Return-Path: Received: from [24.25.9.103] (HELO ms-smtp-04-eri0.southeast.rr.com) by logan.com (CommuniGate Pro SMTP 4.2) with ESMTP id 363143 for flyrotary@lancaironline.net; Thu, 12 Aug 2004 21:58:51 -0400 Received-SPF: none receiver=logan.com; client-ip=24.25.9.103; envelope-from=eanderson@carolina.rr.com Received: from EDWARD (cpe-069-132-183-211.carolina.rr.com [69.132.183.211]) by ms-smtp-04-eri0.southeast.rr.com (8.12.10/8.12.7) with SMTP id i7D1wHSH000566 for ; Thu, 12 Aug 2004 21:58:19 -0400 (EDT) Message-ID: <001501c480d9$03991880$2402a8c0@EDWARD> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: DeltaT Coolant was : [FlyRotary] Re: coolant temps Date: Thu, 12 Aug 2004 21:58:22 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0012_01C480B7.7C3D9E40" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2800.1409 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2800.1409 X-Virus-Scanned: Symantec AntiVirus Scan Engine This is a multi-part message in MIME format. ------=_NextPart_000_0012_01C480B7.7C3D9E40 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable 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. =20 So with Tracy's 30 gpm flow of water =3D 240 lbs/min. Since 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 world, but at = least I didn't do it in Stones and Furlongs {:>) Ed The Columbia Encyclopedia, Sixth Edition. 2001.=20 =20 British thermal unit=20 =20 =20 abbr. Btu, unit for measuring heat quantity in the customary = system of English units of measurement, equal to the amount of heat = required to raise the temperature of one pound of water at its maximum = density [which occurs at a temperature of 39.1 degrees Fahrenheit (=B0F) = ] by 1=B0F. The Btu may also be defined for the temperature difference = between 59=B0F and 60=B0F. One Btu is approximately equivalent to the = following: 251.9 calories; 778.26 foot-pounds; 1055 joules; 107.5 = kilogram-meters; 0.0002928 kilowatt-hours. A pound (0.454 kilogram) of = good coal when burned should yield 14,000 to 15,000 Btu; a pound of = gasoline or other . =20 Ed Anderson RV-6A N494BW Rotary Powered Matthews, NC ----- Original Message -----=20 From: DaveLeonard=20 To: Rotary motors in aircraft=20 Sent: Thursday, August 12, 2004 8:12 PM Subject: [FlyRotary] Re: DeltaT Coolant was : [FlyRotary] Re: coolant = temps 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=20 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 =3D 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=20 Q =3D BTU/min of heat removed by coolant mass flow Assuming 30 GPM coolant flow =3D 30*8 =3D 240 lb/min mass flow. = specific heat of water Cp =3D 1.0 Solving for DeltaT =3D Q/(W*Cp) =3D 2400/(240*1) =3D 2400/240 =3D = 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) =3D 2400/168 =3D 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_0012_01C480B7.7C3D9E40 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Right you are, Dave
 
Below  is one semi-official definition = of BTU in=20 English units.  1 BTU is amount of heat to raise 1 lb of water 1 = degree=20 Fahrenheit.  
 
So with Tracy's 30 gpm flow of water =3D 240 = lbs/min. =20 Since 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=20 new metric world, but at least I didn't do it in Stones and Furlongs=20 {:>)
 
Ed
 
The Columbia Encyclopedia, Sixth=20 Edition.  2001.
 
British = thermal=20 unit
 
 
abbr. Btu, unit for measuring heat quantity in the customary = system=20 of English = units of=20 measurement, equal to the amount of heat required to raise the = temperature of one pound of water at its maximum density [which = occurs at=20 a temperature of 39.1 degrees Fahrenheit (=B0F) ] by 1=B0F. = The Btu may=20 also be defined for the temperature difference between 59=B0F and = 60=B0F. One=20 Btu is approximately equivalent to the following: 251.9 calories; = 778.26=20 foot-pounds; 1055 joules; 107.5 kilogram-meters; 0.0002928 = kilowatt-hours.=20 A pound (0.454 kilogram) of good coal when burned should yield = 14,000 to=20 15,000 Btu; a pound of gasoline or other .
 
 
 
 
 
 
 
Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, NC
----- Original Message -----
From:=20 DaveLeonard=20
Sent: Thursday, August 12, 2004 = 8:12=20 PM
Subject: [FlyRotary] Re: DeltaT = Coolant=20 was : [FlyRotary] Re: coolant temps

Ed, are those units right.  I know that = the=20 specific heat of water is 1.0 cal/(deg = Celsius*gram).  Does=20 that 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=20 to get rid of 2369 BTU/Min through your coolant/radiators.  I = rounded=20 it off 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=20 2400/(240*1)  =3D  2400/240 =3D 10 or  your delta T = for the=20 parameters specified should be around 10F
 
Assuming a 50/50 coolant mix with a Cp  = of 0.7=20 you would have approx 2400/(240 *0.7) =3D 2400/168 =3D 14.2F so I = would say you=20 do not 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=20 Powered
Matthews, NC
 
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