solar array or geothermal energy as a source. But I have seen such a system before, it was something with expanding gas and solar heat, and cooling it back down again in order to generate electricity. Some students project as I can remember....
insAneTunA 2 years ago
I will try to look it up for you, but it has been some time ago that I have seen it here on youtube. If I can't find it here I will search for it on google. I will let you know if I have it.
I see your point, and I can imagine that it's efficient.
insAneTunA 2 years ago
In an ideal exercise we are 100% efficient.
Are you casting doubt on a COP 2.0? If so WHY?
PulseFuelNerd 2 years ago
I am not sure if I understand what casting doubt means, but I have no doubts about your integrity. I could not find the video I mentioned before, I'm sorry for that. but I did find another interesting page.
Gas - Wikipedia, the free encyclopedia
insAneTunA 2 years ago
please see ideal gas laws, p1 t1 v1 and pv=nrt, please go learn what all that means, and then come back for comments.
Thank you for your co-operation in this matter.
d3adp001 2 years ago
I believe in self expression. It is great that we are not all the same. We have thoughts, opinons and points of view.
Please do not naysay, unless you also give support for your view.
I am surprised, that so many, are so inept
Put some thought into your post. I do when I create a video!
Thanks, PFN
PulseFuelNerd 2 years ago
Don't take credit.
Your logic is wrong and unreal.
You don't know every aspect.
Look at Sterling Engines; none have ever output more power than input.
Boyntonstu 2 years ago
Why not take credit?
How is my logic wrong and unreal?
On which aspect am I unaware?
Wow, dude, go play in traffic- was your stirling thought meant to give some form of evidence, like this is any kind of valid argument?
I am asking you to bring it up a notch, before you even come to the table.
PulseFuelNerd 2 years ago
Boyn, I have studied stirlings and other heat engines extensively. can you point to how the mental exercise put forth is different than a stirling engine. Please review those before you come back.
d3adp001 2 years ago
You started with 0 and ended up with 0. One cycle, not two.
CarbideTip 2 years ago
CT, i'm not sure of your meaning: 0 and 0.
What does one cycle not two refer to?
PulseFuelNerd 2 years ago
Are you assuming removing x is free?
No matter how you dump x to the environment, conversion rate still apply. Some ways are better than others but not OU or a gain of 1.
CarbideTip 2 years ago
this is an ideal system mental exercise, therefore conversion and losses dont exist. If you can not do that, then your particpation in this excerise will be greatly inhibited.
The heat in a heat pump moves out of the heated and cooled coils by the gradiant created.
the removal of the excited state is for free, the differencial created allows it to happen.
Come on guys try to keep up.
d3adp001 2 years ago
What good is the exercise? Nothing is ideal. "Removal of the excited state" is NOT free. We still have conversion rates with passive or mechanical. Yes you gain higher efficiency from the environment and see higher cop because of that but even at .000001% loss is still a loss.
Love to be with you on this D3 but I need better convincing.
CarbideTip 2 years ago
I am not trying to convince you of anything. mental excerises and mathmatics can be ideal f=ma e=.5mv^2 and so on are all ideal.
What happens when 1000joules is placed into a volume of 250ml of ideal gas. Tell me about pressures. and if that ideal gas is contained in a cylinder, with a piston and the pressure before in the 250ml is 1atm, and the pressure on the backside of the piston is 1 atm. Assume perfect insulator for cylinder.
red or blue, its just a choice.
d3adp001 2 years ago
CT, the point of the exercise is to get people to think, so they can realize that they are not bound by COP 1.0.
Square world thinkers are bound by COP 1.0 thinking. They can not acknowledge greater COP because it simply does not exist in their world.
Round world thinkers have followed the trail of bread crumbs to discovery and realize that greater than 1.0 COP is possible.
If you are open minded, the facts are there.
Red or Blue Pill, Bro?
PulseFuelNerd 2 years ago
Can you pretend?
The real world has losses and can hide the greater half of this output.
In this ideal exercise, We can identify the full 2.0 COP. The goal is to lead to discovery and personal knowledge.
Only then is there motivation to persist past the losses and failures, to build a device that achieves greater than 1.0 COP
If you can not get it ideally, you will not likely build it for real.
What is maximum COP in your thinking?
What COP are you convinced of currently?
PulseFuelNerd 2 years ago
interesting points
cool70200 2 years ago
Thanks, stay hungry!
PulseFuelNerd 2 years ago
This would be a more efficient electricity generator but in no way would it be twice the energy out than in.
Rufe0 2 years ago
okay, but WHY?
PulseFuelNerd 2 years ago
Rufeo, your comment is completely moot, and void until you can give the mechanical reason why, preferably with some form of mathmatics to go along with it.
d3adp001 2 years ago
Why?For instance the piston moves with a force of say 10. You capture some of this with a generator say 5 (not even getting into losses) the rest of the 5 is used to actually push the atmosphere backwards and out the cylinder. The atmosphere then pushes back with 5 which your generator can capture 2.5. So total input energy is 10 and output is 7. This is better than the normal 5 that would be captured normally from the initial expansion. But it can never be twice the more than the original input
Rufe0 2 years ago
If you could capture 100% of the pistons energy it would not move.
Rufe0 2 years ago