Return-Path: Received: from [24.25.9.102] (HELO ms-smtp-03-eri0.southeast.rr.com) by logan.com (CommuniGate Pro SMTP 4.1.8) with ESMTP id 2883884 for flyrotary@lancaironline.net; Tue, 09 Dec 2003 17:20:52 -0500 Received: from o7y6b5 (clt78-020.carolina.rr.com [24.93.78.20]) by ms-smtp-03-eri0.southeast.rr.com (8.12.10/8.12.7) with SMTP id hB9MKlfP011801 for ; Tue, 9 Dec 2003 17:20:49 -0500 (EST) Message-ID: <003b01c3bea2$2fa619e0$1702a8c0@WorkGroup> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Air Density at altitude Date: Tue, 9 Dec 2003 17:17:07 -0500 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0038_01C3BE78.462FDA40" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2800.1106 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2800.1106 X-Virus-Scanned: Symantec AntiVirus Scan Engine This is a multi-part message in MIME format. ------=_NextPart_000_0038_01C3BE78.462FDA40 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable =20 Ed Anderson wrote: Heat transfer equation Q =3DW*DeltaT*Cp, with W =3D mass flow down by = 30%. So to get rid of the same Q of heat (and since Cp doesn't change = that much) it would appear that means the delta T term would need to = increase by 30% for Q quantity to remain the same. But, I don't know = exactly how a 79Deg colder incoming air would affect the Delta T term. BIll?? Ed, the equation above represents the heat *picked up* by the air, and = is directly proportional to the Delta T that you can achieve in the air = stream. So, if mass flow is down 30%, for equal heat rejection, you = would need the air to change temperature by an offsetting amount. Easier = to use actual numbers rather than %.=20 For example, if W =3D 100, DeltaT =3D 30 F, and Cp =3D 0.25, then Q ~ = 750. Reduce W to 70, and to get Q ~ 750 you would need a deltaT of 42.8 F. = There fore lowering the inlet temperature can increase the temperature = carrying capability of the air flowing through the radiator. However, there is another complicating factor, and that is the = transfer of the heat from the water to the air. Here you need to use the = log-mean delta T, to see if you can get the heat *into* the air. =20 Thanks, Bill Its nice to have someone with some knowledge of thermodynamics to = keep me straight. K&W also points out the log curve in the temperature gradient in the = core from the walls to the center of the air passages through the core. ED ------=_NextPart_000_0038_01C3BE78.462FDA40 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
 
Ed Anderson wrote:

Heat transfer equation Q = =3DW*DeltaT*Cp, with W =3D=20 mass flow down by 30%. So to get rid of the same Q of heat (and since = Cp=20 doesn't change that much) it would appear that means the delta T term = would=20 need to increase by 30% for Q quantity to remain the same. But, I = don't know=20 exactly how a 79Deg colder incoming air would affect the Delta T=20 term.

BIll??

Ed, the equation above represents the = heat=20 *picked up* by the air, and is directly proportional to the Delta T = that you=20 can achieve in the air stream. So, if mass flow is down 30%, for equal = heat=20 rejection, you would need the air to change temperature by an = offsetting=20 amount. Easier to use actual numbers rather than %.

For example, if W =3D 100, DeltaT =3D = 30 F, and Cp =3D=20 0.25, then Q ~ 750.

Reduce W to 70, and to get Q ~ 750 = you would need=20 a deltaT of 42.8 F. There fore lowering the inlet temperature can = increase the=20 temperature carrying capability of the air flowing through the=20 radiator.

However, there is another = complicating factor,=20 and that is the transfer of the heat from the water to the air. Here = you need=20 to use the log-mean delta T, to see if you can get the heat *into* the = air.
 
 
Thanks, Bill
 
    Its nice to have = someone with=20 some knowledge of thermodynamics to keep me straight.
K&W also points out the log curve = in the=20 temperature gradient in the core from the walls to the center of the = air=20 passages through the core.
 
ED
 
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