Boost vs Revs: Comparing Internal Engine Stresses

[Williampreza]

Let's say you had two engines that both made 280 HP. One is boosted and makes peak HP at 7200 RPM and the other is N/A and makes equivalent power by revving to 8000 RPM.

The FI engine is making it's power numbers through higher cylinder pressures and therefore greater torque, while the N/A engine is making power through lower torque numbers at a greater frequency (The power of the Maths!)

Assuming identical internals, which engine is working harder/wearing out faster?

I'll be honest, I'm mostly interested in the forces affecting the rotating assembly specifically the connecting rods, but I don't want to limit any productive discussion.

Answers involving physics and math would be most useful.

[Yoshoobaroo]

As long as you rule out anomalous events like detonation, the higher revs will kill the motor faster than boost.

The force on the rods and crank due to cylinder pressure are trivial compared to the inertial forces of a piston and rod going up and down 133 times PER SECOND. The bearings need to withstand the force of the piston traveling around 25 mph, and reversing its direction in 1/133th of a second. It's incredible these things even work.


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[Teseo]

What about the stress on con rods?

[Yoshoobaroo]

Originally Posted by Teseo

What about the stress on con rods?

They're in the same load path, so the same applies


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[Spuds]

If the internals we're identical, they would be woefully inadequate for one or the other, or moderately inadequate for both.

By inadequate, I mean not optimized, leaving either significant power on the table or being more likely to break under the given conditions.

The answer is therefore: It depends on which situation the engine was originally designed for.

PS, don't wanna math...

[ls1ac]

I think you should stop thinking of HP and look at the torque curve, work load and stress.

[Williampreza]

Originally Posted by ls1ac

I think you should stop thinking of HP and look at the torque curve, work load and stress.

Yeah, that's exactly what I was going for, but I didn't want to hand-feed it.

So, the question becomes, if I'm hammering on one end of a metal rod with a solid base at the other end and with 183.8 ft/lbs of force at a rate of 8000 blows per second, is that going to bend the rod faster or slower than hammering on it with a force of 202.4 ft/lbs at a rate of 7200 times per second.

[Spuds]

Originally Posted by Williampreza

Yeah, that's exactly what I was going for, but I didn't want to hand-feed it.

So, the question becomes, if I'm hammering on one end of a metal rod with a solid base at the other end and with 183.8 ft/lbs of force at a rate of 8000 blows per second, is that going to bend the rod faster or slower than hammering on it with a force of 202.4 ft/lbs at a rate of 7200 times per second.

Well, if you really want answers...

... Ft/lb is not a unit of force, lb is a unit of force. Combustion forces are more distributed than hitting something with a hammer, unless you have some very thick and soft foam on the end lol. And you are vastly oversimplifying the model.

However, to answer that specific question you need a basic understanding of material properties. Steel has what is called a fatigue limit, which is the highest amount of stress a material can endure for an unlimited number of cycles. So as long as your (steel) parts do not exceed that amount of stress, they will last indefinitely.

What is most likely going to cause problems for the situation you care about are the bearings. They operate properly by creating an oil film between two parts. That oil film is maintained by oil pressure. Because fluids act on solids via pressure, you need enough pressure and area (PxA=F) to counteract the forces created by both combustion AND inertia. Oil pressure depends on, among other things, the design of the oil pump. The oil pump design usually increases pressure as rom increases up until cavitation occurs, then the output pressure decreases. So that's one reason why I said it depends on original design.

Another reason is because the balance of combustion and inertial forces are different between the two situations you were talking about. Worst case forces = combustion/compression force vectors + inertial force vectors

Inertial: F=ma. The higher rpms, the higher acceleration. If you have higher forces, you generally need more structure of the same material to withstand it (keeping under the fatigue limit), which in turn makes a bit more force, requiring just a tiny bit more material... Anyway, all this force needs to be supported by the bearings, meaning you need more oil pressure or area. Both of those will further increase forces but let's not go down that rabbit hole too far.

Combustion/compression: this is simpler in theory as it starts out as just linear force on the piston. BUT to withstand extra combustion force, you need stronger (heavier) pistons, connecting rods, etc. Which also adds to inertial forces. Which brings us down the rabbit hole again.



TLDR:. You are asking a question that can only be 'answered' by years of R&D by hundreds of engineers. There is no definitive answer to your question because it all depends on knowing the exact engine parameters and doing a lot of analysis. If you asked specifically about the FA20 for example, there would be an answer, but I'm not going to do the analysis to figure it out lol.

[Irace86.2.0]

E=1/2mv^2 which means the energy of motion goes up dramatically with increasing velocity.

Consider trying to double torque at a given rpm or going from 8psi to 16psi versus trying to add 1000 or 2000rpms to a 7k redline (not doubling anything); it should be obvious which will disintegrate an engine easier.

[Irace86.2.0]

Originally Posted by Williampreza

Yeah, that's exactly what I was going for, but I didn't want to hand-feed it.

So, the question becomes, if I'm hammering on one end of a metal rod with a solid base at the other end and with 183.8 ft/lbs of force at a rate of 8000 blows per second, is that going to bend the rod faster or slower than hammering on it with a force of 202.4 ft/lbs at a rate of 7200 times per second.

Kinda bad analogy. The increase in repetitions aren’t the problem with increasing rpms like how you describe it like banging on metal. Increasing rpms doesn’t lead to more wear or a gasoline car that revs to 7k wouldn’t last 250k miles like a turbo desiel that only revs to 3k. The problem is that increased revs mean increases in the forces needed to accelerate and decelerate the piston so many more times per second.

[cueball89]

Check out the Turbocharging Performance Handbook by Jeff Hartman. I seem to recall a formula and an example he talks about, something to the tune of engine wear increases exponentially with rpm.

[bfrank1972]

Originally Posted by Spuds

Well, if you really want answers...

... Ft/lb is not a unit of force, lb is a unit of force. Combustion forces are more distributed than hitting something with a hammer, unless you have some very thick and soft foam on the end lol. And you are vastly oversimplifying the model.

However, to answer that specific question you need a basic understanding of material properties. Steel has what is called a fatigue limit, which is the highest amount of stress a material can endure for an unlimited number of cycles. So as long as your (steel) parts do not exceed that amount of stress, they will last indefinitely.

What is most likely going to cause problems for the situation you care about are the bearings. They operate properly by creating an oil film between two parts. That oil film is maintained by oil pressure. Because fluids act on solids via pressure, you need enough pressure and area (PxA=F) to counteract the forces created by both combustion AND inertia. Oil pressure depends on, among other things, the design of the oil pump. The oil pump design usually increases pressure as rom increases up until cavitation occurs, then the output pressure decreases. So that's one reason why I said it depends on original design.

Another reason is because the balance of combustion and inertial forces are different between the two situations you were talking about. Worst case forces = combustion/compression force vectors + inertial force vectors

Inertial: F=ma. The higher rpms, the higher acceleration. If you have higher forces, you generally need more structure of the same material to withstand it (keeping under the fatigue limit), which in turn makes a bit more force, requiring just a tiny bit more material... Anyway, all this force needs to be supported by the bearings, meaning you need more oil pressure or area. Both of those will further increase forces but let's not go down that rabbit hole too far.

Combustion/compression: this is simpler in theory as it starts out as just linear force on the piston. BUT to withstand extra combustion force, you need stronger (heavier) pistons, connecting rods, etc. Which also adds to inertial forces. Which brings us down the rabbit hole again.



TLDR:. You are asking a question that can only be 'answered' by years of R&D by hundreds of engineers. There is no definitive answer to your question because it all depends on knowing the exact engine parameters and doing a lot of analysis. If you asked specifically about the FA20 for example, there would be an answer, but I'm not going to do the analysis to figure it out lol.

Great post

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[Lantanafrs2]

Who cares if motors are cheap enough? Anything mechanical will eventually break. I prefer less moving parts.

[cueball89]

A little information from ARP about load and rpm.

Fastener Load
The first step in the process of designing a connecting rod bolt is to determine the load that it must carry. This is accomplished by calculating the dynamic force caused by the oscillating piston and connecting rod. This force is determined from the classical concept that force equals mass times acceleration. The mass includes the mass of the piston plus a portion of the mass of the rod. This mass undergoes oscillating motion as the crankshaft rotates. The resulting acceleration, which is at its maximum value when the piston is at top dead center and bottom dead center, is proportional to the stroke and the square of the engine speed. The oscillating force is sometimes called the reciprocating weight. Its numerical value is proportional to:

It is seen that the design load, the reciprocating weight, depends on the square of the RPM speed. This means that if the speed is doubled, for example, the design load is increased by a factor of 4. This relationship is shown graphically below for one particular rod and piston.