# Unsteady Air Conditioner

#### Clausius2

Science Advisor
Gold Member
I am interested on understanding the unsteady period of functioning of an air conditioner system, with a traditional refrigerant, let's say R-11.

Imagine there is a room which initially is at the same temperature than the external one, i.e. $$T_i(0)=T_e$$. Just when I switch on the system, the evaporator, condenser and refrigerant are at the same temperature $$T_e$$. I really don't see "the movie", I mean, how the refrigerant (which I think will be in liquid state) can be compressed and change its state, and how the evaporator can cause an evaporation if it is initially at the same temperature than ambient (there won't be any heat flux to absorb because both things are at the same temperature), and how the condenser can condense if it is at the same temperature than external atmosphere.

The steady regimen is very easy, but I don't know how the machine starts.

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#### everneo

The refrigerant in the condenser would always be in gaseous state once the liquid state refrigerant pass through the expansion valve. At start when Ti = Te, there would not be any difficulty for the compressor to suck in the gas from condenser and compress it.

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#### Cliff_J

Science Advisor
Clausius2 - its the pressure difference applied to the refridgerant, think triple point chart and it should make sense how even at 10C above ambient a sufficient low pressure or high pressure will cause the respective phase change.

Edit - Just re-read your post. Ok, so there must be some defined startup time when the system has to build up the pressure differential. Maybe Russ or someone intimately familiar with the exact workings will have a better answer.

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#### faust9

Clausius2 said:
I am interested on understanding the unsteady period of functioning of an air conditioner system, with a traditional refrigerant, let's say R-11.

Imagine there is a room which initially is at the same temperature than the external one, i.e. $$T_i(0)=T_e$$. Just when I switch on the system, the evaporator, condenser and refrigerant are at the same temperature $$T_e$$. I really don't see "the movie", I mean, how the refrigerant (which I think will be in liquid state) can be compressed and change its state, and how the evaporator can cause an evaporation if it is initially at the same temperature than ambient (there won't be any heat flux to absorb because both things are at the same temperature), and how the condenser can condense if it is at the same temperature than external atmosphere.

The steady regimen is very easy, but I don't know how the machine starts.
The short answer is one step at a time.

First, you have a misconception: the refrigerant will be in a two phase condition at some pressure depending on the total mass of R11 in the system and the initial temperature of the system. There will be a liquid region at the lowest point and a gaseous region at the highest point.

Switch the compressor motor on and the gaseous portion will get compressed (remember, Liquids are are relatively incompressible so the device which moves a liquid is called a pump not a compressor). Here, I think a T-s diagram might help:
http://www.qrg.northwestern.edu/thermo/design-library/refrig/Ts.gif

The system would be somewhere on the P_low curve initially. It shoulod be somewhere near point 4. The compressor raises the pressure (assume it's an adiabatic process because a real process works in the same way there is just a shifting of 1) to the superheated region (point 1). Work was done on the initial fluid so even if the system and ambient are all initially at the same temperature a transfer of heat can still occure.

The compressor transformed the two phase gas into a super heated gas so the R-11 now needs to be cooled to a saturated liquid. In this step heat is rejected to the environment(point 2).

There is now a high pressure fluid on one side of a throttle valve. This high pressure fluid flows through the throttle valve to a low pressure region (the compressor inlet is on this side of the valve---the motive force on the system is the DP across the compressor) and in doing so transforms into a low pressure low temperature two phase fluid(point 3).

At this point the working fluid can be easily boiled. The boiling process(change of phase) readily absorbes heat energy from the environment(point 4).

This system would not work very well if the entire system was sitting in a room by itself because the heat absorption portion and heat rejection would essentially cancel each other out; however, the added heat from the I^2R losses and mechanical losses in the compressor and electric motor would add additional heat to the environment. If you seperate the heat rejection fluid(outside air, sea water, etc) and the heat absorption fluid(inside air) then you'll be able to cool one fluid while heating up the other fluid.

The process is driven by the DP across the compressor though and that's where you should start when thinking about this sort of thing.

Hopefully I'm not too far off with my description---it's been about 6 years since I've worked with refrigeration cycles.

#### quark

Good explanation by faust9. There may be problems if you consider the control space as source and sink as well. Similar problem will be to keep open the door of a refrigerator which ultimately results in heating up of the space inside the refrigerator. Technically, the room will no more be a sink and its temperature increases due to the compressor work, how small it may be. The increased room temperature(or sink temperature) increases the compressed gas pressure(because saturation pressure at the condensing temperature increases) and finally this results in compressor breakdown, if not assisted by pressure switches.

But if you mean to have the source and sink different but maintained at same temperature, no problem will occur. That is the very principle of a refrigerator, to remove heat from a low temperature source to a high temperature sink. The process is well explained by faust9.

#### Clausius2

Science Advisor
Gold Member
faust9 said:
First, you have a misconception: the refrigerant will be in a two phase condition at some pressure depending on the total mass of R11 in the system and the initial temperature of the system. There will be a liquid region at the lowest point and a gaseous region at the highest point.

.
Why? I cannot see why. Why is it in a two phase condition?. Assuming there is an equilibrium temperature $$T_e\approx 30ºC$$ and an equilibrium pressure at rest of approximately $$P\approx 2 bar$$ the entire refrigerant is in liquid state. So that pumping it wouldn't cause a great internal heating and so condensing process makes no sense. How can you proof your statement about the two phase condition? Which is the equilibrium pressure when the AC remains switched off?.

The rest of your explanation, although it is an excellent one, is an explanation of the well known steady state.

Thanks.

#### Ivan Seeking

Staff Emeritus
Science Advisor
Gold Member
I'm pretty rusty on this stuff but may know the answer. I'm still trying to get the question straight though.

First of all, there is the volume occupied by the liquid. Are you assuming that that volume of liquid equals the volume of the system?

#### Clausius2

Science Advisor
Gold Member
Ivan Seeking said:
I'm pretty rusty on this stuff but may know the answer. I'm still trying to get the question straight though.

First of all, there is the volume occupied by the liquid. Are you assuming that that volume of liquid equals the volume of the system?
I am not assuming nothing. I am searching the truth, how it really works. I don't know what happens when charging the AC system. I don't know if the refrigerant is in liquid state when charging, and if all volume is filled with liquid. I think we need Russ, or some guy who works with this stuff.

#### Ivan Seeking

Staff Emeritus
Science Advisor
Gold Member
Clausius2 said:
I am not assuming nothing. I am searching the truth, how it really works. I don't know what happens when charging the AC system. I don't know if the refrigerant is in liquid state when charging, and if all volume is filled with liquid. I think we need Russ, or some guy who works with this stuff.
No, the entire volume is not filled with liquid. I think that's what's creating the confusion. Liquid is found in or after the condenser and in the liquid resevoir. The rest flashes over to a gas. The system can be charged with either a liquid or a gas, but always to the proper amount such that liquid will never reach the compressor. Think about what happens when we start with an empty system. After being evacuated we start of add freon. This can be either a liquid or a gas, but as soon as the freon enters the evacuated system, the low pressure causes the freon to flash to a gas. As we continue to add more refrigerant, the pressure in the system rises, and pretty soon we have enough pressure so that the gas begins to turn to a liquid. Effectively, we continued to fill until the liquid level reaches the proper level. In practice, we usually add the freon as a gas with the system running. The low pressure on the suction side allows for quicker filling.

Russ is certainly the expert but there was a time that I was factory certified for AC.

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#### Ivan Seeking

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I should add that there is also liquid up to the evaporator or expansions valve, but in all of the systems that I've worked on, with nothing running, gravity kept the liquid on the liquid side of the system; with those components located [physically] lower than the gas side components.

#### faust9

Clausius2 said:
Why? I cannot see why. Why is it in a two phase condition?. Assuming there is an equilibrium temperature $$T_e\approx 30ºC$$ and an equilibrium pressure at rest of approximately $$P\approx 2 bar$$ the entire refrigerant is in liquid state. So that pumping it wouldn't cause a great internal heating and so condensing process makes no sense. How can you proof your statement about the two phase condition? Which is the equilibrium pressure when the AC remains switched off?.

The rest of your explanation, although it is an excellent one, is an explanation of the well known steady state.

Thanks.
Where did you get those numbers from? I can prove my statement by looking at the T-s, P-T and a P-v diagrams. The numbers you have above are for a supercooled liquid. When you compress a supercooled liquid--again, assuming it is adiabatic---you move up to intersect the new pressure curve(on a T-s diagram) which keeps you in the supercooled region still.

If you draw a control volume around the refer system at some point in the system you will find a vapor and at another you will find a liquid. I don't know what proof you want for this---it's just how the processs works. The process begins with a moist vapor i.e. the refrigerant is in the bell portion thus x is defined. Use a T-s and a P-v diagran and see what happens to the process if you were to begin at the point you described (30C, 200Kpa).

For the refer plant to work, a change of phase must occure. For a change of phase to occure, the refer plant must have an initial condition such that the expansion of the R11 resulting from boiling can occure in a constant volume system (refer plants are constant volume systems so if the initial condition was as you described a change of phase would result in a rapid pressure increase within the system). For the above to occure there must be a pre-existing vapor region and a liquid region within the refer plant.

The other option would be a variable volume refer plant. This type of system (I don't know of any variable volume plants though I'm not a HVAC guru) could allow one to start with a liquid filled system.

I'm sorry if my explaination is not clear. I was out in the sun all day at a party. I'll repost tomorrow after a few hours of sleep if you want. The key too understanding this---or any thermodynamic process--- it to look at the graphs. Look at a P-v diagram and a T-s diagram and see what happens when you move from one region to another. Use some basic assumptions like the refer plant is a constant volume system, the throttle valve is a constant h device, the compressor is a constant s device and there is no headloss in the system(no pressure drop across a heat exchanger). The diagrams are the key to understanding what is going on when the system is in steady state or in a transient---the system starts at point 4 from my previous post BTW.

Hope this helped a little.

#### Clausius2

Science Advisor
Gold Member
First of all, thanks for your effort, and hope you had a nice sunny day in your party!!!!.

faust9 said:
Where did you get those numbers from? I can prove my statement by looking at the T-s, P-T and a P-v diagrams. The numbers you have above are for a supercooled liquid. When you compress a supercooled liquid--again, assuming it is adiabatic---you move up to intersect the new pressure curve(on a T-s diagram) which keeps you in the supercooled region still.
My figures come directly from the P-h diagram of R-11. And yes, it is supercooled liquid.

The key too understanding this---or any thermodynamic process--- it to look at the graphs. Look at a P-v diagram and a T-s diagram and see what happens when you move from one region to another. Use some basic assumptions like the refer plant is a constant volume system, the throttle valve is a constant h device, the compressor is a constant s device and there is no headloss in the system(no pressure drop across a heat exchanger). The diagrams are the key to understanding what is going on when the system is in steady state or in a transient---the system starts at point 4 from my previous post BTW.

I think the key of our disagreement is just the point 4 of your diagram. You are assuming the point 4 is just a moisture. Now, when you compress it surely you will increase temperature and then heat flux will be enhanced at the condenser to exhaust it and condensing the gas. On the contrary I don't think the same. I think point 4 may/might/could be in the supercooled region if the pressure and temperature of equilibrium are there. I mean, if I switch off my AC system and go to your party (damn it! you didn't invited me!) for two hours and then I come back and switch on the AC system again, why would the point 4 on moisture region?. I could (depending on what the hell are pressurre and temperature of the refrigerant when I boost the compressor) have an entire volume of liquid inside the pipeline if pressure and temperature of boosting are in the supercooled region. Then, the compressor won't compress anything and won't increase the thermal level of the fluid. We need someone who gives us a figure about where is the refrigerant in the P-h diagram when the AC system spends some time turned off.

By the way, the point 4 is never inside the moisture region. I don't know who draw that diagram, but if you don't want to cause a failure in the compressor, point 4 must be in the superheated region.
---------

Ivan, thanks for your effort also. I cannot see why is not going to be the entire refrigerant at the same temperature and pressure and so at the same state. Once the AC is turned off, there is no interaction which causes such difference (with exception of internal and external temperature of the room which we will assume are the same after a large time, and with exception too of the height difference, which we assume to be negligible in an AC system for one floor of a building). Also, I think you have developed an example of an AC system being charged with freon:

Ivan said:
but as soon as the freon enters the evacuated system, the low pressure causes the freon to flash to a gas.
Why? Not all fluids passing an expansion valve or entering into a low pressure region are vaporized or flashed to gas, if the temperature is the adecuate in the supercooled region. At least it is what P-h diagram for R-11 says.

#### Ivan Seeking

Staff Emeritus
Science Advisor
Gold Member
Clausius2 said:
Also, I think you have developed an example of an AC system being charged with freon:
IIRC, Freon is just a brand name. But I worked with...hmmm...R11, R12, R22, and R111, as I recall.

Why? Not all fluids passing an expansion valve or entering into a low pressure region are vaporized or flashed to gas, if the temperature is the adecuate in the supercooled region. At least it is what P-h diagram for R-11 says.
I will grab my AC Bible and check the charts a little later. Intuitively it makes sense to me but I may not be hitting on the right language or detail. Or, maybe I just never really thought about it! :tongue2:

#### Averagesupernova

Science Advisor
Gold Member
O.P. I guess I'm not seeing why all the details are needed here. Air conditioning and refrigeration that you describe are simple absorbtion principles. It doesn't really matter where you start from, when you evaporate a fluid by lowering the pressure on it, it absorbs thermal energy. This energy shows up again when its pressure is raised. If the temp of the high pressure side is higher than ambient, the ambient absorbs this energy. Net flow of thermal energy is obvious.

#### Ivan Seeking

Staff Emeritus
Science Advisor
Gold Member
Clausius2, the system pressure follows the ambient temp as a saturated vapor. I keep thinking that this is the point that you're missing. Or are you asking how it can exist as a saturated vapor/liquid mix? I can only think to say, how can it not unless we fill the system completely with liquid, or release enough so that the pressure begins to drop? I can't seem to put my finger on the problem any better than that, for the moment.

#### quark

There are two problems with the assumption that liquid can exist in the entire volume of a refrigeration cycle. First, as the specific volume of gas is much more higher than liquid, there won't be any vaporisation(or very less) in the cycle and this ultimately leads to the damage of the compressor. This is taken care by vacuuming the system during charging as suggested by Ivan and care is taken in a way that vapor enters the system. Refrigerants like R123 exist in liquid state at room temperature and these are charged into condenser while the compressor is running, thus creating both liquid and vapor phases.

Secondly, when the system is running, there will be liquid phase in the condenser upto TXV, liquid+vapor in the evaporator and vapor in the suction line. When you stop the system, there will be equilibrium pressure which is well above the saturation temperature of the liquid. So the refrigerant can exist in liquid state all over. But we have to remove latent heat from already existing vapor to make it into liquid. As the latent heat of any liquid is much more higher than the specific heat, it requires, in our case, 191kgs of liquid (and 1K temperature rise) to condense 1kg of vapor(latent heat at 2bar is 173.2kJ/kg K and specific heat of liquid at constant pressure is 0.90219kJ/kg K). So only ambient air should absorb this heat but as soon as the temperature of the gaseous refrigerant equals that of ambience, there will be no more heat transfer.

When you start the system, there is a time delay for compressor. This is to ensure that vapor exists in the system, otherwise liquid may enter into the system(particularly with DX type evaporators) and ultimately results in liquid stroke. With flooded type evaporators, the gas take off is always from top.

#### russ_watters

Mentor
No, sorry, Clausius, I didn't go to Live8
faust9 said:
Switch the compressor motor on and the gaseous portion will get compressed...
faust9 gave a good detailed answer and others provided good info, but to reiterate, the simplest answer is really that everything starts with the compressor - the compressor is what is doing all of the work (in the refrigerant cycle, that is). Since a compressor and a pump are pretty much the same thing, the compressor starts the process and gets the freon circulating and pressurized. Even if the pressures aren't quite right, they'll be close enough and once you get some flow through the expansion valve at high pressure, its going to expand and cool. The fact that all the initial temperatures are ambient actually makes it cool pretty quickly, since the temperature of the refrigerant entering the expansion valve is already considerably cooler than once the system is stable.
Clausius2 said:
Why? I cannot see why. Why is it in a two phase condition?
You know about propane (and other similar fuels), right? Propane is a gas at room temperature and pressure but a liquid at slightly higher pressure. In the tank, you start with a liquid and enough boils off to reach that critical/equilibrium temperature where the pressure of the gas keeps the rest of the liquid a liquid. Same idea with refrigerant.
Assuming there is an equilibrium temperature and an equilibrium pressure at rest of approximately the entire refrigerant is in liquid state.
I think what you may be missing with this issue (as Ivan alluded to, so I'm again just reiterating) is the fact that the volume is constant and the piping isn't "full". The refrigerant simply can't all be a liquid unless there was empty space in the tubing - and if there was, then the pressure would be too low (a vacuum) for it to remain a liquid.

But there is another issue:
Ivan, thanks for your effort also. I cannot see why is not going to be the entire refrigerant at the same temperature and pressure and so at the same state.
Well, think of propane again. Also, consider the phase diagram. There is a line where you get both a liquid and a gas at precisely the same temperature and pressure (think of ice water too). By constraining the volume of the system, you force the system to naturally go to that equilibrium condition.
Why? Not all fluids passing an expansion valve or entering into a low pressure region are vaporized or flashed to gas, if the temperature is the adecuate in the supercooled region.
I'm not sure where exactly you are looking on that p-h diagram - in order for the refrigerant to not be vaporized (again, it can be partially vaporized), it would have to be either extremely cold already or not undergo much of a change in pressure.

#### Clausius2

Science Advisor
Gold Member
Ok. I have looked into your post, and quark's one too. I think now I understand it. You know, sometimes I pretend to be an stubborn guy for you to insist in the same arguments, and so I understand it much better.

Thanks everyone for replying here. An interesting discussion.

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