Prodigy Performance 3.6 Turbo DIY Install

[UselessPickles]

One last attempt to explain from a different direction why CFM capability of the supercharger/turbo is completely irrelevant to my analysis of RIPP's claim of 40% gains at 1800 rpm. Please actually read through this and put some effort into understanding it before responding.


CFM = Cubic Feet Per Minute

That's a rate of flow of a volume of air per unit time. A volumetric flow rate. That's it. Volume/time. Pressure, mass of air, or number of molecules is completely independent of flow rate.

What is the flow rate of the Pentastar engine at 1800 rpm? For simplicity, we will ignore extra volume in the ceiling of the combustion chamber, valve overlap, etc. Let's just go by the displacement.

3.6 L total displacement, but each cylinder requires 2 rotations of the engine to intake and exhaust 1 cylinder full of air.

So (3.6 L) * (1800 rpm) / 2 = 3240 L/minute

Again... this is just volume of air. Pressure, temperature, etc., are all unspecified. It does not specify the AMOUNT of air flowing through the engine. Just the VOLUME. The AMOUNT of air depends on the volume, pressure and temperature.

So what's the volumetric flow rate of the engine at 1800 rpm with 0.3 psi boost? At 0.6 psi boost? At 37 psi boost? The answer is always 3240 L/minute. The difference is the pressure and temperature of the air, which results in a different AMOUNT of air, but the same VOLUME.

Even if there's some inefficiency with cylinder filling (intake valve isn't open long enough to equalize pressure between cylinder and manifold, for example), the end result will still be the same VOLUME of air, but at a slightly lower pressure than manifold pressure (the ECU has lookup tables of volumetric efficiency multipliers to deal with this).

What if you have a supercharger capable of flowing up to 98,000 L/min at 58 psi, and it's producing 23 psi at 1800 rpm? What's the volumetric flow rate through the engine now? Still 3240 L/minute! The supercharger does not change the volume of the cylinders. And it does not change the engine speed. Therefore it cannot possibly change the volumetric flow rate of air through the engine. It only compresses air, changing its pressure and temperature.

Therefore, two different forced induction systems producing the same boost pressures and intake air temperatures on the same engine at the same engine speed cannot possibly be operating at different volumetric flow rates (CFM, or whatever units you care to use), no matter what their peak boost or peak CFM ratings are.

[KaiserBill]

One last attempt to explain from a different direction why CFM capability of the supercharger/turbo is completely irrelevant to my analysis of RIPP's claim of 40% gains at 1800 rpm. Please actually read through this and put some effort into understanding it before responding.


CFM = Cubic Feet Per Minute

That's a rate of flow of a volume of air per unit time. A volumetric flow rate. That's it. Volume/time. Pressure, mass of air, or number of molecules is completely independent of flow rate.

What is the flow rate of the Pentastar engine at 1800 rpm? For simplicity, we will ignore extra volume in the ceiling of the combustion chamber, valve overlap, etc. Let's just go by the displacement.

3.6 L total displacement, but each cylinder requires 2 rotations of the engine to intake and exhaust 1 cylinder full of air.

So (3.6 L) * (1800 rpm) / 2 = 3240 L/minute

Again... this is just volume of air. Pressure, temperature, etc., are all unspecified. It does not specify the AMOUNT of air flowing through the engine. Just the VOLUME. The AMOUNT of air depends on the volume, pressure and temperature.

So what's the volumetric flow rate of the engine at 1800 rpm with 0.3 psi boost? At 0.6 psi boost? At 37 psi boost? The answer is always 3240 L/minute. The difference is the pressure and temperature of the air, which results in a different AMOUNT of air, but the same VOLUME.

Even if there's some inefficiency with cylinder filling (intake valve isn't open long enough to equalize pressure between cylinder and manifold, for example), the end result will still be the same VOLUME of air, but at a slightly lower pressure than manifold pressure (the ECU has lookup tables of volumetric efficiency multipliers to deal with this).

What if you have a supercharger capable of flowing up to 98,000 L/min at 58 psi, and it's producing 23 psi at 1800 rpm? What's the volumetric flow rate through the engine now? Still 3240 L/minute! The supercharger does not change the volume of the cylinders. And it does not change the engine speed. Therefore it cannot possibly change the volumetric flow rate of air through the engine. It only compresses air, changing its pressure and temperature.

Therefore, two different forced induction systems producing the same boost pressures and intake air temperatures on the same engine at the same engine speed cannot possibly be operating at different volumetric flow rates (CFM, or whatever units you care to use), no matter what their peak boost or peak CFM ratings are.

Whatever, you want to believe is fine with me. However, your first formula is wrong-- your volume at 1800rpm is always 3.6L at 5400 rpm the engine still is 3.6L at 150,000rpm still 3.6L! Now CFM is also directly related to the more important number--- lbs/Ft3 measuring air-density. That formula looks like this take CFM x .076Lbs/ft3 and poof that is the magic number. I tend to use CFM because it, well, works just as long as you remember to change it over to Lbs/ft3 and then remember that is per minute. Thusly, even a compressor running at .6 PSI operating at X Lbs/ft3 will create less power than a similar unit doing .6PSI with X+2 Lbs/ft3.

Hence my Vortech examples-- at any given Manifold pressure the V-2E has the potential to produce more horsepower than its similar brother the V-2SCI unit. That little 100CFM difference if it is constant across the entire units performance band produces an extra 7.6Lbs/Ft3 of air mass per minute!!! which if you say rough equates to about 30-50hp at any rpm range for any given boost. If that is a constant 100CFM increase. Now it might not be, but it could be. Which would make your analysis totally meaningless if you didn't take that extra air-mass into account for each given boost level.

So here is the general rule of thumb!!!! And it isn't supper accurate, but if you are making 125 HP at flywheel/ by 10.5 = 12.5hp per Lb/min. So if you are at 1800 rpm and you are pulling in an extra 2.2lbs/min of air (which can then be converted over to CFM, LFM, KG/CM3 or ML/CM or whatever you like) you roughly get a 27.5 hp which is pretty damn close to a 40% increase in power if go from 75HP at 1800 rpm to about 102.5HP: So what it is about a 37 percent increase in power? And it only required an addition 2.2lbs/min of air mass to reach the cylinders.

Until you know exactly how much air/mass is reaching the cylinders for a given boost pressure will not tell you anything of importance. Because guess what moving more air at any pressure means the potential for more air-molecules to be found in that specific volume of gas. Damn that air density.

Also on a side not I think you are misunderstanding Pressure Ratio on the Turbo chart. It doesn't measure the ratio of boost produced for a given volume of exhaust- it measures the ratio of compressione from absolute pressure at the inlet of the turbo to the outlet of the turbo. So lets say you have a 1.25:1 Pressure Ratio at 55,000rpm (impeller speed) means that you out pressure is 1.25 times that of the absolute pressure reading of the inlet side. Most turbos never get much more than 3.0 to 3.5 to 1 ratios. But that is pretty damn impressive. That's why staging turbos can get really insane pressures rapidly.

[UselessPickles]

You have completely failed to comprehend nearly every explanation and formula I have provided. Your responses completely misrepresent what I have said, or state irrelevant information/facts/etc.

For example:

your first formula is wrong-- your volume at 1800rpm is always 3.6L at 5400 rpm the engine still is 3.6L at 150,000rpm still 3.6L!

I have no idea what point you're trying to make here. Your statement does not at all back up your claim that my formula is wrong. Your statement is not in disagreement with my formula. I challenge you to prove that my formula for estimating volumetric flow rate of the engine is incorrect... with actual explanation of the physical processes and use of complete formulas as I have.

Now CFM is also directly related to the more important number--- lbs/Ft3 measuring air-density. That formula looks like this take CFM x .076Lbs/ft3 and poof that is the magic number. I tend to use CFM because it, well, works just as long as you remember to change it over to Lbs/ft3 and then remember that is per minute. Thusly, even a compressor running at .6 PSI operating at X Lbs/ft3 will create less power than a similar unit doing .6PSI with X+2 Lbs/ft3.

Again, a clear indication that you do not actually understand the relationship between pressure, volume, mass, and density of a gas. Two different devices both feeding air into the same engine at the same engine speed with the same amount of pressure in the same atmospheric conditions and same temperature cannot possibly be providing air at different densities (Lbs/ft3) it is absolutely physically impossible.

Density (Lbs/ft3) is directly proportional to pressure. All else equal (same engine, same rpm, same atmospheric conditions, same intake temp, etc). The only way one device could provide denser air is to provide air at a higher pressure. If both produce the same pressure, then they are both providing air with the same density and same flow rate.

Also on a side not I think you are misunderstanding Pressure Ratio on the Turbo chart. It doesn't measure the ratio of boost produced for a given volume of exhaust

You're making a HUGE assumption that I looked at (and misinterpreted) a turbo compressor map to determine the exhaust:boost pressure ratio. I never said anything about the compressor map. Where did you come up with this assumption?

If you must know, I used my knowledge of how the wastegate works combined with the rated boost level of the wastegate spring used in the kit (rated based on a 1:1 exhaust:boost pressure ratio), compared to the actual amount of boost that the spring allows in this turbo kit.



Anyway, it is very clear that you do not have a solid foundation of the physics involved to properly discuss this subject. Please stop. Seriously, this is going no where. Go take a physics class. Learn the Ideal Gas Law, and truly understand it. When you come back and re-read your posts, you will see how completely incorrect and/or irrelevant some of your arguments are. I truly do not intend this as an insult. I'm encouraging you to learn to understand things in a formal/logical/mathematical way that will allow you to more completely/formally express your ideas in terms of math.

I'm also not pulling a "I'm right, you're wrong, no matter what you say". I like to be proven wrong, because I like learning new things. You are, however, failing to provide any complete logical explanations to back up any of your claims. You have made several technically correct statements (by themselves), but have failed to bring them together with any consistency to prove/disprove any particular relationship. You say my formulas are wrong. Please, explain what's wrong with them. Show me the correct formulas. " take CFM x .076Lbs/ft3 and poof that is the magic number" does not count. Start with complete formulas (like the ideal gas law, formula for volumetric flow rate, formula for mass flow rate, etc) and combine/rearrange them to mathematically prove a relationship you are trying to communicate.

[gbaumann]

Two different devices both feeding air into the same engine at the same engine speed with the same amount of pressure in the same atmospheric conditions and same temperature cannot possibly be providing air at different densities (Lbs/ft3) it is absolutely physically impossible. .

So then regarding the dyno results you're trying to reconcile it's looking like an erroneous claim or a different calibration. Question, does Prodigy use Diablosport to generate their calibration files or are they working independently with the CMR software? I wonder how different each mfg's calibration file can be. You'd think that Prodigy, Magnuson, RIPP have all gotten their hands on the others' calibration files and are looking at them.

[Timmy]

Density (Lbs/ft3) is directly proportional to pressure. All else equal (same engine, same rpm, same atmospheric conditions, same intake temp, etc). The only way one device could provide denser air is to provide air at a higher pressure. If both produce the same pressure, then they are both providing air with the same density and same flow rate.

Good explanation Pickles! I kept scratching my head with what was being written in the other posts regarding CFM... As an example... The idea that you could push 1000 CFM @ 1 PSI vs. 1500 CFM @ 1 PSI through the same pipe, what the!?! Did someone not play with a garden hose as a kid enough to know that to obtain an increase in CFM at the same pressure would require an increase in the diameter of the garden hose!!!

For those who didn't get to play with a garden hose as a kid, here's the real formula.

cfm = area of pipe * sqrt (2*Pressure/density)

The idea that one turbo could deliver higher CFM at the same pressure, on the same engine, at the same RPM versus another turbo is just silly. That goes against the laws of physics (as previously stated by said poster named Pickles.)

[UselessPickles]

What do you say to letting me take the post and start a new thread. I only hesitate because you specifically state "not worth its own thread".

If anyone could chart it out pickles, its you. What do you say? Everyone loooooves to O.D. on charts and graphs.

I just didn't feel like committing to fleshing out the beginnings of a complete FI comparison post at the time. Next time I'm looking to kill some time in the evening, I'll start a new thread with the intent to continue to update the first post as a basic comparison guide.

Maybe this is a dumb question, but if RIPP and Mag indicate they are producing 8 PSI of boost with std pulley, and 11 PSI of boost with High Elevation pulley, why does your chart (or their data) only show Magnuson with a max of 6 PSI?

I took the boost data straight from Magnuson's own dyno chart (link in my post above was incorrect, but is now fixed). This is a chart that Ross posted on the wrangler forum, identifying it as a dyno chart from Magnuson:




If you know of any more recent data about that supersedes that chart, please point me to it so I can update my data.

BTW, this is from Magnuson's web site:

In testing, this kit demonstrated between an 80 - 100 HP and 60-70 lb-ft increase at the wheels at just 6 psi.

[UselessPickles]

to obtain an increase in CFM at the same pressure would require an increase in the diameter of the garden hose!!!

Since water is incompressible (in practicality, at least), but air IS compressible, the garden hose is a bit oversimplified.

With water, there IS a direct relationship between volumetric flow rate (e.g., CFM) and mass flow rate (e.g., lbs/ft^3).

With air, the relationship between volumetric flow rate and mass flow rate depends on density, which depends on temperature, pressure and volume.

To make things more confusing, in the world of engine dynamics, they speak of the volumetric efficiency of the engine. It is often explained as the percentage of "volume" of the cylinder that gets filled with air. As I discussed before, the volume of the displacement of the cylinder never changes. So volumetric efficiency is a bit of a misnomer (but not entirely incorrect, either, as I'll explain later). It's really a ratio of the AMOUNT (mass) of air that gets into the cylinder compared to the AMOUNT of air that *would* fill the cylinder *if* it were allowed to reach full equilibrium with the intake manifold.

At 50% volumetric efficiency (number chosen arbitrarily to make my later example easier to comprehend), it doesn't mean that the cylinder is only filled 50% of the of the way with air. That's impossible, because whatever AMOUNT of air enters the cylinder will expand to fill the entire volume.

But if you think of it from the perspective of the intake manifold, the term "volumetric efficiency" actually makes sense: 50% efficiency means that 50% of the cylinder's volume worth of air in the manifold has filled the cylinder. For the Pentastar, a cylinder is 3.6L / 6 = 0.6L. Let's say the manifold pressure is 10 psi (easy number to work with). At 50% volumetric efficiency, that means 50% of 0.6L = 0.3L worth of the manifold's 10 psi air enters the cylinder. That AMOUNT of air will fill the entire 0.6L in the cylinder (when the piston is at the bottom of the stroke). Volume are inversely proportional. That means when that 0.3L of the 10 psi air expands by double to fill the 0.6L cylinder, the pressure will be cut in half to 5 psi* (see disclaimer below). The air in the cylinder now has 50% the pressure of the manifold, AND 50% the amount (mass) of air that would have filled the cylinder if it were allowed to completely "fill" with air from the manifold.

So because of the relationships between volume, pressure, density and mass of air, it can really be thought of as either volumetric efficiency (from the perspective of the manifold) OR mass filling efficiency. It all works out mathematically the same, and will produce the same result in the calculation to determine the amount of air in the cylinder.

*Disclaimer: I simplified a bit for the explanation. When that 0.3L of 10 psi air expands into the 0.6L cylinder, the pressure may not become exactly half, because I believe the temperature of the expanding gas will decrease as well. This gets into some complications that I don't understand. So some of my previous statements in earlier posts may be incorrect about the relationship of how much pressure gets into the cylinder, but the concepts would be correct if you worked in terms of mass rather than pressure.

[KaiserBill]

Good explanation Pickles! I kept scratching my head with what was being written in the other posts regarding CFM... As an example... The idea that you could push 1000 CFM @ 1 PSI vs. 1500 CFM @ 1 PSI through the same pipe, what the!?! Did someone not play with a garden hose as a kid enough to know that to obtain an increase in CFM at the same pressure would require an increase in the diameter of the garden hose!!!

For those who didn't get to play with a garden hose as a kid, here's the real formula.

cfm = area of pipe * sqrt (2*Pressure/density)

The idea that one turbo could deliver higher CFM at the same pressure, on the same engine, at the same RPM versus another turbo is just silly. That goes against the laws of physics (as previously stated by said poster named Pickles.)

No, genius air expands when it enters the cylinder head!!!! This why Roots Lobe Units are called Blowers-- because they develop very very low pressure all the time! So, since no engine runs at 100% volume metric efficiency throughout its entire rpm range-- you can squeeze in more air into the cylinders just by increasing the volume metric flow of the air reaching engine. Remember as the air expands to fill in the open space it will slightly cool. This is where Roots Lobe units really have problems because of the way they create pressure and volume they actually tend to heat the air up more at low rpm then they do at high rpm. But when it comes to volume metric efficiency across a broad RPM range the RL system is hard to beat. Now, with a more air friendly system like the Centrifugal unit Vortech uses the air at low pressures is relatively cold and dense-- the the more you add the greater air density in the cylinder head (up to a point). The point is this at .6 PSI there is a good bet that the Vortech unit can move enough air into the cylinder head to achieve a noticeable increase in the volume metric efficiency of the engine without requiring the engine to need extra manifold pressure to compress more air into the cylinder head. So basically, you might go form say 91% volume metric efficiency to 95% efficiency thus allowing you to burn more fuel creating a mere 30-40hp extra. Physics is just fine... All the fun laws of Boyle's, Charles', and the kinetic theory of heat are intact and happily loving each other.

What, I said was Manifold Pressure doesn't necessarily have a linear relationship between CFM and Boost Pressure! You can very well have two different turbos running at the same engine speed, on similar engines developing two radically different CFM outputs. Why do you think they make so many different units that fit essentially the same size displacement engines??? Is it because they just want to sell people two different turbos for no reason at all? No, it is because different sorts of performance goals require different types of units and peak CFM outputs for specific manifold pressures.

Then I proved to you people with a link to two Vortech V-2 mode Units that outwardly look identical with only minor differences in specifications that have 100CFM difference in peak output. And this translates to 50HP extra at peak performance all while maintaining a maximum boost of 17PSI. So, yeah, would like to try to stating again how wrong I am.

http://www.vortechsuperchargers.com/page.php?id=30157

http://www.vortechsuperchargers.com/page.php?id=30008



My entire point since day one is that if you look a boost graph and decide that some company is misrepresenting their claims on this data alone-- you will probably look foolish. There are just so many more factors that come into play with adding forced induction to an engine. If you don't know how the engines are setup and what goals each kit has-- then it is hard to say that one claim is preposterous because I've got Prodigy's turbo kit and it cannot do it. It might not be designed to do it. And that is just bummer for you. Personally, I think they put out boost graphs because they know most people buying kits aren't well enough versed in forced induction theory to really know what they are looking at.. So yeah you see 10PSI or 11PSI or 22PSI and think wow that is going to be so awesome in my car, truck, or jeep. And not knowing what exactly your engine is doing to start with gives you almost no way of knowing how effective that kit will be on your engine in the long run.

[UselessPickles]

I believe in physics

[UselessPickles]

Yes... my assertion is that either the claim is exaggerated, or RIPP just has a much better calibration at low rpms than Prodigy, or a mix of both. The difference in gains is too much to be explained by differences in efficiency of the RIPP vs Prodigy systems themselves.

Prodigy does their own tuning with the CMR software.

I do remember talking to either RIPP or Prodigy at one point, and they said they worked closely with Diablosport to develop some proprietary tuning capabilities in the CMR software so that they could properly account for boost in the calibrations. All the FI manufacturers work closely with Diablosport for CMR support. I don't know if there's some special calibration capabilities that are available to some manufacturers and not others, so it's tough to know whether it would even be meaningful to compare calibrations between different manufacturers.

I'm curious about whether the CMR software gives them control over the variable valve timing. I was just reading about how ideal cam timing is different at low rpm for a turbo, as compared to NA, to help the turbo spool more quickly.

Anyway, Prodigy is working on an improved tune. I know that low RPMs is an area they will put some effort into, because that have received feedback from multiple people about low RPMs feeling sluggish. It's possible that low RPM just was not a priority for Prodigy previously, so the hadn't yet put effort into optimizing it.