How to predict the exit temperature of an air compressor?

[MysticDream]

TL;DR Summary

Seeking to understand the behavior of air in a positive displacement reciprocating piston compressor.

It’s been said that an air compressor could be “idealized” as a reversible adiabatic process in which case calculating the properties of the gas at the exit would be simple. The problem is, it rarely seems to be the case, in practice, that the process is reversible so how useful is this “idealization”?

I understand that in an irreversible process the work done on the gas can be calculated by the volume change and external force (or pressure). I’m trying to understand how to calculate what this force would be with an air compressor piston running at a high rate of speed, say 1800 or 3600 rpm driven by an electric induction motor. My understanding was that a reversible adiabatic process must happen slowly. It seems that in order for a compressor to maintain a constant angular velocity it must be doing more work on the gas than is necessary before it reaches its final target pressure for delivery (per cycle) making the process irreversible. I’ve thought it may depend on how the shaft is driven. For example, maybe the load on the induction motor driving the piston varies throughout one cycle. If this is true, would it then behave differently if there were a large flywheel attached to the driveshaft?

How could one design a positive displacement cylinder/piston air compressor that is truly reversible? Would it have to run very slowly?

 

[Filip Larsen]

Sounds like you could benefit from reading up on the Carnot cycle.

 

[Chestermiller]

To analyze a compressor operating irreversibly, one would usually employ computational fluid dynamics to include irreversibly of viscous dissipation and conductive heat transfer within the gas.

 

[jack action]

Wouldn't it just still be a polytropic process, with a polytropic index varying slightly from $\gamma$ (ratio of heat capacities)?

 

[MysticDream]

That I don’t know. I should have clarified that I’m assuming the process is adiabatic with no heat exchanger. I’m trying to understand the interaction between only the piston and gas which I suspect depends on it’s speed and how it’s driven. Would a faster moving piston (higher rpm) naturally do more work on the gas because of the nature of a constant angular velocity crank acting on a volume of gas? Because it has to maintain a speed or torque to overcome its final delivery pressure, it seems it would do far more work on the gas than is necessary during its initial compression for each cycle. I’d imagine a truly reversible adiabatic process varying in driving torque or speed throughout one cycle, perhaps slowing down during the initial compression or at least decreasing in torque. It seems the only way for that to be possible mechanically is to have a very slow moving piston. A truly reversible process would make a compressor most efficient because the mass flow rate for the target pressure would be the highest. The temperature would not raise as much due to excess work being done during an irreversible process.

 

[MysticDream]

To analyze a compressor operating irreversibly, one would usually employ computational fluid dynamics to include irreversibly of viscous dissipation and conductive heat transfer within the gas.

Yes, I understand and that is a valid point, but before getting to that, I’m trying to understand how the piston motion affects the gas negating any heat transfer or heat from friction. How could a piston be driven to make the process truly reversible, or at least closer to it?

 

[MysticDream]

Sounds like you could benefit from reading up on the Carnot cycle.

I’m familiar with it. The Carnot cycle is reversible.

 

[Chestermiller]

To make the process more reversible and adiabatic, use lubrication on the piston surface, use an insulating material for the cylinder wall, and move the piston very slowly to avoid viscous dissipation in the gas.

 

[jack action]

I have difficulty following your statements. Can you show a source, a graph, or calculations for these statements:

My understanding was that a reversible adiabatic process must happen slowly.

It seems that in order for a compressor to maintain a constant angular velocity it must be doing more work on the gas than is necessary before it reaches its final target pressure for delivery (per cycle)

Because it has to maintain a speed or torque to overcome its final delivery pressure, it seems it would do far more work on the gas than is necessary during its initial compression for each cycle.

Furthermore, the load cannot be constant: the pressure increases, and the torque will increase as well during the compression process, no matter the piston speed. And if you have a typical slider-crank mechanism, the piston speed will not be constant either; with a constant angular velocity or not. The piston must revert direction at the ends, implying the piston must stop at least two times during the cycle.

A truly reversible process would make a compressor most efficient because the mass flow rate for the target pressure would be the highest.

What is the meaning of "efficient" here? Because the mass flow rate cannot increase the thermodynamic efficiency.

In addition, you seem to want to slow the process to make it "truly reversible" and now state "the mass flow rate would be the highest". It appears as a contradiction.

 

[russ_watters]

@MysticDream I think you do understand the overall issues relatively well, though I'm not completely sure where you want to go with this (if there is anywhere to go). For the overall issue, check "Thermodynamics, an Engineering Approach", by Cengel & Boles, sect 7.11 "Minimizing the Compressor work":
https://netedu.xauat.edu.cn/sykc/gcrlx/content/jcjs/Engineering Thermodynamics.pdf

We have just shown that the work input to a compressor is minimized when the compression process is executed in an internally reversible manner....

Obviously one way of minimizing the compressor work is to approximate an internally reversible process as much as possible by minimizing the irreversibilities such as friction, turbulence, and nonquasi-equilibrium compression. The extent to which this can be accomplished is limited by economic considerations. A second (and more practical) way of reducing the compressor work is to keep the specific volume of the gas as small as possible during the compression process. This is done by maintaining the temperature of the gas as low as possible during compression since the specific volume of a gas is proportional to temperature. Therefore, reducing the work input to a compressor requires that the gas be cooled as it is compressed...

To have a better understanding of the effect of cooling during the compression process, we compare the work input requirements for three kinds of processes: an isentropic process (involves no cooling), a polytropic process (involves some cooling), and an isothermal process (involves maximum cooling)....

It is interesting to observe from this diagram that of the three internally reversible cases considered, the adiabatic compression (Pvk constant) requires the maximum work and the isothermal compression (T constant or Pv constant) requires the minimum. The work input requirement for the polytropic case (Pvn constant) is between these two and decreases as the polytropic exponent n is decreased, by increasing the heat rejection during the compression process. If sufficient heat is removed, the value of n approaches unity and the process becomes isothermal. One common way of cooling the gas during compression is to use cooling jackets around the casing of the compressors.

Some of your specific questions:

How could one design a positive displacement cylinder/piston air compressor that is truly reversible? Would it have to run very slowly?

Yes: big and slow allows better cooling during compression. Obviously there are practical/economic problems with that.

For example, maybe the load on the induction motor driving the piston varies throughout one cycle. If this is true, would it then behave differently if there were a large flywheel attached to the driveshaft?

A flywheel won't change anything if the RPM is close to constant. The electric motor varies torque, which effects motor efficiency a little.

 

[jack action]

Yes: big and slow allows better cooling during compression.

How can there be cooling in an adiabatic process? That is what he said and why I might have trouble following his thought process:

I should have clarified that I’m assuming the process is adiabatic with no heat exchanger.

 

[russ_watters]

How can there be cooling in an adiabatic process? That is what he said and why I might have trouble following his thought process:

You're right, I missed that. If it's adiabatic it won't matter how fast it is, the result should be the same. In most normal compressed air applications you'll eventually lose the heat somehow though. So yeah, there might be conflicting thought processes here.

 

[jack action]

An isothermal compression will always be more desirable than an isentropic one. This is why we prefer compression in multiple stages with intercoolers. Having perfect isothermal compression at high speed is very difficult. But isothermal is not adiabatic.

In the following image, the "real single-stage compression" is what I was referring to when talking about a compression process with a polytropic index different from the heat capacities ratio.

 

[MysticDream]

What is the meaning of "efficient" here? Because the mass flow rate cannot increase the thermodynamic efficiency.

In addition, you seem to want to slow the process to make it "truly reversible" and now state "the mass flow rate would be the highest". It appears as a contradiction.

What I mean is that if our goal for a compressor is to have compressed air at it's highest pressure and lowest temperature, meaning a higher density, for a given cycle rate (rpm), a reversible adiabatic process is more efficient in converting the work to compressed air. Say you have reed valves that only open at the target pressure. If the process was irreversible, the temperature would be higher for the same delivery pressure and the density lower as more work was converted to heat. This lowers the mass flow rate for the same cycle rate (rpm). The goal is to "compress" air, not just pressurize it. You could pressurize air by simply heating it which is not what we want. We want to compress air with minimal amount of temperature increase which happens with a reversible adiabatic process.

In the isothermal case, you may have a compressor that is more effective for it's size, and maybe that is what you intended to point out, but the intercooler is wasting energy so I don't see how it could be more efficient.

Would you disagree with any of my statements here? Let me know if I'm mistaken.

 

[MysticDream]

I have difficulty following your statements.

I apologize if if I haven't been able to describe what I'm trying to understand. Perhaps it would be easier if I asked a question like this:
There is a large flywheel spinning with a large mass and more than enough energy to continue for a while. I engage a clutch coupled to a cylinder/piston compressor head. Will the air be compressed reversibly or irreversibly?

The reason I'm asking is I suspect this large rotating flywheel would cause a force on the gas similar to the common example we hear about gas being compressed irreversibly in a cylinder with a heavy weight on a piston being released suddenly. And yes, I do understand your point about the actual piston not moving at a constant speed but more sinusoidal, however my question is more about how to determine how this motion acts on the gas, whatever the motion is.

Is it always the case that the work done on the gas is reversible because the only work done is what is required to compress the gas? Or is it the case that when more energy is available at the driveshaft, be it in the form of velocity or torque, that more work will be done on the gas and compress (or pressurize) it irreversibly?

 

[russ_watters]

What I mean is that if our goal for a compressor is to have compressed air at it's highest pressure and lowest temperature, meaning a higher density, for a given cycle rate (rpm), a reversible adiabatic process is more efficient in converting the work to compressed air...

Would you disagree with any of my statements here? Let me know if I'm mistaken.

But as we have been discussing, adiabatic means higher temperature not lower temperature. Efficiency is higher if there is cooling (not adiabatic). I feel like you think by slowing the compression down you can achieve a lower temperature even with adiabatic compression(no cooling). You can't. The thing that slower compression (or multiple stages) gives you is time to do the cooling.

 

[MysticDream]

But as we have been discussing, adiabatic means higher temperature not lower temperature.

Yes, higher than with cooling. But I’m comparing adiabatic processes, reversible and irreversible. The reversible case will result in a lower temperature for the same delivery pressure.

I don’t understand why cooling is more efficient. You’re not reusing the heat to compress more gas. You’re throwing it away. Yes, you could compress more gas for the same size and rpm, but I’m talking about efficiency in energy conversion from kinetic to compressed gas, not kinetic to heat. I understand your point about a slower process being able to cool more effectively if an inter cooler is used.

 

[russ_watters]

Yes, higher than with cooling. But I’m comparing adiabatic processes, reversible and irreversible. The reversible case will result in a lower temperature for the same delivery pressure.

A little maybe. I don't think these losses are very significant. And why would you want to do this? What's the ultimate goal? If the ultimate goal is more efficient compression, this isn't the way to do it.

I don’t understand why cooling is more efficient. You’re not reusing the heat to compress more gas. You’re throwing it away.

It's because the heat itself is interfering with compression.

 

[jack action]

but the intercooler is wasting energy so I don't see how it could be more efficient.

I disagree with that.

The work done is the area under the curve on the P-V diagram, which is clearly smaller for the isothermal case, considering the same volume ratio:

the common example we hear about gas being compressed irreversibly in a cylinder with a heavy weight on a piston being released suddenly.

What does the heaviness of the weight, or how sudden the motion is, have to do with the irreversibility of the system?

Is it always the case that the work done on the gas is reversible because the only work done is what is required to compress the gas?

No. In a reversible isentropic compression-expansion, the gas also heats up and cools down, giving back everything it took, heat- and work-wise.

In a reversible isothermal compression-expansion, the same phenomena occur, except the heat is transferred to and from the surroundings.

What would make the process irreversible in our cases is heat gains or losses (due to friction, for example).

Or is it the case that when more energy is available at the driveshaft, be it in the form of velocity or torque, that more work will be done on the gas and compress (or pressurize) it irreversibly?

I really don't understand this question. First, velocity has to do with power, not energy or work. Second, how does doing more work relates to the concept of irreversability? Irreversibility is about losses.

 

[MysticDream]

I disagree with that.

The work done is the area under the curve on the P-V diagram, which is clearly smaller for the isothermal case, considering the same volume ratio...

The pressure is higher in the reversible adiabatic case. Are you saying there is essentially no difference, one is not more wasteful than the other? If so, I understand your point. Still though, I suspect it matters how the gas was compressed which leads to the next question...

What does the heaviness of the weight, or how sudden the motion is, have to do with the irreversibility of the system?

This may be what I need clarification on. My understanding was that if there is a perfectly insulated cylinder and piston with a volume of gas in it and there is a heavy weight on the piston (with stops), and it is released suddenly, the gas compression will be irreversible because it was not compressed in a manner where it is only slightly out of equilibrium during its motion. This excess work leads to heat gain. Is this heat gain from what you're referring to as friction, like internal friction between molecules? The more the difference between the internal pressure and external initially, the faster it's compressed, correct? That is how it has to do with the irreversibility of the system.

I imagined a flywheel storing energy and it's crank and connecting rod mechanism being like the heavy weight on the piston. It puts more force on the piston than is necessary and every compression cycle is way out of equilibrium initially. Therefore, I assumed a compressor with a flywheel would be compressing the gas irreversibly.

 

[MysticDream]

And why would you want to do this? What's the ultimate goal?

To compress gas most efficiently at the lowest exit temperature (stagnation).

It's because the heat itself is interfering with compression.

Yes, I understand that part, but it seems to me when comparing a reversible adiabatic process to an isothermal one, one isn't more efficient than the other. They are both reversible.

 

[MysticDream]

To make the process more reversible and adiabatic, use lubrication on the piston surface, use an insulating material for the cylinder wall, and move the piston very slowly to avoid viscous dissipation in the gas.

This "viscous dissipation" may be what I missed. Is this the cause of heating during irreversible adiabatic compression? If so, what are the limits here and how is it measured? You say move the piston very slowly, but how slow? Is the velocity of the piston movement simply relative to the imbalance of forces? In other words, the greater the imbalance between internal and external force, the greater the rate of compression and the more irreversible it is?

 

[Chestermiller]

This "viscous dissipation" may be what I missed. Is this the cause of heating during irreversible adiabatic compression?

If, by heating, you mean a temperature increase, in adiabatic reversible compression, the temperature also rises. Viscous dissipation results in additional "heating" in irreversible processes.

If so, what are the limits here and how is it measured?

It can be calculated using computational fluid dynamics.

You say move the piston very slowly, but how slow?

It depends on the system.

Is the velocity of the piston movement simply relative to the imbalance of forces? In other words, the greater the imbalance between internal and external force, the greater the rate of compression and the more irreversible it is?

How do you know what the internal force is in an irreversible compression or expansion?

 

[MysticDream]

If, by heating, you mean a temperature increase, in adiabatic reversible compression, the temperature also rises. Viscous dissipation results in additional "heating" in irreversible processes.

Yes, I meant heating in addition to the normal temperature increase during reversible adiabatic compression. Thanks for the clarification.

It can be calculated using computational fluid dynamics.

I know I can look this up and get into it but if you could provide an equation or direct me to a good source on the subject, I'd appreciate it.

It depends on the system.

Do you mean the geometry of the system, like the diameter and stroke distance of the piston? Is a larger piston generally more reversible for the same volumetric flow rate, because it can run at a lower rpm and the piston travels at a lower maximum velocity?

How do you know what the internal force is in an irreversible compression or expansion?

That is a good question. I know my initial force before compression would be atmospheric pressure for a normal air compressor. I'd like to have a better understanding of this.

 

[Chestermiller]

Yes, I meant heating in addition to the normal temperature increase during reversible adiabatic compression. Thanks for the clarification.

I know I can look this up and get into it but if you could provide an equation or direct me to a good source on the subject, I'd appreciate it.

Do you mean the geometry of the system, like the diameter and stroke distance of the piston? Is a larger piston generally more reversible for the same volumetric flow rate, because it can run at a lower rpm and the piston travels at a lower maximum velocity?

That is a good question. I know my initial force before compression would be atmospheric pressure for a normal air compressor. I'd like to have a better understanding of this.

In my judgment, if you really want to understand all this fundamentally, you need to learn about Newtonian fluid mechanics and heat transfer. This is covered in the best book I know of, Transport Phenomena, by Bird, Stewart, and Lightfoot. Transport phenomena include momentum transfer (Newtonian fluid mechanics), heat transfer (conduction and convection), and mass transfer (diffusion). Both fluid dynamics and heat transfer are part of the picture on the microscope in a cylinder of gas undergoing and adiabatic reversible expansion or compression.

 

[jack action]

Are you saying there is essentially no difference, one is not more wasteful than the other?

I'm saying one requires less work than the other.

The reason it doesn't affect the efficiency of an engine (I'm assuming there are heat and expansion processes afterward) is that the heat input is relatively the same (I think it's even less for the isothermal case, not sure) to get the same work output (but higher maximum pressure and temperature in the case of isentropic compression) AND you must reject heat after the expansion. In the isothermal case, instead of rejecting all the heat between the expansion and compression process, you reject some during the compression process as well. So you are wasting heat that would be wasted anyway.

In the case of a compressor with just an expansion process and no way to recuperate the heat loss, the compressor will require extra work or heat to function.

Is this heat gain from what you're referring to as friction, like internal friction between molecules?

Yes. And this is also what @Chestermiller refers to "viscous dissipation":

All fluids, except superfluids, are viscous, meaning that they exert some resistance to deformation: neighbouring parcels of fluid moving at different velocities exert viscous forces on each other.

So, yes, the velocity of the motion should affect those internal forces.

I imagined a flywheel storing energy and it's crank and connecting rod mechanism being like the heavy weight on the piston. It puts more force on the piston than is necessary and every compression cycle is way out of equilibrium initially. Therefore, I assumed a compressor with a flywheel would be compressing the gas irreversibly.

I guess that would be true on the compression side but you will get the opposite effect on the expansion side: for the same average angular velocity, the flywheel slows down the expansion process, leading to smaller viscous forces. I can't tell if one compensates for the other in all cases.

 

[Lnewqban]

But I’m comparing adiabatic processes, reversible and irreversible. The reversible case will result in a lower temperature for the same delivery pressure.

Just two practical considerations:
Adiabatic and reversible compression will not deliver the performance that is needed from a real air compressor.
Trying to save in weight, as well as in machining and material costs, the manufacturer will try to use as little metal and as high rpm's as possible.
That works against internal volume, heat exchanging surfaces, lubrication and slow internal air flows.

I don’t understand why cooling is more efficient. You’re not reusing the heat to compress more gas. You’re throwing it away.

Yes, the user will be throwing away energy in the form of wasted heat for the whole life of the air compressor.
The main causes are that the safe temperatures at which metals and lubrication can work is relatively low, compared to what the compressed air could reach with enough motor power (shaft torque and rpm's).

Based on the manufacturing versus costs compromises slightly explained above, the manufacturers will not tend to be generous in heat exchanging surfaces, less crazy flow velocities, lower rpm's, more complex valves, or about any other investments.
Simply, their main interest in in sales and profit, rather than in means to conservate energy in industrially and commercially compressing air.

 

[MysticDream]

In my judgment, if you really want to understand all this fundamentally, you need to learn about Newtonian fluid mechanics and heat transfer. This is covered in the best book I know of, Transport Phenomena, by Bird, Stewart, and Lightfoot. Transport phenomena include momentum transfer (Newtonian fluid mechanics), heat transfer (conduction and convection), and mass transfer (diffusion). Both fluid dynamics and heat transfer are part of the picture on the microscope in a cylinder of gas undergoing and adiabatic reversible expansion or compression.

Thanks for the recommendation.

Off hand, do you have an estimate of about what max piston velocity starts to make the process significantly irreversible, or a velocity that could be approximated as reversible? Obviously, a shorter stroke would produce a slower max velocity for a given volume.

 

[MysticDream]

I'm saying one requires less work than the other.

The reason it doesn't affect the efficiency of an engine (I'm assuming there are heat and expansion processes afterward) is that the heat input is relatively the same...

In the case of a compressor with just an expansion process and no way to recuperate the heat loss, the compressor will require extra work or heat to function.

Yep, understood.

Yes. And this is also what Chestermiller refers to "viscous dissipation":

So, yes, the velocity of the motion should affect those internal forces.

Thanks for clarifying.

I guess that would be true on the compression side but you will get the opposite effect on the expansion side: for the same average angular velocity, the flywheel slows down the expansion process, leading to smaller viscous forces. I can't tell if one compensates for the other in all cases.

Interesting point I hadn't considered.

 

[MysticDream]

Just two practical considerations:
Adiabatic and reversible compression will not deliver the performance that is needed from a real air compressor...

Simply, their main interest in in sales and profit, rather than in means to conservate energy in industrially and commercially compressing air.

Yes, good points and understood. The size and cost have to be practical. I'm asking these questions to have a better theoretical understanding of the process. I appreciate your input.

 

[Lnewqban]

I'm asking these questions to have a better theoretical understanding of the process.

That is excellent!

I recommend that you study centrifugal compressors, as their internal compression process is much less brutal than the one happening in reciprocating machines.

Please, see:
https://lindbergprocess.com/2020/01/centrifugal-vs-reciprocating/

 

[MysticDream]

I recommend that you study centrifugal compressors, as their internal compression process is much less brutal than the one happening in reciprocating machines.

I've studied them a bit but will more. Thanks.

 

[MysticDream]

In my judgment, if you really want to understand all this fundamentally, you need to learn about Newtonian fluid mechanics and heat transfer. This is covered in the best book I know of, Transport Phenomena, by Bird, Stewart, and Lightfoot. Transport phenomena include momentum transfer (Newtonian fluid mechanics), heat transfer (conduction and convection), and mass transfer (diffusion). Both fluid dynamics and heat transfer are part of the picture on the microscope in a cylinder of gas undergoing and adiabatic reversible expansion or compression.

I've bought the pdf and there is a ton to info to cover which may take a long time. Is it possible for you to narrow the last question down for me? I know I should have to work for it, as I'm sure you have, but any help would be appreciated. I'm wanting to get a general idea of what an approximate reversible process would look like with the velocity of a piston in a cylinder and what the scale looks like as it increases. For example, would it be "near" reversible with a piston peak velocity at 1 m/s and then if doubled in speed it becomes 50% irreversible?

 

[BillMcC]

That I don’t know. I should have clarified that I’m assuming the process is adiabatic with no heat exchanger. I’m trying to understand the interaction between only the piston and gas which I suspect depends on it’s speed and how it’s driven. Would a faster moving piston (higher rpm) naturally do more work on the gas because of the nature of a constant angular velocity crank acting on a volume of gas? Because it has to maintain a speed or torque to overcome its final delivery pressure, it seems it would do far more work on the gas than is necessary during its initial compression for each cycle. I’d imagine a truly reversible adiabatic process varying in driving torque or speed throughout one cycle, perhaps slowing down during the initial compression or at least decreasing in torque. It seems the only way for that to be possible mechanically is to have a very slow moving piston. A truly reversible process would make a compressor most efficient because the mass flow rate for the target pressure would be the highest. The temperature would not raise as much due to excess work being done during an irreversible process.

When it is a piston-type compressor, the head of the compressor usually has some provision for dissipating heat. The lines leaving the compressor sometimes have fins. Most piston compressors are two-stage compressors: One large cylinder compresses the incoming air to some percentage of the final second-stage piston pressure, which keeps changing as the reservoir, the compressor storage tank raises in pressure.