Expansion or compression -- which is more energy efficient?

[T C]

TL;DR: I am describing here a thought experiment where power consumption for expanding and compressing has been compared side by side to find out which one consumes more energy.

I want to share a thought experiment here. Suppose, we have a cylinder with a frictionless piston added to it. Enclosed in it is 1 gm-mole of a diatomic gas at 1 kg/cm^2 pressure and having volume of 0.0224 m^3. The temperature is 27°C i.e. 300°K.

Now, in the first scenario, the gas has been expanded adiabatically by pulling the piston outwards and the volume have been increased to 0.0672 m^3 i.e. three times to its initial volume. As the process is adiabatic, the temperature will fall to -80°C or 193.2°K. That's the temperature of the gas after expansion.

Now, in this scenario, without going into complex equations, we can assume that the cylinder is being pushed against atmospheric pressure i.e. 1 kg/cm^2. We know that the increase in volume and by multiplying both, the amount of energy consumed in the process has been found to be 448 J. and, kindly note that this is the highest amount of energy consumption. Actually the power consumption would be less as the pressure inside isn’t zero but will gradually decrease. But, for the sake of simplicity, let’s consider the power consumption to be this for now.

Now, in the second scenario, the same gas has been compressed to 1/3rd of its original volume. In that case, the temperature will rise to 465.54°K or 192.39°C. The power consumption is 3440 J in the whole process. Now, the hot has been cooled to 27°C i.e. 300°K and is being released to atmospheric pressure i.e. 1 kg/cm^2 pressure. The fall in temperature in that case would be the same as the first scenario as described before i.e. -80°C or 193.2°K. That's the temperature of the gas after being released.

In short, we get the same level of cooling with both the cases. But in the first case, the power consumption would be much less in comparison to the second. Want to know others opinions in this regards.

 

[russ_watters]

Without checking your math, please note that your first case doesn't consume/input mechanical work it outputs mechanical work. The gas pushing on the piston causes the expansion. The piston is not being pulled away from the gas (unless you are also doing work against the atmosphere...).

 

[T C]

(unless you are also doing work against the atmosphere...)

That's what I have suggested. The pressure inside is less while outside it's higher and that's atmospheric.

 

[jack action]

Now, the hot has been cooled to 27°C i.e. 300°K and is being released to atmospheric pressure i.e. 1 kg/cm^2 pressure.

It is not clear which processes (isochoric, isobaric, isentropic, isothermal) you are using to cool and depressurize the fluid in the second scenario. One could assume an isochoric cooling followed by an isothermal expansion, but you seem to talk about opening a valve between the cylinder and the atmosphere to equalize the pressure, and this would further cool the air coming out with an isentropic process, in addition to changing the mass of fluid present inside the cylinder.

Also, could you show your math so that we can see how you got your numbers, and we don't have to guess.

 

[T C]

It is not clear which processes (isochoric, isobaric, isentropic, isothermal) you are using to cool and depressurize the fluid in the second scenario.

Hope the first scenario is clear to you. The second scenario that I have described is like that. The compression is isenthalpic/adiabatic and you can use the adiabatic process equations here for 1 gm-mole of gas. Not very hard I think. After being compressed, the gas is being cooled in isobaric way. And, at the end, it has been released in an isenthalpic way. Hope it's clear now.

Also, could you show your math so that we can see how you got your numbers, and we don't have to guess.

For the first process of the second process? The first process isn't that complicated. Just think of an adiabatic/isenthalpic process where the gas is forcefully expanded to 3 times its initial volume. I think you can do the math by yourself.
The second process have been clearly described and I am sure you can do the math by yourself.

 

[jack action]

the gas is being cooled in isobaric way

That is also an isobaric compression, requiring more work.

And, at the end, it has been released in an isenthalpic way.

Do you mean like an isothermal expansion? Where both work is released and heat added?

The first process isn't that complicated.

Well, I'm glad it is, then it will be easy for you to do the work and show it to us. Just use the LaTeX notation to ease the reading of your equations.

A P-V diagram might be helpful as well.

I am sure you can do the math by yourself.

I'm sure I can too. It is just easier for me - or anyone else - to read your work and look for errors or misinterpretations than for me to make the whole work and hand it freely in a post, just to find out it is not what you meant because we misunderstood each other.

People want to help you after you did the work, not do it for you (even if you swear you already did it).

 

[T C]

First, I must admit my mistake. The calculated power for the first process that I have said before is wrong. It’s not 448 W but rather higher amount. And I am calculating it below:

Let’s calculate the process in this way:

The piston is moving against atmospheric pressure i.e. 101325 N/m²
of force per unit area and the change in volume is (0.0672 m³– 0.0224 m³) i.e. 0.0448 m³.

Therefore, the power consumption by the piston for moving against atmospheric pressure to increase the volume is (101325 X 0.0448) Joules i.e. 4539.38 Joules.

But, we also have to calculate the work done by the gas inside as its volume has increased to three times its initial volume and its temperature has fallen.

Considering the process to be adiabatic and using the adiabatic process formula of TV^γ-1
, the final temperature after the completion of the expansion process is T₂ = T₁(V₁/V₂)^γ-1; the final temperature has been found to be 300°K (initial temperature) X (1/3)^.4 (considering γ = 1.4).

Therefore, the final temperature is 193.31°K.

That means the fall in temperature is 106.68°K.

Therefore, the work done by gas inside the cylinder is 5 X 4.2 X 106.68 J i.e. 2240.28 Joules.

This amount has to be subtracted from the initial power consumption calculation for working against atmospheric pressure i.e. 4539.38 J.

At the end, the net power consumption has been found to be (4539.38 - 2240.28) Joules i.e. 2299.1 Joules.

That is sufficiently less than the 3440 Joules power consumption.

The reason behind such less power consumption during the expansion lies in one fact. And that is during the compression process, the work done on the gas increases its temperature and that heat has to be released for getting the necessary cooling.

But, during expansion, the work done by the piston have been stored as potential energy of the expanded gas and there no increase in temperature occurs during process and that means no wastage of energy as heat as part of the process itself.

 

[T C]

And for the second scenario:
It's basically the same process that we have observed in market available cooling machines like refrigerators or air-conditioners. First, the gas is compressed adiabatically to 1/3^rd of its initial volume.
And, the rise in temperature of the gas in this process can be found by using the same adiabatic process formula that has been used in the previous post i.e. TV^γ-1.
Therefore, the final temperature after compression will be, if the initial temperature is 300°K:

T₂= T₁(V₁/V₂)^γ-1
= 300°K (initial temperature) X (3)^.4(considering γ = 1.4)
= 465.5536 °K​

therefore, the power consumption for the compression is = 5X4.2X(465.5536 - 300) J =3465 J
Now, the compressed gas has been cooled in isobaric and that means the work done on it has been dissipated as heat.
After cooling, the compressed has been released to atmospheric pressure level and by that process the fall of temperature will be same like the first scenario i.e. 106.68°K as in the adiabatic process, the relation between volume and temperature is independent to initial and final pressure level. If the change in volume is the same, the fall in temperature would be the same.
In addition to that, a section of the used energy in the first scenario can be recovered as a large section of the energy spent has been stored as potential energy of the expanded gas. During the cooling process, the temperature and pressure of the expanded gas will rise in isochoric manner. But even if the temperature of the expanded gas will rise to the same temperature as the surrounding atmosphere i.e. 300°K; its pressure will remain sufficiently less than the atmospheric level. That means when the piston will be released, the atmospheric pressure will simply push piston inwards compressing the gas and a part of the energy spent can be recovered in this process.​

 

[jack action]

Now, the compressed gas has been cooled in isobaric and that means the work done on it has been dissipated as heat.

The heat loss in an isobaric cooling process is not equal to the work done on the ideal gas. Instead, the heat transferred ($Q$) is related to the change in internal energy ($U$) and the work done ($W$) by the equation $Q = \Delta U + W$, where the work done is associated with the constant pressure condition.

That would mean that some of the energy loss due to heat transfer came from the work input during the previous isentropic compression.

 

[T C]

The heat loss in an isobaric cooling process is not equal to the work done on the ideal gas. Instead, the heat transferred (Q) is related to the change in internal energy (U) and the work done (W) by the equation Q=ΔU+W, where the work done is associated with the constant pressure condition.

Some part of the internal energy lost isn't my concern here. My point is all the work done has been dissipated as heat. What you have said is some internal energy also is lost in this process, but that doesn't change the basics of what I have said.

 

[jack action]

From the OP:

In short, we get the same level of cooling with both the cases. But in the first case, the power consumption would be much less in comparison to the second. Want to know others opinions in this regards.

If I understand correctly, you seem to claim that both scenarios give the same amount of cooling for different work inputs.

But in the first scenario, there is no heat loss, and in the second one, there is a massive heat loss. This energy can be used for something else; it had to come from somewhere.

That is why if you did a proper analysis of the two scenarios, you would see exactly where the energy goes. Scenarios one and two do not yield the same results, as you state in the OP ("same level of cooling").

 

[T C]

If I understand correctly, you seem to claim that both scenarios give the same amount of cooling for different work inputs.

Yes.

But in the first scenario, there is no heat loss, and in the second one, there is a massive heat loss. This energy can be used for something else; it had to come from somewhere.

Because the energy spent to expand has been stored as potential energy of the expanded gas and I have repeatedly told that it can be recovered. That will further reduce the net power consumption of the process.
In the second scenario, there is a massive heat loss as the work done on the gas has been dissipated as heat and that energy comes from mechanical/electrical energy used to drive the compressor. In the first scenario, the source of energy is the same but a good section of the energy has been stored as potential energy inside the expanded gas.

That is why if you did a proper analysis of the two scenarios, you would see exactly where the energy goes. Scenarios one and two do not yield the same results, as you state in the OP ("same level of cooling").

Why not? The level of change in volume is the same in both the cases and that means the level of cooling should be the same. In the second scenario, huge amount of energy is wasted as it has been dissipated as heat. Even in the second scenario, some more energy spent can be recovered if the compressed gas is used to rotate a turbine with added dynamo. But that will make the whole system too bulky and costly. In fact, in the second scenario, the level of cooling can be lower as the potential energy of compressed gas is added to it as heat after being released. If that has been used to rotate a turbine to generate electricity, the level of cooling would be lower and will be equal to same as the first scenario.

 

[jack action]

and I have repeatedly told that it can be recovered.

All energy in the second scenario can be recovered as well.

In the second scenario, huge amount of energy is wasted as it has been dissipated as heat.

It is only wasted if you don't recuperate it somehow. (To heat something, for example.)

Even in the second scenario, some more energy spent can be recovered if the compressed gas is used to rotate a turbine with added dynamo.

There you go.

But that will make the whole system too bulky and costly.

Well, if you don't need to heat something and you don't need to rotate a turbine, why would anyone spend the energy in the first place? Obviously, the complexity of your machine should reflect the complexity of your needs.

In the end, no matter what path you choose to go from state A to state B - pressure, volume, and temperature -wise - doesn't affect the properties of the gas in both states. But the path you select will depend on what you need the gas to do in between those two states. No path is right or wrong, or "better" than any other.

You could add a third scenario where you do an isochoric cooling to your final temperature, followed by an isothermal expansion, and get the same result with even less work required.

But it is unclear what your restrictions are and, thus, how to evaluate which scenario is best. Why, in the second scenario, do you compress the gas if, in the end, you want an expansion? Why compress it to 1/3 of its volume? Why not 1/10, or 1/100, which would make your comparison even worse if we assume all heat losses are wasted?

 

[T C]

All energy in the second scenario can be recovered as well.

Simply impossible. While compressing, the rise is temperature is 165°. After being cooled and expanded, the fall in temperature is 106°C.

It is only wasted if you don't recuperate it somehow. (To heat something, for example.)

Such attempts are rare and conversion of electricity into heat is in itself a wastage. Heat stays at the lowest level of exergy while electricity at the highest.

 

[T C]

Well, if you don't need to heat something and you don't need to rotate a turbine, why would anyone spend the energy in the first place? Obviously, the complexity of your machine should reflect the complexity of your needs.

For cooling. That's how almost all refrigerators and airconditioners around the world work.

In the end, no matter what path you choose to go from state A to state B - pressure, volume, and temperature -wise - doesn't affect the properties of the gas in both states. But the path you select will depend on what you need the gas to do in between those two states. No path is right or wrong, or "better" than any other.

That may be theoretically true, but in practice these can make big differences.

You could add a third scenario where you do an isochoric cooling to your final temperature, followed by an isothermal expansion, and get the same result with even less work required.

How can I get cooling with an "isothermal" expansion?

But it is unclear what your restrictions are and, thus, how to evaluate which scenario is best. Why, in the second scenario, do you compress the gas if, in the end, you want an expansion? Why compress it to 1/3 of its volume? Why not 1/10, or 1/100, which would make your comparison even worse if we assume all heat losses are wasted?

That's how most of the cooling machines around the world work. First, gas is compressed and become hot in this process. The compressed gas is then cooled and thus releasing the work done together with a part of its enthalpy into surrounding. After being released, its temperature falls and it become colder and is used for cooling.

 

[T C]

In the end, no matter what path you choose to go from state A to state B - pressure, volume, and temperature -wise - doesn't affect the properties of the gas in both states. But the path you select will depend on what you need the gas to do in between those two states. No path is right or wrong, or "better" than any other.

The ultimate gain for us is the same, but that doesn't mean that the processes are the same. So, this logic isn't applicable here. In the first scenario, we get cooling with a expanded gas. In the second scenario, the cooling gas has the same pressure as the initial level but has lower temperature.

 

[Chestermiller]

TL;DR: I am describing here a thought experiment where power consumption for expanding and compressing has been compared side by side to find out which one consumes more energy.

I want to share a thought experiment here. Suppose, we have a cylinder with a frictionless piston added to it. Enclosed in it is 1 gm-mole of a diatomic gas at 1 kg/cm^2 pressure and having volume of 0.0224 m^3. The temperature is 27°C i.e. 300°K.

A pressure of 1 kg/cm^2 corresponds to $0.98\times10^5\ Pa$ pressure. From the ideal gas law, for 1 mole gas at 300 K, the ideal gas law gives a volume of $$\frac{nRT}{P}=\frac{(0.08206)(300)}{1.0}=0.0246\ m^3$$
For an ideal gas, the volume of 1 mole of gas at 1 atm and 273.16 K comes out to be $0.0224m^3$. So you underestimated the initial volume because the actual temperature in this problem is higher than the standard temperature.

Now, in the first scenario, the gas has been expanded adiabatically by pulling the piston outwards and the volume have been increased to 0.0672 m^3 i.e. three times to its initial volume. As the process is adiabatic, the temperature will fall to -80°C or 193.2°K. That's the temperature of the gas after expansion.

This assumes that the adiabatic expansion is reversible. During a reversible adiabatic expansion, the pressure decreases as the volume increases and the gas cools. So it is incorrect to use a constant pressure of 1 atm in the analysis. Pulling tension on the piston in this case results in a decrease in the imposed pressure on the gas. So you did the calculation for case 1 incorrectly.

Now, in this scenario, without going into complex equations, we can assume that the cylinder is being pushed against atmospheric pressure i.e. 1 kg/cm^2. We know that the increase in volume and by multiplying both, the amount of energy consumed in the process has been found to be 448 J. and, kindly note that this is the highest amount of energy consumption. Actually the power consumption would be less as the pressure inside isn’t zero but will gradually decrease. But, for the sake of simplicity, let’s consider the power consumption to be this for now.

The decrease in internal energy U for this case of ideal adiabatic expansion should be exactly equal to the work done by the piston.

$$\Delta U=nC_vR\Delta T=(1)(2.5)(8.314)(193-300)=-2224\ Joules$$

 

[T C]

The decrease in internal energy U for this case of ideal adiabatic expansion should be exactly equal to the work done by the piston.

C_v = 1.5? As far as I know, it's 5.
And while calculating the the power consumption during the expansion process, I first calculated the work done by the piston against atmospheric pressure and then deducted the work done by the gas while expanding. Am I wrong in this calculation?

A pressure of 1 kg/cm^2 corresponds to 0.98×105 Pa pressure. From the ideal gas law, for 1 mole gas at 300 K, the ideal gas law gives a volume of nRTP=(0.08206)(300)1.0=0.0246 m3
For an ideal gas, the volume of 1 mole of gas at 1 atm and 273.16 K comes out to be 0.0224m3. So you underestimated the initial volume because the actual temperature in this problem is higher than the standard temperature.

That means more power consumption during the expansion but does that altered how the ultimate power consumption has been calculated?

 

[jack action]

You will have to use clearer language to make sure everyone follows. For example:

In the first scenario, we get cooling with a expanded gas.

You specified in the OP that the process is adiabatic in the first scenario; therefore, there is no heat exchange, no cooling. If by "cooling" you mean "a lower temperature of the fluid", then I don't understand why you would increase the temperature in the second scenario just to decrease it after, because that seems to be your end goal.

You talk about refrigeration cycles, but their goal is not to lower the internal temperature of the fluid, but to lower the temperature of the surroundings, hence the need for "cooling", i.e. heat exchange.

That means more power consumption

You talk about power (Watts), yet you actually always refer to work (Joules).

 

[T C]

You specified in the OP that the process is adiabatic in the first scenario; therefore, there is no heat exchange, no cooling. If by "cooling" you mean "a lower temperature of the fluid", then I don't understand why you would increase the temperature in the second scenario just to decrease it after, because that seems to be your end goal.

After expansion, the gas/working fluid will become colder as it has been done adiabatically. Now, after the expansion, heat from the surrounding can be enter the gas and that's cooling. And for cooling purpose, you need a fluid of colder temperature to do that. I just want to use an unconventional way to generate the cold fluid.

You talk about refrigeration cycles, but their goal is not to lower the internal temperature of the fluid, but to lower the temperature of the surroundings, hence the need for "cooling", i.e. heat exchange.

The can't generate cooling without lowering the temperature of the working fluid. In present day split air conditioners, the compressor compresses the fluid and after that it has been carried to unit outside the room fitted with a blower. The blower will blow air around the tubes containing the hot, compressed air to cool them down to atmospheric temperature level. That's how a part of the internal enthalpy of the working fluid has been released into surrounding atmosphere. After that, the compressed air has been resent to the unit inside the room and suddenly released. That produce the cooling effect and that cool working fluid will suck heat from the air inside the room and thus cooling it. That's how refrigeration cycles work.

You talk about power (Watts), yet you actually always refer to work (Joules).

Sorry! I should use energy instead of power.

 

[jack action]

Now, after the expansion, heat from the surrounding can be enter the gas and that's cooling.

The cooling you are talking about is an isochoric process (a vertical line on the PV diagram, going up). You failed to mention this.

Then what? How do you bring back your fluid to its initial state to repeat the cycle? Or do you just discard the heated fluid and replace it with a new one?

But all of that still doesn't explain why, in the second scenario, you would heat the fluid before dropping it to the desired temperature.

A refrigeration cycle is like your first scenario, at least, the first part of the cycle. The second part is that you have to recompress the heated fluid (which heats it some more) and then cool it back to its original state. Then you have a complete closed cycle.

 

[Chestermiller]

C_v = 1.5? As far as I know, it's 5.
And while calculating the the power consumption during the expansion process, I first calculated the work done by the piston against atmospheric pressure and then deducted the work done by the gas while expanding. Am I wrong in this calculation?

That means more power consumption during the expansion but does that altered how the ultimate power consumption has been calculated?

sorry;. If R=2 calories/mole-K, then 2.5R= 5 calories/mole-K. In our case,, R=8.314 Joules/mole-K, and 2.5 R = (2.5)(8.314) Joules/mole-K. See my corrected answer in Post#17.

 

[Chestermiller]

In Post #17, I calculated the change in internal energy U in scenario 1. In this post, I will should that this matches the work done by the ideal gas in this case.

The equation for the pressure in terms of the volume for both cases reads: $$P=P_0\left(\frac{V_0}{V}\right)^{\gamma}=P_0\left(\frac{V}{V_0}\right)^{-\gamma}$$ The differential of work along the reversible path is $$dW=PdV=P_0V_0\left(\frac{V}{V_0}\right)^{-\gamma}d\left(\frac{dV}{V_0}\right)$$Integrating this between the initial and final states gives $$W=\frac{P_0V_0}{\gamma-1}\left[1-\left(\frac{V_0}{V}\right)^{\gamma-1}\right]$$where V is the final volume.

I leave it up to you to substitute in this equation and show that the change in internal energy is equal to the work..

 

[russ_watters]

Usually thermodynamic cycles are described with simple statements of the involved processes. @T C you are leaving out steps in your cycles, leaving us to guess what they are. Based on what you've said, it sounds like you intend your second cycle to be an open, compressed-air refrigeration cycle, which could work like this (there are alternatives/variations):

1. Air is drawn into a cylinder (not a thermodynamic process).
2. The air in the cylinder is compressed adiabatically, heating it.
3. The air in the cylinder is cooled back to the starting temperature.
4. The still pressurized air is then released back into the room, cooling it with an isenthalpic expansion which therefore also cools the room.

Energy requirements aside, the problem with this cycle is in step 3 or rather that the entire cycle happens inside an enclosed room. You can't cool the air back to its starting temperature unless you are doing a heat heat exchange with an outside reservoir at that same temperature. Otherwise the simplified result of this cycle is that your mechanical input energy is heating the room. Is heat exchange with an outside reservoir at the same temperature what you intend?

 

[T C]

Then what? How do you bring back your fluid to its initial state to repeat the cycle? Or do you just discard the heated fluid and replace it with a new one?

I don't need to bring back it to its original state. However enclosed and tightly sealed a place is, heat from surrounding atmosphere will always leak in and will heat up the fluid again and everything has to be started from the beginning.

But all of that still doesn't explain why, in the second scenario, you would heat the fluid before dropping it to the desired temperature.

The fluid is being compressed and it heats up in this process. In fact, if it has been compressed adiabatically, it will consume less energy than a adiabatic process. But, isothermal processes are slow and complex and this will increase the cost of machinery.

A refrigeration cycle is like your first scenario, at least, the first part of the cycle. The second part is that you have to recompress the heated fluid (which heats it some more) and then cool it back to its original state. Then you have a complete closed cycle.

Not the first, but the second scenario is a cooling cycle. No cooling machine in the world is now working on "expansion first" process as described in the first scenario.

 

[T C]

1. Air is drawn into a cylinder (not a thermodynamic process).
2. The air in the cylinder is compressed adiabatically, heating it.
3. The air in the cylinder is cooled back to the starting temperature.
4. The still pressurized air is then released back into the room, cooling it with an isenthalpic expansion which therefore also cools the room.

Wrong! What I have tried to describe is a simple cooling cycle used by modern day refrigerators, air conditioners or similar kind of cooling machines that use compressors for cooling.
What I want to show is that if we use direct expansion (the first scenario) instead of compression based cooling used in present day compressor based cooling machines, the energy consumption will always be less with the direct expansion method.

 

[T C]

In Post #17, I calculated the change in internal energy U in scenario 1. In this post, I will should that this matches the work done by the ideal gas in this case.

The equation for the pressure in terms of the volume for both cases reads: $$P=P_0\left(\frac{V_0}{V}\right)^{\gamma}=P_0\left(\frac{V}{V_0}\right)^{-\gamma}$$ The differential of work along the reversible path is $$dW=PdV=P_0V_0\left(\frac{V}{V_0}\right)^{-\gamma}d\left(\frac{dV}{V_0}\right)$$Integrating this between the initial and final states gives $$W=\frac{P_0V_0}{\gamma-1}\left[1-\left(\frac{V_0}{V}\right)^{\gamma-1}\right]$$where V is the final volume.

I leave it up to you to substitute in this equation and show that the change in internal energy is equal to the work..

As it's an adiabatic process, there more simple formulas for calculating the work done by the gas simply by using the fall in temperature and multiplying that with C_v and afterwards multiplying the result with 4.2 to get the work done in Joules. You yourself have used that before and also calculated the work done to be 2224 Joules in the Post#17.
Now, what I have done is that, I first calculated the amount of work done by the piston to push against atmospheric pressure by simply multiplying the change in volume with atmospheric pressure. After that, I calculated the work done by gas inside (due to adiabatic expansion) and subtracted that from the previously calculated energy consumption by the piston for working against the atmospheric pressure.
Am I wrong?
And, at the end, what I want to say is how hardly anyone may try to calculate the energy consumption, at the end the energy consumption in the first scenario will be much less than the energy consumption in the second scenario.