Here's several interesting strategies for reclaiming some of the normally wasted heat ENERGY equivalent HP that's released from the fuel's chemical ENERGY in an internal combustion engine. Of course the 7.3L PSD already reclaims much of the heat ENERGY equivalent HP in its exhaust flow as turbine drive shaft HP to power its turbo to make boost to supply the engine with additional airflow so the idea in the first link of using an exhaust gas driven turbine to power an alternator probably won't provide much additional electrical power if it's done post turbo.

Here's a quote from the second link... "Rathgeber said simple physics prevents a conventional combustion engine from ever exceeding one-third efficiency. A motor with an output of 200 kilowatts (kw) generally creates 400 kw thermal energy as a waste by-product." ...but he should've said "gasoline" internal combustion engine because for my "diesel" 7.2L C7 and the 7.3L PSD about 65% of the heat ENERGY equivalent HP that's released when the fuel's chemical ENERGY is combusted is converted into piston power stroke HP and about 57% of this piston power stroke HP is converted into FWHP so that the overall conversion efficiency at the flywheel is (0.65)(0.57)=0.37 or 37% and not the 33% maximum quoted for a gasser!

Here's some HP numbers for my 7.2L C7 when it's producing its maximum 300 FWHP at a 28 psi boost which I calculated from the data on its factory dyno sheet. There's 812 HP released as heat ENERGY equivalent HP when the fuel's chemical ENERGY is combusted and this 812 heat ENERGY equivalent HP produces 525 piston power stroke HP while 287 HP is lost as rejected heat ENERGY HP to the coolant and out the exhaust! About 40 HP worth of the heat ENERGY equivalent HP going out the exhaust is used to power the turbo to make the 28 psi boost!

Of the 525 piston power stroke HP a "whopping 195 HP" is needed to provide the pumping loss HP during the compression strokes due to the high 28 psi boost. In general the 7.2L C7 needs more boost to make the same FWHP at 2,200 RPM that a 7.3L PSD can make at 2,900 RPM with less boost! Some of this 195 compression stroke pumping loss HP winds up as "heat of compression HP" that's lost to the coolant and the rest is used to aid the combustion.

This leaves 525-195=330 HP remaining as the net piston power stroke HP but piston-cylinder wall friction, rotating friction, and parasitic losses require 30 HP so this leaves a "grand total" of 300 HP at the flywheel. Since (300)/(812)=0.37 this means that 37% of the fuel's 812 chemical ENERGY HP is converted into 300 FWHP when the engine is operated at full power but does that also mean that 812-300=512 HP worth of heat ENERGY equivalent HP is being wasted and can therefore be reclaimed?

Well as I'll post later on this thread that's where things get interesting because according to the laws of "Physics" there's a "cost of doing business HP" associated with "producing HP" using a conventional internal combustion engine and exactly what this "cost HP" is determines the maximum amount of so-called "waste heat ENERGY equivalent HP" that can potentially be reclaimed by these and other strategies that are being proposed to increase fuel efficiency.

Being a retired EE I personally like the idea which was recently proposed (seriously) that all we need to do to solve the ENERGY crisis is to install alternators on each wheel of an all electric car and use those to keep its battery charged up on the fly so you never have to stop and plug in to recharge!

Efficiency Strategies Target Waste Heat...

New efforts to improve fuel economy by squeezing the most efficiency from existing internal combustion engines are focused on the exhaust pipe. One company, for example, proposes a space-efficient, direct-drive turbo generator. Good for hybrid as well as gasoline-fueled vehicles, the exhaust-based system could supply enough electric power to eliminate the need for an alternator. Another automaker hopes to convert wasted engine heat into usable power by applying NASA technology. A modified, radioisotope thermoelectric generator, placed between the car's engine coolant and its exhaust gas manifold, could — in theory — reclaim enough thermal energy to reduce fuel consumption more than 10% in a full-size sedan.

 

I've thought about an exhaust driven alternator for some time. Never had the time or patience to go through with it.

 

To make things run off of exhaust gas work more efficiently, wouldn't that lead to a smaller tail pipe to increase the pressure? What effect will the increased back pressure have on engine efficiency? Especially since there seems to be a movement towards lighter engine components which may have a lower melting point. Sounds more geared for the eco friendly 4 cyl. no power crowd.

It's interesting to read the diesel has almost double the efficiency of a gasser, but typically yields only a 20% or so increase in fuel economy. I know, different BTU content in fuels and all, it just sounds like it should be a larger difference.

 

That is an interesting read Gene. I have wondered why some type of generator could not be put on the wheels to charge the batteries on an electric car like you mentioned.

 

Very interesting Gene, subscribing, not raising my hand yet...

 

In the 1950's Wright made 18-cylinder aircraft radial piston engines that used turbines in the exhaust geared back to the crankshaft. It had a conventional centrifigal supercharger built in. A 3350 cu. in. "Turbo Compound" engine put out 3,400 hp (later 3,700 hp at reduced reliability), of which several hundred were from the "power recovery turbines". It's fuel consumption was rated around 0.38 lbs/hp/hr while cruising about 2,500 hp.

For the extreme 1950's attempt to recover every last HP per drop of fuel, google the Napier Nomad engine. It combined a horizantally-opposed 12-cylinder supercharged (by bleeding air from the axial compressor section of the jet) 2-cycle DIESEL engine with a jet engine (with afterburner) to recover waste heat from the diesel. It was rated at 3,570 hp., presumably including thrust from the afterburner. It's fuel consumption was rated at 0.33 lbs/hp/hr while cruising around 2,248 hp.

By comparison, contemporary small air-cooled aircraft engines were consuming about 0.50 lbs/hp/hr.

Gene, did you ever calculate your fuel consumption in terms of lbs/hp/hr? Also, are you familiar with brake mean effective pressure, which is the combustion (cylinder) pressure applied to each piston head? BMEP is handy in calculating power output. The Convairs I used to fly had a BMEP gauge that actually measured the torque reaction pressure on the propeller reduction gearing.

 

I'm relatively sure that (most?) hybrids do use the vehicles inertia while slowing down to help recharge the batteries. I know the Prius does -- when you let off the accelerator, you can see power reverse direction on the fancy display from batteries --> motor --> wheels to wheels --> motor --> battery. I guess they're locking the electric motor inline and using it as a generator. Pretty clever, if I do say so myself...

 

The so-called crisis about global warming, the diminishing supplies of conventional ENERGY sources, the need for alternative ENERGY sources, and the need for increased efficiencies when using our ENERGY sources is one of my favorite topics because there's so many poorly thought out and even completely nonsensical solutions being proposed to solve it!

To illustrate the sad state of affairs I'll paraphrase a line from a TV debate on the subject where a nonbeliever in global warming was asked... "What's your problem is it ignorance or apathy?" and his reply was... "I don't know and I don't care!" ...well I have the luxury of looking at these problems from a rather detached analytical vantage point because I've got no offspring to worry about and the wife and I will most likely be dead before things really come to a head! My quote below came from this same TV debate...

...and the above is an example of a "completely nonsensical solution" because it represents a "perpetual motion machine"! That's not to say that using a DC generator to reclaim some of a vehicle's kinetic ENERGY when it needs to slow down or come to a stop isn't a good idea!

I shouldn't have even mentioned electric cars because I wanted to concentrate on diesels but perhaps a good way to begin a discussion of how a diesel and its turbo are both "heat engines" and therefore have to live with the efficiency constraints imposed by the principles of "Thermodynamics" is to first discuss "electric motors" which work differently than "heat engines" and therefore have higher efficiencies for converting electrical ENERGY into mechanical HP than diesels have for converting diesel fuel chemical ENERGY into mechanical HP.

Of course comparing efficiencies based on inputs of electrical ENERGY versus chemical ENERGY is "apples and oranges" and advocates of electric cars don't seem to spend much time worrying about where the source of the electrical ENERGY input to their electric cars is coming from and if it's ultimately coming from a source of chemical ENERGY like in a coal or oil fired power plant then the overall efficiency of electric cars is much much lower than for diesels or gassers!

In an electric motor current generates magnetic FORCE fields on the stator and the armature which push against each to produce a FORCE on the armature that acts at a perpendicular DISTANCE from the electric motor's output shaft and since TQ=FORCExDISTANCE this produces a TQ to turn the electric motor's output shaft. The losses due to electrical resistance in the wires and due to friction and parasitic drag in the moving parts mean an electric motor is less than 100% efficient however it can convert more than 90% of the electrical input ENERGY into mechanical HP.

Look at the HP vs RPM (yellow) and the Efficiency vs RPM (green) curves below for an electric motor and note that the Efficiency continuously increases as the HP increases and that the electric motor Efficiency continues to increase as the RPM increases to values higher than the RPM for maximum HP.



Now look at the curves below for my 7.2L C7...



For a diesel the Efficiency is proportional to the inverse of the BSFC=Brake Specific Fuel Consumption lb/hr/HP and since the BSFC reaches a minimum at 1,400 RPM where the peak TQ occurs the diesel Efficiency is a maximum there and I'll discuss later why this is the case because it doesn't really have anything to do with TQ per se!

As the RPM increases above 1,400 RPM the HP increases but the BSFC also increases which means the diesel Efficiency decreases and at higher RPM above the RPM for maximum HP the BSFC increases even more rapidly which means the diesel Efficiency decreases more rapidly at higher RPM.

I'll eventually get around to answering questions and commenting on replies but first I want to give some more background information on how diesels and their turbos are both "heat engines"!

 

Gene Cummins has a 14litre that they call a steam engine, its all about using wasted energy, do a search it's a engineering marvel you will like it.

 

A "heat engine" converts heat ENERGY into mechanical WORK and WORK is defined by WORK=FORCExDISTANCE but unlike the definition for TQ=FORCExDISTANCE in order for WORK to be produced a FORCE must "move" through a DISTANCE that's "parallel" to the FORCE whereas for TQ to be produced a FORCE is "applied" at a DISTANCE that's perpendicular to a shaft and no "motion" ever occurs between the applied FORCE and the perpendicular DISTANCE.

If a heat ENERGY generated FORCE pushes a piston a given DISTANCE then the WORK=FORCExDISTANCE and since HP is the TIME rate for doing WORK the HP~(FORCExDISTANCE)/(TIME) and since VELOCITY=(DISTANCE)/(TIME) you get HP~FORCExVELOCITY and for a diesel we're talking about a combustion generated FORCE and a piston VELOCITY.

The picture below illustrates the so called 1st law of thermodynamics which says that heat ENERGY can be converted into mechanical WORK and conversely that mechanical WORK can be converted into heat ENERGY. The first part of the picture shows that an external FORCE can be applied to move a piston a given DISTANCE and thereby do mechanical WORK on the fill gas inside a cylinder by compressing it to a higher pressure and temperature.



The second part of the picture shows that an external source of heat ENERGY can be applied to the cylinder to heat the fill gas and cause its pressure and temperature to increase and then the resulting FORCE pushes the piston a given DISTANCE to produce mechanical WORK and the resulting HP that's produced is the TIME rate for doing the WORK and that's proportional to the product of the piston FORCE and the piston VELOCITY.

Since heat ENERGY can be converted into mechanical WORK and mechanical WORK can be converted into heat ENERGY the units for mechanical WORK and heat ENERGY are the same and in English units WORK and ENERGY are measured in ft-lb. In English units it's easy to get a calibrated "feel" for what 1 ft-lb of ENERGY is because all you have to do is lift a 1-lb weight a 1-ft DISTANCE in a vertical direction and you've done 1 ft-lb of mechanical WORK and you've increased the potential ENERGY of the weight by 1 ft-lb. You'll gain an exceptionally good "feel" for just how much 1 ft-lb of potential ENERGY is if you drop the weight to produce 1 ft-lb of kinetic ENERGY and then let that 1 ft-lb of ENERGY interact with your foot.

In the picture below the orange/yellow colors represent the heat ENERGY that pushes the piston down to produce power stroke HP and spins the turbine to produce compressor HP to supply the engine with an enhanced airflow.



The diagram below shows a thermodynamic reservoir model for a heat engine. As indicated in the diagram a heat engine uses input heat ENERGY from a high temperature reservoir, it converts some of this heat ENERGY into mechanical WORK, and then rejects the left over heat ENERGY to a lower temperature cold reservoir.

As shown in the diagram the mechanical WORK, W, 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).



An idealized heat engine works on a Carnot cycle which has an Eff=1-(Tc)/(Th), and this implies that if Tc=0 R =-460 F the theoretical maximum Eff=100%. But the only place to find very cold ambient temperatures for a cold reservoir to reject heat to is in deep space and on earth the minimum Tc is typically Tc=530 R=70 F or higher. Also a Carnot cycle requires a very slow operating cycle so it's not applicable for higher RPM internal combustion engines.

For a diesel the mass of hot gas that's produced from the combustion of fuel acts as the high temperature reservoir in the diagram. Some of the input heat ENERGY from this high temperature reservoir produces an expansion pressure and this pressure produces a FORCE which moves the piston through a given DISTANCE to produce mechanical piston WORK and the TIME rate at which this mechanical piston WORK is produced gives the piston power stroke HP.

For a diesel the conversion of heat ENERGY into piston power stroke HP is about 65% efficient because some of the input heat ENERGY is absorbed by the metal surfaces of the combustion chamber and carried away by the coolant, and some of the input heat ENERGY is expelled from the combustion chamber as a mass of hot exhaust gas, and for the diesel these loss mechanisms constitute the lower temperature cold reservoir in the diagram.

Then to get the net FWHP you've got to subtract from the piston power stroke HP the pumping loss HP that's required to produce the intake, compression, and exhaust strokes, and the HP that's required to overcome friction and parasitic losses.

For a turbocharger the mass of rejected hot exhaust gas acts as the high temperature reservoir in the diagram. Some of the input heat ENERGY from this high temperature reservoir produces an expansion pressure which causes a FORCE that moves the turbine blades through a given angle to produce mechanical WORK, and the time rate at which this mechanical WORK is produced gives the turbine HP.

For a turbo the conversion of heat ENERGY into mechanical WORK is less than 50% efficient because some of the input heat ENERGY is absorbed by the metal surfaces of the up-pipe and turbo case and some flows past the turbine blades and out the tailpipe and is then absorbed by the ambient air mass, and for the turbo these loss mechanisms constitute the lower temperature cold reservoir in the diagram.

So if you put all of this thermodynamics stuff together into a computer model that can be calibrated from factory engine dyno data which includes fuel flow, boost, ambient temperature and pressure, etc... you can generate the following graphs. In the first graph below I converted the normally used BSFC into diesel Efficiency %.



The second graph below should be read from right to left for the CS=Compression Stroke, and then from left to right for the PS=Power Stroke. The lower red, green, pink, and blue curves are for the CS, and the upper set of curves is for the PS. Note that at 27* ATDC the peak PSHP for each piston is 1,025 HP!

If you compare my previously described calibration method for what 1 ft-lb of ENERGY is to the fact that the CAT C7 injects 4,060 ft-lb of chemical ENERGY into a single combustion chamber to produce the mechanical HP during a single 180* power stroke of a single piston you'll get a feel for how much ENERGY is required to produce a 1,025 peak HP and a 350 HP averaged over the 180* of a single power stroke!



The chart below is an "accounting budget" which shows how the 4,060 ft-lb of fuel chemical ENERGY is "spent" to make HP at the pistons, flywheel, and rear wheels. Later I'll discuss how some of the 1,435 ft-lb of rejected heat ENERGY that's leftover when the piston PSHP is produced, is used to power the turbo.

 

The above RPM numbers are for my 300 FWHP 7.2L C7 but if you add about 600 RPM to these numbers you'll get the ones for a mildly tuned 300 FWHP 7.3L PSD where the peak TQ occurs at about 2,000 RPM and the maximum HP occurs at about 2,800 RPM.

So for both engines the HP maximum occurs about 800 RPM above the RPM for maximum Efficiency which is also the RPM for peak TQ and the question at hand is does the RPM for maximum Efficiency really have anything to do with TQ per se?

Also what determines the specific RPM for maximum Efficiency and the specific RPM for maximum HP and why are they different by about 600 RPM for the same displacement engine?

Does the answer have anything to do with comparing a 6 cylinder engine to a 8 cylinder engine?

What are the peak TQ and maximum HP RPM values for the 6 cylinder 5.9L and 6.7L Cummins and how do they compare with the ones for a mildly tuned 300 FWHP 7.3L PSD?

What are the peak TQ and maximum HP RPM values for the 7.3L OBS PSD, 6.0L PSD, 6.4L PSD, and the 6.2L, 6.5L, and 6.6L Duramax engines and for any other diesels with comparable displacement?

Is anyone even interested in the answers to the above questions?

If they are please help provide them and help do the research to get the RPM values for peak TQ and maximum HP for the various engines above and also be on the lookout for any data concerning fuel flow vs HP at a given RPM.

For example on my C7 the factory dyno sheet gave GPH fuel flow vs RPM and HP which allowed me to calculate the amount of fuel chemical ENERGY injected during each power stroke. For my C7 at 1,400 RPM for peak TQ the fuel injected for a single piston's power stroke is 4.75x10^-5 gal/Inj=4,805 ft-lb/Inj, and at 2,200 RPM for maximum HP the fuel injected for a single piston's power stroke is 4.015x10^-5 gal/Inj=4,060 ft-lb/Inj.

I think the answer as to why there's a specific RPM for maximum Efficiency and maximum HP has to do with the nature of the combustion process versus the piston velocity. That means I'll need the Stroke for each of the above engines in order to calculate their piston velocities!

 

I am!! Thank You. FTE class in session!

 

Well considering that you run a "6637" and you're "still interested" in what "I've got to say" I guess I'll keep posting on this thread even if it might only be for an audience of one!

I also see you've changed to "single-shot" injectors and since I've seen it posted many times that "single-shots" make more HP than "multi-shot" injectors do I'll offer up some reasons why in general just the opposite might be true but it requires programming the fuel flow for "multi-shot" injectors to make higher HP not to meet lower EPA emissions requirements!

I'll discuss how fuel flow determines HP and derive some basic equations that relate various measures of fuel flow to FWHP=Flywheel HP and to the PSHP=Power Stroke HP that's generated by a single piston. First I'll consider a diesel engine to be a "black box" where a given fuel flow enters at one end and a some FWHP comes out at the other end and then I'll discuss the internals of how combustion in an individual cylinder produces some PSHP.

The total fuel flow into an engine can be quantified in terms of the VFF=Volume Fuel Flow gal/hr or in terms of the MFF=Mass Fuel Flow lb/min. To convert the VFF gal/hr to MFF lb/min multiply the VFF by (DFD) where DFD=Diesel Fuel Density lb/gal to convert from gal to lb and divide by (60) to convert from hr to min and this gives... MFF={(VFF)(DFD)}/(60) lb/min. To convert from MFF lb/min to VFF gal/hr divide the MFF by (DFD) to convert from lb to gal and multiply by (60) to convert from min to hr and this gives... VFF={(MFF)(60)}/(DFD) gal/hr.

If you consider a diesel engine to be a "black box" and define the BTU=British Thermal Unit heating value of the HP equivalent chemical ENERGY that's stored in each gallon of diesel fuel as the DFE=Diesel Fuel Energy Btu/gal and define the net efficiency for converting the fuel's HP equivalent chemical ENERGY into FWHP as the TEF=Thermodynamic Efficiency Flywheel which is a ratio from 0 to 1 then the FWHP is given by...

FWHP={(VFF)(DFE)(TEF)}/(2,545) HP

In this equation the term (VFF)(DFE) or (gal/hr)(Btu/gal) gives the Btu of fuel HP equivalent chemical ENERGY that's released during combustion per hr and multiplying this by (TEF) gives the Btu of ENERGY per hr that's available at the flywheel and dividing by (2,545) converts this flywheel ENERGY btu per hr into FWHP, and that's all there is to deriving an equation it's just several step-by-step arithmetic calculations combined into one final result where symbols are used as place holders for numbers!

If you plug some numbers into this FWHP equation like VFF=1 gal/hr, DFE=130,000 Btu/gal for #2 diesel, and TEF=0.37 you get FWHP={(1)(130,000)(0.37)}/(2,545)=18.9 HP which is the reference number I've used before for converting each 1 gal/hr of fuel flow into FWHP!

If you substitute VFF={(MFF)(60)}/(DFD) gal/hr for VFF in the above FWHP equation and combine the constants you get...

FWHP={(MFF)(DFE)(TEF)}/{(DFD)(42.42)} HP

If you use MFF=1 lb/min and DFD=7 lb/gal for #2 diesel fuel at room temperature you get FWHP={(1)(130,000)(0.37)}/{(7)(42.42)}=162 HP. So each 1 lb/min of fuel flow produces 162 FWHP and as a double check 1 lb/min is 60 lb/hr and dividing by 7 lb/gal gives 8.57 gal/hr and multiplying this by 18.9 FWHP for each 1 gal/hr of fuel flow gives 162 FWHP!

Of course the above equations for FWHP are rather "simplistic" because they assume you've got sufficient airflow to actually combust the fuel's chemical ENERGY and release its heat ENERGY equivalent HP and if the airflow isn't adequate to do this you'll get some black smoke coming out the tailpipe and the actual FWHP will be less than what's predicted by these equations!

To derive a more comprehensive equation for FWHP you need to consider the AFR=Air Fuel Ratio which is defined by AFR=MAF/MFF where MAF=Mass Air Flow lb/min and the total mass airflow lb/min going into an engine's air filter is given by...

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

...where VE=Volumetric Efficiency 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.

Since it requires 15 lb of air to combust 1 lb of fuel an AFR=18 provides a factor of 18/15=1.2 or a 20% "excess airflow" and this is typically adequate to essentially combust all the fuel and avoid black smoke coming out the tailpipe, however it requires an even higher "excess airflow" or EGR or a combination of both to meet the 2004 and later NOx emissions spec.

One strategy for controlling fuel flow is to program the fuel flow to match the airflow in an attempt to maintain a constant AFR and this requires a MFF=MAF/AFR lb/min or since MFF={(VFF)(DFD)}/(60) lb/min it requires a VFF={(MAF)(60)}/{(AFR)(DFD)} gal/hr. Since the MFF lb/min and the VFF gal/hr are both directly proportional to the MAF lb/min the fuel flow needs to be controlled using a MAF sensor or as a function of RPM, AAP, BP, and MAT using the MAF equation and this requires that various sensor inputs be feed into an ECM. Some amount of fuel flow also needs to be programmed as a function of TP=Throttle Position in order to get the whole fueling process started and stopped otherwise the MAF would never change from whatever its current value was.

One can also consider the total VAF=Volume Air Flow ft^3/min going into an engine's air filter and just like for fuel flow one can use the AD=Air Density lb/ft^3 to convert between the mass airflow lb/min and the volume airflow ft^3/min. To convert the VAF ft^3/min to MAF lb/min multiply the VAF by (AD) to convert from ft^3 to lb and this gives... MAF=(VAF)(AD) lb/min. To convert from MAF lb/min to VAF ft^3/min divide the MFF by (AD) to convert from lb to ft^3 and this gives... VAF=(MAF)/(AD) ft^3/min.

Since the density of fuel is approximately constant the VFF gal/hr and the MFF lb/min are equally useful measures for quantifying the fuel flow but for air the density varies greatly as a function of pressure and temperature and is given by... AD={(2.70325)(P)}/{(T+459.67)} lb/ft3 where P=Pressure psi and T=Temperature F. This leads to much confusion when using the VAF ft^3/min measure for quantifying the airflow.

For example the numerical value of the MAF lb/min going into an engine's air filter is the same as the numerical value of the MAF lb/min at each location between the air filter and the intake manifold at which point the total MAF lb/min divides approximately equally between the cylinders.

On the other hand the numerical value of the VAF ft^3/min going into an engine's air filter is less than the numerical value of the VAF ft^3/min coming out of the air filter and going into the turbo, the numerical value of the VAF ft^3/min coming out of the turbo is less than the numerical value of the VAF ft^3/min going into the turbo, and the numerical value of the VAF ft^3/min coming out of the intercooler is less than the value going into the intercooler!

So if using the VAF ft^3/min to quantify airflow is so confusing why not just stick to using the MAF lb/min measure and be done with it? Well for starters you also need to know about the VAF ft^3/min in the CAC tube for example in order to determine the AFV=Air Flow Velocity ft/min in the tube because the AFV=(VAF)/(CSA) ft/min where CSA=Cross Section Area ft^2 and you need to know the AFV ft/min in order to determine the pressure drops across the various air inlet components.

For air flowing through a straight section of pipe the PD=Pressure Difference psi from one end of the pipe to the other end depends on the AD lb/ft^3 and on the AFV ft/min and the PD psi is given by... PD=(K)(AD)(AFV)^n(L/D) psi ...where K is a friction term which includes the surface roughness of the pipe's inner wall, and n=1 for laminar flow, n=2 for turbulent flow, and n has Intermediate values from 1 to 2 for transitional flow in between these two extremes, and (L/D) is the Length-to-Diameter ratio of the pipe. If you solve this equation for AFV ft/min then an applied PD psi across a Length of tube with a given Diameter determines the AFV ft/min and the Diameter determines the CSA ft^2 and this combined with the AFV ft/min determines the VAF ft^3/min airflow in the tube.

You also need the AFV ft/min to calculate pressure drops such as the PD psi across the air filter element, the intercooler, and across the intake valve so you can determine the VE. The VE is needed to determine the CAP=Cylinder Air Pressure psi from the MAP=Manifold Air Pressure psi where MAP=AAP+BP psi and CAP=(VE)(MAP) psi. In my next installment I'll discuss how the CAP psi along with the amount of fuel that's injected into the cylinder and the piston motion dynamics determines the PSHP=Power Stroke HP that's generated by a single piston.

 

To analyze what's happening inside each cylinder you can define the MFI=Mass Fuel Injected lb/inj as the lb weight of fuel that's injected into a cylinder to produce a single power stroke. For each cylinder a power stroke occurs once every 2 revolutions of the crankshaft so there's an injection of MFI 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 total MFF=Mass Fuel Flow lb/min equal to... MFF={(1/2)(Nc)(MFI)(RPM)} lb/min ...and solving for MFI gives... MFI={(2)(MFF)}/{(Nc)(RPM)} lb/inj.

This equation... MFI={(2)(MFF)}/{(Nc)(RPM)} lb/inj ...says the total MFF lb/min going into an engine divides equally among the Nc cylinders and if the MFF lb/min remains constant the lb weight of fuel that's injected into each cylinder decreases inversely proportional to RPM.

Likewise you can define the VFI=Volume Fuel Injected gal/inj as the gal volume of fuel that's injected into a cylinder to produce a single power stroke. For each cylinder a power stroke occurs once every 2 revolutions of the crankshaft so there's an injection of VFI gal 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 total VFF=Volume Fuel Flow gal/hr equal to... VFF={(30)(Nc)(VFI)(RPM)} gal/hr ...and solving for VFI gives... VFI={(VFF)}/{(30)(Nc)(RPM)} gal/inj.

This equation... VFI={(VFF)}/{(30)(Nc)(RPM)} gal/inj ...says the total VFF gal/hr going into an engine divides equally among the Nc cylinders and if the VFF gal/hr remains constant the gal volume of fuel that's injected into each cylinder decreases inversely proportional to RPM.

Since MFF={(VFF/60)(DFD)} lb/min and VFF={(60)(MFF)}/{DFD} gal/hr where DFD=Diesel Fuel Density lb/gal you also have... MFI={(VFI)(DFD)} lb/inj and VFI={(MFI)}/{DFD} gal/inj. The factor (60) converts from hr to min and vice versa so it's needed in the MFF and VFF conversions but the MFI and VFI are both measured per injection so they only have the (DFD) factor to convert between lb and gal.

In the graph below I plotted the equation... VFI={(VFF)}/{(30)(Nc)(RPM)} gal/inj ...on the same graph with my C7 engine dyno curves for VFF gal/hr, FWHP, and FWTQ. As you can see the FWHP curve tends to follow the VFF gal/hr version of the fueling curve while the FWTQ curve tends to follow the VFI gal/inj version of the fueling curve.

The reason the FWHP curve flattens out at a maximum of 300 HP and then decreases at higher RPM is primarily due to the VFF gal/hr fueling curve. Due to its 2 intake valves per cylinder the C7 has plenty of excess airflow to support higher FWHP at higher RPM and the motorhome version of the same engine is programmed for 330 FWHP at 2,400 RPM.

On the other hand even though the motorhome version is programmed for 330 FWHP at 2,400 RPM it's also programmed for the same 860 lb-ft FWTQ at 1,400 RPM as my truck version has so the motorhome program only adds more fuel at higher RPM at about 1,900 RPM and higher which means the VFI gal/inj version of the fueling curve remains unchanged in the vicinity of its maximum at 1,400 RPM.



There're several versions of an ICP sensor "pressure box" available for the C7 and since they fool the ECM into increasing the ICP the fuel curve is increased at all RPM values and this increases the FWHP at lower RPM as well which means the maximum FWTQ is also increased since TQ~HP/RPM.

In general since... FWHP={(VFF)(DFE)(TEF)}/(2,545) HP ...the shape of the FWHP vs RPM curve depends on the shape of the VFF gal/hr version of the fueling curve as well as on the shape of the TEF curve for the net thermodynamic efficiency at the flywheel. The TEF curve decreases somewhat as the RPM increases above the RPM for maximum FWTQ and stock injectors impose a fuel flow limitation at higher RPM and both of these effects combine to limit the FWHP at higher RPM.

Look at the graph above and note at 2,200 RPM the VFF=15.9 gal/hr and using this in the FWHP equation gives... FWHP={(15.9)(130,000)(0.37)}/(2,545)=300.5 HP ...so at 2,200 RPM the TEF=0.37 or 37% because the factory dyno sheet gives 300 FWHP at a VFF=15.9 gal/hr.

If you substitute VFF={(30)(Nc)(VFI)(RPM)} gal/hr in FWHP={(VFF)(DFE)(TEF)}/(2,545) HP you get... FWHP={(30)(Nc)(VFI)(RPM)(DFE)(TEF)}/(2,545) and since... FWTQ={(FWHP)(5,252)}/(RPM) lb-ft ...you get...

FWTQ={(61.9)(Nc)(VFI)(DFE)(TEF)} lb-ft

...and this equation says the FWTQ is proportional to (VFI)(DFE) or the (gal/inj)(Btu/gal) which is the Btu/inj of fuel chemical ENERGY that's released during each combustion event. At the RPM where the VFI gal/inj is a maximum the maximum amount of fuel chemical ENERGY is released and this in turn causes a maximum cylinder pressure which pushes on the piston with a maximum FORCE which produces a maximum TQ on the crankshaft.

Look at the graph above and note at 1,400 RPM the VFI=4.7x10^-5 gal/inj and using this in the FWTQ equation gives... FWTQ={(61.9)(6)(4.7x10^-5)(130,000)(0.38)}=862 lb-ft ...so at 1,400 RPM the TEF=0.38 or 38%.

Since I've been talking about FWTQ I want to make sure I don't inadvertently contribute to any of the "TQ myths" that are shared by many diesel enthusiasts like for example... TQ produces the HP in their diesel engine, TQ is measured at their rear wheels on a dyno, TQ gets them up a hill quickly, TQ wins them a drag race, and TQ makes their diesel better than a gasser!

Even if you can come up with a good reason why having more RWTQ is helpful, and applying more tire FORCE to the ground to pull a stump is the only one that comes to mind, remember that with an appropriate choice of tranny and diff gear ratios you can make your RWTQ as large as you want no matter what your FWTQ is! On the other hand the RWHP is equal to the FWHP minus the loss in the driveline independent of the gear ratio and the RWHP is what's important for every aspect of truck performance that involves motion.