Bad mpg?

[Antonm]

This is because a 6.2 filled to a 14.7 afr (for example) will make more power than a 5.3 L engine filled up at 14.7 afr.

Is this a statement you are willing to stand behind? Its totally wrong, but that hasn't stopped you before.

If you put 10cc of fuel (or really any given amount of fuel within the range each engine can efficiently burn) in either engine (part throttle stuff where talking here) and mix in air to the same AFR, then guess what,,,,, both engines will be making the same power.

The only slight difference would be any difference in efficiency between the engines (which a 5.3 and 6.2 having the same design, they have pretty similar efficiencies).

So that statement ,yet again, demonstrates a gross conceptual error in your part.

And now you go blathering on about transmission downshifts and other crap. The facts are a 5.3 and 6.2 will get very similar (talking second significant digit differences) in fuel milage if driven under the same conditions, moving the same mass. Even the stupid EPA fuel milage test they put on the window sticker shows this.
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[blanchard7684]

Is this a statement you are willing to stand behind? Its totally wrong, but that hasn't stopped you before.

If you put 10cc of fuel (or really any given amount of fuel within the range each engine can efficiently burn) in either engine (part throttle stuff where talking here) and mix in air to the same AFR, then guess what,,,,, both engines will be making the same power.

The only slight difference would be any difference in efficiency between the engines (which a 5.3 and 6.2 having the same design, they have pretty similar efficiencies).

So that statement ,yet again, demonstrates a gross conceptual error in your part.

And now you get blathering on about transmission downshifts and other crap. The facts are a 5.3 and 6.2 will get very similar (talking second significant digit differences) in fuel milage if driven under the same conditions, moving the same mass. Even the stupid EPA fuel milage test they put on the window sticker shows this.
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See previous posts and the linked material. There is an equation for power that literally shows the relationship I’m referencing.

 

[Antonm]

https://view.officeapps.live.com/op/view.aspx?src=https%3A%2F%2Fwww.webpages.uidaho.edu%2Fmindworks%2FIC_Engines%2FLectures%2FLecture%252007%2520-%2520Relationships%2520Between%2520Parameters.pptx&wdOrigin=BROWSELINK

See equation on slide 4.

nv is VE, Vbdc is engine displacement, AFR is air/fuel ratio.

For the power required to execute part load manuever, it is same because it is essentially the same mass.

The difference is engine displacement. The argument at hand is if the 6.2 can do this at higher AFR vs 5.3.

Based on that equation, if you assume VE is constant, the 6.2 can have higher AFR and make same power because the Vbdc term is increased.

So moving the mass up an incline, or into headwind, or tip in acceleration, for instance, the 6.2 can reach the required P on lhs of equation at higher AFR than 5.3 assuming throttle position is constant.

But, if 5.3 increases Ve via throttle, it can reach the required power. However throttle position and AFR are linked together inversely. So the 5.3 has to move to a lower BSFC area on its BMEP vs BSFC curve. This is in rel world represented by the fuel tables and shift schedule.

So the smaller engine will have a harder time keeping fuel consumption minimized in part-load conditions.

This is my point.

OMG are you a moron. Throttle position alone does not, repeat not, decide what the AFR is all by itself, the PCM and its ability to change the amount of fuel delivered gets a vote too in what the afr is .

You can have the same afr at just about any throttle position if the tuner decides that's what they what. And the factory calibrations have it as lean as they can safely get away with for CAFE standard / catalytic reasons.

Yeah during tip -in throttle they PCM commands more fuel, otherwise the engine would bog (basically the PCM is performing the same function as a carburetor accelerator pump did back in the olden days), but the OEMs want that AFR back to as lean as they can get it, as quickly as they can get it, for those all important fuel mileage and catalytic converter longevity reasons, so it doesn't stay rich for long.

And BTW, your link doesn't work.



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[Antonm]

See previous posts and the linked material. There is an equation for power that literally shows the relationship I’m referencing.

Could you post a copy or screenshot of this equation? That way I can show you how you're mis-interpreting it , well, assuming even applies at all that is.
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[blanchard7684]

OMG are you a moron. Throttle position alone does not, repeat not, decide what the AFR is all by itself, the PCM and its ability to change the amount of fuel delivered gets vote too in what the afr is .

You can have the same afr at just about any throttle position if the tuner decides that what they what. And the factory calibrations have it as lean as they can safely get away with for CAFE standard reason.

Yeah during tip -in throttle they PCM commands more fuel, otherwise the engine would bog (basically the PCM is performing the same function as a carburetor accelerator pump did back in the olden days), but the OEMs want that AFR back to as lean as they can get it, as quickly as they can get it, for those all important fuel mileage and catalytic converter longevity reasons, so it doesn't stay rich for long.

And BTW, your link doesn't work.

View attachment 447596

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OMG are you a moron. Throttle position alone does not, repeat not, decide what the AFR is all by itself, the PCM and its ability to change the amount of fuel delivered gets vote too in what the afr is .

You can have the same afr at just about any throttle position if the tuner decides that what they what. And the factory calibrations have it as lean as they can safely get away with for CAFE standard reason.

Yeah during tip -in throttle they PCM commands more fuel, otherwise the engine would bog (basically the PCM is performing the same function as a carburetor accelerator pump did back in the olden days), but the OEMs want that AFR back to as lean as they can get it, as quickly as they can get it, for those all important fuel mileage and catalytic converter longevity reasons, so it doesn't stay rich for long.

And BTW, your link doesn't work.

View attachment 447596

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Link fixed.

Let me summarize.

Power = efficiency factors x Heating value of fuel x VE x displacement x rpm/(afr x constants).

So a 5.3 makes less power than a 6.2 at the same afr.

slide 4.

This is a rearranged formula from Heywood text on combustion engines.

 

[blanchard7684]

OMG are you a moron. Throttle position alone does not, repeat not, decide what the AFR is all by itself, the PCM and its ability to change the amount of fuel delivered gets vote too in what the afr is .

You can have the same afr at just about any throttle position if the tuner decides that what they what. And the factory calibrations have it as lean as they can safely get away with for CAFE standard reason.

Yeah during tip -in throttle they PCM commands more fuel, otherwise the engine would bog (basically the PCM is performing the same function as a carburetor accelerator pump did back in the olden days), but the OEMs want that AFR back to as lean as they can get it, as quickly as they can get it, for those all important fuel mileage and catalytic converter longevity reasons, so it doesn't stay rich for long.



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OMG are you a moron. Throttle position alone does not, repeat not, decide what the AFR is all by itself, the PCM and its ability to change the amount of fuel delivered gets a vote too in what the afr is .

You can have the same afr at just about any throttle position if the tuner decides that's what they what. And the factory calibrations have it as lean as they can safely get away with for CAFE standard / catalytic reasons.

Yeah during tip -in throttle they PCM commands more fuel, otherwise the engine would bog (basically the PCM is performing the same function as a carburetor accelerator pump did back in the olden days), but the OEMs want that AFR back to as lean as they can get it, as quickly as they can get it, for those all important fuel mileage and catalytic converter longevity reasons, so it doesn't stay rich for long.

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No where in this post does it even remotely suggest throttle position is the only input for injector output.

MAF, engine speed, pedal position, cyl head temp, 02 readings, etc etc...all are inputs.

The engine is trying to reach a power output that is commanded, not an afr.

I'm sure you know enough to know that you can't just have any afr at any engine load or any engine speed. The limitations of knock or fuel vaporization apply.

Again the "tuner" is trying to make the engine reach a certain torque level.

 

[blanchard7684]

Is this a statement you are willing to stand behind? Its totally wrong, but that hasn't stopped you before.

If you put 10cc of fuel (or really any given amount of fuel within the range each engine can efficiently burn) in either engine (part throttle stuff where talking here) and mix in air to the same AFR, then guess what,,,,, both engines will be making the same power.
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What burns or combusts is the mixture. one engine has a smaller amount of the mixture than the other.

 

[B-train]

Ok since I’m so clueless you tell me:

How is it the 6.2 can get nearly identical mileage as the 5.3 in the same suv with same gear and same mass?

Since your knowledge is so vast and impressive it should be easy.

I have offered a plausible explanation with facts and solid theoretical basis.

Your turn…

I would say that a 6.2L can get the same MPG because it's not working as hard to move the same mass. It's also true for the old 8.1L in the 2500 avalanche. I know of people who got almost 20 mpg cruising steady on road trips because the motor could just loaf along and didn't need extra fuel to keep up with the load. There are all kinds of variables obviously, but when you compare torque maps the higher torque will always win.

I've had 6.2L denalis for 15 years now and I will say that they get the same, and sometimes better, than the 5.3L yukon I used to own prior to that. Yes, they are the newer generation, but the power output when fed the right fuel is phenomenal. Training the right foot is key.....

 

[Antonm]

Link fixed.

Let me summarize.

Power = efficiency factors x Heating value of fuel x VE x displacement x rpm/(afr x constants).

So a 5.3 makes less power than a 6.2 at the same afr.

slide 4.

This is a rearranged formula from Heywood text on combustion engines.

I think maybe when you fixed the link you directed it to the wrong paper. Because nowhere in the paper you linked (the paper done by the EPA titled "Characterizing Factors Influencing SI Engine Transient Fuel Consumption for Vehicle Simulation in ALPHA"), does it ever even mention the term air-fuel ratio in it,,, anywhere , not even ctrl+f could find that term in the downloaded pdf, so its not like I skimmed over it.

Here's a copy past of the slide 4

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[blanchard7684]

I think maybe when you fixed the link you directed it to the wrong paper. Because nowhere in the paper you linked (the paper done by the EPA titled "Characterizing Factors Influencing SI Engine Transient Fuel Consumption for Vehicle Simulation in ALPHA"), does it ever even mention the term air-fuel ratio in it,,, anywhere , not even ctrl+f could find that term in the downloaded pdf, so its not like I skimmed over it.

Here's a copy past of the slide 4
View attachment 447605
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This link below is what I was referencing. slide 4.

https://view.officeapps.live.com/op/view.aspx?src=https%3A%2F%2Fwww.webpages.uidaho.edu%2Fmindworks%2FIC_Engines%2FLectures%2FLecture%252007%2520-%2520Relationships%2520Between%2520Parameters.pptx&wdOrigin=BROWSELINK

One of the many links I've provided.

 

[Antonm]

What burns or combusts is the mixture. one engine has a smaller amount of the mixture than the other.

lol, Nope, just think about that for a minute.

If driving along at 55mph takes say 70 hp to do, then either engine (the 5.3 or the 6.2) will be injecting the amount of fuel that nets them 70hp, so both engines will have the same amount of the fuel in the cylinders and both will need the same amount of air to burn that fuel

Yes that will be slightly higher percentage open on the smaller engines throttle body, but the same quantity of air and fuel will be in both engines if both are producing the same cruising speed horsepower.
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[blanchard7684]

lol, Nope, just think about that for a minute.

If driving along at 55mph takes say 70 hp to do, then either engine (the 5.3 or the 6.2) will be injecting the amount of fuel that nets them 70hp, so both engines will have the same amount of the fuel in the cylinders and both will need the same amount of air to burn that fuel

Yes that will be slightly higher percentage open on the smaller engines throttle body, but the same quantity of air and fuel will be in both engines if both are producing the same cruising speed horsepower.
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My contention was at-the-same-afr.

Same afr, the smaller displacement engine will make less power.

Equation on slide 4 shows this in previous link.

 

[Antonm]

This link below is what I was referencing. slide 4.

https://view.officeapps.live.com/op/view.aspx?src=https%3A%2F%2Fwww.webpages.uidaho.edu%2Fmindworks%2FIC_Engines%2FLectures%2FLecture%252007%2520-%2520Relationships%2520Between%2520Parameters.pptx&wdOrigin=BROWSELINK

One of the many links I've provided.

And just as I suspected, you are in fact interperting it wrong.

Here's the slide four you think proves what you're talking about. That bottom equation for power does have afr in it, but notice where that comes from (see red boxes in equation above it).

Now lets look at the definition of terms slide in that same presentation;

So, that AFR input is part of how the mass flowrate of fuel is determined,,,pretty sure I've mentioned earlier that power comes from the fuel,,, your eureka moment equation is using AFR to determine how much fuel has been added.
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[Antonm]

My contention was at-the-same-afr.

Same afr, the smaller displacement engine will make less power.

Equation on slide 4 shows this in previous link.

Wrong, that equation is using AFR to determine the amount of fuel added. Not exactly advanced math skills needed to figure that out, pretty sure a C student in a ninth grade algebra class could figure that one out.

What else you got?

Oh BTW, that incorrect link to the EPA paper about their modeling software was interesting, not really relevant to this argument conversation, but an interesting read none the less. Shows how much money (our taxpayer money) the EPA is willing to waste though.

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[blanchard7684]

And just as I suspected, you are in fact interperting it wrong.

Here's the slide four you think proves what you're talking about. That bottom equation for power does have afr in it, but notice where that comes from (see red boxes in equation above it).

View attachment 447606


Now lets look at the definition of terms slide in that same presentation;

View attachment 447607


So, that AFR input is part of how the mass flowrate of fuel is determined,,,pretty sure I've mentioned earlier that power comes from the fuel,,, your eureka moment equation is using AFR to determine how much fuel has been added.
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It is algebraic rearrangement of terms.

which changes nothing in terms of validity.

 

[Antonm]

So let me get this straight,,,, you're saying that the equation posted in your "slide 4" that I copy pasted above is not using AFR just to determine the mass of fuel added. Is that what you're saying a statement you are willing to stand behind?
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[blanchard7684]

And just as I suspected, you are in fact interperting it wrong.

Here's the slide four you think proves what you're talking about. That bottom equation for power does have afr in it, but notice where that comes from (see red boxes in equation above it).

View attachment 447606


Now lets look at the definition of terms slide in that same presentation;

View attachment 447607


So, that AFR input is part of how the mass flowrate of fuel is determined,,,pretty sure I've mentioned earlier that power comes from the fuel,,, your eureka moment equation is using AFR to determine how much fuel has been added.
...

Even looking at just the fuel mass flow, it says the same thing:

70hp ( your example) requires a given fuel flow.

Ok so make that a given in the formula.

The higher displacement ( numerator) can have a higher afr ( denominator) and still make the equation true— ie match the required fuel flow.

This means exactly what is shown in rearranged equation.

It is just a substitution of terms which is a valid mathematical move.

The higher displacement can make 70 hp with higher afr.

Put another way, smaller displacement can’t make as much power at same afr.

 

[blanchard7684]

So let me get this straight,,,, you're saying that the equation posted in your "slide 4" that I copy pasted above is not using AFR just to determine the mass of fuel added. Is that what you're saying a statement you are willing to stand behind?
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So let me get this straight,,,, you're saying that the equation posted in your "slide 4" that I copy pasted above is not using AFR just to determine the mass of fuel added. Is that what you're saying a statement you are willing to stand behind?
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“Just”?

There are many other terms, not just afr.

Afr is of course mass of air divided by mass of fuel.

A fuel mixture ( afr) is required to make combustion occur. It isn’t just the fuel. The air oxidizes the fuel and this happens at the right pressure, temp, and mixing.

This is a valid rearrangement of terms to show how power depends on afr and displacement ( other variables too)

 

[Antonm]

Even looking at just the fuel mass flow, it says the same thing:

70hp ( your example) requires a given fuel flow.

Ok so make that a given in the formula.

The higher displacement ( numerator) can have a higher afr ( denominator) and still make the equation true— ie match the required fuel flow.

This means exactly what is shown in rearranged equation.

It is just a substitution of terms which is a valid mathematical move.

The higher displacement can make 70 hp with higher afr.

Put another way, smaller displacement can’t make as much power at same afr.

UM, no , not even close.

Lets review. Your linked power equation shows that power is equal to a bunch of efficiencies (slide of terms coming up if needed) and fuel flow.

The red box terms in the the numerator of the the mass flowrate of fuel equation are in the numerator of the power equation, and the blue boxed terms in the denominator of the mass fuel flowrate equation and in the denotator of the power equation.

For ease of reference, lets compare the terms in the red and blue colored boxes to the other respective colored box,,, spoiler alert,,they're the same.

So that leaves these terms in the numerator of the power equation( that you linked).



So lets examine what those remaining terms in that numerator are again;

Its a bunch of efficiency terms and another fuel term,,,so your power equation says power is equal to the various efficiencies, times a fuel heating term, times fuel mass flowrate.

So, if you change any one of the terms in either the red or blue boxes, you also change the mass flowrate of fuel,,,you know fuel,,,where the power comes from.
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[Antonm]

“Just”?

There are many other terms, not just afr.

Afr is of course mass of air divided by mass of fuel.

A fuel mixture ( afr) is required to make combustion occur. It isn’t just the fuel. The air oxidizes the fuel and this happens at the right pressure, temp, and mixing.

This is a valid rearrangement of terms to show how power depends on afr and displacement ( other variables too)

If you are maintaining a given power level (like say that 70 hp cruising example) and change any of the terms in either the red or blue boxes above, then you are changing fuel mass flowrate.

Easy example, take P=X/Y, if you're not changing P (its steady at 70 hp cruising) and you move Y down, then X will have to also come down proportionally to maintain P at the set value.

Lets substitute some easy numbers in there to further demonstrate. Lets say 0.5 = 10/20,,, if we change the 20 in the denominator to 10 by itself ,the equation would no longer be true, so the 10 in the numerator would have to change proportionally (to 5 in this case) to keep the equation true, and equal to a value of 0.5.

I should really be charging you for the math tutoring.
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