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Reverse Carnot - Refrigeration Cycle problem

by leonida
Tags: carnot, cycle, refrigeration, reverse
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leonida
#1
Mar7-12, 08:45 PM
P: 10
Hi guys, I have difficulties understanding the what I need to do in this problem. I have a ideal refrigeration cycle. first part is standard calculations, work of compressor and such. Results are Temperature range between -23 and 60 degrees C. Second part is the one that i do not understand. it says: "Assume an expanded control volume where cooling space is at 0 degrees C and the heat rejection space is at 20 degrees C and with the dead state of 20 deg C and 100 kPa. Do a second law analysis of the system." What am i supposed to do here? Is this a cycle in a cycle that runs between o and 20 deg C? or is it a simple ideal reverse Carnot that fits inside the dome? or something else...


thanks
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Andrew Mason
#2
Mar7-12, 10:13 PM
Sci Advisor
HW Helper
P: 6,679
Quote Quote by leonida View Post
Hi guys, I have difficulties understanding the what I need to do in this problem. I have a ideal refrigeration cycle. first part is standard calculations, work of compressor and such. Results are Temperature range between -23 and 60 degrees C. Second part is the one that i do not understand. it says: "Assume an expanded control volume where cooling space is at 0 degrees C and the heat rejection space is at 20 degrees C and with the dead state of 20 deg C and 100 kPa. Do a second law analysis of the system." What am i supposed to do here? Is this a cycle in a cycle that runs between o and 20 deg C? or is it a simple ideal reverse Carnot that fits inside the dome? or something else...


thanks
The "heat rejection space" is simply the hot reservoir (that heat flows into) and the cooling space is the cold reservoir (that heat flows out of). The "dead state" is simply the state at which the system would be at equilibrium with the surroundings so no heat would flow out of or into the system. I am not sure from the problem how that really differs from the hot reservoir.

It is not clear from this what is meant by "expanded control volume". It might help to give us the entire first problem.

AM
leonida
#3
Mar8-12, 06:42 AM
P: 10
AM Thanks for your quick response.

I will post the text of the problem below, and doing the first part of it is not a big deal. I am confused with the same thing you are, 2nd part of it - "expanded control volume". What does that mean?

text of the problem:
A refrigerator uses refrigerant-134a as the working fluid and operates
on the ideal vapor-compression refrigeration cycle except for the compression process.
The refrigerant enters the evaporator with a quality of 30 % and leaves the compressor at 60 C
Compressor work is 0.45 kW.
a) The mass flow rate of the refrigerant, the condenser pressure, and the COP of the refrigerator are to be determined.
b) Assume an expanded control volume where cooling space is at 0 degrees C and the heat rejection space is at 20 degrees C and with the dead state of 20 deg C and 100 kPa. Do a second law analysis of the system

Andrew Mason
#4
Mar12-12, 02:26 PM
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P: 6,679
Reverse Carnot - Refrigeration Cycle problem

Quote Quote by leonida View Post
AM Thanks for your quick response.

I will post the text of the problem below, and doing the first part of it is not a big deal. I am confused with the same thing you are, 2nd part of it - "expanded control volume". What does that mean?

text of the problem:
A refrigerator uses refrigerant-134a as the working fluid and operates
on the ideal vapor-compression refrigeration cycle except for the compression process.
The refrigerant enters the evaporator with a quality of 30 % and leaves the compressor at 60 C
Compressor work is 0.45 kW.
a) The mass flow rate of the refrigerant, the condenser pressure, and the COP of the refrigerator are to be determined.
b) Assume an expanded control volume where cooling space is at 0 degrees C and the heat rejection space is at 20 degrees C and with the dead state of 20 deg C and 100 kPa. Do a second law analysis of the system
"Expanded control volume" must just refer to the volume of refrigerant (vapour) that is in thermal contact with the cold reservoir. The cold reservoir (the air being refrigerated) is being kept cooled to 0 C.

Refrigerant enters the control volume at less than 0C and leaves this space at or close to 0 C. The refrigerant then enters the compressor. It leaves the compressor as a liquid at 60C and is then placed in thermal contact with the hot reservoir (the heat rejection space) which is at 20C. It leaves the heat rejection space at close to 20C. It is then put through the throttle valve causing it to expand/evaporate and cool. It then again enters the control volume and the cycle repeats.

Does that help?

AM
leonida
#5
Mar12-12, 02:37 PM
P: 10
Andrew, thanks again for your help.

To me it does make sense what you have described. But how do i proceed with the 2nd law analysis? I do have dead state at 20 deg C and 100 kPa. How am i supposed to use this information? Restricted Dead State is used for Exergy calculation on system as a whole. Do I use the 0 and 20 deg C as dead states to calculate waste energy for evaporator and condenser as separate equipment? Using 0 deg for evaporator and 20 for condenser? In that case how should i analyze valve?
Andrew Mason
#6
Mar12-12, 02:54 PM
Sci Advisor
HW Helper
P: 6,679
Quote Quote by leonida View Post
Andrew, thanks again for your help.

To me it does make sense what you have described. But how do i proceed with the 2nd law analysis? I do have dead state at 20 deg C and 100 kPa. How am i supposed to use this information? Restricted Dead State is used for Exergy calculation on system as a whole. Do I use the 0 and 20 deg C as dead states to calculate waste energy for evaporator and condenser as separate equipment? Using 0 deg for evaporator and 20 for condenser? In that case how should i analyze valve?
Second law analysis looks at entropy. Draw a diagram of Temperature v. entropy for a unit mass of refrigerant for each part of the cycle. What is happening to entropy in the different stages? I think that is what they are asking.

AM


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