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"Series-sequential" turbos

Discussion in 'Turbo Tech Questions' started by rsimmons, May 21, 2024.

  1. bbi_turbos

    Dec 8, 2021
    I ran your input numbers and got...

    369K coming out of lp turbo.
    470K coming out of high pressure turbo.

    That's 377°F. If it's 70° day outside, subtract 70 from 377= 307° left to cool. Air 2 air is AT MOST 75% efficient, so 307×.75= 230.25 drop. 377° intercooler inlet temp - 230.25° drop = ~147°F air coming out of intercooler, or 337K.
    337K÷288K= 1.17, sqrt of 1.17 = 1.08.
    1.08x62lbs/min= 67.
    Finally 67÷3.0pr= 22lbs/min.
  2. rsimmons

    May 14, 2024
    Okay I've gone back through everything now and things are moving towards matching what you've told me. I was pretty surprised when my MFR out of the intercooler suddenly matched the lb/min from the beginning. I still have a few questions though to make sure I'm doing this right.

    First the first step, I get the CFM of the engine in NA form and then most the calculations seem to be based on that. I do still need to find Wa though to determine PR and mass flow through the low pressure turbo right?

    While I was able to match the flow rate out of the intercooler the two compressor stages were off a little bit, i.e. NA flow rate = 23.8 lb/min, stage 1 =24.8 lb/min, and stage 2 = 27.9 lb/min should these numbers all match or does it only matter for the final coming out of the intercooler?

    All of these calculations are based on ideal conditions with the exception of 95% VE. I've tried crunching the numbers with some assumed losses as well, such as 1psi drop across air filter and 2 psi drop through intercooler and piping, and local baro of 14.5 psi. The ideal PR is 3.0 but when I factor in losses I get an actual of 3.4. If I use 3.4 the 2 stages get a little closer to matching, but post intercooler dips a bit. NA flow = 23.8lb/min, stage 1 = 24.5 lb/min, stage 2 = 27.2 lb/min, and post intercooler = 23.6 lb/min. Should I be assuming these losses or am I over complicating things?

    The last one is kind of a dumb question I feel I should already know the answer to, but I've never really had to think about it since I've never had a question of having an additional temperature rise after a single turbo. When plotting the low pressure turbo, I obviously have a few points in the low revs that land well below where the map start (not in surge, just below the last speed line), since the compressor isn't really operating down there can you just assume 100% efficiency for temp calculations for those points? I would think that if its not actually creating boost, its probably not making much heat either.
  3. bbi_turbos

    Dec 8, 2021
    That's all exactly normal, your seeing how turbos compress the air in stages, your basically watching the density increase.
    For shits and gigs you can run those numbers as a single turbo, you'll see it in one step instead of 2.
    Also fyi, the same equation works for the exhaust, then you can plot on turbine maps.
    rsimmons likes this.
  4. bbi_turbos

    Dec 8, 2021
    Forgot to answer the second part of your question.

    Definitely not 100% efficiency but the pressure ratio is so low that it isn't adding alot of heat. Heat is a percentage of boost, if your boosting 2% then there isn't much heat either. So I guess it could look like it was close to 100%, me myself I would run the numbers how it actually is. BUT even our best numbers aren't 100% accurate, the whole idea of all of this is to get "close enough" where only some minor tweaking is necessary. It gets you away from building a random setup and then endlessly guessing and trying stuff.
    It comes down to the variables and assumptions that have to be made, atmospheric conditions, ve % changing, turbine efficiency etc.
    The more data you can log, egt, drive pressure, pre and post temps from compressors and ic etc, the better and faster you can dial in your setup.
  5. rsimmons

    May 14, 2024
    Thanks with all your help with this. I think I have a decent plan now. The turbo sizing changed from what I was originally thinking. To stay under 30 psi and have decent efficiency it now looks like a G40-900 for primary and GT2252 for secondary.

    You said the same equation works on turbine maps, but if I try using anything from here the numbers make no sense in relation to what's on a turbine map. I've never been able to find any useful information on how to read them. Every source I find says its a complicated process, but never actually attempts to explains it. Do you know of anything that would actually be helpful in learning this? Seems like they wouldn't be out there if nobody knew how to read them. haha
  6. bbi_turbos

    Dec 8, 2021
    If computing the compressor side takes up 2 lines of paper, computing the turbine side takes up the whole paper. The compressor side is just matching up to the engines air demand, pretty easy.
    The turbine side needs you to know the exhaust conditions (flow, pressure, and heat) so you can figure out how much energy the turbine needs. Which is dependent on what the compressor needs (pressure ratio, efficiency, ambient air temp and pressure).
    Yes, it's complicated.

    GENERALLY with compounds figure the turbines expansion ratio the same as pressure ratio. I'll give a quick example for ya.

    So if exhaust temp is 1500°f /1088k,
    1088÷288= 3.78, sq/rt 3.78= 1.94
    1.94× 62lbs/min= 120. 120÷3 = 40lbs/min exhaust flow, at 1.76 expansion ratio. That point on the turbine map needs to be on a line of the turbine/ housing that you picked. If it's above the line, it means the turbine is too small for that flow point, wastegating needed.
    If that point is below the line, the turbine is to large, is flow capacity exceeds what your trying to do.

    We know the hp turbo has a pr rise of 1.76, so we'll just assume a 1.76 drop in exhaust drive pressure. 3.0÷1.76= 1.70 left over for the lp turbo.
    We also know that there's a temp drop thru the hp turbine, figure 300°f, so 1,000°f exhaust at a 1.7 pressure ratio feeding the lp turbine.
    1,000f/811k÷288k =2.82, sq/rt 2.82= 1.68
    1.68×62lbs/min= 104, 104÷1.7= 61lbs/min.
    Lp turbine is 61lbs/min @1.7 expansion ratio.

    I didn't check this against anything, I just ran the numbers so you could get an idea of how the equations work. You'll have to use your experience to know your egts. Don't rely on the turbine numbers TOO much since your only using basic numbers in the equations. A detailed "work out" of the numbers is needed to nail down the turbine side. You have enough to use it as a guide, and with some instrumentation you could dial in your setup really quickly.
    You'll be surprised how big of a lp turbine you'll need, try not to have to gate the lp charger. A wastegate adds expansion ratio, and expansion ratios multiply across the exhaust stages just like boost does. So a 2psi increase here can make for 16psi in the manifold.
    Last edited: Jun 9, 2024
  7. rsimmons

    May 14, 2024
    Okay, your assumptions for the lp turbo are pretty close to my estimates, so I come out with the same numbers that you do. I know you said the lp turbo may need to be surprisingly large, but when I look at the map, that just doesn't seem feasible.

    Let's go with the 61lb/min at 1.7 ratio, and plot that on the map for the G40-900. Surely a turbo that size on a 2.3L is plenty to make around 630 hp, however 61 lb/min is off the map entirely. The highest value to land on a line in the map is about 38 lb/min with the largest 1.19 A/R housing. I've heard that Garrett turbine maps are different than most, but not sure why. Is this where my issue lies, or am I missing something. The smallest turbine in Garrett's line up where 61 lb/min at 1.7 lands on the map is in their 55 series turbos which can support in the range of 2000 hp with 110+mm turbines. Surely it doesn't need to be THAT big, right??
  8. bbi_turbos

    Dec 8, 2021
    Those numbers were assuming a 1:1 drive to boost ratio. Typically 1.2:1 is a good compromise between streetability and performance. Going higher will give you a quicker reacting/ spooling setup, going lower will give you the most hp/psi.
    If boost is 3pr, then the exhaust is 3×1.2= 3.6pr in the manifold. Run the numbers again and you'll see the flow is lower.

    Garrett typically does their maps in total-to-static pressure. Static is atmospheric pressure, total is atmospheric pressure PLUS the speed of the gas. Gas has mass, which has momentum, and so means it contributes to turbine power also.

    You have to realize the lp turbo doesn't know it's feeding a 2.3L engine. It's feeding a 2.3L engine that is being fed by a turbo, which makes it look like a bigger engine. If the hp turbo is doing 1.76pr, then the lp turbo thinks it's feeding (2.3L×1.76pr=4) a 4L engine. What the usual "gotcha" moment is with compounds, is that since the pr is split, the lp turbo has to move that airflow at a pretty low pressure ratio.
  9. rsimmons

    May 14, 2024
    Okay, if I'm doing this right it makes things a lot closer and brings it down to 51lb/min around 2.2 PR. The map for the G40-900 with 1.19 A/R is around 45lb/min at this point. That would just mean the lp turbo needs to be gated, but 5-6lb/min shouldn't be too substantial, right?
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