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.
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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 .

 

[Antonm]

P

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.

You also can't make true either.

Here's a statement I will stand by,,, You are a clueless moron, you like troll, and its freaking hilarious watching you grasp at straws and make things up.

Its almost like watching a political debate were the crazies just make up/ say whatever dumb crap they want that makes no sense. But this is funnier because hopefully you're not in charge of making policy/ law like the clueless politicians.
..

 

[Antonm]

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 .

The argument was you keep wrongfully saying that engines of different displacements can't make the same power at the same AFR. Remember the statement you stand by even though you accidently deleted it.

The only thing, repeat, the one and only thing, that could make a larger engine use less fuel than a smaller engine moving the same mass at the same rare would be if the larger engine happened to be more efficient at that given load than the smaller engine.

Considering both the 5.3 and 6.2 are very similar in design, i would expect their efficiencies' to also be very similar, any differences would likely be in the second significant digit range, and could go either way.
...

 

[blanchard7684]

The argument was you keep wrongfully saying that engines of different displacements can't make the same power at the same AFR. Remember the statement you stand by even though you accidently deleted it.

The formula we have been dissecting literally says this in plain mathematics.

 

[Antonm]

The formula we have been dissecting literally says this in plain mathematics.

Again , no its does not, you just keep imagining, or misunderstanding, or wishing, or whatever that it does.

If that were true, then the solution to getting more MPG would be for everyone to drive huge displacement engines with as many cylinders as possible.

If the math says what you think it does (which it doesn't, but lets just say), and bigger meant better fuel economy, then were does it end , if 6.2L better why 8.0L or 11.0L or 22.0L, do you think all those would get better fuel milage too,,,or does that just sound ridiculous (like the other stuff you been posting?
...

 

[blanchard7684]

You also can't make true either.

Here's a statement I will stand by,,, You are a clueless moron, you like troll, and its freaking hilarious watching you grasp at straws and make things up.

Its almost like watching a political debate were the crazies just make up/ say whatever dumb crap they want that makes no sense. But this is funnier because hopefully you're not in charge of making policy/ law like the clueless politicians.
..

Look..

I can tell this is all new information for you.

It isn’t for me.

I am a practitioner of this information on a regular basis.

I found open source material (that I already knew) and provided it for discussion and your edification.

I can’t post IP.

I will make an apology for being prickly up thread.

However you have kept up incessant ad hominem, insults to my intelligence, mental well being, academic achievement, and others. Some were actually amusing and funny.

Yet I’ve kept this discussion factual. I’ve spent an inordinate amount of time attempting to educate you.

I can tell you have a hang up about this discussion.

That is ok.

Take some time and regroup.

I’m ok. You are ok.

Even though you have lobbed some nasty comments in my direction I don’t hold it against you.

This isn’t 101 level stuff. If it was,then oem’s like Gm wouldn’t have an army of engineers working these engines.

Cheers

 

[blanchard7684]

Again , no its does not, you just keep imagining, or misunderstanding, or wishing, or whatever that it does.

If that were true, then the solution to getting more MPG would be for everyone to drive huge displacement engines with as many cylinders as possible.

If the math says what you think it does (which it doesn't, but lets just say), and bigger meant better fuel economy, then were does it end , if 6.2L better why 8.0L or 11.0L or 22.0L, do you think all those would get better fuel milage too,,,or does that just sound ridiculous (like the other stuff you been posting?
...

Reductio ad absurdum

The difference we are looking for is in a small range of fuel economy.

You are plenty smart enough to know that if one of these variables goes way up then it will change the output of the dependent variable.

No where have I said that the 6.2 gets better economy in all possible driving scenarios.

There is a specific set of conditions that it can.

 

[Antonm]

Look..

I can tell this is all new information for you.

It isn’t for me.

I am a practitioner of this information on a regular basis.

I found open source material (that I already knew) and provided it for discussion and your edification.

I can’t post IP.

I will make an apology for being prickly up thread.

However you have kept up incessant ad hominem, insults to my intelligence, mental well being, academic achievement, and others. Some were actually amusing and funny.

Yet I’ve kept this discussion factual. I’ve spent an inordinate amount of time attempting to educate you.

I can tell you have a hang up about this discussion.

That is ok.

Take some time and regroup.

I’m ok. You are ok.

Even though you have lobbed some nasty comments in my direction I don’t hold it against you.

This isn’t 101 level stuff. If it was,then oem’s like Gm wouldn’t have an army of engineers working these engines.

Cheers

If you are a "practitioner" then you need more practice (like a lot more), because you have some very basic concepts and math fundamentals wrong.

There was a time back in the early 2000's where I taught math and physics to young adults, this discussion has reminded me of some of my more ,,, interesting,, students and how some people refuse to accept when they're mistaken.

But cheers to you as well, this has at least made the last few days less boring if nothing else.
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[djnice]

Entertaining. I don't have time to read it all today. Did you factor the 6.2 has higher compression?

 

[blanchard7684]

The half point in CR would benefit the 6.2. The otto cycle thermal efficiency is highly dependent on compression ratio:

3.5 The Internal combustion engine (Otto Cycle)

3.5 The Internal combustion engine (Otto Cycle)

web.mit.edu

r is compression ratio, and gamma exponent is ratio of specific heats for air (Otto cycle is an "air-standard" thermodynamic cycle for SI 4 stroke engines on gasoline)

 

[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.
...

Let's examine this a bit further.

Taking this view, we can further rearrange the power formula: P = n x Q x fuel flow rate.

The opposing contention is: if power is the same then fuel flow rate has to be the same. Therefore 6.2 and 5.3 make same power at same afr.

What this analysis is missing is that the power equation (P = n x Q x fuel flow rate ) is true for the same engine 5.3 vs 5.3 or 6.2 vs 6.2.
But comparing 6.2 vs 5.3 the main difference is the displacement. The flow rate equation for fuel even shows a direct proportionality to displacement. The 6.2 will draw more air and require more fuel for a given afr.

Here is why.

Fuel flow rate is dependent on afr. There is only a specific range of afr where a combustible mix can occur.

Take 14.7.

We will assume air density, in cylinder, is same between 5.3 and 6.2.

It may be different at extremes of engine operation, but not enough to alter the outcome of the discussion.

For ease of illustration let's say the air density in cylinder is 1 kg/L.

A 5.3 will have 5.3L x 1kg/L = 5.3 kg of air. To get 14.7 afr, then the fuel will have to be 5.3kg/14.7 = 0.36 kg fuel. 5.3 kg air/ 0.36 kg fuel = 14.7.

A 6.2, following same method, will have 0.42kg of fuel.

Since this is over the same time scale this is also the flow rate of fuel.

You can see then that the 5.3 will have less fuel than 6.2 at same afr.

So...with less fuel in this equation P = n x Q x fuel flow rate, it will not have the same power, P, as a 6.2 at the same afr.