Subcooling Finale

In the last article I covered why flash gas is generated leaving the metering device when the pressure drops from the value in the liquid line down to the evaporator pressure. In this example I assumed there was no subcooling present so as to make a baseline example to compare with. In other words, only enough heat was removed to ensure that 100% of the refrigerant was in liquid form. The example was R404a at 280 psig which has a saturation temperature of~ 112 F. It is still saturated but it is right at the 100% liquid line for a pressure of 295 psia. We will use psia as this is the pressure scale the pressure-enthalpy diagram uses. The energy of the refrigerant is ~52 btu/lb leaving the condenser and entering the metering device.

Now here is where it gets interesting. Due to all of the factors we looked at in the last article, a hard truth is that saturated liquid at a lower pressure cannot hold as much energy and stay a liquid as can liquid at a higher pressure. In fact, saturated liquid at -10F and a pressure of 38 psia only contains ~9 btu/lb. Remember that this is the only amount it can hold if it is 100% liquid at saturation. If you add any more heat, you no longer have 100% liquid and this is exactly why when we drive 100% liquid with 52 btu/lb through a metering device and it exits at 38 psia and -10F, we end up with more than 50% flash gas! The simple reason is that there is excess energy in the warm high pressure liquid leaving the condenser compared to the low pressure saturated condition we want in the evaporator. So where does this extra 52-9 = 43 btu/lb go? If you said into making flash gas then you are correct.

To prove this, we need to look at the difference in energy per pound between saturated 100% liquid and saturate 100% vapor at 38 psia. The liquid holds 9 btu/lb while the vapor holds 89 btu/lb. The difference is 89-9 = 80 btu/lb. To calculate how much flash gas we will need to produce to accommodate the extra energy of the liquid leaving the condenser, simply divide the difference in energy of the high and low pressure liquids (43 btu/lb) and the difference in energy between the saturated liquid and vapor at 38 psia (80 btu/lb). Doing the math we end up with (43/80) x 100. We multiply by 100 to put the answer in a percentage format. The answer is ~ 54%. You can verify this by using a pressure enthalpy chart and plotting the point leaving the condenser and dropping straight down till you intercept the evaporator pressure and temperature. The curved lines called refrigerant quality should verify this answer as you should be almost exactly between the 50 and 60 percent lines (Or the decimal equivalent.).

Now we can probably all agree that entering something designed to turn liquid into vapor and absorb heat in the process probably works better when there is more liquid to turn into vapor compared to when there is less. Once you run out of liquid, the show is pretty much over. So entering the evaporator with less than half a tank is not something we would normally choose to do unless energy was free, which it is most definitely not.

The 2nd law of thermodynamics is pretty clear (And absolute.*) that there is no free lunch, so if you want to absorb more energy in the evaporator, you basically have 2 choices, either pump more refrigerant through it or reduce the energy the refrigerant has when it enters. The first choice will increase capacity but not improve efficiency very much if at all. The second will improve efficiency and capacity but will not happen unless we remove more energy from the refrigerant or reduce the compression ratio. Luckily, just about all condensers remove more heat than just enough to produce 100% saturated liquid. AC condensers produce liquid that has had enough energy removed to subcool the liquid anywhere from 6-12 degrees, with the majority somewhere in between.

You may be thinking, great, now I know why they make the condenser coils on high efficiency units so huge. It is to remove more heat and produce more subcooling. Well, you would at least be partially correct in that a larger condenser surface will remove more heat per degree of temperature difference between the refrigerant and the air. However, these high efficiency units actually produce less subcooling for a given set of indoor and outdoor conditions than a lower efficiency system. The reason being is that for a unit with an adaptable metering device such as a TXV, the amount of subcooling is determined by the temperature difference between the refrigerant and the air. A larger condenser requires less of a temperature difference so will produce less subcooling compared to a smaller (Or dirty) condenser. A larger condenser saves energy by allowing the pressure and temperature of the refrigerant to be closer to the evaporator pressure and temperature and this reduces the compression ratio. It also reduces the amount of flash gas for a given condenser leaving condition due to the smaller difference between the maximum liquid energies of the high and low pressure sides. This also tells you why it is such a good idea to let the high side pressure drop when the outside temperature is cooler, as long as there is enough pressure difference to feed the metering devices.

So given this situation, how can we remove more energy from the refrigerant if we cannot do it in the condenser? The answer is to do it outside of the condenser. In fact it pretty much has to occur outside of the condenser because it has to occur with 100% liquid only going through a sensible temperature drop. What you need is a separate heat exchanger and a source that can absorb heat from it. There are many ways to achieve this and I have seen everything from a suction - liquid heat exchanger to reheat coils for warming up the air after it has been over cooled to remove moisture. I often think that those supermarkets you go into that are so damn cold it is hard to tell the difference between the inside of the cooler and the rest of the store would be a great place to try this. In fact, this is a great example to use so let's look a little deeper.

In the summer in areas with humidity outdoors, lowering humidity inside the store is absolutely critical in supermarkets. This is done to avoid water pooling on the floor around coolers and freezers and other nasty issues resulting from excess moisture in the air. The downside is that normal AC units often need to cool the air down considerably more to tackle humidity when it is high outside and this can result in space temperatures in the mid 60's in some places. The solution to over cooling the whole space is to rewarm the air once it leaves the evaporator. You can over cool the air to dry it, either all of it or more economically (And more common these days), only a part of it and then heat it back up so that it is slightly warmer but a lot drier. However, burning fuel to reheat the air you just cooled is an expensive task, not to say counter intuitive. But what if there was a way to reheat the air and actually improve the efficiency of the refrigeration or AC system at the same time? If you have been reading up to this point than you probably know where this is going. Why not run the liquid line of the parallel rack (Or the AC unit's own liquid line for that matter.) into a heat exchanger downstream of the evaporator and increase subcooling while also warming up the supply air so it has a lower RH value and is a lot more comfortable sensibly? Air leaving a dehumidification coil at 52 F and 90% RH, heated up to 60 F through the reheat/subcooling coil will condition the space by keeping humidity in check and provide a moderate amount of sensible cooling which is primarily the main goal for these systems, especially if it is more humid than hot outside.

If you go back to near the beginning of the previous article on subcooling you will see the part about how the condenser empties the bucket of heat so that you can add it back in the evaporator. And since as far as the evaporator is concerned, the condenser can not remove as much heat compared to what is possible to add in the evaporator and as a result, flash gas is produced. But when we subcool the liquid we are removing additional heat from the liquid that can then be used as additional energy space to add heat to in the evaporator. The fact is that the more you subcool the liquid, the more heat you can absorb in the evaporator with the exact same amount of refrigerant being pumped by the compressor. Let's do some examples to make the point and finish off.

In our original example, liquid leaving the condenser with no subcooling contains 52 btu/lb. The specific heat of liquid R404a at 294 psia is ~.43 btu/lb/F, so for every degree we cool the liquid below it's saturation temperature (Subcool it), we remove an additional .43 btu's per pound. If we take a typical subcooling value of 7 degrees and use our example, we will remove an additional 7 X .43 =3 btu/lb so the liquid leaving our example condenser with 7 degrees of subcooling will be at 112-7 = 105F and will contain 52-3 = 49 btu/lb. Using this to calculate surplus energy, we have 49 - 9 = 40 surplus btu/lb between liquid at 105F at 285 psia and -10F liquid at 38 psia. The amount of flash gas we will produce leaving the metering device is (40/80) x 100 = 50%. So by adding 7 degrees subcooling we have dropped our flash gas generated from 54% to 50% and allowed each pound of refrigerant to absorb 3 additional btu/lb.

Now, let's add a subcooler that increases subcooling by another 20F like in our supermarket example so we then have a total of 27 degrees of subcooling. Our liquid temperature will be 112 - 27 = 85F and our energy per pound will be 52-(27 x .43) = 40.5 btu/lb. To calculate the energy difference between our subcooled liquid and the energy of -10F saturated liquid @ 38 psia, we have 40.5 - 9 = 31.5 btu/lb. Flash gas percentage generated = (31.5/80) x 100 = 40%. As well, the amount of heat the refrigerant can absorb in the evaporator is 88.5 - 40.5 = 48 btu/lb. Compare this to our refrigerant that had no subcooling and it could only absorb 88.5 - 52 btu/lb = 36.5 btu/lb (These values do not include super heating of the vapor).

Now to finish, let's look at how much refrigerant the compressor has to pump each hour for each ton of cooling produced in the evaporator. If one ton = 12 000 btu/hr, with no subcooling we would need to pump 12000/36.5 = 329 lbs/hr. With 27 degrees of subcooling, we only need to pump 12000/48 = 250 lbs/hr. That is a pumping reduction of (329-250)/329 x 100 = 24 percent. Not too bad for keeping us all from freezing to death in the supermarket.

* Arthur Eddington was a brilliant guy who lived in the late 19th to mid 20th centuries. He sums up the fact that the 2nd law of thermodynamics is absolute (It has withstood every test put at it, from making whiskey to the destruction of information in evaporating black holes.) when he stated: “The law that entropy always increases is the supreme position among the laws of Nature. If someone points out to you that your theory is in disagreement with Maxwell's equations - then so much the worse for Maxwell's equations. If it is found to be contradicted by observation - well, these experimentalists do bungle things sometimes. But if your theory is found to be against the Second Law of Thermodynamics then I can give you no hope; there is nothing for it to do but collapse in deepest humiliation.” If you want to read more about this interesting polymath, here is a link. https://en.wikipedia.org/wiki/Arthur_Eddington