A Heat Pump Is an Overunity Device?

Mary Rose

Open systems. A closed system is defined when a fixed volume is under study. There can be mass transfers as well as energy transfers across the boundary.

boncho
. . . it says a closed system can be defined when a fixed volume is under study. An open system is the opposite. The first sentence is defining the difference from closed to open . . .

So, you’re saying an open system is not defined by a fixed volume under study?

Sorry, the previous example I sourced is very poorly worded. And appears to maybe be a typo, where "Closed" is perhaps interchanged with "Open", or just varying interpretation on "fixed". My immediate reply is based on the English principles rather than scientific.

For a better description:

1- Closed System :- The system whose boundary does not permit the
transfer of mass between the system and its surrounding i.e. its mass is
constant and only energy being allowed to transfer across the boundary.

2- Open System :- is defined as a region is which the mass is not
necessarily constants besides , the mass as well as energy transfer cross its
boundary .

*

The proper way of defining an open system would be a "control volume".

Types of System

Control Mass System (Closed System)

Its a system of fixed mass with fixed identity.
This type of system is usually referred to as "closed system".
There is no mass transfer across the system boundary.
Energy transfer may take place into or out of the system.

Control Volume System (Open System)

Its a system of fixed volume.
This type of system is usually referred to as "open system” or a "control volume"
Mass transfer can take place across a control volume.
Energy transfer may also occur into or out of the system.
A control volume can be seen as a fixed region across which mass and energy transfers are studied.
Control Surface- Its the boundary of a control volume across which the transfer of both mass and energy takes place.
The mass of a control volume (open system) may or may not be fixed.
When the net influx of mass across the control surface equals zero then the mass of the system is fixed and vice-versa.
The identity of mass in a control volume always changes unlike the case for a control mass system (closed system).
Most of the engineering devices, in general, represent an open system or control volume.

www.nptel.iitm.ac.in...

You can say either is fixed volume, but one is fixed (as in control volume an arbitrarily assigned boundary) and that the other is fixed in that it is fixed in its properties no matter how arbitrary the boundary is so long as the boundary is set within the closed system.

Does that make sense?

*

If you calculate the grand potential in an open system where N represents molecules, the variable N is not fixed.

So if you weren't confused before you can be more so now.

reply to post by Mary Rose

It's none of your business where I get my information from.

Say what?

reply to post by DenyObfuscation


Of course I'm not talking about linking to sources for outside quotes that I post.

I'm talking about judging my sources and making denigrating comments about them that if directed to a member, would be against the terms and conditions.

Going into a temper tantrum about public figures is not conducive to civilized debate.

reply to post by Mary Rose

I'm talking about judging my sources and making denigrating comments about them that if directed to a member, would be against the terms and conditions.

Do YOU not do that?

Going into a temper tantrum about public figures is not conducive to civilized debate.

Who did that?

How about the topic of this thread? Is that just another springboard to your game "Slander the Scientists"?

Where do you stand at the moment regarding the question in this thread's title?

Mary Rose

boncho
Heat is converted to work (or vice versa). There is no way you can change this. Heat is calculable, and so is work.

But is heat calculable and work calculable taking the fact that all "working mediums" don’t behave with the same efficiency into consideration, or not?

Am I making any sense?

Well in the case of a heat pump heat is your only consideration. Were not attempting to make the heat do anything just transfer it from one location to another. A heat pump at its most basic is nothing more then a gate. Because of this when we look at it in classical electrodynamics it looks like heat magically appears. We see the power input we even see the heat it generates from the compressor we see the work being done gas being compressed. And through some miracle it seems like energy freely shows up. Well its not the energy is in the air itself even air at 32 degrees has heat energy we can transfer. Now the catch is if we tried to use this heat to perform work we would quickly see it has insufficient energy. This is why heat pumps need a power source. In classical electrodynamics youll actually see a negative energy input in the cycle. This does not happen when we look at it in physics because we know in physics for example entropy never allows any system to return to its starting point meaning nothing can be transfered even with 100 percent efficiency much less more.

DenyObfuscation
How about the topic of this thread? Is that just another springboard to your game "Slander the Scientists"?

I criticize mainstream science for good reason.

I don't slander an individual.

Here is what has been going on: Challenges to my posts about statements made by independent, self-taught individuals are filled with ridicule of the individual. This is not only an indirect ridicule of me, it is a fallacy of reason, because the topic is not the individual it is the subject matter in question.

DenyObfuscation
Where do you stand at the moment regarding the question in this thread's title?

As far as I'm concerned, a heat pump in a temperate climate, an "energy superstar," is an overunity device, the asinine reasoning behind "overunity doesn't exist" notwithstanding.

Mary Rose
But is heat calculable and work calculable taking the fact that all "working mediums" don’t behave with the same efficiency into consideration, or not?

Re-reading my post, I think I could have worded it with more clarity.

Let me try again: When heat and work are calculated, is the fact that "working mediums" don't behave with the same efficiency taken into consideration as part of the equation? Or should I say formula?

An example with an explanation of which symbol represents it would be helpful.

Mary Rose
As far as I'm concerned, a heat pump in a temperate climate, an "energy superstar," is an overunity device

typical of woo woo "science", change the meaning of a term to fit your distorted view of reality!

reply to post by Mary Rose

As far as I'm concerned, a heat pump in a temperate climate, an "energy superstar," is an overunity device, the asinine reasoning behind "overunity doesn't exist" notwithstanding.

That's funny. Perhaps The Weather Channel should add a overunity forecast to their line-up.

Now seriously, how can you even begin to criticize the reasoning when overunity is an undefined term?

ETA: I forgot about "energy superstar". The way you keep repeating it implies to me that you think it really does mean something important and relevant to this topic. What does "energy superstar" mean exactly, to you?

reply to post by dragonridr


Going back to my statement:

Mary Rose
. . . Carnot believed: Heat itself is converted to work; therefore, all "working mediums" behave with the same efficiency.

He posted temperature-pressure charts for some industrial refrigerants. The charts illustrated the fact that pressure change in relation to the temperature range is different; that is, each compound is shown to be behaving differently. They don't all produce the same pressure increase as the temperature rises.

He said that this is common knowledge among HVAC technicians but that physicists seem to be ignoring this obvious fact.

Can you take the above example of the compounds used for industrial refrigerants and show some math for it which will illustrate how the pressure change in relation to the temperature range is accurately reflected as not a constant?

Mary Rose
Can you take the above example of the compounds used for industrial refrigerants and show some math for it which will illustrate how the pressure change in relation to the temperature range is accurately reflected as not a constant?

The ideal gas law PV=nRT provides a good approximation in many cases, but there is more accurate math which takes the material into account, like this:

en.wikipedia.org...


While it's an improvement over the ideal gas law, it's not a perfect model either:

Above the critical temperature the van der Waals equation is an improvement of the ideal gas law, and for lower temperatures the equation is also qualitatively reasonable for the liquid state and the low-pressure gaseous state. However, the van der Waals model is not appropriate for rigorous quantitative calculations, remaining useful only for teaching and qualitative purposes.

If you scroll further down in that link there is more advanced math.

Mary Rose
reply to post by DenyObfuscation

 

Of course I'm not talking about linking to sources for outside quotes that I post.

I'm talking about judging my sources and making denigrating comments about them that if directed to a member, would be against the terms and conditions.

Going into a temper tantrum about public figures is not conducive to civilized debate.

edit on 11/26/13 by Mary Rose because: Clarify

It's not a temper tantrum, I think it is fair to everyone in this thread for them to know you are parroting information you've picked up in Lindemann's "Open system thermodynamics" DVD that he sells for $20 while claiming its worth $300.

That in fact, would also violate T&C if a member tried to pull it.

In any case, it isn't even science, it is technically philosophies of science. (Poor one at that). And I must say, I just don't feel like it's my "lucky day".

Quite frankly, I don't think anyone has ever offered this much real information on the subject of Thermodynamics, in one package before. So, this must be your lucky day!

The difference between Lindemann and legitimate engineers, scientists, etc… Is that any engineer with a break out idea, can usually pick up funding. Hell, I have funded projects with a chance for far less realized gains than what some revolutionary device could offer. And scientists get funding, while I admit there are many cases this is surrounded in controversy, there are just as many examples of people like Haisch et al, getting hundreds of thousands in grants and other funding to run their projects.

So the idea that some super mafia is going around sabotaging all his work is just BS. And it's clear his "educational" DVD has done nothing but try and obfuscate any actual understanding you might have on the subject.

reply to post by Mary Rose

I criticize mainstream science for good reason.

I don't slander an individual.

You have slandered "mainstream science" as a whole many times, with absolutely no evidence presented to back your opinion. On the other hand, we haven't "Slandered" Lindemann, instead, we have simply pointed out his modus operandi and backed it up with enough examples.

Mary Rose

Mary Rose
But is heat calculable and work calculable taking the fact that all "working mediums" don’t behave with the same efficiency into consideration, or not?

Re-reading my post, I think I could have worded it with more clarity.

I doubt it.

In any case, I have a question for you. Since you believe heat pumps are over unity. I would just like to ask, how much heat makes it into your house compared to all the heat that the sun created? (This is setting those arbitrary boundaries very far now).

I ask you this because you refuse to accept heat and energy as two different things. And when speaking about efficiency you are actually describing something very different.

It has been stated numerous times. Heat pumps are not heat engines.

If work in leaves you with heat, what does… work in - heat - work out leave you with? Is it still over unity?

Your argument is basically stating that science in general needs to redefine its terms because you don't like it. Not because of experimental data, or empirical evidence, simply because energy and heat are the same variable to you and Lindemann, so therefore, why not rewrite everything?

Is that a fair assessment?

Hey OP questions to ask is how are you using your heat pump ? Is it your main source for heat over the winter ? If it is then I am sure you have some type of auxiliary heat like duct heaters or some type of element that heats the air as it is passing through.

Do you live in the Northeast ? The heat pump is not going to be so efficient in the cold winters.
Once it reaches 32 degrees or lower there is very little heat out, or BTU's, to extract from the environment. Thus making your system run longer to achieve your desired temperature.

Change your filters!!!! Dirty filters and even dirty coils ( the thing that your filter protects ) will have a sever impact on how your system runs. Clog up coils cant perform their duty to the air flow if they are covered with dust/debris/gunk and what not.

R-410A sucks. Noone liked it when it first came out... noone likes it now and the only people who do like it are the ones who are selling this whole green movement. They are higher pressures than R-22 and don't pick up as many BTU's per lb of refrigerant compared to the "old" R-22. But hey its "safe" for the environment !

Was your unit sized properly for the home ? Many contractors out there don't even measure the home and calculate if the tonnage matches the size of the home. Many of them just do change outs. They come in and see what you have and say yes we can get you a brand new one more modern and efficient and just swap out the old with the new leaving the same tonnage size without knowing if any additions were made to the home since it was built. Just be careful. Always ask how he came up for this size of a unit for the home and let him show you how it was calculated. A poorly sized unit will cost you money!

I am sure there are other things out there to mention but this is all I can think of now.

Mary Rose

Mary Rose
But is heat calculable and work calculable taking the fact that all "working mediums" don’t behave with the same efficiency into consideration, or not?

Re-reading my post, I think I could have worded it with more clarity.

Let me try again: When heat and work are calculated, is the fact that "working mediums" don't behave with the same efficiency taken into consideration as part of the equation? Or should I say formula?

An example with an explanation of which symbol represents it would be helpful.

Ok let me see if i can get you to see something. There is a flaw in your interpretation so let me start by saying a heat pump performs no work! A heat pump transfers energy think of an electric wire but instead of electricity were transferring heat. So lets say it takes 1 electrical energy to say transfer 5 heat energy in our Model.And for simplicity sake will say 1 heat energy is the same as 1 electrical in reality its about 1/3 but any way. So the simple math would then tell us that for every unit of energy we put in we get 5 out correct? This is what your seeing as over unity but its not its simply how much energy we had to use to transfer the energy.And if we remove heat to power a generator it becomes an open system. Just like leaving your refrigerator door open meaning compressor runs constantly generating more heat and friction until eventually you need more energy then you are making.

Now theres another confusion i see for you so let me give you a quick example your looking at things wrong.A light bulb is about 5 percent efficient at creating light so if our purpose is to create light well a standard bulb sucks. And it would rate at 5 percent efficiency but now if we say its a heating device this changes the whole game.Since 95 percent of the energy goes to heat we have a device that is 95 percent effective as a heater. Now as we know there are far more effective ways to heat a room then a light bulb energy wise its wasteful. This is what your doing with a heat pump your using it for something its not meant to do. And people are telling you its 95 percent efficient because they are showing you the wrong statistics for what your trying to do because math wise sounds like the light bulb is the best heater ever invented.

So what your being sold on is a lie because there only telling you half the story. So what you believe to be over unity isnt since as i said a heat pump doesnt create heat it moves it. So as i said measure the efficiency of a heat pump makes it look like its creating something it not the energy is all ready there.A heat pump just allows us to transfer energy in reverse of nature.

reply to post by dragonridr


The video that I was watching was not on heat pumps.

It was on thermodynamics. More specifically open system thermodynamics.

I took the example of heat pumps out of it to start this thread.

The whole history of the understanding of heat was presented and interpreted.

Mary Rose
reply to post by dragonridr

 

The video that I was watching was not on heat pumps.

It was on thermodynamics. More specifically open system thermodynamics.

I took the example of heat pumps out of it to start this thread.

The whole history of the understanding of heat was presented and interpreted.

Correction:

The whole history of the understanding of heat was presented and interpreted. misinterpreted.

Mary Rose

Keywords: Equilibrium vs. Non-equilibrium



I remember Bearden talking about equilibrium and, instead of "non-equilibrium," he uses the term "disequilibrium" here. Also, I heard him say ". . . When it doesn’t recover symmetry almost immediately, very quickly . . . when it does not minimize disequilibrium effect, particularly if it stabilizes disequilibrium structures and dynamics . . . " So maybe symmetry restores equilibrium and asymmetry is associated with disequilibrium.

Now, going back to Bearden's article that I posted in reference to the 100 units in results in at least 64 units of loss:

Mary Rose
“On Extracting Electromagnetic Energy from the Vacuum” by Thomas E. Bearden

There is another important paragraph:

It is known in particle physics that there can be no symmetry of a mass system without the incorporation of the active vacuum interaction, yet this too is missing from classical electrodynamics. Symmetry implies nonobservables, and asymmetry implies observables. So every observable mass system, being asymmetrical a priori, must be accompanied by nonobservables interacting with it, else it can have no symmetry (or equilibrium). Yet classical electrodynamics continues to assume equilibrium and symmetry in observable systems without incorporating the active vacuum. Everywhere we examine classical electrodynamics, we find non sequiturs of first magnitude. This alone should be a compelling reason for the scientific community to assign the highest priority, ample funding, and the best theoreticians to the sorely-needed revision of electrodynamics from the foundations level up.

Mary Rose
There is another important paragraph:

Symmetry implies nonobservables, and asymmetry implies observables.

Just why do you think that is a valid comment?