Is IC Piston Engine Really Less Than 1% Efficient?

I don't see what the problem is when you consider that the 37bhp @ 5000 rpm is the WOT performance of the vehicle as it approaches top speed. That's when the acceleration is falling off, and the air resistance is consuming most of the work being produced. Sure it sucks! But who drives like that? (oops... that's the wrong kwestion)

Too many things being considered here to try to analyze the entire system down to one number. The energy is leaving the system as losses that are hard (not impossible) to measure. Ever stand next to even a small vehicle as is whips by at 120kph? You'll know where the 'energy' went.

In your posting near to the end, you sated that the amount of petrol being used was0.02cc for every fourth cycle. but you have not multiplied your figures by 4 because the engine dose not fire for every stroke. This is why these engines are called 4 stroke engines, unless it is a 2 stroke, then it would fire every stroke.

Also it would be interesting if you could take this calculation one step further and tell us how many miles to a U.K. gallon this car moves.( 1 U.K. gallon =5 liters).

1 UK Gallon = 4.546 Liters. Being off by .546 liters may not seem like much but when you are talking about 4.5 liters it's over 10%.

Thanks Del, I was completely lost in all the units, angles and ratios, but I can cope with energy in to work out. GA

Why don't you get back on your space-ship and blast off into space using your plasma jet or whatever. Down here on earth, we still need levers and other mechanical components

Finally I did lots of research on the subject raised by you. I located some good web site links.

http://www.fueleconomy.gov/FEG/atv.shtml

http://jagadees.wordpress.com/2007/10/20/ic-engine-efficiency/

www.evworld.com/library/energy_numbers.pdf

http://www.kajul.org/EnergyBlogEN.php?Art=5

I once again am putting down my computation where I know the mistake is in taking air: Fuel ratio in volume rather than mass.

People may criticize, argue, use bad language etc etc Ultimately it appears that efficiency is not more than 15 % as per published data itself. The world needs a better solution- may be even dump IC engines. Today many are working on tinkering and making 10 to 20%(on the 15%) improvements- which is not worth the money spending. We need a disruptive technological approach.

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.

3. Recommended Air to Fuel ratio is about 14.7 to 20 to 1 by mass and not by volume.

4. Hence Fuel intake is 796 x 1/(1+20) = 37.9 cc.

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

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

7. 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.

8. It has 3 cylinders.

9. It generates 39.5 bhp at 5500 rpm.

10. Hence 37.9 cc x 3 cylinders x (5500 rotations/2) in one minute = 0.312675 x 10^6 cc of petrol is consumed in one minute or 0.312675 x (10^6)/ 60 = 0.0.5211x (10^5) cc per sec. = 0.0.05211x (10^2) liter

11. Taking petrol has 33,000 kilo J per liter (got this from website), 0.05211 x 10^2 liters of petrol is consumed in 1 sec. Hence 33 x(10^6)x 0.05211 x 100 = 171.97 x 10^6 joules of energy is input.

12. 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.

13. Hence output x 100/input in %= 2.9455 x(10^4) Joules x 100/171.97 x (10^6) Joules. = 1.71% only !!!!!

14. So efficiency figure has moved up just about 2%.

15. The error may be in taking air to fuel ratio based on volume and not on mass.

16. Web site search gives an indication that petrol engine efficiency is around 15% Fuel to wheel only. This further deteriorates.

17. I as electronic engineer never had thought this to be as low as 15 % granted mistakes in calculation. I think world needs a better engine (need not be Internal Combustion type) which does not degrade the environment.

M S DIVEKAR, I don't mean to be insulting, but reading your posts is really giving me a head-ache.

"3. Recommended Air to Fuel ratio is about 14.7 to 20 to 1 by mass and not by volume.

4. Hence Fuel intake is 796 x 1/(1+20) = 37.9 cc."

cc (cubic centimetres, cm³) is a unit of volume. What you have stated at 4. would only be true if the fuel were the same density as air.

Please re-read the replies you have had from others regarding this matter.

No doubt about what you or I have said. I do not have info about air: fuel ratio.

So our calculations are wrong I admit. But People keep on arguing without pointing out mistake with back up data- which is annoying too. I definitely wish to know how much of fuel is consumed per second - for that very specific condition of Maruti Car 796 cc chamber size, delivering 39.5 BHP at 5500 rpm as claimed by the manufacturer. Bore size = 68.5mm, stroke length 72 mm, Compression ratio 8.7:1, I can give whatever data published in the manual. But I am trying to understand. DD James calculated using bore size etc and came to conclusion that eff was less than 1 % I have taken a different route. People can find mistakes - but let us and the whole world know how exactly in reality some one is getting 15% efficiency.

I do not want unnecessary arguments- I want facts and figures. Why should people flare up? That is not a solution. Ultimately I have found that cars are still not very efficient and the world needs a new engine. I am a scientist- I work towards a solution, by understanding current situation / technology (loop wholes in this technology) - that is all. Even after pointing out so many mistakes no-body can show me how we can make a car with 50 to 80% efficiency. I am looking for that answer.

Mr Divekar,

I believe I have the answer. I have built a working model of a new green engine which substantiates almost all of my expectations. It would be a pleasure to explain the working of this engine to you.

Somehow I missed this post earlier. It sounds like you'd like to explain the working of your engine only to Mr. Divekar. Virtually all of us here would be very interested in knowing about your engine. Please describe it!

Hi MS,

You say: Why should people flare up? That is not a solution.

A reason people flare up when presented with extraordinary claims by charlatans are many. First, such claims, if widely believed, support pseudo science, of which there is already too much. As pseudo science gains a stronghold, society suffers. Well- intentioned people have lost a great deal of money (that might have gone to uses that would benefit society) by investing in pseudo-science scams. Pseudo science, to the extent that it is believed, promotes gullibility and squelches analytical and critical thinking. This too is bad for society: we need more critical thinking, not less, because as a world we face some very serious problems.

Invariably, charlatans will resort to insult when logic fails to support their point. This demeans and devalues discourse.

As is obvious to anyone with a basic understanding of engines, the original poster's claim that engines are less than 1% efficient is completely wrong: a simple calculation of the energy output of an engine vs the energy value of the supplied fuel shows that the 25% efficiency figure that has been used as a rule of thumb for old-tech engines is correct*. So, when the math is so completely wrong, engineers are understandably offended -- we make our livings with math, and don't like to see mathematical garbage presented as valid. It is as if someone walks up to a brain surgeon and says: "An axe is just as good as a scalpel; why do you insist on the silly idea of using a scalpel?"

The same is true regarding physics, the underpinnings for much of what we do. When we see physics trashed, we can find it disheartening. In my daughter's school, her physics teacher is just barely adequate to the task, and science education has generally gone downhill in the US, so many of us are understandably sensitive to seeing pseudo-physics presented as physics.

My personal interest is in keeping the signal-to-noise ratio relatively high at CR4, because then it is a useful, fun site. If instead it is populated by pseudoscience promoters, the site loses its value to me -- there are already plenty of pseudo science sites around, and not enough sites that are focused on sound science.

Some of us are also concerned about fraud and scams, promoters of which show up at CR4 periodically. To the extent to which we fail to denounce fraud and scams, we tacitly permit their existence, so again, you can see how this could be an emotional issue.

Hi Blink,

I really liked your S to N ratio analogy. That is really cool!!

If only some people would listen more and pontificate less what great strides might be made towards gaining a clearer understanding of what has already been achieved and therefore help educate towards what paths might be available for solving the challenges that lie ahead. CR4 is one such forum if used appropriately.

Although IC engine technology has been advancing spasmodically, (I choose my words carefully) there now appears to be some really serious science being devoted in improving efficiency in the realms of the high speed light weight automotive product. It will be interesting to see how the '5 stroke' unit comes along and laser ignition looks very impressive. These are physics/applied engineering based initiatives similar to the solid first principles approach so ably demonstrated by the likes of Rudolph Diesel over a century ago.

Best wishes,

Massey.

I really liked your S to N ratio analogy. That is really cool!!

Thanks!

I recently came across a thread in Physorg, regarding using oxyhydrogen (known to the ill-informed as Browns gas, HHO, etc) to "clean up" coal burning powerplants. In that thread the S/N ratio was about .5. In a Popular Mechanics thread re a test of an "HHO" booster (in which it was shown the the unit had no effect at all, just as theory would predict) the S/N ratio was perhaps .3. for every 10 bad posts there were about 3 pretty good ones (this is more gut feel than actual count).

Fortunately, the S/N ratio on CR4 tends to be pretty high, especially if you include the witty-but-not-strictly-on-topic responses.

Dear "Guest" - as far as I can determine, Massey's first post (to this thread) is #11; it sets out what I believe are completely factual numbers and conclusions. They clearly indicate that the claim in DDJames' original post is wrong. Many posts have shown the corrections to DDJames' bogus "math" and failures to use proper units, in addition to quite bizarre ideas of how much fuel is used (not all of those are from the OP, though). Massey's first post got 3 "Good Answer" scores; the ONLY thing I can find wrong is transposition of a comma and a space - nothing whatever to do with content. Until and unless you, DDJames, or someone else can set down clear, comprehensive, CORRECT, complete, and convincing examples to the contrary, the efficiency of IC engines remains at 25 - 35%, nowhere NEAR "less than 1%". Instead, egregious errors are rife in the posts of the one-per-centers, and have been pointed out time after time; none of these have undergone a full set of corrections acknowledged by the one who posted.

This thread has a surprising number of Good Answers and Almost Good Answers for a long one (a very short thread may have only a single response, and if it gets a GA, that's 100%, so a long thread with high percentage is unusual). I don't recall seeing a single one given to any of the one percent believers. There is a reason.... As Massey said, Q.E.D.

Hi M S Divekar

I don't know if this will help:

Stoichiometric ratio = 14.7 Kg of air for every 1kg of fuel.

Then a square engine with a 3" bore and 3" stroke will have a volume of

3" = 7.62cms

Volume = 3.81 ² x 3.14 x 7.62 = 14.56 x 3.14 x 7.62 = 347.32 cc

1 cc of air weighs 0.00125gms therefore mass of air in cylinder = 0.00125 x 347.32 =

0.43 gms therefore weight of fuel needed = 0.43 / 14.7 = 0.029gms

1 l of petrol weighs 720 gms, 1cc of petrol weighs 0.72 gms . 0.001 cc = 0.00072gms

So 0.00072 x 40.2 = 0.029 gms = 0.04cc

The 3 x3 piston engine needs 0.04 cc approx. for every power stroke.

First let me thank you James for that real input. There was no point in going on finger pointing as many have done unnecessarily.

Let me compute again using some of the data given by you specifically for 796 cc Maruti engine.

· Mass of air in 796 cc = 0.00125 grms/ cc x 796 = 0.995 grms

· Mass of fuel – assuming 14.7 : 1 rato at this condition of maximum power = 0.995/14.7= 0.0677grms

· Density of petrol = 720 grms / 1000 cc = 0.72gr/cc.

· Hence 0.0677 grms of petrol = 0.0677/0.72 = 0.094 cc.

· Fuel contains energy in terms of 33,000 KJ/ liter.

· Hence for 3 cylinders rotating at 5500 rpm, we need 0.094 cc x 3 x 5500/2 = 775.5 cc in one minute or 775.5/60 = 12.925 cc in one second.

· Hence energy input = 12.925 X (10^-3) liters x 33 x (10^6) Joules / liter = 426.525 X (10^3) Joules in one sec

· Max power delivered is 39..5 hp at 5500 rpm

· 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.

· Hence efficiency = 2.9455x (10^4) x 100 / 426.525 X (10^3) = 6.90%

· If Air toFuel ratio is 20 : 1 instead of 14.7, efficiency will be 9.39% for this specific model of car.

· I wonder whether the exercise was worth it.

· I think mechanical engineers – who made me- an electronic engineers do all this calculation – should find some other solution. Junk the IC engine first.

"If we all worked on the assumption that what is accepted as true is really true, there would be little hope of advance." -- Orville Wright

It's often a good idea to have a quick look over one's calculations to make sure everything looks "reasonable".

" ... 775.5 cc [of fuel] in one minute ..."

Does that look reasonable to you? 46½ litres in one hour?

I think you did not read through the whole thing carefully. In a cylinder of 796 cc, fuel drawn in at a time is just 12.925 cc in one second when 5500 rpm/60 revolutions have been completed !!! It is 775.5 cc or or 0.775 liter in one minute when engine is running at maximum power of 39.5 hp.

Why do you jump to wrong conclusion too fast !!

"You progress not through improving what has been done, but reaching toward what has yet to be done." -- Kahlil Gibran

"I try to view the challenges in my life not as annoyances, but as confirmations of fortitude." -- Oprah Winfrey

"I know the price of success: dedication, hard work and an unremitting devotion to the things you want to see happen." -- Frank Lloyd Wright

So I wish to see this IC engine junked

Come ON! THINK!

13cc per second * 3600sec/hr gives you a fuel consumption of 46,800 cc/hr, or 46.8 liters per hour. 46.8 liters is well over 12 gallons! A three-hour trip would require a 37 gallon fuel tank, and if you arrived, you'd be on empty!

This is a very small engine at 0.8 liters displacement (796cc rounded off). At this rate, a 4 liter engine would use 46.8/.8*4=234 liters per hour, or almost 62 gallons per hour. A three hour trip would require a 190 gallon fuel tank!

Admittedly, you aren't going to be running the engine at maximum power the full time, so these numbers are exaggerated, but with such a small engine, trying to keep up with today's freeway traffic would require a large fraction of that power, and therefore a large fraction of that amount of fuel.

I really feel you are taking things too far to prove an non-existent point.

Being an electronic engineer does not condone making mistakes in calculations and units.

It has been earlier also pointed out that just check up the consumption with actual, and see whether they approximately match or not.

@15Kmpl - usual 800CC car of old model (newer ones have 20+ KMPL). Say @40KMPH - one liter will last 15/40*60 = 22 minutes. = 1320 secs

This will be @39 BHP = 39 * 1320 * 746 = 38.4 MJ ?

So the efficiency = 100% ? (38.4/39 = 0.985 = 98.5%)

But we all know it is not.

Where is the fallacy?

_{BTW: I am not mechanical engineer either, nor automobile or things similar. (That's why i took the 0.02 of DDJ on face value)}

1. The displacement of engines is normally given as the total of all its cylinders, so you must divide by the number of cylinders to get the displacement of one cylinder, or NOT multiply by the number of cylinders later on.

2. Anyone suggesting elimination of a technology must suggest something he/she considers better. What is your suggestion? It must be something that can carry 2-5 people plus their 'stuff' in comfort in any weather for 300-500 miles, renew its energy supply economically in 5 minutes at a relatively convenient location, and repeat...

What Del is suggesting is a very normal and simple approach. Just remember, there's more than one way to skin a cat (sorry Del...I couldn't resist). There is nothing wrong with doing a detailed analysis of every component in the system, but that can be time consuming and more prone to error(s).

Hi this is for M.S Divekar

Stoichiometric ratio = 14.7 Kg of air for every 1kg of fuel.

Then a square engine with a 3" bore and 3" stroke will have a volume of :-

(3" = 7.62 cms)

Volume = 3.81 ² x 3.14 x 7.62 = 14.56 x 3.14 x 7.62 = 347.32 cc

1 cc of air weighs 0.00125gms therefore mass of air in cylinder = 0.00125 x 347.32 =

0.43 gms therefore weight of fuel needed = 0.43 / 14.7 = 0.029gms

1 l of petrol weighs 720 gms, 1cc of petrol weighs 0.72 gms . 0.001 cc = 0.00072gms

So 0.00072 x 40.2 = 0.029 gms = 0.4cc

The 3 x3 piston engine needs 0.04 cc approx. for every power stroke.

Taking this a bit further. 39.5 bhp @ 5500 rpm. There are 5500/2 or 2500 power strokes per minute. 2500 x 0.04 = 100cc petrol/min

100cc = 1/10 l therefore 100cc = 3,300,000J : 39.5 bhp = 29455.15J

Therefore efficiency = (29455.15/3300000) x 100 = 0.89% . Is there an explanation for this result ? Some help.

39.5bhp = 29455.15 joules Watts.

100CC petrol per minute- convert into per second and you will have the answer.

And as I said the answer is still too much (about 53%)

We do not get that efficiency in the non CC (combined Cycle) Gas Turbines, except in the larger ones we manufacture.

There is still something wrong about the data.

And as far as I know, the critical Fuel Air ratio (14.7) is at the ignition stage - ie when the air is compressed at about 9bar. There the density is obviously a bit more. Though due to the temperature the density will not be as high as that of room temperature compressed air.

Guest you go from gloating to quite clever to deceitful. Go away till you can put a name to your posts! Please!!

"I did NOT rigorously check all your math, but "39.5 bhp = 29455.15J" can't possibly be correct!

the bhp is a unit of power, while the J is a unit of energy. Energy is the product of power and time, so the power must be multiplied by a time. since the previous paragraph mentioned fuel in cc/min, that time might be one minute. Now what conversion factor did you use to convert from bhp-min to Joules?

This calculation is correct. 2.9455 x 10^4 watts for 1 second also means 2.9455 x 10^4 Joules as 1 watts ec = 1 joule

· 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.

It has been thoroughly checked. No mistakes. I am not looking for minor improvements in eff. I am looking for a new engine with better tahn 50 to 70% eff. We are no where near this.

"Do not follow where the path may lead. Go instead where there is no path and leave a trail." -- Ralph Waldo Emerson

"Determine that the thing can and shall be done, and then we shall find the way." -- Abraham Lincoln

"Mystery creates wonder and wonder is the basis of man's desire to understand." -- Neil Armstrong

OK, I took the time to verify those numbers. Yes, the numbers are correct, but the units are NOT.

Any value of power can NOT equal any unit of energy. The equation should read either:

39.5 bhp-sec = 2.95 x 10^4 Joules (an energy equation)

or

39.5bhp = 2.95 x 10^4 Joules/sec (a power equation).

BTW, the result can not have more digits of precision than the least precise original value.

Also, try using a spell-checker. It would help you to avoid such things as "watts ec".

d.k warner

OK! If you still feel like quibbling: Lets convert to seconds:

39.5 bhp = 29455.15J/min = 29455.15/60 J/s = 490.919 J/sec

AND 3,300,000 J/min = 3,300,000/60 J/s = 55,000 J/sec

Therefore efficiency = (490.9 / 55000) x 100 = 0.89 %

Personally I think it's time for the "Pants on the ground" song.

P.S. I think you should follow it up more rigorously, it's an important issue that affects all of us!

AND 3,300,000 J/min = 3,300,000/60 J/s = 55,000 J/sec DDj,

What ARE you talking about?

I've been reading all this "mullarky" from the beginning, and thought you would have grasped the sense of it before this.

Where did you get the figure 3,300,00 J from? What is your understanding of it?

Stu

Try to follow the thread Stuey, don't make like an expert if you can't read!

Excellent job done BUT, BUT let me know how volume per cylinder is 254.57 cc. I am getting

· . It must be noted that bore dia is 68 mm, stroke length is 72 mm. Hence volume per cylinder is = pi x d^2 x s = pi x 6.8^2 x 7.2 cc = 1045.9 cc. Now there may be other issues which restrict the volume to 796 cc as declared by the manufacturer. Hence it is 796 cc per cylinder and not for all cylinders put together.

Hence 45 % efficiency is also divided by 3 and hence is near 15% only.

Basic math/geometry PLEASE!!

Radius or Diameter?

Volume of cylinder= (Pi x R^2 x H)

OR

Volume of cylinder= (Pi x D^2 x H) /4

I recall seeing a statement that this is a 3 cylinder engine with a 0.8 liter (total) displacement rating. My table numbers seem to support these numbers. This is not difficult math!

OK. You are right- I goofed up. So the efficiency is near 46% but again why did you or whoever scale it down by 50% again? Any reasons?

"Curiosity has its own reason for existence" -- Albert Einstein

"Do not follow where the path may lead. Go instead where there is no path and leave a trail." -- Ralph Waldo Emerson

"Nothing great was ever achieved without enthusiasm." -- Ralph Waldo Emerson

"Let me tell you the secret that has led me to my goal: my strength lies solely in my tenacity." -- Louis Pasteur

I am considering the practical example of an IC engine as used in a personal transportation vehicle. Mechanical losses in the drive train and rolling resistance will consume several horsepower. If 37 hp is available at the crankshaft, I estimate only about 20-25 hp will be available to move people (and the air). Remember, at sustained highway speeds the primary energy use is just overcoming air resistance (drag). If you look back at my 1000 mile trip post #49 , this was a REAL example which indicated typical energy conversion around 22%.

On wiki alone, there are numerous pages describing heat engines, their thermodynamic efficiencies and theoretical limits. Suggest starting here...

http://en.wikipedia.org/wiki/Internal_combustion_engine#Energy_efficiency

If you are just looking at IC engines alone, the shaft horsepower is what you want. Published energy conversion efficiencies can range up to about 45%. If you are debating the pros/cons of IC engines, these numbers can be useful. If you are planning to design a mass-market engine for practical use, I would start by using use a more conservative efficiency value.

I decided to verify my calculation vis-a-vis the table given by you. Pl see some discrepancy I noticed.

I am just quoting from the table given by you.

Fuel mass per stroke = 0.065 grams / power stroke.

Fuel density = 0.72 grams/cc

Fuel volume per power stroke = 0.047 cc/ power stroke as given. But = 0.065g/0.72g/cc= 0.0903 cc

Manf engine rating = 5000 rpm

Power stroke per rev = 0.5

Fuel use = 116.63 cc / mnt = 0.047 *5000 *0.5 = 117.5 cc. BUT = 0.0902 x5000 x 0.5 = 225.5 cc / mnt

Volume conversion = 1000 cc/ liter

Fuel use per mnt = 0.117 liters /mnt as given . BUT = 0.2255 liters/mnt

Theoretical fuel energy = = 33x 10^6 J/ liter

Time conversion = 60 sec / mnt

Theoretical Fuel energy = 64143.84 J/ sec or watts = 33*(10^6) *0.117/60 . BUT = 33*(10^6)* 0.2255/60 = 124025 J/sec or watts

Energy unit conversion = 745.7 w / hp

Theoretical Fuel energy = 86.02 hp as given = 64143.84/745.7 =, BUT = 124025/745.7 = 166.32 HP

Manufacturer's engine rating = 39.5 HP

Engine efficiency = 0.46 as given = 39.5/86.02. BUT is actually = 39.5/ 166.32 = 0.2375 or 23.75% ONLY.

So all the following computations carry this error. Hence efficiency is not 45 % but only about 24% or so. It is still better than 1 % we started with, but still far below my expectations of 60 to 80% at the shaft - as explained by you

"March on. Do not tarry. To go forward is to move toward perfection. March on, and fear not the thorns, or the sharp stones on life's path." -- Kahlil Gibran

"Let me tell you the secret that has led me to my goal: my strength lies solely in my tenacity." -- Louis Pasteur

"Keep away from people who try to belittle your ambitions. Small people always do that, but the really great make you feel that you, too, can become great." -- Mark Twain

Hi! Its me again and I don't want to sound boring BUT it seems to me that while mjb 1962853 has converted fuel into J/sec he has left the horsepower calculation at 39.5 hp/min! So there is a huge difference in his result! Here are my results using the same cylinder bore and stroke dimesnions:

Bore = 6.84 cm , stroke = 7.2cm

Volume of cylinder = pi x r^^2 x h = 3.42 ^^2 x 3.14 x 7.2 = 264.4 c c

Energy density of air = 0.00125 gm/cc

Energy density of air in cylinder = 264.4 x 0.00125 = 0.33 gms

Stoichiometric ratio = 14.7 : 1

Needed density of petrol in cylinder = 0.33/14.7 = 0.022 gms

Density of petrol = 0.72gms/cc

Density 0.001 cc petrol = 0.00072

Volume of petrol in cylinder = 0.035 cc with density 0.00072 x 30.5 = 0.021 gms

Manufacturers figures = 39.5 bhp @ 5000 rpm

Volume of petrol used in 1min = 0.035 x 2500 = 87.5 cc

1 litre petrol = 33,000,000 J

87.5 cc petrol = (33,000,000/1000) x 87.5 = 2887500 J = 3872 bhp

39.5 x 745.7 = 29455.1

(29455.1 / 2887500) x100 = 1.02 % efficiency.

And again :

(39.5 / 3872) x 100 = 1.02 % efficiency.

Here is how mjb 1962853's results should look !

Volume of petrol used 1 min = 224.5 cc

Volume of petrol used in 1 sec = 224.5 / 60 = 3.75 cc

1 litre of petrol = 33,000,000 J

1 cc of petrol = 33,000 J

3.75 x 33,000 = 123 733.5 J = 166 hp/sec

AND 39.5/60 = 0.65 hp/sec

(0.65 /166) x 100 = 0.4% efficiency.

Also he is right 50% efficiency does sound unrealistic!

Unfortunately, you don't seem to be as easily educated as Mr. Divekar.

A power ratio is just as valid when both intake and output are converted to horsepower (as mjb did), as it is when both intake and output values are both converted to Watts (Joules per second), or kiloWatts, or any other valid power units.

Your use of terms does not favor logical progression through the calculations. For example, I quote: "Energy density of air = 0.00125 gm/cc". The value 0.00125 g/cc is the density of air, not the energy density. As far as an IC engine is concerned, the air contains NO energy, and if it did, the value would have to be given in Joules/cc (or any other unit of energy/cc).

"Energy density of air in cylinder = 264.4 x 0.00125 = 0.33 gms". The product 264.4 cc* 0.00125g/cc does indeed give a result of 0.33g, but that value is neither Energy, nor density; it is mass. BTW, the abbreviation for gram, both singular and plural, is "g". "gms" means "gram-milli-seconds"

I'll skip over to your worst infraction: "87.5 cc petrol = (33,000,000/1000) x 87.5 = 2887500 J = 3872 bhp". You start with a volume (87.5 cc), have a couple of energy quantities in the middle, and end up with a power. Volume is NEVER equal to energy, and energy is NEVER equal to power.

87.5 cc petrol contains (33,000,000/1000) x 87.5 = 2887500 J of energy. You MUST include a time to convert energy to power, and you show none.

Once you use the correct terms for quantities, use (and show) correct units for all values, and use true equations, you just might find out that mjb was right, after he corrected his single error.

Hi! dkwarner,

I said at the beginning of this post that there would be a lot of red faces and a lot of anger and disappointment. Before shooting of about time etc., just have a look at mjb's post where he specifically quotes 33Mj/l ! Whats wrong with that? You can also see that he has given the fuel energy in Joules/ sec and has taken the manufacturers value of hp/min. PLEASE just check your facts before going off ! To give an exact quote from his post, so that maybe you'll understand. He says : 123734.26/Joules/or watts sec and then converts this into horsepower as 165.93 hp/sec (please note if it has'nt already caught your attention). He then proceeds to divide manufacturers figures for hp/min (1.e 39.5 bhp/min) by the theoretical hp/sec and comes up with an efficiency of 24% and according to him that is about right. These are the facts that you are going on trying to shove in my face!!!

dkwarner

I am giving you a chance to redeem yourself. OK, energy density of air was a slip up, it should have read plain 'density of air'. Take the final result. 1.02% at peak efficiency is the figure I come up with for the IC piston engine. mjb's figures come out as 0.4% efficiency. So there you have it. Like it or not and I suspect you don't like it !!

"Needed density of petrol in cylinder ..." HUH? Since you give a value in "gms", which I suspect you intend to represent grams ("g"), this would be mass, not density. In any case, you go through four steps, and conclude that 0.022 gms [sic] equal 0.021 gms [also sic].

You later come to the bizarre conclusion that the energy from 87.5 cc of gasoline / petrol equals 3872 brake horsepower, though there is no conceivable way in which BRAKE hosepower can be involved. (see http://en.wikipedia.org/wiki/De_Prony_brake to learn what the term means).

Then you have the term "hp/sec" - could you please explain what it means?

Horsepower is already energy-per-unit-time by definition. You do not divide by 60. The per-unit-rate-of-change of horsepower or

d(hp)/dt

which does have units of [horsepower/second] is not used or needed anywhere in this series of calculations.

You really do need to study your units and how to use them. I made an error in the one table by not thoroughly checking my units, so mistakes do occur. However, you do not seem to understand the fundamental properties of mass, density, energy, and power. I'm not trying to be rude, but it is critically important to understand these properties before trying to analyze more complex systems.

IMHO this horse(power) is long dead, so I'm moving on to the next topic that catches my attention. Sorry, couldn't resist the parting pun

mjb

Horsepower is already energy-per-unit-time by definition. You do not divide by 60. The per-unit-rate-of-change of horsepower or

Just look at your table, that is exactly what you have done. You have given a figure of 224.97 cc /min . Since 1 cc = 33,000 J Then it follows that 224.97 cc = 7424010 J you have next divided this figure by 60 to get 123734.26 J/sec You have further divided this figure by 745.7 to get 165.9 hp/sec ! And to top of all your marvellous calculations you have divided 39.5 which is the manufacturers figure for hp/min (i.e 39.5hp) by your figure for theoretical hp/sec and have announced an efficiency of 24% !! Really great, further instead of admitting this mistake, you are spouting off about mistakes that I have supposedly made, just wonderful. Keep it up!!!!

I welcome a site moderators input as I do not believe I have violated any forum protocols or been rude or insulting in any way. This is an Engineering forum and most here would have hoped this thread would provide some clear answers for you and other curious readers. Here is my last attempt to correct your calculations. If you are not able to check your work by properly understanding and monitoring the units, then it is unlikely you will ever get realistic answers.

Bore = 6.84 [cm] , stroke = 7.2 [cm]
Volume of cylinder = pi x r^2 x h = 3.14159 x 3.42[cm] x 3.42[cm] x 7.2[cm] = 264.6 [cc]
Density of air = 0.0012 [g/cc] (at 20 deg C)
Mass of air in cylinder = 264.6 [cc] x 0.0012 [g/cc] = 0.318 [g]
Stoichiometric ratio = 14.7 : 1
Needed mass of petrol in cylinder = 0.318 [g] / 14.7 = 0.0216 [g]
Density of petrol = 0.72 [g/cc]
Volume of petrol in ONE cylinder = 0.0216 [g] / 0.72 [g/cc] = 0.030 [cc]
** Volume of petrol used in THREE (x3) cylinders in ONE power cycle = 0.090 [cc]**
Manufacturers figures = 39.5 [hp] @ 5000 rpm [1/minute]
Volume of petrol used in 1min = 0.090 [cc] x 5000/2 [1/minute] = 225 [cc/minute]
1 liter petrol = 33,000,000 [J/liter]
225 [cc/minute] x 33,000,000 [J/liter] x 1/1000 [liter/cc] = 7,425,000 [J/minute]
7,425,000 [J/minute] x 1/60 [minute/s] = 123750 [J/s]
123750 [J/s] = 123750 [Watts]
123750 [Watts] x 1/745.7 [hp/Watt] = 166 [hp]
39.5 [hp] / 166 [hp] x 100 = 23.8%
39.5 [hp] x 745.7 [Watts/hp] = 29455 [Watts]
29455 [Watts] / 123750 [Watts] x 100 = 23.8%


And again...

Volume of petrol used 1 min = 225 [cc/minute]
Volume of petrol used in 1 sec = 225 [cc/minute] x 1/60 [minute/s] = 3.75 [cc/s]
1 liter of petrol = 33,000,000 [J/liter]
1 cc of petrol = 33,000 [J/cc]
3.75 [cc/s] x 33,000 [J/cc] = 123750 [J/s] = 166 [hp]
39.5 [hp] / 166 [hp] x 100 = 23.8%

These are reasonable estimates which are supported by a large body of published theoretical and empirical work. I wish you good luck in your endeavors.

Yes, you did indeed find an error in mjb's table. He multiplied instead of dividing. That makes it agree with the fuel intake of my post 104. The discrepancy between your 0.0903 cc and my 0.087 cc is from 2 factors: 1. I took into account the fact that the piston can not aspirate its full volume (I estimated 750 of the 794 cc - I have done no calculations to support or refute this estimate). 2. I used a density of 0.70 g/cc. The density of course changes according to the formulation. In fact here in California, both of our values are probably low, since they always include some ethanol, with a density of around 0.78 g/cc.

Congratulations for sticking around long enough to finally arrive at reasonable values!

Dick

I'm definitely NOT an IC engine expert, but I am a physicist. Unless I am greatly mistaken, in a normally-aspirated engine, the only thing causing the air/fuel mixture to enter the cylinder is the difference in pressure between atmospheric and the cylinder. All fluids take time to flow. 5000 RPM is 83.3 RPS, so one revolution takes 1/83.3=0.012 seconds. There is no appreciable 'vacuum' in the cylinder until the piston has moved down significantly, so we have something less than half a revolution for intake. So the intake time is something like 0.005 seconds, with some variation depending on valve timing. 5 or 6 milliseconds is nowhere near enough time for the pressure inside the cylinder to equalize with the outside, so the pressure in the cylinder before compression must be less than atmospheric. Quantities of gasses must always be corrected for pressure, so the STP (Standard Temperature and Pressure) quantity of fuel/air mixture aspirated must be less than the displacement of the cylinder.

As I indicated, I don't know how much less. It will clearly depend on such things as valve size, valve lift, valve timing, cam shape, and both engine and atmospheric temperatures, so I don't think there can be a known ratio. If any of the engine experts knows an empirical value, I'd love to hear.

Your last statement is clearly the most important one!

Dick

Typically, for an engine in good mechanical condition, running at idle, 15 to 22 inches of Hg is an acceptable range.

Also, a "smooth" reading, without much variance, is as important.

Measured where? I suspect you are referring to manifold pressure, but...

Alright, my first posting here. I've read all this stuff going on. I used to be an engineering technician before I became a cop.

I smell it from all this far away. SCAM.

Yes, and I used caps for SCAM for emphasis.

You gave up way too easy for someone who was so passionate about his idea.

So, what's your play? How much do you want for your idea to save the world from itself?

Come on, son. Confession's good for the soul. Give it up.

I assume you're referring to #112, where DDjames is acquiescing to M S DIVEKAR's onslaught.

I wondered about that ... is DDJ going to deliver a coup de grâce? Or is he going to give us all the opportunity to invest in the Glorious New Future? Or both?

I sit with bated breath.

[BTW - welcome to CR4 - enjoy!]

You may well be right, although I haven't yet heard what the new alternative to the IC engine is, so what might be 'sold'.

I actually think it's more a case of 'a little knowledge is dangerous". A couple of people who have had just enough physics background to fool the truly ignorant into believing they have come up with something new..

... and I second JDG's welcome to CR4!

Dick

Welcome from me too.

Though DD wanted me to go away (he obviously does not like guests, who questions hosts) Still can not miss the chance to second (rather third) the welcome.

Come aboard and Enjoy.

The table provided by mjb1962853 makes matters clear. I think we can close this topic. The efficiency is not as bad as less than 1% but may be 45% - but for some unknown reasons brought down to 26%.

May be another debate can be initiated (as this thread has gone too long) - as to how to improve this and take it to near 80% or more if possible.

"Challenge is a dragon with a gift in its mouth. Tame the dragon and the gift is yours." -- Noela Evans

You apparently have never heard about, nor studied the conclusions of a gentleman named Carnot.

There is one supreme universal (Timeless) law - at least we Hindus believe in - CHANGE IS PERMANENT.

So all man made laws also are subject to change with time. Carnot's law - in implementation can be altered by programmability and electronics. I will not dispute PV /=T constant, but we can always find new ways of implementing it !!! I have that in mind. I understand Physics and laws- but only taking a fersh look - that comes under the Timeless law - CHANGE IS PERMANENT.

May be you can also be an inventor- if you take my words - do not simply accept. They invented long back - with certain assumptions. Question teh assumptions and you will find a new answer.

"Mystery creates wonder and wonder is the basis of man's desire to understand." -- Neil Armstrong from CR 4

"Education's purpose is to replace an empty mind with an open one." -- Malcolm S. Forbes- from CR4

Okay, flap your unaided arms really, really hard. And really, really believe in it. Soon you will succeed in flying. Or maybe not.

MS,

who are you to close anything?

What's with all the quotations?

All the folk you quote actually knew something of which they spoke.

From all the preceding, it seems you don't. Not about reciprocating IC engines as least. - as to how to improve this and take it to near 80% or more if possible.

NOT me! I rather instruct people who actually want to learn something.

Only a dunderhead would make a statement like that. What do you think we all (engine designers, engine companies et al) have been doing for the last 100odd years.

When you get to know something of physics and chemistry you will know that there are some things that are never going to happen.

'Bye.

Stu

Just to make some sense of this thread.

Question paper

All question are compulsory. Q9 contains 100% of marks.

....

Q9: Prove that 2/10 = 2

Music Student: Out of Tune

Art Student : There is no pattern.

Accountant: Cridit and debit don't match.

Philosophy : What can't be can't be.

Religion: It is not destined to be.

Medicine: If you have a controlled group of 10 and you give the medicine to two and placebo to the rest, then the two will .

Engineer: Too simple

2/10 = Two/Ten = Two/ Ten = wo/en

We know from A=1, B=2.. W=23, O = 15, E=5 and N=14

so WO/EN = (23+15)/ (5+14) = 38/19 = 2

QED

Moral: Engineers never bother about "what is the answer", they only ask "What answer you want".

So which realm of incompetence is your daily engagement?

Stu.

It may be better if the admin closes this thread. It is really going nowhere, and any amount of maths is not going to convince the OP about the facts.

Unfortunately these threads will remain a part of history to confuse some young mind in future

So how do those big Diesel electric generators manage to get about 35% or more electrical KWH out of every 100 KWH of fuel energy?

Seems to me that if they do periodic lab test burns of the fuel to determine its exact combustion energy content (bomb calorimeter or similar testing) and the end result is that the in use gen set puts out 35% of that available as tested fuel energy as real measurable electrical power wouldn't that mean that they are probably right about their estimated engine and generator system efficiencies?

So does that mean that Massey 726 has over unity gen sets or more likely someone else can't do basic math?

As has been stated some time ago (energy out / energy in = system efficiency).

All of the mathematical theories in the universe are useless if reality does not agree with them.

I will happily believe your numbers over many others!

Once again guys like you have proven reality does not bow down to theory!

I agree, although with a small proviso:

Theory, appropriately applied and with valid math, usually comes extremely close to predicting reality. For example, I have predicted vehicle performance (and every manufacturer does this too) for clients based only on theory and math -- the vehicle itself does not yet exist. Any manufacturer can get within a couple percent of reality this way. Toyota did not throw some engine parts in a bag, shake it up, and have the Prius engine pop out.

This process, using appropriate theory and appropriate math to predict performance, led us to be able to send a rocket to the moon without having already been there. It's what enabled the Wright brothers to participate in the development of powered flight.

The problem with the central premise of this thread is not one of theory versus practice, it is one of completely ludicrous pseudo science in combination with completely incorrect math. There is little in DDJames presentation that is supported by the theories found in texts on combustion engines.

Real theory very strongly supports Massey's contention just as real theory strongly supports Honda's contention that their home co-generation units are 85% efficient, provided you have a use for all the hot water generated.

To be blunt, only nitwits and charlatans seriously argue that engines are less than 1% efficient -- there is no plausible theory to support such a contention.

Why believe Massey's numbers? Maybe he's making it up, and DDJames is telling the truth. The reason I believe Massey is because theory supports what he says.

Joe Walsh was right. You can't argue with a sick mind.

OP might have a point. The torque figures for Maruti 800 are 59Nm @ 2500 rpm. This is not per second or per minute, therefore must be per revolution of shaft. Hp developed at this rpm would then be 59 x (2500/60) = 59 x 41.6 = 2458.3 W/s = 3.29 hp. But manufacturers use Torque x rpm x 1/5250 = 59 x 2500 = 147500/5250 = 28 hp.

Similarly 37.75Nm @ 5500 rpm = 37.75 x 5500 / 5250 = 39.5 hp. Why use rpm instead of rps ?

Torque does NOT involve "per second", nor "per minute", nor "per revolution". The only purpose of the 2500 rpm is to tell at what engine speed it develops the peak torque. Horsepower peak will typically be at a bit higher rpm for most engines; it will NOT be "horsepower per second", nor "horsepower per minute", nor "horsepower per revolution". Any so-called math based upon such invalid units is bogus and wrong.

Only 'point' OP has is to grind the point (ok edge) of his own axe. His maths is deeply flawed and his "physics" is nonsense.

Don't go there.

Sorry. No, the OP has no valid point re this. For clarity on torque, try Hyperphysics.