Supercharger Question
#23
05-26-2009, 01:17 AM
#26
05-26-2009, 08:30 AM
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I am not sure where you picked up the pressurized air is cooler thing... but that source needs to be double checked man...
The reason why boosted engines run intercoolers and alky injection are to drop the intake charge temp. thus creating a denser fuel to air mixture with better combustion properties.
This is also partly the reason why turbo chargers and superchargers get extremely hot and drastically increase under hood temps...
but I can see where you would see it the other way also. take for instance, letting pressurized air out of a compressor... its damn cold.
Not beating on ya man, just that the heat thing is pretty common knowledge among the boosted boys...
The reason why boosted engines run intercoolers and alky injection are to drop the intake charge temp. thus creating a denser fuel to air mixture with better combustion properties.
This is also partly the reason why turbo chargers and superchargers get extremely hot and drastically increase under hood temps...
but I can see where you would see it the other way also. take for instance, letting pressurized air out of a compressor... its damn cold.
Not beating on ya man, just that the heat thing is pretty common knowledge among the boosted boys...
#27
05-26-2009, 10:19 AM
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#28
05-26-2009, 11:42 AM
#29
05-26-2009, 12:38 PM
It's really quite simple, Gay-Lussac's Law: P1/T1 = P2/T2.
P being pressure and T being temperature. For simplicity's sake we'll use some arbitrary numbers rather than something more realistic. Let's say P1 is 6 and T1 is 2 so we have:
6/2 = P2/T2 -> 3 = P2/T2
Here if we raise the pressure so P2 is 12 we get:
3 = 12/T2
Basic arithmetic, here T2 must equal 4 (12/4 = 3). So as pressure rises temperature does too. Going the opposite way, let's leave P1 and T1 the same and make P2 = 3:
3 = 3/T2
T2 has to be 1, as pressure drops so does temperature. You can plug in numbers for T2 and solve for P2 and you'll find it works the other way as well, it's a direct relationship.
When you let air out of a compressor you are lowering the pressure, that's why it feels cold.
Hopefully this helps if anyone's still confused.
P being pressure and T being temperature. For simplicity's sake we'll use some arbitrary numbers rather than something more realistic. Let's say P1 is 6 and T1 is 2 so we have:
6/2 = P2/T2 -> 3 = P2/T2
Here if we raise the pressure so P2 is 12 we get:
3 = 12/T2
Basic arithmetic, here T2 must equal 4 (12/4 = 3). So as pressure rises temperature does too. Going the opposite way, let's leave P1 and T1 the same and make P2 = 3:
3 = 3/T2
T2 has to be 1, as pressure drops so does temperature. You can plug in numbers for T2 and solve for P2 and you'll find it works the other way as well, it's a direct relationship.
When you let air out of a compressor you are lowering the pressure, that's why it feels cold.
Hopefully this helps if anyone's still confused.
#30
05-26-2009, 01:19 PM
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