Bad mpg?

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Antonm

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If I take 5.3 L at a specified pressure , it will also have a specific mass.

Then if I put this into a 6.2 L container, the pressure will decrease by Boyle’s law.

Um, that's exactly what I said above ,that you said was wrong. But I'm glad to see you now agree.

Here let me post a pic below of where I said that exact thing, then you red underlined it and said it was wrong before you edit the post when you change your mind again.

clueless guy gas press vs volume.png

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Antonm

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When the throttle is actuated it is changing the effective flow coefficient through the entire induction system. It is the main limiting pressure drop, but not all of it.

Again there is pressure restriction from the intake box, throttle, intake manifold, intake runner, and intake valve.

The mass of air will be different, the volume will be different, and the volume inside the cylinder will be different.

True there are other restrictions beside the throttle body,,,,but they are not the limiting or controlling factor at part throttle operation.

The intake tract in a 5.3 is capable of flowing the required air to make at least 355hp. At part throttle conditions, where the engine is making less power than full rated, (which is what we've been talking about the entire time), you just open the throttle body more to get the correct amount of air into the cylinders for that part throttle power.
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blanchard7684

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Um, that's exactly what I said above ,that you said was wrong. But I'm glad to see you now agree.

Here let me post a pic below of where I said that exact thing, then you red underlined it and said it was wrong before you edit the post when you change your mind again.

View attachment 447947
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I can see you didn't read or bother to interpret my post: I clearly said in english language Boyle's law does not apply in your specific, contrived, scenario.

I even underlined it for you as a reference.

You have un convincing case that based on boyle's law the pressure at BDC in 5.3 is higher than in 6.2.
 
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Antonm

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If I take 5.3 L at a specified pressure , it will also have a specific mass.

Then if I put this into a 6.2 L container, the pressure will decrease by Boyle’s law.

All this does is prove Boyle’s law, it doesn’t prove your original point about pressure at BDC being higher in 5.3.


2) it takes a given mass of air. Remember you can get stiochiometric balance in any size cylinder, with different masses of fuel and air, which create different levels of power when combusted, at same afr.

You literally contradict yourself in the same post only a few sentences apart.

You say above "If I take 5.3 L at a specified pressure , it will also have a specific mass. Then if I put this into a 6.2 L container, the pressure will decrease by Boyle’s law."

So you acknowledge that if you put the same volume of air in different size cylinder the pressure will be different (you even wrote this, see above).

Then you seem to acknowledge that its takes a given mass of air to burn a given mass of fuel at stochiometric. Since both the 5.3 and 6.2 make the same part throttle power at cruising speeds , then they would need that same mass of air to that.

But even though you say you understand it takes the same mass of air, and that different size cylinders will have different pressures if the have the same mass of air trapped in them,,,, but you till don't seem to understand how pressure at bdc could be different,,, do you really not see the contradiction in you own statements/ your own logic here?

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Antonm

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I can see you didn't read or bother to interpret my post: I clearly said in english language Boyle's law does not apply in your specific, contrived, scenario.

I even underlined it for you as a reference.

You have un convincing case that based on boyle's law the pressure at BDC in 5.3 is higher than in 6.2.

So last night, before you edited your post, your agreed with P1 * V1 = P2 *V2 ,,, do you no longer agree with equation?
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Antonm

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This stems again from how the induction system is configured.

Just think: an 80 mm throttle, 1.9 intake valves, with 5.3L pulling air through them will move less air mass than a 6.2 L engine with 87 mm throttle and 2.16 intake valves.

At wide open throttle, making 100% rated power, yes. At part throttle (the topic we've been discussing, unless you're now trying to abandon that like you did the afr discussion), then the mass of air is regulated down by the driver via the throttle body so that the appropriate mass is passed to make the desired power.

Assuming that the desired part throttle cruising power level is less that the rated output of both engines, then both intake tracts can flow that amount of air, yes the throttle body will have to be open slightly more on the smaller/ more restrictive intake engine, but the outcome is they will both be flowing the same cubic feet of air to make the same part throttle cruising power.
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Antonm

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2) it takes a given mass of air. Remember you can get stiochiometric balance in any size cylinder, with different masses of fuel and air, which create different levels of power when combusted, at same afr.

True, and if you want to move the same mass (the vehicle) at the same velocity (say that 70mph example used above), then you need to make the power to do that.

And to make that same power you'll need the same amount of fuel,,,,,and to burn that same amount of fuel you'll need the same amount of air to burn that fuel at the same ratio.

So both a 5.3 and a 6.2 will be ingesting the same amount of air when cursing at a given power level.
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Antonm

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Lets look at what your own equations have to say about pressure at bottom dead center (I'm going to assume you believe them because you posted them.

First lets look a the power, , but do remember that the power equation has the fuel flowrate equation directly carried forward into it (remember the blue and red boxed terms earlier)

So what happens to power if only displacement is lowered, but Press at bdc doesn't change.

cluelss guy pbdc press 1.png


Holly crap, power would go down (so the vehicle would start slowing down).

Now why is power going down I wonder,,,

cluelss guy pbdc press 2.png


Oh, its because fuel flow is going down.

So assuming its possible for both a 6.2 and a 5.3 Tahoe to drive down the road, side by side, with both doing 70mph next to each other, then they have to maintain the same cursing power output.

So it ends up looking like this in reality, with pressure at bottom dead center going up when you trap the volume of air needed to burn the same amount of fuel in that smaller cylinder ( of course the revers is also true if you wanted to start with the smaller engine and go to the larger one)

We need power to remain constant between the two engine so both vehicles maintain the same velocity (i.e.. both driving down the road at 70mph).


clueless guy power constant2.png


And BTW, fuel flow will be the same as well.
clueless guy power constant1.png


The value of this pressure change a bottom dead center changes could be shown with p1*v1=p2*v2, but since you recanted/ edited your agreement to that basic equation, I just went with the equations you posted (assuming you agree with them), to illustrate the same point.

So I assume now you'll flip- flop to an efficiency or something to try an explain how your ridiculous assertions could have even a grain of truth to them.
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blanchard7684

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You literally contradict yourself in the same post only a few sentences apart.

You say above "If I take 5.3 L at a specified pressure , it will also have a specific mass. Then if I put this into a 6.2 L container, the pressure will decrease by Boyle’s law."

So you acknowledge that if you put the same volume of air in different size cylinder the pressure will be different (you even wrote this, see above).

Then you seem to acknowledge that its takes a given mass of air to burn a given mass of fuel at stochiometric. Since both the 5.3 and 6.2 make the same part throttle power at cruising speeds , then they would need that same mass of air to that.

But even though you say you understand it takes the same mass of air, and that different size cylinders will have different pressures if the have the same mass of air trapped in them,,,, but you till don't seem to understand how pressure at bdc could be different,,, do you really not see the contradiction in you own statements/ your own logic here?

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You can not discount the effect of acoustic design on the different induction systems.

It will alter pressure at bottom dead center.

Which negates your Boyle's law concept.
 

blanchard7684

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Lets look at what your own equations have to say about pressure at bottom dead center (I'm going to assume you believe them because you posted them.

First lets look a the power, , but do remember that the power equation has the fuel flowrate equation directly carried forward into it (remember the blue and red boxed terms earlier)

So what happens to power if only displacement is lowered, but Press at bdc doesn't change.

View attachment 447950

Holly crap, power would go down (so the vehicle would start slowing down).

Now why is power going down I wonder,,,

View attachment 447951

Oh, its because fuel flow is going down.

So assuming its possible for both a 6.2 and a 5.3 Tahoe to drive down the road, side by side, with both doing 70mph next to each other, then they have to maintain the same cursing power output.

So it ends up looking like this in reality, with pressure at bottom dead center going up when you trap the volume of air needed to burn the same amount of fuel in that smaller cylinder ( of course the revers is also true if you wanted to start with the smaller engine and go to the larger one)

We need power to remain constant between the two engine so both vehicles maintain the same velocity (i.e.. both driving down the road at 70mph).


View attachment 447952

And BTW, fuel flow will be the same as well.
View attachment 447953

The value of this pressure change a bottom dead center changes could be shown with p1*v1=p2*v2, but since you recanted/ edited your agreement to that basic equation, I just went with the equations you posted (assuming you agree with them), to illustrate the same point.

So I assume now you'll flip- flop to an efficiency or something to try an explain how your ridiculous assertions could have even a grain of truth to them.
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Then everytime displacement increases, power drops, at a given afr, given engine speed, given throttle position.

Is this what you want to stick with as conclusion?

Because it is contradictory to all known reality.

I guess the best way to increase power, or maintain power at any afr, engine speed, is to reduce displacement.
 
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Antonm

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You can not discount the effect of acoustic design on the different induction systems.

It will alter pressure at bottom dead center.

Which negates your Boyle's law concept.

So, are you now refuting the very equations you posted and saying they're wrong now?

If the air wasn't getting in the cylinder (for whatever reason) , we'd know that because we directly measure the O2 concentration of the exhaust ( remember the whole thing that started this, the O2 sensor discussion when you tired trolling on Marky Dissod but ended up just showing how little you actually know about the topic), we also measure the mass flowrate of air engine the engine via the mass airflow meter , then confirm what going on in the manifold via the manifold absolute pressure sensor. And we've been doing this on production cars for the last 30 years, so not exactly new tech.

So we know, for certain, how much air is going in.
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blanchard7684

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So, are you now refuting the very equations you posted and saying they're wrong now?

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Going down this path of pressure at bottom dead center, I originally indicated that this pressure would be close to the same (not enough to overcome displacement).

Therefore there is no contradiction of the equation.
 

Antonm

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Then everytime displacement increases, power drops, at a given afr, given engine speed, given throttle position.

Is this what you want to stick with as conclusion?

Because it is contradictory to all known reality.

lol, you can't be this dumb can you?

Yes , every single time a given quantity of air (like say the quantity needed to burn a given amount of fuel at stochiometric ratio) is trapped in a cylinder, if the size of that cylinder is increased in volume, the pressure in that now larger cylinder will go down, and it will do that every single time in a repeatable, measurable, and predictable fashion.

That's exactly what p1*v1=p2*v2 states actually.

So are you now renaming "Boyles law" to "Boyles suggestion that somehow doesn't apply when blanchard 7684 doesn't want it to" ? If you are, maybe makeup an acronym for it, as that's to long to write out very time.
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blanchard7684

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lol, you can't be this dumb can you?

Yes , every single time a given quantity of air (like say the quantity needed to burn a given amount of fuel at stochiometric ratio) is trapped in a cylinder, if the size of that cylinder is increased in volume, the pressure in that now larger cylinder will go down, and it will do that every single time in a repeatable, measurable, and predictable fashion.

That's exactly what p1*v1=p2*v2 states actually.

So are you now renaming "Boyles law" to "Boyles suggestion that somehow doesn't apply when blanchard 7684 doesn't want it to" ? If you are, maybe makeup an acronym for it, as that's to long to write out very time.
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So displacement cuts power.

got it.

AntonM: The only replacement for displacement, is less displacement.

Also AntonM: Boyle's law says I should decrease displacement to increase power.
 

Antonm

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Going down this path of pressure at bottom dead center, I originally indicated that this pressure would be close to the same (not enough to overcome displacement).

Therefore there is no contradiction of the equation.

LOL, your own equations show that can't be true, if pressure at bottom dead center stayed the same , power and fuel flow would change.

Basic math, if you change the value of one term in an equation, then the the value of the solution changes. In this power and fuel consumption would change if only volume at bottom dead center changed (see arrow analysis of your equations above).
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blanchard7684

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LOL, your own equations show that can't be true, if pressure at bottom dead center stayed the same , power and fuel flow would change.

Basic math, if you change the value of one term in an equation, then the the value of the solution changes. In this power and fuel consumption would change if only volume at bottom dead center changed (see arrow analysis of your equations above).
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No I got ya...to increase cylinder pressure, to increase power, or maintain power, you must reduce displacement.

I'll be sure to try this on my next project car.

I'm sure it will work great.
 

Antonm

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So displacement cuts power.

got it.

AntonM: The only replacement for displacement, is less displacement.

Also AntonM: Boyle's law says I should decrease displacement to increase power.

I posted this explanation above , apparently you didn't read it (or are to stupid to understand it, right now I'm leaning toward the latter but lets continue anyway, maybe there's hope yet).

So I'll do it again for you here,,,now pay attention this time

If you Increase displacement, but want to maintain the same part throttle cruising power output, then pressure at bottom dead center will go down (because you trapped the same volume of air in lager cylinder and Boyles suggestion say so).

Lets see that in your own equations;

clueless displ up1.png




Now, lets see what happens if you take in more air in that larger cylinder, (because taking in more air in that larger cylinder is THE ONLY WAY pressure at bottom dead cylinder could remain the same).

clueless displ up3.png


Oh look, power goes up (we don't want power to go up because they we be accelerating vice maintaining our cruising speed but ahhh its going up).

Wonder why power went up;
clueless displ up4.png

Oh, its because fuel flow went up, that's why power is going up.

Now repeat with me,,,,
-power comes from the fuel
-to burn that fuel we need air.
-to make the same power in two engine with the same efficiencies, we have to burn the same amount of fuel.
-to burn the same amount of fuel at the same afr, we need the same mass of air.
-Boyles law is a law and not a suggestion, so if we trap the same mass of air in two different size cylinders the pressures will be different in those cylinders, even though both cylinders contain the same amount of air.

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blanchard7684

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I posted this explanation above , apparently you didn't read it (or are to stupid to understand it, right now I'm leaning toward the latter but lets continue anyway, maybe there's hope yet).

So I'll do it again for you here,,,now pay attention this time

If you Increase displacement, but want to maintain the same part throttle cruising power output, then pressure at bottom dead center will go down (because you trapped the same volume of air in lager cylinder and Boyles suggestion say so).

Lets see that in your own equations;

View attachment 447955



Now, lets see what happens if you take in more air in that larger cylinder, (because taking in more air in that larger cylinder is THE ONLY WAY pressure at bottom dead cylinder could remain the same).

View attachment 447957

Oh look, power goes up (we don't want power to go up because they we be accelerating vice maintaining our cruising speed but ahhh its going up).

Wonder why power went up;
View attachment 447958
Oh, its because fuel flow went up, that's why power is going up.

Now repeat with me,,,,
-power comes from the fuel
-to burn that fuel we need air.
-to make the same power in two engine with the same efficiencies, we have to burn the same amount of fuel.
-to burn the same amount of fuel at the same afr, we need the same mass of air.
-Boyles law is a law and not a suggestion, so if we trap the same mass of air in two different size cylinders the pressures will be different in those cylinders, even though both cylinders contain the same amount of air.

....

Less displacement increases pressure at BDC, so power goes up...because Boyle's law says so.

I got it

Don't worry. I'm already shopping for a smaller displacement platform.

Comedian is your future calling.
 

Antonm

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No I got ya...to increase cylinder pressure, to increase power, or maintain power, you must reduce displacement.

I'll be sure to try this on my next project car.

I'm sure it will work great.

Now you're trying to hopeless spin crap because you know you've been proven wrong, (several times so far) and are tying yet again to grasp at straws by taking things out of context.

Look again and see if I said anything even remotely like that stupid statement of yours above ,,, please quote that.
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Antonm

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Less displacement increases pressure at BDC, so power goes up...because Boyle's law says so.

I got it

Don't worry. I'm already shopping for a smaller displacement platform.

Comedian is your future calling.

Yet another thing you're pulling out of your dumb ***, please quote where I said anything even remotely like that.

Are you perhaps color blind and can't see the directional arrows I put on your own equations ,or are you just to stupid to understand them?

You apparently think you're smarter than Robert Boyle (the creator of Boyle's law) or think he and the entire scientific community for the last few hundred years have gotten in wrong, because in your mind you've disproven this law in a forum posting.
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