ProdriveMS

Big-Den

You're comparing power below the peak to the peak. In that case, friction and vibration have a tiny effect on power compared to cylinder filling. The way I understood it, TopEnd was describing two same displacement engines at their power peak. One at 6000rpm, and one at 12,000 and the question was whether power will double with the doubling of rpm. To that question, the answer is no, because friction and vibration increase as the square of rpm.

Al

Big-DEN

Another example.

OK lets do peak to peak.

A combination that is efficient to 4250RPM vs one that is efficient to 8500 RPM. Lets make TQ peaks 3200 and 6400 to make things closer to "fair".

Combination that is efficient to 4250RPM has a E7 like head, small diameter tubular header, and camshaft designed to optimize its efficicny down low.

One thats efficient to 8500RPM has properly sized head, properly sized header, and camshaft designed to optimize its efficiency up higher.

If thats not a fair example, lets do a 5000 and 10000 RPM examples. INduction/exhaust and camshaft each time decide to optimize engines operation over its designed range.

Not trying to knock your knowledge. I just think theres too many cases which go against "HP will not be double at double the RPM because of frictional and pumping losses"

ProdriveMS

If some had stated that their 400hp car would do a drag limited 200mph, then asked if they doubled it to 800hp, could the car do 400mph with the proper gearing, the answer would be no. Because drag increases as the square of mph. Well, vibration and friction inside an engine increase as the square of rpm just like aerodynamic drag. The first example is fine and everyone would agree it wouldn't happen, so why the confusion on the second? If you imagine a graph of y=x^2, it looks like a parabola. That graph descibes the friction inside your engine. If you look at 0, the graph starts out flat and increase slowly at first, but eventually it will shoot off to the heavens. The friction penalty from 0 to 5,000rpm is not as great as it is from 5,000 to 10,000, or 10,000 to 20,000. Will the engine make more power? Yes. But not double the power. If you were able to keep cylinder filling, thermodynamic efficiency, etc. the same, friction would still get you. It is the fundamental reason. That is why in high revving engines, designers pay very close attention to reducing as much friction as possible. The slightest reduction yields big gains up top.

Al

Big-DEN

We all know double the HP in a car will not double its top speed, unless where talking about very low speeds.

But I believe my example of 5,000 and 10,000 was good. The most efficient 5,000 RPM wonder compared to the most efficient 10,000 RPM wonder with similar bottom end, the 10,000 RPM wonder will be able to have greater than 2x HP potential than that most efficient 5,000 RPM wonder.

Now the difference between the greatest 10,000 RPM wonder of a given displacement and bottom end vs the greatest 20,000 RPM wonder of a given displacement and bottom end is a different question.

So the new question. Say 2.4L f1. Can/induction/exhaust for greatest potential to 10,000 RPM VS the greatest cam/induction/exhaust for greatest potential to 20,000 RPM. Is the 20,000 RPM going to be more than 2x from higher air speeds and stronger resonant tuning effects ( greater natural supercharging to say ), or is the greater friction and pumping losses going to make it less than a 2x affair.

We all can agree on my 5,000 greatest potential vs 10,000 greatest potential on given displacement that the 10,000 RPM optiimized will be greater than 2x over the 5,000 RPM optimized situation.

In higher RPM 10,000 vs 20,000 RPM does friction and pumping losses overcome the greater VE's availabled?

ProdriveMS

Even at low speeds, it wont be exactly double. It may be 1.99999x at low speeds and go down from there at an increasing rate. It all depends on where you are on the graph.
I might have missed it. Did you give an example where this happened? Did you build two motors where this happened? If that was the case, then I believe that the 5000rpm engine wasn't truely optimized for that rpm. Or the 10,000rpm motor was optimized a little better. If the friction penalty was 5hp at 5,000 rpm, it would be 25hp at 10,000. Probably within the range of error- depends on what the friction penalty really is.


The intake pressure wave will be greater at 20k rpm vs 10k if the piston is moving faser. The exhaust wave depends primarily on the energy of combustion. It wont necessarily rise with rpm. The intake wave wont increase at a rate of 2x piston speed though. Even assuming that piston speed doubles from 10k to 20k which it wont(as revs rise with each redesign, stroke is reduced to keep piston speed within the mechanical limits of the bottom end), the increase in intake pressure wave isn't enough to overcome the frictional losses which are increasing at a faster rate.

Air speeds aren't 2x faster. Ports are sized to give maximum velocity at the airflow needed without going sonic. When airflow requirement increases as max revs rise for each redesign, the port size increases to deliver more air while keeping velocity close to the same as before. Great questions Den.

Al

Big-DEN

On the 5,000 vs 10,000 example. Take 5.0L fords. Pick the two best examples of each and use that for a comparison. The 10,000 RPM affair is going to be more than 2x than that of the 5,000 RPM affair. IE: You will not get better than 400 engine HP by 5,000 RPM out of a 5.0L ford. HOwever you will get greater than 800 engine HP or so by 10,000 RPM out of a 5.0L ford designed to do this.

Between 10,000RPM and 20,000RPM I cannot comment.

Just wanted to add the disclaimer, none of us have to physically build anything to prove it, we can take the best examples of either situation that others have done and use it for the comparison.

ProdriveMS

Ok. I don't think the 5k rpm Ford is really optimized to run at that rpm. I would bet that if as much development went into the 5k motor as goes into the 10k motor, the power difference between the two would decrease.

That's fine. I just wasn't sure what examples you were referring to.

Alin

Big-DEN

Prodrive.

The 5,000 RPM and 10,000RPM where just two RPM points, it can be 6,000 and 12,000 as well.

5,000 RPM and 10,000 RPM are an example we can pull real world from.

You can optimize a motor for operation up to 5,000 RPM.

You can optimize a motor for operation up to 10,000 RPM.

Just depends on what your trying to do.

EdmontonSS

Quikin: Thanks for the article, it added some insight that never occured to me before... About the boost "assisting" on the intake stoke... And the fact that frictional losses remain constant, but burning twice the fuel. I guess that with those two factors in perspective, it may be possible to double engine output with 14.7 or so PSI of boost...

Big-DEN: I have to very respectfully disagree with your position that an engine spinning twice the RPM, all other factors being equal, will MORE than double horsepower, due to the air velocity of the charge filling the cylinders. If I get higher air velocity at higher RPM, allowing higher horsepower, I want to know...

Why does my engine "nose over" past 6500 RPM?

My ports are not going "sonic" they are probably well over 220 cc on my little 346. Also the 305 in my truck likes to "nose over" at about 5 k RPM... With your theory, our engines (any engine) should make an expodential horsepower curve... But they don't.
Sure port velocity goes up as RPM increases, but "air" has momentum... A lot of it... relative to the speed of events in an engine... And the airflow, at least in our piston, poppet valve engines, must stop and start over 50 times a second over 6000 RPM. My personal belief is that no 10,000 RPM engine will make double the power of a 5,000 RPM engine, if both use the same technology and are optimized for their respective peak power RPMs.

EdmontonSS

Big-Den: You were also metioning resonant tuning effects. Sure resonant tuning creates high and low pressure points in the manifolds. Pressure is "potential" energy that will assist in supplying the "kinetic" energy to get the air moving, but there is still a huge "kinetic" energy requirement to get the air to fill a cylinder and subsequently "stop" 50 times a second. The "kinetic" energy from the air "stopping" against the closed poppet valve is what creates the pressure "wave" for the next cylinder. Resonant tuning does make an engine more efficient at the RPM the resonant tuning is tuned for, but it is not "free energy" by any means. Air has mass, and the energy to move that air twice as quickly IS expodential. (E=mv2)

Big-DEN

EdmontonSS,

I am fine with you respectfully disagreeing. But its going to be best if the actual technical facts come out.

find your best example of say a 5.0L ford - or make it even a 350 chevy - fully optimized for operation to 5000 RPM with power peak centered right at 5000RPM. Any choice of headers any choice of heads, any camshaft and optimized CR for this.

NOw take that 5.0L ford - or make it even a 350 chevy. fully optimize it for operation to 10,000 RPM with power peak centered right at 10,000 RPM. find the best examples.

The best examples on the 10,000 RPM will more than double the HP produced by the 5,000 RPM example.

To make it even more easy to find information, turn it down to 9,000 RPM and 4,500 RPM.

Again, the cases you find that are fully optimized for operation to 9,000 RPM are going to be more than double the HP for a situation that peaks at 4,500 RPM.

You think you going to have more than 350HP out of a 5.0L ford at 4,500 RPM?

What about more than 700HP at 9,000 RPM? Theres many cases with 600HP being found out of 5.0L ford before 7,500 RPM... 700HP is going to be reached before 9000 RPM on optimized combo.

Now on 10,000 RPM vs 20,000 RPM I dont know, but these examples I'm pulling out I am confident I can fully prove it. I don't really want to waste the time.

EdmontonSS

Big-DEN: I have a feeling that no matter how much searching is done, nobody will find an example to prove you wrong. I think this is simply due to the fact that no-one goes and builds an all-out 4500 or 5000 RPM engine. I think that any racing has a displacement limit via rules, practicality, or otherwise... And with a limited displacement, people are going to build for as high an engine speed as possible. I don't think anyone has built a multi-thousand dollar 5000 RPM motor... I know I wouldn't... I don't feel your "test" is valid, but I will take it all back if you can find such an example yourself, or a good research paper. Possibly some "engine master" engines might provide some insight as they were built for large average power over a specific RPM band (in an attempt to keep costs reasonable, I believe). But again, these engines were far from optimized for a specific RPM...

topend

Would u say the power would be within 10% of the doubled power???
Would the engine consume the air/fuel mixture equivalent of 2x the power???

Big-DEN

Thing about it.

The different RPM ranges have increasing and higher and higher VE ( Volumetric Efficiency ) or cyllinder "overstuffing" per say.

The 20,000 that these F1's are at are nearing 140 VE or 140%!

At around 10,000 RPM your doing in the 120% range

And some good 6000-8000 RPM can exceed 100% to 110% or so.

Theres charts for this.

And the guys are right, frictional and pumping losses increase like that of aerodymics, by the square!

Richiec77

iTrader: (3)

oh yeah. I almost forgot this part too since aerodynamics was brought up.

The crank is now going to have to spin 2x as fast. That creates an aerodynamic drag on the rotating assy also. So now the engine also has to fight against that drag also. I'm guessing it would be safe to say X^2 factor also since it would be aero-based drag.

It's also worse in some design's over others. Example being the LS1 vs Ford 351. Narrow skirt vs Wide skirt

You know top-end, I'd have to assume That the air-fuel amount burned increases linearly, but measured power would be lower since the items above would be using that power. If power was increased linearly to RPM, more A/F would have to be burned to keep the power in line.

Super Speed

iTrader: (8)

A few stickers here and there will get you there in no time...

DarricksZ28

This thread is still going?? This should be moved, as it makes no sense even to discuss it.

Big-DEN

Darricks.

So let me guess what you think.

Boost or manifold overpressure is free so exactly 14.7 PSI on a boost pressure guage == double a engine output

And that an engine built to be most efficient at 5,000 RPM if built to be most efficient at 10,000 RPM will not more than double the HP due to fricitional losses increasing faster than increased VE.

DarricksZ28

In theory 14.7psi+atmo would yield "double" the atmo alone because an engine moves air, and the more it moves the more fuel can be burned the more power will be made. But we don't race in theory, so no it won't exactly, it will take more than 14.7psi. It depends on numerous things. Mainly compressor efficiency, air density and heat added. All this has been stated already.

Will an engine make 2x the power if you spin it 2x the peak RPM? NO. If that were the case then peak wouldn't be where it is. The engine would disintegrate. There are 4 variables you can change to increase power in a gas engine. Airflow, displacement, RPM and volumetric efficiency. Of those four variables, the increasing of RPM yields the least increase in power alone along with displacement. Airflow and VE are the best ways to increase specific power output. It's silly, if you have any understanding of the internal combustion engine, to argue that. Increase airflow, double the airflow you double power while maintaining a given volumetric efficiency. Increasing RPM effectively decreases volumetric efficiency, which is why the torque curve is not linear to redline. Increasing RPM gives less time to fill the cylinder properly, light it off and excavate it. It also increases windage in the crankcase, increases piston side loads and beats the **** out of bearings. Not to mention the harmonics. It would be counter-productive and stupid to even attempt "doubling" power output in such a manner when your application/project isn't limited by displacement, head design/type, etc. To me this isn't a topic for this forum because it makes no sense to even discuss it.

ProdriveMS

Den- it wont. The only thing you have to support your position is a poor example because nobody builds a max effort 5000 rpm engine. What you are comparing is a warmed over engine that happens to peak at 5k vs a 10k properly developed race engine. The physics behind why it cant happen have been laid out already. The strength of the intake pulse increases with piston speed, but piston speed doesn't necessarily double from the 5k to 10k motor. Even if I give you that piston speed is doubled, the energy contained in the pulse hasn't doubled. Meanwhile friction, vibration, not to mention windage which, I completely forgot about until it was mentioned has, increased as the square of rpm. The VE of an engine is not determined by its rpm. Intake air velocity is roughly a factor of piston speed and port size assuming the proper cam is used. Our 600cc FSAE engine got 130% VE pulling through a 20mm restrictor. The VE reflects how well developed the engine is, not how fast it revs. If you threw $250 million at a 5k rpm engine like F1 does, it would hit 140%VE as well. You are comparing apples to oranges.

I used to think that before I read Prof Blair's book. But it is actually the piston moving downward with the intake valve open that creates the negative pressure wave. The example he gave was to try and exhale with your mouth closed. Build pressure with your lungs, then open your mouth. You will get a "pop" sound. That is a positive pressure wave you just heard. Next try to inhale while keeping your mouth closed. Build a vacuum, then open your mouth. You will hear a "huh" sound. That is the negative pressure wave exiting your mouth. In an engine, the piston is your lungs and the valves are your mouth. The positive perssure wave is the exhaust pulse and the negative pressure wave is the intake pulse. In an intake tract, the negative pressure wave travels away from the valve, to the bellmouth where it reverses into a positive pressure wave and starts travelling back toward the intake valve. When it hits the open intake valve, it reverses again as a negative pressure wave and the process starts over. Positive perssure waves will continue to cram air and fuel into the cylinder as long as the intake valve remains open. Everybody tunes off the second wave though.

Al