Mailing List lml@lancaironline.net Message #15502
From: Lorn H. Olsen <lorn@dynacomm.ws>
Sender: Marvin Kaye <marv@lancaironline.net>
Subject: [LML] Re: RAM-air induction
Date: Wed, 25 Sep 2002 12:17:21 -0400
To: <lml>
Villi Seemann <villi.seemann@nordea.com>:
.
.
If we have a RAM air opening of 3 inch diameter, it has an area of appx 7
sq.in ( 1/2  *  3^2  *  PI ) . This means, if we did not have any inlet
losses, back pressure etc, the RAM air would enter our plenum chamber with
.
.
Regards
Villi H. Seemann
Senior Engineer
Infrastructure Network
Phone  (+45) 3333 2101
FAX      (+45) 3333 1130
CellPhn (+45)2220 7690

Villa,

Thank you for the very clear and understandable post. However, I believe that you have a problem in that the area listed above sould be: PI * r^2. With a 3 inch diameter opening, the wrong calculation leads to an answer 2 times as large as it should be.

eg.      1/2  *  3^2  *  PI = 14.14
   and   PI * 1.5^2         =  7.07
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