Laws Governing Workings of the IC Piston Engine

Hey !! Can someone please give me my password so that I can log in as DDjames !! Moan

PPS Bruno dies at the stake (Har! Har! without saying a word !! )

Hi Guys !!!! Found my password for DDjames, sorry for any heartburn. DDjames

OK. So it seems that there are nil comments on my radical new post!!!(pseudonym Prospero) But just think about what it means, you will be stunned out of your minds. B'cos each of the laws referred to are immutable. i.e., they have to apply, there's no getting out of it!

And where is the problem ?

Did anybody mention laws you consider erroneous ? Which ones ?

Do not forget the thermodynamic laws since you write about an Internal Combustion engine (or have I misunderstood ?).

nick name

The very person!! Do the maths. It's pretty frightening!

I still do not understand where is the problem, please elaborate!

Are you building a space ship?

∫dF•J ??

As far as I know impulse M*v or J*ω and force is F=d(M*v)/dt or T= d(J*ω)/dt.

You can write the other way M*V=∫F*dt fot instance.

Hi Nick, What I am trying to point out is that the Internal Combustion piston engine is subject to the Laws of physics just like anything else.

In the initial phase energy is added to the process in the form of combustion of the compressed fuel/air mixture. The pressure inside the cylinder then rises to about 500 psi ( 35 Kg/cm2 ) the final force applied to the piston head depends upon the area of the cylinder. An 80mm dia bore will result in a pressure of 1758.4 Kgf being exerted on the piston head during the power stroke. So the first process is a process of addition.

The next three processes are to do with reduction. Take for instance the impulse. At 6000 rpm a four cylinder 4 stroke engine actually has 12000 power strokes. So the time allowed for each power stroke is just 1/200 of second or 0.005 seconds. Since Impulse = Force x time (i.e., the product of the force and the time for which it is exerted.) = 1758.4 x 0.005 = 8.792 Kgm.

When a force is applied to a lever attached to an axle (as for instance a spanner) at any angle other than a right angle the force (torque) it exerts, is proportionately reduced. As the piston moves down the cylinder it exerts force on the crank shaft at a continuously changing angle. It is possible to work out the exact figure by calculating the force exerted at each crank angle but this is time consuming ( anyone who likes, is welcome to try ) , a rough approximation would be that it averages out to the value of Sin 10 . So 8.79 x 0.173 = 1.52 Kgm.

Further the length of the lever determines the final amount of torque that is exerted . This is represented by the throw of the crankshaft and is generally accepted as being half of the stroke ( which in turn is closely related to the bore) . Taking this example to be a square engine gives 0.04 x 1.52 = 0.06Kgm or 0.06 x 9.8 = 0.9 Nm. So the inescapable conclusion is that the initial force of 1758.4 Kgf exerted on the piston head results in a torque of only 0.6 Nm being exerted on the crank shaft.

There is no need to take this personally or to get hot under the collar, I am not saying that this is necessarily the case, what I am trying to state is that the application of these physical principles to the working of the Internal Combustion piston engine seems to yield this result. To sum up an initial force of 1758.4 Kg exerted on the piston head yields a final torque of 0.6Nm at the crank shaft.

But empirical results show that the actual horsepower developed is much greater how is this possible ? Maybe torque is calculated as torque per second. Then if the engine is running at 6000 rpm as stipulated we have 0.6 x 200 x 6.28 x 6000 / 60000 = 75 Kw = 100 hp. Which seems reasonable, what is of interest is the tiny amount of torqe generated at each power stroke, just 0.6 Nm, is this possible ?

How do you get from torque to horsepower?

What do you get when you multiply torque times rotational speed?

Hi TVP45,

Check out the following link: Torque

You can find the relevant information under "Relationship between torque, power and energy."

I wasn't looking for information. I was trying to get you to think about these things.

Here's another interesting fact. The pressures found in a rifle barrel or a cannon turn out to be in the region of about 35,000 psi. or about 2460 kg/cm**2. Compare this with the 500 psi (35 kg/cm**2) developed in the power stroke of an internal combustion piston engine. And yes the same impulse figures and calculations quoted in my earlier post (#13) also apply to rifles and cannons. Any comments ?

Welcome to mechanics and machine design !! It seems to me that you discover some aspects you never were confronted with in the past. It is OK, I like when somebody makes an effort to come to new knowledge. The only suggestion I have is for you first to look better at details and try to avoid mixing notions. An IC (or any other machine with a similar design basic concept) is analyzed in a bit different way since the geometric parameters of the energy transmitting linkage are not constant over the time. You oversimplify things so that your results are wrong. Same with what you give as example : cannon barrel. Pressure behind the bullet is not constant so that the initial pressure is important but it is not the only aspect. Proof of it is the evolution of barrel length over time and also evolution of the propeller combustion velocity. Of course the basics are the same in an engine "barrel" or in a "cannon" cylinder. Other parameters determine the result which is different function of chosen goal. It goes even further same laws are valid for rockets or submarines. To make your ideas clear make a sketch and send it with explanations. Till now not everything is already clear. At least for me.

Hi Nick

I appreciate the amount of thought you had put into your comment and also the sentiment. Even though there is a lot of truth in what you say, there are still a few pieces missing. Take the concept of torque it is a rotational or turning force. When dealing with a car engine, torque (application of force) arises during the power stroke. So power is calculated as torque x 6.28 x rpm / 60000 (divide this by 746 to get bhp). Yet as I had pointed out in a 4 cylinder 4 stroke engine there are 12000 power strokes at 6000 rpm, so obviously only half the torque being generated is being accounted for by this formula, yet it (the formula) is Universally used. If the torque generated is taken per second (i.e., 12,000/60 = 200) and then multiplied by the rest of the equation it seems to give a more accurate result. But in this case the torque generated per stroke would be only 0.6 Nm. What does a torque of 0.6 Nm mean in actual terms, what if it is applied 200 times a second and so on. These are the questions I had in mind.

Um... correct me if I am wrong, but I think there are only 6000 power strokes at 6000 rpm. Four stroke or four cycle means intake, compression, power and exhaust. Then the cyle is complete. I believe that you may be thinking of the camshaft speed, it is 12,000 rpm when the crankshaft is spinning at 6000 rpm. Remember, it take two full revolutions of the crankshaft to complete the combustion cycle.

Um... correct me if I am wrong, but I think there are only 6000 power strokes at 6000 rpm

Hi Sethtech355,

I appreciate your input. Your idea that there is only one power stroke per every two complete revolutions of the crankshaft in ALL 4 stroke engines seems to be a common misconception, even among engineers, and is precisely the reason for this post. The reality is far different. Think about it, not from the viewpoint of the 4 stroke cycle but from the point of view of the rpm involved and the number of cylinders. So in a 4 cylinder 4 stroke engine : -

1) Cylinder 1 power stroke going down.

2) Cylinder 2 compression stroke going up

3) Cylinder 3 intake stroke going down.

4) Cylinder 4 exhaust stroke going up.

(Note these can appear in a different order but the important thing to keep in mind is that all of these operations happen simultaneously with the power stroke i.e., compression, intake exhaust.)

Now what happens during the power stroke ? The piston moves from TDC to BDC i.e., one half turn of the crankshaft, when the power stroke reaches BDC in cylinder 1 in the given example then the compression stroke in cylinder 2 has just completed, so the next cycle (i.e when power stroke in cylinder 1 has been completed and is at BDC and compression stroke in cylinder 2 has been completed and is at TDC is a power stroke in cylinder 2 moving the crankshaft another half turn. So in sum one power stroke from TDC to BDC in the first half of the cycle in cylinder 1 and one power stroke from TDC to BDC in the second half of the cycle in cylinder 2. So for each complete rotation of the crankshaft there are two power strokes for every full rotation of the crank shaft. Thus in a 4 cylinder 4 stroke engine at 6000 rpm there are in fact 12000 power strokes taking place.

I forgot to add that there will be 3000 combustion events at 6000 rpm in a four cycle engine. There will be 6000 combustion events in a two cycle engine in 6000 rpm.

Hi again,

In a two stroke engine the process of intake of fuel is combined into the downward power stroke and the process of exhaust is combined into the upward compression stroke. So in a single cylinder two stroke engine there is one power stroke for every full revolution of the crank shaft. The interesting thing in this configuration is that as cylinders are added the number of power strokes per crank-shaft revolution goes up accordingly. Thus one cylinder , one power stroke per crankshaft rotation, two cylinders two power strokes per rotation, three cylinders, three power strokes per rotation etc., Thus a 4 cylinder two stroke engine running at 6000 rpm would perform 24,000 power strokes per minute !

I never really thought of it that way. You have a good point. Well done.