Return-Path: Received: from bewersdorff.com ([192.220.83.90] verified) by logan.com (CommuniGate Pro SMTP 4.1) with SMTP id 2511317 for flyrotary@lancaironline.net; Mon, 04 Aug 2003 17:26:30 -0400 Received: (qmail 23915 invoked by uid 21338); 4 Aug 2003 21:26:29 -0000 Received: from unknown (HELO rapunzel) ([216.101.149.124]) (envelope-sender ) by 192.220.83.90 (qmail-ldap-1.03) with SMTP for ; 4 Aug 2003 21:26:29 -0000 From: "Marko Bewersdorff" To: "Rotary motors in aircraft" Subject: RE: [FlyRotary] Re: Cooling? Date: Mon, 4 Aug 2003 14:26:01 -0700 Message-ID: MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_06A1_01C35A94.546BDAC0" X-Priority: 3 (Normal) X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook IMO, Build 9.0.2416 (9.0.2910.0) X-MIMEOLE: Produced By Microsoft MimeOLE V6.00.2600.0000 In-Reply-To: Importance: Normal This is a multi-part message in MIME format. ------=_NextPart_000_06A1_01C35A94.546BDAC0 Content-Type: text/plain; charset="US-ASCII" Content-Transfer-Encoding: 7bit I just plunked temperature into the equation, no pressure change. As long as your plane doesn't read this email and you keep it in the dark about the theory it should just work fine. I think the only issue might be more drag then needed - theoretically. re Marko -----Original Message----- From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net]On Behalf Of Al Gietzen Sent: Monday, August 04, 2003 10:29 AM To: Rotary motors in aircraft Subject: [FlyRotary] Re: Cooling? doesn't that formula depend on absolute temperature like kelvin or so? where zero degree is minus 459 degree F or minus 273 Celsius. the ratio of expansion should be something like: inlet air: 20 degree Celsius (68F) or 293 Kelvin outlet air: 50 degree celsius (122F) or 323 Kelvin then 323 over 293 for a change in temperature gives then the Volume change of more like 1.2 there is some additional expansion due to the pressure difference. Regards Marko Marko; You got me there. You're right, it needs to be the ratio of absolute temp. That gives a ratio of inlet to outlet for the 80 F in and 135 F out of about 1.1. Also for the case you give; 323/293 is 1.1; not 1.2, or did you factor in some additional volume change for pressure. Gees .. you mean now I have to go back and rework my rad outlets? Al ------=_NextPart_000_06A1_01C35A94.546BDAC0 Content-Type: text/html; charset="US-ASCII" Content-Transfer-Encoding: quoted-printable
I just=20 plunked temperature into the equation, no pressure = change.
 
As=20 long as your plane doesn't read this email and you keep it in the dark = about the=20 theory it should just work fine. I think the only issue might be more = drag then=20 needed - theoretically.
 
re
 
Marko
-----Original Message-----
From: Rotary motors in = aircraft=20 [mailto:flyrotary@lancaironline.net]On Behalf Of Al=20 Gietzen
Sent: Monday, August 04, 2003 10:29 AM
To: = Rotary=20 motors in aircraft
Subject: [FlyRotary] Re: Cooling?=20

 

 

doesn't that formula depend on absolute = temperature=20 like kelvin or so? where zero degree is minus 459 degree F or minus = 273=20 Celsius.
the ratio of expansion should be something like:
inlet = air: 20=20 degree Celsius (68F) or 293 Kelvin
outlet air: 50 degree celsius = (122F) or=20 323 Kelvin
then
323 over 293 for a change in temperature = gives
then=20 the Volume change of more like 1.2

there is some additional expansion due to = the pressure=20 difference.

Regards

Marko

Marko;

You got = me=20 there.  You’re right, it needs to be the ratio of absolute = temp.=20  That gives a ratio of inlet to outlet for the 80 F in and 135 F = out of=20 about 1.1.  Also for the case you give; 323/293 is 1.1; not 1.2, = or did=20 you factor in some additional volume change for pressure. =20

Gees .. = you mean=20 now I have to go back and rework my rad outlets?

Al

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