For every 1 PSI = to how much HP/Tq increase?
11-06-2007, 11:06 AM
#11
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There are way too many variables to this equation- Alot of it has to do with the efficiency of the intake ports/cam when under boost and parasitic loss of the FI device. It is not uncommon to see much higher than 7% per psi with a turbo. There is a guy at theturboforums putting over 620hp to the ground with a stock Explorer longblock (220hp at the crank stock) at 16psi with a MP T70.
I theory, you can only make 7% per psi, but that is assuming that the ports of said engine are flowing linear with boost, but that doesnt happen in real life. In real life, FI can overcome "bad" port design and actally make them more efficient, throwing that linear 7% cure off in the favor of more power.
I theory, you can only make 7% per psi, but that is assuming that the ports of said engine are flowing linear with boost, but that doesnt happen in real life. In real life, FI can overcome "bad" port design and actally make them more efficient, throwing that linear 7% cure off in the favor of more power.
11-06-2007, 11:55 AM
#13
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11-06-2007, 12:55 PM
#14
It also provides diminishing returns as boost gets higher. Aside from having to keep your IAT's the same, you won't see a linear 6-7% increase in power as boost goes up.
Take for example an engine that makes 400hp N/A. Add 14.7psi of boost and you get a theoretical 100% power increase. So you're now at 800hp, which is believable. If you add another 14.7psi of boost it probably won't net you another 100% power increase, because you'd then be making 1600hp which seems a bit high. I don't think it's an exponential curve necessarily, but it's just not one linear 6.8%/psi slope.
Take for example an engine that makes 400hp N/A. Add 14.7psi of boost and you get a theoretical 100% power increase. So you're now at 800hp, which is believable. If you add another 14.7psi of boost it probably won't net you another 100% power increase, because you'd then be making 1600hp which seems a bit high. I don't think it's an exponential curve necessarily, but it's just not one linear 6.8%/psi slope.
It isn't a doubling of density and power each atmosphere it's an addition of the base (e.g. not 100% increase at each atmosphere, but +100% of NA for each atmosphere). This comes from Boyle's Law as I recall.
Try it again this way:
NA = 400hp
14.7psi of boost = 800 hp (believable)
29.4psi of boost = 1200 hp (believable imo - if not a little low, but that may be because to get to 29.4 you need to add meth and/or racing fuel which changes the base power)
300%/29.4 psi = 10%/1psi (which is close anyway to the theoretical yield).
11-06-2007, 01:18 PM
#16
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Their is just too much involved in it.
Like stated X number of psi will result in Y number of HP and TQ.
this is based on A the NA numbers of the motor your working with.
the factors of X, Y, and A can be changed by,
B type of camshaft
C type of intercooler
D type of boost system
F type of fuel and the amount delivered.
G type of heads and port flow.
H amount of original compression before boost.
The start of 6 to 7% per pound of boost is a good place to start.
now add to that all the tweaking you can do to raise the percentage
To 8% and on to 9% and so on.
a centrifuge does not make the same percentage rate as a turbo
The stock cam can be used but the after-market cams are built to meet your needs.
stock compression is fine but you can run more boost if you lower it by 1
and the efficiency of the intercooler, air to air, air to water, or a chemical intercooler that adds to the fuel.
It's just too much to play with to figure out where you should be, only to strap it down on a dyno with a jaw dropping WTF.
the best thing to do is take what you know and build the best system possible.
Like stated X number of psi will result in Y number of HP and TQ.
this is based on A the NA numbers of the motor your working with.
the factors of X, Y, and A can be changed by,
B type of camshaft
C type of intercooler
D type of boost system
F type of fuel and the amount delivered.
G type of heads and port flow.
H amount of original compression before boost.
The start of 6 to 7% per pound of boost is a good place to start.
now add to that all the tweaking you can do to raise the percentage
To 8% and on to 9% and so on.
a centrifuge does not make the same percentage rate as a turbo
The stock cam can be used but the after-market cams are built to meet your needs.
stock compression is fine but you can run more boost if you lower it by 1
and the efficiency of the intercooler, air to air, air to water, or a chemical intercooler that adds to the fuel.
It's just too much to play with to figure out where you should be, only to strap it down on a dyno with a jaw dropping WTF.
the best thing to do is take what you know and build the best system possible.
11-06-2007, 01:55 PM
#18
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But this doesn't work anyway.
It isn't a doubling of density and power each atmosphere it's an addition of the base (e.g. not 100% increase at each atmosphere, but +100% of NA for each atmosphere). This comes from Boyle's Law as I recall.
Try it again this way:
NA = 400hp
14.7psi of boost = 800 hp (believable)
29.4psi of boost = 1200 hp (believable imo - if not a little low, but that may be because to get to 29.4 you need to add meth and/or racing fuel which changes the base power)
300%/29.4 psi = 10%/1psi (which is close anyway to the theoretical yield).
It isn't a doubling of density and power each atmosphere it's an addition of the base (e.g. not 100% increase at each atmosphere, but +100% of NA for each atmosphere). This comes from Boyle's Law as I recall.
Try it again this way:
NA = 400hp
14.7psi of boost = 800 hp (believable)
29.4psi of boost = 1200 hp (believable imo - if not a little low, but that may be because to get to 29.4 you need to add meth and/or racing fuel which changes the base power)
300%/29.4 psi = 10%/1psi (which is close anyway to the theoretical yield).
Many people feel it works like this:
X+(X*PSI*6.8/100)= Power at a given boost with X being the starting power number. Lets say X=400, so at 14.7psi we have 800hp. People will then assume: 800+(800*14.7*6.8/100)= new power at 29.4psi when it should really be 400+(400*29.4*6.8/100), because you should always work with the base power figure.
11-06-2007, 02:00 PM
#19
Right, that's basically the second half of what I was trying to say. I've talked to a lot of people that think that 7% keeps compounding on the previous power number. That's all I was trying to say I gues. I can explain better with algebra.
Many people feel it works like this:
X+(X*PSI*6.8/100)= Power at a given boost with X being the starting power number. Lets say X=400, so at 14.7psi we have 800hp. People will then assume: 800+(800*14.7*6.8/100)= new power at 29.4psi when it should really be 400+(400*29.4*6.8/100), because you should always work with the base power figure.
Many people feel it works like this:
X+(X*PSI*6.8/100)= Power at a given boost with X being the starting power number. Lets say X=400, so at 14.7psi we have 800hp. People will then assume: 800+(800*14.7*6.8/100)= new power at 29.4psi when it should really be 400+(400*29.4*6.8/100), because you should always work with the base power figure.
