It has taken me far too long to realize the true nature of PV=nRT. I struggled with how to relate it to a free expansion, where the piston is allowed to travel faster than the expanding gas. Instead, and it may be obvious to many, PV=nRT as related to an expansion, describes the maximum affect possible for a given volume of heated gas in real world conditions. 100cc of 300k gas at 1 bar, becomes 200cc of 600k gas at 1 bar. Along the way it did work on the atmosphere, and in fact, all the work it could ever possibly do in a real engine.
The buffer pressure represents the starting base load that also sets the conditions for inside the power cylinder. If the starting conditions remain the same, yet more load is added externally(10lbs of weight), the gas will not be able to expand to twice its volume. While the gas will still expand to 1 bar at a smaller than 200cc volume, and lift the extra weight, this is an inefficient use of heat.
The extra load, presented in weight, took away a large percentage of the overall energy available in the expanding system.
We can remove the buffer pressure entirely and replace it with weights(mass). If the piston is one square inch we can use a 15lb weight to replace the buffer pressure. It is now much easier to see the work that was done in the initial expansion.
Let's now make the initial starting condition 100cc of 300k gas at 10 bar with a 150lb weight(in place of a 10 bar buffer pressure). If the external load is still only 10lbs, it is now only a fraction of the overall energy available in the expanding system. That means only a fraction of the temperature is required, and the added gas density makes heat transfer more effective.
Added compression ratio, by going beyond the zero point, will certainly allow an atmospheric or low pressure engine to make more power and appear more efficient. But the added efficiency of compression in an external combustion engine is only an illusion. The extra expansion and compression work is wasted in the pursuit of higher power output, when higher buffer pressure and much lower temperatures can be used instead. That is not to say it's a wasted effort, as high output atmospheric engines are likely to be adopted first.
There was no need to substitute the buffer pressure with a 150lb weight, other than to show that along with satisfying the initial 10lb load, we also now have a 150lb oscillating mass. If that sounds inconsequential, just put your finger under it.
How to determine ideal load?
Re: How to determine ideal load?
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You, in one post, have introduced enough questions to necessitate an entire year of thermodynamics and another year of physics. LOL. It is true that it is easier to ask questions, or present interesting conundrums, than to answer them.
When trying to understand PV=nRT it is best to ignore buffer pressure. That is best done by assuming the device is in the vacuum and free fall of space where the buffer pressure is zero and weight is zero. Mass is still 68 kg (150#). That leaves PV=nRT to apply to just the internal working gas as it should always be.
Buffer pressure concerns, just add confusion, and require added modeling variables and complications. Basically, if buffer pressure is added to one part of a cycle, it must be subtracted from an equal and opposite side of the full cycle.
Heat engines must be completely analyzed by modeling the entire cycle. Ignoring the complete cycle will ignore opposing processes, leading to very large misunderstandings. A reciprocating mass leads to accelerations in opposing directions. One way will add energy and the other will subtract. Totaled up, it will be zero energy. The forces will be larger for larger masses, for the same accelerations. The energy sum will come out zero.
Just to help in regard to, free expansion, from, expansion with work. Again think of space, no buffer pressure.
Adiabatic expansion with work involves the gas pushing the piston to expand. All the work the gas does is put into the piston. The piston increases in kinetic energy equal to the work the gas does, and loss of gas energy/temperature. That is true no matter the mass of the piston. Larger mass slower acceleration. That produces maximum adiabatic temperature drop.
In a free adiabatic expansion, some external force has yanked the piston out so fast it has briefly left the gas behind. The gas, doing the same amount of work, does work to itself. All the work goes back into the kinetic energy of the gas. We call that temperature or internal energy U. Since real gas has mass, size, intermolecular forces of attraction and repulsion, and viscosity, some of that energy is lost, we call that a gain of entropy. Most of that gain goes back into the gas as friction and contributes to the constant temperature. However some doesn't, so real gasses experience Jouel-Thomson 'cooling'/temperature drop, as in the Lindy or Gorrie process. Some T-S phase diagrams show data on this phenomenon. This is a very small temperature drop when compared to expansion with work.
Both of these are zero heat processes, adiabatic processes.
Expansion into a buffer pressure, atmosphere, puts work into the buffer, or atmosphere, warming them, and slowing the gas. The gas temperature drops more than for a totally free expansion. Hence why Lindy process works so well here on Earth.
This leads to a further look.at PV=nRT, or how I like to write it here PV=MRT. The M being total working gas mass, n being that same amount in 'ick' moles. Looking at how it relates to Work, heat, and physical units along with modeling assumptions. W=F•d=P∆V, and not W<>∆PV
First: A little trick in physics/engineering, to check our formulating is to check our 'units'. Example N Newtons equals mkg/sec^2, M mass=kg, J Jouels=N•m=m^2kg/sec^2, etc... in the S.I. metric units of measurement. As Matt has pointed out it is best not to mix measurement systems. Hence the S.I., System International.
Modeling processes with PV=nRT requires assumptions. The equation is valid for all real gasses. The problem is everything becomes a variable dependent on every other variable. That is why the assumptions made will limit the usefulness to certain areas of reality.
The equation is valid even into the area of liquefaction phase changes. The problem there is both R and n change for both liquid and gas. So that equation becomes rather useless and is discarded in favor of steam tables, phase diagrams, and more complex equations.
Using algebra, that equation can be rearranged so any single variable can appear alone on the left side. Such as: P=nRT/V. That allows us to calculate one variable from knowing the other values, just by plugging them in. Other equations can be substituted in for variables. Such as W=P∆V goes to P=W/∆V. Substituting in for P we get,
W/∆V=nRT/V
or
W=nRT∆V/V.
Or how work relates to PV=nRT
Note: ∆V=V1-V2, and R will be slightly different and T potentially a lot different, for each. T is a function of P and V.
Another modeling assumption is that M, P, are constant or averaged for the entire inside volume of the engine as V changes, etc... Matt has been pointing out that chopping up the gas mass into small pieces that move around, gives us insight as to how the gas is more realistically reacting. This would ultimately evolve into cutting it up into many many little pieces and computer modeling it using what is call finite element analysis, FEA.
I don't want to get into FEA, but you can easily see where Matt's going with it. The big problem I see in the future of FEA is that it leads to modeling mixing, fluid dynamics, and turbulence, etc... As a professor of mine said, it's sometimes best to close your eyes to all the little details and go with simple approximations. We can live with the constant error.
Now I want to mention that weigh is not mass. Units please:
Weight is force, N Newtons, Ma, kg•m/sec^2
Mass is just kg.
Scales on Earth are calibrated to our gravity to indicate mass kg, and force/weight pounds. That's right there are two 'pounds'. Pounds Force, and Pounds Mass. They differ by a factor of 32.2 . It is very difficult to tell the difference. I avoid them whenever possible. Sorry.
Determining how much work and heat rejection is necessary for a complete cycle can be done several ways. But it will boil down to efficiency is restricted by Th and Tc for both ECE and ICE. The difference is, for a Stirling Th and Tc have no connection to compression ratio, and for an ICE compression ratio dictates Th and Tc.
Buffer pressure and reciprocating mass have little effect on efficiency. Power to weight or size yes, efficiency no.
If that didn't help, please ask for specific clarification. I covered a lot of ground.
.
You, in one post, have introduced enough questions to necessitate an entire year of thermodynamics and another year of physics. LOL. It is true that it is easier to ask questions, or present interesting conundrums, than to answer them.
When trying to understand PV=nRT it is best to ignore buffer pressure. That is best done by assuming the device is in the vacuum and free fall of space where the buffer pressure is zero and weight is zero. Mass is still 68 kg (150#). That leaves PV=nRT to apply to just the internal working gas as it should always be.
Buffer pressure concerns, just add confusion, and require added modeling variables and complications. Basically, if buffer pressure is added to one part of a cycle, it must be subtracted from an equal and opposite side of the full cycle.
Heat engines must be completely analyzed by modeling the entire cycle. Ignoring the complete cycle will ignore opposing processes, leading to very large misunderstandings. A reciprocating mass leads to accelerations in opposing directions. One way will add energy and the other will subtract. Totaled up, it will be zero energy. The forces will be larger for larger masses, for the same accelerations. The energy sum will come out zero.
Just to help in regard to, free expansion, from, expansion with work. Again think of space, no buffer pressure.
Adiabatic expansion with work involves the gas pushing the piston to expand. All the work the gas does is put into the piston. The piston increases in kinetic energy equal to the work the gas does, and loss of gas energy/temperature. That is true no matter the mass of the piston. Larger mass slower acceleration. That produces maximum adiabatic temperature drop.
In a free adiabatic expansion, some external force has yanked the piston out so fast it has briefly left the gas behind. The gas, doing the same amount of work, does work to itself. All the work goes back into the kinetic energy of the gas. We call that temperature or internal energy U. Since real gas has mass, size, intermolecular forces of attraction and repulsion, and viscosity, some of that energy is lost, we call that a gain of entropy. Most of that gain goes back into the gas as friction and contributes to the constant temperature. However some doesn't, so real gasses experience Jouel-Thomson 'cooling'/temperature drop, as in the Lindy or Gorrie process. Some T-S phase diagrams show data on this phenomenon. This is a very small temperature drop when compared to expansion with work.
Both of these are zero heat processes, adiabatic processes.
Expansion into a buffer pressure, atmosphere, puts work into the buffer, or atmosphere, warming them, and slowing the gas. The gas temperature drops more than for a totally free expansion. Hence why Lindy process works so well here on Earth.
This leads to a further look.at PV=nRT, or how I like to write it here PV=MRT. The M being total working gas mass, n being that same amount in 'ick' moles. Looking at how it relates to Work, heat, and physical units along with modeling assumptions. W=F•d=P∆V, and not W<>∆PV
First: A little trick in physics/engineering, to check our formulating is to check our 'units'. Example N Newtons equals mkg/sec^2, M mass=kg, J Jouels=N•m=m^2kg/sec^2, etc... in the S.I. metric units of measurement. As Matt has pointed out it is best not to mix measurement systems. Hence the S.I., System International.
Modeling processes with PV=nRT requires assumptions. The equation is valid for all real gasses. The problem is everything becomes a variable dependent on every other variable. That is why the assumptions made will limit the usefulness to certain areas of reality.
The equation is valid even into the area of liquefaction phase changes. The problem there is both R and n change for both liquid and gas. So that equation becomes rather useless and is discarded in favor of steam tables, phase diagrams, and more complex equations.
Using algebra, that equation can be rearranged so any single variable can appear alone on the left side. Such as: P=nRT/V. That allows us to calculate one variable from knowing the other values, just by plugging them in. Other equations can be substituted in for variables. Such as W=P∆V goes to P=W/∆V. Substituting in for P we get,
W/∆V=nRT/V
or
W=nRT∆V/V.
Or how work relates to PV=nRT
Note: ∆V=V1-V2, and R will be slightly different and T potentially a lot different, for each. T is a function of P and V.
Another modeling assumption is that M, P, are constant or averaged for the entire inside volume of the engine as V changes, etc... Matt has been pointing out that chopping up the gas mass into small pieces that move around, gives us insight as to how the gas is more realistically reacting. This would ultimately evolve into cutting it up into many many little pieces and computer modeling it using what is call finite element analysis, FEA.
I don't want to get into FEA, but you can easily see where Matt's going with it. The big problem I see in the future of FEA is that it leads to modeling mixing, fluid dynamics, and turbulence, etc... As a professor of mine said, it's sometimes best to close your eyes to all the little details and go with simple approximations. We can live with the constant error.
Now I want to mention that weigh is not mass. Units please:
Weight is force, N Newtons, Ma, kg•m/sec^2
Mass is just kg.
Scales on Earth are calibrated to our gravity to indicate mass kg, and force/weight pounds. That's right there are two 'pounds'. Pounds Force, and Pounds Mass. They differ by a factor of 32.2 . It is very difficult to tell the difference. I avoid them whenever possible. Sorry.
Determining how much work and heat rejection is necessary for a complete cycle can be done several ways. But it will boil down to efficiency is restricted by Th and Tc for both ECE and ICE. The difference is, for a Stirling Th and Tc have no connection to compression ratio, and for an ICE compression ratio dictates Th and Tc.
Buffer pressure and reciprocating mass have little effect on efficiency. Power to weight or size yes, efficiency no.
If that didn't help, please ask for specific clarification. I covered a lot of ground.
.
Re: How to determine ideal load?
Thanks for such a thorough response. I'll have to think this over until next week.
As for the mass, think about a ball falling to the ground. It is capable of bouncing back to nearly where it was dropped from, now add the extra force of a power stroke and it should be easy to see where this can lead. The mass must be allowed to bounce off a stiff stopper at the end of its travel like in this engine. https://youtube.com/shorts/rV2ra56iJ6Q? ... gSJncGq16P
As for the mass, think about a ball falling to the ground. It is capable of bouncing back to nearly where it was dropped from, now add the extra force of a power stroke and it should be easy to see where this can lead. The mass must be allowed to bounce off a stiff stopper at the end of its travel like in this engine. https://youtube.com/shorts/rV2ra56iJ6Q? ... gSJncGq16P
Re: How to determine ideal load?
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The crankshaft essentially is a bounce mechanism. In kinematics it is called harmonic motion. Sinusoidal free reversal of reciprocating motion.
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The crankshaft essentially is a bounce mechanism. In kinematics it is called harmonic motion. Sinusoidal free reversal of reciprocating motion.
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