Internal combustion engine efficiency

[physior]

hello

we know that there is 42% limitation in the efficiency of internal combustion engine, thermodynamically calculated by Carnot and we can never overcome this

can someone explain me in few simple words why this is true?

also, does this apply to the energy from internal combustion to generate electricity via a electrical generator?

thanks!

 

[DrClaude]

we know that there is 42% limitation in the efficiency of internal combustion engine, thermodynamically calculated by Carnot and we can never overcome this

That number actually depends on the temperatures at which the engine is operating. There is also another thermodynamic cycle, the Otto cycle, which represents an ideal internal combustion engine and has an even lower upper limit on the efficiency.

can someone explain me in few simple words why this is true?

I can't give you a thermodynamics course online. The simplest answer is that a cyclic engine requires a hot source at T_h and a cold sink at T_c, and the Carnot cycle is the most efficient thermodynamics engine working between these two temperatures and satisfying the 1st and 2nd laws of thermodynamics. Other cycles will lead to the production of additional entropy, meaning that more heat has to be dumped in the cold sink, leaving less heat to do actual work.

also, does this apply to the energy from internal combustion to generate electricity via a electrical generator?

You will have additional losses when converting the work into electricity, so the overall efficiency is going to be lower.

 

[tonyxon22]

I think you can separate your question in two parts:

1) Why is the Carnot Efficiency the limit and can never be overcome?
The Carnot heat engine is an idealization that does not lose heat to the surroundings or work due to friction. You get the top Carnot efficiency considering that all the heat available when burning a combustion product is transformed into work. The Carnot theorem is a relation between the temperature of the “hot source” (the flame of the combustion) against the temperature of the “cold sink” (the environment where you are dumping the residuals from the process). In ideal conditions, the temperature of combustion of a particular fuel is fixed. On the other hand, you can put your engine in the most cold environment possible, let’s say somewhere near the north pole where you have the lowest temperatures on earth. This would give the best Carnot efficiency possible. Actually, the colder the surrounding environment, the better the efficiency of a heat engine, but here on Earth that has a limit. (Note: remember that to calculate the Carnot Efficiency the temperature has to be set on Kelvin, so no such thing as a “negative temperature” is possible).
In the real life however, not all the available temperature difference is transformed into energy. In a real heat engine many things happen to keep you away from the Carnot Efficiency, for example you cannot be sure of a perfect combustion; most of the heat released by the combustion is taken away in the refrigeration circuit; some work is lost in friction in the engine’s parts in movement, etc. So you can never beat the Carnot Efficiency which doesn’t take any of that into account.

2) does this apply to the energy from internal combustion to generate electricity via a electrical generator?
Here you are mixing to things. On one hand there is transformation of the chemical energy stored in the fuel into mechanical energy. For that, the engine releases the chemical energy in the form of heat, which then is used to generate work. As we have discussed the top efficiency for this process would be the Carnot Efficiency. On the other hand, you have the transformation of the already obtained mechanical energy into electrical energy through the generator. In this case the efficiencies are very high (up to 98% if I recall correctly).

To sum it up, the most important looses of efficiency occur in the process of getting the available chemical energy in the fuel and transform it into mechanical energy. After that, the mechanical/electrical transformation is very efficient.

 

[physior]

thanks for your reply

is there an zero-prerequisities tutorial that starts from zero knowledge and builds up to Carnot circle?

I did that in school but I remember nothing

wikipedia articles don't help much

 

[tonyxon22]

There is a "Thermodynamics for Dummies" book... not to call you a "Dummy" or anything like that hehe.I have a lot of "for Dummies" myself. I haven't check that out, but those but are usually very good for that kind of learning.

 

[DrClaude]

is there an zero-prerequisities tutorial that starts from zero knowledge and builds up to Carnot circle?

HyperPhysics can be a good place to start: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heaeng.html#c1

 

[physior]

I gave a go to this guy:
http://www.learnthermo.com/T1-tutorial/ch03/lesson-A/pg02.php

but he says that if we increase pressure, we decrease internal energy! is that true? I thought we increase it, as we pressurize it!

 

[DrClaude]

On the other hand, you can put your engine in the most cold environment possible, let’s say somewhere near the north pole where you have the lowest temperatures on earth. This would give the best Carnot efficiency possible. Actually, the colder the surrounding environment, the better the efficiency of a heat engine, but here on Earth that has a limit.

I disagree here. The "cold sink" temperature in the internal combustion engine corresponds to the lower temperature of the burnt fuel mixture after expansion ("exhaust"). This temperature will be much higher than the ambiant temperature. Again, look at the Otto cycle for something closer to the actual workings of an internal combustion engine, while still being "ideal".

 

[SteamKing]

hello

we know that there is 42% limitation in the efficiency of internal combustion engine, thermodynamically calculated by Carnot and we can never overcome this

can someone explain me in few simple words why this is true?

also, does this apply to the energy from internal combustion to generate electricity via a electrical generator?

thanks!

It's not clear for which internal combustion cycle there is a 42% limitation on efficiency. You should cite some sources for this figure.

There have been large slow-speed diesel engines constructed to power ships which exceed this 'limit', and there are also Brayton cycle engines (better known as gas turbines) which have also had better efficiency.

http://en.wikipedia.org/wiki/Thermal_efficiency

http://en.wikipedia.org/wiki/Diesel_engine

http://www.power-technology.com/features/feature1084/

 

[physior]

no, no gas turbines, we are talking about internal combustion engines

 

[SteamKing]

no, no gas turbines, we are talking about internal combustion engines

Technically, gas turbines are internal combustion engines. They don't use pistons in cylinders like SI or CI motors, though.

 

[tonyxon22]

I disagree here. The "cold sink" temperature in the internal combustion engine corresponds to the lower temperature of the burnt fuel mixture after expansion ("exhaust").

Yes, you are right! It was a misconception from my side. However, I still think that efficiency of internal combustions engines increases with lower temperature of the environment. Is that wrong?

 

[Hybrid_engine]

Seems like people are talking efficiency based on heat only. The real bases for efficiency is brake specific fuel consumption. Horsepower per hour for a given quantity of fuel. Based on heat is very misleading. A standard gas engine rated 25% efficiency will have less than 1/8th the horsepower output per hour for the same quantity of fuel in a VRE engine. Currently a classified design. My point is heat and actual power output in horsepower can be missleading

 

[cjl]

Seems like people are talking efficiency based on heat only. The real bases for efficiency is brake specific fuel consumption. Horsepower per hour for a given quantity of fuel. Based on heat is very misleading. A standard gas engine rated 25% efficiency will have less than 1/8th the horsepower output per hour for the same quantity of fuel in a VRE engine. Currently a classified design. My point is heat and actual power output in horsepower can be missleading

You'll have to provide some sources for your claims if you want anyone to take you seriously, especially the claim that a "classified VRE engine" (whatever that is) can make 8 times the horsepower on the same fuel flow rate as a 25% efficient gas engine (this is impossible, by the way).

 

[russ_watters]

Seems like people are talking efficiency based on heat only. The real bases for efficiency is brake specific fuel consumption. Horsepower per hour for a given quantity of fuel. Based on heat is very misleading. A standard gas engine rated 25% efficiency will have less than 1/8th the horsepower output per hour for the same quantity of fuel in a VRE engine. Currently a classified design. My point is heat and actual power output in horsepower can be missleading

Horsepower per hour is a meaningless unit. The horsepower output at a given set of conditions does not vary over time.

I share cjl's skepticism, but note that PF is not a place for discussion of personal theories, if that is what this is.

 

[Hybrid_engine]

Specific fuel consumption based on constant horsepower output.

 

[Hybrid_engine]

You'll have to provide some sources for your claims if you want anyone to take you seriously, especially the claim that a "classified VRE engine" (whatever that is) can make 8 times the horsepower on the same fuel flow rate as a 25% efficient gas engine (this is impossible, by the way).

How it that impossible? My comparison is based on mechanical power output. You base your idea of impossible on the limited knowledge you possess. If you really think the standard crankshaft engine is the best idea then you are lost in the past.

 

[Hybrid_engine]

8 times the power is 8 times the useable mechanical power. Same quantity of fuel and pressure inside the combustion chamber. 8 times the torque. Torque and rotation equals horsepower something we all know.

 

[Nidum]

http://www.ttengines.com/DCGT2.html

 

[cjl]

How it that impossible? My comparison is based on mechanical power output. You base your idea of impossible on the limited knowledge you possess. If you really think the standard crankshaft engine is the best idea then you are lost in the past.

Because the efficiency is the proportion of the chemical energy input that is transformed to usable mechanical output. 25% efficiency means that 25% of the input energy is turned into usable work at the output shaft. 8 times as much power with the same input fuel would mean that the output shaft was being driven with twice as much energy as the input fuel had, which is clearly impossible (and claiming it is possible constitutes a claim of a perpetual motion machine, which is against the rules of this forum). I'm not claiming modern engines are perfect - far from it in fact, but an engine's efficiency will always be <100%.

 

[Hybrid_engine]

You are making assumptions not based on my reply. Most engineers understand that in a standard crankshaft engine design the pressure on the piston is transmitted to the crankshaft as torque. Due to the inefficiency of the design, most of the time the pressure is more than the torque in lbs exerted on the crank. However there is a point where the torque exceeds pressure as we all know due to the leverage of the offset of the crank journal. None of us will argue with the law of leverage, centripetal, centrifugal forces. My point is the way the energy is transferred to the crankshaft is very inefficient. What I am saying is there are ways to transfer the given energy from combustion 8 times more efficiently than currently used in standard combustion engines. No over unity needed. Since no knowledgeable engineer will argue with all the published papers that relate to this subject, " crankshaft leverage characteristics" it is well known by all that due to the variability of leverage due to the design offset of the crank journals and rotation of a crankshaft relative to the single plane of motion of the piston that the design efficiency is limited to all forces that act on the rotation of the crankshaft. Which include the negative impact of the g forces of changing the piston direction to the opposite direction as it goes up and down. Not forgetting the negative impact pressure from compression has on the total of forces both positive and negative which equals torque produced by the entire mechanism of combustion including friction.

 

[cjl]

That's just a bunch of slightly technical sounding word salad with no substance or substantiation. You really haven't provided any useful claims or evidence here, and it's also diverging pretty far from the original topic.

 

[Hybrid_engine]

Reduced to layman's terms leverage from the offset of the crank journal transmits combustion pressure into torque based on the distance of offset from centerline. Long bar, fulcrum, heavy weight, less pressure than total weight to move weight based on leverage percentage. Crankshaft same principle, combustion pressure, offset "bar" crank journal, Rotation. Total torque based on all forces. Centrifugal and Centripetal forces based on distance and speed are not to be forgotten.

 

[Hybrid_engine]

hello

we know that there is 42% limitation in the efficiency of internal combustion engine, thermodynamically calculated by Carnot and we can never overcome this

can someone explain me in few simple words why this is true?

also, does this apply to the energy from internal combustion to generate electricity via a electrical generator?

thanks!

Who said 42% efficiency is the limit? Efficiency based on what factors?

 

[cjl]

It's clearly not the absolute limit, since some slow speed 2 stroke turbodiesels are in the ~50% range right now.

 

[Hybrid_engine]

It's clearly not the absolute limit, since some slow speed 2 stroke turbodiesels are in the ~50% range right now.

The EmmaMaersk engine runs very slow 100 rpm typically. Its high efficiency is from the long stroke. Using as much entropy as possible and leverage from offset. My question is what defines efficiency? Would specific fuel consumption and power output be a better way to determine efficiency? If you consider an engine 50% efficient that uses a crankshaft and offset journals, what would happen to the efficiency if the offset leverage was constant, not variable as the journal rotates? What would the torque gain be? Also eliminating the negative g forces from the piston changing direction at high speed? What would the gain in torque be with double the offset? Anyone that knows the answer please reply

 

[cjl]

My question is what defines efficiency? Would specific fuel consumption and power output be a better way to determine efficiency?

Brake specific fuel consumption (for a given fuel) is directly related to efficiency. It's just a different way of presenting the same number. Power output is irrelevant to efficiency, since efficiency is a ratio of useful output to input energy. Most extremely efficient engines are very large, however, since there are favorable scaling factors that reduce losses as engine size increases. As for the rest of your post, I have no idea what you're trying to say.

 

[Nidum]

As for the rest of your post, I have no idea what you're trying to say.

+1

 

[Hybrid_engine]

Brake specific fuel consumption (for a given fuel) is directly related to efficiency. It's just a different way of presenting the same number. Power output is irrelevant to efficiency, since efficiency is a ratio of useful output to input energy. Most extremely efficient engines are very large, however, since there are favorable scaling factors that reduce losses as engine size increases. As for the rest of your post, I have no idea what you're trying to say.

 

[Hybrid_engine]

Are you aware of the ability to transmit more torque to the crankshaft than pressure inside the combustion chamber? Due to leverage at a given point of rotation? Have you studied any research on torque and combustion pressure relative to crankshaft degrees of rotation and cylinder pressure at intervals of all degrees of rotation with a specific fuel in a standard crankshaft combustion engine? If not I understand your difficulty understanding what I am saying

 

[Hybrid_engine]

In the real world of combustion engine design, power output is the most important factor compared to quantity of fuel.

 

[cjl]

In the real world of combustion engine design, power output is the most important factor compared to quantity of fuel.

Power output per quantity of fuel (per time)= BSFC = efficiency. It's the same thing. I really don't understand why you're trying to make the distinction.

 

[cjl]

OK, I'll show you why I don't understand what you're saying:

Are you aware of the ability to transmit more torque to the crankshaft than pressure inside the combustion chamber?

Torque and pressure are different. They use different units. They cannot be directly compared. The statement "transmit more torque to the crankshaft than pressure" means absolutely nothing. I could similarly ask if you were aware of the ability to make your electric motor go faster by transmitting more voltage than RPMs. Yes, crankshaft torque is related to cylinder pressure, but you really can't directly compare the two, and the relation is complex.

Due to leverage at a given point of rotation? Have you studied any research on torque and combustion pressure relative to crankshaft degrees of rotation and cylinder pressure at intervals of all degrees of rotation with a specific fuel in a standard crankshaft combustion engine?

I am quite familiar with the thermodynamics and general principles of combustion engines. I still don't have the faintest clue what you are trying to get at here. Yes, combustion pressure as a function of crankshaft rotation is important (and things like mixture ratios, charge stratification, fuel injection type/location, ignition location, and spark timing are very important to this), but you really aren't saying anything of use here.

 

[Hybrid_engine]

Power output per quantity of fuel (per time)= BSFC = efficiency. It's the same thing. I really don't understand why you're trying to make the distinction.

My point is, if you say an engine is 50% efficient with given quantity of fuel and power output, and a completely different engine design has 4 times the power, torque and horsepower available at the drive shaft with same quantity of fuel, the way efficiency is calculated needs to be changed to real world horsepower and torque numbers for a predetermined quantity of fuel in each fuel category.

 

[Hybrid_engine]

OK, I'll show you why I don't understand what you're saying:
Torque and pressure are different. They use different units. They cannot be directly compared. The statement "transmit more torque to the crankshaft than pressure" means absolutely nothing. I could similarly ask if you were aware of the ability to make your electric motor go faster by transmitting more voltage than RPMs. Yes, crankshaft torque is related to cylinder pressure, but you really can't directly compare the two, and the relation is complex.I am quite familiar with the thermodynamics and general principles of combustion engines. I still don't have the faintest clue what you are trying to get at here. Yes, combustion pressure as a function of crankshaft rotation is important (and things like mixture ratios, charge stratification, fuel injection type/location, ignition location, and spark timing are very important to this), but you really aren't saying anything of use here.

 

[Hybrid_engine]

Real world useful power is very relative to combustion pressure and the mechanism of transfering the pressure to the power shaft. That is the whole point in combustion engines. Transferring pressure into rotational force or direct pressure.

 

[cjl]

My point is, if you say an engine is 50% efficient with given quantity of fuel and power output, and a completely different engine design has 4 times the power, torque and horsepower available at the drive shaft with same quantity of fuel, the way efficiency is calculated needs to be changed to real world horsepower and torque numbers for a predetermined quantity of fuel in each fuel category.

No, if an engine is 50% efficient with a given fuel flow rate, and a different engine has 4 times the power with the same fuel consumption, the second engine is impossible. The first engine is extracting 50% of the available energy from the fuel as useful work (this is the definition of 50% efficiency), so the second engine (in order to make 4 times the power) would need to extract twice as much energy as the fuel had. This is clearly not possible.

*Minor caveat, just to prevent future nitpicking: most ICE thermal efficiencies use the lower heating value of the fuel for their "fuel energy content" number. Technically, if you could condense the water vapor out of the exhaust and extract the latent heat of vaporization with some kind of thermodynamic cycle/engine on the exhaust, you could extract more energy from the fuel than this lower heating value would indicate, which could (if you fiddle with the numbers just right) allow for a maximum theoretical efficiency of >100% if you're using the standard formula for efficiency of shaft work out divided by the lower heating value of the fuel burned. In real engines though, the water vapor in the exhaust is a gas, and therefore the use of the lower heating value is correct.

 

[Hybrid_engine]

Everything you mention about timing, spark, air fuel ratio are what impact internal pressure. The mechanism of transferring the pressure is the engine design. To calculate torque you need to add the positive pressure and subtract the negative pressure transferred to the crank. Actually what you mentioned can be left out of the equation. Only need to measure pressures produced and calculate the leverage factors and mechanical efficiency of the engine. What you mentioned has impact on pressure increase or decrease. Not needed to calculate useable torque or horsepower. With exact same airfuel ratio and fuel and temperature in two different engines you get different power output as we all know. My point is, there exist more efficient transfer of energy, based on mechanical advantage, centrifugal and centripetal force, friction, pressure, all determine the final power output.

 

[Hybrid_engine]

No, if an engine is 50% efficient with a given fuel flow rate, and a different engine has 4 times the power with the same fuel consumption, the second engine is impossible. The first engine is extracting 50% of the available energy from the fuel as useful work (this is the definition of 50% efficiency), so the second engine (in order to make 4 times the power) would need to extract twice as much energy as the fuel had. This is clearly not possible.

*Minor caveat, just to prevent future nitpicking: most ICE thermal efficiencies use the lower heating value of the fuel for their "fuel energy content" number. Technically, if you could condense the water vapor out of the exhaust and extract the latent heat of vaporization with some kind of thermodynamic cycle/engine on the exhaust, you could extract more energy from the fuel than this lower heating value would indicate, which could (if you fiddle with the numbers just right) allow for a maximum theoretical efficiency of >100% if you're using the standard formula for efficiency of shaft work out divided by the lower heating value of the fuel burned. In real engines though, the water vapor in the exhaust is a gas, and therefore the use of the lower heating value is correct.

 

[Hybrid_engine]

You are assuming that the combustion pressure produced can only equal double the power if 100% efficient when compared to an engine you call 50% efficient. When you truly understand efficiency and mechanical output you will realize the engine you call 50% efficient is not delivering 50% of the torque and power potential of the fuel. It makes sense to you only if you look at the energy transfer system in a fixed design. Design an engine with constant offset and leverage factor, quadruple the leverage factor, reduce friction with air or magnetic bearings, eliminate g forces from directional changes, designed to give the optimum compression ratio for any fuel, use centrifugal and centripetal force to the maximum advantage in the given space. Is this impossible?

 

[cjl]

You are assuming that the combustion pressure produced can only equal double the power if 100% efficient when compared to an engine you call 50% efficient. When you truly understand efficiency and mechanical output you will realize the engine you call 50% efficient is not delivering 50% of the torque and power potential of the fuel. It makes sense to you only if you look at the energy transfer system in a fixed design. Design an engine with constant offset and leverage factor, quadruple the leverage factor, reduce friction with air or magnetic bearings, eliminate g forces from directional changes, designed to give the optimum compression ratio for any fuel, use centrifugal and centripetal force to the maximum advantage in the given space. Is this impossible?

Combustion pressure and details of the engine are irrelevant - the efficiency is defined as energy out/energy in. You can look at the whole system as a black box and not care about the details as far as an efficiency calculation is concerned - you just have to know the fuel flow in, and the shaft power out. Acceleration from directional changes isn't inherently lossy either - the amount of energy lost depends on the details of the linkages and the movements.

Also, what the heck do you mean by "leverage factor"? You can increase the leverage by increasing the stroke, but this also increases the piston velocity for a given RPM, which tends to decrease maximum RPM (this can be good for efficiency though, depending on the details of the engine). You really should use more standard terminology, and be clearer with your descriptions (though it sounds like you really don't know much about internal combustion engine design).

 

[russ_watters]

Locked pending moderation.
[edit]
I don't really have anything to add to cjl's excellent responses, so I'll just reiterate for emphasis:

Efficiency is simply the output power (mechanical work) divided by the input power (chemical energy per unit time) and is the same whether expressed as a percentage or brake specific fuel consumption. If a typical engine is 25% efficient and a claimed alternate design gets 8x the power output for the same fuel consumption it must be 25%*8=200% efficient, an obvious violation of conservation of energy.

Hybrid_engine, it appears to me that you are getting bogged down in details about internal combustion engines that you don't understand and allowing your misunderstandings to cause you to lose sight of the simple issue of how efficiency and conservation of energy work. Instead you should be using conservation of energy as a guiding principle to help focus your understanding of these other concepts. And you will need to start from scratch and learn quite a lot of thermodynamics before you can attempt to design your own new type of engine.

This thread will remain locked.