Your chit isn't even running Joe. But neither is Gene's as he got rid of it. I take your advice very seriously as I do Genes, not because you are from the same school, but you both bring something to the table. I don't have to accept anybody's chit, like the TN hoopla that was all the buzz, but I like information. I can choose what I accept. I have purposely left the vendors name out of this because that was such a pittiful situation.

I may not always agree with Gene, but he is one fart smellow. I look forward to more posts. You don't have to read them.

BTW, what are you lacking in the 7.3 model Gene?

 

First statement is 100% true with me!! I think I even mentioned it in another thread some time ago. Also, I'm still tuning into see if Gene can hit the maximum word count per post!!!

 

I'm going out on a limb here, but I think this post and especially that last statement are the kinds of things that drives Gene nuts. I'm purposely staying out of the line of fire in this thread, only because I've been down the road a number of times already. To me, experience *and* numbers matter. To Gene, it's all about the numbers no matter what. The problem I've always had is that when you have experience that disagrees with numbers Gene has, he dismisses it. I gave up arguing. Have fun -- I'll keep reading.

 

Wow... time for a coffee break!

 

I'd say easily EarnestEugene.

 

What's the difference between the turbo in the picture below and the turbos in the posts above where the compressor housing outlet is directly connected to the turbine housing inlet with a hose to convert the turbo into a self spooling turbo jet configuration?



Well my answer is that the direct connection hose has been replaced by an intercooler and a diesel engine! However if you bypass the intercooler and block the crankshaft so it can't turn and block open all the intake and exhaust valves and use the injectors for their intended purpose of supplying a fuel flow (to simulate the "torch" that "Pocket" said can't spool a turbo) then you've got a turbo jet configuration and once you start this up using the procedure I described in my previous post to "Tenn" the turbo should spool on fuel flow alone just like the ones in the videos! So I'd say yes a "torch" can definitely spool the turbo!

On the other hand when a diesel engine is in its normal running configuration as is illustrated in the picture above the "diesel engine" itself acts like a "volume airflow throttle" in that the engine controls the volume of air that's allowed to flow from the outlet of the compressor housing to the inlet off the turbine housing! The engine accomplishes this volume airflow control by varying its RPM and a given engine only flows a volume of air that's equal to one-half of its "total displacement " once for every two rotations of its crankshaft!

The resulting mass of air that flows into the engine's intake MAFi for a given volume airflow depends on the air density in the intake manifold and that in turn depends on the BP and MAT in the intake manifold. A given MAFi airflow into the engine's intake is combined in the cylinders with some injected fuel mass in an amount that depends on the mass air-fuel-ratio AFR and this means the resulting mass of gas that flows out of the exhaust MGFe is MGFe=(1+1/AFR)MAFi and for a AFR=20 this gives a MGFe=(1.05)MAFi so the exhaust MGFe is 5% larger than the intake MAFi.

As I've shown below when you calculate the ft-lb of heat ENERGY KEh in a Wg=1 lb mass of "stationary air" at various gas temperatures Tg ranging from 85 F at the compressor inlet to 1,250 F at the turbine inlet and compare that amount of stationary air KEh with the mass flow ENERGY KEf in a Wg=1 lb mass of "flowing air" at various mass flow velocities MPHf ranging from 200 MPH at the compressor inlet to 600 MPH at the turbine inlet you find that in all cases the stationary air heat ENERGY is much larger than the mass flow ENERGY due to the forced movement of the air!

Now the above result might not seem too surprising for the exhaust gas flow MGFe because a lot of heat ENERGY is added to the intake flow MAFi by the combustion process and since much of that combustion heat ENERGY isn't used to produce power stroke HP it's rejected as heat ENERGY coming out the exhaust gas flow MGFe.

However for the intake airflow MAFi someone (especially Pocket) might expect that for an intake air temperature of only 85 F and a 200 MPH mass airflow velocity that the resulting mass flow ENERGY due to the forced movement of the air would be larger than the stationary air heat ENERGY!

Not only is this not the case but as shown in the calculations below when you take the ratio of the stationary air heat ENERGY to the mass flow ENERGY KEh/KEf you see that at the compressor inlet the KEh/KEf=43,573.6/1,337.2=32.6 times more heat ENERGY than mass flow ENERGY, at the turbine inlet KEh/KEf=136,775.2/12,034.9=11.4 times more heat ENERGY than mass flow ENERGY, and at the at the turbine outlet KEh/KEf=108,774.3/334.3=325.4 times more heat ENERGY than mass flow ENERGY so the relative amount of heat ENERGY vs mass flow ENERGY in the cooler intake airflow is 32.6/11.4=2.9 times higher than in the much hotter gas flow going to the turbine!

A given weight of gas molecules Wg is just like an equal weight of any other substance in that if you force it to "flow" or move at a given MPH velocity it has a kinetic ENERGY given by... KEf={(Wg)(MPHf)^2}/(29.913) ft-lb ...and in the case of intake airflow MAFi the compressor wheel needs to supply this ft-lb of ENERGY to force the air to flow into the compressor inlet and the TIME rate of providing this airflow ENERGY is the air flow HP or AFHP that needs to be supplied to the compressor wheel by the turbine!

I'll explain how to calculate the AFHP in a future post because it's rather complicated so for now just consider that whatever this AFHP requirement is for achieving a given intake airflow MAFi it needs to be supplied by the turbine and to account for bearing friction the turbine driveshaft HP or TDHP needs to be somewhat larger than the AFHP that's required for a given intake airflow MAFi.

In the calculations below I chose an exhaust gas flow MGFe=60 lb/min which is based on about the largest possible MAFi=57.1 lb/min airflow that a GTP38R turbo can produce and it's also a convenient number to use since it corresponds to an gas flow of 1 lb/sec and HP is defined as a given ft-lb/sec! So by calculating the difference in total ENERGY of each 1-lb of exhaust gas that flows into and out of the turbine per sec you get the TDHP for a MGFe=60 lb/min which corresponds to a MAFi=MGFe/(1+1/AFR) or for an AFR=20 to a MAFi=(60)/(1+1/20)=57.1 lb/min!

And the answer is TDHP=50.9+21.3=72.2 HP and the portion of the TDHP that's derived from heat ENERGY=50.9/72.2=0.705 or 70.5% and the portion of the TDHP that's derived from mass flow ENERGY=21.3/72.2=0.295 or 29.5%. I've got a pretty good handle on the portion of the TDHP that's derived from heat ENERGY because of measured EGT data but the portion of the TDHP that's derived from mass flow ENERGY is less certain and here I used an educated guess of 600 MPH into the turbine and 100 MPH leaving the turbine.

So lets use round numbers and assume that a TDHP=72.2 HP can supply the compressor with a AFHP=70 HP which implies a HP transmission efficiency of 96.95% and for a direct shaft connection this seems reasonable. Now lets assume an AFR=20 and ask the question is a AFHP=70 HP high enough to force a MAFi=57.1 lb/min airflow through the intake plumbing and into the cylinders for combustion so that it can become a MGFe=60 lb/min exhaust gas flow?

Since the calculation of TDHP=72.2 HP assumed that an exhaust gas flow of MGFe=60 lb/min exists if the resulting AFHP=70 HP isn't high enough to supply a MAFi=57.1 lb/min airflow through the intake then you don't get an exhaust gas flow of MGFe=60 lb/min and you have to begin this whole iterative calculation process again which is why computers are so handy!

In the equations below I show two different ways to calculate the stationary air heat ENERGY. The first equation uses Tg to directly get KEh and the second approach first calculates the random MPHr and then the KEh. These equations can provide additional insights as to how a turbo works and that will also be a topic in a future post!
++++++++++++++++++++++++++
KEh={(80)(Wg)(Tg+459.67)} ft-lb

MPHr=(48.92){(Tg+459.67)^0.5} MPH

KEh={(Wg)(MPHr)^2}/(29.913) ft-lb

KEf={(Wg)(MPHf)^2}/(29.913) ft-lb
++++++++++++++++++++++++++
Tg=1,250 F & Wg=1 lb of stationary air heat ENERGY

KEh={(80)(Wg)(Tg+459.67)}={(80)(1)(1,250+459.67)}= 136,773.6 ft-lb

MPHr=(48.919){(1,250+459.67)^0.5}=2,022.71 MPH

KEh={(Wg)(MPHr)^2}/(29.913)={(1)(2,022.71)^2}/(29.913)=136,775.2 ft-lb
------------------------------------------------
Wg=1 lb of flowing air @ MPHf=600 MPH mass flow ENERGY

KEf={(Wg)(MPHf)^2}/(29.913)={(1)(600)^2}/(29.913)=12,034.9 ft-lb
------------------------------------------------
KEh/KEf=136,775.2/12,034.9=11.4 times more heat ENERGY than mass flow ENERGY!
+++++++++++++++++++++++++++
Tg=900 F & Wg=1 lb of stationary air heat ENERGY

KEh={(80)(Wg)(Tg+459.67)}={(80)(1)(900+459.67)}=10 8,773.6 ft-lb

MPHr=(48.919){(900+459.67)^0.5}=1,803.82 MPH

KEh={(Wg)(MPHr)^2}/(29.913)={(1)(1,803.82)^2}/(29.913)=108,774.3 ft-lb
-----------------------------------------------
Wg=1 lb of flowing air @ MPHf=100 MPH mass flow ENERGY

KEf={(Wg)(MPHf)^2}/(29.913)={(1)(100)^2}/(29.913)=334.3 ft-lb
-----------------------------------------------
KEh/KEf=108,774.3/334.3=325.4 times more heat ENERGY than mass flow ENERGY!
++++++++++++++++++++++++++
MGFe=60 lb/min=1 lb/sec
-----------------------------------------------
Difference in heat ENERGY is 136,775.2-108,774.3=28,0000.9 ft-lb

HP=(ENERGY)/{(TIME)(550)}=(28,0000.9)/{(1)(550)}=50.9 HP
------------------------------------------------
Difference in mass flow ENERGY is 12,034.9-334.3=11,700.6 ft-lb

HP=(ENERGY)/{(TIME)(550)}=(11,700.6)/{(1)(550)}=21.3 HP
------------------------------------------------
Total TDHP=50.9+21.3=72.2 HP

TDHP derived from heat ENERGY=50.9/72.2=0.705 or 70.5%

TDHP derived from mass flow ENERGY=21.3/72.2=0.295 or 29.5%
+++++++++++++++++++++++++++
Tg=85 F & Wg=1 lb of stationary air heat ENERGY

KEh={(80)(Wg)(Tg+459.67)}={(80)(1)(85+459.67)}=43, 573.6 ft-lb

MPHr=(48.919){(85+459.67)^0.5}=1,141.68 MPH

KEh={(Wg)(MPHr)^2}/(29.913)={(1)(1,141.68)^2}/(29.913)=43,574.1 ft-lb
-----------------------------------------------
Wg=1 lb of flowing air @ MPHf=200 MPH mass flow ENERGY

KEf={(Wg)(MPHf)^2}/(29.913)={(1)(200)^2}/(29.913)=1,337.2 ft-lb
----------------------------------------------
KEh/KEf=43,573.6/1,337.2=32.6 times more heat ENERGY than mass flow ENERGY!

 

How do you unsubscribe from a thread...?

 

After all those calculations, Gene still can't account for why a turbo can produce only 5 psi of boost while running 1200 degrees EGT's.

Gene is getting closer. What's funny is he has to start accounting for fueling amounts, airflow, gas density, etc to get closer. Beginning of all of this, heat energy according to him was all that spooled a turbo. After some calculations, a little more than 76% heat energy and slightly more than 23% mass flow spooled the turbo. Now after his last round of calculations, 70% of heat energy and 30% of mass flow spools a turbo. Notice that each time these numbers change, he is accounting for more and more of what is actually happening inside of a diesel engine.

What Gene is forgetting in all of this is fueling quantities vary, and are not a set stoichiometric mix in a diesel.

He is still not accounting for exhaust gas density. Fueling and airflow combined with RPM's will greatly affect this in an engine. You have low airflow, higher fueling, and low RPM's, you get a ton of heat, but you lack the density, pressure, and airflow to spool a turbo to peak boost pressures (again, this is why he can't account for 1200 degrees on the pyro, and only 5 psi of boost, meaning that heat alone can't spool a turbo... also meaning, that heat would account for less than 50% of spooling a turbo).

He is also not accounting for the piston action against the exhaust gas to push it out of the combustion chamber. Hmmm, hot, dense, pressurized exhaust gas being pushed buy a huge piston at an incredible rate of speed. That's going to have a lot of velocity behind it. Oh, you can have hot, not dense, low pressure exhaust gas being pushed by a huge piston at slower speeds (lower RPM's), and what do you get? High EGT's and low boost.

My torch analogy remains intact. Because it doesn't incorporate airflow, pressures, air density, etc. Exactly how heat energy alone doesn't account for that either.

Gene is also missing out on other facts. The main event of combustion is the pinnacle of heat energy and heat exchange. Heat transfers to the pistons, block, heads, etc. When the exhaust valve is opened, the piston pushes the gas (yes pushes, like an air compressor) creating a large amount of air velocity. The gas is still hot, but very quickly cooling off since it's no longer burning. Heat alone won't help the turbo spool, because that gas fresh out of the engine must have a great deal of density (mass, pressure, and temperature combined) in order to spool the turbo. The greater the density, the more your turbo will spool up.

Gene is trying to equate a certain exhaust temp to a specific rate of air intake through the compressor, just like earlier in this thread he claimed he can estimate horsepower based on a certain EGT. However, it doesn't work that way. You can have 250 hp at 1200 degrees EGT's, or you can have 600 hp at 1200 degrees. You can have 5 psi of boost at 1200 degrees, or you can have 25 psi of boost at 1200 degrees. Again, he can't explain why you can have higher EGT's with lower boost pressures. I keep repeating this because he continues to ignore it. Once he finally addresses this very specific issue correctly, he will understand exactly how heat energy actually works on a turbo.

Just because you have heat, doesn't mean the turbo is spooling up. Just because you have heat, doesn't mean you are making power. It takes much more, and anyone who knows anything about how an engine works will understand this. Flow, density, surface area, and heat all combine.

 

Look I'm not sure just what your objective is here in constantly telling everyone that my analysis is flawed and that it doesn't take this and that into account and that it can't correctly predict this and that but at least this time around you did wait a few minutes until after I actually posted some analysis before saying it's wrong which is a big improvement over last time when I had to stop my post in mid sentence due to the wife and I said to stay tuned for the next installment but before I could even post that next installment you chimed in to tell everyone that no matter what that installment said it couldn't possibly predict EGT vs altitude whereas just the opposite is true and in due time I'll explain how it does just that!

Your above claim that... "Beginning of all of this, heat energy according to him was all that spooled a turbo" ...is either a blatant attempt to put words into my mouth or a sign of forgetfulness on your part! Now here's some exact quotes of what was said at the beginning of this thread. This entire exchange which is getting more distasteful to me by the minute as I type this all started because you said...

...if you'd only ask me to explain further and give additional details instead of flatly stating that I was wrong period I would've gladly done so! When "big poppa" asked for more details I gave them...

Note that in the first paragraph I said...

..."exhaust gas heat ENERGY that's primarily responsible for spinning the turbine"

...and in the last paragraph I said...

..."supercharger is powered totally by crankshaft HP whereas a turbocharger is powered primarily by the residual heat ENERGY in the exhaust gas..."

...now does my use of the word "primarily" in both my opening and closing statements sound to anyone else except Pocket like the words he claims that I said...

..."heat energy according to him was all that spooled a turbo"

...which are the words he's now trying to put into my mouth???

Now for some general comments on my efforts to model a turbocharged diesel engine... my current model is "steady state" only which means the engine needs to be operated at a constant load and at constant values of RPM, BP, EGT, MAT, MAFi, MGFe, etc... and since towing approximates a "steady state" operation and since I've got 10 years of towing data with a 7.3L I've got a good handle on what my model can correctly predict!

The way modeling works is that you first need to understand the underlying Physics, derive equations, make calculations, and then compare the predictions with measured data! To even suggest that Physics and equations can't predict something is absurd on its face because we just happen to live in a "physical world" and it's the Physics of the turbo and engine that combine to "generate" the various readings of EGT, BP, RPM, etc... that people "see" as a physical "reality" on their various gauges!

Here's what I said on another truck forum...

...and the reasons listed a few common things that many diesel people "think" are true but actually aren't true at all! I wish more members on this and other truck forms would take a greater interest in wanting to learn how things actually work and I bend over backwards to try and explain what I know about a topic but like for example when I used a swarm of bees to explain the difference between stationary air heat ENERGY and mass flow ENERGY I didn't get any feedback so I've got no idea if everyone understood it or if no one got it at all?

In general when anyone tries to give simplified explanations of complicated problems there's a universal law that says the more "simplified" the explanation is the more technically "incorrect" it is! That's why I really get annoyed when I try to explain things in ways that most can understand only to be told that my explanations are incomplete, flawed, and wrong!

In my next post I'll start giving my explanations of the Physics which govern the intake airflow and the equations to calculate the AFHP. Since the airflow characteristics of the intake plumbing from the air filter to inside the cylinder are quite different from those of the exhaust plumbing for the exhaust gas flow from inside the cylinder back out to the turbine these equations essentially provide a double check on each other.

In closing I like a good argument as well as the next guy and even though several members have cautioned others not to get into one with me because they can't win that's not true because I've lost a few and I don't mind losing one bit because when you learn something new after arguing about it usually gets firmly lodged in your brain and stays there for future use! On the other hand it's not likely you'll win one with me by wearing me down because I've got more time to devote to it than probably anyone else on this forum!

Well here's a post script because I just noticed a few more of Pocket's comments about what I need to include in my turbo-engine model to get the correct results and those comments made me realize that he's missing a fundamental point concerning the "elegance" of my analytical approach which uses a change in ENERGY in a given TIME to calculate the TDHP which is all you need to know to get a 100% correct answer!

As an example on another forum I've been told repeatedly that my drag race model can't possibly give the correct results because it doesn't include "key" parameters that other models use and in fact my model doesn't even consider the truck's acceleration! Well several have posted their MPH trap times and then I run my model and calculate their RWHP vs RPM and it agrees with their dyno sheet which of course they keep secret until I first post my results and they still don't believe me!

Well I won't give the details here except to say that you most definitely don't need to know or to calculate the truck's acceleration to correctly calculate it's ET because HP is the TIME rate of change in ENERGY and to make your MPH increase from MPH1 to a higher value MPH2 in a given T sec you need to apply an average RWHP={(W)(MPH2^2-MPH1^2)}/{(T)(16,452.6)} to the track over that T sec interval. So achieving low ETs is all about "applying RWHP to the track" and nothing else needs to be taken into account!

 

I know what you mean.

Now that Fordziilla01 has all of this good, bad, and useless information. I hope he makes the right decision. Please let this go guys. If some of you are that much of a expert on this matter, heck you should start manufacturing turbo's, you would kick the competitors asz. JFYI I have been working with numbers, and the actual data on projects all of my life. And my conclusion is numbers and formula's are theory until the data is fact.
And what is the facts that Fordzilla ask Which Turbo. I would think it would take more opinions than a few equations that most don't understand and really don't give a chit about. I know I don't have to keep checking this post. But I cant keep from not seeing the feud.
I'm done.

 

This quote says what one item is that I need to get started...

I'm going to use the map for a GTP38R to define the turbo compressor performance and since that's what you've got maybe you'll do some test runs for me? I also need the valve timing for a 7.3L camshaft and I could also use your injector info. I need to know the amount of fuel per injection at max HP and if Jody will tell you your injection timing curve at WOT I'll use that as well.

For my C7 model below I had factory fuel data on their dyno sheet but I had to do some trial and error fuel flow and timing curves and pick the one that gave a reasonable EGT response like is shown below. The steady decline in the KE curve from 30* ATDC to 105* ATDC is because the heat ENERGY in the combustion gas continues to be lost to the surroundings and is still being used to generate PS HP. At C*=105* the exhaust valve begins to open and this causes a sharper decline in KE from 105* to 180* ATDC because now N is decreasing as some of the molecules escape into the exhaust manifold to power the turbo.



BTW people seem to be missing the key point that my model uses the "difference" between the pre and post turbo EGT and it calculates the exhaust gas flow from the AFR, BP, PPM, etc. When you lug an engine it does increase the pre turbo EGT but since the post turbo EGT also increases the BP remains low!

That's a hint so you'll install a post turbo EGT gauge and then with your CFM gauge you can put together a table like before to see how accurately my model predicts your measurements!

 

Click "User CP" on the blue tool bar at the top of the page, then look on the left. Down a little ways is "Subscribed Threads". Click "list subscriptions", then check the box for the thread you want to unsubscribe. Scroll down to the "Selected Threads" drop box and click it. It says "Delete subscription" or something similar.

 

7.3L
Compression Ratio: 17.5:1
Bore: 4.11 inches
Stroke: 4.18 inches
Oops no rod length, and I do not know how to delete

 

Izzy, you da man. Thank you!
Reps for the help.

 

I think there is only one person in this thread who reads every word of Genes posts, and that is Curtis. Hell, i cant even get thru the first sentence of Genes posts and im moving on to the next..