I've been putting out some not quite correct information, which technically speaking means it's WRONG! I've always said that it's all about MAF because it's the lb/min of air flow that determines your HP, however, I just learned something new by playing with my engine model. There's a little more to the MAF vs HP story, and having EQUAL values of MAF does NOT yield EQUAL values of HP, it all depends on how the MAF is created. It turns out that running higher operating temps is even MORE detrimental to your HP than I've been claiming. I was thinking you can compensate for higher temps by pushing the turbo harder to get back the lost MAF due to higher temps, and that would get you back to your original HP, but it does NOT work quite that way!

The bottom line is that the MAF depends on the CAP=Cylinder Air Pressure, and the CAP depends on the MAP=Manifold Air Pressure via the VE=Volumetric Efficiency which depends on RPM and on BP, and if that's all there were to the story then all MAF would be equal no matter how it was created, and would yield the same HP. For example, if you had RPM=2200, BP=20 psi, MAF=27.4 lb/min, RWHP=245, and your IAT increased, or you turned on the A/C and its waste heat reduced the cooling of the charge air going through the IC, and this reduced your MAF and RWHP, you could push the throttle a little harder to increase the BP, to increase the MAP, to increase the CAP, to increase the MAF back to its original value of MAF=27.4 lb/min, and this would get back your original RWHP=245 at RPM=2200.

But it doesn't work out quite this way, and in the new example I give you need to increase the BP from 20 to 27 psi to get back to a MAF=27.4 lb/min, which was reduced because of increases in several operating temps, and you only get back to a RWHP=237 at MAF=27.4 lb/min instead of your original RWHP=245 at MAF=27.4 lb/min.

It turns out that the TE= Thermodynamic Efficiency of the engine for converting the heat energy from the fuel into HP, depends on the CAP, and the TE also depends on RPM and on BP! So the increase in BP at a given RPM gets you back to the same MAF as before, but the TE is now a little lower at the higher CAP, so not quite as much HP is produced by the same lbs of air and fuel as before.

Wife says I've got to go, but I'll post what I've got for now. First two pics are diagrams of the Previous & Current Conditions, and the other two pics are graphs of the same conditions.

 

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So I guess you are saying that you were more right than you thought? Just didn't know it?

 

So based on the BP info you just obtained is there a optimum CAP for maximum RWHP? Can the wastegate or a different wastegate be used/set to keep the CAP at the optimum CAP all the way through the power band? If i read that right???

On a side note doesnt the IAT input effect the BP? So with a tuned wastegate and accurate IAT measurements you could achieve the optimum CAP via BP?

 

This post started out as an addition to the "tail wags the dog" thread to show what happens at higher operating temps. For the "previous condition" with lower temps, BP=20 psi gave IVAF=365 CFM & CMAF=27.4 lb/min. The "current condition" with higher temps requires a BP=24 psi to get back to the previous IVAF=365 CFM, but even though you have the same CFM as before, you now have only a CMAF=26.4 lb/min, which is 1 lb/min less than the "previous condition" with BP=20 psi & IVAF=365 CFM. To get back to the previous CMAF=27.4 lb/min, the BP has to be increased to BP=27 psi, and I thought that would get back to the same HP as in the previous case.

Then I decided to label the RWHP on all the diagrams and graphs for the various conditions, and when I checked my model I was surprised to find I was getting less HP for the current case even though it has the same CMAF as the previous case? I started digging into my model, and realized that the TE was less at the current condition vs the previous condition, and that's why the HP was less even though the CMAF was the same.

The 2 cartoon pics below illustrate the thermodynamics involved in the conversion of (BTU) heat energy into mechanical work (HP) for a diesel. The pic of the piston going up and down relates to the graph of pressure, volume, and temp, to help visualize what the V1, V2, V3, Pa, Pb, Pc, Pd, and Ta,Tb,Tc,Td on the graph are for the 4 strokes. Note that gamma is a property of the air fuel mixture, the Rc is the PSD compression ratio, the Re is the expansion ratio which is determined by the volume increase at the start of the power stroke during the fuel injection and combustion process, and the TE=Thermodynamic Efficiency of the engine for converting the BTU heat energy from the diesel fuel into HP is a complicated function of a whole bunch of stuff!

For the PSD, Rc=17.5, and from the pics it looks like Re is about (1/3)Rc=(1/3)17.5=5.8. If you put gamma=1.4, which is the value for air only, Rc=17.5, and Re=5.8 into the equation for TE, you get TE=58% as the ideal efficiency for the conversion cycle. Now you have to factor in friction, the variation of Re with RPM and BP, the an angle (momentum arm) of the crankshaft during the peak combustion pressure for converting the expansion force to torque, etc... to complete the process.

At BP=0 psi & RPM =2000, which is the torque peak for my early 99, I'm coming up with a maximum value of TE=38%. In my model this maximum TE is tweaked lower as a function of RPM and BP, based on some calculations and estimates using some data I've found. Larger exhausts and upgraded turbos which effect back pressure, intake mods, and any number of other factors determine the exact TE for a given engine. I'm not claiming to have all of this exactly correct (numerically speaking) in my model for my early 99, but I think it's correct enough to explain why the same value of CMAF gives different values of HP due to the differences in TE for the different operating conditions that give the same CMAF. I'm definitely open to any suggestions in this thermodynamics area of engine modeling.

 

This time I'll talk in terms of air molecules to see if that's any more intuitive. The # of molecules in a mole of any substance is Avogadro's number of 6.022x10^23. A mole of dry air weighs 28.97 g, which is 0.0638 lb. Therefore, the # of air molecules per lb is (6.022x10^23)/(0.0638)= 9.44x10^24 molecules/lb.

The standard equation for air density is AD={(2.70325)}{(AP)/(AT+459.67)}, where AD=Air Density, lb/ft3, AP=Air Pressure, psi, and AT=Air Temperature, F, and multiplying this by (9.44x10^24 molecules/lb) gives AD={(2.55x10^25)}{(AP)/(AT+459.67)}, where now AD is the # molecules/ft3. If you plug AP=14.7 psi & AT=32 F into this molecular AD equation, you get AD=7.6x10^23 molecules/ft3, and those who read my post on how you can breathe in outer space might recall that this was my starting off value for the standard molecular density at sea level.

If you plug AP=14.7 psi & AT=70 F into the molecular AD equation, you get AD=7.1x10^23 molecules/ft3, so the net effect of increasing the temp from 32 F to 70 F is to decrease the # of air molecules in each ft3 of volume by about 7%. This occurs because at the higher temp the molecules have a higher random velocity due to their higher thermal kinetic energy, and this increases the average separation between molecules so that fewer molecules fit into each ft3 of volume.

I'm repeating my table of basic air flow equations below, and anywhere you see (2.70325) you can substitute (2.55x10^25) and convert to the # of molecules. For example the equation for CMAF={(0.1285)(RPM)}{[(2.70325)(BP+AAP)]/(MAT+459.67)}{(VEF)} lb/min becomes CMAF={(0.1285)(RPM)}{[(2.55x10^25)(BP+AAP)]/(MAT+459.67)}{(VEF)} molecules/min, and so on. In the previous vs current conditions discussed above, the CMAF of 27.4 lb/min becomes 2.6x10^26 molecules/min.

If you look at the above pic of the single cylinder engine, when the piston is at BDC the pink region is the CV=Cylinder Volume, ft3, and for the PSD, CV=0.0322 ft3. For every 2 rotations of the crankshaft the CV fills with air molecules during the intake stroke, they're compressed during the compression stroke, mixed with fuel and combusted to produce the power stroke, and then expelled during the exhaust stroke. For every 2 rotations of the crank, 1 CV of air flows into & out of, and therefore through a single cylinder engine, and 8 CV's flow through the PSD.

The # of molecules that fill the CV during each intake stroke determines the molecular flow, and this depends on the CAD which is determined by the MAD and the VE, and the densities depend on the pressures and temps, and everything can be traced back to the conditions at the air filter inlet, the heating caused by the turbo, and the cooling provided by the IC. However, the cylinder volume flow only depends on how many CV's fill and empty per min, and this only depends on the # of cylinders and the RPM, and for the 7.3 L, 8 cylinder PSD this is gives a CVAF={(0.1285)(RPM)} which for RPM=2200 is 283 CFM.

The molecular flow for a given CVAF is CMAF={(CVAF)}{(CAD)}, where CAD is in molecules/ft3. In both the previous condition at BP= 20 psi, and the current condition at BP=27 psi, the CAD=0.0967 lb/ft3=9.1x10^23 molecules/ft3, and (283)(0.0967)=27.4 lb/min, and (283)(9.1x10^23)=2.6x10^26 molecules/min.

So for both the previous (BP=20 psi) and current (BP=27 psi) operating conditions, 2.6x10^26 molecules/min flow into the air filter inlet, through the turbo and IC, through the manifold, boots, and plenums, and into the cylinders. However, due to the TE effect already discussed, combusting the same # of air molecules with the same amount of fuel at BP=20 psi produces a RWHP=245, but only a RWHP=237 at a BP=27 psi.

Well even if this wasn't any easier to understand, at least you know how many air molecules fit into a ft3 volume as a function of pressure and temp, and this will be useful if you read my upcoming post on how operating in the mountains at high altitude effects the performance of the turbo and engine.

 

To recap, the Previous Condition: IAT=70 F, TIAT=80 F, A/C Heat To IC=90 F, requires a BP=20 psi at RPM=2200 to give a CMAF=27.4 lb/min and a RWHP=245, which is the requirement for me to tow my 5er up a 3% grade at 71 MPH.

The Current Condition: IAT=90 F, TIAT=110 F, A/C Heat To IC=160 F, requires a BP=27 psi at RPM=2200 to give a CMAF=27.4 lb/min, but due to a lower TE at the higher BP, this same CMAFF only gives a RWHP=237, which is 8 HP less than required to tow my 5er up a 3% grade at 71 MPH.

I checked my model, and came up with several options to tow a 3% grade at the higher temps associated with the Current Condition.

Option 1 is to downshift to 3rd gear and climb the 3% grade at 71 MPH as before, but this requires a BP=29 psi at RPM=3100 which gives a CMAF=34.2 lb/min and a RWHP=245, which is the requirement for me to tow my 5er up a 3% grade at 71 MPH. However, even on its best day with the Banks Big Head my truck could only make a BP=27 psi, so this option is only a theoretical one for me.

Option 2 is to stay in 4th gear and slow down to 65 MPH, which reduces the required RWHP by about 40 to RWHP=205, and now a BP=19 psi at RPM=2000 gives a CMAF=23.0 lb/min and a RWHP=205, which is the requirement for me to tow my 5er up a 3% grade at 65 MPH.

Option 3 is to downshift to 3rd gear and slow down to 50 MPH, which reduces the required RWHP by about 110 to RWHP=135, and now a BP=12 psi at RPM=2200 gives a CMAF=21.2 lb/min and a RWHP=135, which is the requirement for me to tow my 5er up a 3% grade at 50 MPH.

Option 4 is to downshift to 2nd gear and slow down to 30 MPH, which reduces the required RWHP by about 180 to RWHP=65, and now a BP=5 psi at RPM=2000 gives a CMAF=16.2 lb/min and a RWHP=65, which is the requirement for me to tow my 5er up a 3% grade at 30 MPH.

In terms of both types of efficiencies (VE&TE), 2000 is the most efficient RPM, and operating at RPM=2000 at the lowest possible BP provides the best overall efficiency.