Fair enough, Gene. Your clarification is much appreciated regarding the "F250 curve" comments.

I just wanted to make sure you were not under the impression that I took those two points as being "solid" data... but they are interesting, none the less.

I'm hoping to get a full day before to long to visit Joe Mondello's testing lab to get a more formalized set of tests on the 6637, clean, dirty, with clean cover, with dirty cover, my cover, Outerwears cover... all with repeated runs for validation of results and as many vacuum levels as possible to span a full 0-30" WC chart.

My plans include the full range of tests on the clean 6637 with no cover, and a reduced number of points across the entire vacuum range for the remaining tests to identify any curve shifts which may be measureable for these alternate 6637 configurations.

Shoot, I might even go ahead and let them see if they can collapse both a clean and a dirty 6637 element (just for giggles) to see what vacuum and airflow are required for that condition.

 

As long as we're talking (at least sometimes) about air filters and air flow, I figured I'd remind everyone of this thread. I know Hastings air filters have not been a part of the conversation, but it's still interesting to watch. https://www.ford-trucks.com/forums/8...his-video.html

 

A spinning head.

 

Just tells me that we need to install a simple vacuum gauge to monitor the filter restriction.

 

OK Gene. The pleats on the 6637 are 1 7/16" deep and there are 191 of them.

 

Since small bites are easier for most people to chew I'll be posting my "Dyno Analysis" one mouthful at a time! I'm using a GTP38R compressor map as one tool to analyze the dyno data that Izzy gave to back up his "dyno proven" claims for a 6637 ! This approach also gives everyone an opportunity to focus on each particular aspect of the analysis and to make any comments and/or give a critical review of it. At the very end I'll pull everything together in a summary and discuss the results and the conclusions that I draw from them. The 3 runs below are the ones that I'm analyzing and the first step is to plot the operating point for each dyno run on a GTP38R compressor map.

1) run with the tymar netted:
1st run on #2=413.7rwhp

2) run with AFE stage2 netted:
1st run on #2=407.0rwhp

3) run with open element netted:
1st run on #2=424.9rwhp

The first step is to determine the FWHP that's required to produce the measured 424.9 RWHP on run 3 without an air filter because the FWHP determines the minimum fuel flow and this in turn determines the minimum airflow that's required and that in conjunction with the RPM and the BP determines a baseline operating point on the turbo compressor map. After that operating point has been established then the potential effects on RWHP due to increases or decreases in airflow which result from the use of different air filters can be evaluated.

Since I'm analyzing dyno data where the 424.9 RWHP with no air filter is the baseline and where the 413.7 RWHP with the 6637 is only 2.6% lower than the no air filter baseline and the 407.0 RWHP with the AFE is only 4.4% lower than the no air filter baseline and the 413.7 RWHP with the 6637 is only 1.6% higher than the 407.0 RWHP with the AFE then clearly I need to be concerned with every possible detail that might cause even the smallest percentage change in the resulting FWHP that's required to produce a given dyno HP.

So let's start with the fact that a dyno doesn't actually measure a truck's RWHP to begin with! The dyno measures the DRHP=Dyno Roller HP and the way I define terms the RWHP=Rear Wheel HP is equal to the truck's RAHP=Rear Axel HP because the RAHP is applied to the truck's rear wheels by the rear axel half shafts and this is where the "solid" mechanical connection to the flywheel terminates. The tires are between the rear wheels and the road and in the process of applying the RWHP to the road the tires flex and this absorbs some of the RWHP as RRHP=Rolling Resistance HP so that not all of the RWHP gets applied to the road or to the rollers on a chassis dyno. For a dyno run the RWHP=DRHP+RRHP.

I modified this equation... RRHP={(GCW)(MPH^1.3)}/{150,000} ...which gives the RRHP for the entire truck as it rolls down the road at a given MPH to one that can be used in the dyno analysis by converting MPH to engine RPM assuming a stock SRW truck in 3rd gear (direct ratio) with a 3.73 diff and stock 265 75R 16 tires and this gives... RRHP={(RAW)(RPM^1.3)(5.57x10^-8)} ...where RAW=Rear Axel Weight. I use this equation to relate the RWHP to the DRHP as... RWHP=DRHP+{(RAW)(RPM^1.3)(5.57x10^-8)}. I use a RAW=4,000 lb so that for example at a 2,800 RPM the RWHP=DRHP+{(4,000)(2,800^1.3)(5.57x10^-8)}=DRHP+6.7 HP so if a dyno run gives a maximum HP reading at 2,800 RPM then 6.7 HP needs to be added to that dyno measurement in order to get the RWHP.

In general the FWHP that's lost in the transmission, the driveshaft bearings, and the differential due to the heating of these driveline components which is caused by friction and fluid cavitations is given by the DLHP=Driveline Loss HP. The DLHP depends on the FWTQ load that's placed on the driveline and on the rpm of the driveline which for the direct gear that's normally used during a dyno run is the same as the engine RPM. This DLHP can be written as... DLHP=(C1)(FWTQ)(RPM)(RPM)^0.5 ...where (C1) is a constant and this DLHP equation can be simplified to... DLHP=(C2)(FWHP)(RPM)^0.5 ...where the new constant (C2) includes the factor (5,252) for converting {(FWTQ)(RPM)} to (FWHP).

The equation... DLHP=(C2)(FWHP)(RPM)^0.5 ...says that the FWHP which is lost in the driveline is proportional to the amount of FWHP that's being transferred from the flywheel to the rear wheels and it's also proportional to the square root of the RPM rate at which the HP transfer is occurring. This means the RWHP=FWHP-DLHP=FWHP-(C2)(FWHP)(RPM)^0.5 and dividing both sides of this equation by FWHP you get... {(RWHP)/(FWHP)}={1-(C2)(RPM)^0.5}.

The DLE=Drive Line Efficiency is defined as DLE={(RWHP)/(FWHP)} and it's a ratio between 0 to 1 and {(1-DLE)(100)}% gives the percentage of the FWHP that's lost in the driveline. So DLE={1-(C2)(RPM)^0.5} and C2 can be written as C2=(1-DLE*)/(RPM*)^0.5 where DLE* is the DLE at the RPM=RPM* where RPM* is the maximum governed RPM for the engine. This gives... DLE={1-(RPM/RPM*)^0.5(1-DLE*)}.

My measurements on my old F350 are consistent with a DLE*=0.80 at RPM*=3,400 and using these values gives... DLE={1-(RPM/3400)^0.5(0.2)} ...and I use this equation to relate the RWHP to the FWHP as... FWHP=RWHP{1-(RPM/3400)^0.5(0.2)}^-1. At 2,800 RPM this gives FWHP=RWHP{1-(RPM/3400)^0.5(0.2)}^-1=RWHP{1-(2800/3400)^0.5(0.2)}^-1 so that FWHP=(1.2217)RWHP. So if on a given dyno run the maximum FWHP occurs at 2,800 RPM then 18.15% of the FWHP is dissipated as heat in the driveline and only 81.85% shows up at the rear axel during that dyno run.

Clearly the exact RPM at which the various DRHP values occurred is an important factor to consider in my analysis but I can't consider it in a straight forward way because no graphs of DRHP vs RPM are available! So instead I'll analyze a range of RPMs from 2,400 to 3,200 across the maximum HP band for a PSD.

When a given MFF=Mass Fuel Flow lb/min combusts with a given MAF=Mass Air Flow lb/min the "chemical ENERGY" that's stored in the diesel fuel is released as "heat ENERGY" and a given HEHP=Heat Equivalent HP is produced and depending on the thermodynamic efficiency for transforming the HEHP into a piston motion HP a given percentage of the HEHP is transformed into PSHP=Power Stroke HP and then after accounting for friction and parasitic losses which both depend on RPM a given PSHP produces a given FWHP.

In the table below I used a simplifying assumption of a constant net TE=Thermodynamic Efficiency of 37% for converting HEHP into FWHP and I calculated the actual MAF that's required to produce the baseline dyno run of 424.9 DRHP.

DRHP+RRHP at RPM =RWHP/DLE=FWHP/18.9=GPH....MFFx16.5=MAFx13.95/14.7=MAFcorr

424.9+6.1 HP at 2,600=431.0/0.8251=522.4/18.9=27.6....3.22x16.5=53.1 x13.95/14.7=50.4 lb/min

424.9+6.7 HP at 2,800=431.6/0.8185=527.3/18.9=27.9....3.26x16.5=53.8 x13.95/14.7=51.1 lb/min

424.9+7.4 HP at 3,000=432.3/0.8121=532.3/18.9=28.1....3.28x16.5=54.1 x13.95/14.7=51.3 lb/min

424.9+8.0 HP at 3,200=432.9/0.8060=537.1/18.9=28.4....3.31x16.5=54.6 x13.95/14.7=51.8 lb/min

So to produce the nominal 530 FWHP that's required to get the measured 424.9 DRHP you need to have a "minimum" MAF=51 lb/min and the reason I say "minimum" is because I used an AFR=16.6 where... AFR=Air Fuel Ratio is given by AFR=MAF/MFF ...and this means there's only a 10% excess of air to combust with the fuel so the fuel air mixing needs to be good enough to combust 90% of the available air in order to combust all the fuel.

This assumed AFR will cause some smoke and it will "stress" an air filter in the sense that even small reductions in airflow will reduce the AFR enough so that not enough fuel gets combusted to make the measured 424.9 DRHP. If I assumed a larger AFR like an AFR=20 then the required MAF would be larger but there'd be a sufficient amount of excess air so that minor changes in airflow due to different air filters wouldn't change the amount of fuel combusted and there'd be no change in FWHP.

Now we need to plot actual engine MAF capabilities on the GTP38R compressor map and see if we get the desired MAF=51 lb/min in the RPM power band at BP values that are achievable and this is where the "fun" begins! I've given all this before so I'll just cut to the chase. An engine can flow a MAF that's given by MAF={(VE)(CID)(RPM)(AAP+BP)}/{(MAT+459.67)(1,278.46)} lb/min where VE=Volumetric Efficiency is a ratio from 0 to 1, CID={(Nc)(Pi/4)(Bore)^2(Stroke)} in^3, Nc=Number of Cylinders, RPM=Crankshaft revs/min, AAP=Atmospheric Air Pressure psi, BP=Boost Pressure psi, and MAT=Manifold Air Temperature F. For a 7.3L PSD use CID={(Nc)(Pi/4)(Bore)^2(Stroke)}={(8)(Pi/4)(4.11)^2(4.18)}=443.65 in^3, and then the MAF equation for a PSD becomes...

MAF={(VE)(RPM)(AAP+BP)}/{(MAT+459.67)(2.8817)} lb/min.

For this initial cut I'll assume a constant VE=0.85. The actual VE has a maximum at about 2,000 RPM and it decreases for higher RPM and for higher BP. The same is true for the net TE which I also assumed to be a constant value of TE=0.37. These simplifications mean the initial results will both "overestimate" the engine's actual MAF capability and will "underestimate" the required MAF!

The next step is to understand how BP=Boost Pressure psi relates to the TPR=Turbo Pressure Ratio because TPR is the parameter that's related to MAF on a turbo compressor map and you 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.

For this baseline I'm assuming no air filter and an intercooler with no restriction whatsoever so that AFPD=0 and ICPD=0 no matter what the CFM airflow is. Now you have... BP={(AAP)(TPR-1)}={(AAP)(TPR)}-AAP ...so that ...(AAP+BP)={(AAP)(TPR)} ...and of course the MAP=(AAP+BP)={(AAP)(TPR)} ...and the bottom line here is that to increase MAP you need to increase the TPR and/or move to a lower altitude!

If you substitute (AAP+BP)={(AAP)(TPR)} into the equation... MAF={(VE)(RPM)(AAP+BP)}/{(MAT+459.67)(2.8817)} lb/min ...you get MAF={(VE)(RPM)(AAP)(TPR)}/{(MAT+459.67)(2.8817)} lb/min ...and the key to understanding this equation is to realize that the variables MAF, RPM, TPR, and MAT are all interconnected and that if you change RPM for example the MAF, TPR, and MAT also change!

The GTP38R compressor map is referenced to a TIAT=Turbo Inlet Air Temperature F of TIAT=85 F and to a TIAP=Turbo Inlet Air Pressure psi of TIAP=13.95 psi which corresponds to a "standard day" altitude of 1,509 ft and in this baseline I'm assuming the dyno runs were done at these standard reference values. In other words the dyno was located at 1,509 ft above sea level where the AAP=13.95 psi and the AAT=Ambient Air Temperature F was AAT= 85 F and the open turbo inlet ingested this 85 F air.

Since the turbo is open and doesn't have an inlet tube the TIAP=AAP=13.95 psi and the ambient air temp is 85 F so TIAT=85 F. This gives... MAF={(VE)(RPM)(AAP)(TPR)}/{(MAT+459.67)(2.8817)}={(0.85)(RPM)(13.95)(TPR)}/{(MAT+459.67)(2.8817)}={(4.1)(RPM)(TPR)}/{(MAT+459.67)} lb/min so that...

MAF={(4.1)(RPM)(TPR)}/{(MAT+459.67)} lb/min

Say for example you want to plot the point RPM=2,800 and TPR=2.00 on the GTP38R compressor map. That means you have to calculate the MAT that results when the turbo is generating a compression ratio of TPR=2.00. If you follow this example through from top to bottom you'll understand the "Physics" of how much a turbo heats the air as it compresses it and how much the IC=Intercooler cools the air before it's ingested by the engine.

TOATI=Turbo Outlet Air Temperature Ideal, F
TOATI={TIAT+459.67}{(TPR)^0.286}-459.67, F
TOATI={85+459.67}{(2.00)^0.286}-459.67 F
TOATI={544.67}{1.219}-459.67=204.3 F
TCE=Turbo Compressor Efficiency, ratio, TCE=0.70
TOATA=Turbo Outlet Air Temperature Actual, F
TOATA=TIAT+{TOATI-TIAT}/{TCE}, F
TOATA=85+{204.3-85}/{0.70}
TOATA=85+170.4=255.4 F
ICETI=Intercooler Exchanger Temperature Increase, F, ICETI=5 F
ICEAT=Intercooler Exchanger Air Temperature, F
ICEAT=AAT+ICETI, F =85 F+ 5 F=90 F
ICEE=Intercooler Exchanger Efficiency, ratio, ICEE=0.75
MAT=Manifold Air Temperature, F
MAT=TOATA-{TOATA-ICEAT}{ICEE}, F
MAT=255.4-{255.4-90}{0.75} F
MAT=255.4-{124.1}=131.3 F

Now use that MAT=131.3 F in the equation... MAF={(4.1)(RPM)(TPR)}/{(MAT+459.67)}={(4.1)(2,800)(2.00)}/{(131.3+459.67)}={(22,960)}/{(591)}=39.0 lb/min! Now look at the GTP38R compressor map below and find the value TPR=2.0 on the left vertical axis and follow it to the right until it hits the "X" for the X=2,800 RPM line that I hand plotted on the map and "oops" you see a MAF=40.2 lb/min which is 3% larger than the 39.0 lb/min value we just calculated! So Pocket was right all along and my equations and graphs are all wrong after all!

Well all this actually proves is that graphs plotted by hand aren't accurate enough to analyze a several percent difference in air filter performance which is why the next installment of this analysis will feature computer generated graphics. If you look at the GTP38R compressor map you'll see a dotted "surge line" and a dotted "efficiency line" that runs through the center of the efficiency island and I transferred these two reference lines onto my computer generated graphics as reference lines for comparison with the computer generated data that's plotted.



I hand plotted enough TPR vs MAF points at 2,800 RPM and at 3,200 RPM so that your eye can connect the X's and the W's but you certainly can't do fined grain analysis from a plot like this!

However the plot is good enough to show that you can get the MAF=51 lb/min that's required for the 527 FWHP in the top table at 2,800 RPM with only a BP=25 psi! That CFM scale below the MAF scale indicates the CFM that corresponds the MAF at the standard conditions shown where the AD=0.0692 lb/ft^3. So for example at a MAF=45 lb/min the CFM=650 ft^3/min and that was gotten from CFM=MAF/(0.0692).

Now don't be thinking your PSD flows MAF like the RPM vs BP lines on this plot indicate because this plot is based on a number of "optimistic" assumptions but on the other hand as I'll discuss in the next installment it's also based on a "pessimistic" assumption that most drivers don't encounter while just hot rodding around town!

 

Well, I can't read all of that right now, but I will say that Marty/Maintain is pulling 60lbs of boost with his Hypermax through one of the dirtiest 6637s I have ever seen. Food for thought. I will get back to your post a bit later Gene.

 

Let me try to undo the confusion I've caused when I said...

...as you can see from my table...

...I incorrectly used the "corrected MAF" of MAFc=51 lb/min instead of the "actual MAF" of MAFa=54 lb/min which of course is the airflow that's actually required to get the nominal 530 FWHP!

When I generated this table I was planning on analyzing dyno runs done at sea level so that the MAFa=54 lb/min that's actually required to get the nominal 530 FWHP needed to be corrected to a "reference MAF" value of MAFc=51 lb/min so that it could be plotted on a standard reference GTP38R compressor map and then compared to the standard reference engine demand MAFd versus RPM and BP lines on the GTP38R compressor map.

Then I got distracted with chores for a few days and when I returned to the analysis I'd decided to simplify the initial cut and postpone a discussion of "corrected values" by assuming the dyno runs were done at the reference altitude of 1,509 ft where the "corrected values" are the same as the "actual values" but I forgot to delete the last column of my table so I glanced up at it and grabbed the wrong MAF value!

So if a dyno run is done at the reference altitude of 1,509 ft then you plot MAFa=54 lb/min (which is the MAF that's required to get the nominal 530 FWHP at any altitude) on the GTP38R compressor map but if a dyno run is done at sea level with an open turbo inlet then you plot MAFc=51 lb/min on the map.

Now that this can of worms has been opened let me try to explain the reason why you need to use "corrected values" on a turbo compressor map. 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) Since different turbo OEMs specify their maps for different standard "reference" values at the turbo inlet I'll use TIAPr and TIATr to note these standard "reference" values at the turbo inlet.

If you look at a particular TPR vs MAF operating point on a compressor map you'll see a "corrected" or a "reference" value for MAF which I'll call MAFr and this is the MAF the compressor generates at the stated TPR when the turbo inlet sees an actual TIAPa=TIAPr and an actual TIATa=TIATr. For a Garrett compressor map TIAPr=13.95 psi and TIATr=85 F but some other OEM might use TIAPr=14.7 psi and TIATr=60 F as the standard reference conditions for their maps.

So in general how do you use MAFr values from a compressor map to calculate the actual MAFa values for actual operating conditions where the TIAPa and TIATa are different from the standard reference values TIAPr and TIATr for which the map was measured? First I'll give the "conversion" equation then I'll discuss why it contains a term T^0.5 or "square root" of T instead of just T as would be the case if you did a straight forward correction for AD=Air Density at the turbo inlet.

MAFa={(TIAPa)/(TIAPr)}{(TIATr+460)/(TIATa+460)}^0.5{MAFr}

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 AAT=85 F. Well if the air filter inlet is open to the atmosphere and involves no temperature increase then the TIATa=85 F=TIATr so no correction is needed due to temperature. If the R=Restriction of the air filter plus the Restriction of the turbo inlet tube is R=20.76" H2O which is R=0.75 psi then the TIAPa=AAP-R=14.7-0.75=13.95 psi=TIAPr so no correction is needed due to pressure. And this is why referencing a compressor map to TIAPr=13.95 psi and TIATr=85 F works to the convenience of a typical user!

For the "open turbo" baseline there's no air filter or turbo inlet tube so TIAPa=AAP=14.7 psi and the correction for pressure is MAFa={(TIAPa)/(TIAPr)}{MAFr}={(14.7)/(13.95)}{50}={(1.0538)}{50}=52.7 lb/min. If the dyno run was done at sea level and at an AAT=60 F the total correction for both pressure and temperature is... MAFa={(TIAPa)/(TIAPr)}{(TIATr+460)/(TIATa+460)}^0.5{MAFr}={(14.7)/(13.95)}{(85+460)/(60+460)}^0.5{50}={(1.0538)}{(1.048)}^0.5{50}={(1. 0538)}{(1.0238)}{50}={(1.0789)}{50}=53.9 lb/min.

So here's why there's a square root of temperature involved as opposed to a straight forward correction for AD=Air Density at the turbo inlet...

Actual Turbo Inlet MAF={Actual Turbo Inlet AD}{Actual Turbo Inlet CFM} ...MAFa

Reference Turbo Inlet MAF={Reference Turbo Inlet AD}{Reference Turbo Inlet CFM} ...MAFr

TIAD=Turbo Inlet Air Density lb/ft3 is given by... TIAD={(2.70325)(TIAP)}/{(TIAT+460)} lb/ft3 ...where TIAP=Turbo Inlet Air Pressure psi and TIAT=Turbo Inlet Air Temperature F

Turbo Inlet AD Correction={(TIAPa)/(TIAPr)}{(TIATr+460)/(TIATa+460)} ...a straight forward correction for AD

Compressor Wheel RPM Correction={(TIATa+460)/(TIATr+460)}^0.5 ...involves kinetic temperature at blade tips

Turbo Inlet CFM Correction={(TIATa+460)/(TIATr+460)}^0.5 ...Inlet CFM is proportional to Wheel RPM

Turbo Inlet MAF Correction={Turbo Inlet AD Correction}{Turbo Inlet CFM Correction}

Turbo Inlet MAF Correction={(TIAPa)/(TIAPr)}{(TIATr+460)/(TIATa+460)}{(TIATa+460)/(TIATr+460)}^0.5

Turbo Inlet MAF Correction={(TIAPa)/(TIAPr)}{(TIATr+460)/(TIATa+460)}^0.5

Actual Turbo Inlet MAF={Turbo Inlet MAF Correction}{Reference Turbo Inlet MAF}

MAFa={(TIAPa)/(TIAPr)}{(TIATr+460)/(TIATa+460)}^0.5{MAFr}

So to do a conversion from MAFr to MAFa you in fact do a straight forward correction for density which depends on a pressure ratio and on a temperature ratio and then this density correction combines with a temperature only correction of Turbo Inlet CFM which depends on the square root of a temperature ratio so that the net correction to MAF which is the product of the density correction and the CFM correction depends on the square root of a temperature ratio!

The correction to Turbo Inlet CFM which is proportional to Compressor Wheel RPM results from a "black magic" correction to Compressor Wheel RPM which involves only the square root of the temperature at the leading edges of the compressor blades and it has to do with the dynamic compressibility of air as a function of its kinetic temperature.

In my next installment which hopefully won't be followed by a correction post I'll give some real world practical examples of how to use compressor maps to determine engine FWHP capabilities.

 

Well I've been busy building a "new and improved" computer model for a PSD that's leveraged off the model I spent a year developing for my CAT C7. I calibrated my C7 model using factory dyno data on my specific engine which included a complete list of parameters such as HP, TQ, RPM, BP, GPH, and reference temperature and pressure. Over the past year I've been collecting data using my fancy but expensive unit that mounts on my dash and reads all the engine and transmission data from the SAE J1708/J1587/J1939 data link connector and comparing that data to predictions from the model.

I'm pressed for time so I'll post mostly a "picture outline" of how I got the results and I'll fill in more details and answer any questions as my travel schedule and cell connection permit. I've got lots of graphs so I'll spread this over several posts. To recap here's the dyno runs that I replicated with my new PSD dyno model for the purpose of investigating how a given AF=Air Filter restriction effects the measured DRHP=Dyno Roller HP.

1) run with the tymar netted:
1st run on #2=413.7rwhp

2) run with AFE stage2 netted:
1st run on #2=407.0rwhp

3) run with open element netted:
1st run on #2=424.9rwhp

The key to understanding how airflow in a turbo diesel is effected by AF restriction is characterizing the amount of TDHP=Turbine Drive HP that's required to produce a given MAF=Mass Air Flow lb/min. The hot exhaust gas that's generated by a given fuel flow produces a given TDHP which is the input HP to a turbo's turbine wheel. This TDHP powers the turbo's compressor wheel so that it can produce the AFHP=Air Flow HP that's needed to pull a given VAF=Volume Air Flow ft^3/min=CFM through the restriction of an air filter element and then pressurize this ingested airflow and push it through the restriction of an 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).

The diagram below shows the "Physics" of AFHP and how I derived this equation for it... AFHP={(PD)(VAF)}/{(TCE)(229.17)} HP...



...and if you apply this AFHP equation to the CFM airflow through a length of straight pipe you get the graph below...



...and the way to read these curves is to follow the 1,200 CFM line to the right until it intersects the solid pink 4" curve and then go straight down to the dashed pink 4" curve at a 20" H2O restriction and then over to the right hand scale to read 5 HP. Likewise at 800 CFM you see 3 HP at a 17.5" H2O restriction for the green 3.5" curves, and at 700 CFM you see 4.5 HP at a 30" H2O restriction for the blue 3" curves. These AFHP numbers give the TDHP that needs to be applied by the exhaust gas to the turbine wheel in order for the compressor wheel to generate the indicated CFM airflows.

If you divide by 30 you get the results for the required TDHP to generate various CFM flows in a 1 ft straight section of turbo inlet tube but since an actual turbo inlet tube has bends and doesn't have a perfectly smooth inner wall dividing by 10 instead of 30 is probably about right. This means for example that a 1 ft long 4" diameter turbo inlet tube has a 2" H2O restriction at 1,200 CFM and a 0.5" H2O restriction at 600 CFM.

If you apply this AFHP equation to the CFM airflow through the entire air inlet system leading to the cylinders of a turbo diesel engine you get... Turbo AFHP={(BP+ICPD+AFPD)(TOVAF)}/{(TCE)(229.17)} HP ...and on my graphs comparing a no AF baseline with different AF restrictions this Turbo AFHP is noted as TDHP. The TOVAF=Turbo Outlet VAF depends on turbo outlet temperature and pressure which varies as a function of RPM and engine load which determine the BP value. My new PSD model calculates this TDHP so that airflow can be analyzed in terms of a constant TDHP as well as in terms of a constant TPR=Turbo Pressure Ratio which is the parameter used on a turbo compressor map.

I'm using the following two gasser turbo compressor maps to illustrate the effect that VE has on the so called "engine demand lines" that are plotted on turbo compressor maps to determine mass airflow MAF for a given turbo in terms of the TPR parameter. These lines should actually be called "engine accept lines" because the engine doesn't "demand" a given MAF but rather for a given BP and MAT which determine the air density in the intake manifold an engine will "accept" the MAF that's "allowed" by its CID, VE, and RPM. An engine flows a CID volume of air once for every 2 revolutions of its crankshaft and the air density in the intake manifold determines the mass of air that flows through the engine.

For the ideal case of VE=100% the "engine accept lines" on the compressor map below are nicely spread across the map and have a uniform spacing between them. Note the blue dots for BP=0 at the beginning of each line at the bottom of the map where TPR=1 because if you connect these blue dots you've got the airflow versus RPM line for a non-turbocharged engine which was the case for my race engine and as you can see the airflow increases from 80 CFM at 3,000 RPM to 215 CFM at 8,000 RPM which means that if the VE=100% the engine will "accept" a CFM airflow that's proportional to its RPM.



Now look at those same blue dots on this map below where the VE now decreases by an ever increasing amount as the RPM increases. Now you see why I ruined more than one cylinder head while porting and polishing it in an attempt to maximize its VE!



Of course even with a turbo the "engine accept lines" are now all now bunched together between the maximum of the efficiency island and the surge line so what does this map tell you when you read a post that says VE isn't important in a diesel because unlike a gasser the diesel has a turbo that can push as much airflow as it wants to into the engine?

What actually happens is that as the TDHP increases and the compressor wheel pushes harder trying to make the engine "accept" more airflow the BP increases in an "attempt" to increase the airflow but at higher BP the actual airflow achieved is far less than being proportional to the increasing BP or said another way the VE decreases for increasing BP just like the VE decreases for increasing RPM but because of a different reason. For increasing RPM the intake valves aren't open long enough to allow more airflow at a given BP and for increasing BP turbulence limits airflow. For a sufficiently high BP "choked" flow develops at the intake valves and you get no additional flow at all for BP above that limit.

My PSD model has equations that "predict" the decrease in VE due to increases in RPM and BP and the reason I know about the PSD VE as a function of RPM and BP is because I've made CFM vs RPM and BP measurements on my early 99 F350 and compared them to the similar measurements I got from Tenn on his 99.5 truck. I got the picture below from PS ? in NJ and it shows the huge difference in the intake plenums between the early 99 and 99.5 engines.



Even though the 99.5 plenum has a much larger hole to let the air in it still has a flattened out design and based on "dusted" engine tear down reports the outer 4 cylinders are still starved for air! I'm always amazed at how much obsession there is with the perceived effect that AF restriction has on airflow while no one even discusses the potential payoff in upgrading to one of the several versions of larger intake plenums. There's always a tradeoff involved between AF restriction and dirt filtration but investing in upgraded plenums is a win-win deal because they also distort less and thereby allow operation at higher BP.

With this as background the next installment will include the graphs from my new PSD dyno model which I'm still in the process of uploading. Since I had these two maps on my photo site from a long time ago I'll use them as examples of how engine operating points are plotted on a compressor map. The map below isn't for a GTP38 turbo but the BP vs MAF data is form my old early 99 F350.



On the Banks map for a 99.5 GTP38 turbo I plotted some points from my earlier 99.5 PSD model to see how well they agreed with the surge line prediction on the Banks map which I'd previously compared to the surge line on a Garrett map.

 

Yes the 99.5 is larger. However, "upgrading" to a different aftermarket plenum design has already proven not to gain anything at all on 99.5-03 PSD's. The actual restriction is in the design of the heads, not the plenums.

Wide Open Performance to date has the highest power record on the dyno for a 7.3L at 871 hp, and to date also has the quickest E/T for a 7.3L at 10.86 seconds in the quarter mile. What's interesting to note is the plenums that they sell and also use on their engines.

Here is the link to their plenums: http://www.wideopenperformance.net/product.aspx?ID=90

If you notice, they are completely flat, and not even raised up at all like the stock plenums on both the early 99's and the 99.5+ engines. The difference is the thickness of the metal, and it allows you to put a higher torque clamp on to hold the boots in place. I put this set on my truck as well, because I crushed one of my plenums after tightening up the boots. Overall, technically these would decrease the flow of air, if you want to get down to it. However, the plenums are not as restrictive as you might think, and these aftermarket ones flow more than enough air, just like the stockers. By the way, these plenums are also being used on their dragster that they are finishing up.

So yes, the plenums have already been addressed in the aftermarket world a long time ago, and have already been proven that they are not the restriction that people once thought they might be. Rather, it's the design of the heads themselves that is the major restriction for airflow, and you have to spend some big bucks to fix that issue.

 

I'm wondering how the "dusted" motor proves the outer cylinders are starved for air...

Are you saying the "lack of equal dust" on the outer cylinders means they're starved for air?? I would say that an explanation for this is the dust will be pulled into the closest two cylinders first, leaving less for the rest. I don't think that means less air gets there, as in the fuel starvation issue at the 6 & 8 injectors. Besides, wouldn't that show in a cylinder contribution test?? I haven't seen that test run just yet, but I would think an "air starvation" situation would reveal itself there.

 

I'm giving the GTP38R compressor map and the 3 computer generated versions which I'll call graphs #1, #2, and #3 all together so you can reference the computer maps to the GTP38R map. Note the GTP38R map contains a dotted "surge line", a dotted "efficiency line" that runs through the center of the efficiency island, and a 110,000 rpm "speed line" in the upper right hand portion of the map. On the last 2 computer versions of the GTP38R compressor map I needed to switch the horizontal and vertical axes to make automated plotting easier but I transferred all three of these reference lines onto the computer generated graphics as reference lines.

I'm also listing these 4 basic equations together so I can refer to them by their number...

1) MAF={(0.347)(VE)(RPM)(AAP+BP)}/{(MAT+459.67)} lb/min

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

3) TPR={BP+AAP+ICPD}/{AAP-AFPD}

4) TDHP={(BP+ICPD+AFPD)(TOVAF)}/{(TCE)(229.17)} HP









In computer graph #1 the "engine accept lines" are plotted on the map for the ideal case of VE=100% and they have a uniform spacing between them just like in the first gasser example map I gave before. For a 7.3L PSD these "engine accept lines" are determined by equations #1 and #2. You chose values for RPM and TPR and then calculate BP from #2 and plug that into #1 and then use the whole mess of turbo and IC equations I gave before to calculate the MAT and then plug the MAT into #1 and solve #1 for the MAF value which corresponds to the RPM and TPR values.

In computer graphs #2 and #3 the BP values on the right hand scale are read as follows. First look along any RPM line of interest like the "blue" line for 2,800 RPM and find a MAF and TPR point like the point MAF=55 lb/min and TPR=3.1 and then go straight up to the "red" BP line that runs from the bottom left to the top right and move across to the BP scale and see BP=27.5 psi. Now look at the "blown up" graph #3 and see the same point. Now start at the desired MAF=55 lb/min and go over to the "blue" line for 2,800 RPM and then down to TPR=3.1 and then go straight up to the "red" BP line and over to the right to get BP=27.5 psi. As you can see from the compressor maps all of the interesting action for high HP dyno runs occurs at higher values of MAF, TPR, and BP which is why I did the "expanded view" graph #3.

Graphs #2 and #3 are identical except for the "expanded view" in graph #3 and they were both generated from my latest PSD model that includes updates for the VE equations that calculate VE as a function of RPM and BP, the turbo equation to account for heat loss in the case of the turbo, the RRHP that's dissipated in the tires and doesn't get to the dyno rollers, the RPM dependent driveline loss, and several other improvements that I'll discuss at a later time.

Here's some examples to illustrate the different ways my model calculates airflow by predicting the performance of the same truck when it operates at different altitudes. In all 5 cases there's a no AF open turbo inlet that ingests 85 F air and the only parameter that changes is the AAP. Case #1 is at the reference altitude of 1,500 ft, cases #2 and #3 are at sea level, and cases #4 and #5 are at a 5,100 ft altitude.

1) AAP=13.95 psi, RPM=2,900, TPR=2.50, BP=19.9 psi, MAF=47.3 lb/min, AFCFM=683 ft^3/min, TDHP=46.9 HP

2) AAP=14.70 psi, RPM=2,900, TPR=2.50, BP=21.1 psi, MAF=49.9 lb/min, AFCFM=684 ft^3/min, TDHP=49.5 HP

3) AAP=14.70 psi, RPM=2,900, TPR=2.35, BP=18.9 psi, MAF=47.3 lb/min, AFCFM=549 ft^3/min, TDHP=44.1 HP

4) AAP=12.20 psi, RPM=2,900, TPR=2.50, BP=17.3 psi, MAF=41.2 lb/min, AFCFM=680 ft^3/min, TDHP=40.8 HP

5) AAP=12.20 psi, RPM=2,900, TPR=2.95, BP=22.5 psi, MAF=47.2 lb/min, AFCFM=780 ft^3/min, TDHP=54.0 HP

In case #2 if you maintain the same TPR the BP and MAF increase but the AFCFM remains the same and the TDHP that's required to move the additional MAF increases. In case #3 if the TPR is adjusted lower to get the same MAF as before then this original MAF can be gotten at lower values of BP, AFCFM, and TDHP. I haven't shown this point but you could also operate at the original TDHP=46.9 HP instead of at the original TPR=2.50 and this gives a MAF=48.5 lb/min a TPR=2.425, and a BP=20 psi.

In case #4 if you maintain the same TPR the BP and MAF decrease but the AFCFM remains the same and the TDHP that's required to move the lower MAF decreases. In case #5 if the TPR is adjusted higher to get the same MAF as before then this original MAF can be gotten but it requires higher values of BP, AFCFM, and TDHP.

When you operate at lower altitudes the only thing you have to worry about is how much you want to let off the throttle and still have equal or even better performance without working the turbo any harder but at higher altitudes you have to apply considerably more throttle just to recover your original performance and this works the turbo harder by requiring higher values for both the TPR and the TDHP turbo parameters.

If you look at computer graph #3 you'll see that the 3,000 RPM line hits the maximum 110,000 rpm speed line at a TPR=3.3 and that the 2,400 RPM line hits the maximum 110,000 rpm speed line at a TPR=3.5. So I decided to do some DRHP vs RPM computer runs at TPR values of 3.5, 3.2, 2.9, and 2.6 to see how closely the predicted DRHP matched the no AF dyno run with a measured DRHP=424.9 HP and the graph below is from those computer runs.



As you can see the above blue curve for a TPR=3.2 is dead on and it leaves the potential to push the turbo a little harder if needed so I used a TPR=3.2 as the no AF baseline and generated the more detailed DRHP and DRTQ graph below.



In the next installment I'll add the 6637 and AFE air filters and see how their performance compares with the no AF baseline above! Just like I did in the above altitude example I can compare performance at a constant TPR and at a constant TDHP.

 

Now if you look at the above curves the only way for the "analytical jury" to render a completely accurate verdict is for me to take the time (which I'll eventually do) to hand enter into my new PSD dyno model 30 data points between a 0" to 30" H2O restriction for the lower curve I drew through the "3 Donaldson points" and call that the "6637 Donaldson curve" and then do the same thing for the combination of the upper two curves by combining the "gold" AFE curve with the line I drew through the "2 data points F250 posted" and for lack of a better name calling that combined curve the "6637 F250 curve" and if he wants me to rename that curve after myself that's fine with me because being childless I've never had the honor of having anything named after me.

As you can see from my previous computer graph #2 for TPR=1.0 the CFM=433 and for TPR=4.0 the CFM=1,011 so what I did instead of hand entering the exact restriction data from a 0" to 30" H2O restriction for the "6637 Donaldson curve" was to let my computer automatically generate a "linear" H20 restriction from a 0" H2O restriction at TPR=1.0 (CFM=433) to a 30" H2O restriction at TPR=4.0 (CFM=1,011) and as you can see from the "30" in the "pink" cell in my PSD model data entry table below I call the point at TPR=4.0 the "AFPD=Air Filter Pressure Drop At 1,000 CFM, Inches H2O".



In the CFM vs H2O restriction graph above this automated procedure is basically like drawing a line between the points 1,000 CFM at a 30" H20 restriction and 400 CFM at a 0" H20 restriction and calling that the estimated "6637 Donaldson curve" which I'll just call AF#1.

When you enter "15" into that "pink" cell in my PSD model data entry table the computer generates a "linear" H20 restriction from a 0" H2O restriction at TPR=1.0 (CFM=433) to a 15" H2O restriction at TPR=4.0 (CFM=1,011) and in the CFM vs H2O restriction graph above this automated procedure is basically like drawing a line between the points 1,000 CFM at a 15" H20 restriction and 400 CFM at a 0" H20 restriction and calling that the estimated "6637 F250 curve" which I'll just call AF#2.

Of course when you enter "0" into that "pink" cell you get a 0" H2O restriction between TPR=1.0 and TPR=4.0 and this is the no AF baseline which I'll just call No AF. In the two (long awaited) graphs below I didn't actually think to label them with AF#1 and AF#2 but the actual 30" H20 and 15" H20 labels are included along with the actual H20 restriction at the various airflow operating points.

The only input parameter that changed in my PSD model data entry table between the runs for the no AF baseline which is the upper solid curve in each comparison graph below and the runs for the AF#1 and AF#2 which are the lower dashed and dotted curves in each comparison graph is the 30, 15, and 0, entries in the "pink" cell for air filter restriction.

After everyone has had a chance to digest these results and come to their own conclusions I'll post my own thoughts on what they might or might not mean as well as a fundamental reason why using the DRHP from an inertial dyno run or the ET from a drag strip run aren't very accurate ways of determining if a "real world" performance gain has been achieved due to some particular mod!

 

This will be the first in a series of post mortem posts...

Exhibit A

DRHP 424.9, TPR 3.20, BP 29.2, MAF 57.8, TDHP 70.3, NAF Baseline

DRHP 424.9, dyno run #3) with no air filter

Conclusions... Model and dyno run agree exactly for the no air filter baseline


Exhibit B...

AF Restriction 23.5" H2O @ 809 CFM & 407.6 DRHP

DRHP 407.6, TPR 3.35, BP 28.4, MAF 56.0, TDHP 70.5, AF @ equal TDHP

DRHP 407.0, dyno run #2) with AFE air filter

Conclusions... Since the point (23.5" H2O @ 809 CFM) falls exactly on the red "6637 Donaldson curve" and the AFE dyno run gave almost the same DRHP as the model did for the "6637 Donaldson curve" this implies that the gold "advertised AFE curve" is a hoax and that the AFE actually performs like the red "6637 Donaldson curve" in a dyno test!


Exhibit C...

AF Restriction 11.2" H2O @ 819 CFM & 415.5 DRHP

DRHP 415.5, TPR 3.25, BP 28.6, MAF 56.7, TDHP 69.8, AF @ equal TDHP

DRHP 413.7, dyno run #1) with 6637 air filter

Conclusions... Since the point (11.2" H2O @ 819 CFM) falls only 80 CFM below the gold "advertised AFE curve" and 270 CFM above the red "6637 Donaldson curve" and the 6637 dyno run gave almost the same DRHP as the model did for the estimated gold "advertised AFE curve" this implies that the 6637 actually performs close to the gold "advertised AFE curve" in a dyno test!

 

Exhibit A...

AF=00.0" H2O @ CFM=ALL, DRHP=424.9, ..........MAF=57.8, ..........BP=29.2, no AF baseline

AF=11.0" H2O @ CFM=810, DRHP=412.2, -2.9%, MAF=56.1, -2.9%, BP=28.0, -1.2, @ equal TPR

AF=22.0" H2O @ CFM=785, DRHP=399.9, -5.9%, MAF=54.4, -5.9%, BP=26.7, -2.5, @ equal TPR

Conclusions... When comparing an AF with no AF at equal TPR the % decrease in DRHP due to the AF is the same as the % decrease in MAF because there's more than enough fuel to combust all of the available air! How I got these delta changes in BP below is discussed later but here's the answer...

For delta AFPD=11" H2O the delta BP={(3.2)(11/27.68)}=1.27 psi

For delta AFPD=22" H2O the delta BP={(3.2)(22/27.68)}=2.54 psi


Exhibit B...

AF=22.0" H2O @ CFM=785, DRHP=399.9, ...........MAF=54.4, ......... BP=26.7, higher restriction AF baseline

AF=11.0" H2O @ CFM=810, DRHP=412.2, +3.1%, MAF=56.1, +3.1%, BP=28.0, +1.3, @ equal TPR

Conclusions... When comparing a lower restriction AF with a higher restriction AF at equal TPR the % increase in DRHP due to the lower restriction AF is the same as the % increase in MAF because there's more than enough fuel to combust all of the additional air! How I got this delta change in BP below is discussed later but here's the answer...

For delta AFPD=11" H2O the delta BP={(3.2)(11/27.68)}=1.27 psi

Regarding Exhibit A and Exhibit B delta change in BP... since the BP is given by... BP={(AAP)(TPR-1)}-{(TPR)(AFPD/27.68)}-(ICPD) psi ...and since the TPR is constant and to first order the ICPD remains unchanged then the delta change in BP is given by... delta BP={(TPR)(delta AFPD/27.68)}.

Since the heat generated by the turbo only depends on TPR and TIAT and since neither one of these has changed the MAT remains the same and since the MAF is given by... MAF={(0.347)(VE)(RPM)(AAP+BP)}/{(MAT+459.67)} lb/min ...you see that the only parameter that's changed is the BP so that at a given RPM any increases or decreases in MAF due to changes in AF restriction are due solely to the effect delta AFPD has on delta BP in the above equation!

As I'll show in my next post things are more complicated when you compare at equal TDHP because now both the BP and the TPR change and the change in TPR results in a change in turbo generated heat so that the MAT changes and then the changes in MAF are due to both the changes in BP and in MAT!