[CliffordK (OP)]

I was looking at this vapor pressure chart in Wikipedia.  Sorry it only has a few substances, but the idea is the same. 

http://en.wikipedia.org/wiki/Vapor_pressure


Looking at, say the difference in vapor pressure between -10°C (below freezing), and +20°C (room temperature).

If you choose, something like Propane (BP is -42°C), one goes from a vapor pressure from about 2.5 ATM to about 8.5 ATM, or a difference of about 6 ATM.

However, if one chooses a refrigerant like Diethyl-Ether, one goes from about 0.15 ATM up to about 0.6 ATM. 

So...
Theoretically, running the diethyl-ether in a vacuum, rather that under pressure, one should be able to force the phase change with an absolute difference of only about 0.5 ATM.

It would seem as if one should be able to design the diethyl-ether refrigerant to use less energy.  Although, I know that a vacuum pump in air, seems to have to work quite a bit to achieve a low pressure.

[SeanB]

You can use ether as a refrigerant, but a bad side effect is that any leaks will tend to allow oxygen into the system, which will make for a extremely explosive reaction.

[CliffordK (OP)]

You can use ether as a refrigerant, but a bad side effect is that any leaks will tend to allow oxygen into the system, which will make for a extremely explosive reaction.

Good Point.
Heat, Pressure, a Volatile Organic, Plus a risk of adding Oxygen.
May not be the best combination.

One could, of course, choose heavier halocarbons.

My point was that the higher boiling point substances tended to have less of an overall pressure differential change in vapor pressure between the different temperatures, as long as they are operated at a vacuum to force the boiling/evaporation phase.

Of course, the freezing/triple point of a substance would be another limit.

[Bored chemist]

I'm not certain, but I think the big issue is that with ether you have a relatively low pressure in the system- so there's not much mass of ether so it can't carry much heat.
Also I'm pretty sure you need to consider the latent heat of evaporation and other parameters.
It's complicated.

[CliffordK (OP)]

I'm not certain, but I think the big issue is that with ether you have a relatively low pressure in the system- so there's not much mass of ether so it can't carry much heat.

Thanks BC,
That is very insightful. 
However, one could potentially design the system to only use the liquid phase for heating/cooling.

[diagram=658_1]

Also I'm pretty sure you need to consider the latent heat of evaporation and other parameters.
It's complicated.

I was thinking that the latent heat of evaporation was a bad thing.  But, perhaps it is actually a good thing, as it means more heat is removed when evaporating, especially if one is using the liquid phase for heating/cooling as above.

[SeanB]

Refrigeration relies on a phase change to operate. You have a liquid that is hot and under high pressure, which then has the pressure reduced somehow ( either a capillary tube or an expansion valve) so that the boiling point is now lower than the temperature the liquid is at. It then evaporates, and draws the energy required for this phase change from the environment around it. This makes that side cooler. The cold gas is then compressed by some method ( turbine, vane pump, piston, scroll or other method) to give you a high pressure hot gas ( has had energy added to it) that is then allowed to cool to a lower temperature below it's boiling point and condense to a liquid.

Using the expansion to power the compression is not going to add much to the efficiency, and adds a lot more complexity as well. It will not be useful, if it had any benefit it would have been used in some application by now, as refrigeration has been around for nearly a century. There are still units that are in operation that have been running untouched for close to three quarters of a century.

[CliffordK (OP)]

Ok,
So, what I've described will work.
But, it would be more or less a direct energy transfer based on the latent heat of vaporization.

Whereas, using the vapor phase, one can multiply the heating/cooling effects due to the enormous expansion between liquid/gas phases.

[Geezer]

Ok,
So, what I've described will work.

Alas, it won't. You'll simply end up having to do more work with the motor than the work produced by the expander, so you'll be wasting energy. If there was a net gain, you'd be breaking the law. You cannot avoid paying the entropy tax, and for that reason, you can only multiply heating and cooling effects by a number that is less than one.

Multiplying heating and cooling means that you are trying to make heat flow from a cold place to a hot place without doing the necessary work. The work (energy) required to push heat "uphill" is always greater than the amount of heat (energy) transferred. Trying to avoid that is like trying to make time run in the opposite direction.

Homer had something to say about this.

[CliffordK (OP)]

I didn't mean to imply that the system would work with no energy added.  Obviously, even the vapor pressure will be higher on the warm side than on the cold side, so the vacuum pump still needs power.  I guess that does also mean more pressure on the warm return side to recover some of that added energy.  But, at least some energy would need to be expended, especially if the goal is to remove either warmth or cold from the system.  Otherwise the system would tend towards iso-thermal and stop.

We've already discussed that one could make ice by putting water into a vacuum chamber...  this is more or less using the same concept.

[Geezer]

I didn't mean to imply that the system would work with no energy added.  Obviously, even the vapor pressure will be higher on the warm side than on the cold side, so the vacuum pump still needs power.  I guess that does also mean more pressure on the warm return side to recover some of that added energy.  But, at least some energy would need to be expended, especially if the goal is to remove either warmth or cold from the system.  Otherwise the system would tend towards iso-thermal and stop.

We've already discussed that one could make ice by putting water into a vacuum chamber...  this is more or less using the same concept.

No, but you're still trying to get "free energy". It's a heat pump, so you can't get work from the thermal gradient because you are pumping heat "uphill". The energy to transport the heat is all coming from the motor, so if you produce work from the heat difference that results from work done by the motor, because of entropy, you'll get less work from the expander than you could have got directly from the motor.

I missed the bit about making ice with a vacuum. If the idea is to lower the pressure above the water to make it freeze, that does not make sense to me. The freezing temperature of water does not increase as pressure decreases.

[CliffordK (OP)]

I missed the bit about making ice with a vacuum. If the idea is to lower the pressure above the water to make it freeze, that does not make sense to me. The freezing temperature of water does not increase as pressure decreases.

You're in luck.
It seems as if TNS also has a bit of amnesia.

Is vacuum-drying of clothes feasible?

Fortunately, Google doesn't seem to suffer from the same Amnesia problems that TNS seems to.  []

http://webcache.googleusercontent.com/search?q=cache:LRoS6I3LC4gJ:www.thenakedscientists.com/forum/index.php%3Ftopic%3D25084.0+&cd=1&hl=en&ct=clnk&gl=us&client=firefox-a

"How did the discussion get to ice?"
Because some git posted a video of what happens when you connect a vacuum pump to a container of water.
[...]
Have a look at this.

Once the water freezes the vapour pressure drops even further. It's a good way of getting the clothes cold, but not so good at getting them dry.

Unfortunately You-Tube doesn't like my router.  So, I'm taking this on faith.

The same idea as the human body sweating.

A low vapor pressure causes the substance to evaporate. 
The Latent Heat of Vaporization is essentially transferred from the liquid to the vapor.
And, one ends up with ICE.

[Geezer]

Ah, but then you are doing work on the water vapor when you compress it to remove it (and heat) from the vacuum chamber.  If you inject water into a vacuum, there is no heat loss.

[Bored chemist]

" The freezing temperature of water does not increase as pressure decreases."
What's your second guess?
At best, it depends on what temperature / pressure you start at.

[CliffordK (OP)]

The freezing temperature of water does not increase as pressure decreases.

What's your second guess?
At best, it depends on what temperature / pressure you start at.

Ahh...
Yes.
The phase diagram of Water.


For pressures around 1 ATM, Temperatures around 0 to +100 °C
As the pressure increases, the freezing point decreases.
As the pressure decreases, the freezing point increases.

Although, the actual difference in freezing point of water between pressures of 101KPA (1ATM) and .6KPA is only 0.01 °C

The point is that by forcing a liquid to evaporate under low pressure, you can decrease the temperature of the liquid, potentially to the point of turning it into a solid.  However, that operation wouldn't be particularly efficient.  Perhaps an energy efficiency of 1:1.  One needs to also use the rarefied vapor phase to further improve heating/cooling efficiency, to higher efficiency ratings, perhaps 2:1 up to 5:1.

My original question was whether one could use the lower vapor pressure of the heavier liquids to one's advantage.  Operating a refrigeration unit under vacuum rather than under pressure, and thus having a lower difference in vapor pressures between the cold and hot liquid phases.

BC's comments were helpful. 
1: The gas produced under a vacuum would be at a lower density.  It might be cooler, but may not have high enough density for effective heat transfer.
2: The latent heat of vaporization may be a problem.  While it does lead to cooling of the liquid phase, such a process likely is not too energy efficient compared to cooling with the rarefied vapor phase.

I never intended to imply a goal of ZERO energy consumption in this process, but was rather exploring potential efficiency gains through the use of lower pressures.

I do find it interesting that one can get 2 to 5 times the heat gain by using a heat pump vs direct electric resistance heating, but that is a different topic.

[Geezer]

Although, the actual difference in freezing point of water between pressures of 101KPA (1ATM) and .6KPA is only 0.01 °C

Precisely. BC was using a highly pedantic accurate thermometer.

[Geezer]

I'm not saying you are trying to get the process to work with zero energy input, but I don't believe that adding the expander can do anything to reduce the energy input from a thermodynamic point of view.

[Geezer]

I think I figured it out.

The "heat pump" (call it a refrigerator, reverse heat engine or whatever) is a closed system. The mass of working fluid is constant. The only variables are the ambient temperatures surrounding the hot and cold heat exchangers, and the amount of heat pumped.

Thermodynamics insists that the net amount of heat leaving the system must equal the net work done on the system - no exclusions, no rain checks.

Changing the working fluid will affect the heat transfer rate to and from the heat exchangers, and it will change the work done by the compressor, but it can't alter the relationship between the work input and the heat transferred without breaking the second law.

Therefore, different fluids will influence the amount of heat that can be pumped for a given set of conditions and equipment, but they cannot alter the thermal efficiency of the cycle for that set of conditions. That was previously determined by thermodynamics.

[CliffordK (OP)]

 [ Invalid Attachment ]

Ok, here is the general heat pump (or AC/Refrigerator) schematic.

You talk about thermodynamics.  However, this can effectively BREAK the law of thermodynamics...  well, at least crack it a bit. []

The reason is that one can treat the ambient temperature as an infinite heat sink.

In a closed system, yes, it would obey thermodynamics, and you would have a net energy loss, primarily due to friction and systemic inefficiencies. 

However, in an open system, using the ambient heat sink, one gets an extreme increase in efficiency.

The Cold tank, and the vapor coming off of it will be essentially isothermic.  The temperature will be determined by a few things including the vapor pressure above the tank and whether it is allowed to reach equilibrium.  The lower practical limit on the temperature would be the triple point of the liquid.

The thermal efficiency of the heat pump has to do with the multiplicative factor of the ambient heat that is added.

If 1cc of liquid at 290K --> 10cc of vapor.
Essentially the ideal gas law would give you that Volume is proportional to Temperature, and you get the equivalent of the 10cc at 29K, or the energy transfer equivalent of that.

If you heat that back up to ambient temperature, you get 10cc at 290K --> 1cc at 2900K.  Whew!!!

Hmmm, somehow this has to be symmetrical.  The heat gain per cc would be (290-29)*10 + 290 = 2900K.

So, that is the heat equivalent that you are adding at ambient temperature in the cold radiator...  and would need to be removed in the warm radiator.

Perhaps rather than thinking in degrees, I should consider this in calories.  So, that 1:10 evaporation cycle essentially transferred 2610 Calories from the environment into the system.

Now, you will astutely notice that I left out part of the Ideal Gas Law  PV=nRT.  And, the reason is that at the liquid/vapor boudary, one gets a disjoint in the PV relationship.  Compressing and rarefying an "Ideal" gas like Helium would be much less efficient.

I'll have to re-read about the Latent Heat of Vaporization, as that will likely come into play.  While technically that energy is not lost, it is one of the aspects that gets transferred directly from the cold side to the hot side, and thus would likely not give you any "gain".

As far as whether this 2610 calorie gain from the environment could actually generate more power than what it took to generate it...  I don't know.  Since you are adding energy from the environment, you're not truly breaking any laws of physics. 

That is beyond the question of this topic anyway.  I was simply asking if higher Boiling Point molecules would be better for heat pumps or refrigeration.

[Bored chemist]

An unscientific answer is that if higher boiling liquids worked better, the manufacturers would hardly have chosen things like SO2 and NH3 in the first place.

Incidentally, WRT the 0.01C change in freezing point, it's not size that's important- it's which direction it's pointing that matters.

[Geezer]

 Nice try, but what you are describing is still a closed system. The only things that cross the sytem boundaries are heat and work. The mass of the working fluid remains constant. The net work must be equal to the net heat.
 
You are not adding energy from the environment to your system. You are doing work to transport energy from one part of the environment to a different part of the environment. There is no net change in the energy in the universe outside the heat pump.