Design a Refrigeration System

Design and analyze a refrigeration system. In this article, we will look at how to design and analyze a refrigeration system. We'll cover the theory behind an ideal vapor compression cycle, so your performance would be slightly different from the real world scenario.

Scroll down for the YouTube Video Tutorial on Cooling Design and Analysis.

Here we have our basic refrigeration cycle. The main components are the compressor, the condenser, the expander and also the evaporator.

The compressor compresses the refrigerant and pushes it through the system. The condenser rejects unwanted heat from the system. The expansion valve expands the refrigerant. The evaporator absorbs the unwanted heat that has collected in the building, it also produces cooling, which goes into the building.

To design and analyze a refrigeration system, we want to know what the thermodynamic properties of the refrigerant will be in our four key components.

Point 1: between the evaporator and the compressor.

Point 2: at the outlet of the compressor.

Point 3: at the outlet of the condenser, before entering the expansion valve.

Point 4: just after the expansion valve, before it enters the evaporator.

For each of these points we need to know the temperature, entropy, pressure, and enthalpy of the refrigerants.

We will plot our system on two graphs to do this. Notice the gray line that forms a dome over both. Anything to the left of this is where the coolant is a liquid. At any point below the dome, the refrigerant is part vapor, part liquid. Anything on the right hand side of the line is where the superheated steam will be. If the point touches the line on the left, it is a saturated liquid. If it touches the right line, it is saturated steam.

The blue lines mark the performance of our system. From the graphs above, we can see that point 1 will be low temperature, it will also be low pressure, and it will be saturated steam. Point 2 that we can see will be much higher pressure. It's also a high temperature and it's in the superheated region, so it's a superheated vapor. Point 3 that we can see is still high pressure, its temperature has dropped a bit, and it is also on the saturated liquid line, so it will be a saturated liquid.

We can also see that at point 4, right after the expansion valve, when the refrigerant has expanded, the refrigerant will be at a much lower pressure and temperature. It's low pressure, low temperature and it's between the dome. It will therefore be a liquid-vapor mixture.

We will use some acronyms:

T is going to be for temperature.

P is going to be for pressure.

H is going to be for enthalpy.

S is going to be for entropy.

X will be the quality of the coolant. If it is a liquid it is a 0, if it is a vapor it is a 1. We use it to determine the distance between the liquid and the vapor of the refrigerant when it is in the dome region.

Point M is the mass flow of refrigerant

Now there are many ways to start designing one of these systems. Knowing the cooling load you want to achieve would probably be a very good starting point.

Coolant at point 1

We'll start with the compressor. Let's say I have a compressor capable of pushing seven kilograms of refrigerant through the system per second.

From the data of the compressor manufacturers, we can see that it is capable of producing 1200KPA pressure and also needs 320KPA suction pressure. So we start using these values.

We will enter this data in a table.

Therefore, we know that the suction pressure is 320 KPA and it is a saturated vapor at the saturation line (Point 1). We just need to find out what their values will be in the thermodynamic properties of the refrigerant. For this example, we will simply use R134a.

You can get the properties of R-134a refrigerant by clicking here Then scroll down to the saturated vapor refrigerant tables. So we want to find 320 KPA.

Take note of the values highlighted above. These under saturated steam conditions when the refrigerant is at 320 kPa for temperature, volume, enthalpy, and entropy. These are all the values for point 1.

Coolant at point 2

Now we will look for the properties of the refrigerant in state 2.

Now back to the refrigerant tables, this time we need to know two reference points to find the values. At the moment we only know what the pressure is (1200 kPa) and we know that it is superheated steam. Since we are doing this as an ideal cycle, we can assume that the compressor is isotropic and that means that the entropy at point 2 is equal to that at point 1. (0.9301 kJ/kg.K)

So we copy the entropy value from point 1 to point 2. Now we have two data points so we can look in the tables of superheated steam to find the enthalpy and temperature.

Find the superheated refrigerant tables. We are looking for 1200 kPa.

Now the entropy we are looking for is 0.9301 kJ/kg.K. But looking at this list, we can't find it, it's between two values. So to find it, just use linear interpolation.

Watch the YouTube tutorial to see how to calculate this in Excel.

Coolant at point 3

Since this is an ideal system, there is no resistance to flow, so there will be no pressure drop. In the real world, there will obviously be a small pressure drop, but since this is ideal, we can assume that the pressure at point 3 is the same as at point 2.

Now we have the pressure and we know that it is a saturated liquid and that means we can use the tables for saturated liquids to find out the temperature, enthalpy and entropy.

Going back to the property tables, locate the tables for saturated refrigerant and find 1200 kPa. We know that it is a saturated liquid. So we want the columns of saturated liquid. So copy the values for temperature, enthalpy, and entropy.

These are all our properties for point 3.

Coolant at point 4

Now we need to know what the properties of the refrigerant at point 4 will be. This is a bit trickier because it is in the vapor dome. So it's part liquid, part vapor. We still don't know exactly how much vapor or liquid it is. But it's okay because we can get ahead.

We know that the temperature will be the same as at point 1 and the enthalpy will remain constant through the expander. (Look at the graphs to see this.) That means we can use the enthalpy of state 3. So we can plug in those numbers for point 4.

We also know what the pressure is because it will be equal to state 1 and that means we just have to figure out what the entropy is. The way we do this is by finding the quality of the coolant. So what part of liquid or vapor do we have?

For this we return to the tables of saturated refrigerants. We move down until we find the 320 kPa line. Next, we want to copy the entropy, enthalpy, temperature, and pressure of the saturated liquid and vapors. Paste them in Excel, it will be easier.

We find the quality using the above formula. I have already filled in the values. where X is the quality and H is the enthalpy. We already have the value of H4, because it was equal to the enthalpy at point 3. We also know what HF and HG are because we have them here in the table. HF is the saturated liquid and HG is the saturated vapor value.

When we plug the numbers into the formula, you'll see that it's in the form of a decimal number, because it's a ratio, which is about 32%.

We now use this value to determine what the entropy will be in state 4, and we do so using the formula below.

Therefore, S4 equals 0.4436 kilojoules per kilogram per Kelvin.

Therefore, all the properties of the refrigerant have been determined.

Work done by the compressor.

Let us determine the amount of work done by the compressor. We can do that using this formula below.

For this, the enthalpy of state 2 and state 1 and the mass flow rate of the refrigerant are used. From the formula we can see that the compressor is doing 82.29 kilowatts of work on the system.

Evaporator Cooling Load

We can also calculate the cooling load on the evaporator using the formula below.

It's almost exactly the same formula. We just use the enthalpy of point 1 and the enthalpy of point 3. Using this, we can see that it provides 402 kilowatts of cooling.

condenser heat rejection

We can also calculate the heat rejection of the condenser and we can see it in the formula below.

Again, a very simple and similar formula. We can see that the capacitor rejects 485 kilowatts.

Note that this is higher than the cooling provided and this is because we also need to get rid of the heat produced by the compressor. So you can add them to check if your fingers are correct.

If they are not equal, then you did something wrong and you need to go back and take a look at some of your numbers.

coefficient of performance

We can calculate the efficiency of the system, or the coefficient of performance, using this formula below.

This formula is very simple, it is just a work by work relationship. This example, our system has a coefficient of performance of 4.89. This means that for every kilowatt of electricity you put in, you'll get 4.89 kilowatts of cooling, which is very efficient.

Here are all our properties for the system.

If you want to calculate the temperature of the air leaving the evaporator, you can follow this tutorial here.

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