Bad mpg?

[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.
...

 

[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.
...

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.
...

 

[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.

...

 

[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.
...

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?
...

 

[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?
...

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?
...

“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 each 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.
...

 

[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.
...