Thermodynamics: Phase-change refrigeration

Click For Summary
SUMMARY

The discussion focuses on thermodynamics in phase-change refrigeration, specifically addressing entropy, equilibrium, and energy loss in cooling systems. The reversed Otto cycle is identified as the basis for the pV-diagram, highlighting that this cycle experiences an increase in entropy due to irreversible processes. It clarifies that steady state and equilibrium are not synonymous, as steady state can exist without thermal equilibrium. Additionally, energy loss in cooling systems is attributed to friction in compressors and valves, flow turbulence, and inefficiencies in energy conversion, with motor efficiency averaging around 80%.

PREREQUISITES
  • Understanding of the reversed Otto cycle in thermodynamics
  • Knowledge of entropy and its implications in refrigeration cycles
  • Familiarity with the concepts of equilibrium and steady state in thermodynamic systems
  • Basic principles of energy conversion efficiency in refrigeration systems
NEXT STEPS
  • Research the implications of irreversible processes in thermodynamic cycles
  • Study the principles of entropy in relation to refrigeration and heat engines
  • Explore the differences between thermal equilibrium and steady state in fluid dynamics
  • Investigate methods to improve energy efficiency in refrigeration systems
USEFUL FOR

Engineers, thermodynamics students, and professionals involved in HVAC systems, refrigeration technology, and energy efficiency optimization will benefit from this discussion.

Bonulo
Messages
46
Reaction score
0
I'm working on a project about cooling and heating, and have a couple of questions needing answers/clarification:


Entropy in refrigeration:
The pV-diagram for the phase change refrigerator is that of the reversed Otto cycle. This cycle contains two adiabatic and two isochoric processes. The last two being irreversible. Thus the cycle has an entropic increase - since only a Carnot cycle conserves entropy, right? But when is the maximum value of entropy for the refrigeration reached - if it is reached at all?

Equilibrium and steady state in a cooling system:
Is equilibrium and steady state condition in a cooling system the same? I have understood steady state as the condition where the thermodynamic cycle is in equilibrium (e.g. when a refrigerator has been turned on for a small amount of time).


Energy loss in cooling systems:
How does energy loss occur in cooling systems? I think friction in the compressor (and maybe the extension valve) accounts for most of the lost energy from the power input, when work is done on the surroundings and not the working substance (e.g. freon). But flow turbulence of the substance plays a part too. But are there other energy loss sources of importance?


:smile:
 
Physics news on Phys.org
Bonulo said:
I'm working on a project about cooling and heating, and have a couple of questions needing answers/clarification:Entropy in refrigeration:
The pV-diagram for the phase change refrigerator is that of the reversed Otto cycle. This cycle contains two adiabatic and two isochoric processes. The last two being irreversible. Thus the cycle has an entropic increase - since only a Carnot cycle conserves entropy, right? But when is the maximum value of entropy for the refrigeration reached - if it is reached at all?

A forward (heat engine) or backward (refrigerator) Carnot cycle does not conserve entropy of the surroundings. \Delta S = 0 for a Carnot cycle only if you run the forward cycle followed by the backward cycle (and if you assume that you use the output work of the forward cycle as the input work of the reverse cycle, the entropy of the universe does not change).

A refrigerator cycle will result in a decrease in entropy of the surroundings.

Consider the entropy of the surroundings. (Since entropy is a state function, there is no change in entropy of the gas during one complete cycle because the gas returns to exactly the same state at the beginning of every cycle).

\Delta S = \int_1^2 dS + \int_2^3 dS + \int_3^4 dS + \int_4^1 dS

ds = dQ/T

dQ > 0 if the flow of heat is into the reservoir (the hot reservoir) and dQ<0 if heat flow is out of the reservoir (cold).

Since the net heat flow is from a colder to a hotter reservoir, the entropy change around a full cycle is more negative than positive. So maximum entropy for the surroundings is at the beginning of the cycle.

AM
 
Bonulo said:
Equilibrium and steady state in a cooling system:
Is equilibrium and steady state condition in a cooling system the same? I have understood steady state as the condition where the thermodynamic cycle is in equilibrium (e.g. when a refrigerator has been turned on for a small amount of time).
A steady state condition can, and often does, exist without thermal equilibrium. Any time you have gas moving with kinetic energy it is not in thermal equilibrium. Any time you have heat flowing, there is no thermal equilbrium. But a cooling system that has the same volumes and rates of gas being compressed/expanded per unit time and produces the same rate of cooling, it is steady state, but heat is flowing all the time and gas is moving all the time, so none of the gas is in thermal equilibrium.

Energy loss in cooling systems:
How does energy loss occur in cooling systems? I think friction in the compressor (and maybe the extension valve) accounts for most of the lost energy from the power input, when work is done on the surroundings and not the working substance (e.g. freon). But flow turbulence of the substance plays a part too. But are there other energy loss sources of importance?
There are different ways in which energy is lost in a refrigerator cycle. There is energy being expelled as heat, which is energy loss. But that is exactly what you are trying to achieve, so it is not a problem. There is also energy lost in converting input energy to useful thermodynamic work. I think the question may be asking how thermodynamic work compares to input energy.

In a refrigerator input energy consists of electricity used to run a compressor. The efficiency of the motor is about 80% in converting electrical energy into thermodynamic work (compressing gas). There are also friction losses in circulating coolant. There is also friction in the compressor pump which creates further losses. There is also the waste of electrical energy in running the fridge light!

AM
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 13 ·
Replies
13
Views
4K