Help Me With Some Turbo Related Maths

[diamondjo]

Hey guys

I've been trying to figure out what the equivalent displacement of a turbocharged engine from a naturally aspirated engine of the same type would be.

Now I know we're not talking apples for apples here (turbo lag, air/fuel temperature, frictional losses, etc) but some sort of rule of thumb would be good.

Here's the way I figure it.

If you have container full of gas at a pressure of 1 atmosphere (14.7 psi), that means that you have twice the number of molecules in your container than in the equivalent area in the open air.

My car runs a stock RB25DET (2.5 Litre) which gets 7psi (0.47 atmospheres) at full boost.

So 2.5L x 1.47atm = 3.67 Litres on full boost (This is in a perfect world where we don't have any pesky laws of thermodynamics)

This would seem to follow because the stock RB25DE produces 140kw at the crank, whereas a RB25DET produces 184kw.

So (184kw / 140kw) x 2.5L = 3.29 Litres

While not the same result, we're in the same ballpark.

If we assume the only difference between the RB25DET and the RB25DE is that one has a hairdryer, maybe we can account for the shortfall by calling it losses due to thermodynamics. So maybe we can come up with a thermodynamic loss coefficient by dividing what we think the equivalent displacement should be, versus what it is in the real world:

3.29L / 3.67L = 0.89

Here's what I think it all means:

Displacement (in litres) x Boost (in atmospheres) x thermodynamic loss coefficient = equivalent naturally aspirated displacement (when on full boost)

eg: 2.5L x 1.47atm x 0.89 = 3.27L

So if I wanted the equivalent of a 5 Litre donk and I didn't want to do anything else to the car but adjusting boost and I didn't mind crapping my turbine through my exhaust pipe, I'd need to be running 33psi.

What do you guys think?

[sh@un]

One of the major flaws in your maths is the assumption that an n/a production engine like an old 5 litre Holden or Ford engine will fill each cylinder with the equivalent amount of air as the per cylinder displacement on each induction stroke. It doesn't. To do that, it would have to have 100% volumetric efficiency at the revs in question.

The pumping losses through the filter, inlet tract, ports, and past the valves will bring it down to around 80-90%(?) [don't quote me], which is why you have vacuum in the intake on atmo road cars. Race car V8s with short velocity stacks/ram tubes and multi-throttles combined with big ports/valves and huge duration cam shafts can get up to 110% V.E or more, but that's hardly relevant to road use engines (shocking economy, undriveable below 3000 or more rpm, terrible emissions).

Also, and most importantly, the whole point of a turbocharger is to harness heat energy from the engine that would have otherwise been wasted, and to convert that back into extra charge density for the intake air, thereby increasing the efficiency of the system again. Of course you are right in saying that adding an extra atmosphere of pressure won't result in twice the charge density (due to the density reduction when heat is added to the air while being compressed), but with good intercooling, you will see at least 90%, or even 95% in the best examples, of the charge density retained.

And the final factor is the ability of a turbocharged engine (esp. one with a larger than standard turbo) to hold on to more torque at high revs (due to much, much better volumetric efficiency), and make more power that way.

As you would know- There are plenty of Skylines that make heaps more power on pump fuel than 5L Commodore OHV motors, with nowhere near 33psi.

[diamondjo]

An awesome response, thanks for that!

My little quip about 33psi = 5L was sort of a joke, but it was really interesting to hear your reasons why this wouldn't be the case.

Am I close though in saying that for the same engine, with the same setup on the same car - with the only variable being the aspiration, that my RB25DET is almost like a naturally aspirated 3.3Litre RB?

[sh@un]

Using your thermo co-efficient as the multiple (.89), it's probably closer to a 3.8L n/a motor (assuming an 85% V.E for that engine- could be less or more though, of course).

There are just too many variables- look at an F430: 4.3 litre, 490hp, and meets Euro IV emission standards. You could have won an F1 GP with that less than 40 years ago.

[joeyjoejoejuniorshabadoo]

[JET-25L]

geez way too many numbers and too much maths reminds me of maths b in school. Good to see someone knows what they're talking about though!

[sh@un]

Thought of something else last night too

The RB engine is a more efficient design to start with, so you have to also take into account

1) The bore is usually smaller than a big V8- Therefore having less blow-by (more combustion pressure retained)

2) The stroke is usually shorter- Less frictional losses due to piston side-loading and better angular dwell characteristics (improves combustion efficiency)

and

3) The OHC multi-valve set-up of the engine naturally adds efficiency in both 4V air flow ability (not so important in a forced induction environment, but still contributes), and less frictional and inertial losses through a superior valvetrain design

I'm done

[mad082]

in a theoretical wonderland every atmosphere of pressure you put in (using the 1 sized turbo) you double your horsepower. however in the real world this isn't true.

also, comparing the rb25de to the det doesn't take into account the compression difference. you would have to compare the power of a rb5de to that of a rb25de+t.

and then you could only really compare it to a 5.0L rb motor (if you were to scale up a rb25 so it is 5.0L). to do it any other way is silly. just look at the bmw m3 motors. they put out about 236kw stock and they are a natro 3.2L. if you scale that down, that is about 184kw for 2.5L, so the rb25det is equal to a natro 2.5L.

[Beer Baron]

also what if you change turbochargers? a small turbo at 1 bar may give you 180kw. a larger turbo still producing only 1 bar may give you 250kw... there is no accounting for this in your equation. this is because you are measuring forced induction using a measure of the resistance in the inlet tract with no accounting for the actual volume of air being pumped in.

[mad082]

exactly.

and you have to take into account differences in bore vs stroke. most v8's are made to be long stroke motors to give more torque, not high revving high hp motors.

[djr81]

Want to know something scary?

The output of an engine is very closely linked to the airflow into it - commonly measured in lbs/min.

In fact in terms of hp produced per lb/min airflow there is very little difference between a new motor & an old motor, a 2V head or a 4v head, short stroke or long.

So rather than try & align it all to discplacement it is much easier to align it to airflow....

[sh@un]

that's the easiest way to summarize it really-

the more charge density you can cram into the cylinders within a given period of time, the more fuel you can add, the more torque you will have, and the more power you can make.

within reason of mechanical and chemical limits, of course

[Pal]

my head hurts

[sh@un]

*hands Eric a panadol

[diamondjo]

Haha, I'm with Pal, my brain hurts!

Thanks for all that guys it's been really interesting.

I'm going to do some reading on volumetric efficiency now

For those of you playing at home, here's a brief definition:

http://en.wikipedia.org/wiki/Volumetric_efficiency

[bombtrack]

Is it just coincidence that almost all the avatars in this thread are Simpsons characters(or at least Matt Groenig-esque toons)?

Great thread though, very interesting

[sh@un]

make your own here!

www.simpsonsmovie.com

[PM-R33]

Is it just coincidence that almost all the avatars in this thread are Simpsons characters(or at least Matt Groenig-esque toons)?

Great thread though, very interesting

Hahaha thats what i was thinking aswell.

But yeah i agree, great thread!

[sxc33]

Ummm. 7.

Interesting read lol

[diamondjo]

Okay, here's another question. Why does the standard gauge under the tacho measure from -700 mm Hg to +700mm Hg? When you're just cruising, obviously it's sitting down at -7. Presumably the turbo isn't going backwards and creating a vacuum. So what, if anything, does a negative pressure reading mean?