Return-Path: Received: from fed1mtao03.cox.net ([68.6.19.242] verified) by logan.com (CommuniGate Pro SMTP 4.1.8) with ESMTP id 2883654 for flyrotary@lancaironline.net; Tue, 09 Dec 2003 13:52:42 -0500 Received: from BigAl ([68.107.116.221]) by fed1mtao03.cox.net (InterMail vM.5.01.06.05 201-253-122-130-105-20030824) with ESMTP id <20031209185241.XRJL28419.fed1mtao03.cox.net@BigAl> for ; Tue, 9 Dec 2003 13:52:41 -0500 From: "Al Gietzen" To: "'Rotary motors in aircraft'" Subject: RE: [FlyRotary] Re: Air Density at altitude Date: Tue, 9 Dec 2003 10:52:53 -0800 Message-ID: <000e01c3be85$a72937d0$6400a8c0@BigAl> MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_000F_01C3BE42.9905F7D0" X-Priority: 3 (Normal) X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook, Build 10.0.4024 Importance: Normal In-Reply-To: X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2800.1165 This is a multi-part message in MIME format. ------=_NextPart_000_000F_01C3BE42.9905F7D0 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: 7bit > Heat transfer equation Q =W*DeltaT*Cp, with W = 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 Anderson > Wouldn't the percentage change in DeltaT (DeltaDeltaT?) have to be measured from absolute 0? An 80degree change would then represent...what?... about 20%? No; the heat transfer is directly proportional (roughly) to the temp drop from rad surface to airstream, all else being the same. So it you have 80F air and 180F rad at low altitude (100 delta), then 50F air would give you about 30% more heat rejection; ignoring the reduced conductivity of the lower pressure air. Most of the temp drop is from the wall to the free steam air temp. Al ------=_NextPart_000_000F_01C3BE42.9905F7D0 Content-Type: text/html; charset="us-ascii" Content-Transfer-Encoding: quoted-printable

> 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 Anderson

Wouldn't the percentage change in DeltaT (DeltaDeltaT?) have to be measured from absolute 0? An 80degree change would then represent...what?... = about 20%?

No; the heat transfer is directly proportional = (roughly) to the temp drop from rad surface to airstream, all else being the same. = So it you have 80F air and 180F rad at low altitude (100 delta), then 50F = air would give you about 30% more heat rejection; ignoring the reduced = conductivity of the lower pressure air. Most of the temp drop is from the wall = to the free steam air temp.

<= /font>

------=_NextPart_000_000F_01C3BE42.9905F7D0--