Theoretical Max of an IC Engine

[TexanJohn]

I am a car enthusiast and have absolutely no engineering educational/knoweldge, so keep that in mind.

If an internal combustion engine revolutions per minute increase simply by opening the throttle, I assume that this is a function of the pressure differential between the air outside the motor and the low pressure area created within the motor (assume we are discussing a naturally aspirted engine). Assume the parts could withstand the forces (pistons, pins, rods, bearings, springs, etc). What would happen if I just left the throttle open? The RPM's would continue to increase obviously, but at what point would they max? Or would they? I assume at some point other restrictions regarding the intake manifold, cam design & valve events, etc. would take over. But would the pressure differential come close to zero? My assumption is yes based on experience, but I am not for sure what would happen with the RPM's.

Just a mind exercise that I haven't worked out in my head.

 

[russ_watters]

ICE's aren't quite my specialty, but...

The pressure differential is more or less fixed as a function of the piston/combustion chamber geometry. It doesn't decrease unless you put less fuel/air into it, but it doesn't ever go to zero (remember, the engine geometry pressurizes the air/fuel mixture, then it combusts). Deviating from that too much, destroys your efficiency - I'm not really sure of how much they deviate, though. But yes, in an unloaded engine, a small increase or decrease in pressure at compustion will cause it to accelerate or decelerate.

Now consider your car - if you step on the gas when its in neutral, the rpm will zoom up until it gets to the governor that pulls off the gass. If there were no governor, the engine would keep revving-up until the engine overheated and siezed.

 

[Cliff_J]

Ok, you've got a few questions in there.

Yes, the pressure inside and outside the manifold would be nearly the same until the RPMs meant that the volume of air being inhaled by the motor would cause the small throttle opening to restrict the airflow and create another vacuum inside the manifold, all because the air can't flow fast enough through the opening to equalize the pressure.

Second, the RPM limit would be in reality limited by the valve springs and their inability to close the valves and keep them closed for a good seal at high RPM. In Formula1 racing, they use air pressure in addition to springs to hold the valves shut, and you can bet they are super lightweight too. In layman terms, this is called "valve float" because the valves aren't following their normal movement and are sort of floating somewhere in between the places they should be.

If the valvetrain could be designed to handle any RPM, then at some point the volumetric efficiency (ability to fill the combustion chamber) starts to fall off and continually does so until the engine is no longer going to make enough power to spin any faster. This theoretical limit might be 20,000 RPM or higher, difficult to say and it would depend on design a lot but some motorcycle engines and Formula 1 are able to make power at those RPMs.

After the valvetrain, the connecting rods are the likely point of faliure. If you double the RPM, you quadruple the force because it goes up by the square of the difference. So at 20,000 RPM, the forces are 16 times greater than at 5,000 RPM. Better have some strong metal and some very lightweight pistons.

 

[TexanJohn]

Assume the parts won't break. They are all 4 times as strong as carbon fiber and 4 times lighter, and also that we have infinite cooling capacity.

The engine couldn't rev forever. I agree that VE would eventaully fall so much that the engine couldn't speed up any more. The inlet to the manifold would become a restriction at some point. I was just curious if there was a way to guess, discuss, or model what the limit would be, or if somebody had already thought the whole thing out. When they built the first 4 cycle, did they know to limit the incoming air, or did it just rev to the point of failure the first couple of times?

The camshaft design would also become a limiting factor as it affects the ability to allow air to enter the cylinders. I guess I was just wondering if it was possible to say that a 350" V-8 small block (in theory) would only rev to X rpms given an air restriction (probably based on inlet and camshaft; camshaft limitations would be hard to determine) of Y.

Thanks for the responses.

 

[Cliff_J]

The first engines actually had goveners on them to prevent over-reving. It was probably a result of destroying a few in R&D efforts.

Anyways, the early ones didn't fire on every compression cycle, and its odd hearing them run because the RPMs hunt up and down as it revs up and the govener kills the spark and they slow back down.

On some of the stationary crankshaft/moving cylinder engines used on early aircraft between WWI and WWII, the spark was the only means of controlling RPM. So you had spark maybe every 5-6 compression cycles at idle, and then full power.

Odd to think what was deemed acceptable when fuel was cheap and the air was clean...

It would be tricky to model even the popular 350cid chevy unless you had a good data set on the VE of the engine across the RPM range. You could maybe ballpark the figures with some flowbench results that were integrated over time against intake runner length, but it would be some pretty sophisticated CFD to scale 28.8 inches of water up to whatever vacuum the cylinder has at valve opening and so on. Pretty dynamic stuff occurring, empircal data might be the easiest by far.

Here's a new Renault F1 engine, 2.4L V8. The rumors are these engines will be developed to go to 22,000RPM to compensate for the loss of 2 cylinders off last year's motor spec. Still, impressive stuff.

http://video.google.com/videoplay?docid=-7393939630149644203&q=F1

 

[Gigabyte111]

assuming the engine in quiestion is a four stroke

it seems to me that the only limit on the max RPM of an IC engine would be at the point where the time it takes a piston to depress, and return to it's origional position (i will call that C in my mathmatical equation) is equal to the time it takes the gas fuel mixture to explode ( E in my equation)

so since D=R*T (distance = Rate * time)

and Maximum revolutions is when C=E

so if we assume that the travel of the pistion is 3 inches (6 inches to complete it's trip and it takes .01 seconds to explode the fuel mixture

then r=d/t or r=6/.01
therefore R=600 or 600RPS or 36000 RPM

so for the given engine specifications, and assumed burn time, the maximum RPM is 36000RPM



OR AT LEAST THAT IS WHAT I THINK

 

[TexanJohn]

http://video.google.com/videoplay?docid=-7393939630149644203&q=F1

Check out this homemade model engine running on alcohol. Definitely the coolest, toughest model engine I have ever seen. The video is awesome.

http://weberprecision.com/

 

[TexanJohn]

assuming the engine in quiestion is a four stroke

it seems to me that the only limit on the max RPM of an IC engine would be at the point where the time it takes a piston to depress, and return to it's origional position (i will call that C in my mathmatical equation) is equal to the time it takes the gas fuel mixture to explode ( E in my equation)

so since D=R*T (distance = Rate * time)

and Maximum revolutions is when C=E

so if we assume that the travel of the pistion is 3 inches (6 inches to complete it's trip and it takes .01 seconds to explode the fuel mixture

then r=d/t or r=6/.01
therefore R=600 or 600RPS or 36000 RPM

so for the given engine specifications, and assumed burn time, the maximum RPM is 36000RPM



OR AT LEAST THAT IS WHAT I THINK

I like your thought process. Agreeing to your assumptions as well as that 36,000 RPM was obtainable (i.e. the engine will stay together), my question would be this: Would some form of pressure equilibrium (due to flow constraints) be reached before then?

 

[Cyrus]

I would say the following would happen:

As your engine speed increases, (RPM goes up), the pistons and valves have to move faster. Now, you want to assume that they can withstand the stresses in such operation, so we don’t have to worry about fatigue or stress failure. Even with this assumption, your idea that the pressure differential goes to zero is wrong. Every time the piston goes down, no matter how long it takes, it will create a vacuum. The reason is because all the valves are closed during the down stroke, so the only thing faster RPM’s does is decrease the amount of time the piston takes to create that vacuum. Now, the piston HAS to create the SAME amount of vacuum no matter how fast or slow the RPM because the piston’s displacement is the same amount on the down stroke, no matter how fast it’s going. This means your assumption is HIGHLY based on the fact that you have PERFECT valves that can keep up and give you seals in instantaneously small time limits. The next consideration is your maximum RPM. Given your perfect situation, you have a PERFECT air/gas mixture. Then when you compress this and ignite it should combust with the BEST power output compared to any other ratio. This will cause the pistons and the crank shaft to accelerate. Over time it will get faster and faster, but, at some point the crank shaft will get to a speed that exactly matches the maximum power output of the piston. The reason is because even if you have a perfect mixture and perfect combustion, there is always SOME upper limit to that value (your piston can only move to ONE Top Dead Center value no matter what, which means you can only get so much compression with your given perfect air mixture ratio). So after the crank shaft gets up to this ideal speed, no matter what you do, you can never go faster, because you NEVER have any ability to output MORE power, because your internal geometry has not changed. Now if you take reality into account, you would find a MUCH lower value that you are limited to. Back to your question, the pressure differential NEVER goes to Zero. It just has less time to exchange flow between the low pressure air in the cylinder and the high pressure air in the atmosphere (because it spends less time with the valves open as RPM goes up), which means LESS air by volume can enter as it speeds up for that small instant in time in which the intake valve is open. This means less power as RPM goes up. If you adjust your mixture by leaning it out as your amount of air entering decreases so you get back to your new ideal fuel/air ratio due to your reduced air flow caused by increased RPM and decreased intake valve time being opened, I *think* you would find that your power output goes down. Because you are now compressing LESS volume of gas and LESS volume of fuel, which means you overall have LESS chemical energy available in the SMALLER amount of fuel. This is also because, again, you can only compress your gas so much, and we are already assuming we can compress it to its ideal limit.

 

[brewnog]

Generally, in practice, maximum cyclic speeds are determined (through engine geometry) by maximum mean piston speeds, which are in turn limited by gas flow resistance, and inertia of moving parts. For small automotive spark ignition engines these are in the region of 15m/s, but this is being pushed up constantly by increasing volumetric efficiency, lightning rotating and reciprocating parts, and increasing strength (this is one of the reasons high performance engines will run with a forged steel bottom end, and strengthened valvetrain). Race engines are obviously higher than this, but also go wrong a lot more.

 

[Gigabyte111]

Every time the piston goes down, no matter how long it takes, it will create a vacuum. The reason is because all the valves are closed during the down stroke, so the only thing faster RPM’s does is decrease the amount of time the piston takes to create that vacuum.

not entirely true, because if as you say the pistion creates a vaccum, then there would be nothing to push the pistion on downward stroke, thus no running engine. the way it works is this, the Gas/air mixture is ignited. I am no expert in chemisty but what i understand is that when you turn a liquid into a gas, it expands greatly. much in the way that a gun goes off except in that case it is a solid transforming into a gass that shoots the bullet. now what i was saing was that if a pistion takes less time to complete its down stroke and begin its upstroke then it takes for the gas fuel mixture to finish expanding. this means that the pistion will begin to encounter more and more resistance from the still expanding air and will soon reach a point that the pistion can not compres the air any more and the rpm increase will cease.


on another thought, if my math is not correct and the engine really could rev for ever, what would happen if the incoming air hit suppersonic speeds?

 

[turbo]

I have done a lot of tuning on a fairly high-performance carbureted 4-stroke (my old H-D Wide-Glide with heads reworked by Dave Perewitz, steep cams, S&S Super E carb, exhaust pipes with anti-reversion inserts, low restriction air cleaner, etc) and there are a lot of things to take into account.

1) Dense (cool) incoming air with well-atomized fuel optimizes power. Hot exhaust pipes help scavange exhaust through the exhaust ports for more power. So you want cool incoming air and nice hot exhaust pipes. Some people ensure this by wrapping the pipes with insulating tape.

2) When the exhaust gas exits the exhaust pipe (especially if it is lightly baffled or unbaffled) it creates an acoustic reversion pulse that opposes the clean scavanging of the next pulse of exhaust gas, limiting throughput, so there must be some way of suppressing reversion pulses.

3) As mentioned before, the fuel should be finely atomized for optimal burning (short burn-time). This is a problem with a butterfly-carb like the S&S, which operates most happily at full throttle. The solution for this (for me) was a Yost power tube - an insert that sits over the main jet and creates an asymmetry that atomizes the fuel as it exits the venturi and enters the carb throat.

4) Another concern is smoothness of the head porting. You might think that polishing the ports would make the fuel/air and exhaust move more quickly, but this is not the case. Small-scale roughness on the port walls results in micro-turbulence (acting like a layer of tiny bearings) that allows the bulk of the air/fuel mixture and exhaust to move far more quickly - a key to power.

As you can see, there are a lot of limiting factors that can prevent a 4-stroke engine from "running away". Knowing that there are power addicts like myself that will try to minimize these limiting factors, Harley makes sure that their ignition systems incorporate rev limiters. Of course, we circumvent those, too, but that's another story. There are more limiting factors, including valve timing (cams), ignition timing, etc. It is quite possible to make a 4-stroke engine exceed design parameters and self-destruct, but it will not simply rev up more and more at full throttle until the engineering issues are addressed.

 

[Cyrus]

not entirely true, because if as you say the pistion creates a vaccum, then there would be nothing to push the pistion on downward stroke, thus no running engine.

Do you understand how an engine works? From what you just wrote I take it that you do not. An engine's piston creates a vacuum (P<<Atm) on the down stroke and a compression (P>>Atm) on the upstroke.

im no expert in chemisty but what i understand is that when you turn a liquid into a gas, it expands greatly.

No, that's wrong. That is NOT what your engine is doing. When you ignite the gas in an engine, it is vaporized into the mixture of air. This ignition causes a rise in temperature which by Pv=RT causes a rise in pressure and pushes the piston down. I don't think you understand the workings of an engine.

Why don't you just open a standard general chemistry book and look at the change in enthalphy for combustion of Propane and Oxygen:

$$C_3_H_8 (g) + 5O_2 \rightarrow 3CO_2(g)+4H_2O(l)$$

$$\Delta H_{rxn}^o = -2220 kJ$$

So if this engine is ideal, and it always has 100% combustion, with an efficiency of $$\eta_{eff} = 1$$, and the air/fuel ratio is always perfect, then as your RPM goes up, there is less time the valves remain open, and there is less volume of air flow into the cylinders, so you have to have a lower proportion of fuel to keep your ratio just right, which means less moles of reactants, which means less power out put.

Here is another implication that just crossed my mind. Even though the power per single combustion goes down as RPM goes up, you have more RPM, and thus more combustions per unit time with the higher RPM range than you do with the lower RPM range. So there is probably some trade off where you are optimal, which is why you see a power v rpm curve of an engine peak out at around 1/2 to 3/4 of full RPM and go back down. Some tradeoff inbetween is the optimization value. Where your amount of energy per unit time, (Energy * RPM) is the most compared to any other set of values. This would mean RPM does not increase limitless, because as RPM goes up, power goes down at a faster rate per unit RPM. So overall power goes down.

 

[Gigabyte111]

" If you put a tiny amount of high-energy fuel (like gasoline) in a small, enclosed space and ignite it, an incredible amount of energy is released in the form of expanding gas. "-http://auto.howstuffworks.com/engine3.htm"
^
well according to this| site i am right and your wrong.

and also as even more proof, get a teapot with a top that can seal very well, and on the spout, atach a LARGE baloon. once that water has boiled away you will have a baloon at least 15 times the size of the original volume of water.

 

[Cyrus]

Why don't you go back to your link and look at the 4-stroke diagram:

When the piston reaches the top of its stroke, the spark plug emits a spark to ignite the gasoline. The gasoline charge in the cylinder explodes, driving the piston down. (Part 3 of the figure)

http://auto.howstuffworks.com/engine4.htm

Your engine is not being powered by liquid changing state to gas. It is due to the BURNING of the gas. The chemical composition changes, releasing energy and causing rising temperature. Your tea pot example is not what is happening in your engine. (That is going from water to steam, no exothermal energy release)

I would recommend a good book on cars, "How your car works" by Sam Julty. You will enjoy reading it.