Determine the coefficient of performance of this cycle

AI Thread Summary
In an ideal refrigeration cycle with a condensing temperature of 40°C and an evaporation temperature of -20°C, the coefficient of performance (COP) can be calculated using the formula C.O.Pc = TL/(TH-TL), resulting in a value of 4.2. The discussion highlights the importance of understanding the working fluids, R12 and ammonia, and their properties, as they influence performance metrics like COP. Participants emphasize the need to refer to thermodynamic tables and T-S diagrams to determine enthalpy values and temperatures at various points in the cycle. The conversation also touches on practical insights from real-world applications, such as the effects of refrigerant levels on system performance. Overall, mastering these concepts enhances comprehension of refrigeration cycles and their efficiencies.
Nemo's
Messages
69
Reaction score
0

Homework Statement


In an ideal refrigeration cycle, the temperature of the condensing vapour is 40oC and the temperature during evaporation is -20oC. Determine the coefficient of performance of this cycle for the working fluids; R12 and ammonia.

Homework Equations


C.O.Pc = TL/(TH-TL)

The Attempt at a Solution


C.O.Pc= ((-20+273)/(40--20))
= 4.2
I don't understand how the working fluid would affect the results of the C.O.Pc (where c stands for carnot efficiency)
 
Physics news on Phys.org
Carnot tells what is the best possibly achievable
i think your question, why does the refrigerant make a difference
lies in the Molier charts for real refrigerants
here's more than any mortal could want to know
https://www.downloads.siemens.com/download-center/Download.aspx?pos=download&fct=getasset&id1=8359

this looks like a good worked out example
http://www.saylor.org/site/wp-content/uploads/2013/08/BolesLectureNotesThermodynamicsChapter10.pdf
see example 10-1
upload_2016-5-7_9-52-4.png


old jim
 
Last edited by a moderator:
jim hardy said:
Thanks you for the helpful material you attached.
From the second attachment I concluded that COPR=(h1-h4)/(h2-h1) so I was able to get h1 from the tables at -20οC. for R12 h1=178.73 and similarly I got h2=74.59. However,I don't know the temperature at 3 or 4 so I can't get h4. What should I do?
 
Nemo's said:
Thanks you for the helpful material you attached.
From the second attachment I concluded that COPR=(h1-h4)/(h2-h1) so I was able to get h1 from the tables at -20οC. for R12 h1=178.73 and similarly I got h2=74.59. However,I don't know the temperature at 3 or 4 so I can't get h4. What should I do?
The temperature at 4 is going to be the same TL as in the ideal Carnot cycle.

Do like Jim suggests, and look at the worked example he attached in Post #2.
 
Nemo's said:
However,I don't know the temperature at 3 or 4 ... What should I do?

1. Go back to original question
Nemo's said:
In an ideal refrigeration cycle, the temperature of the condensing vapour is 40oC and the temperature during evaporation is -20oC.
Ideal cycle, eh ?
That's what is represented in the T-S diagram above,
Think about it
working fluid enters evaporator as liquid-vapor mix at saturation pressure and leaves as saturated vapor at same pressure, as stated by phrase "4-1 constant pressure heat addition" and as shown by proximity to saturation line on the T-S diagram ...
that sounds to me like constant temperature
What do you think from looking at the chart ?

in their ideal cycle chart they don't show any superheat so i'd assume saturation temperature on both ends of the evaporator, just different quality.
Isn't that what is the "x" is here , fraction of vapor ?
upload_2016-5-7_17-35-23.png


2. so i think if you know t at 1 you know t at 4. What do you think ?

It is helpful to remember refrigerators work by heat of vaporization
and imagine yourself inside a tiny submarine riding along through the pipes
I think that'll help you visualize .

3. so - what goes on in the condenser ? Heat of vaporization is rejected...
First it cools the superheated compressor discharge , point 2 to saturation line,
then it rejects heat of vaporization at constant pressure again as stated, that's saturation line to other side of saturation curve point 3. Between the two sides of the saturation line it is just giving up heat of vaporization, and that has to be at saturation temperature... I think you probably know T at 3...

4. Print yourself several copies of that T-S chart and think your way around the circuit,
writing on each segment your own words about what is going on. When you get the words boiled down to a concise summary and it "feels right" you will feel an upwelling sense of accomplishment. Probably you'll improve on it after a night's sleep.Just a couple days ago i charged a friend's car airconditioner.
When they're low on freon they can't liquefy in the condenser
so no liquid enters the evaporator
meaning there can't be much absorbtion of heat in the evaporator because there's no vaporization going on in there anymore
and the air comes out of the dashboard vents warm. Point 1 is in superheat.
It is enlightening to feel the air get cooler as the compressor's suction pressure comes up , which is counterintuitive ...
but what's happening is it's coming out of superheat and is approaching saturation pressure for the cabin air temperature.

In a real airconditioner they want a small amount of superheat at evaporator outlet to assure the compressor doesn't ingest incompressible liquid
and they want a small amount of subcooling at condenser outlet to assure you've given up all the heat of vaporization
but those practicalities aren't shown on the ideal diagram.

We learn more by doing than by reading about doing
$49 buys a set of inexpensive refrigeration gages for your car
and summer is just around the corner...
You'll quickly observe that the compressor discharge line is the hottest point in the system
and you'll get a visceral "feel" for the power of evaporation .
Not to mention you'll gain sudden popularity...old jim
 
Last edited:
Back
Top