R123 thermophysical properties and COP calculation
R123 thermophysical properties and COP calculation
(OP)
Hi there, i'm looking for some help to calculate the COP of a chiller (low pressure chiller) based on the pressure of the refrigerant that i have inside the Evaporator and the Condsenser.
the problem is i'm getting a very high COP when i'm doing the calculation, please find below the procedure:
1) h1= enthalpy of the Vapor after the evaporator (based on the presuure inside the evaporator pe)
2) h2= enthalpy of the vapor after the compressor( based on the pressure inside the condensor Pc and entropy s2 as s1=s2)
3) h3= enthalpy of the liquid refrigerant after the condensor( based on the pressure of the refrigerent)
4) h4=h3
pe=0.043 MPa
pc=0.152 MPa
COP=(h1-h4)/(h2-h1)
this is how we used to calculate the COP at the university, following the same procedure and we used to do it for R134 not for R123.
is there something that i don't know about R123 properties!
Regards
the problem is i'm getting a very high COP when i'm doing the calculation, please find below the procedure:
1) h1= enthalpy of the Vapor after the evaporator (based on the presuure inside the evaporator pe)
2) h2= enthalpy of the vapor after the compressor( based on the pressure inside the condensor Pc and entropy s2 as s1=s2)
3) h3= enthalpy of the liquid refrigerant after the condensor( based on the pressure of the refrigerent)
4) h4=h3
pe=0.043 MPa
pc=0.152 MPa
COP=(h1-h4)/(h2-h1)
this is how we used to calculate the COP at the university, following the same procedure and we used to do it for R134 not for R123.
is there something that i don't know about R123 properties!
Regards





RE: R123 thermophysical properties and COP calculation
but you mayuse the wront teperatures. Refrigerant temeprature in condenser will be higher than ambient. the UA will determine how much higher.
in the evaporator refrigerant will be colder than then fluid. Again, UA will determinehow much colder.
You also need to include compressor efficiency.
Now you know how to make a chiller more effficient: larger heat exchanger surface and more efficient compressor.
RE: R123 thermophysical properties and COP calculation
Regards
RE: R123 thermophysical properties and COP calculation
RE: R123 thermophysical properties and COP calculation
RE: R123 thermophysical properties and COP calculation
Your S1=S2 assumption is the biggest factor that makes your COP so high. That's a perfectly efficient compressor. Look up a chart with the isentropic efficiency of that compressor at the operating point you're interested in. All of its heat goes into the vapor. If the compressor is hermetic, the motor heat goes in there too. HerrKaLeun mentioned this in his post.
Another reason for high COP is that R123 is one of the very best refrigerants in terms of work needed to get the proper lift for HVAC needs. Look here: Good charts in here even though off-topic.
Also, give this document a glance.
Wait for someone else to explain the superheat question.
Best to you,
Goober Dave
Haven't see the forum policies? Do so now: Forum Policies
RE: R123 thermophysical properties and COP calculation
But previously
So it would seem that it's because you are using saturation properties, not actual properties.