How much to convert to VXR 500
Discussion
I've absolutely no doubt it would be safe, but I just wanted to avoid doing the trip twice, (getting lifts etc). Tho' I do like the idea of coming home to the s/c car, kind of reminds me of picking up my new VXR6 on the day after I returned from holls two years ago...........Also, I'm guessing the second trip wouldn't take quite as long as the first! (LOL!).
Compressed air = increased temperature.
You may recall Boyles law and Ideal gas laws from school chemistry!
http://en.wikipedia.org/wiki/Ideal_ga
Its impossible to have forced induction without increased heat. IMO an intercooler is always preferred.
You may recall Boyles law and Ideal gas laws from school chemistry!
http://en.wikipedia.org/wiki/Ideal_ga
Its impossible to have forced induction without increased heat. IMO an intercooler is always preferred.
skyblue465 said:
What's the turnaround on the s/c conversion? I'd like to have mine done in about 6-7 wks when I get back from my snowboarding holls.
I was hoping to have it done within a day? cos it's a bit of a trek to Greens for me. Do we have to lose the stut brace? or is it not in the way?
If we had it very first thing, we could lend you a car to go and explore for the day, and then it should be ready last thing, as long as its just an s/c. Anything additional would take two min.
greens vauxhall said:
skyblue465 said:
What's the turnaround on the s/c conversion? I'd like to have mine done in about 6-7 wks when I get back from my snowboarding holls.
I was hoping to have it done within a day? cos it's a bit of a trek to Greens for me. Do we have to lose the stut brace? or is it not in the way?
If we had it very first thing, we could lend you a car to go and explore for the day, and then it should be ready last thing, as long as its just an s/c. Anything additional would take two min.
That's exactly the answer I was hoping for, the only addition may be bunging in the my straight through center section, don't 'spect that would take long tho'. I'll call soon to make arangements, who should I ask for?
ringram said:
Compressed air = increased temperature.
You may recall Boyles law and Ideal gas laws from school chemistry!
http://en.wikipedia.org/wiki/Ideal_ga
Its impossible to have forced induction without increased heat. IMO an intercooler is always preferred.
You may recall Boyles law and Ideal gas laws from school chemistry!
http://en.wikipedia.org/wiki/Ideal_ga
Its impossible to have forced induction without increased heat. IMO an intercooler is always preferred.
I was under the impression that the efficiency of the compressor was directly related to heat generated.
100% efficient = no heat, the poorer the efficiency, the more heat is generated, if the compressor is of a more efficient design, then it would generate less heat?
Gelf VXR said:
ringram said:
Compressed air = increased temperature.
You may recall Boyles law and Ideal gas laws from school chemistry!
http://en.wikipedia.org/wiki/Ideal_ga
Its impossible to have forced induction without increased heat. IMO an intercooler is always preferred.
You may recall Boyles law and Ideal gas laws from school chemistry!
http://en.wikipedia.org/wiki/Ideal_ga
Its impossible to have forced induction without increased heat. IMO an intercooler is always preferred.
I was under the impression that the efficiency of the compressor was directly related to heat generated.
100% efficient = no heat, the poorer the efficiency, the more heat is generated, if the compressor is of a more efficient design, then it would generate less heat?
Correct!
Interesting discussion, I have to agree with Ringram.
Surely the efficiency of the compressor is a measure of the volume of air at a given pressure that can be created for a given amount of horsepower in. Any time you take a volume of gas and compress it into a smaller space, hence increase its density and pressure, there is a finite mathematical equation defining the increase in temperature that will occur. This is basic physics and surely non - negotiable? As I understand it (which may be wrong!) using a charge cooler or intercooler will increase the engine's output at a given boost by effectively increasing the density of the charge. So not using a cooler would require higher boost pressure to get the same power?
Surely the efficiency of the compressor is a measure of the volume of air at a given pressure that can be created for a given amount of horsepower in. Any time you take a volume of gas and compress it into a smaller space, hence increase its density and pressure, there is a finite mathematical equation defining the increase in temperature that will occur. This is basic physics and surely non - negotiable? As I understand it (which may be wrong!) using a charge cooler or intercooler will increase the engine's output at a given boost by effectively increasing the density of the charge. So not using a cooler would require higher boost pressure to get the same power?
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