Forced Induction Supercharger/Turbocharger

Understanding flow characteristics

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Old Jul 22, 2004 | 09:18 PM
  #1  
Rodrigues's Avatar
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Understanding flow characteristics

I have an intercooler with the spearco 2-230 cores http://www.turboneticsinc.com/liquid.html, which spearco rates to 1500cfm. Now lets say I get a T88, which flows aprox 1700cfm... what happens? Seems like a lot of people are getting away with tiny intercoolers... heres a pic of mine.


http://kak.com/camaro/intercooler/DSCN0741.JPG
http://kak.com/camaro/intercooler/DSCN0740.JPG

Last edited by Rodrigues; Jul 22, 2004 at 09:22 PM.
Old Jul 22, 2004 | 09:45 PM
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Re: Understanding flow characteristics

The intercooler will explode since there is nowhere for the extra 200 cfm to go. Just kidding.

I'm not sure if they're rating it based on restriction or cooling capacity. Either way, that's way bigger than the Vortech Aftercooler and people run 9's with that.

Mike
Old Jul 22, 2004 | 10:30 PM
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Re: Understanding flow characteristics

Think its possible to push over 1,200 at the crank through an intercooler like this?

Originally Posted by engineermike
The intercooler will explode since there is nowhere for the extra 200 cfm to go. Just kidding.

I'm not sure if they're rating it based on restriction or cooling capacity. Either way, that's way bigger than the Vortech Aftercooler and people run 9's with that.

Mike
Old Jul 22, 2004 | 10:36 PM
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Re: Understanding flow characteristics

1500cfm is measured at a pressure drop (1.5 psi i believe). so if you try to flow more through it, you will get a greater pressure drop. 1500cfm is a nice size IC.
Old Jul 22, 2004 | 10:44 PM
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Re: Understanding flow characteristics

Is the pressure drop usually pretty linear vs initial boost?
Old Jul 22, 2004 | 10:52 PM
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Re: Understanding flow characteristics

Pressure drop is proportional to the square of flow. So, if you plan on flowing a TRUE 1700 cfm, then you could find the pressure drop by:

dP = 1.5 * (1700/1500)^2 = 1.93 psi
Old Jul 22, 2004 | 11:53 PM
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Re: Understanding flow characteristics

According to the graphs of the t88 its able to push a peak of 120 lb min, so 120/.069= 1739 peak. Now lets assume a 2 psi drop. So the more pressure i push through, a larger percent is being lost to pressure drop. So even if I push exactly 1500cfm, im still losing 1.5 psi.... not too shabby only losing a half psi or so for 200 more cfm imo.
Originally Posted by engineermike
Pressure drop is proportional to the square of flow. So, if you plan on flowing a TRUE 1700 cfm, then you could find the pressure drop by:

dP = 1.5 * (1700/1500)^2 = 1.93 psi
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