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

  • #1
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.
 
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Answers and Replies

  • #2
247
2
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
 
  • #3
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

I am asking
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?
 
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  • #4
33
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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.
 
  • #5
Thanks ExNihilo for you input.

I have a few problems because...

My first engine had a compression ratio of approx 12.5, the spark plugs would only last 3-4 days before they would carbonize and short out, I lowered the compression and the plugs lasted 2 months. I have video evidence see http://www.youtube.com/watch?v=XnwJuNbvULY"

My first engine had a really strange timing curve when it was a high compression engine the engine used to knock when the rpm was above 3000 rpm this was under no load to stop the knocking I had to reduce the total timing to 3-4 degrees before top dead center so what I have found with my 2nd engine the vacuum advance is 20 degrees that brings my first engine to 16 degrees after top dead center but the piston is moving away not getting closer so I guess you could half the value. I still have the computer and the program I used to control my engine timing and I have plotted there output the results are here http://ampair.tripod.com/Graph/My_first_engine.html" [Broken]

1. This is the result of my first program was a 4 ms delay from the trigger there was no power from my engine.

2. After a few nights work no knocking sound.

3. This is a later program after I lowered the compression to 8:1
On this graph you will notice that the peak timing is reached by 1400-1500 rpm then it swings negative above 1500 rpm then dropping suddenly at 3000 rpm to about 20 deg btdc for the rest of the range and this is with a compression ratio 8 to 1.

4. After a while I got the feel for the timing the blue line is a guess below 3000 rpm.

I never tested the fuel consumption for my first engine.
My second engine the efficiency falls off the more load you put on it I do not think that is normal but on the other hand when my engine was in good condition I think I used to get 45 mpg (imp) doing 80 km/h on the hwy but back then never tested the city cycle, now have tested the city cycle and have a record of 45 mpg (imp) ....Explane, I am at a loss for that one, I have evidence at http://203.161.71.130/forum/topic.asp?TOPIC_ID=9"
If you go here you can see more results and test methods http://203.161.71.130/forum/forum.asp?FORUM_ID=4"
I know my methods are a little out but you would have to compare the stats to my results.

The Stats for the run is as follows
Idle 13.99 % its a bit low cause the cops did not pull me up this time! but still stopped for 4 smokes.
Accelerating time 17.32%
Test run time 99.5 mins
So please an explanation how by tweaking my engine I can go from 59.1% to 78.4% better than a normal engine what I am saying is I got a 32.6% increase on my own test results, If I increase the compression I will still get better results because it will lower the timing and I have proved that as carbon builds up the timing goes negative so compressive ignition losses get less, and from what I understand the pressure build up from the flame front happens at top dead center I must have changed that some how.
From memory a normal engines peak and rough idle rpm will happen at about 20 deg btdc my engine 40 deg btdc see http://203.161.71.130/forum/topic.asp?TOPIC_ID=14" note there are 2 marks on my pulley 20 deg apart eg so when the second mark gets to the end of the timing plate its 40 deg btdc.
 
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  • #6
56
1
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.
 
  • #7
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.
see http://www.youtube.com/watch?v=bKc9xwYV7ks"
 
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  • #8
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Hi smokingwheels,

In my opinion you are too focused on a particular factor, which is the fuel consumption I suppose and overlooked many other factors between a random average 2L 4 cylinders car and you specific setup. Your measure of gas consumption in a city run is polluted by too many factors which makes it unreliable. For example: tires pressure, winds, traffic light, aero dynamic drag, etc. And just to add some absurdity, I would say if I use an average untuned car, doing a city run, on a downhill portion. I will certainly end up consuming less fuel than your test.

If you really want to know the efficiency, then measure it right that the output of the engine. Put a certain qty of fuel, lets say 0.5L, then load your engine and measure how much energy it outputs. Then make the ratio output/input. The difficulty here is how to measure accurately the output energy. Can you get in touch with a school or lab which has a power bench to help you to make such a measure?

Nevertheless, the quick and approximative answer is that, your engine cannot reach 78% efficiency. Here is the Wikipedia article: http://en.wikipedia.org/wiki/Thermal_efficiency" [Broken]. Read sections "Carnot efficiency" and "Automobiles: Otto cycle". The idea of 78% someone told you was probably the maximum theoretical efficiency of the Carnot cycle. And this, only during the power stroke (please remember that you must supply energy to the other 3 strokes). Let's focus on the power stroke. To reach 78% you still needs a lot of conditions. To reach such a value, your engine much reach Tmax at least 1063 degrees Celcius. The Aluminium melting point is 660 degrees Celcius. Are you absolutely certain you reached that temperature? Hopefully your cooling system is flawless. Next, this assumes that the Tmin at the end of the power stroke is 20 degree Celcius. You know well that it is not true at all (the exhaust pipe right next to the engine is very hot).

It's is understandable you are excited being on something. But the sad reality is that whatever you do, you will never beat 35%, especially for a gasoline engine (Wikipedia article "for example the average automobile engine is less than 35% efficient"). Efficiency here means output/input. Not the one from "power stroke only". Read in particular the sections "Engine cycle efficiency" and "Other inefficiencies" of the Wikipedia article above. Unlike the digital world, I am afraid that there will be no outstanding revolution in the area of IC engine.

In case you don't have access to the measurement equipment of a school or lab. How about a down to Earth experiment?

1- Car1 = yours. Car2 = a reasonably well maintained car having the same engine cylinder volume.

2- Put the same quantity of gasoline in each car. Now, make a long trip together until one car runs out of gas. Of course the objective of the race is to cover maximum distance so it is in the driver best interest to have a smooth driving style.

3- Let's assume Car2 runs out of gas first. Do you think that yours will cover more than double the distance? Because I am sure Car2 will have less the 30% efficiency and Car1 has a "supposed" 78% efficiency.

Do that test and let us know the results. Good lucks.
 
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  • #9
Smokingwheels,

There might be people on this forum that can help you: http://ecomodder.com/
You seem to be extremely focused on engine efficiency. I know a few little tricks. Let me know what all you have modded. Here are some suggestions other than what you seem to have already done:
1. polish the combustion chamber: you want the air to be hot so it expands as much as possible. If you polish the piston top and cylinder dome to a mirror finish not only do you reflect heat radiated as light back to the air you decrease the surface area of those parts and therefore their absorbtion of heat.
2. Change to a higher temp thermostat: This doesn't show improvements on every car. Some cars have coolant going through the intake manifold. This heats it up to optimize fuel vaporization, combustion, and emissions.
3. change to an underdrive pulley. This doesn't increase combustion efficiency, just decreases parasitic load. You can also just eliminate accessories you don't use. AC, power steering, etc.
4. change to iridium spark plugs: this might help keep them in service a little while longer.
5. side gap your spark plugs: this can actually decrease spark plug life but increases power
6. key your spark plugs. Basically get a spark plug keying kit and install them so the open end of the spark plug faces the intake valves
7. wrap your exhaust with heat shield.
8. You CAN increase your compression ratio and decrease engine knock by injecting water to cool the engine. There are plenty of kits online. Basically it injects the water as a fine mist and cools hot spots that can cause preignition/detonation/knock
9. Cylinder head groves might help clean up your combustion a little depending on the shape of your cylinder head dome http://somender-singh.com/content/view/7/31/ [Broken]
I've known some people to get good results from this, and some got almost none

Synthetic oils everywhere! Engine, transmission, differential if separate from transmission

then there is aerodynamics. This is usually where most of the improvement in gas mileage comes in.
 
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  • #10
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.

Vespa71 thanks for your input.
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???

EngineExhausttemp.jpg


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.
 
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  • #11
russ_watters
Mentor
20,963
7,585
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/SPRING/thermodynamics/notes/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%
 
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  • #12
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 http://www.engr.colostate.edu/~allan/thermo/page1/page1f.html" [Broken]
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 http://www.youtube.com/watch?v=5CK_5KCKBUw"
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.
 
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  • #13
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 http://203.161.71.130/forum" in the fuel consumption testing section.
 
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  • #14
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.
 
  • #15
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.
 
  • #16
33
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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?
 
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  • #17
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??????
 
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  • #18
That's amazing as a result. How about the road test using 2 cars I proposed in Post #8 above?

I do have data but its a little shaky and no proof.

My car when engine was new and when driven at 80km/h in overdrive 4.18 L/100km.
My EFI car when driven at 80km/h in overdrive 6.33 L/100km

My next job is to work out the trip travel times and the estimate cc/min and measure rpm.
 
  • #19
Im learning slowly...

I compared the fuel consumption of a normal engine and my engine using the fuel required for no load and the fuel required for load and compared the thermal efficiency of the power stroke only.

Normal engine output 4.334 kw 1730 rpm consumes approx 96.24 cc/min
I then worked out how much fuel goes up the the other 3 cycles and arrived at 48.25 cc/min that leaves 47.75 left to do work in the power stroke.

Does anyone know if there are graphs of this on the net?

47.75cc/min / 48.25cc/min = 0.988637 which is a ratio.
Calculated thermal efficiency on the power stroke.
62kj/kw
4.334kw = 286kj
47.75cc/min = .04775 * 34.8 = 1.6617 mj = 1667 kj
So 286 / 1667kj = .0156 = 17.1% power stroke thermal efficiency at 1730 rpm and 4.334 kw.

Modified engine output 4.334 kw 1730 rpm consumes approx 42.61cc/min
I then worked out how much fuel goes up the the other 3 cycles and arrived at 29.16 cc/min that leaves 13.45cc/min left to do work in the power stroke.

13.45cc/min / 29.16cc/min = 0.461248 which is a ratio of load to no load.
Calculated thermal efficiency on the power stroke.
62kj/kw
4.334kw = 286kj
13.45cc/min = .01345 * 34.8 = .468 mj = 468 kj
So 286 / 468kj = .5542 = 61.1% power stroke thermal efficiency at 1730 rpm and 4.334 kw.

What I did next is divide the efficiency percentage to find the ratio
61.1% / 17.1% = 3.57 times better at that particular load/rpm in the power stroke.
So maybe the best or ideal % is 68.4% - 1 stroke of friction.

I think the highest theoretical ratio is about 4 eg raise compression to the maximum level where you still have control. not very practical though but it would prove a point.

One thing I have tested a long time ago is my engine is approx 10% efficient at full throttle.
Are there any graphs for load and thermal efficiency on the net because I think I have move the line around a bit.

I am currently thinking this explanation for the increase in efficiency in the power stroke.
I have treated the power stroke like in electronics Ohms law, when you double the voltage eg double the distance, with the same load resistor eg kw output you get 4 times the power eg the thermal efficiency or work of the power stroke but not more torque. I know the problem is much more complex that but I only have a theory on it at the moment.

So what I am saying if you can go double the distance on the same fuel you efficiency must be 4 times better somewhere??

I will work on my first run in a few days where I did a rough figure of 4.11L/100km my reference is at approx 110 km/h so I have to adjust the 9.41 L down a little.
I have a reference from my EFI car its rated at 7L/100km on the HWY and I have done a trip at 80km/h. My results probably wont be that accurate though.

I'm open to your thoughts....
On the lighter side...
I am interested in ignition system that fires a spark and I can also feed music into it and listen from the ignition coil....
 
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  • #20
cjl
Science Advisor
1,892
461
How are you determining your engine's output power? You keep using that 4.334 kW figure, but how was it measured?
 
  • #21
How are you determining your engine's output power? You keep using that 4.334 kW figure, but how was it measured?

I calculated it from the ADR rules the city mileage test the dyno setting is 6.7kw at 80 km/h and a dynamic load for acceleration which I ignored then assumed a 30% increase for drivetrain losses. I worked out my average speed then then calculated the load for 40.95 km/h. I know my methods are rough but I dont have any money to do it the right way I guess this shows in my results.

My first test on my test track was 40.13 mpg (imp)
then 41.97 mpg (imp)
then 42.44 mpg (imp)
and the last test was 45 mpg
So that's a 12% increase on the same distance track.
I made small improvement between the tests the raw data is available at "ftp://203.161.71.130/Engine%20Data/Raw%20Data/"[/URL]
 
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  • #22
cjl
Science Advisor
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That seems like that could be your error. Honestly, I doubt that your estimation is at all correct, and I'd be more willing to believe that it was wrong than that you're getting >60% efficiency.

(Oh, and just so you know most car engines have their best efficiency at high throttle at around 2500-3000 RPM. You can find actual, accurate efficiency charts if you search for "Brake specific fuel consumption")
 

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