# Exploration of a refrigeration installation in the entropy chart

# Introduction

The objective of this exploration is to make you discover the cycle of a refrigeration installation in the (T, s) thermodynamic entropy chart.

It completes guided exploration (S-M3-V9), where the cycle was presented, with explanations on its settings and its representation in the (h, ln (P)) chart.

# Loading the model

We will now study the ejector refrigeration cycle.

## Load the model

Click on the following link: Open a file in ThermoptimOpen a file in Thermoptim

You can also:

- either open the "Project files/Example catalog" (CtrlE) and select model m6.3 in Chapter 6 model list.
- or directly open the diagram file (refrig_light.dia) using the "File/Open" menu in the diagram editor menu, and the project file (refrig_light.prj) using the "Project files/Load a project" menu in the simulator.

# Cycle plot in the (T,s) entropy chart

## First step: loading the water (T,s) entropy chart

Click this button

You can also open the diagram using the "Interactive Diagrams" line in the "Special" menu of the simulator screen, which opens an interface that links the simulator and the diagram. Double-click in the field at the top left of this interface to choose the type of diagram desired (here "Vapors").

Once the diagram is open, choose "R134a" from the substance menu, and select "(T,s)" from the "Chart" menu.

## Second step: loading a pre-recorded cycle corresponding to the loaded project, the layout of which has been previously refined in order to be more precise

Click this button

You can also open this cycle as follows: in the diagram window, choose "Load a cycle" from the Cycle menu, and select "refrig_lightEnThin.txt" from the list of available cycles. Then click on the "Connected points" line in the Cycle menu.

## Cycle analysis

Point 1 slightly superheated compared to saturated steam is placed on the 2 bar isobar.

Irreversible compression results in an increase in entropy. Cooling with outside air has two stages: desuperheating (2 - 3a) in the steam zone and condensation along the horizontal line segment (3a - 3).

Isenthalpic throttling (3 - 4) leads to an increase in entropy, point 4 being located in the liquid-vapor equilibrium zone.

The useful effect, the production of cold in the evaporator, corresponds to the horizontal which passes through point 4, except for the slight overheating which leads to point 1.

## Identification of some characteristic points in the chart

What is the point in the liquid zone?

What is the point in two-phase zone?

What is the point with the highest temperature?

# Comparison with the reverse Carnot cycle

η = T2/(T1 - T2)

The reverse Carnot cycle is the one that leads to the best COP. It is given by this formula, T1 and T2 being the temperatures of the hot and cold sources (T1 > T2), expressed in K and not in °C.

The comparison with the reverse Carnot cycle will be done in four stages:

1. display of temperatures of external sources in the entropy chart
2. display of a first reverse Carnot cycle
3. display of a second reverse Carnot cycle
4. comparison of the project cycle and the reverse Carnot cycle

## Première étape

## First step: the value of the temperature of the cold compartment (-5 °C) is displayed on the chart

Click this button

You can also open this cycle as follows:

- in the diagram window, choose "Load a cycle" from the Cycle menu
- and select "refrigTColdsource.txt" from the list of available cycles.

Note that the refrigerant evaporates at a temperature lower than that of the cold compartment

What is the temperature difference (R134a - cold compartment)?

### The value of the outside air temperature (18 ° C) is displayed on the chart

Click this button

You can also open this cycle as follows:

- in the diagram window, choose "Load a cycle" from the Cycle menu
- and select "refrigTairExt.txt" from the list of available cycles.

Note that the refrigerant condenses at a higher temperature than the outside air.

What is the average temperature difference (R134a - outside air) during the condensation stage?

## Second step

### Loading of the reverse Carnot cycle relating to the refrigerant, that is to say corresponding to a cycle evolving between the extreme temperatures of the refrigerant. Please note, point D has no physical reality

Click this button

You can also open this cycle as follows:

- in the diagram window, choose "Load a cycle" from the Cycle menu
- and select "refrigCarnot.txt" from the list of available cycles.

### Calculation of the reverse Carnot COP

To calculate the reverse Carnot COP, determine the values ​​of the hot and cold temperatures in Kelvin, and apply the formula above.

Enter the temperature value of point 4 in Kelvin

Enter the temperature value of point 2 in Kelvin

Calculate the COP of this reverse Carnot cycle and enter it

## Third step

### Loading of the reverse Carnot cycle relative to external sources, that is to say corresponding to a cycle evolving between the temperatures of the two external sources.

Loading a second reverse Carnot cycle

Calculate the reverse Carnot COP of the cycle delimited by points a, b, c and d.

By comparing the value found for this reverse Carnot cycle with the previous one, you can judge the impact of irreversibilities due to temperature differences.

Click this button

You can also open this cycle as follows:

- in the diagram window, choose "Load a cycle" from the Cycle menu
- and select "refrigCarnotSources.txt" from the list of available cycles.

Enter the temperature value of point d in Kelvin

Enter the temperature value of point c in Kelvin

Calculate the COP of this reverse Carnot cycle and enter it

### Clearing the second cycle of reverse Carnot

We have displayed the two reverse Carnot cycles which could be considered a priori for the refrigerator, but only that relating to the refrigerant can serve as a reference in the present case because of the need to take into account a temperature difference between the refrigerant and external sources. For the rest of the scenario, we erase the reverse Carnot cycle relating to external sources

Click this button

You can also delete the different cycles that have become unnecessary by following these steps: Display the Cycle manager from the Cycle menu. Click on the "Update the cycle table" button, and deselect the rows that you do not want to appear on the chart.

## Fourth step: comparison of the project cycle studied and the reverse Carnot cycle

You will now be able to compare the project cycle studied with the reverse Carnot cycle relating to the refrigerant.

Let DeltaThot be the difference between the high temperature of the reverse Carnot cycle and that of the outside air.

Let DeltaTcold be the difference between the low temperature of the reverse Carnot cycle and that of the cold compartment.

### Analysis of temperature differences

temperature differences between the refrigerator and external sources

Technologically speaking, the heat exchanger having a finite dimension, the working fluid cannot be at the same temperature as the external sources, which constitutes a first difference with the reverse Carnot cycle.

DeltaThot is almost 3 times higher than DeltaTcold DeltaThot is almost 2 times higher than DeltaTcold DeltaThot is close to DeltaTcold DeltaThot is almost 2 times lower than DeltaTcold DeltaThot is almost 3 times lower than DeltaTcold

### Taking into account temperature differences due to desuperheating

Due to the need to desuperheat the refrigerant before condensing it, the cycle has a "horn" between points 2 and 2a which explains why DeltaThot is superior to DeltaTcold

What is the value of the difference T2 - T3a ?

What is the value of the difference TD - Outside air

What is the value of the difference between the cold compartment and Tc?

### Taking into account compression irreversibilities

Compression can be assumed to be adiabatic, but not isentropic, given the irreversibilities that take place in the compressor, which induces a new deviation with the reverse Carnot cycle.

To answer the question, open the compressor screen, enter an isentropic efficiency equal to 1, and recalculate the process

What would be the outlet temperature of isentropic compression?

### Taking into account expansion irreversibilities

In theory, it would be possible to achieve an expansion close to the isentropic, but the technological reality is different, and this for three reasons:

1. expansion of a two-phase mixture is done, unless special precautions are taken, with poor isentropic efficiency
2. the work involved remains very small
3. there is no suitable and inexpensive expansion machine, especially for small capacity refrigeration machines.

What is the increase in entropy due to expansion?

# Conclusion

This exploration allowed you to familiarize yourself with the representation of the cycle of a refrigeration installation in the entropy chart and to compare it with the reverse Carnot cycle.