# Overall efficiency of cascaded reverse Carnot engine & a Rankine turbine

• TMT
In summary, the reverse Carnot engine achieves a higher efficiency when the heat is rejected to the steam boiler and then collected from the ambient.
TMT
Assuming a reverse Carnot engine (Heat Pump Achieve 353K after compression and 253K after evaporation) is cascaded to Rankine turbine such that heat rejection to a steam boiler and heat is collected from a thermally isolated compartment first (used as condensation by cooling) then from ambient (with 318K temperature) via Heat Exchangers
As far as my knowledge COP(Heating)=QHot/W
=TSink/TSink-TSource
where
TSink=353K Heat rejected to Steam boiler (After compression)
TSource=318K Heat collected from ambient
COP(Heating)=318/(353-318) = 318/40 = 7.95
Again for rankine turbine efficiency = 1 - TCondensed/TVapor
where
TCondensed =253K
TVapor=353K
Turbine efficiency = 1 - 253/353 = 0.283

if Heat pump driven by N Kw
Heat pump deliver N x 7.95 Kw
Turbine can convert this with %28.3 efficiency
Turbine work out = N x 7.95 x 0.238 = N x 2.252 Kw
Now overall system efficiency = EnergyOutput/EnergyInput
= N x 2.252 Kw/N Kw
= 2.252
Achieved Efficiency > 1

Am I wrong? ( excess energy taken from environment not crated from absent)

TMT said:
Assuming a reverse Carnot engine (Heat Pump Achieve 353K after compression and 253K after evaporation) is cascaded to Rankine turbine such that heat rejection to a steam boiler and heat is collected from a thermally isolated compartment first (used as condensation by cooling) then from ambient (with 318K temperature) via Heat Exchangers
As far as my knowledge COP(Heating)=QHot/W
=TSink/TSink-TSource
where
TSink=353K Heat rejected to Steam boiler (After compression)
TSource=318K Heat collected from ambient
COP(Heating)=318/(353-318) = 318/40 = 7.95
Again for rankine turbine efficiency = 1 - TCondensed/TVapor
where
TCondensed =253K
TVapor=353K
Turbine efficiency = 1 - 253/353 = 0.283

if Heat pump driven by N Kw
Heat pump deliver N x 7.95 Kw
Turbine can convert this with %28.3 efficiency
Turbine work out = N x 7.95 x 0.238 = N x 2.252 Kw
Now overall system efficiency = EnergyOutput/EnergyInput
= N x 2.252 Kw/N Kw
= 2.252
Achieved Efficiency > 1

Am I wrong? ( excess energy taken from environment not crated from absent)
Would it be correct to change COP ? to 9.085 and how will that change your example ?
I for one could use some picture examples,

Thank you for sharp eyes 353-318 is 35; not 40 of course
COP became 9.085 as you say
"Turbine work out = N x 7.95 x 0.238 = N x 2.252 Kw" also wrong it must be
"Turbine work out = N x 7.95 x 0.283 = N x 2.252 Kw"
than, overall efficiency became (9.085 x 0.283) => 2.57
which is still > 1
RonL said:
I for one could use some picture examples,
Sorry, I didn't get what you mean?

TMT said:
Thank you for sharp eyes 353-318 is 35; not 40 of course
COP became 9.085 as you say
"Turbine work out = N x 7.95 x 0.238 = N x 2.252 Kw" also wrong it must be
"Turbine work out = N x 7.95 x 0.283 = N x 2.252 Kw"
than, overall efficiency became (9.085 x 0.283) => 2.57
which is still > 1

Sorry, I didn't get what you mean?
Please disregard my question I have never been a good student and now my recall is slowing down very much.
I hope you get some other comments and I'll be working my mind a little as to where all the exchanges take place and what connects them together.
I'll keep looking in.

While it is impossible to convert heat 100% to work, you can use a certain amount of heat, to move an even larger amount of heat somewhere else. It all depends on the relative temperature differences, and the degree of irreversibility.

If you burn a fuel, produce a high-temperature flame (say 2000 kelvins), and transfer all the heat to the surrounding room (at 300 kelvins), the efficiency of supplying heat is 1. However, there is a huge irreversibility in transferring heat across a 1700 kelvin temperature difference. Irreversibility is always a sign that things could have been done more efficiently - an opportunity was missed. And entropy was increased.

On the other hand, if I use a Carnot engine (T_h=2000, T_L = 250 kelvins) to drive a Carnot heat pump (T_h=300, T_L = 275 kelvins), there is no irreversibility, because any heat transfer is accomplished across an infinitesimal temperature difference. The engine efficiency is 0.875, and the heat pump COP is 12.00. So you are able to move 10.5 times as much heat with the heat pump as is supplied to the Carnot engine. Everything being reversible, entropy was not increased. The high-temperature heat was used to full advantage, for the situation.

Cengel and Boles would say that high temperature heat has a higher "quality" than low temperature heat. In this case, the high temperature (high quality) heat supply enables you to efficiently run a Carnot engine (rather than "dumping" it into a low-temperature room), and then run a heat pump very effectively, because it only has to transfer heat from a reservoir 25 kelvins below the room temperature.

russ_watters
First of all thank you for contribution, I confess that I'm not good with thermodynamics.I don't know how irreversibility is calculated.
Randy Beikmann said:
On the other hand, if I use a Carnot engine (T_h=2000, T_L = 250 kelvins) to drive a Carnot heat pump (T_h=300, T_L = 275 kelvins), there is no irreversibility, because any heat transfer is accomplished across an infinitesimal temperature difference. The engine efficiency is 0.875, and the heat pump COP is 12.00.
calculation with your example almost same values with my calculation,
efficiency:1-275/200=0.8625
COP.: 275/(300-275)=11
I'm not sure if your example is appropriate with my concern.
if I don't misunderstood you drive Carnot heat pump with Carnot heat engine in your example But I drive Carnot heat engine with heat supplied by Carnot heat pump
(Notice pumping heat from environment will increases entropy in our concerned system)

Than I'm looking overall efficiency of such setup is worthy or not. The work driven from Carnot heat engine is greater than work supplied to Carnot heat pump.which is surprising me.

As I admitted my thermodynamic knowledge is very poor. I strongly disbelief the result. in first glance it is against to1st law (energy can not be created from noting but efficiency > 1 is saying more amount of energy produced from supplied amount.) but noticing more energy taken from environment to our system and must be count as input energy than efficiency drop down under 1 which is comfortable with my learnings.

Actually my target was cooling the environment as a solution for global warming. which yield a question that "how get rid of the energy gathered from environment".
If I can convert his energy into electricity, than I can easily transmit that energy further places or use it in some other work requirements

This process need a work. Organic rankine cycle turbine will be best of all (since heat pump using environment as heat source, can not build up high temperature but 353K can be achievable.

With this temperature and organic working fluid we can get pressure above 16 Bar. With this pressure we can run turbine and generate electricity.

Is such a kind of setup is worthy or not? is derivable energy greater than supplied energy? then I looked for overall efficiency and found > 1

I need to confirm if my approach is violating any thermodynamic law or not before continuing further.

I will appreciate any contribution

if it is not possible workout with this setup; I like to see violated rules to correct my thermodynamic knowledge​

TMT said:
Actually my target was cooling the environment as a solution for global warming. which yield a question that "how get rid of the energy gathered from environment".
If I can convert his energy into electricity, than I can easily transmit that energy further places or use it in some other work requirements

If this is your end goal, we can forget everything said and jump to the conclusion. Heat pumps can only transfer heat from one reservoir to another. Since your low- and high-temperature reservoirs are both on our planet (and together make an isolated system), this in itself will not change the temperature of the earth. In fact, since real heat engines and heat pumps will never be as efficient as their Carnot versions (which are impossible to build or run), the heat supplied to the engine will require a fuel input, and the overall process will actually increase the Earth's temperature.

To cool the earth, you would have to transfer heat to outer space, which can only be done easily by radiation heat transfer. You could theoretically run a laser and aim it out to space. If you used solar cells to power the laser, your system would send a fraction of the sunlight that hits the cells back out to space. But I suppose you could do that more efficiently with a mirror. And either way, air traffic control and pilots wouldn't be too happy with you!

russ_watters
Randy Beikmann said:
To cool the earth, you would have to transfer heat to outer space,
If you accept Heat is a form of energy, and if you reduce energy content of a hot body, its temperature will drop. it is not necessary establish a contact with colder body to reduce temperature . if we send some atmospheric energy to outer space in some way Earth's temperature will decrease.

I'm aware that Heat pumps can only transfer heat from one reservoir to another.
We are also transferring heat from ambient to a thermally isolated and closed system.
I think you missed some essential points in setup. Yes both reservoirs are on our planet but one of reservoir is thermally isolated which creates temperature differences between reservoirs. (Actually we are talking for three reservoir. )
1. Heat source where heat pump absorbs heat (Ambient; unbounded reservoir T_H);
2. Thermally isolated pressure vessel (where heat delivered & vapor generated Heat sink for Heat pump T_L; also Heat source for heat engine T_H);
3. Thermally isolated pressure vessel used for condensation process by cooling (Heat sink for heat engine T_L)
System briefly explained as heat in first source collected in second source and work done between second and third source
Since heat transferred from ambient to a closed system. (Even amount is very very tiny) ambient temperature will drop .and temperature will increase in closed system.(hence more total energy amount in fixed volume)
If we release this energy back to environment you will be right temperature of Earth will not changed.
But hypothetically assume this closed system send to space what will be happened?
The Earth temperature will stay as before or reduced? (even amount is very very very tiny)
We convert this heat energy first to electricity than create electromagnetic wave and transmit to space. if you accept E.M. wave (like microwave) is carrying energy and penetrated through atmosphere then will sink in deep of space.
Since this energy taken from atmosphere globe going to cool down.
if you use fuel to create E.M. wave you are right some how temperature of Earth will increase.
Since using fuel you are creating extra heat and it is not possible to convert all this heat to E.M. wave some heat energy will stay on earth.
Transmitting this energy to space and cooling Earth with this is not primary issue
can we convert atmospheric low grade heat energy to electricity? is our main concern

TMT said:
If you accept Heat is a form of energy, and if you reduce energy content of a hot body, its temperature will drop. it is not necessary establish a contact with colder body to reduce temperature . if we send some atmospheric energy to outer space in some way Earth's temperature will decrease.

I'm aware that Heat pumps can only transfer heat from one reservoir to another.
We are also transferring heat from ambient to a thermally isolated and closed system.
I think you missed some essential points in setup. Yes both reservoirs are on our planet but one of reservoir is thermally isolated which creates temperature differences between reservoirs. (Actually we are talking for three reservoir. )
1. Heat source where heat pump absorbs heat (Ambient; unbounded reservoir T_H);
2. Thermally isolated pressure vessel (where heat delivered & vapor generated Heat sink for Heat pump T_L; also Heat source for heat engine T_H);
3. Thermally isolated pressure vessel used for condensation process by cooling (Heat sink for heat engine T_L)
System briefly explained as heat in first source collected in second source and work done between second and third source
Since heat transferred from ambient to a closed system. (Even amount is very very tiny) ambient temperature will drop .and temperature will increase in closed system.(hence more total energy amount in fixed volume)
If we release this energy back to environment you will be right temperature of Earth will not changed.
But hypothetically assume this closed system send to space what will be happened?
The Earth temperature will stay as before or reduced? (even amount is very very very tiny)
We convert this heat energy first to electricity than create electromagnetic wave and transmit to space. if you accept E.M. wave (like microwave) is carrying energy and penetrated through atmosphere then will sink in deep of space.
Since this energy taken from atmosphere globe going to cool down.
if you use fuel to create E.M. wave you are right some how temperature of Earth will increase.
Since using fuel you are creating extra heat and it is not possible to convert all this heat to E.M. wave some heat energy will stay on earth.
Transmitting this energy to space and cooling Earth with this is not primary issue
can we convert atmospheric low grade heat energy to electricity? is our main concern

The way you are describing it, reservoirs #2 and #3 are not actually isolated systems. "Isolated system" means that no energy can cross the boundary, and you are transferring heat across the boundaries of both.

In the abstract, you could transfer heat into a thermally insulated pressure vessel, and trap the heat there. But it would need to be an infinite reservoir (or the temperature would soon get very high, killing the COP), and the insulation would need to be perfect. Of course a real system couldn't do this.

Radiating to outer space is the only possible way I see to do this in reality.

You are right #2 and #3 are not actually isolated system. But they behave in this manner for a specific time frame
Randy Beikmann said:
In the abstract, you could transfer heat into a thermally insulated pressure vessel, and trap the heat there. But it would need to be an infinite reservoir (or the temperature would soon get very high, killing the COP)
you are right again but if you consider heat engine is also driving out some of this energy pressure vessel can't build the situation you described. Of course you are right again in real word we can't create a perfect isolation. But say that it is not insulator it is retarder, holding situation for a while. whenever some energy converted to work and this work is grater than externally supplied energy to run heat pump our mission will be completed. we are not claim to perform this conversion without leaks.and losses As 3rd law saying it is impossible prevent loss in any of process dealing with heat. Therefore complete and perfect isolation is impossible.

Randy Beikmann said:
The way you are describing it, reservoirs #2 and #3 are not actually isolated systems. "Isolated system" means that no energy can cross the boundary, and you are transferring heat across the boundaries of both.

In the abstract, you could transfer heat into a thermally insulated pressure vessel, and trap the heat there. But it would need to be an infinite reservoir (or the temperature would soon get very high, killing the COP), and the insulation would need to be perfect. Of course a real system couldn't do this.

Radiating to outer space is the only possible way I see to do this in reality.
In view of radiating to outer space not being a practical option, would there be a point of time separation, that would make some machine transforming heat to mechanical power a worthwhile effort ?

The easiest way to radiate heat back into space is just to reflect radiation from the sun!

Randy Beikmann
russ_watters said:
The easiest way to radiate heat back into space is just to reflect radiation from the sun!
this only prevents further increase of temperature of Earth not cool down it!
But if we could convert some energy (heat) to electricity then we can either store it in chemical compound or convert to potential energy like pumping water to higher altitudes or create E.M. wave and transmit to space. OR UTILISE ON EARTH FOR WORK than man kind will not consume fossil fuels any more then energy content of atmosphere stay constant (almost) since the heat produced from work recycled.

TMT said:
this only prevents further increase of temperature of Earth not cool down it!
No, the Earth heats every day and cools every night. The balance of radiation in and out is what makes Earth's average temperature what it is.
OR UTILISE ON EARTH FOR WORK than man kind will not consume fossil fuels any more then energy content of atmosphere stay constant (almost) since the heat produced from work recycled.
For the good reasons given by others, extracting electricity from environmental heat is very difficult/inefficient. Better bet would be to use solar power...but installing enough solar to get us off fossil fuels is a tall order in itself.

...also should point out that nuclear power adds heat to the atmosphere too...

russ_watters said:
No, the Earth heats every day and cools every night. The balance of radiation in and out is what makes Earth's average temperature what it is.
I'm aware it! the greenhouse effect work as blanket and preventing heating and cooling process rates mismatch. Of course Sun is main heat buildup object and reflection of radiation coming from sun will gratly help balancing heating and cooling process rate But mankind need energy and population growing in geometric speed that increase total energy demand every day (also every living object is another heat generator mankind plus live stocks needed as food is establish considerable heat source) how will you reflect this unbalancing heat like as sun radiation
we must figure out the process to convert the atmospheric heat to electricity regardless its difficulties
In this thread I try to demonstrate it may be possible with a cascaded heat pump with a rankine turbine As I admit my thermodynamic knowledge is limited and I'm trying to learn by myself. If any thermodynamic rule against to my setup I will be very glad to see it.
I am not (even think to) racing my knowledge with some one else but I must see the said at here is matching with my learning
But anyway I appreciate all of guy contribute to this thread at least they are enlightening me

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## 1. What is a cascaded reverse Carnot engine and a Rankine turbine?

A cascaded reverse Carnot engine is a system that consists of two or more Carnot engines connected in series, with the waste heat from the first engine being used as the heat input for the second engine. A Rankine turbine is a type of steam turbine that uses the Rankine cycle to convert thermal energy into mechanical work.

## 2. How does a cascaded reverse Carnot engine and a Rankine turbine improve overall efficiency?

By connecting multiple Carnot engines in series, the waste heat from one engine is utilized as the heat input for the next engine, increasing the overall efficiency of the system. The Rankine turbine also improves efficiency by utilizing the Rankine cycle, which allows for better heat transfer and energy conversion.

## 3. What factors affect the overall efficiency of a cascaded reverse Carnot engine and a Rankine turbine?

The overall efficiency of this system is affected by several factors, including the temperature of the heat source, the temperature of the heat sink, and the efficiency of the individual engines and turbine. Other factors such as the design and maintenance of the system can also impact efficiency.

## 4. How is the efficiency of a cascaded reverse Carnot engine and a Rankine turbine calculated?

The efficiency of this system can be calculated by taking the product of the individual efficiencies of the Carnot engines and the turbine. The efficiency of a Carnot engine is equal to 1 - (Tc/Th), where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir. The efficiency of a Rankine turbine can be calculated using the Rankine cycle efficiency equation.

## 5. What are some real-world applications of cascaded reverse Carnot engines and Rankine turbines?

These systems are commonly used in power generation, such as in steam power plants. They are also used in refrigeration and air conditioning systems, where waste heat from the condenser can be used as the heat input for the evaporator. Additionally, they can be used in combined heat and power systems to simultaneously generate electricity and heat for industrial or residential use.

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