Revisiting Refrigeration Efficiency: Impact of Potential Typo on Calculations

In summary, the conversation revolved around the topic of efficiency and COP (Coefficient of Performance). It was discussed that COP can be greater than 100% and is not limited to 1:1 or 100%. An analogy of using a bucket to transport heat was used to explain the concept of COP. The conversation also touched upon the calculation of work done and specific enthalpy values from tables, as well as the validity of the problem being discussed.
  • #1
yecko
Gold Member
279
15
Homework Statement
A vapor compression refrigeration cycle uses R-22 and follows the theoretical single-stage cycle. The condensing temperature is 48 oC, and the evaporating temperature is -18 oC. The power input to the cycle is 3 kW, and the mass flow rate of refrigerant is 0.1 kg/s. Determine (a) the heat rejected from the condenser, (b) the coefficient of performance, and (c) the refrigerating efficiency.
Relevant Equations
R-22 table
Refrigeration system
1AA7C4B1-3A81-4C65-B3E9-8E6729C1E074.jpeg

Can efficiency be larger than 100%?

I have double checked the values of
enthalpy of point 1,2,3 from R-22 tables.
And I obtained enthalpy of point 4 from the work done given.

Thank you.
 
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  • #2
I can't read your handwriting. What do the i's stand for?

From your tables, what are the conditions at 1,2,3, and 4 ? (T, P, u, h, s)
 
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  • #3
i is enthalpy, same as h as your notation.
(From R22 table)
i1=i2=260.51
i3=397.81
W=m(i4-i3)=3kW
i4=427.81
Ql=m(i3-i2)=13.72
COP= Ql/W=13.72/3=4.57
=>Which is higher than COP(theoretical)= 3.866
Is there anything I have done wrong?
Thank you.
 
  • #4
I'm just having trouble understanding the rationale of what you did.
 
  • #5
I'm having trouble comparing with you because my tables must be different from yours. At point 3, my tables show a specific enthalpy of 242.92 kJ/kg, and at point 1, they show a specific enthalpy of about 106 kJ/kg. So, we show the same difference between these two states. But, it's hard for me to help without the same exact tables as yours.
 
  • #6
46BCAFED-E9CB-4EF0-908C-AF6D270E5834.png
9B6BF36A-FE14-4CC8-A877-A5F0F3E00197.png


My tables are attached here
 
  • #7
yecko said:
Can efficiency be larger than 100%?
Very common question/concern; COP isn't efficiency, which is probably part of the reason why they called it by a different name. Yes, it certainly can be more than 100%. It can be almost anything.

Think of COP this way:
You have a bucket. You can put dirt in that bucket and carry it from one place to another. You want to know how efficient your dirt transportation is. Maybe you measure it in kg of dirt moved per Joule of work. That's not a percentage, but it kind of is an efficiency.

But what if what you are moving is heat (or "cold")? You can fill the bucket with hot water or ice and carry it from one place to another. Then you're moving a certain number of Joules of thermal energy with a certain number of Joules of work, and the amount of heat you move has no relation whatsoever to the amount of energy it took to move it. The ratio can truly be anything. Strictly speaking it isn't efficiency, because efficiency is a fraction of work-in you get to keep. Here, the work-in just plain isn't what you are after.

That's basically what COP is. Depending on the process, there are different thermodynamic rules governing what the COP can be, but it is not limited to 1:1 or 100%.
 
  • #8
yecko said:
View attachment 274320View attachment 274321

My tables are attached here
That 30 kJ/kg seems very suspicious. What would the work have been if the compression were isentropic and the pressure of the superheated vapor exiting the compressor were the equilibrium vapor pressure at 48 C?
 
  • #9
russ_watters said:
Very common question/concern; COP isn't efficiency, which is probably part of the reason why they called it by a different name. Yes, it certainly can be more than 100%. It can be almost anything.

Think of COP this way:
You have a bucket. You can put dirt in that bucket and carry it from one place to another. You want to know how efficient your dirt transportation is. Maybe you measure it in kg of dirt moved per Joule of work. That's not a percentage, but it kind of is an efficiency.

But what if what you are moving is heat (or "cold")? You can fill the bucket with hot water or ice and carry it from one place to another. Then you're moving a certain number of Joules of thermal energy with a certain number of Joules of work, and the amount of heat you move has no relation whatsoever to the amount of energy it took to move it. The ratio can truly be anything. Strictly speaking it isn't efficiency, because efficiency is a fraction of work-in you get to keep. Here, the work-in just plain isn't what you are after.

That's basically what COP is. Depending on the process, there are different thermodynamic rules governing what the COP can be, but it is not limited to 1:1 or 100%.
I mean efficiency (part c of the question), which is COP/ COP_carnot, or Ql/Qh or Tl/Th
 
  • #10
Chestermiller said:
That 30 kJ/kg seems very suspicious.
I am sorry but where do you mean by the 30kJ/kg?
 
  • #11
yecko said:
I am sorry but where do you mean by the 30kJ/kg?
3 KW at 0.1 kg/sec is 30 kJ/kg
 
  • #12
For isentropic operation of the compressor with an exit pressure of about 19 bars (the pressure in the condenser), I estimate an enthalpy change of about 50 kJ/kg (for the ideal compressional work). That should really be the minimum amount of work.
 
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  • #13
Chestermiller said:
For isentropic operation of the compressor with an exit pressure of about 19 bars (the pressure in the condenser), I estimate an enthalpy change of about 50 kJ/kg (for the ideal compressional work). That should really be the minimum amount of work.
The problem is invalid, right?
 
  • #14
It seems that way to me. Maybe the 3 Kw is a typo, and it should have been 5 Kw. How would that affect your answer and your conclusions?
 

1. What is the COP refrigeration cycle?

The COP (Coefficient of Performance) refrigeration cycle is a thermodynamic cycle used in refrigeration systems to transfer heat from a low-temperature source to a high-temperature sink. It involves four main components: evaporator, compressor, condenser, and expansion valve. The cycle operates by compressing a refrigerant gas, which becomes hot and then condenses into a liquid, releasing heat. The liquid then expands through the expansion valve, causing it to cool and evaporate. This process absorbs heat from the surrounding environment, creating a cooling effect.

2. How does the COP refrigeration cycle work?

The COP refrigeration cycle works by using a refrigerant, a substance that can change between liquid and gas states at various temperatures, to transfer heat from one location to another. The cycle begins with the refrigerant in a low-pressure, low-temperature liquid state in the evaporator. The refrigerant is then compressed by the compressor, increasing its pressure and temperature. The hot, high-pressure gas then flows to the condenser, where it releases heat and condenses back into a liquid. The liquid refrigerant then flows through the expansion valve, where it expands and cools, absorbing heat from the surrounding environment. The cycle then repeats, continuously transferring heat from the evaporator to the condenser.

3. What is the purpose of the COP refrigeration cycle?

The main purpose of the COP refrigeration cycle is to provide a cooling effect by transferring heat from a low-temperature source, such as a room or food storage area, to a high-temperature sink, such as the outside environment. This is achieved by using the thermodynamic properties of the refrigerant to absorb heat from the surrounding environment and release it in a different location. Refrigeration systems that use the COP refrigeration cycle are commonly used in refrigerators, air conditioners, and industrial cooling systems.

4. What factors affect the COP of a refrigeration cycle?

The COP of a refrigeration cycle can be affected by several factors. One of the main factors is the type of refrigerant used, as different refrigerants have different thermodynamic properties. The efficiency of the compressor also plays a significant role in the COP, as a more efficient compressor can compress the refrigerant with less energy input. The temperature difference between the evaporator and condenser also affects the COP, with a larger temperature difference resulting in a lower COP. Additionally, the design and maintenance of the refrigeration system can impact its overall COP.

5. What are the benefits of using the COP refrigeration cycle?

There are several benefits to using the COP refrigeration cycle in refrigeration systems. One of the main benefits is its energy efficiency, as it can transfer heat from one location to another using relatively little energy. This makes it a cost-effective option for cooling and air conditioning. The cycle also allows for precise temperature control, making it suitable for various applications, from food storage to industrial processes. Additionally, the use of refrigerants in the cycle can also have environmental benefits, as newer refrigerants are being developed to have lower global warming potential and ozone depletion potential.

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