Refrigeration - Work done by evaporator, condenser, expansion valve

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In vapor compression refrigeration, I know how to calculate the work done by the compressor (enthalpy at outlet - enthalpy at inlet), but I am lost on how to calculate the work done by the evaporator, condenser, and the expansion valve.

For the evaporator and condenser, if I wish to have a certain amount of heat exchange occur between the refrigerant and the air, would I just need to calculate the flow rate needed to attain the right amount of convection and get how much work needs to be done by the fan to generate that much air flow? If not, what should I do?

For the expansion valve, however, I don't know where to get started.

Thank you for the help in advance.
 

Answers and Replies

  • #2
rock.freak667
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Normally your evaporator and condenser will do no work (i.e. no mechanical work is being done by them). They will just facilitate heat exchange which you can get by the same enthalpy in - enthalpy out deal (this equation comes from the steady state-steady flow equation).

Depending on the conditions of the refrigerant, you can get the heat required to cause the conditions and then use that to obtain the mass flow of air required.

Ideally, your expansion valve just expands the volume the refrigerant at the expense of its enthalpy as the refrigerant will go back to the compressor at this stage.


If you have some conditions you'd like to work with, you can post them and we may be able to help with wherever we can.
 
  • #3
256bits
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The expansion valve in a typical home air conditioner is typically a capillary tube. The expansion process here is ideally a constant enthalpy process ( isenthalpic ), where the temperature and pressure of the fluid decreases.

The heat removed from the fluid in the condensor Qcondensor = Qevaporator + Wcompressor.

I usually like to give a reference for you read
http://www.coolingdevice.net/4.html [Broken]
http://web.me.unr.edu/me372/Spring2001/Vapor Compression Refrigeration Cycles.pdf
 
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  • #4
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Technically, there is no "work" done by the condenser or evaporator. Work is the exchange of mechanical energy wheras heat is the exchange of thermal energy. In a condenser or evaporator, heat transfer is occuring, thus thermal energy is being exchanged. Probably no need to worry about the terminology but just know the difference between them.

When you say you want to calculate the work done in the condenser and evaporator, what you actually mean is the energy transferred (in this case, thermal energy i.e., heat). I'm not sure of your conditions, but I'll assume that both evaporation and condensation are isobaric and isothermal. To calculate the energy removed or added during either process, multiply the mass flow rate of the working fluid (the refrigerant) by the change in enthalpy from inlet to outlet.

Hope this helps
 
  • #5
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Thanks everyone, but I think I may have been confused when I first asked this question. If there is no mechanical work done by the condenser or evaporator, how would the energy required to operate the heat exchanger inside them or maintaining them at a certain temperature be calculated?
 
  • #6
rock.freak667
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Thanks everyone, but I think I may have been confused when I first asked this question. If there is no mechanical work done by the condenser or evaporator, how would the energy required to operate the heat exchanger inside them or maintaining them at a certain temperature be calculated?
If you take for example the condenser which rejects heat to the environment, you can get the energy lost by the fluid by using Q = mass flow * change in enthalpy.

This Q will be the same as the heat energy gained by the air (or wherever it is rejecting to)
 
  • #7
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If you take for example the condenser which rejects heat to the environment, you can get the energy lost by the fluid by using Q = mass flow * change in enthalpy.

This Q will be the same as the heat energy gained by the air (or wherever it is rejecting to)
I know that, but I think the point of my question was missed again - what I meant was, how much electricity do I need to supply to the condenser/evaporator to make them function properly by allowing them enough heat exchange? Or can they function without any energy input, and the only energy that I would need to take into account when I calculate how much energy should be supplied to make the refrigeration unit function is the work done by the compressor?

I don't mean to come across as rude - I just wanted to make my question clear.
 
  • #8
256bits
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The condensor and evaporator do not use electricity to operate. They exchange heat with the surroundings.

If you are asking how can you rate what size of fan and motor you need to blow air across the coils or fins then then please verify?
 
  • #9
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I know that, but I think the point of my question was missed again - what I meant was, how much electricity do I need to supply to the condenser/evaporator to make them function properly by allowing them enough heat exchange? Or can they function without any energy input, and the only energy that I would need to take into account when I calculate how much energy should be supplied to make the refrigeration unit function is the work done by the compressor?

I don't mean to come across as rude - I just wanted to make my question clear.
I think you are referring to fan power consumption. For example, fans blowing air over finned tubes (in a condenser) to dissipate a quantity of thermal energy?
 
  • #10
russ_watters
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If that's the case, then the question is very difficult to answer, particularly for the evaporator fan, which depends more on the ductwork system and accessories than on the evaporator itself.
 
  • #11
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If that's the case, then the question is very difficult to answer, particularly for the evaporator fan, which depends more on the ductwork system and accessories than on the evaporator itself.
Is the power consumed by the fan not just a function of rotational speed & fan diameter? For any given fan at constant speed, there will be "x" amount of power consumed. This only changes if the fan speed does. The operating point of the fan will depend on the ducting system though.
 
  • #12
russ_watters
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Is the power consumed by the fan not just a function of rotational speed & fan diameter?
There are probably a good half dozen different basic parameters that affect power consumption, and fan geometry is much, much more complicated than just diameter. Holding all else constant, fan power is a cube function of RPM though.
For any given fan at constant speed, there will be "x" amount of power consumed. This only changes if the fan speed does. The operating point of the fan will depend on the ducting system though.
Basically correct.
 

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