Based on some of the comments in your reply it seems to me that since you apparently don't understand some of the basics of diesel engine "Physics" these concepts probably aren't understood by many others as well! In an attempt to not further confuse these issues by jumbling them all together I'll start with your two "incorrect" quotes above and in future posts I'll address your other replies one point at a time.

Now one problem we've had in our past discussions is that you reply to something I've posted and state that it isn't true and then state that instead something else (usually the opposite) is actually true but you seldom offer up any "explanations" to justify your claims. When I post on this site I try to explain in sometimes "excruciating detail" why what I've posted is correct because the only way to advance everyone's understanding of these topics of interest including mine is to discuss them by giving explanations as to the WHY and the HOW!

At least you do bother to reply to many of my posts which I do appreciate and I hope you don't take offense to my use of the word "incorrect" in referring to your two quotes above and I'm not giving these following details to refute your quotes in an attempt to "win an argument" or to prove "how smart I am" because at age 67 and counting my philosophy of life now is to just get through however much of it I've got left with a minimum of aggravation! As I've said before this is my "hobby" for my personal "enjoyment" and by taking my time to explain to others what I've learned about diesel engines during my past 10 years as a fulltime RVer helps with my understanding of them as well.

So lets start with your claim that... "You can have higher HP with lower EGT's"!

Now I'm assuming by HP you mean FWHP and not piston power stroke HP? I've already explained why and how you can have "equal" but not "higher" piston power stroke HP with lower EGTs by increasing the airflow but that when this extra airflow is compressed it causes a net decrease in FWHP even though the piston power stroke HP is unchanged. Now is there anything in my explanation of this point that isn't clear or that you disagree with?

Everyone knows that when they dial up a higher FWHP setting on their DP tuner they also get higher EGTs to go along with that higher FWHP and there's a fundamental reason why a diesel engine must generate higher EGTs when it produces higher FWHP. A diesel engine is a thermodynamic "heat engine" which converts the heat ENERGY that's released by combusting the fuel's chemical ENERGY into mechanical WORK and the TIME rate of producing this WORK is the FWHP that's produced by the engine. This means the FWHP depends on the TIME rate at which the fuel's chemical ENERGY is combusted.

A diesel engine converts about 65% of the heat ENERGY that's liberated by combusting fuel into piston power stroke HP. Some of the combustion heat ENERGY pushes on the piston to produce power stroke HP and in the process of pushing the piston down some of the combustion heat ENERGY is absorbed by the metal surfaces of the combustion chambers and cylinder walls and is carried away by the coolant.

Towards the end of the power stroke the exhaust valve opens and some of the combustion heat ENERGY is expelled from the cylinders as a mass of hot exhaust gas and it's this rejected heat ENERGY in the exhaust gas that causes a higher EGT when more fuel is combusted to get more FWHP!

So according to the basic laws of "Thermodynamics" the general statement in my earlier post that... "you can't make high HP without also having high EGT" ...is correct but since you claim the opposite that... "You can have higher HP with lower EGT's" you must've had some specific example in mind and I'd like to hear what it is.

For example if you could somehow increase the underlying Thermodynamic Efficiency of the 7.3L engine then in principle you could combust the same amount of fuel and convert more the combustion heat ENERGY into FWHP and have less rejected as heat ENERGY in the exhaust gas and that would give higher FWHP with lower EGTs but that's really comparing the FWHP vs EGT characteristics of two different engines. If you then consider only the higher efficiency version of the 7.3L engine my earlier general statement that... "you can't make high HP without also having high EGT" is still correct.

Now I'll address your claim that... "Heat energy" is NOT what spools a turbo. Take a torch and hold it up to the turbine housing. That's heat energy. Will the turbo spool? No. So what spools up a turbo? It's called "mass flow rate"!

Well on FTE I've posted many times in great detail why the above claim isn't true and I've started a few new threads on this topic including one where I gave reference links to member reports that when they switched to B100 their maximum boost "mysteriously" decreased by several psi and I explained that the decrease in boost was due to the fact that for a given fuel flow B100 releases less combustion heat ENERGY and that since it's the leftover combustion heat ENERGY that's rejected in the exhaust gas that spools the turbo it's normal to have lower boost when running 100% biodiesel!

I also did a detailed post for Clux where we worked through a "thought experiment" together in which a 7.3L engine was powered by an external motor and was spun cold with the ignition off up to 3,400 RPM on a 70 F day and that provided a 25 lb/min MAF airflow through the engine but the turbine was only spun hard enough to produce a 2 psi boost! No heat ENERGY means very little boost is produced no matter how much MAF airflow goes through the engine! In this same thread I also calculated the pre and post turbo EGT difference that corresponds to various boost pressures and showed how you could predict the boost level by measuring this EGT difference and then plugging it into my equation!

I started a thread that discussed turbochargers versus superchargers and I pointed out that the reason why using a turbocharger gives an improved fuel efficiency for an engine is that a turbocharger extracts a lot of the leftover combustion heat ENERGY that's rejected into the exhaust and would otherwise be wasted and just exit the tailpipe and I recently did a similar post which discussed a proposal to use the leftover combustion heat ENERGY that's rejected into the exhaust to spool a turbine to power an alternator thereby reducing the parasitic loss on the crankshaft and improving fuel efficiency!

I also gave links to vendors who sell "heat blankets" to wrap around the turbine housing so that more heat ENERGY is captured inside the housing to spool the turbine and thereby improve the turbo's performance as a Thermodynamic "heat engine"!

But I won't bother to give any links to these previous efforts because no one ever bothers to click on my reference links anyway and it's clear from some of your later comments in this reply where you bring up points like the effect of MAT on MAF as if I never thought to consider them when in fact I'd discussed them in detail in the link I gave on compressor wheel size that you didn't check the link I gave! Now with my slow cell connection I know better than most what a pain it is to go read someone's reference link before responding to their post but I wouldn't dream of lecturing someone on a point that might've already been covered in their link without checking it first!

Well since a picture is supposedly worth a thousand words I'll begin this topic again by referring to the two pictures below. In the first one the orange color represents the combustion heat ENERGY that pushes the piston down to produce power stroke HP and the yellow color represents the leftover combustion heat ENERGY that spools the turbine to produce the compressor wheel HP that's need to supply the engine with an enhanced airflow.




On the right hand side in the second picture below note the caption..."Turbine Wheel (converts exhaust ENERGY into shaft POWER to drive the compressor)"! Whoever wrote this caption got it correct and the exhaust ENERGY that's being referred to is the heat ENERGY from the combusted fuel that's leftover from the power stroke due to the inherent inefficiency of the diesel "heat engine" cycle and some of this heat ENERGY is used to spool the turbine thereby providing the POWER to drive the compressor wheel! Since the unit for ENERGY is ft-lb and the unit for POWER is ft-lb/sec it's the TIME rate at which the leftover heat ENERGY is extracted from the exhaust gas that determines the drive HP to the turbine shaft!



The compressor wheel then converts about 75% of this mechanical turbine driveshaft HP back into ENERGY by increasing the kinetic ENERGY of the incoming air molecules. The vanes on the compressor wheel add some ft-lb of kinetic ENERGY to the incoming air molecules by slinging them from the center of the wheel to its perimeter where the molecules are then forced through the narrow gap shown in the above picture.

Then these higher kinetic ENERGY air molecules are captured by the scroll of the compressor housing and since air pressure is defined as molecular kinetic ENERGY divided by volume which is ENERGY density this is how a turbo compresses the incoming airflow! Then as shown in the first picture the compressed air is pushed from the turbo outlet and through the IC and into the intake manifold to produce the given BP so in essence it's the leftover combustion heat ENERGY that produces the BP which is why for a given fuel flow B100 due to its lower combustion heat ENERGY produces a lower BP!

The exhaust gas heat ENERGY is defined as the kinetic ENERGY of the exhaust gas molecules and for an engine with EGR it's the exhaust gas heat ENERGY that allows the higher kinetic ENERGY exhaust gas molecules to be directly recycled back to the intake and to re-enter the engine under pressure! The only difference between EGR and a turbo is that with a turbo some of the exhaust gas heat ENERGY is first extracted to spool the turbo and then clean pressured air is allowed to re-enter the engine instead of dirty pressured exhaust gas. Of course clean is a relative term that does depend on your choice of air filter!

Regarding your comment... "Take a torch and hold it up to the turbine housing. That's heat energy. Will the turbo spool? No."

Well you've got the torch in the wrong location! If you place the torch inside the exhaust manifold so that the heat ENERGY released from the torch can increase the kinetic ENERGY of the air molecules inside the manifold then these higher ENERGY molecules can transfer some of their increased kinetic ENERGY to the turbine blades and spool the turbo! But to make this work in the steady state you've got to pulse the torch on and off at say a 50% duty cycle and use a valve that closes to let the air heat up and the pressure build to spin the turbine and then open the valve to let in some replacement air as the manifold cools down!

 

All those words... Wow. Here's what he's saying Gene, and my example: By changing turbos, the EGTs go down and the HP stays the same or goes up. Why?

Here's my answer and example: Because the smaller exhaust housing becomes a restriction and causes the EGT to shoot way up farther up the power band. Mine did. Today, I have the van turbo and my EGTs are 200 degrees cooler. I didn't lose any HP -- I'm relatively sure of that, maybe gained some. Only the dyno will tell. Or will it? LOL

 

[quote] I also gave links to vendors who sell "heat blankets" to wrap around the turbine housing so that more heat ENERGY is captured inside the housing to spool the turbine and thereby improve the turbo's performance as a Thermodynamic "heat engine"!

But I won't bother to give any links to these previous efforts because no one ever bothers to click on my reference links. Taken from Gene's last post.[Quote] Gene, I would like to see these links, and hear you therory on the blanket wraps, interesting. Post them up or PM them to me, thanks..

 

Well, I must apologize... I never intended this post to become a pissing match as to who understands the physics better about how the compressed air that turbo forces in the motor affects 100 different variables. It appears this thread has gotten completely out of control. I was just looking for first hand experience and opinions as to the best turbo to purchase for the future plans of my application.

Don't get me wrong everyone, I do very much appreciate the posts that were of value and offered what I was looking for. I not trying to split hairs as to fuel/air/temp mixtures, I just wanted a couple of personal opinions as to the best bang for your buck when it comes to turbos.

 

Well "Ceramic coatings" and "heat blankets" for the exhaust manifold, up-pipes, and turbine housing are supported by thermodynamic theory but I don't claim any credit for this theory because it's much older than even me! An additional advantage of using insulation is to reduce the heat soak in the engine compartment and if you do consider installing insulation remember to allow much more than the normal time to cool the turbo bearing before shutting down!

A popular viewpoint is that the turbine is powered by the pistons pushing the mass of exhaust gas past the turbine blades and the following isn't my theory either but these are my words for trying to explain why it's the exhaust gas heat ENERGY that's primarily responsible for spinning the turbine and the pistons pushing on the exhaust gas play primarily a "blocking" role by keeping the heat ENERGY expansion pressure from moving away from the turbine.

While in the combustion chamber the heat ENERGY expansion pressure of the combustion gas is constrained in all directions except the direction that allows the piston to be pushed down and produce power stroke HP. In the exhaust manifold the heat ENERGY expansion pressure of the exhaust gas is constrained in all directions except the direction that inhibits the piston from being pushed up during the exhaust stroke and the direction that pushes on the turbine.

The heat ENERGY in a given mass of high temperature exhaust gas resides in the high speed random motion of its molecules and this random molecular velocity is several orders of magnitude larger than the velocity of the pistons moving up towards TDC during the exhaust strokes. So relative to the rate at which the hot exhaust gas is trying to expand the pistons are essentially moving up in "slow motion" and it's the rate of exhaust gas expansion due to its heat ENERGY content that rapidly moves past the turbine blades and spools the turbo!

That being said in order to complete an exhaust stroke the piston does have to eventually move up against the inhibiting force of the heat ENERGY expansion pressure in the exhaust manifold and this requires some amount of crankshaft HP but not nearly as much crankshaft HP as would be required if you used a supercharger instead of a turbocharger. That's because a supercharger is powered totally by crankshaft HP whereas a turbocharger is powered primarily by the residual heat ENERGY in the exhaust gas!

I found the links below using the FTE search function but with my slow cell connection it takes too much time to try and find some of the specific references I mentioned so just hunt an peck if you've got the time and bandwidth! Well for some reason after I pasted these into FTE the "linking feature" disappeared? Well now I guess you'll have to search yourself.

A Turbocharger Is A "Heat Engine"... ( 1 2 )
ernesteugene

insulating the turbo an up-pipes?? ( 1 2 3 4 )

Went from 17psi to 22psi when I switched from Biodiesel to Diesel
Ed in Maine

Ceramic Up-Pipes - worth it?
Alpine6

I want to develop an idea i had for a diesel engine... ( 1 2 3 4 5 ... Last Page)
parkland

 

Don't apologize. Some people have an overwhelming need to be "right", and will drown anyone that disagrees with them in theory and data but not much actual experience.
it has been an interesting thread as I too I'm considering turbo options. you just have to block out the BS and add others opinions to your own research to make an educated decision. I will probably choose the Garrett when I jump, but its nice to see the other options.

Barney

 

fordzilla01, this is great, does it sound a little borderline on edge, yes, but these guys are getting to the core of everything so we ALL understand the full makeup of a turbo and its function and the end result, whats good for your application, I think its great, We're all learning alot here. Edit:: Like Farmboy said below, and I agree, can't go wrong with the van or 38R, good stuff. Adam, your going to like that deal...love mine, no overkill here, got plans..
Appreciate it Gene, will do...

 

Agreed!
I've been following this thread and found it rather interesting. I didn't think it'd be that hard to explain which turbo is better, and I'm pretty sure everyone here knows how a turbo works.


Basically it sounds like you have a choice bewteen the 38R or a van turbo.

I upgraded to stage 1 SS injectors with a stock turbo, and the thing smokes like a coal plant. I can barely even accelerate without wiffs of smoke coming out the tailpipe. I have a van turbo that I was going to install, but got in on a great deal for the 38R, so now I'm upgrading my engine to fit that. I use my truck for a fair amount of towing on the farm, and I want the power, but no smoke. The 38R may be a bit of overkill for me, but I'm hoping it'll clean the smoke up and keep my EGT's down.

Good luck with your decision, either way you'll be happy!

 

You assume that everyone runs a certain air/fuel ratio when they mod their trucks. They don't. Matt at Gearhead Performance put down 600 hp on the dyno, and his EGT's barely cracked 1100 degrees when he did it.

Why is it my truck at 350 rwhp runs lower EGT's than it did when I made 300 rwhp? In fact, my truck at WOT runs the same EGT's as it did when it was bone stock, making 215 hp at the rear wheels. At the end of a drag strip run, I'm only reaching 1200 degrees. At 300 hp, I was reaching close to 1500 degrees.

There's my examples. It's not so cut and dry. It's all about airflow (and a bit of fuel atomization too).

Now you keep saying that adding air reduces power. You have to realize Ernest, that at some point, introducing more air doesn't really reduce power. That power levels off and you get lower EGT's without sacrificing all that power. But don't tell that to Ben (owner of Double Overtime), who is in the process of putting on a 3rd turbo. That's right, he'll be running 3 turbos soon. He wants more air so he can make more power.

And don't tell that to Gale Banks and his team running the 7 second diesel. They have so much air and nitrous that they can make a run down the dragstrip without blowing smoke. To make a diesel not smoke, it has to be running lean. And as much power as they put down to make a 7 second pass, they have a lot of air.



Now Ernest, as for the "heat energy".

Heat energy is only part of the equation. In your calculations, you left out many factors. So I'll detail it for you since you like math so much. You might actually enjoy this.

The basic equation for mass flow rate is: M = P x V x A

M is mass flow rate
P is density
V is velocity
A is the flow area

Mass flow rate is the mass of a substance which passes through a given surface per unit of time.

Now, since you are so keen on heat energy, let's look at density first. How do you calculate air density? The formula for that is D = (M x P) / (R x T)

D is density
M is the molar mass
P is pressure
R is the universal gas constant
T is temperature

Take note of "T", that's where your heat energy comes from. Heat helps to calculate air density, so yes heat has an effect on spooling up a turbo. But heat is not doing the work all by itself.

Looking back at the rest of the mass flow rate equation, you have heat built in through calculating air density. On top of that, you also have the air velocity. If you were to take a tank of air, and heat it up, if there is no velocity, then it's not powering or moving any object. That air has to move. Just like a pinwheel, there must be a velocity to spin it. So we must have an opening.

Now we all know that hot air expands, creating pressure, and that expansion will push air towards any opening it can find. Pressure doesn't have to be heat related, but rather it can be volume related. Add more air to the same space, it creates pressure. Now add an opening, and this creates air velocity "V". An example? Your typical air compressor. The air coming out is usually cool, not hot, but it's under pressure because an air pump is adding volume, so once there is an opening, the air rushes out to escape. So your example of a pulsing torch spooling a turbo with heat... well my counter argument is an air compressor. Cool air can do the same thing with pressure. The key is it creates air velocity. An engine compresses.... and an engine creates heat. Heat and volume creates pressure, which in turn creates air velocity when an opening (exhaust valve) is available. Now, when we introduce more air to the intake side, we get more air on the exhaust side. So it's not only the extra heat that is driving that turbo, but there is also more air. That's why you can make extra power, spool a larger turbo, and still run low EGT's.

Lastly, that air velocity can change not only due to temperature and pressure, but also the flow area "A". Let's say you go from the 2" stock sized up-pipes to 3" stock sized up-pipes. What will happen? Well the flow area has changed. Heat remains the same out of the engine, but with the larger up-pipes, you will lose mass flow rate. David Lott at Diesel Innovations experimented with different sized up-pipes, and found that going too large will decrease boost, because the exhaust gas isn't moving at a flow rate to keep the turbo spooled like it should.

So there you have it Ernest, I gave you information with equations, as you requested. Conclusion is that heat energy is not the only driving factor in spooling a turbo. Heat energy is only part of the equation. Now I don't want to leave anything out, and the examples and equations I gave are very very very basic, so that anyone reading this can understand it. However, to truly calculate the flow rate of a gas, the equation is actually much more complicated, including things like compressibility coefficient, friction coefficient, pressure variances from start to finish, etc.

Last example. How do you run an engine without heat? Click to see......

http://www.mdi.lu/english/

The Aircar's engine runs on compressed air. You fill an air tank, and use air pressure to push a piston. Since the air is cool, there is no heat energy that's driving the piston. This is a prime example of mass flow rate in motion.

 

As I've already posted in the link I gave I think Garrett's claim that... "The 88mm GT Compressor Wheel Provides 33% More Flow than the Stock 80mm Wheel" ...is comparing the CFM volume airflow of the two different size compressor wheels at a given compressor wheel rpm and I don't think Garrett's claim has anything whatsoever to do with... "the air temperature at the turbo outlet discharge" ...as you suggested above!

A given compressor wheel flows a given VAF=Volume Air Flow ft^3/min=CFM at a given compressor wheel rpm and in the link I pointed out that the GTP38R 88mm wheel is only 10% larger in diameter and only 21% larger in disk area than the stock GTP38 80mm wheel so that some of the 33% increase in CFM airflow must be due to the differences in the blade design and that the GTP38R compressor wheel appears to be a "glorified" version of a WW=E99 compressor wheel and for sure the GTP38R compressor wheel doesn't look anything at all like a stock L99.5 compressor wheel. So I think the "33% More Flow" claim relates to "CFM volume airflow at a given compressor wheel rpm" and that this increase is due to differences in both the size and the blade design of the GTP38R compressor wheel versus the stock compressor wheel.

I think a question of much interest to many is what happens if you just change from a stock turbo to a GTP38R and keep everything else the same? Some people seem to think they'll get a 33% increase in the MAF airflow going into the engine and that this will increase their FWHP! Well what I think would happen is that if they did get a 33% increase in MAF and if they were previously operating at an AFR of say 18 or higher so that they weren't making any smoke their FWHP would actually decrease! If they were previously operating at an AFR of say 17 or lower and making some smoke their FWHP would increase and the lower their initial AFR was the more the FWHP would increase but if some of the smoke was due to a weak HPOP the increase in FWHP wouldn't be as much as might be expected from their initial AFR!

If we consider a comparison test where we keep everything else the same so that the GTP38R and stock turbos are operated at MAF vs TPR points on their respective compressor maps where the compressor wheel rpm is the same then you'll get a 33% increase in CFM volume airflow at the turbo inlet and if the air density at the air filter inlet remains the same for the comparison test then you'll get a 33% increase in MAF going into the engine. So the question is where are "candidate" MAF vs TPR operating points on the respective compressor maps where the compressor wheel rpm is the same?

Well here's what I've come up with so far. Below is a GTP38R compressor map and as you can see the compressor wheel rpm curves range from 50,000 rpm in the vicinity of MAF=36 lb/min and TPR=1.2 to 110,000 rpm in the vicinity of MAF=60 lb/min and TPR=3.3.



I thought I had a compressor map for a stock turbo that also showed the compressor wheel rpm curves for it but if I did I can't find it on my computer and I can't find one on the internet either! What I did find is the comparison maps below but they don't include the wheel rpm curves.



So for now I've given up on trying to do a comparison analysis based on keeping the compressor wheel rpm the same but since it's clear from the above comparison maps that a GTP38R has the "potential" for supplying more airflow I'll address the more general question of what's required to get a given increase in MAF with a GTP38R turbo while still keeping everything else the same at least as much as is possible and for a reasons I'll give later I'll analyze what's required to get a 30% increase in MAF!

Well it requires more HP (from somewhere) to move 30% more CFM airflow through the restriction of the air filter element and then pressurize this ingested airflow in the compressor and then push it through the restriction of the IC=Intercooler and into the intake manifold to produce a given MAP=Manifold Air Pressure psi which corresponds to a given BP=Boost Pressure psi and then through the intake valves with a given VE=Volumetric Efficiency and finally on into the cylinders to produce a given CAP=Cylinder Air Pressure psi where VE=(CAP)/(MAP).

Also this additional HP (from somewhere) needs to be applied to the turbine driveshaft as TDHP=Turbine Drive HP in an amount that's sufficient to produce a 30% increase in MAF and I claim that this HP (from somewhere) comes from the leftover combustion heat ENERGY in the hot exhaust gas that's generated by a given fuel flow and that the amount of heat ENERGY per sec that's removed from the hot exhaust gas as it flows past the turbine is what supplies the input HP to the turbine wheel so that this TDHP can then power the compressor wheel and supply it with a sufficient amount of AFHP=Air Flow HP to pull and then push a given CFM airflow through the intake plumbing and into the engine!

Well to do a comparison analysis I need to establish a "baseline" and since the "33% More Flow" claim is a comparison to a "stock turbo" I'm going to use a "stock BP=17.2 psi" and then calculate the baseline MAF. For a 7.3L PSD the MAF is given by... MAF={(0.347)(VE)(RPM)(AAP+BP)}/{(MAT+459.67)} lb/min ...I'll also use VE=0.812, RPM=2,900, AAP=14.7 psi, BP=17.2 psi, MAT=126 F ...where this is the "steady state" value of MAT which occurs after the engine has been operated at a steady BP=17.2 psi for about 1 minute or more.

So the baseline MAF is... MAF={(0.347)(VE)(RPM)(AAP+BP)}/{(MAT+459.67)}={(0.347)(0.812)(2,900)(14.7+17.2)}/{(126+459.67)}=44.5 lb/min and a 30% increase in the baseline MAF gives a MAF=57.8 lb/min and this is achieved at a BP=29.5 psi where the VE is 96.7% of it's previous value due to the higher airflow turbulence at the higher boost so the new VE=0.785 and the new MAT is higher because "the air temperature at the turbo outlet discharge" is higher and the new steady state MAT=144 F.

If you plug these new numbers into the MAF equation you get... MAF={(0.347)(0.785)(2,900)(14.7+29.5)}/{(144+459.67)}=57.8 lb/min so we see everything checks and the reason I used a 30% increase in MAF and a "stock BP=17.2 psi" is because they allow me to pick exact values from my computer model!

I plotted the baseline operating point MAF=44.5 lb/min, BP=17.2 psi, TPR=2.30 and the 30% increase operating point MAF=57.8 lb/min, BP=29.5 psi, TPR=3.30 on the comparison map below. If you look at the "official" GTP38R compressor map in the first graph you'll see that the 30% increase operating point MAF=57.8 lb/min, BP=29.5 psi, TPR=3.30 is at the top limit of the map on the compressor wheel rpm curve for 100,000 rpm but on the "promotional" GTP38R compressor map this same 30% increase operating point appears to be well within the map limits!



Well until proven differently I'm considering the "promotional" map to be "advertising hype" and the "official" map to be the actual performance! It's interesting to note that an exact 33% increase in the baseline MAF of 44.5 lb/min is 59.2 lb/min which at 2,900 RPM corresponds to a BP=31.2 psi at a TPR=3.44 and this point is past the 100,000 rpm curve and well off the "official" map! Of course Garrett could define a slightly less challenging baseline than the one I used and then their 33% increase would lie on their "official" map!

So the next project is to calculate the FWHP and GPH fuel flow for the baseline MAF and the 30% increase MAF and the required TDHP=Turbine Drive HP for each case and do some more thinking and see what I can come up with! If anyone has any suggestions, helpful hints, or comments please give them!

 

Well I agree with the first part of this statement but as to the correctness of the last part... "a stock turbo is not" ...as an infamous ex-president once said... "it all depends on what the definition of is...is"!

My analysis shows that for a BP=25 psi the stock turbo is within its compressor map using a stock L99.5 compressor wheel and it's well within its compressor map using a E99 (WW) compressor wheel! Now I've advised many times not to run a stock turbo at a 25 psi BP too frequently but that's to save wear and tear on its bearing not because you're running it outside the bounds of its compressor map! Well since someone might want to know what my definition of "within its map" and "well within its map" is I'll give a calculation below.

The first step is to relate the BP=Boost Pressure psi to the TPR=Turbo Pressure Ratio because the "corrected" MAF=Mass Air Flow lb/min at a given TPR is what defines an operating point on a turbo compressor map. If anyone's interested they can read the details here... https://www.ford-trucks.com/forums/8...ml#post7228722 ...for how I derived the following equation for BP...

BP={(AAP)(TPR-1)}-{(TPR)(AFPD/27.68)}-(ICPD) psi

...where AAP=Atmospheric Air Pressure psi, TPR=Turbo Pressure Ratio and it's the ratio of the Turbo Outlet Pressure to the Turbo Inlet Pressure, AFPD=Air Filter Pressure Drop Inches H20, and ICPD=Intercooler Pressure Drop psi.

From the above equation you see that ICPD trades one-for-one with BP. For example a 0.5 psi reduction in ICPD would provide a an equal 0.5 psi increase in BP so a lower restriction IC appears to provided a lot of bang for the buck but since it's not all that restrictive to begin with the best you can likely do is pick up a few tenth's of a psi in BP!

The effect of AFPD is magnified by the TPR! For example at a TPR=2.85 a AFPD=12.5" H20=0.45 psi causes a 1.28 psi reduction in BP compared to having no air filter at all! But this example is for a 7.3L AIS and if you change to a lower restriction air filter like a large K&N cone you only reduce to about a AFPD=8" H20=0.29 psi and this still causes a 0.83 psi reduction in BP for a net increase in BP you only 0.45 psi as the "bang" for the "penalty" of ingesting all that dirty air!

The effect of AAP is magnified by (TPR-1). For example at a TPR=2.85 a sea level AAP=14.7 psi gives a {(AAP)(TPR-1)}={(14.7)(2.85-1)}=27.2 psi contribution to BP but at a 5,500 ft altitude a AAP=12.0 psi gives only a {(AAP)(TPR-1)}={(12.0)(2.85-1)}=22.2 psi contribution to BP which is a 5.0 psi reduction in BP compared to sea level! This is why towing at altitude is so hard on your turbo because it has to work at a higher TPR to make the same BP that it can more easily generate at sea level!

So lets use the following values... TPR=2.85, AAP=14.7 psi, AFPD=12.5"=0.45 psi, ICPD=0.93 and see what BP we get... BP={(AAP)(TPR-1)}-{(TPR)(AFPD/27.68)}-(ICPD)={(14.7)(2.85-1)}-{(2.85)(12.5/27.68)}-(0.93)=(27.2)-(1.28)-(0.93)=25.0 psi!

So for this example a TPR=2.85 corresponds to a BP=25.0 psi and now we need to calculate the "actual" MAF airflow into the engine for a BP=25 psi and then "correct" that "actual" MAF to the standard reference temperature and pressure using the procedure specified on the Garret website. Here's a few points to keep in mind when using turbo compressor maps...

1) A turbo compressor map specifics only the "compressor performance" independent of the actual "turbine performance" by using a controlled input HP to drive the turbine shaft.

2) The "compressor performance" is measured and plotted on a map for standard "reference" values of TIAP=Turbo Inlet Air Pressure psi and TIAT=Turbo Inlet Air Temperature F.

3) For Garrett compressor maps the standard reference conditions are... TIAPr=13.95 psi and TIATr=85 F ...but some other OEM might use the references TIAP=14.7 psi and TIATr=60 F or some other reference for their maps so you just can't compare one OEMs map with another's map until you do the conversions that I discuss below.

The way most people use a compressor map is to look at a candidate operating point on it at some particular value of TPR and "reference" MAFr and then ask if I operate at that point on the map what will my "actual" MAFa be at that TPR? Well in general the answer is... MAFa={(TIAPa)/(TIAPr)}{(TIATr+460)/(TIATa+460)}^0.5{MAFr} ...and for a Garrett map in particular just plug TIAPr=13.95 psi and TIATr=85 F into this equation along the MAFr from the point on the map and the values for your actual TIAPa and TIATa.

Say for example you have an operating point on a Garrett compressor map that gives a MAFr=50 lb/min and you want to calculate the airflow at sea level where AAP=14.7 psi and say the ambient temperature is 70 F and the inlet airflow temperature increases by 15 F before reaching the turbo inlet. That means TIATa=85 F=TIATr so no correction is needed due to temperature. If at the airflow in question the air filter restriction R just happens to be 20.76" H2O which is 0.75 psi then the TIAPa=AAP-R=14.7-0.75=13.95 psi=TIAPr so no correction is needed due to pressure.

I used my model to calculate the actual MAFa at the values of TPR and RPM shown below and then I corrected the actual MAFa to the reference MAFr using this equation... MAFr={(TIAPr)/(TIAPa)}{(TIATa+460)/(TIATr+460)}^0.5{MAFa} ...and I chose an ambient air temperature so that the TIATa=85 F=TIATr so no correction was needed due to temperature and I only needed to correct for the different air filter restrictions at the different operating points.

The points below are for a TPR=2.85 which gives a BP=25.0 psi.

RPM......MAFa.....MAFr

2,400.....45.9.......44.9=1

2,800.....52.2.......51.1=2

3,200.....57.0.......55.8=3

The points below are for a TPR=2.5 which gives a BP=20.4 psi.

RPM......MAFa.....MAFr

2,400.....41.5.......40.4=4

2,800.....47.1.......45.8=5

3,200.....51.5.......50.1=6

The points below are for a TPR=2.1 which gives a BP=15.1 psi.

RPM......MAFa.....MAFr

2,400.....35.8.......34.6=7

2,800.....40.7.......39.3=8

3,200.....44.5.......43.0=9

I then plotted the 9 points on the Banks "promotional" compressor map which compares the stock L99.5 wheel with the "Banks wheel" which of course is just a repackaged E99 (WW) compressor wheel and that's shown below.



The reason I did this is because the Garrett "promotional" compressor map shown below is wrong as it "promotes" the GTP38R as having the same "surge line" that a stock turbo with a stock L99.5 wheel has!



I'm still looking for my other turbo maps but I did find the map below where years ago I'd overlaid the E99 "surge line" on a GTP38R map and discovered that in the range of interest it had about the same "surge line" as the GTP38R! Based on the Banks "promotional" compressor map If you do the WW mod I'd say you're well within the bounds of the compressor map at BP=25 psi but I don't think anyone should make it a routine practice to push their stock turbo that hard.



I've decided the best way to do the analysis I mentioned in my last post is to take the time to build an engine model for a 7.3L that calculates what's happening inside the cylinder every 1 crankshaft degree during the compression stroke and power stroke including the injection of fuel etc... so that will take some time to do.

Does anyone know the length of a 7.3L rod? I need the length-to-stroke ratio. If someone's rebuilding an engine a measurement of how far the piston descends in the cylinder as the crankshaft is rotated 90* from TDC will also tell me what I need to know. It should go down about 2.4" which is about 57% of the 4.18" stroke!

 

Please remember the parts about... "the air is cool", "no heat energy", and "mass flow rate in motion" when you get to the end of my replies! This time around I'll start with your last comment and work my way back up the list to the first ones including your equations and I'll use them along with a few others to do some calculations which will answer questions like... How much HP do you need to apply to the turbine's driveshaft so that the compressor wheel can generate a given psi boost in the spider? ...and... Where does this turbine driveshaft HP come from and how do you calculate how much of it is available to apply to the turbine's driveshaft?

Due to my cell connection when I clicked on your link all I got was the main screen which showed some tiny golf cart size vehicles and then everything bogged down but I did see enough to comment that promoting compressed air powered vehicles as a solution to our ENERGY problems is another example of blatant opportunism and that site reminded me of my quote at the conclusion of my three part series which begins here... The true story of how HP and TQ were actually discovered...Part 1 of 3 ... http://thedieselgarage.com/forums/showthread.php?t=358 ... where I said ...

Well it turns out that GM gets the bailout not Ford but for very small flywheels a 1 million RPM has been achieved and if you do the math you'll see that using a flywheel to store the ENERGY is much more efficient than using compressed air to store the ENERGY! You could even have a water powered car by using a storage tank on the roof to store potential ENERGY to power a water-wheel motor! As I pointed out in post #8 here... Strategies for reclaiming waste heat ENERGY equivalent HP... https://www.ford-trucks.com/forums/8...ml#post7635931 ...

If you calculate the total amount of heat ENERGY that's stored as combustible chemical ENERGY in a 30 gallon tank full of #2 diesel and compare that to the total amount of electrical ENERGY that can possibly be stored in a battery or to the total amount of air molecule kinetic ENERGY that can possibly be stored in a tank of compressed air or to the total amount of mechanical kinetic ENERGY that can possibly be stored in a flywheel you'll see why diesel powered trucks won't be disappearing from the roads anytime soon but they might eventually be powered by a pollution free "steam engine" where its boiler is supplied with heat ENERGY from a miniature "fission reactor" and when you use E=Mc^2 to calculate the total amount of ENERGY that's stored in 0.1 lb of fissionable material you'll see that all other forms of ENERGY storage pale in comparison to "nuclear ENERGY"!

Well getting back to the topic at hand of "what powers the turbo?" in the first paragraph of the above thread I also said...

...and when you stop and think about it the entire truck including the diesel engine and all of its accessories such as the alternator, the water pump, and the A/C compressor are all actually powered by the heat ENERGY that's stored in its fuel tank because that's the only source of ENERGY that's available! It might seem strange to some but an air conditioner is just a "heat pump" running a reverse cycle and it's truly a "heat engine" even when its prime source of ENERGY comes from the electrical grid!

In general HP is defined as the TIME rate at which any form of ENERGY is converted to mechanical WORK. Independent of the specific details of the actual conversion process that's used if you convert 550 ft-lb of any form of ENERGY into mechanical WORK in a TIME of 1 second you've generated 1 HP!

I use the term "heat ENERGY equivalent HP" to define the "equivalent HP" that's potentially available from a given amount of "heat ENERGY" and each gallon of #2 diesel contains 101,140,000 ft-lb of heat ENERGY! If the conversion process is 100% efficient then each 1 gal/hr of #2 diesel generates {(101,140,000}/{(3,600)(550)}=51.08 HP! Dividing by the factor (3,600) converts the ENERGY consumption from hours to seconds and dividing by the factor (550) converts the ENERGY consumption into units of 1 HP!

Now lets compare a "supercharger" to a "turbocharger" which is normally just called a turbo. The supercharger receives its drive HP "directly" from the crankshaft by mechanical means which involve converting the linear motion of the pistons into a rotating motion of the crankshaft but "indirectly" the drive HP for a supercharger is coming from the TIME rate at which heat ENERGY is being released from the combusting fuel.

The turbo has no direct mechanical connection to the crankshaft so the turbo must receive all of its drive HP by somehow extracting enough ft-lb of kinetic ENERGY per sec from the exhaust gas molecules to provide the HP that's required to spin the compressor wheel so that it can produce a given psi boost at a given CFM airflow. The exhaust gas molecules contain two forms of molecular kinetic ENERGY that can be converted into turbine driveshaft HP.

During the exhaust strokes the pistons push the entire mass of exhaust gas molecules (which each have an individual mass Mg) along the plumbing leading to the turbine with a translational velocity Vt and this gives each of the exhaust gas molecules a translational kinetic ENERGY KEt=1/2MgVt^2.

During the exhaust stroke the MPV=Mean Piston Velocity MPH is the average velocity of the piston as it moves up from BDC to TDC and is given by... MPV={(Stroke)(RPM)}/(528) MPH ... so for a 7.3L at 2,800 RPM you get a... MPV={(4.18)(2,800)}/(528)=22.2 MPH. The 7.3L Bore is 4.11" and if the diameter of the final pipe leading to the turbine housing where the up-pipes join together also has a 4.11" diameter then the translational velocity Vt imparted to the exhaust gas molecules in that section of pipe during the 90* crankshaft rotation of a single piston's exhaust stroke would be the same as the mean 22.2 MPH piston velocity during that exhaust stroke.

The residual heat ENERGY in the exhaust gas which is determined by its EGT causes the molecules to buzz around like a swarm of angry bees in every possible direction with a random velocity Vr and this gives each of the exhaust gas molecules a random kinetic ENERGY KEr=1/2MgVr^2.

If you check out this link... Kinetic Temperature... http://hyperphysics.phy-astr.gsu.edu...ic/kintem.html ...and scroll down to the Molecular Speed Calculation section and use the conversion C=(5/9)(F-32) to see that an EGT=1,250 F is an EGT=677 C and then plug 29 amu for dry air and 677 C into the boxes indicated you'll see that the random RMS velocity of the molecules in the exhaust gas due to their residual heat ENERGY is 2,022 MPH!

This means the amount of random kinetic ENERGY in the exhaust gas (heat ENERGY) is larger than the amount of "piston pushing" translational kinetic ENERGY (mass flow ENERGY) by a factor of (Vr/Vt)^2=(2,022/22.2)^2=8,296. Now just because the exhaust gas molecules contain roughly 8,000 times more random kinetic ENERGY due to the residual heat ENERGY in the exhaust than they do "piston pushing" translational kinetic ENERGY due to their mass flow doesn't prove that it's the heat ENERGY that spools the turbine but my wife says I've had enough fun for now so I'll have to continue this story at a later time...

 

Is there a map around with the Late 99 wheel in the ATS housing? I'm curious now...

Specifically, I wonder (for obvious reasons) what the van turbo map would look like with the late wheel in the ported compressor housing. hhhmmmm...

 

Wow! Where have I been? I have some serious catching up to do.

 

Heat energy alone can't explain why at higher altitude, your EGT's rise, yet your turbo is laggier and you lose power and boost.

Heat energy alone can't explain why you can get a turbo to spool slower at 1400 degrees EGT's, but spool it faster at 800 degrees EGT's.

You need to know mass flow rate. In the altitude example, you have to account for pressure and air density variations at different altitudes. In the EGT example, you have to account for velocity, as an engine at low RPM's with spiking EGT's really high won't spool a turbo nearly as quickly as an engine running high RPM's with lower EGT's.

See, it's a much bigger picture than just heat energy alone.