# Homework Help: Thermodynamics: Phase-change refrigeration

1. Apr 27, 2006

### Bonulo

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?

2. Apr 27, 2006

### Andrew Mason

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

3. Apr 27, 2006

### Andrew Mason

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.

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