Viewing a turbocharger as a variable displacement device - THEORY/DISCUSSION

It's no secret that an engine, any engine, is basically just a glorified air pump. In naturally aspirated form, a four stroke 2 liter piston engine moves approximately 2 liters of air through it with every two revolutions of the crankshaft.

There are outside influences on this, such as volumetric efficiency, temperature, restrictions in either the intake or exhaust, ambient air pressure, and probably a couple others that when factored in, all limit the actual airflow, albeit only slightly. Because different engines inherently have different restrictions, to me it seems more appropriate to designate engines based off of a measured maximum airflow, rather than displacement.

This is where the turbo comes into play. As boost pressure increases, the airflow requirement does also. Theoretically speaking, a heavily boosted 4G63 can have the same maximum airflow requirement as the NA 8.3 liter Viper V-10, or even more.

The reason I bring all of this up is simple. After any Mustang or Camaro or insert car name here owner has been on the receiving end of the Mitsubishi Pimp Hand, the response I get most often is "No fair!, you have turbo!".

My answer has been that "I'm only running 14 lbs. which is only twice the ambient air pressure, and your 350 is nearly three times larger than my engine, so even with the boost, you still have more engine than me." Usually this confuses them enough that they just drive away.

The problem that I'm having is that I know it isn't that simple. The math is much more complicated I think. I know that power output is increased as the cylinder pressures increase. I think this could one of those exponentially decaying functions (thank you Barnes), and if that is the case, there isn't any real purpose to classifying engines based off of an airflow requirement, because two completely dissimilar engines can exist with the same airflow requirement, and make vast differences in power output.

Please discuss.

 

[Barnes]

I think where where things get complicated is that the air volume relationship is not 1 to 1 with respect to power. A 4G63 boosted with 14.7lbs of air may(speculation) have the same volume of air going into it as a 4 liter motor, but it won't have the same dynamic compression ratio, and thus not the same power. This is where thermodynamics comes in. Compression ratio strong affects how much energy is produced in the Otto cycle . The higher the compression ratio, the more efficient the cycle and the more work it does. This is why diesel engines are more efficient than gas engines, they run much higher dynamic compression ratios.

EDIT: Also, the turbocharger acts something like an energy recycler. Energy that would otherwise get dumped out the exhaust is put into the turbine to provide work to the compressor. So that is also another x-factor.

 

[cheekychimp]

But surely your last point only indicates that a turbo engine might be more efficient in recycling that energy. Isn't the reason we have a wastegate, that if that recycled energy were allowed to be continuously fed back into the engine supplementing the power the engine made on each cycle, pressure (and power) would just continue to rise until the engine exploded?

Another issue that I believe 'might' need to be factored in (although I am not sure how, since as you point out the math gets extremely complicated) is that a great deal of fuel injected into a turbo engine is for 'cooling'. The effects may be marginal but I would think your cubic capacity is going to be effected by how much unburnt fuel is taking up space in the combustion chamber and being pumped out the exhaust and/or burnt in the catalytic converter.

 

[Barnes]

My point is that the turbocharged engine recycles some of that energy. A normally aspirated car losses all of that energy. And yes, turbo chargers limit the speed of the turbo charger, therefore we are only 'recycling' a portion of the energy. The point being, this is a factor in comparing N/A cars vs. turbo cars.

 

[Muskrat]

This is just a quick post, so bear with me. Some of this would be a lot easier to explain if I made diagrams:

Everything else being equal (VE, afr, compression, etc..) Power is a function of air flow. The more air the engine pushes through it, the more power it makes. There are two ways to do this: increase displacement (bigger cylinders, or more cylinders, or both), or forced induction.

You might have seen this, because I post it everywhere, but it's a really good base level explanation of turbocharged engines and how they produce power:

Turbo Systems 101, 102, 103

Now lets look at the simplified Otto cycle itself for a 4-stroke engine (which, just to clarify, isn't used for diesel engines):

- We start at TDC with low cylinder volume (V) and low cylinder pressure (p). We'll call this point 1. [it might help to understand this is you mark up the diagram of the simplified otto cycle in the wiki article]

- We then increase V, and p remains the same during the induction stroke of the engine, until we're at BDC. This is point 2. [green line]

- We then compress the contents of the cylinder until we're back at TDC. V decreases and P increases. This is where compression ratio comes into play. This is point 3. [orange line]

- We then induce combustion with a spark. p increases from this, V stays the same. This is point 4. [vertical red line]

- As the cylinder is forced down to BDC, p decreases and V increases. This is the power stroke. Point 5. [rest of the red line]

- The exhaust valves open, letting off any remaining pressure. p decreases to atmospheric pressure V stays. Point 6. Then the piston moves back up to TDC, exhausting the exhaust gasses. Point 7.

If you look at the graph you can see two clear area's enclose by the curve. One by the intake and exhaust potions, and the other by compression, combustion, and power. Each area represents 1 cycle of the piston. a 4 stroke engine has 2 piston cycles per combustion event. The net work produced is the area of the larger curve. The area of the smaller curve is called pumping loss. It's the work required to pump the air in and out of the engine.

Now, lets look at the effect of a turbo on the Otto cycle. There are three:

1) The first is that the intake stroke happens at higher than atmospheric pressure. What this means is that the "pumping loss" area is now a net GAIN in work, instead of a net LOSS.

2) It shifts the entire graph up, pretty much, because everything is now starting at much higher pressure. Only the blue line will remain in the same place, pretty much.

3) It allows the increase in pressure from combustion to be greater, because more air is in the cylinder, therefore more gas can be burned, which releases more energy. This means the vertical red line is much longer, which increases the big area of the curve, our net work output.

How increased compression ration affects the graph is by elongating the compression line. Point 3 is moved further up the y-axis. It can also help to get a more complete burn, which will increase the combustion line.

How this relates to your original question:

If you had two engines with the exact same characteristics, but one was a 4 cylinder turbo, and one was an 8 cylinder. The 8 cyl would have double the power of the 4 cyl (if NA). It's basically doing this cycle 8 times instead of just 4.

So, in order for your 4 cyl turbo to have the same power as the 8, you need the area's enclosed by the turbo curves to be double the area of the NA curve.

Assuming WOT. For an NA engine the curve is constant. For a turbo car it varies with boost pressure. This is why a turbo car's power is more peaky than an NA.

Keep in mind Power = Work/Time. So As RPM increases, and this cycle is completed more often in less time your power goes up.

I hope that sheds some light, and wasn't to much verbal (textage?) diarrhea.

I might go back and edit this for clarity later, but I think you can get idea gist of it.

 

[beaner]

There's so many variables involved it's insane.

I just think of boost as resistance to flow. The psi # is a completely arbitrary figure. It's all about the amount of air in and out, not the psi. If you told me your car has 14.7psi and half the displacement of mine, I can assume your car is piss slow, equal, or something I really really want to go for a ride in. Like 14.7psi on what snail? A t25 or 42r? That's just looking at turbo size.

Understand that 16g @ 20psi on a modified 4g63 will not feel the same as a 16g @ 20psi on a bone stock 4g63. Imagine taking that stock setup @ 20psi, then porting out everything, a better exhaust, better ic & piping, upping the compression, redesigning the manifolds, etc etc etc. The displacement and turbo have stayed the same, but the air flows so much better making it feel just as fast as before but maybe at only 15psi. The real gain of the mods will be felt when the boost is put back to 20psi as before.

That's the way it works out in my head. I'm sure there's someone on here that has forgotten more than I'll ever know and can explain it way better.

 

[AnotherNewb]

To classify an engine in air flow makes no sense. If the final purpose of the engine were to displace a certain amount of air then it would make sense to measure its size in cfm, lb hr, or whatever. However, an engines purpose is to produce torque. It would make more sense to rate an automobile engine in HP or TQ like most outboard marine engines therby giving a clear referance of that perticular engines ability.

 

[curtis]

It does make sense if you think area under the curve into the evaluation.

Given a stack of heads, one stock, one with the best port job theoretically possible and the others ported with different angles on the valves, margin areas, bowls blended etc etc . Lets say all heads have the same cams.

If you flow bench each head at .050, .100, .150 etc all the way to max lift then over lay all plotted points on a graph one head maybe flat and the other might be a straight line from 0 to max lift then the theoretically best head is straight up then flat over to max. One thing that has always killed me is flow bench numbers. People that do this are there to make a living not give away port jobs so the competition is always there and they advertise there best numbers.

Company A might say there head flows 375 at .600 and another might say .400 at .550. Most times the numbers advertised don't fricking matter. Its the max your cam opens that you worry about, anything above that is BS. Then import car cams are measured in the metric system 10.5 lft etc which confuses people even more because they want convert the numbers over and take the word of the shop owner.

How it works is the cam opens once and closes once in each cycle but and this is the big but.... its at the other numbers twice in the cycle so the car with the greatest area under the curve for a port job with everything else being equal will win. This is why you see alot of equally modified cars and there's always one out there that is car lengths ahead.


Adding forced induction or spray only adds to this. Its about the amount of air that can get in and get out and the math isn't that complicated. All the guy with the flow bench has to do is change the specific gravity that the head is measured at. Adding , taking away etc. They all know how to cheat the system to make the heads lie. If you ever have something benched make sure your at the industry standard of 28 inches for a valve in a head or I think its 20 for a carb or T body and open hole, also make sure what your comparing to is at standard as well. Only exception is people that don't preach numbers but do a before and after on the head. If the head is at x in dirty used form then picks up to x + 2% when it comes out of the hot tank then when your cutter is done with it its at x + 30% gain. I wouldn't worry to much and car will take care of business. 20 to 30% increase is a hell of alot on a head at any lift.

Now not turning this into a head thread and not going into "at what feet per second does velocity matter" what about change in pressure at the same velocity etc. When stuff goes turbulent on the short turn radius..... We could debate this sh*t for years.


Plus Barnes doesn't let us get into fluid dynamics anymore. /ubbthreads/images/graemlins/rofl.gif

 

[belize1334]

I've only two things to add.

First, I prefer to think of VE as the n/a rated VE and I'd very much like to try to convince you all to do the same. The advantage is that you can think of the engine and turbo as being decoupled save for the small restriction effects of the turbo assembly. Thought of this way, even at 20psi my engine is still only consuming ~2.0L of air per 2 revolutions. How can this be? Simple. The engine sees only the intake and exhaust ports so the volume of air consumed should be measured at the intake port. The VE measures this volume and compares it to the rated displacement. If I then turbocharge it I can increase the density ratio to say ~2:1 which means that those 2 liters comprise nearly twice the mass of air and hence ~ twice the power. People are continually referring to a turbocharged engine consuming x times more volume because of the turbo but that's just not the case. The TURBO consumes this larger volume since it has to supply the mass of air that the engine is consuming. The engine is consuming it's rated VE in volume but at a larger density. Conservation of mass then dictates what the volume consumed at the compressor must be. Alternatively you can think of the boost pressure as being dictated by the disparity in volumes which necessitates a density ratio and this in turn dictates the boost pressure via the adiabatic process. This is exactly the way people think about superchargers which are positive displacement devices and essentially cram 6liters of air into a 4liter engine.

So, our 2 liters consume about 200cfm n/a. If the turbo is giving a density ratio of about 2:1 then the engine is consuming twice the mass that it would at atmospheric pressure. Since the turbo itself is consuming that air at atmospheric it requires twice the volume so that puts it at about 400cfm. This puts us in 14b territory. E3-16g would be more like 2.5:1 density ratio and hence ~500cfm. I highly recommend that people play around on the stealth316.com turbo pressure applet. It's a neat toy.

The second point I'd like to make is just a recent observation. If you calculate the speed of the air entering the compressor housing of a 16g (~49mm diameter at the inlet) you find that it reaches the speed of sound at almost exactly the stated limit of flow for that turbocharger. For the 14b the diameter is more like 44mm and the flow scales accordingly. Now, when the airflow reaches the speed of sound you can no longer propagate a pressure signal so the pressure differential can't pull in any more air. It would seem that the restriction on flow for these turbos is not the compressor efficiency but rather the cavitation of air at the inlet. It might be worth porting the inlet on one of these and seeing if it ups the upper limit on achievable power.

 

Beaner, you've confirmed one of my points exactly. In your example you've taken the stocker 4G63 and compared it to one that has far less restrictions due to modification. The modified engine is more powerful because "the air flows better", and therefore has a higher airflow requirement. If the classification of these two engines was airflow based, then it would be apparent that one would be more powerful even though they both displace 2 liters.

 

Maybe, considering everything above, we could pose a scenario here. One four cylinder, hell, let's make it a bone stock NA 4G63. Now imagine an 8 cylinder, comprising of two bone stock NA 4G63s stacked up like a straight 8. Because we'd be trying to keep all of the outside variables OUT of the equation, the straight 8 version would have to have two intake manifolds, two exhaust manifolds, two throttle bodies, etc, so as to keep resistance comparable to the four cylinder version. This should effectively make the straight 8 require exactly twice the airflow, and make exactly twice the power of the four. Now we'll shove exactly twice the air into the four cylinder. Would the power equally double?

EDIT: All of this, while assuming fuel is there as needed to maintain proper A/F ratios.

 

[belize1334]

^^ I'm guessing no. The 4-banger won't make as much power at a 2:1 density ratio as the 8-cyl at 1:1. The reason is that you can't run as much timing on the boosted engine since the intake charge is more volatile. Also, the exhaust housing creates a mild restriction on the exhaust side of the engine and thus requires more energy to move the exhaust out of the engine. This has the dual effect placing a slight load on the rising piston during the exhaust stroke (robbing torque) and also reducing the amount of exhaust that is evacuated from the cylinder. Evacuating less exhaust means that the engine will draw in less charge on the intake stroke and thus reduce the n/a VE.

Moral is, doubling the density of the air is less efficient than doubling the displacement and thus the power made will be slightly less. As an example, consider that the brake-specific-fuel-consumption (bsfc) for n/a is typically around 0.45 and for turbo more like 0.6. The units are not terribly important but it's something like (lb/hr)/bhp. Now, if you consider that a n/a car is tuned to around 14.5:1 afr and turbo is like 12:1 under boost, you can get go from fuel consumption to air consumption by multiplying bsfc by afr. For n/a we get brake-specific-airflow to be 0.45*14.7=6.6 lb/hr/bhp and for turbo it's 0.6*12=7.2 lb/hr/bhp. The result is that for the same airflow the n/a car is making more power than the turbo to the tune of about (7.2/6.6) = 109%. That is, if you use the "typical" bsfc as a measure you find that n/a cars make about 8% more power than turbo cars for the same cfm rating at the primary inlet.

 

Going back to my original post,

Quote:


The reason I bring all of this up is simple. After any Mustang or Camaro or insert car name here owner has been on the receiving end of the Mitsubishi Pimp Hand, the response I get most often is "No fair!, you have turbo!".

My answer has been that "I'm only running 14 lbs. which is only twice the ambient air pressure, and your 350 is nearly three times larger than my engine, so even with the boost, you still have more engine than me." Usually this confuses them enough that they just drive away.





I suppose the big motor fella's just can't drive then? I know a Mustang or a Camaro isn't heavier than my overweight Galant.

 

[4thStroke]

The newer Mustangs and Camaros are pushing 2 tons. I think the new Mustang did shed a few pounds though from the previous years.

 

Well, truth be told I'm talking more about the late 90's, early 2000's versions. It seems to be fairly difficult for me to find someone with a newer version of either to want to run 'em. I ran into a SRT-8 300C a couple weeks ago, and he was either full on the accelerator or full on the brakes. Well I caught him at a stop light, and considering the low compression numbers I pulled on my last compression test, I was surprised at how well I kept up. 0-80ish and I was only a car or two back. I know that's a big car, but they still run hard. I also found a new Camaro SS that sounded quite a bit healthier than stock with Diego the last time I saw him, and on a rolling 1st gear start through redline 2nd, (yeah, yeah, short race), it was a surprise again how well I kept up. I guess I suspected that the new "muscle" from the last few years would be a lot more impressive.

 

[gtluke]

This is pointless ricer math.
Especially when I show up in my turbo 5.8L
Run what you brung.

 

[mitsuturbo]

I agree with this guy. ^^^^^^^^

 

Quoting gtluke:
This is pointless ricer math.
Especially when I show up in my turbo 5.8L
Run what you brung.



I thought you were over turbo's?

/ubbthreads/images/graemlins/rofl.gif

Yeah, yeah, you're right. It really doesn't matter unless you win. I can only imagine how bad ass your Stang is, closest I can come up with is a friends ProCharged 347 '85 convertible, dyno'd @ 686 RWHP. Talk about a rush...