HFC-134a Refrigerant & 2nd Law of Thermodynamics

In summary, after analyzing the process of using HFC-134a refrigerant in a refrigeration/heat pump cycle, it appears that the useful output from the three Carnot heat engines exceeds the input from the compressor, indicating a 1.5% deficit on the cold side of the loop. It is possible that the cold section could absorb heat from an outside heat reservoir in order to complete the cycle.
  • #1
Devin-M
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I am trying to reconcile HFC-134a refrigerant with the 2nd Law of Thermodynamics. I shouldn't be able to extract useful work from a single temperature reservoir.

Suppose I use a refrigeration / heat pump cycle with HFC-134a as shown below.

-Compressor adds 29 kJ/kg increasing refrigerant temperature from -60C (saturated vapor) to -18C

-Hot side heat exchanger is divided into 3 sections -- two variable temperature sections and one constant temperature section.

-The log mean temp of hot side section 1 is -24C, temp of hot side section 2 is -30C, and log mean temp of hot side section 3 is -45.3C

-If 3 carnot efficiency heat engines are connected across the 3 hot sections to the -60C cold section, they extract 30.96kJ/kg of useful work.

Since the compressor only adds 29kJ/kg of energy, but the useful work of the heat engines is 30.96kJ/kg, won't this mean there is 1.96kJ/kg energy deficit on the cold side of the loop before the compressor? Can't this cold section absorb heat from an outside heat reservoir that is greater than -60C to complete the cycle?

hfc-134a5-.jpg


https://www.speakev.com/cdn-cgi/image/format=auto,onerror=redirect,width=1920,height=1920,fit=scale-down/https://www.speakev.com/attachments/hfc-134a6-jpg.147707/
 
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  • #2
Devin-M said:
I am trying to reconcile HFC-134a refrigerant with the 2nd Law of Thermodynamics. I shouldn't be able to extract useful work from a single temperature reservoir.

Suppose I use a refrigeration / heat pump cycle with HFC-134a as shown below.

-Compressor adds 29 kJ/kg increasing refrigerant temperature from -60C (saturated vapor) to -18C

-Hot side heat exchanger is divided into 3 sections -- two variable temperature sections and one constant temperature section.

-The log mean temp of hot side section 1 is -24C, temp of hot side section 2 is -30C, and log mean temp of hot side section 3 is -45.3C

-If 3 carnot efficiency heat engines are connected across the 3 hot sections to the -60C cold section, they extract 30.96kJ/kg of useful work.

Since the compressor only adds 29kJ/kg of energy, but the useful work of the heat engines is 30.96kJ/kg, won't this mean there is 1.96kJ/kg energy deficit on the cold side of the loop before the compressor? Can't this cold section absorb heat from an outside heat reservoir that is greater than -60C to complete the cycle?

View attachment 295196

https://www.speakev.com/cdn-cgi/image/format=auto,onerror=redirect,width=1920,height=1920,fit=scale-down/https://www.speakev.com/attachments/hfc-134a6-jpg.147707/
It is not clear to me what is being done here. Can you please provide a schematic diagram of the process? Also, are those log-mean temperatures or log-mean temperature differences?
 
  • #3
Hi,

Where do you get the numerical data for your cycle ? Surely not from the graph ?
Probably not here

##\ ##
 
  • #4
Chestermiller said:
It is not clear to me what is being done here. Can you please provide a schematic diagram of the process? Also, are those log-mean temperatures or log-mean temperature differences?

BvU said:
Hi,

Where do you get the numerical data for your cycle ? Surely not from the graph ?
Probably not here

##\ ##

I'm going to redo my calculations with numbers from the chart you linked. So far, I've got:

Compressor Entrance
Saturated Vapor
Enthalpy: 361.7 kJ/kg
Entropy: 1.80 kJ/(kg) (K)
Temperature: -60C
Pressure: 15.89kPa

Adding 27.1 kJ/kg with the compressor at constant entropy-->

Compressor Exit:
Superheated Vapor
Enthalpy: 388.8 kJ/kg
Entropy: 1.80 kJ/(kg) (K)
Temperature: -20C
Pressure: 70.00kPa
 
  • #5
Devin-M said:
I'm going to redo my calculations with numbers from the chart you linked. So far, I've got:

Compressor Entrance
Saturated Vapor
Enthalpy: 361.7 kJ/kg
Entropy: 1.80 kJ/(kg) (K)
Temperature: -60C
Pressure: 15.89kPa

Adding 27.1 kJ/kg with the compressor at constant entropy-->

Compressor Exit:
Superheated Vapor
Enthalpy: 388.8 kJ/kg
Entropy: 1.80 kJ/(kg) (K)
Temperature: -20C
Pressure: 70.00kPa
The pressures don't seem to match what the diagram shows.
 
  • #6
Chestermiller said:
The pressures don't seem to match what the diagram shows.
I'm trying to keep the compressor input at or below the original value of 29 kJ/kg while using values found in the charts via the link.
 
  • #7
I'm still calculating 1.5% more useful output from the 3 Carnot heat engines than input to the compressor...

For Reference

---

Compressor Entrance
Saturated Vapor
Enthalpy: 361.7 kJ/kg
Entropy: 1.80 kJ/(kg) (K)
Temperature: -60C
Pressure: 15.89kPa

+27.1 kJ/kg Compressor

Compressor Exit
Superheated Vapor
Enthalpy: 388.8 kJ/kg
Entropy: 1.80 kJ/(kg) (K)
Temperature: -20C
Pressure: 70.00kPa

Heat Engine 1
Hot Exchanger Temp Range: Hot -20C Cold -33.83C Variable
Hot Exchanger Log Mean Temp: -26.98C
Cold Exchanger Temp: -60C
Carnot Efficiency: 13.413%
Energy Transfer: 10.6 kJ/kg
Useful Amount: 1.421 kJ/kg

Saturated Vapor:
Enthalpy: 378.2 kJ/kg
Entropy: 1.75 kJ/(kg) (K)
Temperature: -33.83C
Pressure: 70.00kPa

Heat Engine 2
Hot Exchanger Temp Range: -33.83C Constant
Hot Exchanger Log Mean Temp: -33.83C
Cold Exchanger Temp: -60C
Carnot Efficiency: 10.935%
Energy Transfer: 222.1 kJ/kg
Useful Amount: 24.2866 kJ/kg

Saturated Liquid:
Enthalpy: 156.1 kJ/kg
Entropy: 0.82 kJ/(kg) (K)
Temperature: -33.83C
Pressure: 70.00kPa

Heat Engine 3
Hot Exchanger Temp Range: Hot -33.83C Cold -60C Variable
Hot Exchanger Log Mean Temp: -47.16C
Cold Exchanger Temp: -60C
Carnot Efficiency: 5.682%
Energy Transfer: 32.1 kJ/kg
Useful Amount: 1.823 kJ/kg

Hot Exit:
Enthalpy: 124 kJ/kg
Temperature: -60C
Pressure: 70.00kPa

Cold Entrance:
Saturated Liquid
Enthalpy: 124 kJ/kg
Entropy: 0.68 kJ/(kg) (K)
Temperature: -60C
Pressure: 15.89kPa

Total Useful Amount:
27.53 kJ/kg

Useful Output / Compressor Input = 27.53 / 27.1 = 101.5%
 
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  • #8
1.5 % is not so bad when reading from a graph...
 
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  • #9
BvU said:
1.5 % is not so bad when reading from a graph...
I got those values from the link you posted.

BvU said:
 
  • #10
Devin-M said:
I'm still calculating 1.5% more useful output from the 3 Carnot heat engines than input to the compressor...
As you improve your calculations, you reduce your error. We already went through this 6 months ago.
 
  • #11
BvU said:
Hi,

Where do you get the numerical data for your cycle ? Surely not from the graph ?
Probably not here

##\ ##
It looks like this is exactly where the OP got his data.
 
  • #12
Using log mean temperatures assumes the heat capacities are constant. I agree with the others: your calculations have too much inaccuracy.
 
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  • #13
I calculated the energy transfer through heat engine 1 from the difference between the enthalpy at the compressor exit and the enthalpy at saturated vapor...

Devin-M said:
Heat Engine 1
Hot Exchanger Temp Range: Hot -20C Cold -33.83C Variable
Hot Exchanger Log Mean Temp: -26.98C
Cold Exchanger Temp: -60C
Carnot Efficiency: 13.413%
Energy Transfer: 10.6 kJ/kg
Useful Amount: 1.421 kJ/kg

Devin-M said:
Compressor Exit
Superheated Vapor
Enthalpy: 388.8 kJ/kg

Devin-M said:
Saturated Vapor:
Enthalpy: 378.2 kJ/kg
Entropy: 1.75 kJ/(kg) (K)
 
  • #14
The problem is just roundoff error. The entropy of the superheated vapor at -20 should be the same as that of the saturated vapor at -60, 1.8024 rather than 1.80. I used the latter value and got a match on the work.

Here is the calculation for the Carnot engine(s):

Hot Side refrigerant in:
Temperature -20 C
Pressure 70 kPa
Enthalpy 388.8 kJ
Entropy 1.8024

Hot Side refrigerant out:
Temperature = -60 C
Pressure = 70 kPa
Enthalpy = 124.0 kJ
Entropy = 0.6873

Hot Side refrigerant change in enthalpy = ##\Delta H_{hot}=Q_{hot}=-264.8 kJ##
Hot Side refrigerant change in entropy = ##\Delta S_{hot}=-1.1151##

Temperature of cold side refrigerant = T_{cold}=-60 C = 213.2 K

The entropy balance of the Carnot engine working fluid is: $$-\Delta S_{hot}+\frac{Q_{cold}}{T_{cold}}=0$$where ##Q_{cold}## is the heat transferred from the Carnot working fluid to the refrigerant on the cold side. This gives ##Q_{cold}=237.7\ kJ##

The energy balance on the Carnot engine working fluid is:$$-Q_{hot}-Q_{cold}=W=264.8-237.7=27.1\ kJ$$
 
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  • #15
Chestermiller said:
The problem is just roundoff error. The entropy of the superheated vapor at -20 should be the same as that of the saturated vapor at -60, 1.8024 rather than 1.80. I used the latter value and got a match on the work.

Here is the calculation for the Carnot engine(s):

Hot Side refrigerant in:
Temperature -20 C
Pressure 70 kPa
Enthalpy 388.8 kJ
Entropy 1.8024

Hot Side refrigerant out:
Temperature = -60 C
Pressure = 70 kPa
Enthalpy = 124.0 kJ
Entropy = 0.6873

Hot Side refrigerant change in enthalpy = ##\Delta H_{hot}=Q_{hot}=-264.8 kJ##
Hot Side refrigerant change in entropy = ##\Delta S_{hot}=-1.1151##

Temperature of cold side refrigerant = T_{cold}=-60 C = 213.2 K

The entropy balance of the Carnot engine working fluid is: $$-\Delta S_{hot}+\frac{Q_{cold}}{T_{cold}}=0$$where ##Q_{cold}## is the heat transferred from the Carnot working fluid to the refrigerant on the cold side. This gives ##Q_{cold}=237.7\ kJ##

The energy balance on the Carnot engine working fluid is:$$-Q_{hot}-Q_{cold}=W=264.8-237.7=27.1\ kJ$$
I’ve calculated that only edit:237.27kJ/kg is returned to the cold side via the 3 carnot heat engines (since 27.53kJ/kg is extracted as useful work), so to complete the refrigeration cycle where is the other 0.43kJ/kg coming from? If it’s absorbed from the environment how do we reconcile it with the 2nd Law of Thermodynamics, since we shouldn’t be able to convert heat to work with a device that works by extracting heat from a single temperature reservoir (ie the environment).
 
Last edited:
  • #16
Devin-M said:
I’ve calculated that only edit:237.27kJ/kg is returned to the cold side via the 3 carnot heat engines (since 27.53kJ/kg is extracted as useful work), so to complete the refrigeration cycle where is the other 0.43kJ/kg coming from? If it’s absorbed from the environment how do we reconcile it with the 2nd Law of Thermodynamics, since we shouldn’t be able to convert heat to work with a device that works by extracting heat from a single temperature reservoir (ie the environment).
I showed that your 237.37 is inaccurate, and that the correct value is 237.7. You integrated CpdT/T to get the hot side entropy change (assuming constant Cp), but the thermodynamic table clearly gives the overall entropy change directly (which is needed to get the cold side heat return of the engine(s)). You didn't have to integrate to get the entropy change, and, if you had used the values from the table, you would have gotten the correct result.
 
  • #17
I used this as reference for Carnot heat engine 1...

heat_engine_1.jpg
Devin-M said:
Heat Engine 1
Hot Exchanger Temp Range: Hot -20C Cold -33.83C Variable
Hot Exchanger Log Mean Temp: -26.98C
Cold Exchanger Temp: -60C
Carnot Efficiency: 13.413%
Energy Transfer: 10.6 kJ/kg
Useful Amount: 1.421 kJ/kg
 
  • #18
For heat engine 1, I noted the 10.6kJ/kg change in enthalpy between 70kPa -20, and Sat Vap 70kPa -33.83. So 10.6kJ/kg went into the hot side of Carnot heat engine 1. Next I found the log mean of the temperature range (-20C to -33.83C) on the hot side of the heat engine was -26.98C. Next I calculated the carnot efficiency of a heat engine with a -26.98C hot side and a -60C cold side-- and that is 13.413% efficiency. Next I calculated the useful work from the heat engine was 1.421kJ/kg. To find the heat returned to the cold stream I would subtract 1.421kJ/kg from 10.6kJ/kg to arrive at 9.179kJ/kg is returned to the cold stream through heat engine 1.

Can you please point out what stage of the calculation I made my error?
 
  • #19
Devin-M said:
For heat engine 1, I noted the 10.6kJ/kg change in enthalpy between 70kPa -20, and Sat Vap 70kPa -33.83. So 10.6kJ/kg went into the hot side of Carnot heat engine 1. Next I found the log mean of the temperature range (-20C to -33.83C) on the hot side of the heat engine was -26.98C. Next I calculated the carnot efficiency of a heat engine with a -26.98C hot side and a -60C cold side-- and that is 13.413% efficiency. Next I calculated the useful work from the heat engine was 1.421kJ/kg. To find the heat returned to the cold stream I would subtract 1.421kJ/kg from 10.6kJ/kg to arrive at 9.179kJ/kg is returned to the cold stream through heat engine 1.

Can you please point out what stage of the calculation I made my error?
Why did you use the heat capacity and log mean temperatures if you could get the entropy change directly from the table? You do realize that using the log-mean temperature is tantamount to assuming that the heat capacity is constant, right?

The entropy change is 1.8004 - 1.7571 = 0.0433. So the heat returned to the cold stream should be (0.0433)(213.2) = 9.23 kJ/kg. The hot side heat removal was 10.6 kJ/kg. So the work should be 1.37 kJ/kg, and the efficiency should be 12.9%. By the way, the 10.6 is not precise enough to allow us to express the cold side heat removal or the efficiency to 3 significant figures..
 
  • #20
While I try and digest that explanation, what about the simpler case (Heat Engine 2) with constant temperatures on both the hot and cold sides? There's an enthalpy change of 222.1 kJ/kg. Hot temp constant -33.83C and cold temp constant -60C. Carnot efficiency is 10.935%. 222.1 kJ/kg goes into the hot side and 24.2866 kJ/kg comes out as useful energy. 197.81kJ/kg goes back into the cold stream through heat engine 2, correct?
 
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  • #21
Devin-M said:
While I try and digest that explanation, what about the simpler case (Heat Engine 2) with constant temperatures on both the hot and cold sides? There's an enthalpy change of 222.1 kJ/kg. Hot temp constant -33.83C and cold temp constant -60C. Carnot efficiency is 10.935%. 222.1 kJ/kg goes into the hot side and 24.2866 kJ/kg comes out as useful energy. 197.81kJ/kg goes back into the cold stream through heat engine 2, correct?
For that "engine," both methods will give the same answer (of course). However, the calculated values will depend on what one uses for the absolute Kelvin temperature at 0 C (273., 273.2, 273.15, or whatever). It also depends on what value the developer of your table used. If you divide their enthalpy of vaporization by their entropy of vaporization at -20 C, you get a temperature of 253.18 K, implying a value at 0 C of 273.18.

The point of. all this is that (1) the table may not have enough precision to resolve the difference you are looking for and (2) using the log-mean temperatures for engines 1 and 3 results in inaccuracy, by inherently neglecting the effect of temperature on heat capacity.
 
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1. What is HFC-134a refrigerant?

HFC-134a refrigerant is a chemical compound used in refrigeration systems and air conditioners as a cooling agent. It is a type of hydrofluorocarbon (HFC) that does not contain chlorine, making it a more environmentally friendly alternative to older refrigerants like CFCs and HCFCs.

2. How does HFC-134a refrigerant work in a cooling system?

HFC-134a refrigerant works by absorbing heat from the surrounding air or liquid and then releasing it through a process called phase change. When the refrigerant is compressed, it becomes a high-pressure gas. As it expands, it cools down and turns into a liquid, absorbing heat from the surrounding environment. The liquid refrigerant then travels through the system and evaporates, releasing the absorbed heat and cooling the surrounding area.

3. What is the 2nd Law of Thermodynamics?

The 2nd Law of Thermodynamics states that in any energy transfer or conversion, some energy will inevitably be lost as heat. This means that no system can be 100% efficient, and there will always be some energy that cannot be converted into useful work. This law has important implications for refrigeration systems, as it limits their efficiency and requires them to continuously release heat into the environment.

4. How does the 2nd Law of Thermodynamics apply to HFC-134a refrigerant?

The 2nd Law of Thermodynamics applies to HFC-134a refrigerant in two ways. Firstly, as mentioned earlier, the refrigerant must release heat into the environment in order to cool the surrounding area. This is due to the law's requirement for some energy to be lost as heat in any energy transfer. Secondly, the refrigerant itself is subject to the law, meaning that it will eventually degrade and lose its cooling properties over time, leading to the need for replacement.

5. What are the environmental impacts of using HFC-134a refrigerant?

While HFC-134a is considered a more environmentally friendly alternative to older refrigerants like CFCs and HCFCs, it still has negative impacts on the environment. HFCs are potent greenhouse gases, contributing to global warming and climate change. In addition, if the refrigerant is not properly disposed of, it can leak into the atmosphere and contribute to ozone depletion. Therefore, it is important to properly handle and dispose of HFC-134a refrigerant to minimize its environmental impact.

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