Wow, its been a long time since I have written a blog post!
Times fun when you’re having flies …
as frogs are often heard to say.
I have been putting a lot of new content up on the Commissioning Resources website, so that has taken my time. But fairly recently, I had a discussion with a friend who was having a hard time wrapping their head around the coefficient of performance for of a heat pump/refrigeration process and I came up with an analogy that – while not perfect – worked for them and which they found somewhat amusing.
So I decided I would try to resurrect my blog posting activities by sharing it for what’s worth.
The fundamental question was …
It seems like magic that you can get a COP = 4. I’m having a hard time wrapping my head around the fact that you can get 4 units of energy OUT for putting in 1 unit of energy.
The Somewhat Technical Answer
I started out by saying that I thought maybe the key was to think about the compressor as doing work to move energy rather than creating the cooling effect.
In other words, a refrigerant at a saturation temperature/pressure of “X”°F/Y psia will produce “Z” Btu’s of cooling via the phase change that occurs if heat is applied to the evaporator, causing the liquid refrigerant to boil and become a vapor. It’s the energy absorbed by the phase change process produces the cooling.
The amount of energy absorbed per pound of refrigerant as well as the saturation pressure associated with the temperature that the phase change occurs at will be a function of the physical properties of the refrigerant.
In other words, you may need to move “U” pounds of refrigerant A at a saturation temperature/pressure of “X”°F/Y psia, but move “V” pounds of refrigerant B to produce the same refrigeration effect at a saturation temperature/pressure of “X”°F/Y psia.
Once the refrigerant has gone through the phase change, the problem becomes getting rid of the heat by condensing the refrigerant. One way to do that is to move it to a higher saturation temperature and pressure so that you can use some other medium that is cool relative to this new, elevated pressure and temperature to reverse the process and condense the refrigerant.
The compressor accomplishes this for us by compressing the cool vapor from the evaporator. In doing this, it does work on the refrigerant (the pv/J part of the steady flow energy equation) …
… and the amount of work it does can be determined by plotting the cycle on a pressure enthalpy diagram[i].
The work includes the irreversibility losses, i.e. there is a change in entropy. All of this will be specific to the refrigerant that is use, as will the evaporator saturation temperature and pressure relationships.
In addition, you will put more energy into the compressor motor than you get out as shaft power to the compressor because of the losses in the motor. If the motor is cooled by the refrigeration process, then these losses will also show up as heat to be rejected at the condenser.
At the evaporator coil and condenser coil, the energy transfer is 100% efficient; i.e. 100% of the energy removed from the fluid flowing through the evaporator shows up as vaporized refrigerant and 100% of the energy removed from the condenser by the air or water flowing through it shows up as an increase in air or water temperature. But the amount of energy rejected is more than the cooling effect because the compressor energy is also being rejected.
I think my friend kind of knew this all along; he basically alluded to it in what he said when he initiated the discussion. But somehow, my saying it back to him caused the dots to connect. All I really did was mirror back what he already know. That is the power of having a discussion I think.
But at that point, I was on a roll, so I continued with my analogy, which they patiently tolerated. (You, of course, can just stop reading this and I will never know).
Suppose you have a nice little cabin out in the Pacific North West woods next to a very pretty, deep lake that was fed by streams which were fed by melting glaciers.. Most of the time the cabin is quite comfortable, but there is the occasional hot summer day when it would be nice to have some sort of cooling system.
One day, after going snorkeling to see the fish in the lake, you realize that the water towards the bottom of the lake is actually pretty cold, even though the surface water temperature is very pleasant.
That gives you an idea.
You go buy an 800 cfm fan coil unit, install it in the basement of your cabin, and run a pipe from the inlet of the cooling coil out to just below the surface of the lake, then add a vertical extension to it so that when you open the valve to the coil, the head produced by the water level in the lake will cause water to flow through your coil, but the flow will be from the bottom of the lake, where the water is coldest.
You buy a kiddie pool to place under the outlet of the coil to catch the water so it doesn’t flood your cabin. The good news is that you can make 76°F air with this arrangement, which will cool down your cabin, which is at 90°F but very low RH (i.e. the coil is running dry).
The bad news is that the 4.8 gpm it takes to do this adds up and the kiddie pool starts to overflow. So you build a flume and reservoir that allows you to fill a bucket, climb up a ladder 15 feet, and dump the 4.8 gpm into it, which returns it to the lake.
The bottom line is that the system is doing a ton of cooling by changing the temperature of the water going through the coil 14°F.
Natural forces are producing the cooling effect;
- The head created by the difference in the level of the lake and the outlet of your pipe moves the water through the coil to the basement.
- The ability of the water to absorb heat by changing temperature provides the actual cooling effect. Basically, the lake water is your refrigerant; its just doing the cooling with a sensible energy change vs. a phase change.
But to keep your cabin from flooding you need to do some extra work to move the water back to the lake, which involves carrying a bucket of water multiple times from the basement level to the flume level. When you dump the water into the flume, you are above the level of the lake.
This is a bigger elevation change than the difference between the water level in the lake and the water level in the Kiddie pool. But to get the water to flow from your cabin back to the lake, you have to dump it into the flume at the higher elevation.
Bottom line, to keep the system working and keep from flooding your cabin, on average you need to move 4.8 gallons of water through a 15 foot elevation change.
But, of course, the mass of the water is not the only thing you move up the ladder. You also move your own mass and the weight of the bucket. If you do the math with the water horse power equation …
… you discover that the water hp is about 0.018 hp.
But if you convert the gallons of water in the bucket to pounds and add your weight and the bucket weight to it and multiplying it by the 15 foot elevation change, and the number of trips you need to make to keep the basement from flooding, you discover that you are doing 0.087 hp of work or 220 Btus.
If your body was about 25% efficient, you would need to consume a lot of calories to keep this process going[ii].
Since you find the free cooling to be quite desirable on the occasional hot day, but would rather not have to climb the ladder so much, you invent a device that can do that for you using solar cells as a source of power and begin to market your new product.
Changing the Refrigerant
As a result of the success of your invention, you accumulate great wealth and decide to buy a place on the US Virgin Islands so you can spend some of your time there relaxing on the beach, snorkeling, and watching sunsets.
Given the high temperatures and humidity levels, you decide to install your cooling system in one room of your beach house to provide a bit of relief from the heat and humidity, this time using seawater as the refrigerant.
When you commission your system, you discover a number of differences from the system in your cabin.
For one thing, given the humidity in addition to the heat as well as the available water temperature, you realize you probably will need a larger fan coil unit; at one ton, your current model can not dehumidify and only performs sensible cooling. So while it helps, what is really needed is some relief from the humidity in addition to the heat.
But you decide that the sensible cooling is better than nothing, so you continue to commission the system while waiting for your new, larger fan coil unit to arrive. In doing that, you discover that to create the ton of cooling, you need a bit more flow, specifically 4.9 gpm instead of 4.8 gpm.
After investigating and determining that your flow measurement is in fact accurate, you realize that the specific heat of seawater is lower than that of the pure fresh water in the lake by your cabin; 1.00 Btu/lb-°F for the fresh water vs. 0.96 Btu/lb-°F for the seawater.
In other words, it is a different refrigerant and because of its physical properties, you need to move more of it to produce the same refrigeration effect.
You also realize that the reason you seem to float better when snorkeling in the Caribbean is that the density of the saltwater is higher than that of the fresh water in your lake back at the cabin; 62.29 lb/cu.ft. for the fresh water vs. 64.00 lb/cu.ft. for the saltwater.
That means that you have to do a bit more work to keep the system running. More specifically, you find that you are moving 2,509 lb/hr of saltwater up the 15 foot ladder or 0.0874 hp when you add your weight and the bucket into the mix. This is in contrast with the vs. the 2,398 lb/hr you had to move up the ladder in your cabin using 0.0866 hp when you’re the weight of you and the bucket is added in.
Ultimately, you conclude that with a bit of development, you can expand your product line to provide a product suitable for providing relief to owners of USVI beach houses. And what better place to do the development than from the deck of your beach house, over looking the Caribbean.
Thus Ends the Analogy
Hopefully that was more useful than silly.
The idea was to illustrate that the actual refrigeration effect was provided by the refrigerant (the lake water or sea water) absorbing heat. But to reject the heat, work had to be done to move the heat to a location where it could be rejected.
In the case of the initial example it was done by carrying a bucket up a ladder to an elevation that would allow it to flow back to the lake were natural forces (like deep sky effect and evaporative cooling) would cool it back down.
But if you change refrigerant (seawater instead of fresh water), because its physical properties are different (its not as good of a refrigerant as pure water), you end up needing to move more mass to move the heat from the kiddie pool back to the oceans where evaporative cooling and deep sky effect can cool it back down.
Senior Engineer – Facility Dynamics Engineering
Visit Our Commissioning Resources Website at http://www.av8rdas.com/
[i] If you want an example of a pressure/enthalpy diagram, you will find one in this blog post. If you want to understand how to use one in practical terms, Sporlan publishes a very will done technical guide that is well worth reading in my opinion.
[ii] In working on the analogy, I found a really interesting blog post about the efficiency of the human body. The author was looking at biking and walking.
Here is a summary table from the post showing miles per gallon for different activities and energy sources. The difference between food and gas/lard is the energy density of our average diet vs. the energy we would get if all we ate was lard, which was the closest he could come to the equivalent of gasoline in terms of energy density.