Return-Path: Sender: (Marvin Kaye) To: lml@lancaironline.net Date: Tue, 01 Jun 2004 20:20:06 -0400 Message-ID: X-Original-Return-Path: Received: from imo-m15.mx.aol.com ([64.12.138.205] verified) by logan.com (CommuniGate Pro SMTP 4.2b3) with ESMTP id 99910 for lml@lancaironline.net; Tue, 01 Jun 2004 19:10:07 -0400 Received: from Sky2high@aol.com by imo-m15.mx.aol.com (mail_out_v37_r2.6.) id q.46.4fbea405 (14374) for ; Tue, 1 Jun 2004 19:09:30 -0400 (EDT) From: Sky2high@aol.com X-Original-Message-ID: <46.4fbea405.2dee66aa@aol.com> X-Original-Date: Tue, 1 Jun 2004 19:09:30 EDT Subject: Re: [LML] Air Intake volume for IO360 B1F X-Original-To: lml@lancaironline.net MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="-----------------------------1086131370" X-Mailer: 9.0 for Windows sub 910 -------------------------------1086131370 Content-Type: text/plain; charset="US-ASCII" Content-Transfer-Encoding: 7bit In a message dated 6/1/2004 5:43:33 PM Central Standard Time, Charles.Nelson@therasense.com writes: Does anyone know the CFM required for the IO360B1F? Charles, Try this, Each cylinder sucks a load of air every second revolution. So two cylinders suck in air every revolution. So, at WOT, 360/2 = 180 CI per revolution. At 2700 RPM, that is 2700 x 180 = 486000 CI/M or 486000/ (12x12x12) = 486000/1728 = 281.25 CFM In other words, a room, 5' x 7' (with an 8 foot ceiling) would contain the air your engine would need for one minute at sea-level take-off power. Also, at sea level, air weighs .0807/CF so we would need 22.7 pounds of air per minute and 1.5 pounds of fuel (approx 15:1 ratio). This works out to about 1/4 gallon per min or 15 gal/hr. What was that question again? Scott Krueger AKA Grayhawk Sky2high@aol.com II-P N92EX IO320 Aurora, IL (KARR) LML, where ideas collide and you decide! -------------------------------1086131370 Content-Type: text/html; charset="US-ASCII" Content-Transfer-Encoding: quoted-printable
In a message dated 6/1/2004 5:43:33 PM Central Standard Time,=20 Charles.Nelson@therasense.com writes:
<= FONT=20 style=3D"BACKGROUND-COLOR: transparent" face=3DArial color=3D#000000 size= =3D2>

Does anyone know the CFM req= uired=20 for the IO360B1F?

Charles,
 
Try this, Each cylinder sucks a load of air every second revolution.&nb= sp;=20 So two cylinders suck in air every revolution.  So, at WOT,  360/2= =3D=20 180 CI per revolution.  At 2700 RPM, that is 2700 x 180 =3D 486000 CI/M= or=20 486000/ (12x12x12) =3D 486000/1728 =3D 281.25 CFM  In other words, a ro= om, 5' x=20 7' (with an 8 foot ceiling) would contain the air your engine would need for= one=20 minute at sea-level take-off power.  Also, at sea level, air weighs=20 .0807/CF so we would need 22.7 pounds of air per minute and 1.5 pounds=20= of=20 fuel (approx 15:1 ratio).  This works out to about 1/4 gallon per min o= r 15=20 gal/hr.
 
What was that question again?=20
 
Scott Krueger=20 AKA Grayhawk
Sky2high@aol.com
II-P N92EX IO320 Aurora, IL=20 (KARR)

LML, where ideas collide and you=20 decide!
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