Well due to the wife bugging me to get ready for dinner I missed this... "during the 90* crankshaft rotation of a single piston's exhaust stroke" ...and of course an exhaust stroke actually occurs during a "180* crankshaft rotation"! It's also of interest to note that each of the 8 cylinders is undergoing a 180* exhaust stroke during every 720* of crankshaft rotation and that due to the 7.3L firing order... 1 2 (7 3) 4 5 (6 8) ...cylinders 7&3 on the passenger side and cylinders 6&8 on the driver's side are pushing their exhaust gas into the same exhaust manifold during the same 90* portion of their exhaust strokes and that means they're also trying to push their exhaust gas back into each other's cylinders during that overlap period!

And yes I also realize that when the mass flow of exhaust gas exits "the final pipe leading to the turbine housing" and actually enters the turbine housing that its mass flow velocity is greatly increased due to the restriction in the turbine housing prior to the exhaust gas reaching the turbine!

The picture below is the turbine on my early 99 replacement turbo after only 300 miles of use because I had to pull it to replace the L99.5 compressor wheel that Ford uses for rebuilt early 99 turbos with a E99 wheel because the surge when towing with a L99.5 wheel in a E99 compressor housing is even worse than with a stock 99.5 turbo!



The picture below shows the new E99 compressor wheel installed and I personally think it's the safest practice to pull the turbo when changing compressor wheels!



Now the turbine in the first picture needs to receive enough TDHP=Turbine Driveshaft HP from the exhaust gas to supply the compressor wheel in the second picture with the AFHP=Air Flow HP it needs to establish a given MAF airflow into the engine at a given BP!

So here's the bottom line results of my efforts to date... Consider the point #2 I did previously to show that a BP=25 psi is well within a stock turbo's compressor map for RPM=2,800, TPR=2.85, BP=25.0 psi, which gives an "actual" MAF=52.2 lb/min. For this point #2 the "total required" TDHP=58.0 HP and 44.3 HP or 76.4% comes from the residual heat ENERGY in the exhaust gas and 13.7 HP or 23.6% comes from the mass flow ENERGY in the exhaust gas! So based on this result I think it's accurate to say that it's the residual heat ENERGY in the exhaust gas that spools the turbine!

Due to my cell connection I just got around to seeing some of the things posted about me on an earlier page and based on those comments, other snide remarks, and the apparent lack of general interest that's been expressed lately I'm tempted to end my contribution to this thread here and be done with it! However since there might be someone who wants to learn more about the underlying Physics of how turbos actually work I'll at least give my equations and show how I used them to get the above results. Also if any one expresses an interest in how I derived the following equations I'll provide that as well.

There're two different ways to calculate the TDHPr=44.3 HP or 76.4% portion that's provided by the residual heat ENERGY in the exhaust gas. I used the Physics of "heat capacity" to derive this equation...

TDHPr={(1.333)(MAF)(EGT2-EGT1)}/(550) HP

...where EGT2 is the pre-turbo EGT in F, EGT1 is the post-turbo EGT in F, and MAF is the lb/min airflow into the engine. Based on EGT measurements made by me and those reported by Banks and by others for a BP=25 psi with a stock turbo I used EGT2=1,250 F and EGT1=900 F for point #2 operating conditions which give a MAF=52.2 lb/min. Plugging these numbers into "my" equation gives... TDHPr={(1.333)(MAF)(EGT2-EGT1)}/(550)={(1.333)(52.2)(1,250-900)}/(550)=44.3 HP!

Well the MAF coming out the exhaust is actually about 5% larger than the MAF=52.2 lb/min going into the engine because of the 5% fuel mass injected but I'm trying to illustrate a basic point about the effect of EGT on boost and if you believe in "my" equation and you want the highest possible boost from a given turbo then you definitely want the highest possible "non-damaging" pre-turbo EGT and the lowest possible post-turbo EGT because that combination provides the largest possible drive HP to the turbine so that the compressor can make the boost!

Here's another independent way of arriving at this same conclusion. If you take my previous equation for random kinetic ENERGY... KEr=1/2MgVr^2 ...and convert it to "English" units which due to pride in my heritage I try to use whenever possible you get... KEr={(Wg)(MPHr)^2}/(29.9138) ft-lb ...where now Wg is the total weight in lb of all the molecules in a given mass of exhaust gas and MPHr is the MPH random RMS velocity of the molecules in the exhaust gas due to the residual heat ENERGY at a given EGT in F.

The HP is then given by... TDHPr={(KEr2-KEr1)}/{TIME)(550)} HP ...where KEr2 is the pre-turbo ft-lb of random kinetic ENERGY in the exhaust gas and KEr1 is the post-turbo ft-lb of random kinetic ENERGY in the exhaust gas . Then substitute KEr={(Wg)(MPHr^2)}/(29.9138) ft-lb in this HP equation using MPHr2 and MPHr1 to denote pre and post turbine velocities and recognize that the term Wg/TIME lb/sec is just MAF/60 and you get a second independent way for calculating the TDHPr as...

TDHPr={(MAF)/(987,156)}{(MPHr2)^2-(MPHr1)^2} HP

Now go to 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 gives the pre-turbine MPHr2=2,022 MPH.

Now repeat the above for an EGT=900 F which is an EGT=482 C and this gives the post-turbine MPHr1=1,803 MPH and then use these numbers and MAF=52.2 lb/min in "my" second equation and get... TDHPr={(MAF)/(987,156)}{(MPHr2)^2-(MPHr1)^2}={(52.2)/(987,156)}{(2,022)^2-(1,803)^2}=44.3 HP!

So both equations give the same answer as they must because if I bothered to show how I derived the first one based on "heat capacity" you'd see that the same underlying Physics are operative. Well I guess naysayers like Pocket and Izzy will probably claim that my equations don't prove squat but consider these facts... 1) there's a decrease in EGT after the exhaust gas passes through the turbine housing and enters the downpipe... 2) a decrease in EGT means that the heat ENERGY in the exhaust gas decreased... 3) a given decrease in ENERGY over a given TIME means that a given amount of HP was generated.

So based on these facts if this HP that was generated wasn't used to power the turbine to make boost just what the heck was it used for and where did it go? Well I did say that the "total required" TDHP=58.0 HP and that only 44.3 HP or 76.4% comes from the residual heat ENERGY in the exhaust gas and 13.7 HP or 23.6% comes from the mass flow ENERGY in the exhaust gas so here's how to calculate the portion that comes from the mass flow ENERGY in the exhaust gas.

You just take the equation above and replace the subscripts r=random with t=translational and you get...

TDHPt={(MAF)/(987,156)}{(MPHt2)^2-(MPHt1)^2} HP

...and here I used MPHt2=525 MPH for the pre-turbine mass flow velocity and MPHt1=100 MPH for the post turbine mass flow velocity and I again used MAF=52.2 lb/min and this gives... TDHPt={(MAF)/(987,156)}{(MPHt2)^2-(MPHt1)^2}={(52.2)/(987,156)}{(525)^2-(100)^2}=14.0 HP ...which is a tad more than the 13.7 HP or 23.6% that I cited but I had to make an educated estimate of these mass flow velocities whereas with EGT I had real data which I'd measured by moving my EGT probe from its post-turbo location in my Banks downpipe to a pre-turbo location in the driver's side up-pipe.

 

Good luck!!!

 

That's what I was thinking!!!

 

I guess it works better and lowers EGT just is not enough anymore.

 

Well if you watch this video of a turbo "heat engine" in action you might reconsider your claim that it's not heat ENERGY that spools the turbine! If you're in a hurry slide over to the 2 min mark where the real "heat ENERGY" action begins!

http://www.youtube.com/watch?v=l-spt...layer_embedded

 

Interesting that you posted that video, because in that video, something very obvious happens that you didn't take into account on a real life Powerstroke....


heat soak


In all your equations, you try to prove that it's heat that drives a turbo. Now in your latest set of equations, it's 76% heat energy, and 24% mass flow. So you are starting to understand that it's not only heat that contributes. Problem is, mass flow already accounts for temperature (heat). Another problem, is that pesky heat soak issue. So your equations and calculations are incorrect.

Look at how the turbo and the pipe glow. That's where most of your heat energy is going, to heat up the turbine housing and all related piping, and to keep it hot. Your figures as far as what percentage of heat vs mass flow spools a turbo.... way off.


Funny thing happened on the way home. I have this nice long hill climb to my house, and I held my speed so that I was lugging the engine at 1400 RPM's. The entire way up the hill, my EGT's kept increasing from about 800 degrees at the time I hit my terminal speed, to almost 1100 degrees by the time I hit the top of the hill. Same speed, same RPM's all the way up. With more heat, and according to your theory, I should have had more boost as EGT's were rising. Problem is... boost remained exactly the same all the way up. That 300 degree increase in temperatures added exactly nothing to spooling my turbo any further.

Gene, you forgot to mention that during heat exchange, there is loss that happens elsewhere, such as the turbine housing, piping, etc. A turbo is not 100% efficient at converting heat to useful power.

 

WTH did I just watch? Maybe you could tell me what I was looking at. For all I know that was a flux capacitor.

 

Anyone have a link to the write-up on the "flux capacitor" mod........now THATS some funny chit right there.....

 

Since you've made so many "incorrect statements" such as the one above and other "silly remarks" and downright "snide comments" about me personally I think I need to amend my previous statement about you which was...

...to read you "definitely" don't understand some of the basics of "Physics" period!!!

Now as it turns out I'm currently just hanging out waiting for my wife to finish her several month annual visit with her family so I guess spending my spare time correcting all of your many mistakes in this thread is as good a use for it as any especially if in the process of correcting your mistakes I can help others learn some basics as to how and why things work! So let's lets begin with your above "incorrect statement" and as my schedule permits I'll go back in this thread and dredge up some of the others and also explain how and why they're wrong as well.

Here's an analogy for conceptualizing how and why "mass flow" and "temperature" are two totally "independent" attributes of the exhaust gas that powers a turbine and why "mass flow" most definitely doesn't account for "temperature (heat)" as you've claimed! Picture a bear who just stole some honey and there's a huge swarm of bees buzzing around his head as the bear just stands there motionless enjoying his ill gotten honey.

The flight speeds of the individual bees vary from 8 MPH to 12 MPH but they mostly fly in the 9 to 11 MPH range and the average flight speed of all the bees is 10 MPH. The flight trajectories of the individual bees are erratic and constantly changing as bees bounce off the bear's head and collide with each other but collectively all of the bees remain in the vicinity of the bear's head and the bear is just standing still eating the honey.

This bee swarm is analogous to a "stationary mass" of hot exhaust gas molecules and the bee flight speeds represent the gas molecular velocities and the "square" of the "bee's average flight speed" represents the "degree of bee anger" and this is what determines how much "collision pressure" the bear feels when bees smack into his face. For my analogy the "degree of bee anger" represents the "absolute temperature" of the mass of hot exhaust gas and the "heat ENERGY" contained in the mass of hot exhaust gas is proportional to its "absolute temperature"! This illustrates how you can have "temperature" and "heat ENERGY" but no "mass flow"!

Now the bear lumbers away from the crime scene at a walking speed of 1 MPH and the bees continue to swarm around his head. If you were a bee observing from inside the swarm the flight of the bees around you would appear to be unchanged from when the bear was just standing still however to an external observer it's clear that the collective mass of bees is now slowly drifting along with the bear at 1 MPH. This 1 MPH is the "mass flow" velocity and even though the bees might not perceive an increased level of anger when the bear begins to walk away at 1 MPH the bees do have to increase the total "flight ENERGY" of the swarm by a factor of (1/10)^2 or by 1% in order to keep up with the bear.

Now assume the bees calm down some and reduce their average flight speed from 10 MPH to only 5 MPH. Their anger temperature is now decrease by a factor of (10/5)^2=4 and now their "anger ENERGY" is only (5/1)^2 or 25 times larger than their "mass flow ENERGY" whereas before the calm down it was 100 times larger!

Now assume the bees really calm down and they all come to rest on the bear's head so that their absolute anger temperature is now zero! Well the bear is still walking at 1 MPH so the collective mass of bees still have a 1 MPH "mass flow" velocity even though their temperature is now zero and now the "mass flow ENERGY" accounts for 100% of the total instead of only 1% which was the case when the bees were the maddest!

Hopefully the above examples illustrate how "mass flow" and "temperature" are two "independent" attributes and that knowing "mass flow" doesn't tell you anything whatsoever about "temperature" and vice versa that knowing the "temperature" doesn't tell you anything whatsoever about "mass flow"!

Well I guess I've still got time to address this silly comment...

Well I didn't forget anything but you evidently forgot to read this link... Strategies for reclaiming waste heat ENERGY equivalent HP... https://www.ford-trucks.com/forums/8...ml#post7635931 ...which I gave two of my posts ago on this very thread and in that link I explained heat engine efficiency in great detail!

If a turbo or any other heat engine was "100% efficient at converting heat to useful power" which is a silly notion to even suggest then the exhaust gas molecules exiting the turbine housing would have an absolute temperature of zero and they would just fall out of the end of the tailpipe like grains of sand and lie on the road and then someone would have to come along and sweep them up for disposal!

The diagram below shows the thermodynamic reservoir model for a heat engine. As indicated in the diagram a heat engine uses a given amount of input heat ENERGY from a high temperature reservoir Qh and it converts some of this heat ENERGY into mechanical WORK in a given amount of TIME which is the output HP and then the heat engine rejects the left over heat ENERGY to a lower temperature cold reservoir as Qc.

As shown in the diagram the mechanical WORK W that's done by a heat engine is equal to the difference between the input heat ENERGY from the high temperature reservoir Qh and the left over heat ENERGY that's rejected to the lower temperature cold reservoir Qc so that W=Qh-Qc. The efficiency Eff for converting heat ENERGY into mechanical WORK is Eff=(W)/(Qh)=(Qh- Qc)/(Qh)=1-(Qc)/(Qh).



In my previous post above I calculated the ft-lb amount of pre-turbo heat ENERGY in the exhaust gas which is the Qh and the ft-lb amount of post-turbo heat ENERGY in the exhaust gas which is the Qc and the Eff=1-(Qc)/(Qh). Then I took the difference in these two ENERGY values which is (Qh-Qc) and I then divided (Qh-Qc) by TIME to get the drive HP to the turbine.

So in my previous post where I calculated the drive HP that's applied to the turbine I most definitely took into account the conversion efficiency of the turbo and when I finish my detailed 7.3L model I'll be able to estimate how much of the injected fuel ENERGY is used to run the turbo and that'll give a fuel efficiency of the turbo which is different than the heat engine efficiency of the turbo.

You and everyone else should be aware that when I post something I've usually done quite a bit of brain work before hand and I've already done some supporting analysis to back up what I post! So my suggestion to anyone who reads my posts and then feels an immediate and compelling need to tell me how wrong, flawed, and poorly thought out they are is to please initially assume that they're correct and try to learn something from them and then if after thinking it over for 24 hours you're still sure there's something wrong in one of them please let me know by quoting back the specific statements which you think are incorrect and exactly what your objections to them are.

 

Im so lost...

 

It's just a home-made jet engine made from a turbo. There are a bunch of vids out there of similar cobbled together engines.

 

Well if you watched the entire video which I can't do just now due to my cell connection then you know more than me! When I originally found that link last winter I thought it might be a secret Garret test facility and that a group of inebriated engineers were doing a hot bench test on an experimental turbo!

However since I'd found the link on an 18 wheeler's site I concluded that a bunch of truckers had a few too many beers while replacing someone's worn out turbo on a big rig and weren't sure to do with the old turbo and then someone like me suggested turning it into a jet engine!

Basically all you need to do is use a large diameter flexible hose and hook the outlet from the compressor housing to the inlet of the turbine housing and then poke a 1/4" or so diameter metal tube through a slit in the large hose and you've approximated the jet below.



Then just like you start a jet shoot a blast of compressed air and a little fuel through the metal tube and ignite it and from then on the equation... Fuel+Air=Combustion=heat ENERGY ...takes over and you can control the turbine rpm with the furl flow! Don't forget to use a pump to supply some oil to the bearing and let the return oil drain into a bucket or the turbo jet won't last very long.

Here's a link to another setup that I recall shows more details.

 

Apparently I struck a nerve or something.

And I do understand physics. I also have a better understanding of how engines work. I also know that heat alone can't spool a turbo. There is more to the equation that is missing. Problem is, you ignore it. You're pretty good about ignoring something blatantly obvious too, especially when it could possibly contradict you. But whatever, I'm used to that from other past threads too.

Anyway, I'm done with this thread, since Gene is taking this personally.

 

Gene, don't take this personally, but from proven statistics and factual evidence gather on this forum, 96.35% of those viewing this thread just read the first sentence of your posts and then skip to the next post.
We all know that you have immense knowledge of physics and numerous other topics, you just don't have to tell everyone about it every time you make a post. This could have been a great thread about a very simple question, but it continues to be a pissing match between two guys trying to prove who knows more about nothing. The only reason I keep reading it is to see who can write more in one post than the other trying to prove a point.

You have my approval to be pissed at me for poking fun at you, but you know what, I don't care. I've got better things to do than try and get approval or prove myself to the internet.

 

Ding Ding... I prefer to read about what has been done, and what does work... not what should work, and what does not equate...

I know that the 1.00 exhaust housing on a stocker works better then the .84... I know that the 38r works better then the stocker with a 1.00 housing... I know a Modded H2e works better then the 38r...

I know these things because I have ran them, alot...