Does anyone have/know what the BSFC or SFC is for F series Diesels? Prefereably in torque curve format. I found the curves linked below for the cummins and am trying to find it for for the F series diesels. I doing a little research on why the dodges get better MPG and wanted to see if it started with the BSFC.

http://www.turbodieselregister.com/t...1_FEimage1.gif

http://www.turbodieselregister.com/t...1_FEimage2.gif

 

Um... what the heck is BSFC?

 

Brake Specific Fuel Consumption. What it does is measure the engine's fuel consumption based on power produced at the crank. It's how you compare apples to apples on different engines when it comes to fuel consumption. In other words, it's a comparison of efficiency via specific power output.

Take two engines both making 300 hp. Which one makes 300 HP more efficiently? BSFC measures that.

 

Doug, I'm not sure where you'll find BSFC charts for the Powerstrokes. I've never seen them.

 

Learn something new every day! Tried to rep ya Curtis but I'm in rep jail.

 

I don't see where having the BSFC curve in a "torque curve format" is good for anything at all, and certainty not for doing MPG analysis. When I get the time I'll dust off my 99.5 PSD model and see how much that might differ from my newly calibrated CAT C7 model.

The BSFC=Brake Specific Fuel Consumption lb/hr/HP, and the graph below is a plot of BSFC vs RPM for a diesel engine that's comparable to my CAT C7. The "CONSUMPTION" curve is the BSFC lb/hr/HP and as you can see the BSFC curve has a broad minimum in the vicinity where the TQ is a maximum, and the BSFC increases significantly as the RPM increases above 1,800 RPM.

Now it seems that everyone but me always plots the TQ curve along with the BSFC curve, but I think this is confusing because people misinterpret this as a "cause and effect" relationship as if the "maximum TQ" is causing the "minimum BSFC"!

Also if you want to use the BSFC vs RPM curve to analyze MPG you need to plot the BSFC vs RPM curve and the RWHP vs RPM curve on the same graph because TQ has nothing whatsoever to do with MPG, or for that matter with anything else except possibly breaking something in the driveline!



If you define the MFF=Mass Fuel Flow lb/min then the BSFC={(60)(MFF)}/{HP}. In this graph below from my computer model for my C7 the "green" fuel flow curve is given in terms of the VFF=Volume Fuel Flow gal/hr=gph where VFF={(60)(MFF)}/(DFD) gal/hr, where DFD=Diesel Fuel Density, lb/gal, which is typically about 7 lb/gal for #2 diesel at a temperature of 70 F.

Solving the VFF equation for MFF gives, MFF={(VFF)(DFD)}/{60}, and substituting this into the equation for BSFC gives, BSFC={(60)(MFF)}/{HP}={(VFF)(DFD)}/{HP}, or BSFC={(VFF)(DFD)}/{HP}. As seen in my graph at 1,400 RPM where VFF=11.9 gph and HP=230, you get a BSFC={(VFF)(DFD)}/{HP}={(11.9)(7)}/{230}=0.36.

So at each value of RPM the value of the "blue" BSFC curve is the value of the "green" VFF curve multiplied by (7) for #2 diesel, and then divided by the value of the "red" HP curve. So the "blue" BSFC curve depends on the "green" VFF curve and on the "red" HP curve, but not on the TQ curve, which is no longer shown on my graph just to make that point clear!



If you're interested in MPG then the BSFC curve should be plotted with the HP curve because the % grade and the speed on the flat determines the required HP, and you apply the same HP to the ground no matter which gear you use! Of course if you chose a gear that gives too high a MPH that makes the HP requirement higher than the engine can produce at that RPM so you have to downshift to run at higher RPM which gives more HP to meet the demand.

For example the sweet spot for my C7 is in the 1,400 RPM to 1,550 RPM range, and with no headwind on the flat I can hold 58 MPH at 1,450 RPM in 6th gear, but if I get a headwind or much of a grade I have to downshift to 5th gear and go 58 MPH at 1,675 RPM so as to get the additional HP at the higher RPM to meet the demand!

The MFF can also be written as, MFF={(1/2)(Nc)(CFM)(RPM)} lb/min, where Nc=Number of Cylinders and CFM=Cylinder Fuel Mass lb where the CFM is the lb of fuel that's injected into each cylinder for each power stroke. That gives BSFC={(60)(MFF)}/{HP}={(60)(1/2)(Nc)(CFM)(RPM)}/{HP} or BSFC={(30)(Nc)(CFM)(RPM)}/{HP}.

So as you can see from the above equation the BSFC is determined by how much "piston HP" can be extracted from a given CFM lb of fuel that's injected into the cylinder for a single power stroke, and this "piston HP" depends on the "piston FORCE" where the "piston FORCE" is determined by the thermodynamic combustion process which releases the fuel's ENERGY which causes the cylinder pressure to push on the piston, and on the "piston VELOCITY" where the "piston VELOCITY" determines the crankshaft RPM.

If you look at this data table you'll see all the parameters that are entered into the model for determining the BSFC. The neat thing about having a model is that I can change these parameters and determine their effect on HP, BSFC, etc... My model is still a work in progress and it helps me to understand the finer points about the inner workings of dieses engines!



Later I'll post some examples of how calculate MPG from the above types of data.

 

"Now it seems that everyone but me always plots the TQ curve along with the BSFC curve, but I think this is confusing because people misinterpret this as a "cause and effect" relationship as if the "maximum TQ" is causing the "minimum BSFC"!"

Maybe they typically use TQ since TQ is what engines actually produce and HP is just calculated from TQ. (HP=TQ*rpm/5252 IIRC).

So then, does anyone have an opinion as to why the Dodges seem to get better mpg? If it is due to a lower BSFC then does that imply more airflow? Or is it due to the fact that the Cummins is an in-line 6 which produces more torque at the low end and so there may be less gear reduction and hence less gear resistance in the tranny? Another possibility might be the use of variable displacement. I know they do that in hemi's and other gasers but I haven't heard of them doing it in diesels ( i think because of the high compression).

Thanks for the posts guys.

 

Without looking at comparison BSFC charts for Cummins and Powerstrokes, I'm not really sure if that's why Cummins typically get better mileage. It would make sense to explain it that way. Gearing is nearly the same on both Cummins and Powerstrokes, so we can rule that out. Fords are heavier trucks, so that's one area where the Dodge has the advantage, and hence the better mileage. However, even the guys who have swapped a Cummins in the Ford body still have better mileage than Powerstrokes.

So what contributes to engine efficiency? Pretty much everything from engine internals, to injectors, to airflow (intake, turbo, heads, etc), fuel system, and on and on. The 6.4L Powerstoke is actually a more efficient engine than the 7.3L, just the DPF gets in the way so you can't see that. Remove the DPF, and you can make a 6.4L PSD get more than 20 MPG easily. So why is the 6.4L so much more efficient? Well for one, they are now using a Common Rail setup just like Cummins and Dmax engines. Our older PSD's are stuck with the HEUI system. HEUI is reliable, but it's not as efficient. Common rail, for example, injects fuel at higher injection pressures. This results in better atomization, which in turn gives you a better fuel burn, resulting in more efficiency.

Ernest is correct in that you want to use HP and RPM's instead of TQ. It takes a certain amount of HP to maintain a certain speed under certain conditions. BSFC calculations require that you know RPM's and power output to calculate your fuel consumption, which is where the HP calculation comes into play.

 

Pocket has it right,and that's what the engine manufacture's says about the difference between common rail and the old HEUI, and on this comparison is that cummins is 2 less cylinders to feed. That's why my next mod will be that Adrealine HPOP

 

Well said and here's an example using my CAT C7 engine model of how BSFC={(VFF)(DFD)}/{HP} lb/hr/HP can be used to calculate MPG for my Freightliner and why you must relate BSFC to HP in order to get MPG. For now assume that we're dealing with straight summer blend #2 diesel fuel so that each gallon of fuel weighs 7 lb and contains 101,140,000 ft-lb of chemical ENERGY. Since DFD=7 lb/gal the BSFC={(7)(VFF)}/{HP} lb/hr/HP.

The chart below gives the CAT C7 BSFC={(7)(VFF)}/{HP} lb/hr/HP as a function of %Throttle, Boost psi, and RPM. When I'm cruising empty on the flat with no head wind I need between 59% to 62% Throttle to go 59.5 MPH at 1,500 RPM. I know this because in addition to RPM and Boost I have a real-time read out of my TPS! If you look at this chart you see that for RPM=1,500 the BSFC=0.357 at 59% Throttle and 13 psi Boost, and the BSFC=0.358 at 62% Throttle and 14 psi Boost.



Now MPG=MPH/VFF and we know MPH=59.5 but we need to relate the BSFC to HP in order to calculate the VFF from VFF={(BSFC)(HP)}/{7} which then gives MPG=MPH/VFF={(MPH)(7)}/{(BSFC)(HP)}.

The chart below gives the CAT C7 Flywheel HP as a function of %Throttle, Boost psi, and RPM. If you look at this chart you see that for RPM=1,500 the Flywheel HP=97.1 at 59% Throttle and 13 psi Boost, and the Flywheel HP=105.4 at 62% Throttle and 14 psi Boost.



At 59% Throttle this gives MPG={(MPH)(7)}/{(BSFC)(HP)}={(59.5)(7)}/{(0.357)(97.1)}=12.015, and at 62% Throttle this gives MPG={(MPH)(7)}/{(BSFC)(HP)}={(59.5)(7)}/{(0.358)(105.4)}=11.038. So even though the BSFC remains almost unchanged just a slight "nudge" on my throttle decreases my MPG from 12 to 11. Also the above charts for BSFC and Flywheel HP only apply to the conditions in the data input chart shown in my earlier post which includes things like ambient temperature, barometric pressure, etc.

The graph below gives an "overview picture" of the RWHP required to overcome aerodynamic drag and rolling resistance for a number of different trucks. Note that for my Freightliner it requires about 80 RWHP to maintain 60 MPH and this works out to about 100 Flywheel HP like in my example.

Note that an 18 wheeler requires 250 RWHP to maintain 70 MPH on the flat. When they come to a hill and slow down to say 30 MPH they only require 50 RWHP to overcome aerodynamic drag and rolling resistance and this gives them the extra 200 HP to climb the hill!



If you want to read more about aerodynamic drag and rolling resistance check this thread...

How to get your RWHP from a simple road test...
https://www.ford-trucks.com/forums/8...ad-test-2.html

...and at some point I'll be discussing engine efficiency in this thread...

Making more HP or just blowing smoke???
https://www.ford-trucks.com/forums/8...ing-smoke.html

 

The HP in a diesel engine originates from its reciprocating pistons and not from its crankshaft which is where the TQ is defined. The crankshaft, driveline, and powered wheels are just a convenient way of coupling the piston HP to the road.

A fundamental way of defining HP is HP={(FORCE)(VELOCITY)}/{550} ft-lb/sec where the lb FORCE is moving parallel to the ft/sec VELOCITY. A combustion generated lb FORCE pushes on the tops of the pistons, and this lb FORCE makes the pistons move a given ft DISTANCE that's parallel to the lb FORCE, in a given sec TIME, and that means the pistons move at a given ft/sec VELOCITY={DISTANCE}/{TIME}, and this forced motion of the pistons defines a piston HP={(piston FORCE)(piston VELOCITY)}/{550} ft-lb/sec, and a certain percentage of this piston HP is transferred to the truck's rear wheels by the driveline as RWHP, and as the rear tires turn they apply this RWHP to the road in the form of a lb tire FORCE which is a reaction FORCE that's applied at the tire's contact patch with the road, and this tire FORCE moves along the road at the same VELOCITY as does the truck, and by tire FORCE I mean the total tire contact patch FORCE from all of the powered wheels.

Since the truck's rear axel is attached to the truck's frame by suspension members, this tire FORCE generates a reaction FORCE in the truck's frame which is equivalent to pushing the truck at its rear bumper with an external push FORCE vector that has the same magnitude and direction as the total combined tire FORCE vectors from all of the powered wheels. Instead of worrying about what's happening at each tire contact patch with the road you can just imagine that the truck is being propelled forward by a net push FORCE vector applied to its rear bumper and in terms of MPH this push FORCE is related to RWHP as follows... push FORCE={(RWHP)(375)}/{MPH} and the RWHP in turn is equal to the total piston HP minus the driveline loss and by using these relationships you never have to even mention the word TQ and you can solve any truck performance problem of interest.

So in my above example my Freightliner was being propelled by a push FORCE={(RWHP)(375)}/{MPH}={(80)(375)}/{60}=500 lb. But keep in mind that this 500 lb push FORCE is moving at 60 MPH and that's what HP is all about. On the other hand TQ is a static concept that involves no TIME or motion in its definition and it takes a FORCE which moves to perform WORK and HP is the TIME rate of doing WORK.

 

First let me say how much I appreciate you folks bringing your experience and expertise to this discussion. Not to mention Ernest's generosity with regard to posting the fruits of his labors for our reading pleasure. I also apologize for my sometimes late replies. Between the ranch, the rental properties, and work; it sometimes takes a bit for me to get back to the various bulletin boards I visit.

Anyway, through algebraic exercises one can take fundamental relationships and redefine them in terms of the variables needed to analyze a specific application. Ernest provides a good example of this above with his equation “BSFC={(VFF)(DFD)}/HP” which yielded the units of lb/hr/hp. Actually, the fundamental units of BSFC are grams/joule and is time invariant as is the problem we are trying to solve for (mpg). Note the lack of a time variable in either grams per joule or miles per gallon. In fact, if we take Ernest's units of lb/hr/hp they easily reduce to become time invariant. That is, HP = (watts times some constant depending on application)/(time). So Ernest's units=lb/hr/hp=lb/t/hp=lb/t/{(watts)(c)/(t)}=lb/{(watts)(c)}. And so again we are right back to amount of fuel divided by power produced. Since we are actually dealing with time invariant quantities (BSFC, mpg, force, etc) it would seem that using torque (also time invariant) would make sense. Another interesting note is that if you include the time variable in the analysis you end up having to include the work definition. Note that the equation for work (force times distance = f*d ) is identical to the units of torque (ft*lbs).

"The HP in a diesel engine originates from its reciprocating pistons and not from its crankshaft which is where the TQ is defined. The crankshaft, driveline, and powered wheels are just a convenient way of coupling the piston HP to the road."


Here again I would say that certainly you can define or derive a value for HP from the piston dynamics, but it doesn't buy you anything except a lot of unit conversion and generation of more variables. However, I admit that the generation of more variables and equations may actually be desired for a specific application. In fact, this is one trick in math for solving multiple simultaneous equations when you have more variables than equations. That is, if you have more variables than you have equations to solve for, then you can take one of the equations and express one of its variables in terms of the other equations there by creating another equation until you have enough equations to solve for all your variables. There is actually a little more to it than that but thats the jest of it.

So then, it takes a certain amount of Force (a time invariant quantity), not HP to move the vehicle down the road under certain conditions. Thats not to say you can't define or derive a value for HP for the required force. Its just that its not required. The force required is a function of the resistance to motion (another time invariant quantity). The efficiency of generation of the force is what we are analyzing, and that; in its most fundamental form is simply the conversion efficiency of chemical energy to mechanical energy. Which again is a time in variant quantity. We can arbitrarily pick any point between the chemical energy, its resulting combustion force and the force applied to the road to define a conversion efficiency to that point. BSFC seems to typically uses the flywheel as its reference point.

The only reason to use a time inclusive defined quantity is if your choice of measurement necessitates it. Things such as flow rates, calculating HP from accelerations rates etc. all require a time component. However, once the measurement is made the time component will eventually be removed to determine the desired efficiencies.

Here's a link to an interesting discussion on BSFC. The thing I found most interesting was that BSFC is most efficient at WOT.

Browser Warning

 

Whats with the browser warning? The links still seems to work though