Assumed violation of physics - Heat vs. Work

Good evening Dale,

One matter absent from your analysis is the small matter of heat transfer efficiency into and out of the reservoirs.
Your perfect machine analysis is silent in this respect.
 Mentor Hi Studiot, I believe that you are talking about the rate of transfer to the reservoir, not efficiency. Correct? The transfer is passive, so it doesn't use or generate work, so I don't see where efficiency would come in.
 No I'm not talking about rate of transfer. Let us say you want to use the heat pump to heat a room, obtaining the heat from the outside atmosphere ( a pretty typical practical application). 1)In order to extract heat from the air to your working fluid you have to pass the air over some sort of heat exhanger. Probably you require to expend energy in a fan. 2)Having picked up the heat, you have to pump your working fluid around in heat pump and perform the compression and expansion cycles. 3)You now know what I'm going to say. Having pumped your hot fluid into the room you need a second heat exchanger/fan to remove and distribute the heat. I was intending to come round to explaining that the manufacturers only quote the the COP of the heat exchanger using (2) as the work budget. This makes advertising headlines but poor engineering.

Mentor
 Quote by Studiot I was intending to come round to explaining that the manufacturers only quote the the COP of the heat exchanger using (2) as the work budget. This makes advertising headlines but poor engineering.
Oh, I didn't realize that. I thought that the COP was for the whole system, which of course includes all of those other fans and parts that you mention for the work.

Well, all that does is reduce the real efficiency further from the ideal efficiency. Since even an ideal heat engine and heat pump could not gain energy this way, what you mention is just one more way for a real system to perform even worse. Worse than 0 is pretty bad.
 Recognitions: Gold Member Are you guys discussing heat pumps from the 80's? I can buy a heat pump with COP of 6.6 @ +7 degrees C when I want to keep an indoor temperature at 20 degrees C. While it is freezing cold outside, -15 degrees C, the COP is 4.3. The heat pump is said to deliver decent heat, when it is -30 degrees C outside, but the factory guarantee down to -25. It is a Mitsubishi Electric heat pump I have been reading about. Vidar

Mentor
 Quote by Low-Q Are you guys discussing heat pumps from the 80's? I can buy a heat pump with COP of 6.6 @ +7 degrees C when I want to keep an indoor temperature at 20 degrees C.
No, we are talking about ideal heat pumps. An ideal heat pump at those temperatures would have a COP of 22.6.

 Quote by Low-Q While it is freezing cold outside, -15 degrees C, the COP is 4.3.
The ideal would be 8.4.

We are giving the idea the complete "benefit of the doubt" and talking about using theoretically perfect heat pumps. And even then it takes all of the work from the ideal heat engine to run the ideal heat pump.

 Quote by DaleSpam That is a reasonable objection. Rather than getting a specific set of numbers to exactly match, we can solve for the general system and just leave it in terms of variables that you can plug in. Using the same notation as in post 14, the efficiency of an ideal heat engine is again: $$\frac{W_E}{Q_{HE}}=1-\frac{T_{CE}}{T_{HE}}$$ (...)
I'm sorry but personally I think these kinds of calculations are a waste of time and entirely meaningless, irrelevant, misleading and frankly ridiculous.

As I said before, your using Carnot efficiency on a scale of absolute zero. Its nonsensical.

We are talking about an engine running on ambient heat. "Free" heat all around us in the air. A virtually unlimited supply that is renewed on a daily basis by the sun and will never run out until the sun itself burns out.

Lets take a "real world" example. Not really "REAL" just using real numbers in a hypothetical situation.

Lets say I build an experimental engine that runs on hot air and it works.

I set up hot air solar panels to help concentrate the heat for the engine. I even set up a heat pump along with the solar panels to add more heat on cloudy days.

The engine takes off. It draws in hot air and blows out cold air and generates electricity. I even set up duct work to use the cold air from the engine to air condition my house.

Day after day goes by and it keeps running day and night. I notice that the heat pump is redundant as it never goes on. It wasn't needed.

I'm sitting pretty. All the free energy I can use.

One day the wind blows my solar collector over. I didn't even notice because the engine kept running just the same without it. Still producing electricity. But now the air conditioning is way too cold.

Earlier I measured the heat in the solar panels at an average of 200 degrees Fahrenheit.

The engine was always utilizing 100% of all the heat the solar panels could produce and sometimes then some. Exhausting cold air into the house at a comfortable 60 degree Fahrenheit, even on balmy summer days.

Now with the solar panels destroyed I take new measurements. The engine is now drawing in 60 degree Fahrenheit ambient air directly. I'm shocked to see that the exhaust is now negative 30 degrees Fahrenheit. Thirty degrees below zero! No wonder the house was getting so damn cold.

What is the theoretical "Carnot Efficiency" of this little miracle engine ?

Unfortunately, I know, the thing didn't cool the air all the way down to absolute zero and the whole apparatus didn't disappear reducing itself to a single Bose-Einstein Condensate particle in the process so it can't be 100% efficient at removing every last drop of heat so that the exhaust air was colder than the dark side of the moon or deep interstellar space.

I'm a little fuzzy on the math and the conversion between Fahrenheit and Kelvin and all that technical stuff.

Based on the data. The "Th" and "Tc". all that complicated math stuff that's all Greek to me anyway.

What's the Carnot efficiency of this engine before and after the destruction of my solar collector ?

Mentor
 Quote by Tom Booth I'm sorry but personally I think these kinds of calculations are a waste of time and entirely meaningless, irrelevant, misleading and frankly ridiculous.
The calculations are a correct analysis of the scenario. The fact that you don't like the answer doesn't make a bit of difference.

 Quote by Tom Booth As I said before, your using Carnot efficiency on a scale of absolute zero. Its nonsensical.
Then please post a scientific reference that says that you can use another temperature scale for these equations.

 Quote by Tom Booth We are talking about an engine running on ambient heat. "Free" heat all around us in the air.
The thing is that heat engines don't run on heat. They run on differences in heat.

 Quote by Tom Booth I'm a little fuzzy on the math and the conversion between Fahrenheit and Kelvin and all that technical stuff. Based on the data. The "Th" and "Tc". all that complicated math stuff that's all Greek to me anyway.
So I gathered.

 Quote by DaleSpam The calculations are a correct analysis of the scenario. The fact that you don't like the answer doesn't make a bit of difference. Then please post a scientific reference that says that you can use another temperature scale for these equations. The thing is that heat engines don't run on heat. They run on differences in heat. So I gathered.

 Oh, I didn't realize that. I thought that the COP was for the whole system, which of course includes all of those other fans and parts that you mention for the work.
Not totally without justification the heat pump manufacturer will say that he does not (necessarily) manufacture the distribution parts.

An air to water system produces water at a stated temp and COP. What you do with it is then your business.

To transfer heat from the ouside air to the heat pump fluid requires the cold reservoir in the heat exchanger to be be colder than the outside air.
To produce warm air at 20°C require the hot reservoir in the indoor heat exchanger to be warmer than 20°C

Thus neither of these figures can be used for COP calculations.

When making comparisons it should be noted in most cases the work input is performed by electricity.
This is more expensive per joule than some other forms of heating.
So the heat output that derived from the work input will be at higher cost.

 Quote by DaleSpam The calculations are a correct analysis of the scenario. The fact that you don't like the answer doesn't make a bit of difference.
I like the answers fine. If it's understood what it means. most people don't so its misleading.

 Then please post a scientific reference that says that you can use another temperature scale for these equations.
It isn't the scale. It's that actual utilization of supplied heat to such a heat engine in practical terms. The 100% of all the heat collected by the solar panels for example is something different from ALL THE HEAT down to absolute zero.

Carnot efficiency does not describe practical utilization of the supplied heat, it describes how much of all the heat. ALL of it. That means if you are talking about a hot air engine running on heat in the air, the engine has to cool the air to absolute zero to be 100% "efficient". The engine could be working great making all the energy anyone could ever need but the Carnot efficiency numbers would make it look like a dog. A waste of time.

 The thing is that heat engines don't run on heat. They run on differences in heat.
We can debate that issue another time. I don't exactly disagree. Let's just say this engine, based on Tesla's concept can evolve its own temperature difference. an ingenious combined heat engine slash cooling system. Let's not worry about the working principle, lets just see some numbers.
 It is not too unlike this little combined heat engine and cooling system, just scaled up a bit. Different kind of heat engine. Different kind of cooling method. There is no temperature difference until the engine is started, afterwards it maintains its own temperature difference. This little bird is a heat engine running on ambient heat and evaporative cooling. Evaporative cooling is also driven by ambient heat. So energy from ambient heat in the air is being utilized to produce both energy output and supply cooling. Until the engine starts, everything is the same temperature. http://www.youtube.com/watch?v=Rq3K6Ma0wIU This is a little different, yet similar. http://www.youtube.com/watch?v=ARD3ctp80ac Still running on ambient heat and simple evaporative cooling from a damp piece of paper.
 Mentor A device that uses a heat pump to create a temperature difference to drive a heat engine, achieving a positive energy output is a violation of the 2nd law of thermodynamics: a type 2 perpetual motion machine. We don't do perpetual motion machines here. Thread locked.