ok here's the scenario , lets say you run a blower and run into detonation at high rpm. if you put in headers your boost will drop, right?
will the drop in boost get rid of the detonation assuming the same timing as before????

 

the boost will drop but your hp level will stay the same pretty much, so the problem you have will most likely still be a problem. it is not caused directly by air pressure.

 

I believe the more boost/intake backpressure the more heat, the more heat the more detonation.

Increased flow = same amount of air and power w/less heat.

 

so i have one for each answer! <img border="0" title="" alt="[Big Grin]" src="gr_grin.gif" />

any other suggestions?

 

Let's look at the mixture in the chamber right before TDC when it is about to detonate. You will have about the same amount of total mixture there anyway, and at the same temp. why? Let me guide you to Gay-Lussac's Law of ideal gases. The pressure of a quantity of a gas is proportional to the temperature at constant volumn. This will tell you why temps go up with higher pressure. take 2 liters of air at 100 degrees and stuff it into a space that is 1 liter in size for 2 bars of pressure. final temp will be 200 degree. Take 4 liters of air at 100 degrees and stuff it into a space that is 1 liter in size for 4 Bars of pressure. you get 400 degrees. Notice that if you double the pressure you double the temp at the same final volumn like in Gay-Lussac's Law.

I am certain that you will not drop any temp in the mixture in the combustion chamber as the supercharger still has to flow the same amount of air through it thus it does the same amount of work. Don't count on headers to solve your KR issues. You will gain hp with the headers though.

get some alky/water <img border="0" title="" alt="[Wink]" src="gr_images/icons/wink.gif" />

 

</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by AlienDroid:
[QBtake 2 liters of air at 100 degrees and stuff it into a space that is 1 liter in size for 2 bars of pressure. final temp will be 200 degree. Take 4 liters of air at 100 degrees and stuff it into a space that is 1 liter in size for 4 Bars of pressure. you get 400 degrees[/QB]</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Were you going to qualify that with any particular temperature scale?

 

</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by 1dirtyZ:
<strong> ok here's the scenario , lets say you run a blower and run into detonation at high rpm. if you put in headers your boost will drop, right?
will the drop in boost get rid of the detonation assuming the same timing as before???? </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">It really depends on how much boost you are going to reduce it by. I don't think it will be enough to solve your KR problems though.

 

</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by AlienDroid:
<strong> Let's look at the mixture in the chamber right before TDC when it is about to detonate. You will have about the same amount of total mixture there anyway, and at the same temp. why? Let me guide you to Gay-Lussac's Law of ideal gases. The pressure of a quantity of a gas is proportional to the temperature at constant volumn. This will tell you why temps go up with higher pressure. take 2 liters of air at 100 degrees and stuff it into a space that is 1 liter in size for 2 bars of pressure. final temp will be 200 degree. Take 4 liters of air at 100 degrees and stuff it into a space that is 1 liter in size for 4 Bars of pressure. you get 400 degrees. Notice that if you double the pressure you double the temp at the same final volumn like in Gay-Lussac's Law.

I am certain that you will not drop any temp in the mixture in the combustion chamber as the supercharger still has to flow the same amount of air through it thus it does the same amount of work. Don't count on headers to solve your KR issues. You will gain hp with the headers though.

get some alky/water <img border="0" title="" alt="[Wink]" src="gr_images/icons/wink.gif" /> </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Physics in the wrong hands is dangerous <img border="0" title="" alt="[Wink]" src="gr_images/icons/wink.gif" />

You are forgetting a fundemental law of thermodynamics in all of these crazy calculations you do.

Now, I know your not going to believe it coming from me, so I will site a reputable source.

Allen Lockheed, a noted higher-performance engineer and an expert on computer modeling emphasizes a point that you need to consider "The airflow through an engine remains the same whether or not it is supercharged or turbocharged. The power comes from the increase in air-charge density".

Chew on that for a while and you will realize why all the stuff that you put together above means nothing.

Not a flame, just trying to help out.

Good Luck,
Kevin

 

"The airflow through an engine remains the same whether or not it is supercharged or turbocharged. The power comes from the increase in air-charge density".

This is a good point. Flow is the same, as in speed of flow. air-charge density means that there is more mixture per volumn so air flow at the same speed actually gets more air through. so what he said is well understood. but what I'm saying is that when you see a drop in boost pressure, that means nothing by itself. You can't just sit there and assume you the mixture will be cooler or as much cooler as you think it will be.


"Were you going to qualify that with any particular temperature scale? "

probably should have. Should have used thermal energy and not degrees.


Wish I could get some funding to do an experiment on this. <img border="0" alt="[cheers]" title="" src="graemlins/gr_cheers.gif" />

 

Anyone want any more crazy wacked out calculations and theories.

and no need for insults NoGo.

 

I am not trying to insult, just merely pointing out what you had insinuated was incorrect.

And, what you are still insuating is still incorrect.

<strong>
I am certain that you will not drop any temp in the mixture in the combustion chamber
</strong>

Change in air charge density (read more or less combustible gas) will directly effect cylinder temperatures, and thus pre-ignition or detonation. You are making the incorrect assumption that since the compressor is going to be flowing the same amount of air, then your combustion temperatures are going to be the same no matter what you do.

<strong>
but what I'm saying is that when you see a drop in boost pressure, that means nothing by itself
</strong>

A reduction in boost pressure means everything when we're talking pre-ignition and detonation. Boost pressure directly effects the dynamic compression of the motor. The higher the boost, the higher the dynamic compression. The higher the dynamic compression, the higher the chances of detonating and pre-ignition.

<strong>
You can't just sit there and assume you the mixture will be cooler or as much cooler as you think it will be.
</strong>

Sure I can. If you would like, I will use the same law that you did above to show you why the intake tamperature is going to be cooler.

Good Luck,
Kevin

 

</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by AlienDroid:
[QB"Were you going to qualify that with any particular temperature scale? "

probably should have. Should have used thermal energy and not degrees.
[/QB]</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">You can use degrees as long as it's Kelvin. But then the 100, 200 would be way off.