Compression Ratios / Short Block

[Spartarus]

You can make 300 whp on 10 PSI on pump gas and stock compression with great drivability. I dunno what everyone's whining about.

With adequate cooling and the right viscosity, you can support any hp you want. The oil pressure freak out was a bit of a "chicken little" type event.

For an actual racecar, time attack, for instance, it's always been an art form to keep a wild power turbo setup running.

[wparsons]

Lower compression lets you run more boost, but it also requires you run more boost to make the same power. More boost = more heat, meaning you might need to run a more efficient IC.

[Spartarus]

Originally Posted by wparsons

Lower compression lets you run more boost, but it also requires you run more boost to make the same power. More boost = more heat, meaning you might need to run a more efficient IC.

All other things equal, increased charge temperature will necessarily make an air-to-air intercooler more efficient.

I'll ask a few annoying, mythbuster-type questions.

Why does it require you to run more boost? Does it necessarily require you to run more boost? How much boost offsets a drop in compression?

If you had a magical detonation-proof fuel, why would you still need an intercooler? Why would an intercooled turbo still make more power? (Because it would, by the way. That's not a trick question.)

Another. You have 2 engines. In the first, you leave everything alone. In the second, you drop compression and raise boost, maintaining an equivalent effective compression ratio. Both engines have the exact same effective compression. Which one makes more power?

I'll give you a hint. Which engine has more potential energy in each cylinder?

One more. Say you take an engine and raise the compression 1 point. Is there any more air or fuel present to burn? No. So there's no more energy there. In fact, it took even more energy to compress it than before. So, why does it make any more power at all? (Again, not a trick question. It does make more power, just not much.)

Oh, and 1 more. The turbocharger doesn't drive anything. Why does it increase the thermal efficiency of an engine? Why can a supercharger, which robs power in a parasitic way, also increase the efficiency of an engine?

We haven't even addressed cam timing and Miller cycle yet.

[ogrowup]

Originally Posted by Spartarus

Oh I like this one. This one is good.

So, you make high peak HP, and much lower peak torque, and you don't get to high RPM's very often.

You want HP / TQ to be roughly square, and you want a lot more torque.

Horsepower is just a math equation. It comes from torque.
Specifically (Torque x RPM) / 5252 = HP
What that means for you, is you have to hit peak torque below 5252 RPM to have nice square TQ / HP numbers.

Your gears reverse the equation and turn horsepower back into torque at the wheels. (if the output is a constant, and you lower speed, you raise torque.) Torque at the wheels is what accelerates the car.

There are many ways to achieve this, and it depends on many variables.

I'll be honest, The GTX 3076 in that Ptuning kit is a little big for that goal.

You're going to want to retain as much compression as you can, because if you're looking for a "torquey" motor, I guarantee you're also looking for quick transient throttle response. Lower compression slows transient throttle response, both in and out of boost.

People say high compression spools a turbo faster, but that's an erroneous conclusion from a chicken-and-egg scenario. Whatever. To the driver, that statement is true, that's all you need to worry about.

I can continue, but in order to sort out the inputs, like compression, you need to refine your output goals.

Thanks for the intelligent answer...BTW, I left out something- I have a water meth kit. Currently I get 365hp at the wheel and Max boost at 7000rpm. This is my DD, so I don't spend a whole lot of time at 7000rpm.
I spoke to the Ptuning folks and they said I'd be much happier all around with the OEM block, the 12.5:1 CR, and swap out to the new, smaller GTX28. I'll lose a little at the top end, but at the lower end, boost will start around 3300rpm instead of 5000rpm. They had the same misgivings about the lower compression.

[Ultramaroon]

Nobody answered so I'll give it a go. Pulling this out of my ass. Learn me some.

Originally Posted by Spartarus

Why does it require you to run more boost? Does it necessarily require you to run more boost? How much boost offsets a drop in compression?

Why: Because, all other properties being equal, lower static compression ratio also means less air/fuel being pulled on the intake stroke.

Necessary: Maybe not if a bigger CAC is an option.

How much: Lots of math, ideal gas law, not truly adiabatic system, blah blah... Enough so that the air mass you're stuffing in there is equal?

If you had a magical detonation-proof fuel, why would you still need an intercooler? Why would an intercooled turbo still make more power? (Because it would, by the way. That's not a trick question.)

Because lower temp at same pressure equals denser charge - more mass.

Another. You have 2 engines. In the first, you leave everything alone. In the second, you drop compression and raise boost, maintaining an equivalent effective compression ratio. Both engines have the exact same effective compression. Which one makes more power?

I'll give you a hint. Which engine has more potential energy in each cylinder?

The first because more energy is extracted from the cycle.

One more. Say you take an engine and raise the compression 1 point. Is there any more air or fuel present to burn? No. So there's no more energy there. In fact, it took even more energy to compress it than before. So, why does it make any more power at all? (Again, not a trick question. It does make more power, just not much.)

Again, I think it has something to do with efficiency of the cycle but I don't remember for sure.

Oh, and 1 more. The turbocharger doesn't drive anything. Why does it increase the thermal efficiency of an engine? Why can a supercharger, which robs power in a parasitic way, also increase the efficiency of an engine?

Turbo converts wasted heat into work which goes to compressing intake charge, blah blah...

Supercharger? Hmmm... not sure about that. Does it have anything to do with the CAC?

[Ken@PTuning]

Oh now this is fun lol

Originally Posted by Spartarus
Why does it require you to run more boost? Does it necessarily require you to run more boost? How much boost offsets a drop in compression?

The drop in max volume (total displacement) is almost negligible.

You've reduced the thermal efficiency of the system. To get the same amount of output work, you'll need more potential energy (fuel, Q in) and the only way to do that is to increase the amount of air input.

Originally Posted by Spartarus
If you had a magical detonation-proof fuel, why would you still need an intercooler? Why would an intercooled turbo still make more power? (Because it would, by the way. That's not a trick question.)

An increase in pressure increases the temperature. This decreases your specific volume. With an intercooler and removing the extra heat, your specific volumetric flow rate increases. More air means more the ability for more fuel, increasing heat input (Q in).

Originally Posted by Spartarus
Another. You have 2 engines. In the first, you leave everything alone. In the second, you drop compression and raise boost, maintaining an equivalent effective compression ratio. Both engines have the exact same effective compression. Which one makes more power?

I'll give you a hint. Which engine has more potential energy in each cylinder?

The second will make more power. You've reduced the efficiency but there is a larger mass of air and fuel. Hence, there's more potential energy in the system. The engine that can move more air wins (usually).

Originally Posted by Spartarus
One more. Say you take an engine and raise the compression 1 point. Is there any more air or fuel present to burn? No. So there's no more energy there. In fact, it took even more energy to compress it than before. So, why does it make any more power at all? (Again, not a trick question. It does make more power, just not much.)

You've increased the thermal efficiency of the system. You may have increased the work input to compress it but the work output is now much higher as well. It's easier to see when plotted against a P-v or T-s diagram.

Originally Posted by Spartarus
Oh, and 1 more. The turbocharger doesn't drive anything. Why does it increase the thermal efficiency of an engine? Why can a supercharger, which robs power in a parasitic way, also increase the efficiency of an engine?

The turbo uses the wasted energy of the otto cycle since the end of the cycle is at both a higher pressure and temperature than ambient (that differential is your lost potential energy). This means your work input is due to what would have been lost energy. This increases the thermal efficiency of the system (work input stays the same but work output increases, or work input decreases and work output stays the same).

As for the supercharger, part of your output work is used to drive the supercharger. Your total efficiency is reduced (work input increases for the same work output).

[Ken@PTuning]

Originally Posted by ogrowup

Thanks for the intelligent answer...BTW, I left out something- I have a water meth kit. Currently I get 365hp at the wheel and Max boost at 7000rpm. This is my DD, so I don't spend a whole lot of time at 7000rpm.
I spoke to the Ptuning folks and they said I'd be much happier all around with the OEM block, the 12.5:1 CR, and swap out to the new, smaller GTX28. I'll lose a little at the top end, but at the lower end, boost will start around 3300rpm instead of 5000rpm. They had the same misgivings about the lower compression.

Yup, if I don't have to crack open the engine build it (or swap it with a built motor, same thing), I won't. For your power level and goals, you're actually better off with the stock motor.

[Spartarus]

I'm going to start by saying Oh My! An intelligent debate on the internet. Achievement unlocked.

I'm going to follow by saying I saw the thread about the updated Ptuning kit. All I can say is I'm impressed.

OK. On to the fact-straightening exercise, slash debate.

Originally Posted by Ken@PTuning

Oh now this is fun lol

Quite fun.

First. Displacement volume. Both of you addressed this, so I'll address that first.

Originally Posted by Ultramaroon

Why: Because, all other properties being equal, lower static compression ratio also means less air/fuel being pulled on the intake stroke.

Originally Posted by Ken@PTuning

The drop in max volume (total displacement) is almost negligible.

A drop in volume you say? Remember, all we're doing is lowering the piston crown. That's it. We're not destroking the motor. The swept volume, and therefore displacement remains unchanged. The new void at TDC is also present at BDC. In fact, at BDC, the total cylinder volume actually increases. Yes, it is negligible, but if you are scavenging effectively, your "effective" displacement is actually going up, not down. Again, a negligible amount but the misconception is clear and present.

NEXT. Intercoolers. Let's have a thought experiment.

Originally Posted by Ultramaroon

Because lower temp at same pressure equals denser charge - more mass.

Originally Posted by Ken@PTuning

An increase in pressure increases the temperature. This decreases your specific volume. With an intercooler and removing the extra heat, your specific volumetric flow rate increases. More air means more the ability for more fuel, increasing heat input (Q in).

All true. But here's a question. Yes, increasing pressure increases temperature. The opposite is also true. These are idealized as adiabatic processes. Why, then, is the pressure drop across an intercooler not proportional to the temperature drop? How can an "idealized" intercooler be isobaric? Given that pressure and temperature are typically proportional? It's a volume thing. We're putting the V in PV=NRT. An isobaric reduction in temperature leaves a smaller volume with a uniform pressure. In an ideal, lossless, frictionless, free-flowing system, the pressure is uniform throughout.

Next up, thermal efficiency and work! Expansion ratios! This covers the rest of the questions and answers.

Aaaaand I'm leaving it until tomorrow. I'm too tired to type the rest of this. It's been barrels of fun. Stay tuned for another exciting episode. I want to talk about PV diagrams, and Q. No, Otto and Carnot's Q, not Ian Fleming's funny Englishman with the cool toys.

[Ken@PTuning]

Originally Posted by Spartarus

I'm going to start by saying Oh My! An intelligent debate on the internet. Achievement unlocked.

I'm going to follow by saying I saw the thread about the updated Ptuning kit. All I can say is I'm impressed.

OK. On to the fact-straightening exercise, slash debate.



Quite fun.

First. Displacement volume. Both of you addressed this, so I'll address that first.




A drop in volume you say? Remember, all we're doing is lowering the piston crown. That's it. We're not destroking the motor. The swept volume, and therefore displacement remains unchanged. The new void at TDC is also present at BDC. In fact, at BDC, the total cylinder volume actually increases. Yes, it is negligible, but if you are scavenging effectively, your "effective" displacement is actually going up, not down. Again, a negligible amount but the misconception is clear and present.

NEXT. Intercoolers. Let's have a thought experiment.





All true. But here's a question. Yes, increasing pressure increases temperature. The opposite is also true. These are idealized as adiabatic processes. Why, then, is the pressure drop across an intercooler not proportional to the temperature drop? How can an "idealized" intercooler be isobaric? Given that pressure and temperature are typically proportional? It's a volume thing. We're putting the V in PV=NRT. An isobaric reduction in temperature leaves a smaller volume with a uniform pressure. In an ideal, lossless, frictionless, free-flowing system, the pressure is uniform throughout.

Next up, thermal efficiency and work! Expansion ratios! This covers the rest of the questions and answers.

Aaaaand I'm leaving it until tomorrow. I'm too tired to type the rest of this. It's been barrels of fun. Stay tuned for another exciting episode. I want to talk about PV diagrams, and Q. No, Otto and Carnot's Q, not Ian Fleming's funny Englishman with the cool toys.

Man, you're going to make me pull out my text books from college lol.

Yes, you're absolutely correct about volume in the chamber. It would indeed be an increase in volume. Once again though, very small to the point I really wouldn't consider it for anything worth thinking about.

The pressure drop you're referring to in the intercooler is the loss due to friction (a rather large increase in surface area), turbulence, etc. which is of course only a part of the total pressure drop from all the curves, bends, transitions, etc. in your charge air system. I'm sure I have the complex fluid dynamics equations somewhere (I actually had a MathCAD file written to calculate this). If this were a closed system, a temperature change would cause a change in pressure but we're talking about a system that is allowed to change. And don't forget, time is a very important factor in the consideration of what is considered isobaric. The heat rejection in the intercooler is a relatively slow process (when compared to a turbo or the internals of an engine) with ample time for the system to equalize any pressure transitions (Vdot does not change, but vdot does). The turbo however is a very rapid addition of kinetic energy with a very rapid deceleration as it approaches stagnation exiting the outlet.