Why is my refrigeration cycle efficiency over 100% with R-22?

[yecko]

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

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.

 

[Chestermiller]

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)

 

[yecko]

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.

 

[Chestermiller]

I'm just having trouble understanding the rationale of what you did.

 

[Chestermiller]

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.

 

[yecko]

My tables are attached here

 

[russ_watters]

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%.

 

[Chestermiller]

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?

 

[yecko]

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

 

[yecko]

That 30 kJ/kg seems very suspicious.

I am sorry but where do you mean by the 30kJ/kg?

 

[Chestermiller]

I am sorry but where do you mean by the 30kJ/kg?

3 KW at 0.1 kg/sec is 30 kJ/kg

 

[Chestermiller]

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.

 

[yecko]

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?

 

[Chestermiller]

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?