Well it's true that during their dyno runs high HP diesels spew forth a lot of black smoke, and they do the same thing at the drag strip, but does a lot of smoke really mean they're making their maximum possible HP? It seems the Banks "smokeless" diesel does quite well at the drag strip without making any smoke at all!

On the thread... How Much Coal Do You Roll???... someone lamented that they couldn't make smoke with their stock engine so I suggested...

That suggestion was intended to be humorous, but it does illustrate that more smoke doesn't necessarily equate to more HP, and that set me to thinking this is a good topic for one of my technical posts!

The AFR=Air Fuel Ratio given by AFR=MAF/MFF holds the key for unlocking your engine's potential to produce more HP as well as its potential for spewing forth a lot of black smoke! In this AFR equation MAF=Mass Air Flow lb/min and MFF=Mass Fuel Flow lb/min.

The MAF is given by MAF={(VE)(CID)(RPM)(AAP+BP)}/{(MAT+459.67)(1,278.46)} lb/min where VE=Volumetric Efficiency a ratio 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. You can get real time values of your MAT using AE or you can pull to the side of the road after doing my road test procedure and use an "IR gun" to shoot temps of the intake plumbing to get an estimate of MAT.

To illustrate how this equation works I'll use some typical numbers from my PSD engine model. I'll use an AFIAT=Air Filter Inlet Air Temperature F of 80 F and a AFIAP=Air Filter Inlet Air Pressure psi of AAP=14.53 psi which corresponds to an altitude of 315 ft because these values work out to give a TIAT=Turbo Inlet Air Temperature F of 85 F and a TIAP=Turbo Inlet Air Pressure psi of 13.95 psi and these are the "standard conditions" at the turbo inlet when turbo compressor maps are measured.

The MAT values used in the following examples assume the A/C is off and that the ICEE=Intercooler Exchanger Efficiency a ratio 0 to 1 is 0.75 and that the ICEAT=Intercooler Exchanger Air Temperature F is 85 F which is only 5 F higher than the 80 F ambient temperature. So 85 F is the lowest possible outlet temperature from the IC. Running the A/C can heat the IC to 120 F or even higher which is why I'm glad that my Freightliner has its A/C condenser below the IC instead of in front of the IC which is the case for pickups.

The maximum VE=0.830 occurs at RPM=2,000 & BP=0 psi and the VE is reduced at higher values of BP and at both higher and lower values of RPM. To illustrate this maximum VE=0.830 condition assume you're in 4th gear with a 4R100 and going down a slight grade and applying 20% throttle to maintain 70 MPH. For RPM=2,000, BP=0 psi, VE=0.830, my model gives MAT=87 F and these numbers give... MAF={(VE)(RPM)(AAP+BP)}/{(MAT+459.67)(2.8817)}={(0.830)(2,000)(14.53+0)}/{(87+459.67)(2.8817)}=15.3 lb/min.

Now assume you come to a grade and apply 60% throttle to maintain the same road speed, now RPM=2,000, BP=10 psi, VE=0.8136, and MAT=121 F and these numbers give... MAF={(VE)(RPM)(AAP+BP)}/{(MAT+459.67)(2.8817)}={(0.8136)(2,000)(14.53+10)}/{(121+459.67)(2.8817)}=23.9 lb/min.

Now assume you downshift to 3rd gear and apply 100% WOT and now have RPM=2,800, BP=26 psi, VE=0.6975, and MAT=158 F and these numbers give... MAF={(VE)(RPM)(AAP+BP)}/{(MAT+459.67)(2.8817)}={(0.6975)(2,800)(14.53+26)}/{(158+459.67)(2.8817)}=44.5 lb/min.

Now you pull into a rest stop to make some temperature measurements with your "IR gun" and let the MAT cool back down to 87 F. Now you have RPM=600, BP=0 psi, VE=0.6706, and MAT=87 F and these numbers give... MAF={(VE)(RPM)(AAP+BP)}/{(MAT+459.67)(2.8817)}={(0.6706)(600)(14.53+0)}/{(87+459.67)(2.8817)}=3.7 lb/min.

So hopefully you now see what parameters determine MAF and why it varies from 3.7 lb/min at idle RPM to 44.5 lb/min at the RPM for maximum HP. In the next installment I'll cover MFF, AFR, and how AFR effects combustion, EGT, HP, and smoke!

 

YEP ernesteugene most of your equation are correct. But as we know Banks has put in a lot of research and proved that smoke dont' always make power. They may have not figured it the same as you did but it still works.

 

Yep, I double checked em and they look good to me

 

I did a quick Excel spreadsheet with the calculations. They seem to work out but am curious about the MFF portion and the resulting AFR value. Looks like I can't save an xls file here so try http://www.ktech-usa.com/Seven-Three...oke%20Calc.xls

 

Ernestegene, great write up as always, yes smoke rolling is wasted HP and higher EGT's, the trick is get it to burn clean, cooler air and lots of it helps. I hate to say this but Honda has a diesel in testing that burns cleaner than a gas engine.

 

I love your posts Eugene! Always informative!

 

Great post Gene. I wasn't aware that the AC would raise the IC temps so much.

jwhitetail, I think there's a difference in smoke and emissions for this particular conversation, but Gene did bring up Banks truck, which I thought also runs a DPF (which smoke would plug up pretty quick).

 

Someday I'll do a post on my intercooler equations!

I've read rumors that Banks is cheating by using NOS to go fast without smoke, but I haven't seen any reference to him employing a DPF? I'm still in Vegas and the Banks truck is supposed to run in Phoenix in a couple of weeks maybe I should go check it out? When you listen to the videos of his drag runs it doesn't sound like it's got a DPF?

Banks breaks new ground for diesel dragsters...
http://www.bosch-diesel.us/pool/pdf/...eTechGroup.pdf

"...While torque to power ratio for truck diesels when measured in pound feet to horsepower is typically about 2:1 Banks is developing engines that are closer to 1.5:1 ...Lower torque higher horsepower is the goal ...We‘re right at 1,300 hp with the 6.6L Duramax ...The S10 has already achieved quarter mile performance of 7.72 seconds at 179 mph and Banks is aiming for under 7 seconds and 200 mph with a diesel dragster that is under development ...the Bosch diesel system is the key to our efforts. It allows us to have a combustion recipe that is smoke free and highly efficient..."

Well at least one thing is clear from the above. Since he's an engineer Banks knows that acceleration is all about HP so I'm sure he agrees with my quote...

In the world of Physics all aspects of a truck's performance are determined by its ability to rapidly produce some form of ENERGY, and a truck's TIME rate of ENERGY production is determined solely by its HP. The truck that produces the largest TIME rate of increase in its kinetic ENERGY wins the drag race, the truck that produces the largest TIME rate of increase in its potential ENERGY wins the hill climb, and the truck's maximum speed is determined by how much of the truck's kinetic ENERGY is available to transfer to the air molecules in front of it to increase their kinetic ENERGY which is what's required to push them out of the truck's path through the air.

 

Before moving on let me explain how I derived the equation MAF={(VE)(CID)(RPM)(AAP+BP)}/{(MAT+459.67)(1,278.46)} lb/min. The boost gauge is measuring the BP which is the "gauge" pressure in the intake manifold/spider so that MAP=Manifold Air Pressure psi is given by MAP={AAP+BP} psi. The cylinder fill condition at BDC of the intake stroke isn't perfect so the CAP=Cylinder Air Pressure psi is a little less than the MAP and CAP={(VE)(MAP)}={(VE)(AAP+BP)} psi.

So the VE specifics how efficiently the cylinder gets filled with air under various operating conditions that depend on BP and RPM. For a PSD the VE has a maximum value at 2,000 RPM and BP=0 psi. As BP increases the air flow becomes more turbulent and this decreases the filling efficiency so that the VE progressively decreases as BP increases.

Well below 2,000 RPM the intake valves remain open longer which aids cylinder filling, but the air flow velocity is lower and the inertia of the air causes the flow to have a starting and stopping action, and the net effect is a progressive reduction in VE as the RPM decreases below 2,000 RPM due to the air's inertia.

In the vicinity of 2,000 RPM the air flow velocity is higher and the inertia of the air aids filling. When an intake valve opens the built up inertia provides a RAM air effect and pushes more air into the cylinders than you'd get at lower RPM where the flow has more of a starting and stopping action.

However at higher RPM well above 2,000 RPM the intake valves aren't open long enough for the RAM air effect to work well and the net effect at higher RPM is a progressive reduction in VE as the RPM increases above 2,000 RPM due to shorter intake valve opening times.

The VE also relates the CAD=Cylinder Air Density lb/ft^3 to the MAD=Manifold Air Density lb/ft^3 by CAD={(VE)(MAD)} lb/ft^3 where MAD={(2.70325)(MAP)}/{MAT+459.67}={(2.70325)(AAP+BP)}/{MAT+459.67} lb/ft^3. As the piston descends to BDC of the intake stroke the SCV=Swept Cylinder Volume ft^3 is given by SCV={(Pi/4)(Bore)^2(Stroke)} ft^3 and the CAM=Cylinder Air Mass lb is given by... CAM={(CAD)(SCV)}={(VE)(MAD)(SCV)}={(VE)(2.70325)(A AP+BP)(SCV)}/{MAT+459.67} lb.

There's an intake stroke which ingests a CAM={(VE)(2.70325)(AAP+BP)(SCV)}/{MAT+459.67} lb of air into each cylinder once every 2 revolutions of the crankshaft so for an engine with Nc cylinders operating at a given RPM the MAF={(1/2)(Nc)(CAM)(RPM)} lb/min which gives MAF={(VE)(2.70325)(1/2)(Nc)(SCV)(AAP+BP)(RPM)}/{MAT+459.67} lb/min, and using CID={(1,728)(Nc)(SCV)} in this equation gives...

MAF={(VE)(CID)(RPM)(AAP+BP)}/{(MAT+459.67)(1,278.46)} lb/min

...and that's how I derived the MAF equation.

 

Man I've got a headache

 

Your probably right. I remember reading about him testing his exhaust systems with the DPF and even trying them out on a track, but that probably doesn't mean the big tracks, just some local runs for test purposes.

 

Correct. Someone posted a while back that they counted 4 nitrous bottles mounted in the truck. There was also a rumor about the fuel they were using too, but I'm not sure. The fuel part I would have to see to believe. Still, the amount of power they make is impressive. For that I'll give Banks cudos, but it still doesn't make me a Banks fan. Gale needs to stick to racing, since that's what he does best.

At any rate, I've got my truck running smokeless down the track too. The only smoke you'll see from me is while I am staging and trying to light the turbo. Once the tree is lit and I launch, there's no more smoke out the pipe.

If your truck is still smoking at WOT (track or dyno runs), it means you have room for more air. That can be fixed by either properly sizing your turbo(s), or by adding nitrous to the mix. I've also found that injection pressures, fueling tables, and injection timing will add or remove smoke.

The reason for this torque/hp ratio change is simple..... it's RPM's. Our trucks are redlining at a little over 3000 RPM's, and the Banks truck is hitting close to 6000 RPM's. The higher a diesel's RPM's, the closer the torque to hp ratio will become. Also at higher RPM's, several other things happen. One, you'll run less smoke, and two, you actually put less strain on the engine. Imagine the cylinder pressures it takes to make full power at 2000 RPM's. Now make that same power at 4000 RPM's. If you have the same power output at higher RPM's, it means that you'll have lower cylinder pressures at 4000 RPM's verses 2000 RPM's. As Ernest pointed out, power output is based on time. You have more "time" at 4000 RPM's than you do at half the engine speed.

To put this in laymen's terms. Imagine you have a hand ratchet and an air ratchet. Both are capable of outputting 100 ft/lbs of torque to tighten a bolt. Which one will win the race? Your arm, or the automatic air ratchet? Well the air ratchet can apply the same power at a greater speed, so the air ratchet will tighten the bolt first, while your arm is getting tired trying to keep up. It's no contest. That's probably the easiest way I can explain it. Ernest, you think that sums it up for the dummies like me who don't do math? LOL

 

The banks truck does not have DPF, they run NOS because there is no intercooler.

 

Time to get out my gun.......

Banks' Sidewinder S-10 Drag "truck" would smoke just as bad as any other high-HP diesel truck if it weren't for the nitrous oxide...probably worse because it relies so much on the N2O to operate properly. It uses multiple stages of computer-controlled nitrous oxide injection to control smoke during spool, launch, and down the track.

The part about it that absolutely makes me want to kill is that Gale Banks and his cronies are adamant that the truck actually produces less pollution than other trucks that smoke so much. NOT TRUE. The truck produces the same amount of actual pollution as any other truck....and again, probably more because it is able to burn so much more fuel.

Yes, the nitrous oxide allows the truck to burn more fuel and burn it completely. In turn, this produces less VISIBLE smoke than competitors not using N2O.
This is where Gale's uncanny sales pitches come from. Since any Tom, Dick, or Harry can SEE the black smoke from other trucks, they automatically assume that they're "dirty". Gale Banks claims "cleaner" with his Sidewinder "truck" because there is substantially less VISIBLE smoke......but the actual pollution coming from the exhaust is the same or worse (since MORE fuel is actually burned).

Black smoke IS THE REASON we have stupid emission garbage like particulate filters and the programming needed to run the regeneration cycle that USES MORE FUEL than a truck without a DPF. There again. A heavier-polluting truck that APPEARS to pollute less.

Rubbish.

 

Well here's the first installment on MFF, AFR, etc... and several more posts will be required to complete the HP vs smoke story! I'm glad someone's double checking my equations if you find anything that you even think might be an error please let me know and I'll recheck everything yet again.

To recap from the air flow side of the AFR equation we've got two forms of the MAF equation... MAF={(VE)(CID)(RPM)(AAP+BP)}/{(MAT+459.67)(1,278.46)} lb/min and... MAF={(1/2)(Nc)(CAM)(RPM)} lb/min where CAM=Cylinder Air Mass lb is the lb weight of air that's ingested by a single cylinder during the intake stroke. Combining these two equations gives... CAM={(VE)(CID/Nc)(AAP+BP)}/{(MAT+459.67)(639.23)} lb where the term (CID/Nc) is just the SCV in^3.

To see how this equation works for a 7.3L PSD with CID=443.65 in^3 and Nc=8 lets calculate the CAM for this example...

"Now assume you downshift to 3rd gear and apply 100% WOT and now have RPM=2,800, BP=26 psi, VE=0.6975, and MAT=158 F and these numbers give... MAF={(VE)(RPM)(AAP+BP)}/{(MAT+459.67)(2.8817)}={(0.6975)(2,800)(14.53+26)}/{(158+459.67)(2.8817)}=44.5 lb/min."

...which gives... CAM={(VE)(CID/Nc)(AAP+BP)}/{(MAT+459.67)(639.23)}={(0.6975)(443.65/8)(14.53+26)}/{(158+459.67)(639.23)}=0.0039706 lb of air ingested by each cylinder during their intake strokes.

At the risk of confusing CFM with VAF=Volume Air Flow ft^3/min=cfm I'll define CFM=Cylinder Fuel Mass lb as the lb weight of fuel that's injected into a single cylinder during the power stroke. For each cylinder a power stroke occurs once every 2 revolutions of the crankshaft so there's an injection of CFM lb of fuel into each cylinder once every 2 revolutions of the crankshaft. For an engine with Nc cylinders operating at a given RPM this gives a MFF=Mass Fuel Flow lb/min equal to... MFF={(1/2)(Nc)(CFM)(RPM)} lb/min.

Since we normally buy fuel by the gallon and watch our fuel gauge decline by the hour as we tow a heavy load down the road we can convert MFF lb/min to VFF=Volume Fuel Flow gal/hr=gph as follows... MFF={(VFF/60)(DFD)} lb/min and VFF={(60)(MFF)}/{DFD} gal/hr where DFD=Diesel Fuel Density lb/gal which is typically about 7 lb/gal for #2 diesel at room temperature. Putting all this into one equation gives... VFF={(30)(Nc)(CFM)(RPM)}/{DFD} gal/hr, and solving this for CFM gives... CFM={(VFF)(DFD)}/{(Nc)(RPM)(30)} lb.

From the perspective of a single cylinder the AFR={(CAM)/(CFM)}=[{(VE)(CID/Nc)(AAP+BP)}/{(MAT+459.67)(639.23)}]/[{(VFF)(DFD)}/{(Nc)(RPM)(30)}] and rearranging terms and noting that the term Nc cancels out gives... AFR={(VE)(CID)(RPM)(AAP+BP)}/{(MAT+459.67)(VFF)(DFD)(21.308)}

From the perspective of the entire engine the AFR={(MAF)/(MFF)}=[{(VE)(CID)(RPM)(AAP+BP)}/{(MAT+459.67)(1,278.46)}]/[{(VFF/60)(DFD)}] and rearranging terms gives... AFR={(VE)(CID)(RPM)(AAP+BP)}/{(MAT+459.67)(VFF)(DFD)(21.308)} which is the same equation as from a single cylinder's perspective once again demonstrating that the "whole" is equal to the "sum" of its parts, that is until you burn a hole in one of the pistons by trying to make too much smoke!

Lets apply the AFR equation to the... "downshift to 3rd gear and apply 100% WOT" ...example above where we already calculated MAF=44.5 lb/min and CAM=0.0039706 lb. From my PSD engine model I get VFF=21.21 gph which gives MFF={(VFF/60)(DFD)}={(21.21/60)(7)}=2.475 lb/min and CFM={(VFF)(DFD)}/{(Nc)(RPM)(30)}={(21.21)(7)}/{(8)(2,800)(30)}=0.0002209 lb.

So from the perspective of a single cylinder the AFR={(CAM)/(CFM)}={(0.0039706)/(0.0002209)}=17.97 and from the perspective of the entire engine the AFR={(MAF)/(MFF)}={(44.5)/(2.475)}=17.97 and since both approaches agree everything looks good except that this AFR produces some smoke in the process of getting about the maximum possible FWHP from this amount of air flow, and in the next post I'll discuss why this is.

In my next installment I'll discuss the TE=Thermodynamic Efficiency a ratio from 0 to 1 which is the efficiency for converting the TIME rate of release of the fuel's chemical ENERGY into the TIME rate of increase in the heat ENERGY of the cylinder's air mass and how this determines the "piston HP" and how that relates to the FWHP.

I'll also discuss how TE varies with RPM & BP, BSFC=Brake Specific Fuel Consumption lb/hr/hp, how and why CAM effects EGT, including a "Quantum Physics" mechanism that Gale Banks probably doesn't even know about, and eventually I'll derive an equation for FWHP in terms of AFR and explain why some "smoke" is a necessary evil for producing the maximum possible FWHP in a 7.3L PSD.