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What do turbine maps tell us about a turbo?

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Old 10-02-2007 | 03:09 AM
  #16  
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Thanks Hohn;

Typically I always thought of turbines as enthalpy utilization devices, but the corrected gas flow concept is much easier to understand.

Might want to goof with this myself. Mind posting the formulas for corrected gas flow and PR?

FWIW you can work backwards through your calculations. From the engine CFM / compressor side of things. My guess is that you could use it to maybe predict turbine A/R charateristics for a given compressor using Hp as something equal for both devices.

Compressor Hp
Fan Hp = P = ( Q * p ) / ( 229 *u )
P = Power in Hp
Q = Flow Rate in CFM
p = Pressure in psi (actually be better described as delta p as in p2-p1)
u= Efficiency coefficient (from the compressor map)

Jim
Old 10-02-2007 | 12:56 PM
  #17  
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Well, a turbine IS an enthalpy utlization device. Corrected gas flow is what appears on the turbine maps from Garrett, so that's what I went with.

We have to remember that you can't calculate both PR and CGF at the same time-- you have to fix one variable. I think it's more useful to fix PR and calculate gas flow, so that's what I did. The reason you can't calc both at the same time is that as PR goes up, the "corrected" gas flow goes down for the same mass flow. That's why you see the turbine maps level out-- the increasing mass flow at the turbine "piles up" and appears in the form of a higher PR, not as more gas flow.


For example, look how the gas flow changes wildly with PR in the example below:

Turbine inlet temp: 1000F
actual mass flow in: 18lb/min
Turbine inlet pressure: 14.7psi
Corrected mass flow: 28.7lb/min!!

But if I double the PR, watch this:
Turbine inlet temp: 1000ºF
Turbine inlet pressure: 29.4psi
actual mass flow: 18lb/min
Corrected mass flow: 14.37lb/min

I just cut it in half when I double PR.

So, increasing mass flow through the turbine can EITHER increase the mass flowing through it OR it can increase the PR at the turbine-- but not both. WHICH it does depends on how much mass flow, because the turbine map tells us that while the turbine can initially absorb all additional flow without increasing PR, it gets to the point where ALL additional flow goes straight to increasing PR.

The formula is in absolute temp and pressure and such, but here it is (just ideal gas, as far as I can tell), correcting to "standard" conditions.

Corrected mass flow= input mass flow * (sqrt (TurbineInletTemp+460)) / sqrt (518.7) / (turbine PR in absolute pressure).

Justin
Old 10-02-2007 | 01:05 PM
  #18  
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From: Let'r Buck!
Originally Posted by HOHN
That's why you see the turbine maps level out-- the increasing mass flow at the turbine "piles up" and appears in the form of a higher PR, not as more gas flow.
and from earlier

The plateaus in your graph (and what I believe you are refering to in the first quote above) are exhibiting choking behavior.

Please explain to me where you would like to operate on the curve and why.

Thanks!
Old 10-02-2007 | 04:14 PM
  #19  
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From: Cummins Technical Center, IN
Originally Posted by gies
Please explain to me where you would like to operate on the curve and why.

Thanks!
That's a GREAT question, one I'm trying to figure out myself.

It's telling that these turbine maps level out, and that this level area corresponds roughly with a certain level of efficiency. The higher the efficiency (from a larger housing), the more flow it takes before it levels out.

By my reasoning (questionable to others, as you've seen) I'm mostly concerned with where a turbo starts making boost. Boost means that the operating PR of the compressor is greater than 1. So if my compressor is a 1.5PR, then I'm seeing about 7.35psi of boost (half of 14.7psi, which is atmospheric pressure).

What has compressor to do with turbine PR? They roughly correlate. Lots of guys have done tests showing a 1:1 correlation of boost to drive pressure, and in some cases boost is a hair higher.

We can therefore reasonably assume that when the turbine is operating at 1.5PR, the compressor is fairly close to this as well, and will begin showing a little boost. For operating reference, this is about where the stock turbo operates on the hwy with the cruise control set on level ground.

So I look at the 1.5PR point and the 2.0PR points on each curve, because this is the transition from "finally up on boost" to "fully spooled" in my mind. Most of the curves have levelled out (or come close to it) by the 2.0 PR point.

PRs below 1.5 might be somewhat useful, but you'll notice most of the maps don't even start until 1.25

"Choked flow" is oddly enough more related to compressor than turbine mapping, from what I can tell. When the compressor goes off its map, the efficiency plummets and the RPM will skyrocket. This means it takes a LOT more shaft HP to drive the compressor and the restriction of the turbine thus goes way up, and drive pressure soars, too.

Then there's the "choked flow" inthe turbine when the flow goes supersonic, which would also spike the DP by a large amount.

From what I can tell, the turbine maps only concern the turbine wheel and housing in isolation-- completely ignoring the effect of the wheel on the other end of the shaft.

So if we want to know what's going on, we have to look at how they affect each other.

Justin
Old 10-03-2007 | 01:17 AM
  #20  
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So turbine PR is?

PR = ( P1 + P2 ) / P1
P1 = Drive Pressure
P2 = Back Pressure (~ atmospheric depend upon exhaust pipe charactersistics?)

The most desirable DPR (Drive Pressure Ratio) is less than or equal to 1. But realistically, often the DPR is greater than 1 especially with the OEM Holsets.

Hence those Garret turbine maps become meaninful.

FWIW you don't always have to convert temperatures to absolute, because the math cancels itself.

Jim
Old 10-03-2007 | 01:26 AM
  #21  
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Originally Posted by HOHN
Well, a turbine IS an enthalpy utlization device. Corrected gas flow is what appears on the turbine maps from Garrett, so that's what I went with.
Justin
I guess what I meant is that I in my calculations I could never get past trying to calculate basically:

(Delta Enthalpy x Gas Flow Rate) x Turbine Efficientcy = Turbine Hp
Where:
Enthalpy = Massive Headache (for me)

CGF gives you a number to work with based upon real world measureable data (PR) and to a lesser degree engine RPM / Boost - CFM (LBM/MIN). The problem I would think difficult is equating Turbine Hp generation to Compressor Hp consumption. Figure out a way to draw lines plotting points between the Compressor and Turbine maps for a given application and then you have something.

Basically Hp(in / turbine) must equal Hp(out / compressor) or the turbo rpm will not be constant. The Compressor Hp is relatively easy, but the for Turbine Hp I think you might have to delve into enthalpy unless you can derive the change in enthalpy backwards / sideways using CGF.

Jim
Old 10-03-2007 | 01:32 AM
  #22  
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From: Cummins Technical Center, IN
Yes. But keep in mind the all these pressures are in ABSOLUTE pressure-- meaning you have to add the "drive pressure gauge reading" to the atmospheric value.

In your example, P2 would be atmospheric pressure (backpressure in the exhaust can be ignored imo except in severe cases-- like twins with 3" exhaust).

I assume by DPR you mean the ratio of Drive to Boost pressures, correct? If so, then you are correct that we'd want this to be one or less.

The turbine map is deceptively simple. There's a LOT of useful information hiding in those simple little curves.

JH
Old 10-03-2007 | 01:39 AM
  #23  
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Yeah...

DPR = Drive / Boost (absolute pressures)

Hohn, go back and reread my posts, I made some edits as you were typing.

I did not know you were online...

Jim
Old 10-03-2007 | 04:15 AM
  #24  
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From: Cummins Technical Center, IN
Originally Posted by NoSeeUm
I guess what I meant is that I in my calculations I could never get past trying to calculate basically:

(Delta Enthalpy x Gas Flow Rate) x Turbine Efficientcy = Turbine Hp
Where:
Enthalpy = Massive Headache (for me)

CGF gives you a number to work with based upon real world measureable data (PR) and to a lesser degree engine RPM / Boost - CFM (LBM/MIN). The problem I would think difficult is equating Turbine Hp generation to Compressor Hp consumption. Figure out a way to draw lines plotting points between the Compressor and Turbine maps for a given application and then you have something.

Basically Hp(in / turbine) must equal Hp(out / compressor) or the turbo rpm will not be constant. The Compressor Hp is relatively easy, but the for Turbine Hp I think you might have to delve into enthalpy unless you can derive the change in enthalpy backwards / sideways using CGF.

Jim
Jim-- I think you're onto something. You certainly want turbine hp out to equal compressor hp in (minus any hp loss to viscous friction).

I think corrected gas flow works elegantly and simply, so I don't feel the need to calculate enthalpy change or attempt to calculate turbine shaft hp.

Corrected gas flow and PRs will get me all the info I need within acceptable (to me) margins of error. I will try to add examples of this later in this thread (too late tonight).

The ONE unresolved "fly in my ointment" thus far to me is accounting for engine load. Why is it that higher engine loads so dramatically affect spoolup? It is JUST that higher engine loads increase EGT and thus the drive energy to the turbine? Could it be just that simple? Or is there something else more complex that I'm missing?

Anyway, I've satisfied myself that I can figure with reasonable probabability what I want based on corrected gas flow.

For example, let's run a calculation for the Garrett stg 3 on Hammer's truck. He reports that he sees 4-6psi positive boost pressure with the cruise set on the hwy. Let's say that occurs at 1800rpm.

359 cubic inches X 1800rpm / 3456 X .95 (volumetric efficiency)= ~178CFM

Density of air= .075lb/cu ft. So we have 13.35lb/min of airflow going through the engine (assuming it's drawing in 70º air).

Let's say we are running an air/fuel ratio of about 26:1, so we add another half-pound to the mass flow coming out. So let's call mass flow out to be 13.8lb/min. What is the EGT? Who knows? That's a function of timing, engine load, and a ton of other variables that are beyond our needs. We need to ignore the effect of the turbo and guesstimate what EGT might be without a turbo attached. Let's call it 1000º F.

When we "correct" the gas flow based on these figures, we get 23.1lb/min at a PR of 1:1.

Looking at the Turbine Map for a Hammer's GT37 (sorry, no link), we see that the curve starts at 1.25PR. So we go back and re-calc the gas flow correction based on 1.25PR and get 18.5lb/min

Again looking at the turbine map for the GT3788, we see that 1.25PR falls at right around 19lb/min. So our "calculated" guesstimate is within a half-pound of the "real world" example!

Since the operating PR is 1.25, and we know that boost roughly follows drive pressure, where would we be if the compressor also was at 1.25PR? Answer: right around 4psi of boost (14.7psi divided by 4).

So our "calculated" values are VERY close to the real world numbers Hammer is reporting.

This validates my methodology in my mind, at least close enough to be able to say how a given Garrett turbo will spool relative to another, just by looking at the turbine maps. Remember that the maps can only tell you WHEN the turbo starts making boost, not HOW FAST is will spool after that.

But you can bet a Garrett BB turbo spools pretty quick.

Based on my analysis of the turbine maps, I suggested that Hammer try a GT4088 because the turbine map says it should spool 18% later in terms of RPM and load and such. It might be just what he needs, because he says his actually spools too soon and runs out of steam.

Anyway, as I learn more about this, I feel its very eye-opening and wish I would have learned it sooner.

It also makes me want those B-W turbine maps even worse!

Justin
Old 10-03-2007 | 11:05 AM
  #25  
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Based on my observations, the stock Holset's turbine on the '05 attains a 5:1PR, and goes supersonic around 50PSI.

Only with aftermarket fueling, of course.

Wish there was a convenient way to measure shaft RPM; intuition tells me that dangerous RPM spikes are largely prevented by the flow disruptions that accompany driving off the top of the map.
Old 10-04-2007 | 06:44 PM
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Good stuff Hohn, keep working.

The whole "What will the EGT be?" question has bugged me for practically day one. In essence, I think you would need to build a mathematical model / curve based on real world data and tune it based upon your assumptions. Then go see how close you get, for example with your truck your turbo.

If you want to know how to calculate the effects on EGT based upon DPR >1, I think I have that one.

I also think, that if you keep working on your spreadsheet you will get there...

Jim
Old 10-04-2007 | 07:50 PM
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All this is a little over my head but I found this by searching how to read compressor maps on google. Its from bimmerforums.com and thought you might like it.
Pics wouldn't come up for me.

How To: Read Compressor Maps

--------------------------------------------------------------------------------

This guide isn't meant to be an in-depth look at turbo sizing, but rather to give the reader a working knowledge of how to read compressor maps. In other words, to decipher what all those swirling lines mean when choosing a turbo compressor.

Terminology

Compressor - This is the "cold" side of the turbo that sucks in intake air and compresses it for the engine to later combust with fuel.

Turbine - This is the "hot" side of the turbo. Hot exhaust gasses pass through it, expanding and cooling. This expansion spins a turbine wheel that drives the compressor wheel via a shaft. Unfortunately turbo manufacturers don't make turbine maps available to the general public.

Absolute Pressure - This is pressure referenced from a pure vacuum. Most calculations done involving compressors use absolute pressure. Note - 1 atmosphere = ~14.7 psia (Absolute pressure in pounds per square inch) = 0 psig (gage pressure in pounds per square inch). Your boost gauge reads in psig, referenced to local atmospheric pressure.

ONTO THE MAPS

Surge - This is lowest amount of airflow a compressor can supply at a given pressure ratio(getting to that). Any pressure above this at this airflow, the compressor will "gulp" air. This is not good for your turbo, or your power output. Fortunately you have to saddle a pretty huge compressor with a small turbine to really worry about this effect.

Here is a compressor map with the surge line highlighted in red.



On the X-axis(horizontal) you'll notice the mass airflow of the compressor in lbs/min. On the Y-axis there is the Pressure Ratio. Pressure ratio is defined as follows:

Atmospheric Pressure + Boost Pressure = Pressure Ratio
Atmospheric Pressure

So the astute reader will notice a pressure ratio of 1.0 is the exact same as atmospheric. A pressure ratio of 2.0 is equivalent to 1 atmosphere or ~14.7 psig in your intake manifold. Without concrete data proving otherwise, it is always the best course of action to assume the pressure is ambient at the compressor inlet and make note of the pressure drops of the system will in the end cause less horsepower to be produced than the mass flowrate of the turbo would suggest.

The oval shaped rings on the compressor map are efficiency islands. These are regions where the compressor has approximately the same efficiency at compressing the air. Of course, the higher the efficiency the better since the compressor will be introducing less unneeded heat into the charge air. Note that as you go away from the maximum efficiency island, you always go down in efficiency. By the time you're off the map you're usually in the <60% range, which is not a good thing.

The lines that slope from the surge line to the right and down across the efficiency islands are constant speed lines. This would be really useful if you could match up the speed of the compressor to the speed of the turbine and find out its efficiency and mass flowrate for that shaft speed, but since we don't have turbine maps we're kind of at a disadvantage there for picking the ultimate turbo match. The maps used here out of Garrett's publicly available catalog aren't too detailed. Some maps will have much more data like putting RPM values on the speed lines, more efficiency islands etc.

I won't go into the hard equation to calculate the mass airflow of the engine, as it really doesn't gain anybody any further insight into the turbo selection process. The only important things to understand that the big factors in how much mass airflow an engine is consuming are:

Engine Displacement
Volumetric Efficiency(how well the engine breathes)
Pressure at the inlet valves(BOOST!)
RPM

By altering these things(more displacement, cams to increase VE, more boost, more RPMs) you can make the engine combust more air and make more power. I'll be attaching a spreadsheet that makes calculating the airflow of an engine an easy matter. It does over simplify things since it doesn't vary VE by RPM and whatnot, but it is a reasonably close approximation. I use a VE of 90% in most my calculations. It is pretty close to what a modern 4 valve engine gets in high RPMs, and tends to be conservative on less modified engines.

So go ahead and download the spreadsheet and we can look at a compressor map.

Here I'll look at a GT30R turbo on an S52B32 engine with a VE of 90%, displacement of 3.2L and maximum RPM of 7000. For the first go, we'll see what happens at a modest boost level of ~8.7 psi(pressure ratio of 1.6).



How I evaluate compressor maps is to note the airflow at 2000 RPM. Find that on the X-axis and draw a straight line from that point at a PR of 1 to the airflow at 3000 RPM at your desired PR(1.6 in this case). This gives you an idea of how a typical turbo will look when spooling up, and let you know if it's at a risk of surging. From there, the line should stay at that PR all the way to the airflow at redline ~39 lb/min here.

As you can see, surge is not a problem here, but this turbo sure does look a bit too small for this sized engine! It goes off the map just before redline, so that means it is very inefficient at higher revs.

Let's see what happens when we up the boost to ~17.4 psi(PR of 2.2).



No risk of surge due to this being a large engine for the turbo, but boy does it ever get REALLY inefficient at higher revs. Past 6000 RPM it is again off the map.

Let's go to a slightly larger turbo, a GT35R, to see the difference. Same boost of 17.4 psi(PR of 2.2).



Now that is more like it! Notice how the engine spends a good amount of time in the really efficient islands, and the turbo is still at 72% efficiency all the way to redline. I'd think this turbo would be putting out in the 450-500rwhp range at this boost on this engine, and that's probably being a bit conservative. If the VE of the engine was even higher(which it can be), this turbo could still put out even more power. The compressor map also suggests it has a bit more headroom on this particular engine.



I hope that was helpful to everybody, and gives people a start in the right direction on reading compressor maps themselves. If you want to modify the Excel airflow chart I attached, just extend the RPM row and copy the formula in the airflow cells for CFM and lb/min over below the RPM and it should fill in correctly. You can change the displacement, VE and pressure ratio in the parameters section to get an idea of how these change things.


Enjoy boost junkies!
Old 10-06-2007 | 01:17 PM
  #28  
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Hohn, here is a couple of threads might interest you. Not sure where BSFC falls in there.

Diesel Freak:
stoic for diesel 14:1
Smoke threshold ~18:1
useable EGT 28:1
unit of volume will not work
must use units of mass, pounds mass, or kilograms. This take all of the density variations out of the problem
tbuelna
Direct-injected, compression-ignited engines always combust at near stoichiometric, if the engine is designed properly. The fuel injected will only combust when sufficient oxygen is present at the flame front. The fuel will continue to burn as long conditions are suitable within the combustion region (ie. oxygen, fuel and heat). If the available oxygen within the chamber is used up before the available fuel, soot is created. If the heat in the combustion zone is reduced sufficiently, due to expansion of the end gas and slow combustion, combustion halts and the remaining fuel and oxygen are dumped out the exhaust. If more oxygen is pumped into the woking chamber than is necessary for the amount of fuel injected, then engine efficiency is reduced.

The basic nature of a compression-ignited means that it can run "lean" or "rich" without doing damage to the engine. Ultimately, the amount of fuel delivered is dictated by the load on the engine, and not by the amount of air flowing past the throttle as in an SI engine.
Competition Diesel

Engine Tips

Some comments on too much air (IE: twins) and the reduction of Hp (efficientcy) due to pump losses become greater I am assuming. It also might seem to appear why advancement of timing helps as the AFR (Air Fuel Ratio) leans out up around 40:1. Even some NOx control stuff.

Jim
Old 10-06-2007 | 10:45 PM
  #29  
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That's all good stuff, right there.

But I'll disagree with Cliff's assertion that 28:1 is required for useable EGT.

The reality is that you can get wildly different EGT numbers for the SAME air/fuel ratio, depending on injector efficiency, timing, and turbo operating conditions.

We can burn diesel fuel in a lab at different air/fuel ratios and assign different ratios as being certain "points"-- the "smoke point", the "Stoich point" the "EGT" point, etc.

But when we try to burn that fuel in an imperfect engine, more variables are introduced that make hard and fast rules difficult to come by.

For the first part, EGT is almost as much a measure of expansion and speed of combustion as it is air/fuel ratio (AFR). For example, if I take a given diesel engine that runs well at 17:1 compression ratio, I can drop the EGT just by raising the compression ratio-- even with the exact same AFR. The higher compression accomplished two good things-- it increases the speed of combustion by confining the fuel and oxygen molecules to a smaller volume (increasing the likelihood of collision) and expands that combustion products to a proportionally higher volume, reducing the temp of the exhaust products by lowering pressure (Boyle's Law of temp/pressure relations).

So any hard and fast rules about trying to equate a certain AFR with a certain EGT are going to be pretty much impossible dreams.

Garrett uses a 22:1 AFR for calculating for diesels, so that's what I used. the irony of all that is using THEIR stated AFR, they make their own hp claims for their turbos impossible..

Funny, eh?
Old 10-23-2007 | 02:26 AM
  #30  
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Funny, eh?
Yeah ain't it. Smoke and Mirrors.....

DF was likely refering only to the ISB and usable "as in" towing ability?

Not sure.....

One thing that I find not often discussed is the effect of fuel vaporization in the combustion chamber. Hence, a big difference between the Otto and the Diesel Cycles for which the diesel has the efficientcy advantage.

I was not aware of lower EGT higher compression concept. I am guessing that correspondingly the combustion temperature increases though?

Off topic sorry;
Jim



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