Can We Create a More Efficient Engine Than the Carnot Engine?

[person_random_normal]

1 - T1/T2 gives efficiency of a carnot engine working between two temperature sources T1 &T2 and we know that, that efficiency is unbeatable but can there be another engine possible serving the same purpose as that of carnot engine working in an entirely different manner and MORE EFFICIENT than the carnot engine correspondingly working ?

 

[EM_Guy]

1 - T1/T2 gives efficiency of a carnot engine working between two temperature sources T1 &T2 and we know that, that efficiency is unbeatable but can there be another engine possible serving the same purpose as that of carnot engine working in an entirely different manner and MORE EFFICIENT than the carnot engine correspondingly working ?

No. As you correctly said, "the efficiency is unbeatable." This is a direct result of the 2nd Law of Thermodynamics. You can not violate the 2nd Law.

 

[johnbbahm]

1 - T1/T2 gives efficiency of a carnot engine working between two temperature sources T1 &T2 and we know that, that efficiency is unbeatable but can there be another engine possible serving the same purpose as that of carnot engine working in an entirely different manner and MORE EFFICIENT than the carnot engine correspondingly working ?

In regards to extracting work from heat, likely not!
Fuel cells may be able to extract a higher percentage of the potential energy from liquid fuel.

 

[person_random_normal]

the heat engines driving our cars follow the carnot cycle ?

 

[EM_Guy]

the heat engines driving our cars follow the carnot cycle ?

No. They aren't that efficient.

 

[person_random_normal]

i mean to ask are engines in our car follow the carnot cycle or they work differently

 

[johnbbahm]

i mean to ask are engines in our car follow the carnot cycle or they work differently

Gasoline and Diesel engines are heat engines and must follow Carnot's rules.

 

[person_random_normal]

WHY

 

[EM_Guy]

Gasoline and Diesel engines are heat engines and must follow Carnot's rules.

Gasoline and diesel engines are not as efficient as Carnot engines.

 

[johnbbahm]

Gasoline and diesel engines are not as efficient as Carnot engines.

I know, but as heat engines must follow his rules,

 

[person_random_normal]

i could not understand you

 

[JorisL]

WHY

Well the Carnot cycle is quasistatic. This means the power at any given time is identically zero.
So while a Carnot engine has a high efficiency, it delivers 0 Joules of energy per second. This renders it useless as a "powersource" for movement.

There can be found modifications to the Carnot cycle where the power is nonzero in the literature.
The simplest approach I know of is due to Curzon and Ahlborn which is discussed in this article as well http://arxiv.org/abs/1405.2273

I know, but as heat engines must follow his rules,

You should state that their efficiency is bounded from above by the Carnot efficiency.
That's why he made that comment.

 

[EM_Guy]

WHY

"It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work." - Kelvin-Planck statement of the 2nd Law

"It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body." - Clausius statement of the 2nd Law.

These two statements are equivalent. A machine that violates the 2nd Law is a perpetual motion machine. There are no perpetual motion machines, because no machine can violate the 2nd Law.

A reversible process is a process that can be reversed without leaving any trace on the surroundings.

The Carnot cycle is comprised of four reversible processes - two isothermal and two adiabatic. The efficiencies of all reversible heat engines operating between the same two reservoirs are the same. The efficiency of an irreversible heat engine is always less than the efficiency of a reversible heat engine operating between the same two reservoirs.

 

[johnbbahm]

i could not understand you

Gasoline and Diesel engines still must adhere to the 1 - T1/T2 equation, they cannot get better.
https://en.wikipedia.org/?title=Heat_engine

 

[jerromyjon]

Can someone explain how an internal combustion engine and a carnot engine are related? I'm not familiar with carnot engines at all.

 

[EM_Guy]

Jerromyjon,

My advice: First learn the Carnot engine. Then, learn about other types of engines.

Heat engines receive heat from a high temperature source, they convert part of this heat to work (usually in the form of a rotating shaft), they reject the remaining waste heat to a low-temperature sink, and they operate on a cycle. Heat engines usually involve a working fluid to and from which heat is transferred while undergoing a cycle. Internal combustion engines technically do not operate on a thermodynamic cycle. That is the working fluid (combustion gases) do not undergo a complete cycle; new combustion gases are injected into the engine at specific times, and the exhaust gases are thrown out at specific times. Since the working fluid is not operating on a cycle, an internal combustion engine is technically not operating on a thermodynamic cycle. But for the sake of simplicity, you can model an internal combustion engine as a heat engine that works on a thermodynamic cycle.

The Carnot cycle consists of four reversible processes. Note that reversible processes never occur in nature. Rather, a reversible process is an idealization. Reversible processes demonstrate the theoretical limits of the corresponding irreversible processes. Factors that cause a process to be irreversible include friction, unrestrained expansion, mixing of two fluids, heat transfer across a finite temperature difference, power dissipation in electrical resistance, inelastic deformation of solids, and chemical reactions.

The four reversible processes that make up the Carnot cycle include:

1. Reversible Isothermal Expansion. The working fluid at some temperature (TH) is brought into close contact with a source at temperature (TH). The gas slowly expands - doing work on the surroundings. As it does so, it tends to decrease in temperature, but as soon as it decreases an infinitesimal amount (dT), heat flows from the source to the working fluid - keeping the working fluid temperature at (TH).

2. Reversible Adiabatic Expansion. The working fluid is now not in contact with the thermal reservoir, and the working fluid is well insulated. So, no heat is transferred to or from the working fluid (aka - adiabatic process). The gas continues to expand doing work on the surroundings, and as it does so, its temperature drops from TH to TL.

3. Reversible Isothermal Compression. The gas (at temperature TL) is brought into contact with the thermal sink at temperature TL. The gas is compressed by an external force, which is doing work on the gas. This tends to cause the temperature to increase, but whenever the temperature increases an infinitesimal amount dT, heat is transferred from the working fluid to the sink. Thus, the gas temperature is held constant at TL (isothermal).

4. Reversible Adiabatic Compression. The gas is now not in contact with the thermal sink and the working fluid is well insulated (thus adiabatic). The gas is compressed in a reversible manner. As it is compressed, its temperature rises from TL to TH, thus completing the cycle.

You should find and study the PV-diagram of the Carnot cycle.

The efficiency of a heat engine:

$\eta = \frac{Wnet,out}{Qin} = \frac{Qin - Qout}{Qin} = 1 - \frac{Qout}{Qin}$

Based on the Carnot cycle and the Carnot principles, a temperature scale can be defined. The second law of thermodynamics only requires that the ratio of heat transfer from a high temperature reservoir to the working fluid of a reversible heat engine to the heat transfer from the working fluid of the said heat engine to a low temperature reservoir be equal to the ratio of some function of the temperature of the high temperature reservoir to the same function of the temperature of the low temperature reservoir.

$(\frac{QH}{QL})rev = \frac{\phi(TH)}{\phi(TL)}$

Lord Kelvin simply chose to let $\phi(T) = T$ so that

$(\frac{QH}{QL})rev = \frac{TH}{TL}$

where $TH$ and $TL$ are absolute temperatures.

 

[jerromyjon]

Hi and thanks EM Guy for the excellent breakdown! I am an automotive technician who has rebuilt many various engines and transmissions, and while that does not explicitly imply a physical understanding you can probably imagine I have had adequate exposure to engines that "don't run" and trust me when I say the computer systems rarely give you a solid lead as to why that is.

Gasoline and Diesel engines are heat engines and must follow Carnot's rules.

This is what caught my attention, I have heard of heat engines and simply assumed they were similar to a steam engine, where the expansion of the phase change from liquid to gas states provides the energy to do work. I wasn't sure if I was missing something about gasoline engines where an extensive amount of work is done to "remove" heat from the system, while none of that heat energy is actually utilized.

 

[EM_Guy]

Hi and thanks EM Guy for the excellent breakdown! I am an automotive technician who has rebuilt many various engines and transmissions, and while that does not explicitly imply a physical understanding you can probably imagine I have had adequate exposure to engines that "don't run" and trust me when I say the computer systems rarely give you a solid lead as to why that is.

This is what caught my attention, I have heard of heat engines and simply assumed they were similar to a steam engine, where the expansion of the phase change from liquid to gas states provides the energy to do work. I wasn't sure if I was missing something about gasoline engines where an extensive amount of work is done to "remove" heat from the system, while none of that heat energy is actually utilized.

The removal of "waste heat" from an engine (whether it works on an actual thermodynamic cycle or not) is necessary. That heat isn't utilized, because it can't be utilized.

In other words, someone might ask (and it almost sounds like you are asking this), "Why include the reversible isothermal compression (step 3 of the Carnot cycle) and waste all of this heat that is just going to the surroundings? Can't we use that heat to do more work?" The answer is, "No." In the first two steps of the Carnot cycle, two reversible processes are performed which results in work being done on the surroundings. Work producing devices produce the most work when the processes used are reversible processes. To continue the cycle, we need to get the working fluid back to its state at the beginning of the cycle (or we need to dump this working fluid and replace it with new working fluid). To accomplish this, work needs to be done on the working fluid, but in order to optimize the net work done by the working fluid in the cycle, we need to minimize the work done on the working fluid to get it back up to its original state. Working consuming devices consume the least work when the processes used are reversible processes. To complete the cycle, this low pressure, low temperature working fluid needs to be compressed. During the reversible isothermal compression process, heat will be removed from the working fluid and will be "wasted" on the surroundings. There is no getting around this. There are other ways to get the working fluid back up to its original state, but none that would improve the thermal efficiency of the Carnot engine.

 

[jerromyjon]

I'm sorry, I am still not seeing any connection whatsoever between a combustion engine and the carnot cycle, unless it has to do with heat being removed from the cylinders prior to the next combustion cycle.

 

[JorisL]

I'm sorry, I am still not seeing any connection whatsoever between a combustion engine and the carnot cycle, unless it has to do with heat being removed from the cylinders prior to the next combustion cycle.

It's simple. The Carnot cycle is proven to be the most efficient cycle you can possibly get for a cycle running between two heat baths.
For another type of engine, the heat baths are intake of air is cold and exhaust is hot to state it simply.
We know we can never surpass (or reach for that matter) the efficiency $1-T_c/T_h$.

And that is the connection for me. It is a profound one too, there are a lot of possible cycles. None of which are more efficient than this simple cycle.

 

[EM_Guy]

Do you have a book on thermodynamics?

http://www.mechadyne-int.com/vva-reference/part-load-pumping-losses-si-engine
https://www.grc.nasa.gov/www/K-12/airplane/otto.html

The Otto cycle is the idealized cycle of the spark-ignition engine. The combustion process is replaced by a heat addition process. The Otto cycle includes an isentropic expansion followed by a constant specific volume heat rejection. In the actual spark ignition cycle, at the end of the expansion stroke, the exhaust valve opens, and the combustion waste gases are removed. Then, the intake valve opens and the air-fuel mixture is added during the intake stroke. From the point that the exhaust valve opens to the end of the intake stroke, this is simplified and modeled simply as a constant volume heat rejection. Although, some models of the Otto cycle do show reduction and increase of specific volume during the exhaust and intake strokes.

An isentropic process is a constant entropy process. All processes that are both reversible and adiabatic are isentropic.

 

[EM_Guy]

Otto cycle efficiency:

$\eta = 1 - \frac{1}{r^{k-1}}$

r = compression ratio
k = specific heat ratio

This will always be less than the Carnot efficiency.

 

[jerromyjon]

Now it makes perfect sense. Thank you so much!

 

[EM_Guy]

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