Turbo Versus Roots Net Horsepower

[Interested Observer]

Hotrod,
PV=nRT, or, P=nRT/V.   T1/V1 = T2/V2  => P1=P2

Smaller T and smaller V; Pressure is the same.

The pressure is imposed by the turbo, the volume (or density) is automatically adjusted to accommodate the temperature change. 

Suppose we had a magical intercooler with no frictional losses.  What you are proposing is that the pressures on either side of the frictionless cooler are different!  How can that be?  Why wouldn’t they equilibrate?

Dynoroom,
There is a difference between simply asserting a “basic” concept, and explaining it. 

[McRat]

The thing about turbos for racing apps is that they are somewhat self-adjusting, and crank driven stuff is tied into RPM.

This is why a turbo acts like a cam swap, and the crank-driven stuff doesn't.

If your desire is maximum HP at maximum sustainable RPM, with a continuous aggressive climb in the HP curve (can't use it if you can't get there), you use a turbo.  If you need over 60 PSIG worth of boost, you'll need two of them, staged.

That 2003 Hot Rod article appears to be using a My-Little-Pony or Playschool Brand turbo.  A turbo that loses power before 6000 is way to small for a gas engine, or tuned badly.  But even so, in the Hot Rod Article the turbo wins even though it was set up wrong.

[fredvance]

You might get a book by Corky Bell "Maximum Boost and Supercharging" very informative. You can call Corky at 830-438-2890. He is a good guy to talk to. A master of turbocharging.

[38flattie]

You might get a book by Corky Bell "Maximum Boost and Supercharging" very informative. You can call Corky at 830-438-2890. He is a good guy to talk to. A master of turbocharging.

Fred, thank you!

I just had a 45 minute discussion with Corky. He's a very nice guy, and very helpful. He's going to be our 'advisor', on the fuel and turbo setup.

I laid out my goals, and detailed the description of the current engine setup. He thinks we can easily meet my goals, with about 20.5 psi boost, and keep dual turbos in the max efficiency range.

He said that was the easy part-keeping the engine in one piece may be much harder!

[Bob Drury]

  Just to add to the conversation, my friend Chuck Smithfield who is a long time drag racer, and used to come to the salt with LeVan Prothero, built a Blown destroked small block Chevy which set the NHRA record in BB/A about five years ago.
  Chuck put a 1471 blower on top of the "small" small block, and ran it drasticly under driven.
  The reason was to keep the air temp down while still providing the neccesary boost to get the job done.
  Chuck spent about five years making this motor work and it is quite a piece.
  He modified Brodix "Buick" heads with a second set of spark plugs inside the valve covers!
  He was seeing high EGT readings when running the motor with a single MSD front mounted magneto, so he knew that with the spark plugs (in the standard location) showing he was safe, that he was burning fuel in the exhaust exiting the combustion chambers.
  Chuck then added a second ignition (a MSD crank trigger type), installed it in the standard distributor hole and added the second set of spark plugs near the rocker arms.
  The spark plug wires were routed under the blower manifold and through slots in the upper base of the valve covers.
  He ran this second ignition slightly retarded and the results were astounding. After being told by MSD that this setup wouldn't work, he  proved them wrong.
  He not only dropped his EGT readings by over a hundred degrees but also blew the National Record away in BB/A and held it for so long that the NHRA retired the record.
  Incidently using a Lenco five speed Peanut transmission which was manually  lever shifted, He launched at 9,000 rpm, pulled second gear at the tree (9,800) and hit the quarter at 10,000 rpm.
  When I asked him what happened if he hit 10,300 in the lights he smiled and pointed to a pile of well ventilated aluminum oil pans......... Bob

[hotrod]

That is probably true, the large hotside would spool slower and if you walked into the throttle off the line it would take a few seconds to get on boost.
If you did not wind out low gear it might not get on full boost until you were under load in second gear. Turbos don't really come on strong until the engine is under high load (high exhaust flow volume and exhaust port pressures). In drag racing this can be a problem as some turbos will not reach target boost at all in low gear because engine load is never high enough, and bigger turbos can fall out of boost on shifts if they are a bit slow.

There are some things that the tuner can do with electronic engine management to help manage boost. For example retarding ignition timing briefly at midrange rpm can significantly improve turbo spool because the late ignition timing means the mixture is still burning as the exhaust valve is opening. You lose a bit of power from the engine but all that unused energy is now available to the turbocharger hot side turbine to get it spooled up.

Getting the right size balance between the hot side turbine and the cold side compressor is one of the tricks to a good turbo setup. Too small of a compressor won't flow enough air at high enough boost, make it too big and the hot side turbine may not be able to spin it up. Make the hot side smaller and it will spin up faster but might be too small at the top end and strangle the exhaust side of the engine and go into choke flow where no matter how high the exhaust pressure goes it simply cannot stuff enough exhaust gasses through the turbine. This is where the turbo wizards come in and start fiddling with wheel size on both the hot and cold side and the "trim" of the turbo compressor and the hot side turbine to get the right balance.

Since the turbo is essentially a small gas turbine you also have to manage heat so you deliver hot exhaust gases to the turbo without cooking the rest of your engine compartment.

Larry

[hotrod]

Suppose we had a magical intercooler with no frictional losses.  What you are proposing is that the pressures on either side of the frictionless cooler are different!  How can that be?  Why wouldn’t they equilibrate?

Because it is a dynamic system.

In a magical friction free intercooler the pressures would try to equilibration but the volume flow would be where most of the adjustment took place. You would be pushing say 300 cfm of very hot air into the intercooler at one boost pressure  and getting 250 cfm out of air at about the same pressure but much lower temp.

In a real system you get changes in both. There will be  volume drop across the intercooler along with a pressure drop as the temperature goes down.

It is like an electronic circuit. You only see voltage drops across resistors when there is current flowing. Once the current goes to zero all your voltage drops go away too.

Larry

[Interested Observer]

Well, except for friction losses, generally a few percent, the pressures on either side of an intercooler will be the same.  That is obvious from the fact that the cooler core offers little or no resistance to the flow of air, which allows the two sides to equalize across the core.  It has nothing to do with cooling, heating, or the price of pickles.

Sometimes ingrained notions are hard to change.

[dw230]

This got into a P'ing match that I have no knowledge of.

DW

[1 fast evo 2]

Mike, please explain "all boost pressure is NOT equal". Other than the parasitic drag, what is the difference? It seems to me, that 14.2 psi at a set temp, in a 366CID engine, would always be the same. Can you please expound on this?

Turbos are interesting critters. Lets suppose you have two turbos. Your engine needs 600 cfm of air flow at 14.7 psi. One turbo has its sweet spot at 450 cfm at that boost pressure and the other has a sweet spot of 600 cfm at 14.7 psi.

The first might deliver 600 cfm of air at 14.7 psi but its discharge temp is 250 deg F, the second delivers the same 600 cfm at 14.7 psi but its discharge temp is 150 deg F.  That difference in discharge temp is wasted power that came from the exhaust flow of the engine.

The turbo that has the 250 deg F discharge temp might have an exhaust back pressure of 1.8x boost pressure so it is fighting an exhaust pressure of 26.46 psi when the cam is in overlap as it is trying to stuff air in the cylinder at 14.7 psi. Needless to say the exhaust burps back into the cylinder until the exhaust valve is closed contaminating the intake charge. This is why most turbo engines have different cam timing than the same engine with no blower or a mechanical blower, often a rapid closing of the exhaust valves and little overlap.

The properly sized turbocharge might only have an exhaust back pressure of 1.2x boost pressure, putting exhaust pressure at 17.64 psi. This means less blow back into the cylinder as the exhaust valve closes and the engine has to do less work pushing the exhaust out of the cylinder against the turbo back pressure in the exhaust system.

Not only that, if your run both turbos through an ice over water intercooler tank, the very hot turbocharger will lose more pressure due to cooling of the hot air charge, so it will really have to do the work to make perhaps 17 psi at the discharge outlet to actually deliver 14.7 psi at the intake manifold, where the properly sized turbocharger might only need to make 15.2 psi to overcome intercooler losses and still deliver 14.7 psi at the intake manifold.

(these are all just wild guess numbers but illustrate the trade offs you must make as you size the turbocharger.)

Your real power gain is due to the total increase in useful mixture in the cylinder when the valves close, and how much power you can make burning it, minus the power you will use up getting the mixture in there.

In a belt or crank driven blower the energy cost is directly from the crank power output. In a turbo there are still power losses but they are hidden in things like exhaust gas back pressure and the work the engine needs to do pushing the exhaust gases out against that back pressure, and the need to use a short overlap cam to keep from getting excessive exhaust gas contamination in the cylinder due to high pressure exhaust gas back flowing into the cylinder before the exhaust valve closes.

In a mechanical blower you can throw away mixture with over lap and get great scavanging so all the mixture in the cylinder is uncontaminated. In the turbo not always true, although a turbocharger with a huge hotside might have very low backpressure but the trade off is it spools slowly and may be lazy to come on boost. A smaller hotside that spools fast might come into boost violently when the engine comes under load.

In the real world you need to find a middle ground between those extremes. Fast enough spool so you have good and predictable throttle response, but low enough exhaust gas back pressure that you don't throw away too much of your power overcoming that restriction.

That is where the art of turbocharger setup is. Picking the right turbo size and compressor trim to get the right airflow and boost pressure with good throttle response, and a hot side sized to give reasonable spool up and low back pressure.

Then you throw in a few tuning tricks to manage boost by juggling fuel mixtures and ignition timing to bring on boost or soften boost onset at critical rpm ranges then keep all those balls in the air so everything works together in harmony.

Larry

38flattie sorry I didn't get back to this sooner but Larry gave you a very well executed explanation. I agree with everything he has posted above(and later for that matter).

[jl222]

 
  Yeah..Evo I agree with Larry also. I know a lot of intercooler manufactures advertise no pressure drop but they
must be testing with no temp or coolant added.

  We always showed a temp drop across the intercoolers, ''old setup'' temp recorder broke for use on new intercooler.

  Will try to post some temp and pressure records from ''real life'' Bville and El Mirage runs later today.

  We have always taken boost and temp recordings for over 20 yrs now. We tried a new data system for intake

temp out of blower and before intercooler [ + other sensors] but can't get system to work.

  F.A.S.T.---EFI records boost and temp but only after intercooler.

            JL222

                   

 

[jl222]

Hotrod,
PV=nRT, or, P=nRT/V.   T1/V1 = T2/V2  => P1=P2

Smaller T and smaller V; Pressure is the same.

The pressure is imposed by the turbo, the volume (or density) is automatically adjusted to accommodate the temperature change. 

Suppose we had a magical intercooler with no frictional losses.  What you are proposing is that the pressures on either side of the frictionless cooler are different!  How can that be?  Why wouldn’t they equilibrate?

Dynoroom,
There is a difference between simply asserting a “basic” concept, and explaining it. 

  ''In the interest of accuracy'', according to the book [ chemical principles by P.W. selwood] Interested Observers formula
is asx backwards.

   ''Charles' Law---If the absolute temperature T of a gas sample is changed to T2 it will be found that the new volume V2 is related to the old volume V1 by the expression '' 

                                                              V1/T1 = V2/T2 OR  V2=V1 X T2/T1

  ''This is the mathematical expression for Charles Law which may be stated as follows''

  ''The volume of a fixed mass of gas varies directly as the absolute temperature provided the pressure does not change''
 
  Charles' Law is volume 1 divided by temp 1 = volume 2 divided by temp 2, not the temp divided by the volume.

 I really like this book because it shows problems worked out + charts to look at. A little of chart on sample of hydrogen gas

                 temp T1               Volume  V       
                 deg Kelvin K            ml or millimeter
                  218                        420
                  300                        578
                  385                        739
                  450                        868
                  610                       1180 

   By looking at the 300 and 610 temps and their volumes you can see that by cooling from 610 deg to 300 deg the volume
 is half as much
 
   Well I might as well type in some other gas laws copied out of this chemistry book

   Boyles Law

 ''If the pressure P1 exerted by, and on, a gas sample is changed to P2, it will be found that the new volume V2 is related to the old volume V1 by the expression''

                                               P1V1 = P2V1

  '' This is the mathematical expression for Boyles Law which may, in words, be stated as follows;''

     ''The volume of a fixed mass of gas varies inversely as the pressure, provided the temperature does not change.''

    '' Boyle's Law may be written as this''

                                                         V2= V1 X P1/P2

       Chart showing effect of pressure on sample of hydrogen gas at 25 degree C

                    Pressure P                                 Volume V
                     mm mercury                                ml

                     50.0                                         884
                     170                                          250
                     310                                          413
                     498                                           88.8
                     760                                           58.1
                     1000                                         44.1
 
    By looking at the 498 and 1000 pressures and volumes you can see how doubling the pressure shrinks the volume in half

  Its getting late, going to post before I lose this, wil post formula for temp pressure and volume later


    JL222

  p.s. glad I never took chemistry as later chapters or really tough

[38flattie]

JL222, I'd love to see "some temp and pressure records from ''real life'' Bville and El Mirage runs"!

[Interested Observer]

Re: reply #56--
Apparently JL222 skipped algebra as well, since V1/T1=V2/T2 and T1/V1=T2/V2 are fully equivalent to each other and both yield V2=V1xT2/T1.

And he may be pleased to know that both Charles’ and Boyle’s laws are simply particular cases of the more general “Ideal Gas Law”, (PV=nRT), and are easily derived from it. 

[jl222]

 INTERESTED OBSEVER  Problem out of ''Chemical Principles''

  IF the volume of a gas is 150 ml at 298 deg K what would the volume be at 653 deg K assuming the pressure does not change?

                        V2= V1 X T2/T1    SOLUTION   V2= 150 ML X 653/298
                                                                         150ML X 2.19=  328.69 ml  correct


  Interested Obsever's   V2= VI X TI/T2  SOLUTION  V2= 150 ml x 298/653
                                                                             150ml x .456 =  68.45 ml  incorrect go back to class

                      JL222