Return-Path: Received: from fed1rmmtao04.cox.net ([68.230.241.35] verified) by logan.com (CommuniGate Pro SMTP 4.2) with ESMTP id 363435 for flyrotary@lancaironline.net; Fri, 13 Aug 2004 08:45:39 -0400 Received-SPF: none receiver=logan.com; client-ip=68.230.241.35; envelope-from=daveleonard@cox.net Received: from davidandanne ([68.111.224.107]) by fed1rmmtao04.cox.net (InterMail vM.6.01.03.02.01 201-2131-111-104-103-20040709) with SMTP id <20040813124507.DRXQ21610.fed1rmmtao04.cox.net@davidandanne> for ; Fri, 13 Aug 2004 08:45:07 -0400 From: "DaveLeonard" To: "Rotary motors in aircraft" Subject: RE: [FlyRotary] Re: DeltaT Coolant was : [FlyRotary] Re: coolant temps Date: Fri, 13 Aug 2004 05:45:08 -0700 Message-ID: MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_001A_01C480F8.B1312810" 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 V6.00.2800.1441 In-Reply-To: Importance: Normal This is a multi-part message in MIME format. ------=_NextPart_000_001A_01C480F8.B1312810 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: 7bit Mark, Ed took care of that right in the second line "So with Tracy's 30 gpm flow of water = 240 lbs/min" Dave Leonard Ed, Please humor me (a non-engineer) while I ask a dumb question. If it takes 1BTU to raise 1lb of water 1 degree, and you factor in 30 gpm flow to come up with a 2400 BTU requirement for a 10 degree rise for 1 lb of water, where does the number of pounds of water figure into the equation, or do we just ignore that issue? Water is 8.34 lbs/gal, and say you have 2 gallons of coolant, that would be 16.68 lbs. Seems that we would need to multiply the 2400 figure by 16.68 to arrive at a total system requirement of 40,032 BTU/min. What am I missing here? Mark S. At 09:58 PM 8/12/2004 -0400, you wrote: Right you are, Dave Below is one semi-official definition of BTU in English units. 1 BTU is amount of heat to raise 1 lb of water 1 degree Fahrenheit. So with Tracy's 30 gpm flow of water = 240 lbs/min. Since its temperature is raised 10 degree F we have BTU = 240 * 10 * 1 = 2400 BTU/min I know I'm ancient and I should move into the new metric world, but at least I didn't do it in Stones and Furlongs {:>) ------=_NextPart_000_001A_01C480F8.B1312810 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable

Mark, = Ed took care=20 of that right in the second line "So with = Tracy's 30 gpm=20 flow of water =3D 240 lbs/min"

Dave=20 Leonard

Ed,
Please=20 humor me (a non-engineer) while I ask a dumb question. If it = takes 1BTU=20 to raise 1lb of water 1 degree, and you factor in 30 gpm flow to come = up with=20 a 2400 BTU requirement for a 10 degree rise for 1 lb of water, where = does the=20 number of pounds of water figure into the equation, or do we just = ignore that=20 issue? Water is 8.34 lbs/gal, and say you have 2 gallons of = coolant,=20 that would be 16.68 lbs. Seems that we would need to multiply = the 2400=20 figure by 16.68 to arrive at a total system requirement of 40,032=20 BTU/min. What am I missing here?

Mark=20 S.

At 09:58 PM 8/12/2004 -0400, = you=20 wrote:

Right you are,=20 Dave

Below is one = semi-official=20 definition of BTU in English units. 1 BTU is amount of heat to = raise 1=20 lb of water 1 degree Fahrenheit. =

So with Tracy's 30 gpm flow of water =3D 240 = lbs/min. Since=20 its temperature is raised 10 degree F we = have

BTU =3D 240 * 10 * 1 =3D 2400 = BTU/min

I know I'm ancient and I should move into the new = metric=20 world, but at least I didn't do it in Stones and Furlongs=20 {:>)

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