Does the Carnot heat engine law apply to an internal combustion engine?

The law says that the maximum thermal efficiency of a heat engine can be 88%.

If I have and internal combustion engine using 9.8 cc/min instead of 25 to 30 cc/min and the other accepted figure is a 2L 4 cylinder engine for idle is it uses approx 5 HP of fuel just just to idle.

So I am doing the same work as 5 HP of fuel at 30% efficiency but now only using 2.43 HP worth of fuel.
If my maths are correct a unmodified engine uses 105.76% more fuel.

Its been said my engine efficiency was around 78% is this true?

Could someone please do the numbers for me to see if the 78% is correct?

If by chance I raise the compression ratio of my engine it will become more efficient because the ignition phase pressure build up compressive losses will be reduced eg if I have to start the ignition phase after top dead center.

I currently have an engine that needs the timing to be roughly 4 degrees after top dead center at 500 rpm the other point is at about 1427 rpm it is roughly at top dead center my engine has a timing slope 0.0035 per rpm in a 1 demention graph.

I have video evidence online of the 9.8 cc/min if anybody would like to see to confirm my results.
 smokingwheels, The answer to your title, "Does the Carnot heat engine law apply to an internal combustion engine?" is "yes". For the rest of it, ... what are you asking, and "No". fFish

 Quote by Fish4Fun smokingwheels, The answer to your title, "Does the Carnot heat engine law apply to an internal combustion engine?" is "yes". For the rest of it, ... what are you asking, and "No". fFish
Its been said my engine efficiency was around 78% is this true?
ok so a no on this one, so my question is do you think it is not possible?

Then Could someone please do the numbers an explain for me to see what efficiency I have?

Does the Carnot heat engine law apply to an internal combustion engine?

Hi,

I am sorry I can give you only a vague answer. You can elaborate further with reading more documentation. I don't remember exactly the formula to calculate the efficiency of the internal combustion engine. However, I still remember the principles and the values.

- The efficiency depends on the difference of temperature when the power stroke occurs.
T highest when combustion sparks and T lowest when the piston reaches its lowest course.

- To achieve the best efficiency, this Tmax must be increased. Either by using fuel with higher energy output, or more conveniently by increasing the compression ratio. You know that well I am sure. I guess you also know that there is also a practical limit to the increase of this compression either because of mechanic manufacturing or simply because of the resistance of the materials (melting).

Here is the common sense, if a simple trick to increase the compression would increase the efficiency. Don't you think the car manufacturer would do it and use it as marketing advantage? There is max ratio in compression for gasoline engine. Above that, the mix air + gasoline would explode spontaneously which will put the power stoke out of sync.

In short, with the internal combustion engine, it will be very hard to go above 30%. The principle itself relies on loss of heat. The 78% you mentioned could be the % of the "power" stroke alone, the 3 remaining strokes are resistant. The overall efficiency must take in account all 4 strokes. With a lot of skills and maintenance, you can fine tune your engine to claim a few % that I guess 5% max better than the average engine. There is no way you can get 78%.

No saying that these fine tuned engines have a shorter life time because the materials is more stressed. Or because they could be upset with a slight change in parameters (fuel quality, ambient temperature, driving style, etc). When the ideal conditions are not met, these engines are not necessary better than the average engine.

Hope my answer is not too wrong and could give you some hints to dig further.
 Efficiency= Work-out/chemical Energy in, for the system. (apprx 10 kWh/kg for gas and diesel means 10 kWh entered chemically derived heat, gives 3,333 kWh mecanical work at 30% eff.). Your system would, by the discription given, produce lower eff. than normal ic-engine. Max carnot-eff. is deltaT/T-h (temperature difference highest-lowest in Kelvin/celcius, divided by exhaust-temp in Kelvin. Best of luck.

 Quote by Vespa71 Efficiency= Work-out/chemical Energy in, for the system. (apprx 10 kWh/kg for gas and diesel means 10 kWh entered chemically derived heat, gives 3,333 kWh mecanical work at 30% eff.). Your system would, by the discription given, produce lower eff. than normal ic-engine. Max carnot-eff. is deltaT/T-h (temperature difference highest-lowest in Kelvin/celcius, divided by exhaust-temp in Kelvin. Best of luck.
If I had a little money I could measure such things, hey if you know if anybody can donate stuff to my cause PM me Thanks

Acording to common knolage a 4 cyclinder 2 L engine consumes 5 HP 3.68kw of fuel to just idle
So that is roughtly 20.33 cc/min as per a program called turbo calc with no overheads like water pump altnator etc 5-8cc/min???.
My average fuel consumption on my best city run is 23.46 cc/min run for 99.5min with overheads with rough guess the engine was doing 1700 rpm on average (I will have to modify a program to process the Tacho graphs again).
So what I am saying is I am driving my car on what a normal one would consume just to idle.
I have it on video of my engine using 9.8 cc/min of fuel to idle at about 650 rpm and at 750 rpm is calculated to be about 11.??cc/min with overheads.

 Quote by Vespa71 Efficiency= Work-out/chemical Energy in, for the system. (apprx 10 kWh/kg for gas and diesel means 10 kWh entered chemically derived heat, gives 3,333 kWh mecanical work at 30% eff.). Your system would, by the discription given, produce lower eff. than normal ic-engine. Max carnot-eff. is deltaT/T-h (temperature difference highest-lowest in Kelvin/celcius, divided by exhaust-temp in Kelvin. Best of luck.
I have done a crude test and started a thread on another forum to compare results, I can put my test on video and host on youtube if needed.

I took my car for a drive until it warmed up ~195 deg F (thermostat) then I checked the exhaust temperature after letting it idle for a few mins I used my hand and then water.

The outside temperature was 24.8 Deg C, RH 41%, wind SSW 20km/h.

Point 1 in the pic was where water was just starting to boil when sprayed on the pipe approx 1 m of plumbing.
So a 75 degree C above ambient, The engine was approx 90.5 deg C also

Point 2 in the pic was where I could just touch it for about 1 second approx 2.4 m of plumbing.
So a rough guess 55 Deg C

The end of the system at the rear of my car I could hold on to the pipe.
Maybe 40 deg C???

I have started a topic on the speed talk forum see http://speedtalk.com/forum/viewtopic.php?f=1&t=25862
I did it to get some clue on what the normal temp is with other engine.
There is a report of a person has just started an engine for a few mins and he put his hand on the exhaust pipe 10 inches back from the end and burnt his hand.

Ok so If I have lower exhaust temp then my engine is more efficient, I need some one to measure mine one day.
 Mentor I don't think the first response was quite accurate/clear regarding the title. The Carnot heat engine is an idealized heat engine that predicts an upper-bound efficiency for all possible heat engines. The actual cycle is not possible in reality. So it applies only in that loose sense. To get a more realistic idea of the efficiency of a particular thermodynamic cycle, you have to use the efficiency equation of that particular thermodynamic cycle. You can read about it for cars here: http://web.mit.edu/16.unified/www/SP...es/node25.html From that link, the maximum theoretical efficiency of an Otto cycle engine with a compression ratio of 12.5:1 is about 60%. Note, that's the efficiency at the drive shaft of the engine, not the efficiency of the car. A substantial fraction (30-40%) of the power at the drive shaft is lost in the drivetrain and accessories of the car. So an otto cycle car with a 12.5:1 compression ratio can't possibly have an overall efficiency greater than about 40%

 Quote by ExNihilo Hi, - The efficiency depends on the difference of temperature when the power stroke occurs. T highest when combustion sparks and T lowest when the piston reaches its lowest course. - To achieve the best efficiency, this Tmax must be increased. Either by using fuel with higher energy output, or more conveniently by increasing the compression ratio. You know that well I am sure. I guess you also know that there is also a practical limit to the increase of this compression either because of mechanic manufacturing or simply because of the resistance of the materials (melting). Here is the common sense, if a simple trick to increase the compression would increase the efficiency. Don't you think the car manufacturer would do it and use it as marketing advantage? There is max ratio in compression for gasoline engine. Above that, the mix air + gasoline would explode spontaneously which will put the power stoke out of sync. In short, with the internal combustion engine, it will be very hard to go above 30%. The principle itself relies on loss of heat. The 78% you mentioned could be the % of the "power" stroke alone, the 3 remaining strokes are resistant. The overall efficiency must take in account all 4 strokes. With a lot of skills and maintenance, you can fine tune your engine to claim a few % that I guess 5% max better than the average engine. There is no way you can get 78%. No saying that these fine tuned engines have a shorter life time because the materials is more stressed. Or because they could be upset with a slight change in parameters (fuel quality, ambient temperature, driving style, etc). When the ideal conditions are not met, these engines are not necessary better than the average engine.
Thanks ExNihilo

I have found this site at the Colorado uni Internal Combustion Engine Thermodynamics Outline Tools
So I have started learning a bit more and playing with the variables of thermodynamics.
Its been said that if you increase turbulence you increase the flame rate, So if you decrease turbulence you slow the flame rate down which is what I have done or what my tests with sound waves have reveled see Subbie air flow test ported and unported tubes
Some where in the video I show the cigarette papers after a hour or so of AC/DC and various sine waves, The untreated tube paper seem to be battered more than the treated tube.

The Tmax I think has been increased because my exhaust gasses are lower when compared to a normal engine eg my exhaust temp is about 100 deg C 1 m from the discharge (75 deg C above ambient and my engine coolant temp is 90.5 Deg C).

The mix air + gasoline would explode spontaneously and melt things I agree but from the tests from my first engine eg knocking with the timing approx 24 degrees to much advance, but the the problem with that figure is my knocking reference was 3-4 deg BTDC under no load hence I destroyed the bearings in the engine very quickly but never hurt the pistons I confirmed that when I lowered the compression later on..I agree I really messed the power stroke up and disturbed many things I had not clue about.

From my first engine data what little there is I maybe able to run insane compression ratios, I could be wrong but my current engine needs spark plugs 3 heat ranges up to keep the carbon away.

 Quote by russ_watters I don't think the first response was quite accurate/clear regarding the title. The Carnot heat engine is an idealized heat engine that predicts an upper-bound efficiency for all possible heat engines. The actual cycle is not possible in reality.
Yes think this thread has gone a bit off from the title.

But I have done 334.5 km trip at 80km/h on approx 14L of fuel(I will my need my reluctant witness eg x girlfriend) in my 1984 wagon with carby.

Possible errors with fuel consumption is about 2.5L depending on the driveway angle/slope.
I did 334.5 km on 14L of fuel +-2.5 so 16.5 and 11.5
16.5L of fuel = 4.93 L/100km
14L of fuel = 4.18 L/100km
11.5 of fuel = 3.43 L/100km
All figures are below what my more modern EFI 1990 sedan does it is rated a 7L/100km on the hwy and has simlar mass to my wagon. I did a 300km trip at 80 km/h and used 19 L that's about 6.33L/100km. So my worst case is 4.68 L better over the trip. I hope to reproduce the my result one day when I restore my engine's condition.

I guess I am not going to get very far unless I can prove what I did, I accept that.

The overall efficiency of an engine with 12.5:1 compression is 40%. ok is it true that the overall efficiency of a modem day car is 20%?
My current engine compression is 10.5:1 but the normal factory compression is 9.4:1
Why do I have better fuel economy with 1.1 points raised compression eg approx 78% better MPG in the city, I know my tests may be a bit out though you can remove some points for that. see My forum in the fuel consumption testing section.
 If you want to measure efficiency it seems a good way would be to put your engine on a brake dyno. Inertial dyno's would be useless in this case. The real world has to many variables to get foolproof results. (tail wind, slopes, weather, drafting, etc.) In theory what you would do is choose an RPM and run the engine at a certain load for a few minutes, and measure how much fuel went into the engine. You could test at all different RPM's and loads to see which is most efficient. I'm sure you can make some formulas to use for the brake dyno. Not sure how much time on one of these machines costs.

 Quote by michaelwoodco If you want to measure efficiency it seems a good way would be to put your engine on a brake dyno. Inertial dyno's would be useless in this case. The real world has to many variables to get foolproof results. (tail wind, slopes, weather, drafting, etc.) In theory what you would do is choose an RPM and run the engine at a certain load for a few minutes, and measure how much fuel went into the engine. You could test at all different RPM's and loads to see which is most efficient. I'm sure you can make some formulas to use for the brake dyno. Not sure how much time on one of these machines costs.
A dyno ok I rang the Sarich Corporation Limited 25 Mar 2011 and they gave me the dyno figure for the ADR mileage test matching the mass of my car. I used that figure and did a few rough calculations. What I have found is that a normal engine similar to mine uses x amount of fuel just to get to 1791 rpm this amount is more than what my engine uses to drive around the block. I need some one to check my results and calculations before I go announcing that I have something new, I don't mind being corrected but there is not much interest so far. Oh and most of the calculations where done on a mobile phone LOL.

The rough figure for reduction in fuel is 55% but that is with some % error.

 Quote by smokingwheels A dyno ok I rang the Sarich Corporation Limited 25 Mar 2011 and they gave me the dyno figure for the ADR mileage test matching the mass of my car. I used that figure and did a few rough calculations. What I have found is that a normal engine similar to mine uses x amount of fuel just to get to 1791 rpm this amount is more than what my engine uses to drive around the block. I need some one to check my results and calculations before I go announcing that I have something new, I don't mind being corrected but there is not much interest so far. Oh and most of the calculations where done on a mobile phone LOL. The rough figure for reduction in fuel is 55% but that is with some % error.
That's amazing as a result. How about the road test using 2 cars I proposed in Post #8 above?
 What is the indicated thermal efficiency of an engine? Fuel used in total 42.61 cc/min @ 1730 rpm. Fuel used to get to 1730 rpm = 29.16 cc/min no load. Fuel left = 13.45 cc/min to do approx 4.335 kw at 1730 rpm approx 40.95 kph Fuel heat content 34.8 kj/l Approx 62kj / kw According to the graph on http://web.mit.edu/16.unified/www/SP...es/node25.html The max ideal engine efficiency is about 62.81% efficient for a compression ratio of 10.5:1 when you look at the bit map with paint. So I used the fuel used to create the power at 1730 rpm. Work over heat input. 3. The attempt at a solution 4.335 kw/min = 286 kj 13.45 cc/min = 0.01345 L * 34.8 kj = .46806 Mj = 468 kj So 286kj / 468 kj = .611 * 100 = 61.1 % indicated thermal efficiency Or is it this 4.335 kw/min = 286 kj 42.61 cc/min = 0.04261 L * 34.8 kj = 1.482 Mj = 1482 kj So 286kj / 1482 kj = .192 * 100 = 19.2 % indicated thermal efficiency. So how can 19.2 % be right when I am using approx 50% less fuel do I have to add the power output just to get to 1730 rpm? or is it this so roughly there is 7.78 kw friction output on a normal engine to get to 1730 rpm I use 27 cc/min to do that. and I use 13 cc/min to do 4.335 kw output eg the difference between load and no load. A normal engine uses about 47cc/min for 4.335 kw output that does not include the total which according to my guess is 96 cc/min I have no idea how accurate that is but its based on 3.7 kw of fuel just to idle eg 5 HP. Load output 4.335 kw =286 kj Friction output 7.78 kw = 486 kj 42.61 cc/min = 0.04261 L * 34.8 kj = 1.482 Mj = 1482 kj 286+486= 772/1482 kj = .52 * 100 = 52% indicated thermal efficiency. ok does efficiency varie with load is that how is works no one has told me. So a normal engine uses approx 96cc/min for 4.335 kw at the same rpm 4.335 kw/min = 286 kj 96 cc/min = 0.096 L * 34.8 kj = 3.34 Mj = 3340 kj So 286kj / 3340 kj = .085 * 100 = 8.5 % indicated thermal efficiency. Thus normal engine runs 8.5% @ 4.335kw@1730 rpm My engine runs 19.2% @ 4.335 kw at 1730 rpm So 19.2 -8.5 = 10.7 /8.5= 1.258 *100= 128 % better on the thermal efficiency side of things??????

 Tags engine calculation, smokingwheels