Talk:Thermal efficiency

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What in blazes is going on in the thermal efficiency section? E_hp - E_r = 1 ?!?!

So E_hp was greater than one to begin with?! Something needs explanation or changing here! —The preceding unsigned comment was added by 165.124.116.206 (talk) 23:06, August 22, 2007 (UTC)

Internal combustion engines are not "thermal devices" like furnaces or steam engines. Heat is a byproduct of burning fuel to generate pressure. Heat is actually unwanted, thus radiators and so on. External combustion engines are "thermal devices" because they specifically use heat, to generate pressure. —Preceding unsigned comment added by Arantius (talk • contribs) 16:44, 3 January 2008 (UTC)[reply]

Pressure (useful work) is generated by the heating of the air inside the piston of an internal combustion engine. ...so it is a cyclic thermal device, converts heat into work and obeys the 2nd law of thermodynamics. The "unwanted heat" is due to the inherent inefficiency, as stated by the 2nd law. 86.159.124.167 (talk) 15:19, 11 January 2008 (UTC)[reply]

Applications

I undid your revision of the thermal efficiency article due to grammer and the material added being confused and out of place. I'm not sure the Carnot limit applies to Solar cells like you state (do you have any references?) and the bit about detonations just seemed odd to me. I don't think higher temperatures around shockwaves do allow higher efficiencies and while there are large temperature gradients within engines it doesn't seem relavent to the article to me. Andrew.Ainsworth (talk) 13:27, 13 January 2008 (UTC)[reply]

With a little love for the article you could have reinserted my corrections concerning the International System of Units. Grammar: I don't know, maybe you mean style? Relevant vs thinkofthechildren: Do not make the mind of our young readers dull by mentioning only the application the adults mostly look up. After all this article is classified as stub.

Other inefficiencies

Unreferenced statement at the bottom concerning electric vehicle system efficiency("much greater") compared to I/C is incorrect because the analysis is very incomplete. Author completely ignores losses in battery storage cycle, electric motor efficiency, weight of vehicle difference, transmission losses from power plant to battery charger, etc. Full analysis would also extend to difference in efficiencies all the way back to coal extraction and oil extraction, transportation, conversion to useful fuel, energy expense of waste disposal(including transportation, cleanup, etc). I agree it should be more efficient, but not much greater. That yields unrealistic expectations.

And of course there are losses on the internal combustion side that have not been mentioned: the additional energy required to refine petroleum into gasoline as opposed to higher molecular weight fuel oil, the energy cost and evaporative fuel loss of transporting gasoline through the supply chain to gas stations, the energy cost of building, heating, lighting, manning, and maintaining an infrastructure of gas stations that would not be needed with electric cars, the energy cost of building, repairing and maintaining IC engines which are many times more complicated than electric engines, and particularly the energy saved through regenerative braking. But you have a point. I removed the adjective "much", and I will look for sources. --Chetvorno^TALK 07:05, 31 May 2010 (UTC)[reply]

Some conceptual clarification should be convenient for this article

The background idea is that engines are not perfectly lossless. When an engine is NOT perfectly lossless (tipically due to friction), then some losses will occur, consisting generally of (useful) work and non-thermal energy, that will go out the system and get lost, and will get transformed, by means of heat, into a higher surrounding temperature and atoms and molecules motion.

The 2nd law of thermodynamics tells us that these losses will make efficiency decrease, and will make maximum efficiency impossible.

The 2nd law of thermodynamics speaks about entropy and entropy increase, so the losses suffered by the imperfect engine, will entail an entropy increase (in the engine's surroundings). This way, efficency and entropy will be inversely correlated.

If the engine IS PERFECTLY LOSSLESS, then its efficiency will be maximum. And then (?) entropy will be zero within (inside) the engine.

A perfectly lossless engine, is like an "isolated system"; and the 2nd law of thermodynamics tells us that, within (inside) an isolated system, entropy will increase to a maximum.

So we are saying that, within a perfectly lossless or closed engine, both efficiency will be maximum and entropy will be maximum (?).

I reproduce here a fragment from the article:

" This limiting value [...] called the Carnot cycle efficiency [...] is the efficiency of an unattainable, ideal, lossless (reversible) engine cycle called the Carnot cycle. "

Consider the bold highlighted words; lossless condition, and reversibility, are being correlated.

Summarizing: on the article, it is not clear whether efficiency and entropy are being considered equal things or inverse things, and so it is not clear whether inside of a closed, lossless, isolated system, both efficiency and entropy will be maximum, or contrarily efficiency will be maximum while entropy will be minimum (if entropy will be minimum, then, being the engine or system lossless or closed, the 2nd law of thermodynamics is being contradicted).

I think this conceptual issue should be explicited and clarified on the article. --Faustnh (talk) 17:37, 14 June 2009 (UTC)[reply]

Yeah, I think the word 'lossless' to describe the Carnot cycle is confusing. None of the heat energy entering any engine is of course 'lost': it either comes out as work performed on the environment or heat exhausted into the lower temperature reservoir. But if 'losses' are defined as the energy wasted by the engine as exhaust heat, the Carnot cycle is not 'lossless'.

To answer your questions: the Carnot cycle is a reversible cycle. That means that the total entropy of the engine and it's surroundings (the two heat reservoirs and whatever the engine does work on), considered as a closed system, is unchanged by the operation of the engine. What Carnot's theorem says is that this perfect frictionless entropyless engine is not 100% efficient; some of the energy input as heat from the hot reservoir is wasted ('lost') as heat exhausted into the cold reservoir. The amount doesn't depend on the engine but on the temperature of the reservoirs: if the cold reservoir is 2/3 the temperature of the hot reservoir, this 'perfect' engine will be only 33% efficient. Real world engines are irreversible, having irreversible cycles, friction etc. so considered as a closed system (engine + environment) their total entropy increases as they run. For them, even more of the input heat is wasted as heat exhausted to the environment, so they have greater 'losses' and lower efficiency. You could say that as engines become more 'perfect' (reversible), as irreversibilities such as friction are eliminated and they produce less entropy, they become more efficient and have lower 'losses', but even at zero entropy they approach a limiting value of efficiency and losses that can't be improved, the Carnot value, dependent on the temperatures of their environment. --Chetvorno^TALK 10:35, 15 June 2009 (UTC)[reply]

By the way Chetvorno, admitted that Carnot engine is not actually "lossless", but is a bit "open", and, thus, admitted that for such reason that engine will suffer entropy, then I'd like to bring here another question I made at On another strange question about the 2nd law of thermodynamics:

The universe is "lossless"; it is not "a bit open" like Carnot engine. So it should be maximum efficient, and so zero entropic (?) --Faustnh (talk) 14:42, 15 June 2009 (UTC)[reply]

Addition: if 2nd law of thermodynamics says an isolated system tends to maximum entropy, then how can we say an isolated engine is maximum efficient ? --Faustnh (talk) 22:27, 15 June 2009 (UTC)[reply]

conceptual flaws

This article is riddled with conceptual flaws that will lead the reader to false conclusions. example: electric heat pumps may have a thermal efficiency of >100% because they move much more heat than the electric energy consumed. The definition must be extremely concise about what parts of the system are included and excluded in the calculation, and where gains or losses from outside arise. -- 99.233.186.4 (talk) 16:52, 13 November 2009 (UTC)[reply]

The point you made about heat pumps is actually included in the article: " [COP]...can often be greater than 100%. Since these devices are moving heat, not creating it, the amount of heat they move can be greater than the input work." However the article may not be as concise as it could be, as you say. I added the energy balance (1st Law of thermodynamics) equations for heat engines and heat pumps to the article. --Chetvorno^TALK 20:24, 13 November 2009 (UTC)[reply]

100% efficient

Electric resistance room heaters are 100% efficient - all the energy is ultimately dissipated as heat into the room. Where else does it go?

Electric heaters also produce significant amount of photons which might not be absorbed, which creates significant amount of loss.212.156.122.30 (talk) 09:09, 18 March 2011 (UTC)[reply]

The tiny fraction of energy radiated as light photons is absorbed by surfaces and converted to heat. Electric heaters are considered 100% efficient. --Chetvorno^TALK 10:53, 18 March 2011 (UTC)[reply]

Athermal engines efficiency 100%

The efficiency of traditional heat engines is a limit 1/3， 33.33%.^[1]^[2]

Px=Py=Pz=P,

PV=(2/3)NEk =(1/3)mv²N

when 3-Dimension

V²=Vx²+Vy²+Vz²

when 1-D in 3-Dimension

Vx²=3Vx²=3v²=V²

PxV=(1/3)mV²N=(1/3)mVx²N

=(1/3)m3Vx²N

=mVx²N

=2EKN

Px>P

Px=3P=3NRT/V

ηs=A/Q=1 -(T2/T1)

=1 -(T2/Q1)S

entropy S→0, so

(T2/Q1)S→0,

ηs→1

0≤η<<1

Q=3PV

or

Q=U=3PV

η0=A/Q=PV/3PV

=1/3

A=3PV

ηs=A/Q=3PV/3PV

=1

References

^ Deng Yu，Deng Hai，热效率趋向100%的能量利用新途径I:从无序到有序，中国工程热物理学会，工程热力学与能源利用学术会议，1994
^ Deng Yu，Deng Hai，邓宇，邓海，热效率趋向100%的能量利用新途径II:量子化内能的分解及有序化的演变，中国工程热物理学会，工程热力学与能源利用学术会议，1994

Wrong! (Completely and utterly)

In the 'Engine cycles' section, there is this remarkable sentence:

< The higher the compression ratio, the higher the temperature in the cylinder as the fuel burns and so the higher the efficiency. >

but that is exactly what the equation immediately above does NOT say. What it says is that the efficiency depends ONLY on the compression ratio. The 'temperature in the cylinder as the fuel burns' is not relevant. (The same applies to the Brayton cycle). 86.3.108.41 (talk) 23:39, 29 September 2011 (UTC)[reply]

I made an edit to fix this 86.176.165.110 (talk) 12:32, 18 October 2011 (UTC)[reply]

Carnot Efficiency and burning

The statement under the section "Carnot efficiency":

"Carnot's theorem only applies to heat engines, where fuel is burned. Devices that convert the fuel's energy directly into work without burning it, such as fuel cells, can exceed the Carnot efficiency."

is dubious. You can't convert fuel into energy without burning it unless you want to very narrowly define burning as only that which produces fire. As long as something is being oxidized it is being burned, just like our body "burns" calories without fire. TimL • talk 22:15, 19 September 2013 (UTC)[reply]

The distinction is that in a fuel cell the chemical energy is not converted into heat first. A fuel cell operates like a battery, the chemical energy is converted directly into electricity. Carnot's theorem applies only to converting heat to other forms of energy. --Chetvorno^TALK 01:03, 20 September 2013 (UTC)[reply]

Carnot's theorem applies beyond cases where fuel is burned (such as electric heat pumps and refrigerators). But the statement is otherwise correct; if your initial energy source has low entropy, you can convert the energy into work more efficiently than you can with an equal amount of high-entropy thermal energy. If you burned hydrogen fuel to drive a heat engine to run a generator system, you would likely recover much less electrical energy than with a fuel cell. One way of thinking about it is that minimizing losses is equivalent to minimizing entropy generation during the energy conversion process, so any step that creates excess entropy is wasteful. Any irreversible step generates entropy. Burning fuel is an irreversible, entropy-generating step; discharging a fuel cell or a rechargeable battery may be a reversible process, depending on the chemistry. That almost guarantees that it will be more efficient.JB Gnome (talk) 06:49, 21 August 2014 (UTC)[reply]

[1] Deng Yu，Deng Hai，热效率趋向100%的能量利用新途径I:从无序到有序，中国工程热物理学会，工程热力学与能源利用学术会议，1994

[2] Deng Yu，Deng Hai，邓宇，邓海，热效率趋向100%的能量利用新途径II:量子化内能的分解及有序化的演变，中国工程热物理学会，工程热力学与能源利用学术会议，1994

[1]

[2]