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In the very nature of compounding the first stage must be larger than the second on the order nearly equal to the pressure ratio it delivers to the second stage. (In the example I gave it would need to be around twice as "big" or more depending on other conditions)
This is the part I'm most highly interested in. I have heard people talking about size differences between compound turbos of as little as 20% (sizing the little one at 90% and the large at 110% of a single turbo flow), this is the first time I've heard of making the big one double.

Of course it will depend entirely on the application. When I get around to trying it my goal is boost at low rpm's and holding through to the highest rpm's you use in normal driving. I'm after maximum drivability and no smoke.
So for me the larger turbo will only be sized big enough to flow the expected maximum airflow.

Quite different to making a dyno queen where the only concern is power. This is always going to lead you to bigger turbos for their lower exhaust backpressure and hitting good efficiency at the engines maximum flow.
 

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...........................................
You just proved that the large turbo doesn't need to be double the flow of the small one.

You have just sized a turbo system for operation of a set boost level at a fixed speed. Fine for an industrial generator but not what I'm planning to do.

My plan involves around 15psi from the small turbo, compounded by about 7.5psi from the big one to give a total boost of 30psi. Hence a low pressure ratio of 1.27 and difference in turbo size (flow) of about that.

The compromises of getting more boost over a wider rpm range will lead to variation in turbo sizing and in pressure ratios.
For example I don't expect to be running even pressure ratios between the turbos although it probably will occur as a coincidence somewhere in the operation range and it'll be in a rev range that doesn't get visited often.

A small problem with your calcs.

A pressure ratio of 2.25 with an intake at STP and efficiency of 65% will give an outlet temp of 410K and density ratio of 1.61 (1.2kg/m^3 to 1.93kg/m^3).

A pressure ratio of 2.25 sucking on the outlet of the first turbo gives outlet conditions of approx 575K (202 deg C) and a density ratio of 1.61.
 

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Your Foundation ( Block,heads,crank, Rods Etc. Wont Take It. Those 6 Bt,s Twin Turboed Blow Up Constantly. My Machineist Build Em Weekly For These Young Guys And Its A Money Pit For Them. Its Well Beyond There Limits For Durability. I,am Just Tryin To Save You,all Alot Of Grief And $$$$$$$$$$$$$$$$$$$$$$$$$$$
Someone emailed me a few pictures of a 6BT powered pulling tractor which crapped itself under very high boost.

The engine unzipped right around the crankcase.
 

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As for the sake of others hoping to learn about staged compressors, I will have to clarify a few things to stop any possible confusion.

First be aware of this portion:

(Dougal)
"My plan involves around 15psi from the small turbo, compounded by about 7.5psi from the big one to give a total boost of 30psi. Hence a low pressure ratio of 1.27 and difference in turbo size (flow) of about that."
Typo on my part.
Substitute "pressure ratio of 1.27" for "density ratio of 1.27" and it will all become very clear.
 

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Dougal:

I am sorry for trying to help explain how to size a staged turbo system.

There is absolutely no problem with the calculations, I explicitly said it was more complicated due to intercooling (wich I left out for simplicity of explanation).

I can see that my attempt has fallen on deaf ears (or blind eyes as the case may be)

My appologies.

Do as you wish, pretend I had never said a word.
I found an error in your calcs. Specifically the density ratio you gave did not check out. If you can't handle your calculations being checked, then why post them up?

It's the same as claiming facts without being able to back them up with reliable sources.

You're really putting a damper on the warm fluffy atmosphere here.
 

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I guess I just see someone trying to convert a PR to a specific temperature for a specific ambient condition, and then convert that again to another temperature unit, and then use that to get a ratio between the two.

.....
......Then taking the next step of doing a conversion to get your units right. Then further taking a ratio between two numbers already rounded two to three times depending on the route taken.

And after all of that calculating from pressure to temperature, to another unit you come up with a number that is ~0.15 different from mine.
Not at all.

See I didn't do any conversions and didn't lose any rounding errors.
I have a turbo sizing spreadsheet I made which calculates everything to whatever precision the program is capable of using.

I don't have to convert temperatures or pressures. See STP (standard temp and pressure) is the same regardless of units. Start your calcs with 14.7 psi or start with 101.3 kPa, it makes no difference.
Pressure ratios, efficiency, density ratios etc are all dimensionless.

Your density ratio calcs were 10% out. The impact of that could be 10% difference in engine power.
The difference between 400hp and 360. Quite meaningful to some people here.


Of course the interesting upshot of your calcs was something I had missed before.
The difference in turbo sizing is the density ratio created by the first turbo. To use a large turbo which is double the flow of the small, you'd need a density ratio of 2.
Not many systems outside tractor pulling will do that.
 

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With a PR of 2.25, if the density ratio was 1.46 then the compressor was operating at an efficiency of .648 (1.46 divided by 2.25) or simply 65%, the efficiency I stated to have been calculating for.

However, with a PR of 2.25, if the density ratio was found to be 1.61, then the compressor was operating at an efficiency of .715 (1.61 divided by 2.25) or simply 72%.
What are your input conditions? I am using STP (standard temperature and pressure) and yes it is 1.61 at 65% of adiabatic efficiency for a pressure ratio of 2.25.

Input conditions control the output conditions and the density ratio completely. With different input conditions, your output conditions and density will be quite different.

With intercooling between each stage (as you should) the second stage will need to move 819 CFM and the first stage will have to move 1916 CFM.
I have yet to see a staged turbo setup with intercooling between the stages. But it is irrelevant.

The difference in turbo flow is the density ratio feeding the second. Whether this is the result of just one turbo or an intercooled turbo doesn't actually matter.
If you get a density ratio of 2 then twice as big makes sense. But that's an extreme setup, not a typical one.


I will also add that 30psi boost is not at all in the realm of staged compressors. You will likely find that you will be less efficient running compounds for such a dismal pressure ratio. That is a job best left to a single compressor.
If you read my reasons why, it makes perfect sense why I'm not opting for a single turbo.
The reasons are response, being able to use a small variable vane turbo as the little one for maximum low end torque, while compounding it allows it to flow more air so it doesn't overspeed and blow up.

I'm not making a 20" wheeled street truck.

30psi is the starting line, not the finishing post.
 

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With a PR of 2.25, if the density ratio was 1.46 then the compressor was operating at an efficiency of .648 (1.46 divided by 2.25) or simply 65%, the efficiency I stated to have been calculating for.
Is that your actual calculation to get density ratio?

Because that is very very wrong. I suggest you get yourself a thermodynamics text book and look up "brayton cycle" for the easiest solution.


Calculation of density:

1. Start with intake conditions, pressure, temperature, density. Use PV=NRT to ensure you know all relevant conditions. Work in absolute temperature (Kelvin scale), My conditinosa are for temp.293K; pressure 101.3 kPa and air density of 1.2 kg/m^3

2. Find your pressure ratio, in this case 2.25

3. Apply adiabatic compression to find the outlet temperature relating to a 100% adiabatic compression. T2/T1 = compression^(gamma-1/gamma).
Gamma is 1.4 for air. Tout = 369.4K

4. Find the adiabatic temperature rise. This is your adiabatic outlet temp (369.4-273=76.4K) your adiabatic outlet temp.

5. Divide this temp rise by your compressor efficiency (0.65 in this case). This gives you your actual outlet temperature. (76.4/0.65+273 = 410.5K)

6. Use PV=nRT to find the the volume (and density) of the output charge using your pressure ratio and your actual outlet temp and density. Density = 1.93, intake density =1.2
density ratio =(1.93/1.2=1.61)

7. Divide original density by resulting density to get the density ratio. 1.2/1.93 = 1.61
 

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I am interested if you entered 2.25 PR into your template yet with a compressor efficiency of 100%.

The correct density ratio would be 2.25:1, and yet I have no doubt in my mind that your template will not return that answer.

I must be missing something.

:(

Yes you are missing something, something vitally important. The fact that air heats by itself when it is compressed. It's called "adiabatic compression" and it's why bike pumps get hot and turbos blow hot air.

A compressor of 100% efficiency doesn't put any extra heat into the air, but the air still heats when it's compressed.

A 100% efficient compressor at pr of 2.25 has an density ratio of 1.78. When sucking on air at STP.
 

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I understand what adiabatic means, and had wondered if we were to assume that for the outlet temperature (plus that from a given level of inefficiency)?

Either way, when this was started my calculation was 1.46 vs yours of 1.61. I do not see how if we are to assume the turbo production will be adiabatic temp plus that from inefficiency that my solution could be lower than yours at any point with my having placed a 1:1 relationship between 100% efficiency and the Pressure Ratio.

Why at 100% is my assumption giving a number greater than yours and at 65% efficiency yours is giving one greater than mine?

Like I said, I want to understand the discrepency between these numbers so the correct method is used. If I have oversimplified it, then please tell me exactly why.

Thanks
In a nutshell:
You're getting wrong numbers because you're using the wrong method.
I have covered the correct method above in 7 stops. If you do not understand and use that method then you will get the wrong answer everytime.

You can't "simplify" the maths and get the right answer. There is no linear relationship 1:1 or otherwise.
 

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Hey folks,
I was searching around the web for details on a wrangler /4bt and found you guys. Im a diesel fuel injection tech and actually have a recipe for a 400 horse 4bt. Cummins actually has a 321 horse 4bt stock. Once I found out those basic parts the rest was easy.
INJECTORS 0432131753
PUMP 0402744806, This is a A model pump,but you can get these numbers out of a stanadyne, CAV or a VE with the proper tune (I know i customized one hell of a DB2 for a 93 turbo ford 7.3)
For the turbo I would use a schwitzer s200 # 316998 it support's enough fuel to get you in the 400 hp range due to its large turbine size, but the compressor wheel is still s200 with extended tip technology so it will spool pretty quick to. I would stay away from holset's they tend to tear apart when you push them hard enough
What airflow and boost would you need for 400hp and how much smoke are we talking?

Petrol engine rule of thumb was about 10hp for every lb of air/minute. Diesel I think you're closer to 7hp which would mean about 60lb/min airlow.
 

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I like to think in things more along the lines of volume of air not boost. For instance The owner of my shop has a 99 powerstroke all the bells and whistle,ram air intake, 120 hp 3 stage bullydog chip and a 4 inch exhaust, he was getting about 28psi of boost but at 120 he had a lot of black smoke and the seat of the pants feel was the same as at 75 hp with little black smoke.
We slapped in a Garret Powermax ball bearing turbo boost went down to about 26psi (wastegated) but low and behold @ 120 no smoke and mega power. The reason effiency and volume.
Reduced turbine backpressure and/or reduced charge temps would do that. Increasing the engines VE through less retained exhaust. A more efficient turbo compressor gets a lower temp and more air molecules in for the same boost so doesn't work the turbine as hard so exh backpressure is less.

But to get a certain flow through a certain size engine requires a certain amount of boost. That amount of boost is a wide window depending on many factors but it'd be good to know the the ballpark.

Got any links to the compressor map of one of these Schwitzers?
 

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Have a question that has been nagging me and can tell from reading posts, more intelligence and familiarity with diesels,turbos and intercoolers/aftercoolers than I'll ever have. Am curious if a person could use A/C system as a intercooler/aftercooler to get a denser charge of air and lower EGT's. Like an "on demand" system, seems it would be handy pulling steep grades or towing applications. Would the tradeoff be worth the while(hp drawn vs cooler EGTs) or would they pretty much cancel each other out? Input appreciated. Ive heard(not sure if true) some supercharger appl. might be using this method (Eaton, Kenne Bell)
You might get the tiniest bit of net power gain (power gained from cooling - power lost to drive AC system) if you put a lot of work into designing the system.
But it's soo much easier to simply make your intercooler as effective as possible as they don't suck crankshaft power.
 
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