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This is a screenshot of a lab where we were experimenting Vapor-Compression Refrig. Systems.

I am trying to understand what is happening between points 1-2 because I thought that because of max isothermal efficiency the gradient of the line shouldn't be more than the constant entropy line between the two points.

Also I am not sure how to explain the dip between points 5-1.

Refrigerant: R134a Schematics System schematics

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  • $\begingroup$ You are on the right track. This is why we normally condense first and pump a liquid, rather than compress a gas prior to condensation. It's more energy efficient, but not always practical. The compressor is reducing entropy. See here $\endgroup$ – Phil Sweet May 28 '19 at 0:51
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I am trying to understand what is happening between points 1-2 because I thought that because of max isothermal efficiency the gradient of the line shouldn't be more than the constant entropy line between the two points.

Short Answer

Some cooling took place in the compression step between points $1$ and $2$ that isn't shown because lines are easier for a program to draw.

Longer Answer

I think the graph as shown suffers from the fact that the cycle is drawn with straight lines from point-to-point instead of curves that would reflect rational design.

That's not to say that the curve shown is impossible. Any movement on a P-H diagram is possible so long as your machine has a cold sink to dump its waste heat and the machine has energy to perform work. However, in refrigeration examples, practicality means you are limited in your movement on a P-H diagram such that you shouldn't cross to lower temperature isotherms below that of your cold sink's temperature unless you are in an evaporation step. Otherwise you're wasting energy in some way or another. For example, the isotherm barrier you can't cross until the evaporation step is the outlet temperature of your condenser. In the drawing that temperature appears to be $36^{\circ}\text{C}$-ish.

I think you're right to be suspicious of the $1\rightarrow2$ step because if you interpret its path literally, its initial course shows a decrease in entropy (it's P-H path is steeper than the isentropic line) while near $20^{\circ}\text{C}$. In other words, it is initially decreasing in entropy while below the isotherm of the cold sink, $36^{\circ}\text{C}$. In order to decrease in entropy during compression, heat transfer from gas to the environment must simultaneously occur for each incremental increase in pressure during the compression (as mentioned by Phil Sweet). The initial path of the $1\rightarrow2$ line only makes sense if there is a $20^{\circ}\text{C}$ cold sink available to the refrigeration machine which clearly isn't the case. If this were true, then the $2\rightarrow3$ line would ride the $20^{\circ}\text{C}$ isotherm, instead of the $36^{\circ}\text{C}$ isotherm.

That said, it is possible to shift the compression line to the left while never falling in temperature below the $36^{\circ}\text{C}$ isotherm. For example, the ideal refrigeration cycle, a portion of a Carnot cycle, starting from point $1$ would follow an isentropic line (somewhere between the $1.8$ and $1.9$ blue isentropic lines) up until it intersects the cold sink isotherm (isentropic compression), $36^{\circ}\text{C}$. Then, it would ride this isotherm (somewhere between the red $20^{\circ}\text{C}$ and $40^{\circ}\text{C}$ isotherm lines ) up to the black dew point line (note: isothermal compression = compression + cooling) at a point to the left of point $2$. I drew in magenta what this ideal portion of a carnot cycle would look like:

annotation depicting idealized compression step in refrigeration cycle

Real world compressor equipment probably would not be able to follow this ideal curve since it is cheaper to simply have one single-stage compressor instead of many compression stages interlaced with many heat exchangers. Also, riding close to the dew point line is dangerous since formation of incompressible fluids damage most compressor types.

My guess for why the $1\rightarrow2$ line is so wrong is because the software generating point $2$ assumed that some cooling took place during or after the compression step and simply connected point $1$ with point $2$ with a straight line because it's simpler to program drawing a straight line.

Also I am not sure how to explain the dip between points 5-1.

The drop in pressure for the $5\rightarrow1$ step is probably due to friction in the piping between the evaporator and the compressor suction inlet. The rise in enthalpy is probably due to imperfect insulation of said piping (the environment warming the cold refrigerant vapor).

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Are you sure that the diagram is correct because I believe that point 5 should be at the saturation point on the bell shape curve, and the dip should start after that.

You see 4 to 1 is meant to be constant pressure process in evaporator(that absorbs heat and cause cooling) and after the saturation point constant pressure lines have a negative slope, thus causing a dip in the curve.

A standard process that is generally demonstrated in the books has point 5(and point 1 overlapping) at the saturation point and thus no dip is shown in such a case.

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  • $\begingroup$ I took a screenshot of the cycle straight from the output of the software that is used to control the experiment. I had assumed it would be drawing the correct cycle. After calculating the efficiency, it comes out as around 400%. Do you have any suggestions as to what might be causing this? $\endgroup$ – ZiZ Dec 29 '18 at 20:30
  • $\begingroup$ Efficiency is not calculated for refrigeration cycle, instead we calculate coefficient of performance for refrigeration, which is always greater than 1. $\endgroup$ – anshul suri Jan 1 '19 at 21:24

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