Does 8 psi of boost with the air temperature being 65 degrees equal 8 psi of boost with the air temperature being 30 degrees?
I think I know the answer, but i ask because with my stock s/c pulley I was seeing ~8 lbs of boost with the outside temperature being in the mid sixties, now that it's below freezing outside, I'm seeing 11-12 psi of boost.
I bought a smaller pulley based on me only boosting to 8 psi, not 12. I don't want to boost to 15 psi and push the motor that far.
What have you s/c guys done to tune for this kind of change in air temperature?
Do you guys switch back and forth between pulley sizes?
I ask because I don't think it's safe to run a smaller pulley based on a starting point of 11 psi.
Or does outside temperature matter?
How about you tell us what size pulley you want to install.
Hotter air will be less dense at the same level of boost pressure than cooler air. The cooler, denser charge will contain more O2, therefore hold more power potential, and be less prone to detonation. That's why intercooled setups make the same or better power than non-intercooled at lower boost levels.
15.3 @ 89.97mph, 14's on the way?
The increase in boost could be from the charger sucking in all of the cold air...then in the process of compressing, the charger heats up the incoming air to near-usual manifold temps, hence a higher pressure reading. That would be my guess.
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"Youth in Asia"...I don't see anything wrong with that.
Yes, but its in the STOCK calabration, GM takes abinet air temp in to account.
Chris
'02 Z-24 Supercharged
13.7 @102.45 MPH Third Place, 2007 GMSC Bash SOLD AS OF 01MAR08
honestly, just dont press the pedal that far if you're concerned.
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- Sold my beloved J in April 2010 -
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Hotter air will be less dense at the same level of boost pressure than cooler air. The cooler, denser charge will contain more O2, therefore hold more power potential, and be less prone to detonation. That's why intercooled setups make the same or better power than non-intercooled at lower boost levels.
theres the perfect answer
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honestly, just dont press the pedal that far if you're concerned.
whats the fun in that?
have you thought about a water sprayer like the STI's have? or a nitrous/co2?
this way you would only get that extra cool air when u wanted it.
http://www.myspace.com/15102113
Air follows the ideal gas laws...that said you can calculate the new air temp after compression as follows:
(P1*V1)/T1 = (P2*T2)/T2 :: 1 being known or original case; 2 being condition of interest.
If your comparing the same volume of interest (charge pipes) you can reduce the equation to:
P1/T1 = P2/T2
Pressure should be in kpa and Temperature should be used in Celcius or Kelvin
Using this equation with 8psig boost over atm. pressure of (14.7psi or 101.3kpa) @ 70degree F (21.1 C) outside temp the equation will look like this:
(101.3kpa)/(21.1C) = (156.5kpa)/(X) :: X represents the new temperature of interest with 8 psig boost in the intake
X=32.6 degress C = 90.7 degrees F (makes a air compression temperature difference of 20.7 degress F @ 70F outside temp)
Next using another scenario with the same 8psi of boost and an outside temp of 90 degrees F:
(101.3kpa)/(32.2C) = (156.5)/X
X=49.77C or 121.6F which gives a temperature difference after compression of (121.6F - 90F) = 31.6 F which is greater than the 20.7F degrees gained when the outside ambient temp was 70F
So YES, the ambient temperature outside makes a difference to the air temp gained after compression. I hope this helps. Remember this DOESN'T include any heat transfered from hotter surfaces, also known as heat soak.
^ Yep, that's what I was thinking.......thanks for the clarification though.
Anyone running 15 psi with the M62 on an ECO?
^^Yeah, Josh is right. Gotta love physics :-)
2010 Subaru Impreza WRX Limited
1999 Cavalier Z24 Supercharged
1999 Grand AM SE (Beater Car)
1997 GMC Sierra
2007 Honda CBR 600RR
2005 Honda TRX450R
Admiral Jedi wrote:^ Yep, that's what I was thinking.......thanks for the clarification though.
Anyone running 15 psi with the M62 on an ECO?
Airtonics did for about 1000 feet, then he coasted (actually he ran it for months, but that was just when it decided to give up the connecting rods ghost after two strong 103+MPH runs).
Sure Josh, but that still doesn't explain why Chris is seeing more boost. The colder air still has to heat up past what just compressing the air will...
I also believe you are supposed to use Kelvin when making those calculations, otherwise, after pressurizing 8 psi over atm @ -200°C the final temp would be -35.8°K...
As an engineer, I know you see what is wrong with that.
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"Youth in Asia"...I don't see anything wrong with that.
Brian Whalen wrote:Sure Josh, but that still doesn't explain why Chris is seeing more boost. The colder air still has to heat up past what just compressing the air will...
I also believe you are supposed to use Kelvin when making those calculations, otherwise, after pressurizing 8 psi over atm @ -200°C the final temp would be -35.8°K...
As an engineer, I know you see what is wrong with that.
Well, you can see how the heat transfer would be greater from the blower (at temp) to the cooler air (increasing temperature difference will increase rate of heat transfer). You can use that same gas eq to show that the temperature increase would increase the pressure. I doubt if it would be enough to account for a 3-4 psi increase (thermo exam less than 24 hours ago so I don't feel like doing more calcs), but it's a thought.
fortune cookie say:
better a delay than a disaster.
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Sure Josh, but that still doesn't explain why Chris is seeing more boost. The colder air still has to heat up past what just compressing the air will...
This car is intercooled. Hasn't been mentioned, but pressure is most likely measured post cooler. Smaller delta T at compressor results in smaller delta P across the cooler. Is he measuring an increase in pressure, or a decrease in pressure loss? Simple installation of gauge at s/c outlet might help.
-->Slow
Brian Whalen wrote:I also believe you are supposed to use Kelvin when making those calculations, otherwise, after pressurizing 8 psi over atm @ -200°C the final temp would be -35.8°K...
As an engineer, I know you see what is wrong with that.
Actually in the ideal gas law you can use either K or C (1 degree K = 1 + 273.15 degrees C) in the comparison of two identical states(meaning you aren't crossing the freezing point of water) for the temperature since they both share the same unit of measurement. Meaning a 1 degree raise in C requires the same energy as 1 degree raise in K. If I were to compare two conditions one at 32F and another at above the freezing point of water, say 60F, yes I would need the temp to be in K since 32F is equal to 0C and the equation would no longer work.
To be absolutely correct, yes, you want to use only K in the equation but with due caution either can be used.
You do know that -200C is -328F right? - Are you boosting your J on the dark side of Mercury?
OHV notec wrote:Well, you can see how the heat transfer would be greater from the blower (at temp) to the cooler air (increasing temperature difference will increase rate of heat transfer). You can use that same gas eq to show that the temperature increase would increase the pressure. I doubt if it would be enough to account for a 3-4 psi increase (thermo exam less than 24 hours ago so I don't feel like doing more calcs), but it's a thought.
Your right, given that nature of the blower, "a positive displacement pump" it becomes more efficient as the temp drops looking at it in a pure mass basis. Also at lower temp air, like you said above, the air will heat with a great delta T when at temp which given the proper amount of time for the air to heat up on the back side of the pos. displacement pump(expanding air from from a temperature increase can't escape which leads to building pressure) could be the reason for the added boost at colder temps.
However, a 3psi increase @ 32F intake air would mean the air would need to be heated to 102C or 215F which means this definitely plays a role in the added boost but not the entire reason.
Hmm...perhaps its time for another iteration of the ideal gas law; PV=mRT ==> compare two volumes the equation becomes:
(m1*T1)/P1 = (m2&T2)/P2 - this would involve the added mass of the air that would be present in the intake at different temperatures instead of assuming it constant as done previously.
Joshua Dearman wrote:You do know that -200C is -328F right? - Are you boosting your J on the dark side of Mercury?
lol...I mentioned that because Chris did say the ambiant air temp is below freezing.
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"Youth in Asia"...I don't see anything wrong with that.
