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Turbo's 101 Class

JohnnyT

New Member
Sticky this bitch.

What turbo you pick for your car has a direct correlation on how much enjoyment you will receive from it. No one wants to wait through 2/3rd of the powerband for a turbo to spool, and on the other side of the spectrum no one wants a turbo that doesn't flow enough power for their goals and needs.
OEM manufactures realize this, and that is why OEM turbo system have turbos that spool within the first 1/3rd of their powerband, because this is where you spend the most of your driving. Not many people, aside from time attack road racers, spend any serious time near redline.

So first, we need to figure out how much airflow you require. There is a rather simple formula to figure this out, and you only need to do it once (for each boost pressure), then you can plot the figure on various compressor maps.

Pressure Ratio

First, complete the following equation to figure out your Pressure Ratio (PR). Pressure Ratio is defined as absolute outlet air pressure divided by absolute inlet air pressure. This formula is the only you will need to keep figuring if you want to change boost levels. There are a few different variables in here why we use the numbers we do, but without getting into the Ideal Gas Law, we can just use these numbers.

14.7 + Boost Pressure/14.7
For example:
14.7+ 15psi = 29.7/14.7 = 2.0 PR (Round to the nearest tenth)

The 14.7 represents ambient air pressure. Of course, at our altitude, we don't have this much air pressure. At 5000ft, we have 12.4psi of ambient air pressure. So, a more accurate way to calculate this would be the following formula:

12.4+ Boost Pressure/12.4
An example:
12.4+ 15psi = 27.4/12.4 = 2.2 PR (Rounded to the nearest tenth)

As we can cleary see, our pressure ratio is 0.2 higher than without accounting for altitude (which is a good deal). This points out why our turbos need to flow MORE air for a given PSI than at sea level, so they are working harder for a same pressure level than a turbo would be at sea level.

Also, if you have a lot of intake piping, such as a cold air intake with a filter and a large air box, you should also deduct another 1 PSIA from your absolute ambient air pressure. In our scenario, this would mean 12.4psia now becomes 11.4psia. This is to account of for restriction of your intake. You don't have to do this (I'm not in this example), but it will give you a more accurate PR.

Just a quick example, I will not use this again:

12.4 + 15psi =27.4/12.4-1= 2.4 PR

Now, we use the newly figured out Pressure Ratio (the Y axis on the compressor map) to figure out how much airflow you need for your desired boost level. To do this, we need some more information. So plug in the following formula:

(Displacement) (RPM) (Volumetric Efficiency) (Pressure Ratio)/5660

Displacement is easy, use your displacement in liters rounded to the nearest tenth. A Mazdaspeed3/6 would obviously be 2.3. For a rotary, we have to convert to piston math, so a 13B is 2.6L, a 12A is 2.4L, 20B is 4L.
RPM is the redline of your motor. Even if your motor doesn't make power all the way to redline, you need to plug in the highest RPM you're going to see, because even if power doesn't increase with RPM, airflow does. If your redline is 6000rpm, but you don't rev past 5500rpm, then plug in 5500rpm.
Volumetric efficiency is simply because no system is 100% efficient. For most systems, 90% is acceptable for Rotaries, 92-95% for 4-valve piston motors. To input this into the formula, enter it as a whole number, so 90% is entered as 90 (not 0.9).
Finally, the pressure ratio (PR) is the number we just figured out, which is boost-dependent, so it changes depending on what boost and RPM you want to run.

So, for example on a 13B-REW:

(2.6L) (8000rpm) (90%VE) (2.2)/5660 = 727.6

The number we just received is our CFM (Cubic Feet per Minute), which is a volume measurement of a gas, in this case, air.
So now we have an air flow requirement of 727.6 CFM. So what? Of course, turbo compressor maps have air flow in LBS per Minute (Lbs/min) on the X axis. So now, we need to complete yet another conversion to get our CFM into lbs/min.

Formula:
(CFM) (0.069) = lbs/min

And our example:
(727.6) (0.069) = 50.2 lbs/min

One quick way we can calculate power is 1 lb/min of airflow is equal to about 10rwhp. So, for 50.2lbs/min of air, this should be 502rwhp. This is just a ballpark, but it should get you close. Rotaries require more airflow per horsepower than piston motors, so for rotaries the forumla is 7.69 rwhp per 1 lb of airflow per minute. So in our example this would be 386rwhp. This is why turbo sizing for rotaries is so different, and why rotaries seemingly run larger turbos for less power. If you rotary is streetported, add 10% for the horsepower calculator. Bridgeports can add 16-18%.

To take this method one step further and to compensate for our altitude, we loose about 10% power up here. For the 10rwhp per 1/lbs of air in piston math and 7.69rwhp per 1lb of air for rotaries doesn't hold true up here, it's 10% LESS. So 1lb of air is 9rwhp for pistons and 1lbs of air is 6.92rwhp for rotaries. Or just figure your horsepower number using normal math and then subtract 10%.

And now these are the numbers we use to plot on our compressor map. The vertical axis/y axis is the Pressure Ratio, which in our example is 2.2. The horizontal axis/X axis is in lbs/min.
On the compressor map, there are what we can "efficiency islands". The turbo is most efficient in the center of the map. Think of the map as a geographical topographry map, with the center efficiency island being the highest point (like a mountain), and the map getting smaller as it spreads out. You want to be as close to the center as possible on the map to be in the best efficiency range. Too far to the left of the map and you will surge (the turbo is pumping more air than the engine can ingest), and too far to the right and the turbo will become a restriction and starve your motor, also known as the choke line.

So now, let's plot what we just did in our example on a Garrett GT3582R compressor map. The PR is 2.2 and the lbs/min is 50.2 (or 50lbs/min):


As we can see in our example, this turbo (Garrett GT3582R) is perfectly suited for our application. North or South doesn't matter much, as long as we stay within the main efficiency island. This would offer great turbo response and airflow. This turbo would also give us room to grow, let's plug in our formulas to see how much power we could ring out of this motor on our example engine.

Example:
12.4 +18psi =30.4/12.4 = 2.4 PR

(2.6L) (8000rpm) (90%VE) (2.4 PR)/5660 = 811 CFM

(811 CFM) (0.069) =56 lbs/min

Alright, so we went from 50.2 lbs/min of airflow required at 15psi to 56lbs/min of airflow for 18psi. This should equate to 431rwhp in rotary terms, 560rwhp in piston math.

And replotted on the map:


We are off the main efficiency island but still within the means of the compressor. So, let's go for broke and see what we can do with even more boost on our sample motor:

12.4 + 24psi = 36.4/12.4 = 2.9 PR

(2.6L) (8000rpm) (90%VE) (2.9 PR)/5600 = 959 CFM

(959 CFM) (0.069) =66 lb/min

So at 24psi we are pushing 66lbs/min of airflow on our 13B-REW 2-rotor Wankel. This should be 508rwhp on a rotary, and 660rwhp on a piston motor.

Let's plot it:


Basically off the map. We are at the point where the turbo can't supply enough airflow for the motor and could become a restriction. Also, the air is becoming super heated at this point, because the compressor is being over-spun.

So now we know the GT3582R is too small is we want to flow 66+ lbs/min of air on our example motor. So lets plot this on a larger compressor, such as the Garrett GT4294R:


Boom! Basically on the main efficiency island of the GT42R. Again, this is for a rotary, but we can see this turbo would be the best choice here. Although, this doesn't mean the GT35R wouldn't be a good option, although it would be ill advised.

OK, enough rotary stuff, let's show the Mazdaspeed guys some love. In this example, we can even work backwards, starting with what horsepower we want. For this, lets say we want 400whp.

Formula:
400whp/10 = 40 lb/min

40 lbs/min/ 0.069 = 580 cfm

Okay, we know now we need 580 cfm in order to get 400whp on our Mazdaspeed3/6 (the MS6 might need slightly more because of the higher drivetrain loss).

12.4 + 20psi = 32.4/12.4 = 2.6 PR

(2.3L) (6000rpm) (95%VE) (2.6)/5660 = 602 cfm

(602 cfm) (0.069) = 42 lbs/min

So a little strong, but we're pretty damn close. This means, for out alltitude, you would need to run 20psi to get to the right PR. Now we can look at some compressor maps and see which one will give us the best trade offs of power and spool.

Here's a GT3071R:


A pretty good match. We are just a little to the right of the main efficiency island but pretty much spot on. The GT3071R would be a great 350whp-450whp choice for Mazdaspeed cars.

In the last example, we worked backwards by first figuring out how much power we wanted. This is easy because 10rwhp = 1whp, so simply move the decimal point over one spot to the left. 50lb/min of air is 500whp (again, this is a ROUGH estimate, but it will get you close).

We can also do this for a rotary, but as I outlined in an earlier post, the formula is different. Say we want 500rwhp in our single turbo FC.

500rwhp/7.69 = 65 lbs/min

So we know we need 65lbs/min of air to make 500rwhp on a rotary. This applies for any rotary; 12A, 13B, 20B, etc.

This makes it quick and easy to look at compressor maps on the fly. The Pressure Ratio is easy enough to figure out. Then, with a quick glance at the map, you figure out the most power that compressor can support.

Let's look at a GT4088R compressor map:


We can see this turbo flows around 68lbs/min of air at a PR of 2.3. So this turbo can support roughly 680rwhp on a piston motor and 523rwhp on a rotary. Again, this is at a PR of 2.3, so you need to calculate where you will be at for your airflow requirements.

Here is another turbo I am real excited about; the new Garrett GTX3582R. This features Garretts new Billet comp wheel. In case you haven't heard, billet wheels are all the rage now because they are over 40% lighter than traditional/typical cast wheels, which means quicker spool, lower discharge temperatures, and greater power potential. Really, there are NO drawbacks to a billet wheel, besides cost. If you can afford one, GET ONE!

Here is the GTX35R comp map:


We can see this turbo supports 78lbs/min of air at a PR of 2.8! That's 780whp on a piston motor and 599rwhp on a rotary! Of course, this is off the good efficiency islands so you're running on the bleeding edge.

The GTX35R is happiest in a wide range of airflow, around 40-60lbs/min and at PR's of 2.0 through 3.0. That's a pretty broad range to be on the main efficiency island, proving just how flexible this new compressor truly is.

Using our previous example where we fell off the standard GT35R's map, we had 66lbs/min at a PR of 2.9. We were just on the map with the old cast-comp wheel GT35R, but the GTX35R is comfortably supporting those requirements. Go billet!! I would love to map out a Precision 6262 billet (the turbo I am probably going with on my single turbo FD build) but getting a compressor cost a considerable amount of money. Precision is a small company that stuffs proprietary compressors and turbines in Garrett CHRA's and some Garrett/Turbonetics housings (Turbonetics also modified Garrett cores). So, Precision has no compressor maps.

Here is a quick cheat-sheet I came up with for Pressure Ratios for given boost levels. These are the same for all motors, and only vary with ambient air pressure (altitude) and desires boost level (these were all figured with an ambient psia of 12.4 for 5000ft of elevation):

10psi=1.8PR
12psi=2.0PR
14psi=2.1PR
16psi=2.3PR
18psi=2.45PR
20psi=2.6PR
22psi=2.8PR
24psi=2.9PR
26psi=3.1PR
28psi=3.3PR
30psi=3.4PR
32psi=3.6PR

And there you have it! It's not that difficult, and with just a little math you can be selecting the perfect turbo for your application. Post up any questions you have and put up some maps and we can work them out!

Thanks!

We already figured for roughly 400whp on an MS3 you need 42lbs/min of air at a PR of 2.6.

Plotted on the GT3076R:


And on the GT3071R:


We can see the GT3071R will handle it comfortably (on the 3rd main island) and be a good trade off for power and spool. The 76R is going to have more room to grow and will produce lower compressor discharge temps for the same boost level as the 71R, at the cost of a greater boost threshold.

(sorry the comp map pictures a little large, it's the first ones I found)

The GT30 series turbos are really well suited for two main motor types, medium displacement, high revving, low boost applications or small displacement, low revving, high boost applications (such as the MS3). High boost would be over 25psi, because look at the map, it goes to a Pressure Ratio of 4 (37-38psi)! At this level, again, you're on the bleeding edge, but still, this turbo can handle it.

Alright, let's figure out a GT2876R for 400whp on an MS3 and see if it's a good match.

We can also plug in your 11.8psia figure for your higher elevation. We will also need a little more boost.

So:

11.8 + 21psi/11.8 = 2.7 PR

(2.3) (6000RPM) (95%VE) (2.7PR)/5660 = 592 cfm

(592 cfm) (0.069) = 41lbs/min of air

So this gets us 41lbs/min of air required at a PR of 2.7.

Plotted on the GT2876R comp map:
 

Picklz

SUDO Make me a SAMCH
VERY nice, Gonna stick this nao!

edit: okay someone beat me to the sticky, still, great post.
 
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