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.
...
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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