Re-Visiting Carnot’s Theorem

The proof by contradiction used in physics textbooks is one of those arguments that appear surprising, then self-evident, then deceptive in its simplicity. You – or maybe only: I – cannot resist turning it over and over in your head again, viewing it from different angles.

tl;dr: I just wanted to introduce the time-honored tradition of ASCII text art images to illustrate Carnot’s Theorem, but this post got out of hand when I mulled about how to  refute an erroneous counter-argument. As there are still research papers being written about Carnot’s efficiency I feel vindicated for writing a really long post though.

Carnot‘s arguments prove that there is a maximum efficiency of a thermodynamic heat engine – a machine that turns heat into mechanical energy. He gives the maximum value by evaluating one specific, idealized process, and then proves that a machine with higher efficiency would give rise to a paradox. The engine uses part of the heat available in a large, hot reservoir of heat and turns it into mechanical work and waste heat – the latter dumped to a colder ‘environment’ in a 4-step process. (Note that while our modern reformulation of the proof by contradiction refers to the Second Law of Thermodynamics, Carnot’s initial version was based on the caloric theory.)

The efficiency of such an engine η – mechanical energy per cycle over input heat energy – only depends on the two temperatures (More details and references here):

\eta_\text{carnot} = \frac {T_1-T_2}{T_1}

These are absolute temperatures in Kelvin; this universal efficiency can be used to define what we mean by absolute temperature.

I am going to use ‘nice’ numbers. To make ηcarnot equal to 1/2, the hot temperature
T1 = 273° = 546 K, and the colder ‘environment’ has T2 = 0°C = 273 K.

If this machine is run in reverse, it uses mechanical input energy to ‘pump’ energy from the cold environment to the hot reservoir – it is a heat pump using the ambient reservoir as a heat source. The Coefficient of Performance (COP, ε) of the heat pump is heat output over mechanical input, the inverse of the efficiency of the corresponding engine. εcarnot is 2 for the temperatures given above.

If we combine two such perfect machines – an engine and a heat pump, both connected to the hot space and to the cold environment, their effects cancel out: The mechanical energy released by the engine drives the heat pump which ‘pumps back’ the same amount of energy.

In the ASCII images energies are translated to arrows, and the number of parallel arrows indicates the amount of energy per cycle (or power). For each device, the number or arrows flowing in and out is the same; energy is always conserved. I am viewing this from the heat pump’s perspective, so I call the cold environment the source, and the hot environment room.

Neither of the heat reservoirs are heated or cooled in this ideal case as the same amount of energy flows from and to each of the heat reservoirs:

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
           | | | |                         | | | |
           v v v v                         ^ ^ ^ ^
           | | | |                         | | | |
       |------------|                 |---------------|
       |   Engine   |->->->->->->->->-|   Heat pump   |
       |  Eta = 1/2 |->->->->->->->->-| COP=2 Eta=1/2 |
       |------------|                 |---------------|
             | |                             | |
             v v                             ^ ^
             | |                             | |
|----------------------------------------------------------| 
|        Cold source at temperature T_2 = 0°C = 273 K      | 
|----------------------------------------------------------|

If either of the two machines works less than perfectly and in tandem with a perfect machine, anything is still fine:

If the engine is far less than perfect and has an efficiency of only 1/4 – while the heat pump still works perfectly – more of the engine’s heat energy input is now converted to waste heat and diverted to the environment:

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
           | | | |                           | |  
           v v v v                           ^ ^  
           | | | |                           | |  
       |------------|                 |---------------|
       |   Engine   |->->->->->->->->-|   Heat pump   |
       |  Eta = 1/4 |                 | COP=2 Eta=1/2 |
       |------------|                 |---------------|
            | | |                             |
            v v v                             ^
            | | |                             |
|----------------------------------------------------------| 
|        Cold source at temperature T_2 = 0°C = 273 K      | 
|----------------------------------------------------------|

Now two net units of energy flow from the hot room to the environment (summing up the arrows to and from the devices):

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
                              | |                                
                              v v                                
                              | | 
                     |------------------|
                     |   Combination:   |
                     | Eta=1/4 COP=1/2  |
                     |------------------|                            
                              | |                              
                              v v                              
                              | |                             
|----------------------------------------------------------| 
|        Cold source at temperature T_2 = 0°C = 273 K      | 
|----------------------------------------------------------|

Using a real-live heat pump with a COP of 3/2 (< 2) together with a perfect engine …

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
          | | | |                             | | | 
          v v v v                             ^ ^ ^ 
          | | | |                             | | |
       |------------|                 |-----------------|
       |   Engine   |->->->->->->->->-|    Heat pump    |
       |  Eta = 1/2 |->->->->->->->->-|     COP=3/2     |
       |------------|                 |-----------------|
            | |                                 |
            v v                                 ^
            | |                                 |
|----------------------------------------------------------| 
|        Cold source at temperature T_2 = 0°C = 273 K      | 
|----------------------------------------------------------|

… causes again a non-paradoxical net flow of one unit of energy from the room to the environment.

In the most extreme case  a poor heat pump (not worth this name) with a COP of 1 just translates mechanical energy into heat energy 1:1. This is a resistive heating element, a heating rod, and net heat fortunately flows from hot to cold without paradoxes:

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
            | |                                |   
            v v                                ^   
            | |                                |   
       |------------|                 |-----------------|
       |   Engine   |->->->->->->->->-|   'Heat pump'   |
       |  Eta = 1/2 |                 |     COP = 1     |
       |------------|                 |-----------------|
             |                                 
             v                                 
             |                                 
|----------------------------------------------------------| 
|        Cold source at temperature T_2 = 0°C = 273 K      | 
|----------------------------------------------------------|

The textbook paradox in encountered, when an ideal heat pump is combined with an allegedly better-than-possible engine, e.g. one with an efficiency:

ηengine = 2/3 (> ηcarnot = 1/2)

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
           | | |                           | | | |
           v v v                           ^ ^ ^ ^
           | | |                           | | | |
       |------------|                 |---------------|
       |   Engine   |->->->->->->->->-|   Heat pump   |
       |  Eta = 2/3 |->->->->->->->->-| COP=2 Eta=1/2 |
       |------------|                 |---------------|
             |                               | |
             v                               ^ ^
             |                               | |
|----------------------------------------------------------| 
|        Cold source at temperature T_2 = 0°C = 273 K      | 
|----------------------------------------------------------|

The net effect / heat flow is then:

|----------------------------------------------------------|
|        Hot room at temperature T_1 = 273°C = 546 K       | 
|----------------------------------------------------------| 
                             | 
                             ^ 
                             | 
                   |------------------| 
                   |   Combination:   | 
                   | Eta=3/2; COP=1/2 | 
                   |------------------| 
                             | 
                             ^ 
                             | 
|----------------------------------------------------------| 
|       Cold source at temperature T_2 = 0°C = 273 K       | 
|----------------------------------------------------------|

One unit of heat would flow from the environment to the room, from the colder to the warmer body without any other change being made to the system. The combination of these machines would violate the Second Law of Thermodynamics; it is a Perpetuum Mobile of the Second Kind.

If the heat pump has a higher COP than the inverse of the perfect engine’s efficiency, a similar paradox arises, and again one unit of heat flows in the forbidden direction:

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
            | |                             | | |
            v v                             ^ ^ ^
            | |                             | | |
       |------------|                 |---------------|
       |   Engine   |->->->->->->->->-|   Heat pump   |
       |  Eta = 1/2 |                 |    COP = 3    |
       |------------|                 |---------------|
             |                               | |
             v                               ^ ^
             |                               | |
|----------------------------------------------------------| 
|        Cold source at temperature T_2 = 0°C = 273 K      | 
|----------------------------------------------------------|

A weird question: Can’t we circumvent the paradox if we pair the impossible superior engine with a poor heat pump?

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
           | | |                             | |  
           v v v                             ^ ^  
           | | |                             | |  
       |------------|                 |---------------|
       |   Engine   |->->->->->->->->-|   Heat pump   |
       |  Eta = 2/3 |->->->->->->->->-|    COP = 1    |
       |------------|                 |---------------|
             |                                
             v                                
             |                                
|----------------------------------------------------------| 
|        Cold source at temperature T_2 = 0°C = 273 K      | 
|----------------------------------------------------------

Indeed: If the COP of the heat pump (= 1) is smaller than the inverse of the engine’s efficiency (3/2), there will be no apparent violation of the Second Law – one unit of net heat flows from hot to cold.

An engine with low efficiency 1/4 would ‘fix’ the second paradox involving the better-than-perfect heat pump:

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
           | | | |                          | | |
           v v v v                          ^ ^ ^
           | | | |                          | | |
       |------------|                 |---------------|
       |   Engine   |->->->->->->->->-|   Heat pump   |
       |  Eta = 1/4 |                 |     COP=3     |
       |------------|                 |---------------|
            | | |                            | |
            v v v                            ^ ^
            | | |                            | |
|----------------------------------------------------------| 
|        Cold source at temperature T_2 = 0°C = 273 K      | 
|----------------------------------------------------------|

But we cannot combine heat pumps and engines at will, just to circumvent the paradox – one counter-example is sufficient: Any realistic engine combined with any realistic heat pump – plus all combinations of those machines with ‘worse’ ones – have to result in net flow from hot to cold …

The Second Law identifies such ‘sets’ of engines and heat pumps that will all work together nicely. It’s easier to see this when all examples are condensed into one formula:

The heat extracted in total from the hot room – Q1 –  is the difference of heat used by the engine and heat delivered by the heat pump, both of which are defined in relation to the same mechanical work W:

Q_1 = W\left (\frac{1}{\eta_\text{engine}}-\varepsilon_\text{heatpump}\right)

This is also automatically equal to Qas another quick calculation shows or by just considering that energy is conserved: Some heat goes into the combination of the two machines, part of it – W – flows internally from the engine to the heat pump. But no part of the input Q1 can be lost, so the output of the combined machine has to match the input. Energy ‘losses’ such as energy due to friction will flow to either of the heat reservoirs: If an engine is less-then-perfect, more heat will be wasted to the environment; and if the heat pump is less-than-perfect a greater part of mechanical energy will be translated to heat only 1:1. You might be even lucky: Some part of heat generated by friction might end up in the hot room.

As Q1 has to be > 0 according to the Second Low, the performance numbers have to related by this inequality:

\frac{1}{\eta_\text{engine}}\geq\varepsilon_\text{heatpump}

The equal sign is true if the effects of the two machines just cancel each other.

If we start from a combination of two perfect machines (ηengine = 1/2 = 1/εheatpump) and increase either ηengine or εheatpump, this condition would be violated and heat would flow from cold to hot without efforts.

But also an engine with efficiency = 1 would work happily with the worst heat pump with COP = 1. No paradox would arise at first glance  – as 1/1 >= 1:

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
             |                                |   
             v                                ^   
             |                                |   
       |------------|                 |-----------------|
       |   Engine   |->->->->->->->->-|   'Heat pump'   |
       |   Eta = 1  |                 |      COP=1      |
       |------------|                 |-----------------|
                                               
                                               
                                               
|----------------------------------------------------------| 
|        Cold source at temperature T_2 = 0°C = 273 K      | 
|----------------------------------------------------------|

What’s wrong here?

Because of conservation of energy ε is always greater equal 1; so the set of valid combinations of machines all consistent with each other is defined by:

\frac{1}{\eta_\text{engine}}\geq\varepsilon_\text{heatpump}\geq1

… for all efficiencies η and COPs / ε of machines in a valid set. The combination η = ε = 1 is still not ruled out immediately.

But if the alleged best engine (in a ‘set’) would have an efficiency of 1, then the alleged best heat pump would have an Coefficient of Performance of only 1 – and this is actually the only heat pump possible as ε has to be both lower equal and greater equal than 1. It cannot get better without creating paradoxes!

If one real-live heat pump is found that is just slightly better than a heating rod – say
ε = 1,1 – then performance numbers for the set of consisent, non-paradoxical machines need to fulfill:

\eta_\text{engine}\leq\eta_\text{best engine}

and

\varepsilon_\text{heatpump}\leq\varepsilon_\text{best heatpump}

… in addition to the inequality relating η and ε.

If ε = 1,1 is a candidate for the best heat pump, a set of valid machines would comprise:

  • All heat pumps with ε between 1 and 1,1 (as per limits on ε)
  • All engines with η between 0 and 0,9 (as per inequality following the Second Law plus limit on η).

Consistent sets of machines are thus given by a stronger condition – by adding a limit for both efficiency and COP ‘in between’:

\frac{1}{\eta_\text{engine}}\geq\text{Some Number}\geq\varepsilon_\text{heatpump}\geq1

Carnot has designed a hypothetical ideal heat pump that could have a COP of εcarnot = 1/ηcarnot. It is a limiting case of a reversible machine, but feasible on principle. εcarnot  is thus a valid upper limit for heat pumps, a candidate for Some Number. In order to make this inequality true for all sets of machines (ideal ones plus all worse ones) then 1/ηcarnot = εcarnot also constitutes a limit for engines:

\frac{1}{\eta_\text{engine}}\geq\frac{1}{\eta_\text{carnot}}\geq\varepsilon_\text{heatpump}\geq1

So in order to rule out all paradoxes, Some Number in Between has to be provided for each set of machines. But what defines a set? As machines of totally different making have to work with each other without violating this equality, this number can only be a function of the only parameters characterizing the system – the two temperatures

Carnot’s efficiency is only a function of the temperatures. His hypothetical process is reversible, the machine can work either as a heat pump or an engine. If we could come up with a better process for a reversible heat pump (ε > εcarnot), the machine run in reverse would be an engine with η less than ηcarnot, whereas a ‘better’ engine would lower the upper bound for heat pumps.

If you have found one truly reversible process, both η and ε associated with it are necessarily the upper bounds of performance of the respective machines, so you cannot push Some Number in one direction or the other, and the efficiencies of all reversible engines have to be equal – and thus equal to ηcarnot. The ‘resistive heater’ with ε = 1 is the iconic irreversible device. It will not turn into a perfect engine with η = 1 when ‘run in reverse’.

The seemingly odd thing is that 1/ηcarnot appears like a lower bound for ε at first glance if you just declare ηcarnot an upper bound for corresponding engines and take the inverse, while in practice and according to common sense it is the maximum value for all heat pumps, including irreversible ones. (As a rule of thumb a typical heat pump for space heating has a COP only 50% of 1/ηcarnot.)

But this ‘contradiction’ is yet another way of stating that there is one universal performance indicator of all reversible machines making use of two heat reservoirs: The COP of a hypothetical ‘superior’ reversible heat pump would be at least 1/ηcarnot  … as good as Carnot’s reversible machine, maybe better. But the same is true for the hypothetical superior engine with an efficiency of at least ηcarnot. So the performance numbers of all reversible machines (all in one set, characterized by the two temperatures) have to be exactly the same.

Steam pump / Verkehrt laufende Dampfmaschine

Historical piston compressor (from the time when engines with pistons looked like the ones in textbooks), installed 1878 in the salt mine of Bex, Switzerland. 1943 it was still in operation. Such machines used in salt processing were considered the first heat pumps.

How Does It Work? (The Heat Pump System, That Is)

Over the holidays I stayed away from social media, read quantum physics textbooks instead, and The Chief Engineer and I mulled over the fundamental questions of life, the universe and everything. Such as: How to explain our heat pump system?

Many blog postings were actually answers to questions, and am consolidating all these answers to frequently asked questions again in a list of such answers. However, this list has grown quickly.

An astute reader suggested to create an ‘animation’ of the gradual evolution of the system’s state. As I learned from discussions, one major confusion was related to the role of the solar collector and the fact that you have to factor in the history of the heat source: This is true for every heat pump system that uses a heat source that can be ‘depleted’, in contrast to a flow of ground water at a constant temperature for example. With the latter, the ‘state’ of the system only depends on the current ambient temperature, and you can explain it in a way not too different from pontificating on a wood or gas boiler.

One thing you have to accept though is how a heat pump as such works: I have given up to go into thermodynamical details, and I also think that the refigerator analogy is not helpful. So for this pragmatic introduction a heat pump is just a device that generates heating energy as an output, the input energy being electrical energy and heat energy extracted from a rather cold heat source somewhere near the building. For 8kW heating power you need about 2kW electrical energy and 6kW ambient energy. The ratio of 8kW and 2kW is called the coefficient of performance.

What the typical intro to heat pumps in physics textbooks does not point out is that the ambient heat source actually has to be able to deliver that input energyduring a whole heating season. There is no such thing as the infinite reservoir of energy usually depicted as a large box. Actually, the worse the performance of a heat pump is – the ratio of output heat energy and input electrical energy, the smaller are the demands on the heat source. The Chief Engineer has coined the term The Heat Source Paradox for this!

The lower the temperature of the heat source, the smaller the coefficient of performance is: So if you run an air source heat pump in mid-winter (using a big ventilator) then less energy is extracted from that air source than a geothermal heat pump would extract from ground. But if you build a geothermal heat source that’s too small in relation to a building’s heating demands, you see the same effect: Ground freezes, source temperature decreases, performance decreases, and you need more electrical energy and less ambient energy.

I am harping on the role of the heat source as the whole point of our ‘innovation’ is our special heat source that has two components, both of them being essential: An unglazed solar collector and an underground water / ice tank plus the surrounding ground. The solar collector allows to replenish the energy stored in the tank quickly, even in winter, and the tank is a buffer: When no energy is harvested by the collector at ambient temperatures below 0°C water freezes and releases latent heat. So you can call that an air heat pump with a huge, silent and mainentance-free ‘absorber’ plus a buffer that provides energy for periods of frost and that allows for storing all the energy you don’t need immediately. Ground does provide some energy as well, and I am planning to post about my related simulations. It can be visulized as an extension of the ice / water energy storage into the surroundings. But the active volume or area of ground is smaller than for geothermal systems as most of the ambient energy actually comes from the solar collector: The critical months in our climate are Dec-Jan-Feb: Before and after, the solar collector would be sufficient as the only heat source. In the three ‘ice months’ water is typically frozen in the tank, but even then the solar collector provides for 75-80% of the ambient energy needed to drive the heat pump.

Components are off-the-shelf products, actually rather simple and cheap ones, such as the most stupid, non-smart brine-water heat pump. What is special is 1) the arrangement of the heat exchanger in the water tank and 2) the custom control logic, that is programming of the control unit.

So here is finally the series of images of the system’s state, shown in a gallery and with captions: You can scroll down to see the series embedded in the post, or click on the first image to see an enlarged view and then click through the slide-show.

Information for German readers: This post contains the German version of this slide-show.

An Efficiency Greater Than 1?

No, my next project is not building a Perpetuum Mobile.

Sometimes I mull upon definitions of performance indicators. It seems straight-forward that the efficiency of a wood log or oil burner is smaller than 1 – if combustion is not perfect you will never be able to turn the caloric value into heat, due to various losses and incomplete combustion.

Our solar panels have an ‘efficiency’ or power ratio of about 16,5%. So 16.5% of solar energy are converted to electrical energy which does not seem a lot. However, that number is meaningless without adding economic context as solar energy is free. Higher efficiency would allow for much smaller panels. If efficiency were only 1% and panels were incredibly cheap and I had ample roof spaces I might not care though.

The coefficient of performance of a heat pump is 4-5 which sometimes leaves you with this weird feeling of using odd definitions. Electrical power is ‘multiplied’ by a factor always greater than one. Is that based on crackpottery?

Heat pump.

Our heat pump. (5 connections: 2x heat source – brine, 3x heating water hot water / heating water supply, joint return).

Actually, we are cheating here when considering the ‘input’ – in contrast to the way we view photovoltaic panels: If 1 kW of electrical power is magically converted to 4 kW of heating power, the remaining 3 kW are provided by a cold or lukewarm heat source. Since those are (economically) free, they don’t count. But you might still wonder, why the number is so much higher than 1.

My favorite answer:

There is an absolute minimum temperature, and our typical refrigerators and heat pumps operate well above it.

The efficiency of thermodynamic machines is most often explained by starting with an ideal process using an ideal substance – using a perfect gas as a refrigerant that runs in a closed circuit. (For more details see pointers in the Further Reading section below). The gas would be expanded at a low temperature. This low temperature is constant as heat is transferred from the heat source to the gas. At a higher temperature the gas is compressed and releases heat. The heat released is the sum of the heat taken in at lower temperatures plus the electrical energy fed in to the compressor – so there is no violation of energy conservation. In order to ‘jump’ from the lower to the higher temperature, the gas is compressed – by a compressor run on electrical power – without exchanging heat with the environment. This process is repeating itself again and again, and with every cycle the same heat energy is released at the higher temperature.

In defining the coefficient of performance the energy from the heat source is omitted, in contrast to the electrical energy:

COP = \frac {\text{Heat released at higher temperature per cycle}}{\text{Electrical energy fed into the compressor per cycle}}

The efficiency of a heat pump is the inverse of the efficiency of an ideal engine – the same machine, running in reverse. The engine has an efficiency lower than 1 as expected. Just as the ambient energy fed into the heat pump is ‘free’, the related heat released by the engine to the environment is useless and thus not included in the engine’s ‘output’.

100 1870 (Voitsberg steam power plant)

One of Austria’s last coal power plants – Kraftwerk Voitsberg, retired in 2006 (Florian Probst, Wikimedia). Thermodynamically, this is like ‘a heat pump running in reverse. That’s why I don’t like when a heat pump is said to ‘work like a refrigerator, just in reverse’ (Hinting at: The useful heat provided by the heat pump is equivalent to the waste heat of the refrigerator). If you run the cycle backwards, a heat pump would become sort of a steam power plant.

The calculation (see below) results in a simple expression as the efficiency only depends on temperatures. Naming the higher temperature (heating water) T1 and the temperature of the heat source (‘environment’, our water tank for example) T….

COP = \frac {T_1}{T_1-T_2}

The important thing here is that temperatures have to be calculated in absolute values: 0°C is equal to 273,15 Kelvin, so for a typical heat pump and floor loops the nominator is about 307 K (35°C) whereas the denominator is the difference between both temperature levels – 35°C and 0°C, so 35 K. Thus the theoretical COP is as high as 8,8!

Two silly examples:

  • Would the heat pump operate close to absolute zero, say, trying to pump heat from 5 K to 40 K, the COP would only be
    40 / 35 = 1,14.
  • On the other hand, using the sun as a heat source (6000 K) the COP would be
    6035 / 35 = 172.

So, as heat pump owners we are lucky to live in an environment rather hot compared to absolute zero, on a planet where temperatures don’t vary that much in different places, compared to how far away we are from absolute zero.

__________________________

Further reading:

Richard Feynman has often used unusual approaches and new perspectives when explaining the basics in his legendary Physics Lectures. He introduces (potential) energy at the very beginning of the course drawing on Carnot’s argument, even before he defines force, acceleration, velocity etc. (!) In deriving the efficiency of an ideal thermodynamic engine many chapters later he pictured a funny machine made from rubber bands, but otherwise he follows the classical arguments:

Chapter 44 of Feynman’s Physics Lectures Vol 1, The Laws of Thermodynamics.

For an ideal gas heat energies and mechanical energies are calculated for the four steps of Carnot’s ideal process – based on the Ideal Gas Law. The result is the much more universal efficiency given above. There can’t be any better machine as combining an ideal engine with an ideal heat pump / refrigerator (the same type of machine running in reverse) would violate the second law of thermodynamics – stated as a principle: Heat cannot flow from a colder to a warmer body and be turned into mechanical energy, with the remaining system staying the same.

KarnoyiCikl

Pressure over Volume for Carnot’s process, when using the machine as an engine (running it counter-clockwise it describes a heat pump): AB: Expansion at constant high temperature, BC: Expansion without heat exchange (cooling), CD: Compression at constant low temperature, DA: Compression without heat exhange (gas heats up). (Image: Kara98, Wikimedia).

Feynman stated several times in his lectures that he does not want to teach history of physics or downplayed the importance of learning about history of science a bit (though it seems he was well versed in – as e.g. his efforts to follow Newton’s geometrical prove of Kepler’s Laws showed). For historical background of the evolution of Carnot’s ideas and his legacy see the the definitive resource on classical thermodynamics and its history – Peter Mander’s blog carnotcycle.wordpress.com:

What had puzzled me is once why we accidentally latched onto such a universal law, using just the Ideal Gas Law.The reason is that the Gas Law has the absolute temperature already included. Historically, it did take quite a while until pressure, volume and temperature had been combined in a single equation – see Peter Mander’s excellent article on the historical background of this equation.

Having explained Carnot’s Cycle and efficiency, every course in thermodynamics reveals a deeper explanation: The efficiency of an ideal engine could actually be used as a starting point defining the new scale of temperature.

Temperature scale according to Kelvin (William Thomson)

Carnot engines with different efficiencies due to different lower temperatures. If one of the temperatures is declared the reference temperature, the other can be determined by / defined by the efficiency of the ideal machine (Image: Olivier Cleynen, Wikimedia.)

However, according to the following paper, Carnot did not rigorously prove that his ideal cycle would be the optimum one. But it can be done, applying variational principles – optimizing the process for maximum work done or maximum efficiency:

Carnot Theory: Derivation and Extension, paper by Liqiu Wang

How to Evaluate a Heat Pump’s Performance?

The straight-forward way is to read off two energy values at the end of a period – day, month, or season:

  1. The electrical energy used by the heat pump
  2. and the heating energy delivered.

The Seasonal Performance Factor (SPF) is the ratio of these – the factor the input electrical energy is ‘multiplied with’ to yield heating energy. The difference between these two energies is supplied by the heat source – the underground water tank / ‘cistern’ plus solar collector in our setup.

But there might not be a separate power meter just for the heat pump’s compressor. Fortunately, performance factors can also be evaluated from vendors’ datasheets and measured brine / heating water temperatures:

Datasheets provide the Coefficient of Performance (COP) – the ‘instantaneous’ ratio of heating power and electrical power. The COP decreases with increasing temperature of the heating water, and with decreasing temperature of the source  – the brine circuit immersed in the cold ice / water tank. E.g when heating the water in floor loops to 35°C the COP is a bit greater than 4 if the water in the underground tank is frozen (0°C). The textbook formula based on Carnot’s ideal process for thermodynamic machines is 8,8 for 0°C/35°; realistic COPs are typically by about factor of 2 lower.

Heat Pump Performance, from Datasheets

COPs, eletrical power (input) and heating power (output) of a ‘7 kW’ brine / water heat pump. Temperatures in the legend are heating water infeed temperatures – 35°C as required by floor loops and 50°C for hot water heating.

If you measure the temperature of the brine and the temperature of the heating water every few minutes, you can determine the COP from these diagrams and take averages for days, months, or seasons.

But should PF and average COP actually be the same?

Average power is total energy divided by time, so (with bars denoting averages):

\text{Performance Factor } = \frac {\text{Total Heating Energy } \mathnormal{E_{H}}} {\text{Total Electrical Energy } \mathnormal{E_{E}}} = \frac {\text{Average Heating Power } \mathnormal{\bar{P}_{H}}} {\text{Average Electrical Power }\mathnormal{\bar{P}_{E}} }

On the other hand the average COP is calculated from data taken at many different times. At any point of time t,

\text{Coefficient of Performance(t)} = \frac {\text{Heating Power }P_{H}(t))} {\text{Electrical Power } P_{E}(t))}

Having measured the COP at N times, the average COP is thus:

\overline{COP}(t) = \frac {1}{N} \sum \frac{P_{H}(t)}{P_{E}(t)} = \overline{\frac{P_{H}(t)}{P_{E}(t)}}

\overline{\frac{P_{H}(t)}{P_{E}(t)}} is not necessarily equal to \frac{\overline{P_{H}}}{\overline{P_{E}}}

When is the average of ratios equal to the ratios of the averages?

If electrical power and heating power would fluctuate wildly we would be in trouble. Consider this hypothetical scenario of odd non-physical power readings:

  • PH = 10, PE = 1
  • PH = 2, PE = 20

The ratio of averages is: (10 + 2) / (1 + 20) = 12 / 21 = 0,57
The average of ratios is: (10/1 + 2/20) / 2 = (10 + 0,1) / 2 = 5,05

Quite a difference. Good that typical powers look like this:

Measured heating power, electrical power, COP (heat pump)

Powers measured on 2015-02-20 . Two space heating periods with a COP between 4 and 5, and one heating hot water cycle: the COP gradually decreases as heating water temperature increases.

Powers change only by a fraction of their absolute values – the heat pump is basically ON or OFF.  When these data were taken in February, average daily ambient temperature was between 0°C and 5°C, and per day about 75kWh were used for space heating and hot water. Since heat pump output is constant, daily run times change with heating demands.

Results for the red hot tap water heating cycle:

  • Performance Factor calculated from energies: 3,68
  • Average COP: 3,76.

I wanted to know how much powers are allowed to change without invalidating the average COP method:

Electrical power and heating power rise / fall about linearly, so they can be described by two parameters: Initial powers when the heat pump is turned on, and the slope of the curve or relative change of power within on cycle. The Performance Factor is determined from energies, the areas of trapezoids under the curves. For calculating the COP the ratio needs to be integrated, which results in a not so nice integral.

The important thing is that COP and PF are proportional to the ratio of inital powers and their relative match only depends on the slopes of the heating power and electrical power curves. As long as the relative increase / decrease of those powers is significantly smaller than 1, the difference in performance indicators is just a few percent. In the example curve, the heating energy decreases by 15%, while electrical energy increases by 52% – performance indicators would differ by less than 2%. This small difference is not too sensitive to changes in slopes.

All is well.

Solar collector, spring 2015.

Happily harvesting ambient energy.

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Detailed monthly and seasonal performance data are given in this document.