Return-Path: Received: from fed1rmmtao02.cox.net ([68.230.241.37] verified) by logan.com (CommuniGate Pro SMTP 4.2) with ESMTP id 363075 for flyrotary@lancaironline.net; Thu, 12 Aug 2004 20:13:09 -0400 Received-SPF: none receiver=logan.com; client-ip=68.230.241.37; envelope-from=daveleonard@cox.net Received: from davidandanne ([68.111.224.107]) by fed1rmmtao02.cox.net (InterMail vM.6.01.03.02.01 201-2131-111-104-103-20040709) with SMTP id <20040813001237.CMBZ14393.fed1rmmtao02.cox.net@davidandanne> for ; Thu, 12 Aug 2004 20:12:37 -0400 From: "DaveLeonard" To: "Rotary motors in aircraft" Subject: RE: [FlyRotary] DeltaT Coolant was : [FlyRotary] Re: coolant temps Date: Thu, 12 Aug 2004 17:12:38 -0700 Message-ID: MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_004F_01C4808F.91CF0A00" X-Priority: 3 (Normal) X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook IMO, Build 9.0.2416 (9.0.2910.0) In-Reply-To: X-MIMEOLE: Produced By Microsoft MimeOLE V6.00.2800.1441 Importance: Normal This is a multi-part message in MIME format. ------=_NextPart_000_004F_01C4808F.91CF0A00 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: 7bit Ed, are those units right. I know that the specific heat of water is 1.0 cal/(deg Celsius*gram). Does that also work out to 1.0 BTU/(deg. Farhengight * Lb.) ? Dave Leonard Tracy my calculations shows your coolant temp drop is where it should be: My calculations show that at 7 gph fuel burn you need to get rid of 2369 BTU/Min through your coolant/radiators. I rounded it off to 2400 BTU/min. Q = W*DeltaT*Cp Basic Heat/Mass Flow equation With water as the mass with a weight of 8 lbs/ gallon and a specific heat of 1.0 Q = BTU/min of heat removed by coolant mass flow Assuming 30 GPM coolant flow = 30*8 = 240 lb/min mass flow. specific heat of water Cp = 1.0 Solving for DeltaT = Q/(W*Cp) = 2400/(240*1) = 2400/240 = 10 or your delta T for the parameters specified should be around 10F Assuming a 50/50 coolant mix with a Cp of 0.7 you would have approx 2400/(240 *0.7) = 2400/168 = 14.2F so I would say you do not fly with a 50/50 coolant mix but something closer to pure water. But in any case, certainly in the ball park. You reported 10-12F under those conditions, so I would say condition is 4. Normal operation Ed Ed Anderson RV-6A N494BW Rotary Powered Matthews, NC ------=_NextPart_000_004F_01C4808F.91CF0A00 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable

Ed, are those units right. I know that = the=20 specific heat of water is 1.0 cal/(deg = Celsius*gram). Does that=20 also work out to 1.0 BTU/(deg. Farhengight * Lb.) ?

Dave = Leonard

Tracy my calculations shows your coolant temp = drop is=20 where it should be:

My calculations show that at 7 gph fuel burn = you need to=20 get rid of 2369 BTU/Min through your coolant/radiators. I = rounded it off=20 to 2400 BTU/min.

Q =3D W*DeltaT*Cp Basic Heat/Mass = Flow=20 equation With water as the mass with a weight of 8 lbs/ gallon = and a=20 specific heat of 1.0

Q =3D BTU/min of heat removed by coolant mass=20 flow

Assuming 30 GPM coolant flow =3D 30*8 = =3D 240 lb/min=20 mass flow. specific heat of water Cp =3D 1.0

Solving for DeltaT =3D Q/(W*Cp) =3D = 2400/(240*1) =20 =3D 2400/240 =3D 10 or your delta T for the parameters = specified=20 should be around 10F

Assuming a 50/50 coolant mix with a Cp = of 0.7 you=20 would have approx 2400/(240 *0.7) =3D 2400/168 =3D 14.2F so I would = say you do not=20 fly with

a 50/50 coolant mix but something closer = to pure=20 water. But in any case, certainly in the ball park.

You reported 10-12F under those conditions, so = I would=20 say condition is 4. Normal operation

Ed

Ed Anderson
RV-6A N494BW Rotary = Powered
Matthews,=20 NC

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