Expansion or compression -- which is more energy efficient?

[T C]

where A is the area of the piston, and the differential displacement of the piston is ds=dV/A. This is the work that is used in the first law of thermodynamics, and is equal to the decrease in internal energy of the gas (for adiabatic change). It is what you are calling the total "energy consumption."

No. The gas works as it expands and the piston works by pulling atmospheric pressure outwards. The work done by the piston against the atmospheric pressure is positive here while the work done by the gas is negative. It has to be deducted from the work done by the done against the atmospheric pressure.

here is no energy consumption involved in advancing the piston for the case of a massless, frictionless piston. The work done by the gas on one side of the piston is equal in magnitude to the work done on the other side of the piston by the atmosphere and the pulling force

How can you say so? The piston is working against the atmospheric pressure and the gas inside is working on the piston. So get the net energy expenditure, we have to deduct the work done by the gas inside from the work done by the piston by working against atmospheric pressure.

 

[russ_watters]

Like all other cooling cycles, it's closed. It sucks heat from one place and dumps that into atmosphere. And there is no phase change occurring in this process.

The high pressure side of piston will face the atmosphere and there is no poking hole. The enclosed gas will remain inside the cylinder.

So, are these the processes? I really don't like guessing, but....

Gas in the cylinder starts at atmospheric pressure and room temperature.

1. Gas is adiabatically expanded and cooled. That part you at least made clear.
2. Gas absorbs heat from the room, heating it to room temperature, but remaining a lower pressure.
3. Piston compresses the gas, heating it, while heat is transferred outdoors (outdoors and indoors are at same temperature), keeping the gas at room temp (isothermal compression), and ending up at the state it started in.

Is this your cycle?

 

[Chestermiller]

No. The gas works as it expands and the piston works by pulling atmospheric pressure outwards. The work done by the piston against the atmospheric pressure is positive here while the work done by the gas is negative. It has to be deducted from the work done by the done against the atmospheric pressure.

How can you say so? The piston is working against the atmospheric pressure and the gas inside is working on the piston. So get the net energy expenditure, we have to deduct the work done by the gas inside from the work done by the piston by working against atmospheric pressure.

I totally disagree with your interpretation and stand by what I said previously. You should know that I have a lot of practical experience with thermodynamics (>60 years), so I am very confident in what I am saying.

I can help you work your way through this if you are willing to work with me. If so, I would like you to start out by writing down a Newton's law force balance on the piston that includes $P_{atm}$, $P_{gas}$, F (the external pulling force needed to move the piston), and the piston area A.

Thank you.

 

[T C]

So, are these the processes? I really don't like guessing, but....

Gas in the cylinder starts at atmospheric pressure and room temperature.

1. Gas is adiabatically expanded and cooled. That part you at least made clear.
2. Gas absorbs heat from the room, heating it to room temperature, but remaining a lower pressure.
3. Piston compresses the gas, heating it, while heat is transferred outdoors (outdoors and indoors are at same temperature), keeping the gas at room temp (isothermal compression), and ending up at the state it started in.

Is this your cycle?

That's it.

 

[T C]

I totally disagree with your interpretation and stand by what I said previously. You should know that I have a lot of practical experience with thermodynamics (>60 years), so I am very confident in what I am saying.

In the first scenario, I have calculated the net energy consumption by subtracting the work done by the gas from the energy spent by the piston i.e. (4539.38 - 2240.28) J. And now, as per you, it should be (4539.38 + 2240.28) J, right?

 

[Chestermiller]

In the first scenario, I have calculated the net energy consumptioOn by subtracting the work done by the gas from the energy spent by the piston i.e. (4539.38 - 2240.28) J. And now, as per you, it should be (4539.38 + 2240.28) J, right?

There is no net work done on the piston, and, thus, no net energy "spent by the piston." I show this below.

From Newton's 2nd law of. motion applied to the piston, we have $$P_{gas}A+F-P_{atm}A=0\tag{1}$$The term $P_{gas}A$ represents the forward force of the gas on the inside face of the piston, the term F represents the forward external pulling force (needed to advance the piston) applied on the outside face of the piston, and the term $P_{atm}A$ represents the backward pressure force exerted by the air on the outside face of the piston. Notice that, since we are assuming a massless frictionless piston, the net force on the piston is zero at all times during the process. This means that the net of the forces exerted on the faces of the piston is zero.

Next let's consider the work done on the piston (and the net energy consumed by the piston). If we multiply Eqn. 1 by the differential forward displacement of the piston ds and integrate between the Initial and final states of the system, we obtain: $$\int_{V_i}^{V_f}{P_{gas}dV}+\int{Fds}-P_{atm}(V_f-V_i)=0\tag{2}$$The first term on the left-hand side represents the work done by the gas on the inside face of the piston; the second term represents the work done by the applied external force on the outside face of the piston; and the third term (with minus sign included) represents minus the work done by the atmosphere on the outside face of the piston. Eqn. 2 shows that (1) the net work done on the piston by its surroundings is zero and the net work by the piston on its surroundings is zero. Therefore, with regard to the piston, advancing it consumes no net energy.

Let us rearrange Eqn. 2 to obtain:$$\int_{V_i}^{V_f}{P_{gas}dV}=P_{atm}(V_f-V_i)-\int{Fds}\tag{3}$$The term on the LHS represents the work done by the gas on its surroundings (the inside piston face). The LHS represents the work done by the outside face of the piston on the surroundings of the combined system comprising the gas and the piston. You can see that the work done by the gas on its surrounding is exactly equal to the work done by the combined system of gas and piston on its surroundings.

When applying the first law of thermodynamics to a process, the first step in the analysis must be to define, in a precise manner, the specific system that is being analyzed. In the case of scenario 1, there are two simple choices one can make:

1. The gas alone
2. The combination of the gas and the cylinder

Analyzing the process based on either of these choices gives exactly the same results,

CHOICE 1: System = gas alone

In this case, the first law of thermodynamics gives, $$\Delta U=nC_v(T_f-T_i)=-\int_{V_i}^{V_f}{P_{gas}dV}\tag{4}$$
CHOICE 2: System = gas plus piston

In analyzing this case, it is necessary to assume that the massless frictionless piston is thermally insulated such that the internal energy change of the piston is zero. (This is an implicit assumption in case 1 as well). If we write down the equation for the first law of thermodynamics in this case, we obtain $$\Delta U=nC_v(T_f-T_i)=-P_{atm}(V_f-V_i)+\int{Fds}\tag{5}$$Next, combining this with Eqn. 3, we obtain:$$\Delta U=nC_v(T_f-T_i)=-\int_{V_i}^{V_f}{P_{gas}dV}\tag{6}$$

Note that Eqns. 5 and 6 are, as expected, identical..

We have shown here that, for scenario 1,

1. No net energy is consumed in advancing the piston from its initial position to its final position (i.e., no net work is done on the piston)
2. The only energy consumption is the result of expanding the gas. However, even here, no energy is dissipated because the decrease in internal energy of the gas is exactly equal to the work done on the surroundings (the work to push the atmosphere back minus the work expended in using an external tensile force F to pull the piston).

 

[jack action]

Gas in the cylinder starts at atmospheric pressure and room temperature.

1. Gas is adiabatically expanded and cooled. That part you at least made clear.
2. Gas absorbs heat from the room, heating it to room temperature, but remaining a lower pressure.
3. Piston compresses the gas, heating it, while heat is transferred outdoors (outdoors and indoors are at same temperature), keeping the gas at room temp (isothermal compression), and ending up at the state it started in.

That's it.

The hard part of this cycle is to decompress isobarically as it heats up in (2), and then recompress it as the internal heat is transferred out - somehow, outside the box too - isothermally in process (3).

But the real tricky part is trying to do it fast. Isothermal processes are usually really hard to accomplish quickly. For reference, see the reverse Stirling cycle:

The cycle is reversible, meaning that if supplied with mechanical power, it can function as a heat pump for heating or cooling, and even for cryogenic cooling.

At least one company already makes these Stirling cycle refrigerators. They claim their high initial cost can be recuperated over time.

​

 

[T C]

The hard part of this cycle is to decompress isobarically as it heats up in (2), and then recompress it as the internal heat is transferred out - somehow, outside the box too - isothermally in process (3).

What is isobaric decompression? As far as I know, decompression means decrease in pressure.

 

[jack action]

What is isobaric decompression? As far as I know, decompression means decrease in pressure.

It means an increase in volume while keeping the pressure constant. I should of used "isobaric expansion" which is more appropriate.

 

[T C]

It means an increase in volume while keeping the pressure constant. I should of used "isobaric expansion" which is more appropriate.

My aim is to generate cooling. In case of isobaric expansion, the temperature of the fluid must be increased to keep the pressure same while increasing the volume. I have clearly stated at the very beginning that it's adiabatic.

 

[jack action]

My aim is to generate cooling. In case of isobaric expansion, the temperature of the fluid must be increased to keep the pressure same while increasing the volume. I have clearly stated at the very beginning that it's adiabatic.

"Isobaric" and "adiabatic" don't go together. I recall that you replied, "That's it", to the following:

2. Gas absorbs heat from the room, heating it to room temperature, but remaining a lower pressure.

In an isobaric expansion, the heat is transferred from the surroundings to the fluid inside. You cannot go around this. The heat transferred is $Q = C_p \Delta T$, The change in internal energy is $\Delta U = C_v \Delta T$, and the work is $W = R\Delta T$. https://en.wikipedia.org/wiki/Isobaric_process

 

[Chestermiller]

In the first scenario, I have calculated the net energy consumption by subtracting the work done by the gas from the energy spent by the piston i.e. (4539.38 - 2240.28) J. And now, as per you, it should be (4539.38 + 2240.28) J, right?

Here are my calculations for scenario 1.

Given:
Initial Temperature = $T_i=300.2K$
Initial Pressure = $P_i=101325\ Pa$
Number of moles n =1.0

Calculations:
Initial Volume$$V_i=\frac{nRT_i}{P_i}=\frac{(1)(8.314)(300.2)}{101325}=0.02463\ m^3$$
Final Volume$$V_f=3V_i=0.0739\ m^3$$
Final Temperature$$T_f=T_i\left(\frac{V_i}{V_f}\right)^{\gamma-1}=193.4\ K$$
Final Pressure$$P_f=P_i\left(\frac{V_i}{V_f}\right)=21764\ Pa$$
Change in internal energy of gas
$$\Delta U=nC_vR\Delta T=(1)(2.5)(8.314)(193.4-300)$$$$=-2216.\ Joules$$
Work done by gas on its surroundings (the trailing face of the piston)$$W_{gas}=\frac{P_iV_i}{\gamma-1}\left[1-\left(\frac{V_i}{V_f}\right)^{\gamma-1}\right]$$$$=\frac{(101325)(0.02463)}{0.4}\left[1-\left(\frac{0.02463}{0.0739}\right)^{0.4}\right]=2219\ Joules$$ Work done by leading face of piston to push back atmosphere $$W_{atm}=P_{atm}(V_f-V_i)$$$$=101325(0.0739-0.02463)=4992\ Joules$$
Work done by supplemental rod welded to leading face of piston to pull the piston forward$$\int{Fds}=W_{atm}-W_{gas}=4992-2219=2,773 \ Joules$$
These results confirm that

1. the decrease in internal energy of the gas is equal to the work done by the gas on its surroundings (i.e., the trailing edge of the piston)

2. The work required to push back the atmosphere at the leading edge of the piston is supplied by the pushing force of the gas at the trailing face of the piston plus the pulling force of the supplemental rod welded to the leading face of the piston. For the particular parameters of this problem, the supplemental rod supplies roughly 56% of the required work while the gas supples roughly 44%.

 

[T C]

"Isobaric" and "adiabatic" don't go together. I recall that you replied, "That's it", to the following:

I have never mentioned "isobaric". So, no question of mixing up.

In an isobaric expansion, the heat is transferred from the surroundings to the fluid inside. You cannot go around this. The heat transferred is Q=CpΔT, The change in internal energy is ΔU=CvΔT, and the work is W=RΔT. https://en.wikipedia.org/wiki/Isobaric_process

That doesn't mean that if heat enters from outside inside the cylinder, the process in itself become isobaric. To make a process isobaric, much more heat has to be added to the fluid inside.

 

[T C]

Here are my calculations for scenario 1.

Given:
Initial Temperature = Ti=300.2K
Initial Pressure = Pi=101325 Pa
Number of moles n =1.0

Calculations:
Initial VolumeVi=nRTiPi=(1)(8.314)(300.2)101325=0.02463 m3
Final VolumeVf=3Vi=0.0739 m3
Final TemperatureTf=Ti(ViVf)γ−1=193.4 K
Final PressurePf=Pi(ViVf)=21764 Pa
Change in internal energy of gas
ΔU=nCvRΔT=(1)(2.5)(8.314)(193.4−300)=−2216. Joules
Work done by gas on its surroundings (the trailing face of the piston)Wgas=PiViγ−1[1−(ViVf)γ−1]=(101325)(0.02463)0.4[1−(0.024630.0739)0.4]=2219 Joules Work done by leading face of piston to push back atmosphere Watm=Patm(Vf−Vi)=101325(0.0739−0.02463)=4992 Joules
Work done by supplemental rod welded to leading face of piston to pull the piston forward∫Fds=Watm−Wgas=4992−2219=2,773 Joules
These results confirm that

1. the decrease in internal energy of the gas is equal to the work done by the gas on its surroundings (i.e., the trailing edge of the piston)

2. The work required to push back the atmosphere at the leading edge of the piston is supplied by the pushing force of the gas at the trailing face of the piston plus the pulling force of the supplemental rod welded to the leading face of the piston. For the particular parameters of this problem, the supplemental rod supplies roughly 56% of the required work while the gas supples roughly 44%.

You have just confirmed what I have guessed. Whatsoever, this shows that by expansion, cooling can be done with less energy consumption.

 

[Chestermiller]

You have just confirmed what I have guessed. Whatsoever, this shows that by expansion, cooling can be done with less energy consumption.

Cooling by expansion can be done with less energy consumption than by what?

This seems to me to be basically just a standard reversible adiabatic expansion scenario, analogous to removing pebbles from a cylinder and piston oriented vertically. The increase in the potential energy of the pebbles would equal the work done by the gas, and would also equal to the decrease in the internal energy of the gas. I fail to see why you regard this particular expansion scenario as being so special (and providing new understanding).

 

[T C]

Cooling by expansion can be done with less energy consumption than by what?

In comparison to cooling by compression.

 

[jack action]

I have never mentioned "isobaric". So, no question of mixing up.

That doesn't mean that if heat enters from outside inside the cylinder, the process in itself become isobaric. To make a process isobaric, much more heat has to be added to the fluid inside.

You did say part of your cycle is isobaric when @russ_watters said:

Gas in the cylinder starts at atmospheric pressure and room temperature.

1. Gas is adiabatically expanded and cooled. That part you at least made clear.
2. Gas absorbs heat from the room, heating it to room temperature, but remaining a lower pressure.
3. Piston compresses the gas, heating it, while heat is transferred outdoors (outdoors and indoors are at same temperature), keeping the gas at room temp (isothermal compression), and ending up at the state it started in.

Is this your cycle?

The first process is adiabatic;
The second process is isobaric;
The third process is isothermal.

Here is the PV diagram of this cycle:

And you answered in post #34:

That's it.

 

[T C]

The first process is adiabatic;
The second process is isobaric;
The third process is isothermal.

Here is the PV diagram of this cycle:

The gas will remain in lower pressure doesn't mean that the pressure will remain same during the process, but rather the volume will remain same i.e. it's an isochoric (not isobaric) process. And the third process isn't isothermal as the heat will be transferred to outside after being compressed and not during the compression process.

 

[Chestermiller]

In comparison to cooling by compression.

Here's a big surprise: The surroundings doing adiabatic compression work on a gas causes its internal energy and temperature to increase. A gas doing adiabatic expansion work on its surrounding causes its internal energy and temperature to decrease. Isn't that what you are saying?

 

[jack action]

but rather the volume will remain same i.e. it's an isochoric (not isobaric) process.

In an isochoric process, the pressure increases, so it doesn't "remain at a lower pressure".

And the third process isn't isothermal as the heat will be transferred to outside after being compressed and not during the compression process.

Heat is transferred to the outside in an isothermal compression.

That being said, @russ_watters clearly stated the third process he described was an isothermal compression, and you responded "That's it."

Now you MUST do a PV diagram of your cycle (like I did) for us to follow you. What we got now from you is:

1. An adiabatic expansion from 300 K to 193 K;
2. An isochoric heating from 193 K to an unknown pressure (somewhere between states 1 and 2) and temperature (it cannot be 300 K, as the third process isn't isothermal);
3. A third process, which is a compression that is not isothermal. It seems to be adiabatic from your statement (compression without heat rejection). What are the pressure and temperature at the end of this process?
4. A possible fourth process that would be needed to reject the heat from the previous compression, possibly isochoric as well?

To sum it up (referring to states in the following image): adiabatic expansion (2-1), followed by an isochoric heating (1-4), followed by an adiabatic compression (4-3, ends at very high pressure and temperature), followed by an isochoric cooling (3-2) to go back to a temperature of 300 K. This would be a reverse Otto cycle:

 

[T C]

Here's a big surprise: The surroundings doing adiabatic compression work on a gas causes its internal energy and temperature to increase. A gas doing adiabatic expansion work on its surrounding causes its internal energy and temperature to decrease. Isn't that what you are saying?

The fall in internal energy and temperature will only cause when the gas expands as that means doing some kind of work and the only source of energy for the gas to do that work is its internal energy, right?

 

[T C]

Now, let’s calculate how much energy can be recovered during the return stroke in the first scenario:

During the outward motion of the piston, the final volume of the gas is 0.0672 m³ and the temperature is (300-106)°K i.e. 194°K or -79°C.

Now, during the cooling process, heat from the cooling area will enter the gas and its temperature and pressure will rise while the volume will remain the same i.e. it will be an isochoric process.

Let’s assume that the temperature will rise to 300°K.

That means the pressure too will rise following Gay-Lussac’s law.

During the expansion process, the pressure at the beginning is 1 barA and that will fall to (1/3)^1.4 i.e. 0.21479800499241808350276432702957 barA.

Now, as the temperature will rise, the pressure will also rise and at 300°K, the pressure will become (0.21479800499241808350276432702957) X (300/194) barA i.e. 0.45510014855795202093624595280225 barA.

Now, the outside pressure is 1 barA and the pressure inside is 0.33216186339033724253004792839625 barA. If the piston is suddenly released, it will push the gas inwards and its pressure will increase together with its temperature and the volume will decrease.

In such a scenario, the volume will fall to = 0.0672 X (0.45510014855795202093624595280225/1)^(1/1.4) m³ i.e. 0.03058272998309437580691572802831 m³.

Therefore, the decrease in volume is = (0.0672 – 0.03058272998309437580691572802831) m³ = 0.03661727001690562419308427197169 m³.

The force causing this fall in volume is atmospheric pressure i.e. 101,325 N/m².

Therefore, the work done by atmospheric pressure here is = (101,325 X 0.03661727001690562419308427197169) Joules = 3,710.2448844629623713642638575314 Joules.

But, a good section of the work done has been converted into the increase in enthalpy and temperature of the enclosed gas.

Now, at the starting of the process, the temperature of the gas was 300°K and the volume changes from 0.0672 m³ to 0.03058272998309437580691572802831 m³.

Therefore, the rise in temperature is = (0.0672/0.03058272998309437580691572802831)^2/7 X 300°K = 411.03467801462645180436738278542°K.
Here, (γ-1)/γ = 2/7 as γ = 1.4
Therefore, the rise in temperature is = (411.03467801462645180436738278542 – 300)°K = 111.03467801462645180436738278542°K.

That means addition to the enthalpy of the gas inside is = 5 X 4.2 X 111.03467801462645180436738278542 Joules = 2,331.7282383071554878917150384939 Joules.

Now, the remaining work = 1,378.5166461558082701774422423371 Joules.

And this can be recovered during the recompression process that is adiabatic.

Now, the previously calculated net power consumption for doing the expansion was 2773 Joules. And, after recovering 1,378.5166461558082701774422423371 during the recompression process, the net energy consumption during the process has become (2773 - 1,378.5166461558082701774422423371) joules i.e. 1,394.4833538441917298225577576629 Joules.

That is sufficiently less than the power consumption in comparison to the second scenario.
In the second scenario, such recovery can also be possible if the compressed gas is released through a turbine attached with dynamo and in this process, a section of input energy can be recovered. But, that will make the system bulky, costly and expensive to maintain. In short, the cost will outweigh gain in energy input.

 

[jack action]

In light of this new definition of the complete cycle of scenario 1, with my calculations, this is what I get:

Scenario 1:

During the outward motion of the piston, the final volume of the gas is 0.0672 m³ and the temperature is (300-106)°K i.e. 194°K or -79°C. --> isentropic expansion

Now, during the cooling process, heat from the cooling area will enter the gas and its temperature and pressure will rise while the volume will remain the same i.e. it will be an isochoric process.

Let’s assume that the temperature will rise to 300°K. --> isochoric heating

[...]

In such a scenario, the volume will fall to = 0.0672 X (0.45510014855795202093624595280225/1)(1/1.4) m3 i.e. 0.03058272998309437580691572802831 m3.

[...]

The force causing this fall in volume is atmospheric pressure i.e. 101,325 N/m2.

[...]

Now, at the starting of the process, the temperature of the gas was 300°K and the volume changes from 0.0672 m3 to 0.03058272998309437580691572802831 m3.

Therefore, the rise in temperature is = (0.0672/0.03058272998309437580691572802831)2/7 X 300°K = 411.03467801462645180436738278542°K.

--> isentropic compression

[...]

I added an isobaric cooling process that is needed to close the system and restore the fluid to its original state.

	P (Pa)	T (K)	$\rho$ (kg/m³)	W (J/kg)	Q (J/kg)
1	101325	300.0	1.1768
2 isentropic expansion (1:3)	21764	193.3	0.3923	76554	0
3 isochoric heating (to $T_1$)	33775	300.0	0.3923	0	-76554
4 isentropic compression (to $P_1$)	101325	410.6	0.8598	-79371	0
5 isobaric cooling (to state 1)	101325	300.0	1.1768	-31748	111119
			NET:	-34575	34575

Scenario 2:

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.

	P (Pa)	T (K)	$\rho$ (kg/m³)	W (J/kg)	Q (J/kg)
1	101325	300.0	1.1768
2 isentropic compression (3:1)	471722	465.6	3.5305	-118785	0
3 isochoric cooling (to $T_1$)	303975	300.0	3.5305	0	118785
4 isothermal expansion (to state 1)	101325	300.0	1.1768	94591	-94591
			NET:	-24194	24194

In scenario 1, we have to input 34574 J/kg of net work to get a cooling of 76554 J/kg in the isochoric heating process.
In scenario 2, we have to input 24194 J/kg of net work to get a cooling of 94591 J/kg in the isothermal expansion process.

Scenario 2 gives more cooling for less work.

 

[T C]

I added an isobaric cooling process that is needed to close the system and restore the fluid to its original state.

P (Pa)	T (K)	ρ (kg/m³)	W (J/kg)	Q (J/kg)
1	101325	300.0	1.1768
2 isentropic expansion (1:3)	21764	193.3	0.3923	76554	0
3 isochoric heating (to T1)	33775	300.0	0.3923	0	-76554
4 isentropic compression (to P1)	101325	410.6	0.8598	-79371	0
5 isobaric cooling (to state 1)	101325	300.0	1.1768	-31748	111119
NET:	-34575	34575

In the fifth phase, no work is necessary by the external machinery and for me, that's the mot important part. So this 31748 J part can be eliminated from the net energy consumption part. In the fifth/last phase, this can be done simply by passing the fluid through a heat exchanger and that doesn't need any kind big energy input.

 

[Chestermiller]

The fall in internal energy and temperature will only cause when the gas expands as that means doing some kind of work and the only source of energy for the gas to do that work is its internal energy, right?

If I understand you correctly, right for an adiabatic process.