Bad mpg?

[blanchard7684]

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.


View attachment 447618


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

View attachment 447619

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

View attachment 447620

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.
...

The expression says power is proportional to displacement and inversely proportional to afr.

You circled it. Literally.

You are simply working through a valid derivation , term by term. You aren’t changing the validity of the final expression.

Critiquing how the other variables on right hand side will not change this fact.

If I tell you force is equal to mass times acceleration, then redefine accel in terms of momentum ,and redefine mass in terms of density and volume, It doesn’t change the validity of either final expression nor any of the preceding expressions.

This is substitution of terms.

Dancing around this doesn’t change the fact that the final expression says exactly what I said it did.

 

[blanchard7684]

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.

I should really be charging you for the math tutoring.
...

Now we are getting somewhere.

If p is 70 then you can have a 6.2 ( numerator) displacement with higher afr ( denominator) and make the same P as a 5.3 with lower afr.

Hence 6.2 can have better mileage in some situations than 5.3; and for the same afr the 6.2 can make more power than 5.3

 

[Antonm]

The expression says power is proportional to displacement and inversely proportional to afr.

Can you not read or are you just that dumb?

It says nothing of the sort. It says power = efficiency x fuel flow.

I even boxed the terms to show how they compare / help your simple mind grasp them.

Yor not even being a good troll at this point, you're just being dumb.
...

 

[Antonm]

Now we are getting somewhere.

If p is 70 then you can have a 6.2 ( numerator) displacement with higher afr ( denominator) and make the same P as a 5.3 with lower afr.

Hence 6.2 can have better mileage in some situations than 5.3; and for the same afr the 6.2 can make more power than 5.3

Look at the equation again, if you change ANY of the terms in either the blue or red boxes, then you are changing fuel flow assuming a steady power output.

You need some remedial math classes.
...

 

[blanchard7684]

Can you not read or are you just that dumb?

It says nothing of the sort. It says power = efficiency x fuel flow.

I even boxed the terms to show how they compare / help your simple mind grasp them.

Yor not even being a good troll at this point, you're just being dumb.
...

You boxed the very terms that show the inverse relationship between power , afr, and displacement.

But somehow you can’t see the correlation.

You fascinate me.

 

[blanchard7684]

Look at the equation again, if you change ANY of the terms in either the blue or red boxes, then you are changing fuel flow assuming a steady power output.

You need some remedial math classes.
...

If you want to look at it that way that is your choice.

But it does not negate the validity of the final expression, nor does it negate any obvious conclusions that stem from it , such as lower displacement does not make the same power as higher displacement at-the-same-afr.

Mass flow of fuel is a function of afr. It is also a function of displacement. This is clear in the expression that you are stuck on.

Afr and displacement are still inversely related in the equation for mass flow of fuel.

There is no getting around this.

 

[Antonm]

You boxed the very terms that show the inverse relationship between power , afr, and displacement.

But somehow you can’t see the correlation.

You fascinate me.

Because that correlation is not there, that correlation only exists in your imagination.
..

 

[blanchard7684]

P

Because that correlation is not there, that correlation only exists in your imagination.
..

Based on this comment I can tell you are willfully ignoring what is an obvious fact.

This is just like the other thread on disablers where RG23RST handed you the facts on a platter and you still wouldn’t concede the point.

If you are so brazen as to look at a valid algebraic rearrangement of a formula and impudently refuse to see an obvious correlation, there is no point in continuing the discussion.

I can’t make it any simpler.

 

[Antonm]

If you want to look at it that way that is your choice.

But it does not negate the validity of the final expression, nor does it negate any obvious conclusions that stem from it , such as lower displacement does not make the same power as higher displacement at-the-same-afr.

Did you bother to read past slide #4 ?

They do talk about ways to increase power later on. Here's that slide, it even talks about AFR, is says once you go off stochiometric you lose combustion efficiency (which is true).

Just to keep your brain straight (hopefully) the P is no longer a constant in what they're taking about below. They're talking about ways to raise power (P↑) and not about making the same power with two different engines. So if P↑, then other variables can /will change as well (like you're not going to get more power by just adding BDC (boost pressure) with no increase in fuel flow


...

 

[blanchard7684]

Did you bother to read past slide #4 ?

They do talk about ways to increase power later on. Here's that slide, it even talks about AFR, is says once you go off stochiometric you lose combustion efficiency (which is true).

Just to keep your brain straight (hopefully) the P is no longer a constant in what they're taking about below. They're talking about ways to raise power (P↑) and not about making the same power with two different engines. So if P↑, then other variables can /will change as well (like you're not going to get more power by just adding BDC (boost pressure) with no increase in fuel flow

View attachment 447624
...

Sure did and supports the notion that larger displacement can achieve higher fuel mileage in some situations.

Thanks for agreeing.

The other slides show how VE changes things as well.

Key point there is that Ve isn’t just a function of engine dimensions ( your claim)

But I’ll leave that one alone .