2nd Gen Non-Engine/Transmission "Fuel Consumption vs. RPM" question

[RustyJC]

BSFC measures fuel mass per unit of time per brake horsepower, not per RPM as a previous poster stated. For an engine operating at full load, best BSFC normally occurs at the torque peak (or peak BMEP) RPM. If one's foot were to the floor and maximum torque and BHP were being demanded from the engine, total fuel consumption would increase with RPM, but so would BHP, so unless you go back to a BSFC measurement that takes this into account, it really is a meaningless comparison.



The question that (as far as I can tell) hasn't been answered is - What if I'm running at a constant 60 MPH pulling a 5th wheel (i. e. , a constant load situation). Will fuel consumption be better in 6th gear at 2000 RPM or 5th gear at 2600 RPM (4. 10 gears)? This is a constant BHP situation, so to achieve a given BHP, torque output must be higher when running at 2000 RPM than at 2600 RPM. (BHP = [Torque x RPM]/5252) Higher torque means higher BMEP. Higher BMEP generally means better BSFC. The 2000 RPM situation, in addition to higher BMEP, will have reduced frictional losses, so should produce better BSFC and total fuel consumption.



At the end of the day, the only way to be absolutely sure what the RPM/BSFC relationship would be for a given engine would be to produce a 3 dimensional variable torque (BMEP)/variable speed engine BSFC map on a dynamometer.



Rusty

 

[Rman]

Originally posted by RustyJC

BSFC measures fuel mass per unit of time per brake horsepower, not per RPM as a previous poster stated. For an engine operating at full load, best BSFC normally occurs at the torque peak (or peak BMEP) RPM. If one's foot were to the floor and maximum torque and BHP were being demanded from the engine, total fuel consumption would increase with RPM, but so would BHP, so unless you go back to a BSFC measurement that takes this into account, it really is a meaningless comparison.



The question that (as far as I can tell) hasn't been answered is - What if I'm running at a constant 60 MPH pulling a 5th wheel (i. e. , a constant load situation). Will fuel consumption be better in 6th gear at 2000 RPM or 5th gear at 2600 RPM (4. 10 gears)? This is a constant BHP situation, so to achieve a given BHP, torque output must be higher when running at 2000 RPM than at 2600 RPM. (BHP = [Torque x RPM]/5252) Higher torque means higher BMEP. Higher BMEP generally means better BSFC. The 2000 RPM situation, in addition to higher BMEP, will have reduced frictional losses, so should produce better BSFC and total fuel consumption.



At the end of the day, the only way to be absolutely sure what the RPM/BSFC relationship would be for a given engine would be to produce a 3 dimensional variable load/variable speed engine BSFC map on a dynamometer.



Rusty

A fine use of acronyms Rusty!!! Where's my coffee. . i have a headache now

 

[RustyJC]

BSFC = brake specific fuel consumption (lbs/bhp-hr, for example)

BHP = brake horsepower

BMEP = brake mean effective pressure (the theoretical constant unit pressure applied to the piston crown during the power stroke - 211 psig, for example)

RPM = engine revolutions per minute



Did I get all the acronyms defined?



Rusty

 

[klenger]

One of the problems with the original question is doubling the RPM's. Given a low RPM of 1500 and a high RPM of 3000, that's basically the entire operating range of a 24V ISB. It's just not a practical thing to do. A better way to word the question might be "if I increase RPM's 10%, will the MPG go down 10%". From practical experience, it's just not the linear. Rusty worded it better than I can ever understand. So much depends on load, wind, terrain, and even the fuel.

 

[jponder]

I think that is a good question rusty posed about pulling the fifth wheel. I have heard all this about 4. 10 beibg a better rearend and I always thought then just drop from 6th to 5th if you want higher revs. I pulled a D4H once and man I never got out of 4th and I was just clocking the rpms. TDR and Dodge need to post a $1million dollar prize for the best mileage in one of our trucks and I bet we would find all the answers out real quick.

 

[Hohn]

This is a REALLY GOOD thead, and Rusty's post needs to be read carefully.



We also need to understand a few things. One thing that separates diesels from gassers is the variable air:fuel ratio.



What RPM a diesel ends up at is strictly a function of engine output vs load at that RPM. IF the engine makes more power than the load, RPM will rise. If the load is greater than the engine torque, then RPM will fall.



This is why diesels can take hills so much better. As the load rise (from going uphill) the RPM will come down a LOT slower than in a gasser because the torque does NOT decrease with RPM as much. In a CTD, you have torque from 2700 RPM all the way down to 1400. A gasser that has peak power at 2700 loses torque very rapidly as RPMs fall.



Anyway, the variable fuel ratio is the key to diesel economy. This is because we can then use just enough fuel to get the job done- no more (wasteful) and no less (inadequate).



In theory, fuel economy would be optimized at the RPM JUST immediately above lugging. This is from the ENGINE's efficiency component.



The thing is, overall economy also means tire and wind resistance and all that. Both are directly proportional to speed.



Thus, max fuel economy is the lowest RPM the engine can handle in its highest gear. In my truck, that's about 1200 RPM in 6th, which comes out to about 45mph.



Heed the words about BSFC. THAT"S what matters. RPM is irrelevant-- it's just a byproduct of the fueling vs engine load (within the engine's operating range). RPM only plays a role (IMHO) in the engine's ability to burn the fuel that is being injected.



So, the bottom line (I would say) is that MAX mileage is the tallest gear/lowest RPM between 1400 and 2700 that will burn fuel without smoking.



Right?



HOHN