Is IC Piston Engine Really Less Than 1% Efficient?

I have some doubts about the torque figures used thus far, and about the consistency of measurement units. Let me suggest a few grains (or shovels) of salt....

I reply only to your title header. According to the relevant research publications, the efficiency of a modern IC motor ranges between 24 to 30 %. Depends on the motor. This is low enough, but this is what there is.

Hi! Yes, I agree with you and that is precisely the point that I am trying to make ! For years and years I took it for granted that the efficiency of the IC piston engine was around 20% - 25%. because that is what I had been told ! Now, when I have investigated the phenomenon for myself I find that it just cannot be true. The laws applying to levers, which incidentally are almost the oldest physical laws known to man. Say that it cannot be true. It is just simple common sense. The amount of torque generated depends on the length of lever. In the IC piston engine the length of the lever happens to be extremely small ! Add to that the fact that the force to the lever is applied at a continuously changing angle that averages out to an angle of Sin 10, add to that the fact that the exhaust valve has to open 10 degrees to 15 degrees before BDC if the crank shaft is not going to shatter, and the vaunted efficiency of the IC piston engine begins to melt like butter on hot toast.

Hi John,

Yes I am talking about a simple normally aspirated engine, without turbo chargers etc., and as I had tried to demonstrate the calorific value of the fuel used compared to the amount of torque generated turns out to be on the order of 0.7% which is really cause for concern. The question is do we keep on pretending that we are getting efficiencies of 20% - 30% from these engines or do we do something about it ?

Your investigation is too complex. But just to correct

from where did you get that value of 3.6MJ/lt? Petrol has only about 50 (a bit less) KJ/Kg which boils down to almost 65KJ/L

Your values are inflated by a factor of approximately 55.

At this assuming all your calculations are correct, the efficiency will be 0.5*55 = 27% almost. And that is almost the figure expected.

Hi!

Interesting post and I must thank you for correcting me, the actual amount of energy in 1 litre of petrol is anywhere between 33 MJ/l and 36 MJ/l and not as I had erroneously stated 3.6 MJ/l so that means that the efficiency of the IC piston engine works out to 0.07% and not 0.7% as previously stated. Sorry for the mistake and thanks for pointing it out. Your figure of 50KJ/l is as you can see out by something like a factor of 10 ^^3. As for your calculation that my figures are out by a factor of 55, I don't really see how you arrive at that conclusion.

P.S. Check out this site to see the calorific values of different fuels:

http://www.bts.gov/publications/national_transportation_statistics/csv/table_04_06_m.csv

You will find the information under notes if you scroll down.

I am not sure where your 2:1 bore to stroke ratio came from. Its common to see engines with equal bore to stroke numbers and some of the old stationary engines used 1:2 bore to stroke ratios.

The length of the connecting rod and what RPM range the engine will run at determines the much of that ratio.

Sounds like you may need to do a lot more research and homework!

GA.

As I was reading through the posts I was wondering if someone was going to comment on the relationship of piston stroke and crankshaft throw and the fact that the diameter of the piston is not a factor regarding crankshaft throw.

NO! The throw of the crank shaft has to be half the stroke of the piston. The diameter can be any value the designer wants, as long as the connecting rod does not strike the cylinder walls.

Try imagining a cylinder that is 1 ft long and has a diameter of 3 ins. What happens when the piston goes down the cylinder, you have a really long connecting rod but the throw of the crankshaft would have to be tiny wouldn't it ! That is why in an IC piston engine the bore and stroke are closely related. So your stipulation that the diameter of the piston can be any value as long as the connecting rod does not strike the piston walls is really a non sequitur. Ok. Yes, the normal definition of the crank throw is taken as half the stroke but this is closely related to the piston bore as is illustrated by the above example. But equally often the throw of the crank shaft is taken as half the piston bore plus minus 15%. So the basic argument I have made in this thread remains unchanged and if you still have doubts just look up the specifications of as many different car engine makes as you can.

An engine with a 3 inch bore and a 12 inch stroke would need a long connecting rod and considerably greater spacing between the crankshaft center and the cylinder base but its entirely possible to build one and have it work without cylinder wall to connecting rod clearance issues. Basic geometry shows it can work. The greater the distance between the piston pivot point (wrist pin end) and the crankshaft pivot point (rod bearing end)the lesser the deflection at the piston.

The old low speed steam engines often used large strokes with lesser bores to create high torque and use as much expensive energy of the steam as possible in single stage engines.

Here is a better explanation about bore and stoke ratios and the how's and why's behind them. http://www.statemaster.com/encyclopedia/Stroke-ratio#Stroke.2Fbore_ratio

An engine with a 3 inch bore and a 12 inch stroke would need a long connecting rod and considerably greater spacing between the crankshaft center and the cylinder base but its entirely possible to build one and have it work without cylinder wall to connecting rod clearance issues.

What would be the point ! Maybe if you could calculate just how long the connecting rod would have to be it might be really interesting !

"What would be the point ..." ?

IMHO, the point would be to illustrate that your erroneous statements about limitations of the bore/stroke ratio are more examples of your woolly thinking.

I almost dug out my 20 year old LEGO set and built him a demonstrator model! (1/3 scale of course.)

I think that all college students in India (and America now too) should be required to have a LEGO and Erector set lab that is 4 hours a day 5 days a week for two at least two semesters before they can become engineers.

Did you ever use Meccano?

[Edit: Woops! Learn something new every day - seems Erector is merkin for Meccano].

Wooly!!! Just calculate (a) length of connecting rod (b) length of crank throw!

"Calculate"? Depending upon the mechanism used, they can be nearly any relationship you'd ever want - but it's the bore/stroke relationship that was the topic. Connecting rods are typically anywhere from about 1.2 to 1.7 times throw in common practice. Look at the mechanism at http://www.petervaldivia.com/technology/mechanisms/steam-engine.php. Observe that the diameter of the piston is independent of the crankshaft throw or connecting rod length; the identical mechanism could have a rod the size of where it passes through the sealing gland, for example, giving a stroke of, say, 20 times the bore. This is totally true for internal combustion designs as well as steam engines. And yes, "woolly thinking" applies to the post that it commented upon.

Heres a few referance points if you have a problem.

Here is my Minneapolis Moline tractor. As you can see it has a 4.25 inch bore and a 5 inch stroke. http://www.tractordata.com/farm-tractors/000/9/6/961-minneapolis-moline-uts-engine.html

Here is a John Deere model R diesel with a 5.75 inch bore and a 8 inch stroke. http://www.tractordata.com/farm-tractors/000/0/3/34-john-deere-r-engine.html

Here is a list of Ford 460 big block engines. As you can see 4.36 inch bore by 4.3 inch stroke is used. And these engines can run at over 6000+ RPM is built properly. http://www.4wheeloffroad.com/brandpages/ford/131_0404_533_cubic_inch_big_block_ford/index.html

Here the supposedly most efficient gasoline engine in production. The Toyota Prius. As you can see it has a 3.17 inch bore by 3.48 inch stroke. http://www.toyota.com/prius-hybrid/specs.html

I dont know what all else you are talking about but I do know that ratios of far greater than 2:1 bore to stroke are used every where in billions of engines of all makes and designs.

Seems to me someone did not do their homework on practical engine designs used in real life applications.

We are talking at cross purposes here, ofcourse there are square engines and over square and under square engines as your post illustrates. I just can't figure out where you got this 2:1 ratio from!

Here is one. Of course it's not an IC engine, but with a very slow-burning fuel, it could be...

Hey guys!!

DDjames appears to also go by geartrust. See a similar discussion here. Revolutionary Gearbox Designs Disclose Ignoring the Physics Laws .

After non answering questions and getting pinned down on specifics he left that thread.

Hey Guest!!

Look if you don't have the guts to put your name to your opinions, maybe its better if you just keep your opinions to yourself. That's where they'lldo the most good .

Deploying pudding cup bait, ENGAGE

Hey! packrat,

I am talking about your IC piston engine and not some hypothetical gearbox design!

I am talking about your IC piston engine and not some hypothetical gearbox design!

My IC piston engine? You mean the one sitting on the floor of my garage? The 250CID Ford in-line six cylinder? How did you know about it? I thought you were discussing internal combustion engines in general, excuse me while I go close the shutters on the garage window.

Dear Massey,

Before we get into all this talk of bursting bubbles etc., Do you mind if we just check your figures ? OK! You state that the hourly consumption of gas for your reciprocating gas generators is as follows:-

1) Hourly consumption of gasoline = 2830 KW = 2,830, 000 W = 2,830,000 J right !!!

2) One litre of diesel = 37,000,0000 J !!!

Now an average car that gives about 15 Km/l traveling at 60 km/hr uses up one litre

of fuel in 15 minutes right, and that is just with an output of about 15 hp or 11,190 W. You are stating that an engine that produces 1035 KW of power at the terminals uses just a fraction, 7% to be accurate, of 1 litre of fuel to get that kind of output.

Let me tell you sir, this forum has the reputation of being one of the leading engineering forums in the world. Yes, CR4 has a huge reputation, so where are all these facts coming from. Does it make you feel better to think that your reciprocating engines are running at , what was it 95% efficiency. Excuse me, the figure is more like 0.95% efficiency.

I have quickly scanned most of the posts and simply say:

It is unlikely that millions of competent engineers have been wrong on the point you make for so many years, and the more so for those who operate energy in-energy out type plants as Massey76 appears to be.

If you did have a case to make it could only be by taking the fuel cycle back to an earlier part of the cycle - for example when the sun shone on the plants that became the oil that is used in today's engines.

It seems that the core issue/error in your proposition is about how much energy there is, or is not, in 1 litre of fuel - and that you have used the wrong value.

It might be useful to note the wisdom of Occam's Razor as part of reconsidering your case.

.

"Hourly consumption of gasoline = 2830 KW = 2,830, 000 W = 2,830,000 J right !!!"

Think we could be onto something here. What is your definition of the Joule?

Why not cross into Europe and start using the metric system, things would be far clearer to you then, instead of questioning facts that are well known.

joule

   /dʒul, dʒaʊl/ Show Spelled[jool, joul] Show IPA –nounPhysics. the SI unit of work or energy, equal to the work done by a force of one newton when its point of application moves througha distance of one meter in the direction of the force: equivalent to 10⁷ ergs and one watt-second. Abbreviation: J, j

The facts as most of the world know them.

You don't seem to understand the metric system yourself.

Massey,

Good illustration.

GA.

Stu

Hi Stueywright,

Thanks for comment and the GA! Its a funny thing but when you are really involved with something as basic as the energy conversion business and you are using first principles stuff on a daily basis the current topic really gets under the skin!!

Best wishes,

Massey.

DDJames-

There is something very wrong with your calculations, but I do not have the time or the energy to pick through them. I will say that Massey's numbers, based on actual measurements of stationary power plants. are much closer to my own measurements of power produced per gallon of fuel (although I must say, I am a bit envious of Mr. Massey- my numbers are a bit closer to 30%- probably because I am working with smaller or older equipment). In my book, real world measurements trump theory any day.

When considering the "efficiency" of the IC engine used for transportation purposes, how much of the energy consumed is actually used for moving a load through a distance? I suspect it varies significantly from driver to driver, and whether the load is being moved in the city or in a rural environment. I would not be surprised if your typical driver actually got an effective efficiency far less than the engine is capable of, but you would have that same problem with virtually any technology with which you tried to use to replace the internal combustion engine.

You are, I believe, chasing moonbeams...

Ok, lets make this simple, you give me your definition of how many joules there are in 1 litre of petrol and we'll take it from there and compare that to the figures Mr. Massey has given !

Hi cwarner7 11,

Good to hear your views. The plant to which I refer is a 1996 dated GE Jenbacher installation featuring their 320 series engines running at 1500rpm,which employs a 'lean burn' technology with an air to fuel ratio varying from 14 to 17 : 1.

It also uses a computerised ignition timing system(Morse) which through monitoring a pair of seismic detectors (one per bank of cylinders, advance until knocking is detected then retard a 'tad) is able to continuously alter the spark timing to optimise the combustion with respect to load and methane number. Other manufactures did not possess this level of sophistication at that point in time though many do nowadays.

If you are interested you may find the GE Jenbacher website of interest where you will see that their latest products are claimed to have 40% shaft efficiency.

I believe that once laser ingnition technology becomes established that thermal efficiency will rise further.

Best wishes,

Massey.

My old car runs at around 14 Kms/lt

so that is about 36 /14 = 2.6 MJ/km

Let me assume the combustion to be about 60% (though it is maintained well and emission are much below the required norms). The bye products water and CO₂ being stable are at low energy levels.

So the combustion generates say about 1.5 MJ/km I drive. (2.6*0.6)

Out of this a portion is only escaping as heat (engine, drive chain friction, exhaust gases) - I am taking system approach as exchange of energy - wheel frictions are not part of it. And all these are escaping as heat energy.

As per the OP, only 0.05% is actually used. That means at say 40Kmph speed i am generating about 1MW heat output ?

How about running a small thermal powerstation with it?

You really don't have a clue as to what you are talking about do you !!

I have and i am talking in terms of law of conservation of energy. (I assume here the mc² portion is a bit insignificant).

Once I assume that, the energy of the petrol after combustion has to go somewhere?

- Insignificant energy lost in EM. The spark energy is almost totally retained within the chamber? No light emission (except headlight), Just a fraction stored as chemical through alternator and batteries

- mechanical Energy is the output

- left out is the heat energy

- the other loss is due to incomplete combustion of the fuel (that i took to be 60%)

So the

Total Energy = Combustion Losses + mechanical Energy + Heat Losses

= Total Energy* combustion Efficiency + mech Energy + Heat Losses.

1 = Combustion Efficiency + Mech Energy/Total Energy + Heat Loss/ Total Energy

System Efficiency = MechEnergy/Total Energy = 1 - combustion Efficiency - (Heat loss/Total Energy)

0.0005 = 1 - (0.6) - (Heat Loss/Total Energy)

heat Loss = Total Energy * (0.4 - 0.0005) = 0.3995 * Total Energy

= 0.3995 * 36 MJ/L = 14 MJ/lt

If 1 lt runs for say 90 Sec, then the power wasted = 0.15MW - ideal condition as per manual 40 KMPH on level road for best combustion efficiency.

This is an excellent topic raised by Mr James. As an electronic engineer I had been only told by someone that Diesel engines are inefficiecnt Hence earlier I had floated the idea of a 2^N Internal Combustion engine in this forum to improve efficiency. I was not aware of calculations then. This thread is useful from that point of view.

Hey Guy, Found a couple of aberrations in you calcs. No matter. Not the time to go into them now. Just passing by (like the 'noses on the faces of the maids of the court of King Karactakus)

Having swung spanners for 55 years, I'd sure like to get rid of the bastard thing too.

Desperate for a motive device which is cleaner, quieter, easier to fix, and more efficient

But, which one????????

Cheers,

Stu

Hi!

I would sincerely like to know what those aberrations might be. Just take the trouble to write a few lines, on where you thinkImight have gone wrong, I would appreciate it

Hi!

Thanks for taking the time. The figure of 2830 KW that you had attributed to me was actually taken from Mr. Masseys post. Ofcourse if you had an output of 2380 KW for one hour the figure would have to be given in KWH. So 1 J/sec = 1 W and 1W/sec = 1J.

I quote from Mr. Masseys post : "Each engine has an hourly consumption rate of 2830 kW gas and produces 1035kW of electricity at the alternator terminals."

So what am I supposed to make from that ? "An hourly consumption rate of 2830kW gas ??? " Clearly it means that in one hour his engine consumes 2830kW of gas, which you must admit is food for thought, when it is known that 1 litre of diesel as you yourself had stated has an energy of about 37 MJ.

However IF there is a mistake in the calculations that I had made, regarding the torque produced etc., I think it would be more constructive for the moment to discuss that aspect and then maybe we could go on to discuss the thermodynamics part of it.

The IC piston engine is inherently inefficienct and it is never going to improve as has been proved by the introduction, of MPFI, multiple valves, double overhead cams, DFI and so on. With all these improvements one would have expected quantum increases in efficiency that simply have not materialised. I state again that the root cause for this lack of improvement lies in the three factors I had mentioned.

1) Extremely short lever arm as represented by the crank throw.

2) Varying angle of attack of the connecting rod as it comes down the cylinder.

3) Premature opening of the exhaust valve before piston reaches BDC.

Further the manufacturers figures( forthe Maruti Suzuki 800) themselves namely 37 bhp at 5000 rpm show that each power stroke produces just 4 ft lbs of torque. i.e 37 x 550 / 5000 = 4.07 ft lbs. BHP is mechanically measured directly from the crankshaft (flywheel) so there is not much room for error. Take it from there.

All that stuff about short throw, variable angle whatever is part of the inefficiency of the internal conversion engine BUT mainly expressed ultimately as "friction loss". The other main losses are "exhaust heat", "pumping loss" and "heat loss to coolant".

These energy balances have been done a million times and are well known - your mistake will be in something basic like a units conversion/comparison and I really don't feel like going through it all, if you won't, to find it - unless of course you want to put a bet on it !! Would $1000.00 be OK

All that stuff about short throw, variable angle whatever is part of the inefficiency of the internal conversion engine BUT mainly expressed ultimately as "friction loss". The other main losses are "exhaust heat", "pumping loss" and "heat loss to coolant".

______________________________________________

I have used the manufacturers own figures to show that the power produced during the power stroke is woefully inefficient. BHP is calculated directly from the crankshaft output, there is little leeway for error. What I state is just common sense. You say that a short crank throw translates into frictional loss ??? Yet 1 lb of force applied at right angles to a lever that is 1 foot long produces a torque of 1 ft lb if the other end of the lever is attached to an axle. So in this situation the lever is 100% efficient. (output/effort ) x 100 = 100%. Now if the length of the lever is a fraction of a foot, the mechanical efficiency is proportionately reduced. If further to this the pressure is applied to the lever at something other than 90 degrees then the efficiency is reduced even further. What I am trying to show is that in the IC piston engine this reduction in efficiency is so great that the mechanical efficiency of the engine works out at something less than 1% !! That is the case I am trying, rightly or wrongly, to make.

The only problem with your theory seems to be that there is nothing rotating in the IC piston engine. You seem to think that the crankshaft rotates by itself. I think that guest and his crazy calculations are a sight better than your theory.

Great galloping dung beetles, man!

What are you talking about?

The crankshaft is rotating in an ICE. The piston is moving linearly thus power equals force times velocity of the piston. The crankshaft rotates thus power equals torque times angular velocity.

I think we are getting to the nub, here.

Your statement:

" 1 J/sec = 1 W and 1W/sec = 1J "

and other examples of sloppy thinking such as overlooking the difference between hours and seconds^[1], and:

"... averages out to an angle of Sin 10 ..." ^[2]

suggest that perhaps it's time for a refresher course in elementary maths and physics before you resume your postulations.

^[1] #13

^[2] #4

What did you open this thread for?

You've said that it's a heap of crap, we've agreed with you, and now you're complaining??????

Personally, the demise of the IC piston engine can't happen soon enough.

Cheers,

Stu ( Engine Designer)

I need some clarifications pl..

1. The last sentence of para 2 should read (not pressure exerted) -"And the total force generated on this area would be 500 x 3.14 = 1570 lbsf."

2. The catch in the argument hinges around the Sin 10. I did not understand from where this came from? Is it possible to provide some diagrammatic representation for this?

3. You said Maruti has a bore of 68.4 mm and a stroke of 72 mm- while you had argued that stroke length has to be ½ of bore dia !!!

4. For Maruti stroke length being 2.8 in = 0. 233 in ( why was 0.112 in taken instead)?

5. Hence torque of 4.07 ft lbs results in 4.07/0.233 = 17.44 lbsf and not 36 lbf which in turn divided by Sin 10 becomes 100.8lbsf.

6. Then next paragraph you find Torque/ initial force which also has a radius factor in it !!! Torque = Force x Radius. Hence there is some confusion in computation of efficiency.

7. The Maruthi has a fuel intake of 800 cc. So how did you get 0.02 cc?

I do find that you have made a very logical argument, but kindly explain some assumptions like 1) Sin 10, 2) Torue/ initial force, 3) considering 0.02 cc, 4) Initial force of 2846 (how did you arrive at this).

This will give me food for thought.

You said Maruti has a bore of 68.4 mm and a stroke of 72 mm- while you had argued that stroke length has to be ½ of bore dia !!!

4. For Maruti stroke length being 2.8 in = 0. 233 in ( why was 0.112 in taken instead)?

_______________________________________________________________

I think your confusion is justified, since it was a long post and therefore difficult to follow. But first to clear up some of the points you had mentioned:

1) First I had never said that the stroke length has to be half the bore diameter, that would be ridiculous! What I did say was that the throw of the crankshaft has to be half the diameter of the piston bore.

2) What is the throw of the crankshaft ? You understand that the motion produced by an IC piston engine is linear (i.e., the piston moves up and down the cylinder), am I right ? This up and down or linear motion has to be translated into the rotary motion of the crank shaft. This is achieved through the crank throw, which you can imagine as a short lever attached at one end to the crank shaft and at the other to the connecting rod which is in turn attached to the piston.

3) Coming to the Sin 10 part of the post. Ideally in order to produce maximum results from a lever, the pressure has to be applied at right angles. For instance if you are turning a nut with a spanner, you apply the pressure at right angles to the end of the spanner. However, in the IC piston engine this is not possible. The piston is moving down the cylinder perpendicularly so the angle at which the connecting rod is applying pressure to the lever (crank throw) is constantly changing. If averaged out this might equal a value of about Sin 10 which is equal to 0.173. Sin 90 = 1 or unity, so if there is a deviation from this angle of 90 degrees in the pressure applied, the resultant will be less than unity.

4) Lastly the Maruti 800 has a displacement of 800 cc , the fuel intake on each stroke works out to be about 0.02 cc.

If you have understood what I have written so far, maybe we can go into further explanations.

Much of the whole calculation/discussion is indicative of someone who does not understand physical systems particularly well.

______________________________________

And that kind of attitude, in a nutshell, is why we are at an impasse where the IC piston engine is concerned. and again

____________________________________

A short lever or an "unfavourable" angle reduces the force/torque but increases the

"speed" or something else to offset the supposed disadvantage.

____________________________________________________

Wow !! Mind blowing ideas!

There was slight mistake in my computation posted earlier. See updated figure is still worse !!!

1. Piston displacement = 796 cc. Hence petrol intake = 796 cc.

2. This is assumed to be a 4 stroke engine. Hence petrol is drawn in once in every 2 revolutions.

3. It has 3 cylinders.

4. It generates 39.5 bhp at 5500 rpm.

5. Hence 796 cc x 3 cylinders x (5500 rotations/2) in one minute = 6.567 x 10^6 cc of petrol is consumed in one minute or 6.567 x (10^6)/ 60 = 1.0945x (10^5) cc per sec.

6. Taking petrol has 33,000 kilo J per liter (got this from website), 1.0945 x 10^2 liters of petrol is consumed in 1 sec. Hence 33 x(10^6)x 1.0945 x 100 = 3611.85 x 10^6 joules of energy is input.

7. Output = 39.5 bHP. 1 bHP = 745.7 w Hence output = 39.5 x 745.7 w in one second = 2.9455x (10^4) Joules.

8. Hence output x 100/input in %= 2.9455 x(10^4) Joules x 100/3611.85 x (10^6) Joules. = 0.08155% only !!!!!

9. I should be throwing out the Maruti car and every other fuel engine vehicle.

10. Here is food for thought- LOOK FOR ALTERNATIVE= INNOVATE.

I have been analyzing some of the statements made by Mr Tardis at comment 45 .

TARDIS says

"That is not correct. It doesn't matter how long the lever is. As long as there is no friction between the rotating elements the power out will equal power in (I'm only talking about a lever here). Regardless of the moment arm (or length of the lever) the power in will be the product of the applied force the velocity at the location where the force is applied. The power output is the product of the torque and angular velocity of the shaft. So the moment arm is irrelevant."

Power = force x velocity – I accept this in terms of dimensions. Next statement is Power out is product of Torque and angular velocity of the shaft !! Hence Force x radius / sec which means angular velocity has unit of 1/sec only !! Hope my guess is right.

Another statement is by Mr Trevor at 39.

Trevor states

"The sort of loss/inefficiency you are seeking to explain is nothing to do with lever advantage and that sort of concept because at the end of the day (friction and heat loss ignored) power = pressure by volume/sec. A short lever or an "unfavourable" angle reduces the force/torque but increases the "speed" or something else to offset the supposed disadvantage."

Power = Pressure x volume / sec is correct as Pressure = Force / meter^2 and Volume = Meter^3. Hence Power = Force x meter / sec – fits well dimensionally. Hence we are concerned about area enclosed by pressure x volume curve..

Hence it is not power = pressure by volume/sec. Kindly confirm what I have stated above.

"angular velocity has unit of 1/sec only" That is correct. However one must be measuring the angles in radians, and angular velocities in radians/sec. Since the radian itself is a ratio of two distances, the units of those distances cancel; thus the radian has no units, and mathematically radians/sec is 1/sec.

Since engine speeds are commonly measured in RPMs, those values must be divided by 60 to get RPS, and multiplied by 2π (2 pi) to get radians/sec or 1/sec.

"Hence we are concerned about area enclosed by pressure x volume curve.." This is also correct, as long as you don't forget time. The area under the PV curve is work or energy. Divide that by the appropriate time and you have power.

Mr James- see I ahve taken a different route to compute using Maruti manufacturer's data sheets. BUT YOU ARE RIGHT - EFFECIENCY IS AS BAD AS < 1%

1. Piston displacement = 796 cc. Hence petrol intake = 796 cc.

2. This is assumed to be a 4 stroke engine. Hence petrol is drawn in once in every 2 revolutions.

3. It has 3 cylinders.

4. It generates 39.5 bhp at 5500 rpm.

5. Hence 796 cc x 3 cylinders x (5500 rotations/2) in one minute = 6.567 x 10^6 cc of petrol is consumed in one minute or 6.567 x (10^6)/ 60 = 1.0945x (10^5) cc per sec.

6. Taking petrol has 33,000 kilo J per liter, 1.0945 x 10^2 liters of petrol is consumed in 1 sec. Hence 3.3 x(10^6)x 1.0945 x 100 = 361.185 x 10^6 joules of energy is input.

7. Output = 39.5 bHP. 1 bHP = 745.7 w Hence output = 39.5 x 745.7 w in one second = 2.9455x (10^4) Joules.

8. Hence output x 100/input in %= 2.9455 x(10^4) Joules x 100/361.185 x (10^6) Joules. = 0.8155% only !!!!!

9. I should be throwing out the Maruti car and every other fuel engine vehicle.

10. Here is food for thought- LOOK FOR ALTERNATIVE= INNOVATE.

After filling the tank for howmany seconds your car run?

pause to think for a moment before using maths.

Piston displacement = 796 cc. Hence petrol intake = 796 cc.

Hence petrol is drawn in once in every 2 revolutions.

RPM = 5000 = 2500 double revolutions = 2500 Intakes per minute

= 2500 * 796CC per minute

= 1990 lts/minute

With an average tank capacity = 30 lts, your full tank will last only for 1 second (or just less than it)

I am sorry, you really will have to throw your car out with this rate of consumption.

Or are we talking about Boeing ?

"1. Piston displacement = 796 cc. Hence petrol intake = 796 cc."

You're off to a very BAD start! MOST of the intake is AIR! If you filled a cylinder with petrol, your engine would come to a very fast and catastrophic stop (or it wouldn't have started in the first place). Ignition of any fuel requires a fuel/oxidizer ratio within a certain range, and that range does not include infinity (or however else you may define 100/0).

Yes, the old hydraulic lock problem

Mr Divekar,

How can you post such figures, most of this is erroneous, for example:

"1.0945 x 10^2 liters of petrol is consumed in 1 sec."

Since this number is better written as 109.45 litres, how many small car fuel tanks would be suitable!

109 litres per second of petrol! I am so glad that I do not live in your numerical world!

I reckon to need 35 litres for the whole of my weeks driving!

1. Now I can see that I have left out most important data (not available to me) is the fuel to air ratio.
2. Hence entire 796 CC drawn in is not petrol.
3. Only some automobile expert can provide some data - likely figure.
4. I do not see any other mistake in the calculation or approach.
5. I do see point many have made that the car cannot run for few seconds with such high fuel consumption.
6. Hence incorporating Fuel:air ratio will change efficiency figures dramatically.
7. Since I have used only one system of units (SI and not outdated units) and not confused with bore dia, crank shaft throw etc etc, the calculations are easy to understand.
   
8. Anyway the exercise was worth it as I ma trying to see how to push my idea of 2^N Internal combustion engine posted in CR4 sometime back.
   

MS,

You post replies as you might be an expert in auto engineering.

As such, did you not know about the stoichiometric ratio of air to fuel, in the case in point, of petrol?

Cheers,

Stu.

Assuming MS has an intake of 0.02cc/ stroke and there are 5000 power strokes per minute it is 100 cc/minute

or 100/60 cc/sec

or (100/60) /1000 liters/sec

= (1/600) lts per sec

= (1/600)* 36 MJ /sec

= 0.06 MW

= 60 KW

Mech Power output is 37BHP = 27 KW

Efficiency = 27/60 = 45%

All as per OP data (and it looks to be a bit too much, I would have assumed only half of it)

All input data is as supplied by OP (36MJ/lt, 5000 RPM, 0.02cc/stroke, 37 BHP)

Infact as far as I remember there are only 1 power stroke per 2 RPM and that way the efficiency will shoot up to unimaginary level.

So either my calculation is wrong.

Or the input data is GIGO.

Let me recalculate incorporating Air: Fuel ratio.

1. Piston displacement = 796 cc. Hence petrol intake = 796 cc.

2. Manufacturer also states that the compression ratio is 9:1. May be this also means Air : Fuel = 9:1. Liquid cannot be compressed and hence the piston can move all the way up till air is compressed.

3. Hence Fuel intake is 796 x 1/(1+9) = 79.6 cc. I do not know from where 0.02 cc figure is appearing and what is the basis for it.

4. This is assumed to be a 4 stroke engine. Hence petrol is drawn in once in every 2 revolutions.

5. Further some people are confusing stroke to revolutions !!

6. A stroke is equivalent to just 180 degrees or half a revolution in a 4 stroke engine. Hence in a 5500 rpm, fuel is drawn in 5500 / 2 times in a minute.

7. It has 3 cylinders.

8. It generates 39.5 bhp at 5500 rpm.

9. Hence 79.6 cc x 3 cylinders x (5500 rotations/2) in one minute = 0.6567 x 10^6 cc of petrol is consumed in one minute or 0.6567 x (10^6)/ 60 = 0.10945x (10^5) cc per sec. = 0.10945x (10^2) liter

10. Taking petrol has 33,000 kilo J per liter (got this from website), 0.10945 x 10^2 liters of petrol is consumed in 1 sec. Hence 33 x(10^6)x 0.10945 x 100 = 361.135 x 10^6 joules of energy is input.

11. Output = 39.5 bHP. 1 bHP = 745.7 w Hence output = 39.5 x 745.7 w in one second = 2.9455x (10^4) Joules.

12. Hence output x 100/input in %= 2.9455 x(10^4) Joules x 100/361.185 x (10^6) Joules. = 0.8155% only !!!!!

13. So efficiency figure has moved up by just one order, but still less than 1%.

14. I look forward to someone challenging this backed with solid data and not making critical comments.

compression ratio is 9:1. May be this also means Air : Fuel = 9:1.

Are you serious ?

The critical Air Fuel ratio for Gasoline Engines is 14.7:1 (by mass)

remember both are not haveing same density.

Hence Fuel intake is 796 x 1/(1+9) = 79.6 cc. I do not know from where 0.02 cc figure is appearing and what is the basis for it.

The value 0.02 has come from OP, I simply didn't bother to calculate it. It could have been easily calculated based on the above data and the density of air (about 1 Kg/m³ under STP) and converting it to the combustion parameter. Gasoline of course has a Rel density of 0.72 approx (ie 720 Kg/m³)

This is assumed to be a 4 stroke engine. Hence petrol is drawn in once in every 2 revolutions.

right.

The total displacement of the engine is 796cc

Hope this much of solid data is enough.

Are you factoring your air/fuel ratio as mass or volume. It will make a big difference if you are using the wrong one.

First you suggest that 238.8 CC engine at 5500 RPM uses .6567(10^6) = 656700 CC of fuel a minute. Thats wrong , it uses 656700 CC of air and fuel mixed together.

Also I see 33000 apparently referenced as 33 X (10^6). That causes another questionable calculation issue.

If you want to make things simpler and easier to understand stick with whole numbers and don't mix in scientific notation in here and there.

I suggest you recheck your math and what the standard of what and A/F ratio is measured by, volume or mass.

I just want to bring to your attention that I have an old MARUTI (originally SUZUKI of Japan) car. I have the User's manual from which I have reproduced all the data. Unfortunately it does not spell out Air : Fuel ratio. I took it as Compression ratio. Hence my assumption was based on volume and not mass.

What puzzles me is - I still have not seen anyone putting solid data to prove that petrol engine efficiency is not as bad as <1%, but somewhere 45 to 50%. I am arguing with concrete data. I am stating that Engine rating is 796 CC as per manufacturer's User's manual. From where did you bring in 238.8 cc (I have not said).

I wish I had facility to scan and post it on this web site. Do not doubt it. Of course I found from web site that Petrol has a energy value of 33KJ/ liter, but James has said 3.7MJ / liter (which is 10 times less than what I am saying). See I have pasted statement from this link www.cleavebooks.co.uk/dictunit/notes2.htm.

"A litre of petrol contains about 33,000 kilojoules of energy, and a car traveling at 50 kilometres per hour might use 1 litre of petrol in about 10 minutes.
This is a power of 55 kW (or about 70 horsepower). Of course, due to the inefficiencies of the system only about 25% of that is actually delivered to the wheels. (Unfortunately!)
Whereas if 1 litre of petrol were burnt in just 1 second, as in an explosion, then it would produce 33,000 kW of power (over 44,000 horsepower).
Incidentally, it would be almost 100% efficient!"

Other sources for the energy content of fuels include http://en.wikipedia.org/wiki/Gasoline#Energy_content and http://hypertextbook.com/facts/2003/ArthurGolnik.shtml.

They all generally agree on something around 44MJ/kg, which is around 32MJ/liter.

Note that you say 33kJ/liter, whereas your link says 33,000 kJ/liter, which is 33MJ/liter.

AND, your link indicates 25% of the energy delivered to the wheels. Since there are lots of losses between the engine and the wheels, the engine efficiency must be significantly higher.

BTW, regarding that last statement of your quote: "Incidentally, it would be almost 100% efficient!" 100% efficient at performing what task? Blowing you and your car to smithereens?

I do agree with you that some where I have written 33kJ/liter. But I rechecked my calculation where I have taken 33,000 KJ/ Liter. I am reproducing the calculation. The last line about 100% eff is not mine- it is from that web site I gave you link.

I have also seen comment by Mr John DG about - definition of compression ratio. I have really no manufacturer's data about Air: Fuel ratio. Liquid cannot be compressed, only air is compressed. Hence using this logic I have equated compression ratio to Air: Fuel ratio.

Let me recalculate incorporating Air: Fuel ratio.

1. Piston displacement = 796 cc. Hence petrol intake = 796 cc.

2. Manufacturer also states that the compression ratio is 9:1. May be this also means Air : Fuel = 9:1. Liquid cannot be compressed and hence the piston can move all the way up till air is compressed.

3. Hence Fuel intake is 796 x 1/(1+9) = 79.6 cc. I do not know from where 0.02 cc figure is appearing and what is the basis for it.

4. This is assumed to be a 4 stroke engine. Hence petrol is drawn in once in every 2 revolutions.

5. Further some people are confusing stroke to revolutions !!

6. A stroke is equivalent to just 180 degrees or half a revolution in a 4 stroke engine. Hence in a 5500 rpm, fuel is drawn in 5500 / 2 times in a minute.

7. It has 3 cylinders.

8. It generates 39.5 bhp at 5500 rpm.

9. Hence 79.6 cc x 3 cylinders x (5500 rotations/2) in one minute = 0.6567 x 10^6 cc of petrol is consumed in one minute or 0.6567 x (10^6)/ 60 = 0.10945x (10^5) cc per sec. = 0.10945x (10^2) liter

10. Taking petrol has 33,000 kilo J per liter (got this from website), 0.10945 x 10^2 liters of petrol is consumed in 1 sec. Hence 33 x(10^6)x 0.10945 x 100 = 361.135 x 10^6 joules of energy is input.

11. Output = 39.5 bHP. 1 bHP = 745.7 w Hence output = 39.5 x 745.7 w in one second = 2.9455x (10^4) Joules.

12. Hence output x 100/input in %= 2.9455 x(10^4) Joules x 100/361.185 x (10^6) Joules. = 0.8155% only !!!!!

Where do the experts stand after such a heated debate over the issue? Have we all goofed up in our approach to calculations / debate etc or is it really true that efficiency is less than 1%?

I have also seen comment by Mr John DG about - definition of compression ratio. I have really no manufacturer's data about Air: Fuel ratio. Liquid cannot be compressed, only air is compressed. Hence using this logic I have equated compression ratio to Air: Fuel ratio.

Well, I've been sitting back enjoying all of this hilarity. I've just got to chime in.

Compression ratio and air:fuel ratio are two different factors. They do not directly relate as you're trying to do. One is the relationship of max to min travel of the piston in the bore, and the corresponding volumes in the cylinder; the other is the optimization of the fuel and air relationship for combustion. While there can be correlation between these factors, they can be divorced, and one can be altered without affecting the other.

Your logic is faulty.

Let's just get this compression ratio thing out of the way to stop it clouding the issue.

"The compression ratio of an internal-combustion engine or external combustion engine is a value that represents the ratio of the volume of its combustion chamber; from its largest capacity to its smallest capacity. It is a fundamental specification for many common combustion engines."

(Quoted from Wikipedia - not always to be trusted, but in this case 100% correct).

In other words, the volume of the enclosed space at TDC divided by the volume at BDC.

Ok Ill bite assuming you are right what is the alternative you refer to in the last line of your opening post?

Hello mjb1962853

I surrender, a very beautiful post, very clearly stated. I have been convincingly bested. My apologies to this forum. You have proved conclusively that the IC piston engine does indeed possess efficiencies of 25% and more! Interestingly enough though there is still the question of the gyrobus . These buses weighed 11 tons (11,000 Kg) and were powered by a one ton flywheel whose kinetic energy was 6.2MJ. Using this energy these huge buses could travel on a level road at speeds of between 50 kmh and 60 kmh for distances of 6km and in some cases 10 km. Now 6.2 MJ is far from being 33 MJ and 11000 Kg is about the weight of a dozen normal cars. So it does make you think. I am not being contentious, I am just stating the facts as they occurred to me. The fleet of gyrobuses were in service for 8 years at various places around the world.

You concede to this, when Massey's post basically provides the same evidence?

I think you are just playing with us, and I for one will no longer participate.

THINK about the rotating mass of the crankshaft + the gear shafts + the drive shaft + the wheels + the tires; subtract for the friction of bearing surfaces and drive train direction changes;

NOW IMAGINE: A tiny hammer banging on the crankshaft journal, in place of the piston + connecting rod;

The very small FORCE of the gasoline burning / gas expansion / piston push is TRANSLATED into POWER by the continuous action of the TINY FORCE repeated many times over a short period of time.

Therefore, YOUR CALCULATIONS ARE ABSOLUTELY CORRECT, for one firing of one piston in one cylinder!

Even a one cylinder engine contains a flywheel to maintain (conserve) the momemtum of the force, so that it can use some of the power to drive the piston back up to the top of the cylinder, draw it back down during intake, and push it back up one more time while COMPRESSING gas / air mixture.

Therefore, (I believe / surmise / posit ) that the increased power output of the reciprocating engine is obtained through the CO-OPERATIVE INTEGRATION of complex motions into a SINGLE FORCE!

Hi Tomfranprat,

Another very good and interesting post. Yeah! I must say it all adds up. I suppose the 55 bhp that I had used as an example is no laughing matter, just think of 55 above average horses doing the work ! Sorry for the inconvenience and thanks for the sporting post.

I always knew guests do not get GA here .

With all the brain draining calculations (which OP thinks to be crazy, but unfortunately they care not crazier )

Can you really find the mistake, these are simple maths (based on OP's data) and not as complicated as the HHO or quite a few other overunity ones around here.

What happened to the conservation of Energy ?

BTW: what has got torque got to do with it ? (rhyming with TT's grammy winner )

There is some energy created in the stroke 3. Either this energy (pressure) is transferred to the crank shaft, or it exits through the exhaust in stroke 4 (that case it works more like jet propelled car rather than a normal IC)

Can we forget about the torque (of course the total theory is based on it so difficult for OP to forget) but as far as I know the total efficiency calculation is based on the principle of energy (not even power, though that can be used as a direct derivative of energy)

As I mentioned earlier in one of my crazy calculations

Output = Input - losses.

Losses here can only be heat losses - please give some other losses taking place if there are.

Exactly how much heat is getting lost in the car? out of 30MJ/lt assuming a lt runs for 15 km in some 20 minutes it boils down to 25KW heat being generated (at 99.95% conversion of energy to heat)

Exactly how this is getting dissipated without frying the driver?

Why do one has to go round about way to torque and then back to energy?

Losses here can only be heat losses - please give some other losses taking place if there are.

Losses by heat transfer to the cooling system; losses from internal friction (bearings/other rubbing surfaces); losses from incomplete combustion are three.

Heat transfer to cooling system/ friction - heat losses - dissipated/ raises temperature of the system.

Incomplete combustion already taken care in previous post (in fact at least a couple) with a pessimistic value put to combustion efficiency.

The OP is really not bothered about this since he is worried about the energy created in combustion but not converted to mechanical power at the driven axle.

I thing I phrased my clause a bit wrong, sorry for that.

I should have said losses through heat energy (or losses dissipated as heat) you know what I mean -

these will increase system temperature or have heat energy interaction.

As I mentioned earlier the other energy interaction (EM is almost not exixtant in the system boundary)

A bit of sound energy yes but is quite insignificant

Well here's one .. XXXX .. coz I think you made a Good Answer.