Turbine wheel questions

[dieselguy86]

Hi all,

This is a question for the experts, or guys that are/were in the design field. If I'm wrong someone correct me, please.
We all know compressor trim affects the compressors pr capability at any given shaft speed. My question is, does turbine wheel trim correspond with the expansion ratio it is capable of?

I've been working with a guy who has twin turbos on his semi truck (that I built for him, I added a turbo over the factory gt47 turbo). I'm aware that with compounds turbos, the pr "load" should be equally split between both stages. In this case both should be doing 2.25pr each, but he's found its been working better with the low pressure doing 3.3 and the hp 1.5. Trying to make the hp do more immediately makes the exhaust drive pressure climb really high, really quick. With the 3.3/1.5 split drive pressure is close to 1/1 most of the way until the end of the revs where it starts to exceed boost, which I think is to be expected.

I've since gotten ahold of a factory compound turbo and found that their hp compressor has a really low trim of .43, the one were trying to use currently is ~55. So I ordered a billet wheel that I'm going to re profile to achieve the same trim.
The same is true for the factory hp turbine wheel, it's a 76 trim, where all the other "normal" gt42 turbines are 84 trim. So will it be necessary to re profile the gt47 turbine we are using to a lower trim in order to drive the low trim compressor? I'm just wondering what all turbine wheel trim affects, I really don't want to have to re profile turbine wheels.

-Joe

[racket]

Hi Joe

LOL.......you've had a lot of "Views", but nobody wants to tackle this one :-)

If I'm reading correctly , you're running a 5:1 overal comp PR , 3.3 X 1.5 = 4.95 , whereas you'd like a 2.25 X 2.25 .

OK your current 1.5 PR on the high pressure comp won't need much energy from the turb so you can run a fairly "open" turb stage with the largest A/R scroll and little "backpressure" .

But the low pressure turb stage will be required to have considerable backpressure to produce the 3.3 PR , this backpressure needs to be reduced so a larger scroll A/R required .

Without knowing the exact specs of each turbo its hard to know where the problem lies , are there interstage intercoolers as well as aftercooling ??

Cheers
John

[dieselguy86]

Hi John,

You are correct, 2.25 x 2.25pr is the goal. Right now the engine is basically not using the hp turbo. Wouldn't a properly working hp turbo lower the energy density going to the lp, somewhat acting like a larger a/r housing?

Currently the high pressure is a factory stock Garrett gt47, 74mm x 102mm compressor, 92mm x 83 turbine. Right now we're using a 1.42ar housing on it, with option of 1.27 housing.

The low pressure is a gt55, 112/102 turbine, 94mm x 123mm compressor, currently running a 1.24a/r open housing. Next a/r housing up is a 1.4, which I've been thinking about.

Airflow should be around 165lb/min, 2.75lb/sec. But will probably be close to 3lb/sec once we can get the hp turbo to start doing more and raise overall boost to 70psi. The egts can run hot at the high end, which tells me boost and airflow aren't where they need to be yet. My designed itt is 1200°f.

No inter-stage cooler yet, has it but I need to go back down to install it for him. The numbers showed something like a .4 increase in expansion ratio required to compress the hot air vs cooled, which is a big deal when there is another turbo downstream requiring its own exp ratio.

Truck runs good, it's making ~1500hp @ 60psi at 1:1, but its internally eating at me that the turbos aren't doing their equal share. I feel I need to understand why so I can work out future systems.

I have another garrett hp comp from John Deere and guess what, it's 43 trim. My plan now is to trim a billet 74 x 115 comp to 72mm, which raised my question of if I need a low trim turbine to power a low trim comp. I'm going machine out the map width enhancement groove in the housing and insert a solid chunk and taper the inlet to the inducer. I'm wondering if the map groove is helping to shove so much air into the wheel that the turb needs an excessive pressure drop to power it. Also going to make up a flow straightener, I think the gt55s compressor flow is spiraling against the gt47s rotation.

There's alot going on and I'm 1,500 miles away from the problem.

-Joe

[racket]

Hi Joe

I'm running into trouble with the HP turb stage , it has a Corrected Flow of ~57 lbs/min , but my numbers indicate you're pushing ~10% extra through it, so more backpressure , but if its running choked it should be producing more than a 1.5 PR with those high gas velocities.

Then your LP turb stage only has ~65 lbs/min Corrected Flow with the 1.24 A/R and only ~70 lbs/min Corrected with a 1.4 housing , you need the bigger housing to lower the LP comps PR and to increase the PR across the HP turb stage so it can up its comp PR .

I think you need a much larger LP turb stage , the HP exducer area is 5410 sq mms , the LP exducer 8170 sq mms , only ~50% bigger , if the pressure drop across the two stages is to be divided equally its gotta be ~twice as big as the HP.

A GT6041 turb stage with your sorta mass flow has an exducer of 119 mm - 11120 sq mms , heh heh , twice your HP turb exducer area and a Corrected Flow of ~90 lbs/min :-)

Cheers
John

[dieselguy86]

Hi John,

You may remember, but I just happen to have a brand new gt6041 in my shop. Other thing to do would be to add a second gt55, but that's quite a bit of plumbing work.

This is the info I'm missing out on. Would there be any way you could break it down into steps on how you figured all that out? It doesn't even need to be all in one go, even one step a day/ couple days so I could post process it.

I've known that if I could calculate things in the manifold, then I could figure out the rest, but how do you know what speeds the gasses are flowing? How do you calculate the flow potential through the housings when you don't know the area of the throat? How do you correct the flow?

The hp turbo is using a wastegated housing, so does that help the ~10% extra were trying to shove through it?

I already know what your going to say, it's in Rodgers and Cohen's book lol, it's been m.i.a since June/ July when we packed up for the move. Wife handled the house, which is where all my books were at.

I know you're busy doing your build, as well as advising everyone else, so I get it if you don't have the time to do a break down. Don't feel bad to say no

-Joe

[racket]

Hi Joe

I'll do it step by step :-)

Cheers
John

[racket]

Hi Joe

Lets look at your LP comp map www.garrettmotion.com/wp-content/uploads/2018/05/Comp-Map-GTX5533-94mm-1.jpg 165 lbs/min at a 2.25 PR is in the choke region with efficiencies at ~55% at best , with a minimum of 137 C degree rise requiring at least 117 C degree drop through the turb stage , you need that 106 mm inducered GT6041 comp with effic ~73% and only a 103 C deg rise

I'll do the numbers using the GT60

Cheers
John

[racket]

Hi Joe

The GT6041 very comfortably flows 165 lbs/min at 73% effic at a 2.25 PR , temp rise using a 15 deg C day will be 103 degrees , so 118 C out , +273 = 391 K .

A 103 deg comp rise will need ~88 C deg drop through the turb , or a PR of ~1.68 , lets say 2:1 if we add on the exhaust velocity .

Your HP stage comp will be running choked at 165 lbs actual flow , 87 lbs/min Corrected flow at ~60% effic at 2.25 PR with a ~170 C deg rise needing ~146 C deg drop, which will require a PR of 2.25 across a say 74% effic turb stage from ~1073K

Now to get 165 lbs/min through the turb stage that has a Corrected Flow of ~57 lbs/min from a temp of 1073K you'll need to have a ~5.6 PR entering the stage .

If we divide our 5.6 by the 2.25 PR required to power the comp then the interstage PR is 2.48 , well in excess of the required PR across the LP turb stage .

But this is where we really run into problems , the GT6041 has a Corrected Flow of ~90-95 lbs/min with the 1.47 A/R

Corrected flow = Actual flow times sq root of gas temp/standard temp, divided by the PR going in

So Corr Flow = 165 x sq root 923K/288K , divided by our 2.48 PR out of the HP turb

165 X 1.79 div by 2.48 = 119 lbs/min Corrected Flow , still far in excess of the turb stages capacity unless the inlet PR is increased .


LOL.......my heads hurting ..........I'd better stop and check the numbers latter on :-)

Cheers
John

[dieselguy86]

Hi John,

Don't forget, and this might muddy things up some, but we have wastegates on both turbos. So we can dump the excess, with the hp stage im limited to the factory gate, which I think is around 25mm. The lp currently has a 38mm gate, but I can upsize that to 44mm if needed. Wouldn't the addition of "fresh" exhaust out the hp gate raise the gt60 itt, helping to keep the pr down "some"?

Plus, we can live with some drive pressure, acceptable would be ~1.2:1, maybe 1.5 at the very high end.

[racket]

That does help :-)

[racket]

Hi Joe

It could be that your current GT55 wastegate isn't big enough, the turb stage only has a Corrected Flow of ~70 lbs/min , you'd need to be dumping a huge amount of gas out of it .

Are you measuring exhaust pressures before each turbo ??

Cheers
John

[dieselguy86]

Hi John,

No we don't, just in the manifold. I would like to add more though, especially when we get the inter-stage cooler installed.
I bet if we would of had 2 drive pressure gauges we would of noticed the gt55 being too small.

-Joe

[dieselguy86]

Hi John,

How did you figure out what pr was needed in the manifold to squeeze 165lb/min through the hp turb? Only way I can think of is to use the continuity equation in Thomas Kamps book, assuming gas speed would be mach (choked) and rearranging it to solve for pressure. Is there a simpler way?

As I was playing with all the numbers today on another system (for myself) I worked out a crazy gas velocity of 1,922ft/sec thru a turbine. Knowing that was well above mach, I was curious what mach would be at 1,400°, and some online calculators all said 1,820ft/sec, which seems crazy. They also said pressure doesn't matter, but I always thought pressure did change speed of sound, any of that make sense?

-Joe

[racket]

Hi Joe

Yep , you need pitot tube total pressure gauges before each turbo turbine stage, as well as thermocouples .

Speed of sound is temp related , the hotter the faster, ~2,000 ft/sec isn't unusual in a glowing turbo , pressure isn't important .

As for the flow through the HP turb , I checked out the turb map www.garrettmotion.com/wp-content/uploads/2018/05/Turbine-Flow-GT47.jpg for the Gt47with a 1.39 housing and it gave 55 lbs Corrected , so added on a couple more for your bigger housing, I used 800 Deg C - 1073 K as the TIT

Now Corrected Flow = Actual Flow X sq root of TIT/288 all divided by the PR going in

Corrected Flow = 165 X sq root 1073/288 divided by 5

= 165 X 1.93 divided by 5

=63
Now 63lbs/min Corrected Flow is considerably bigger than the 57 lbs/min on the map , so to get 57 lbs/min Corrected I needed to increase the PR going in , basicaly increase the density .

165 X 1.93 /5.6 = 56.8 ..............Corrected now equals the map.

Hope this helps

Cheers
John

[finiteparts]

Hi Joe,

I haven't visited here in a bit and I see that John has answered your questions well.  I had actually looked at this in the past, since my design does require a free-turbine for powering a propeller.  I had started to code up the process, but got sidetracked in the past and decided to jump back into that this morning.  Many of the newer texts have chapters dedicated to the turbomachinery component matching process and I had been using a typical process shown  which takes advantage of the mass flow parameters (MFP) of each component.  Since we can pull the corrected mass flow from the maps, we can quickly calculate the MFP for each turbine then just multiply by the corresponding pressure and temperature ratios to get the required LPT MFP in order to pass the same mass flow as the given HPT.  The equation is shown here:

Now, I generally like to follow the standard SAE station naming convention, so this equation is just a touch off.   Station 3 is the inlet to the HPT, while station 4 is the exit of the HPT and/or the inlet to the LPT. 

I ran the case over a GT47-ish map that I found online:

By pulling off 10 points of the curve, I built an approximate map in my model.  I assumed a constant 69% turbine efficiency for the GT47 turbine, which I know is wrong, but I just thought that I would run something without having to guess how the peak efficiency changes over the map.  I calculate the Tt4 and then calculate the Tt4/Tt3 ratio in each step.

As you can see, the mismatch becomes more pronounced as you go up in turbine expansion ratio because the HPT has taken more work out of the gas flow and left the LPT with lower energy flow (lower pressure and lower temperature).    Now, if your LPT has a smaller corrected flow than what is demanded by the system, you push more work to the LPT because you are taking a larger portion of the pressure drop across the LPT.  Basically, you are not running the HPT at the correct turbine expansion ratio (taking less expansion across it) because of the blockage due to the poorly sized LPT.

I hope this helps to add something to what John has already explained.

Chris