X-Virus-Scanned: clean according to Sophos on Logan.com X-SpamCatcher-Score: 2 [X] Return-Path: Received: from ms-smtp-05.southeast.rr.com ([24.25.9.104] verified) by logan.com (CommuniGate Pro SMTP 5.1.9) with ESMTP id 2061529 for flyrotary@lancaironline.net; Tue, 22 May 2007 13:27:04 -0400 Received-SPF: pass receiver=logan.com; client-ip=24.25.9.104; envelope-from=eanderson@carolina.rr.com Received: from edward2 (cpe-024-074-103-061.carolina.res.rr.com [24.74.103.61]) by ms-smtp-05.southeast.rr.com (8.13.6/8.13.6) with SMTP id l4MHQArW006647 for ; Tue, 22 May 2007 13:26:10 -0400 (EDT) Message-ID: <000b01c79c96$8d4c3e00$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" Subject: Fw: Gear box oil temps Date: Tue, 22 May 2007 13:27:32 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0006_01C79C74.F405AAE0" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.3028 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3028 X-Virus-Scanned: Symantec AntiVirus Scan Engine This is a multi-part message in MIME format. ------=_NextPart_000_0006_01C79C74.F405AAE0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Figures I found on the web indicates 3% is a fairly standard figure used = for a planetary gearbox efficiency. So if that figure is close and you are producing in cruise - say 160 HP- = then the gear box will produce heat with about 3% of the input power. = So 3% of 160 HP =3D 4.8 HP which is mostly converted to heat. I assume the oil is already cooled by the oil cooler before it is fed = to the PSRU. So the temp of the oil going into the gear box might be = around 180F. I am not certain how many GPM of oil Tracy's PSRU flows = but the old Ross flowed around 1 to 1 1/2 pint/minute, but that was = apparently rather marginal. So assume Tracy's flows 8 times that or 8 = pints/minute =3D 1 gal a min. Probably not that high but lower flow = just means the temperature calculation will be higher than in this = example. The Cp of oil is around 0.4 and its mass is approx 7.3 lbs/gallon. 4.8Hp converts to 203 BTU/Min, so we need to find out what temperature = rise that might cause in the oil. Delta T =3D Heat/(Mass flow *Cp) or if we start out with 180F oil (Ti) = then we need to find the final temperature (Tf) . So Tf =3D Ti + = (Heat/(mass flow *cp) Tf =3D 180F + (203/((7.3 *.5) *0.4) =3D 180F + 69F =3D 249F So using this example and assuming I haven't screwed up, I would expect = the oil temps coming out of the gear box to be around 249F. =20 Less efficiency would mean higher temps. More oil flow would lower the = temp and less would increase it. Less power being produced would also = decrease the temperatures. There may be some heat lost through air flow = around the gearbox, but probably offset by heat from the engine through = the spacers and plate. In my case, I would likely producing around 80 HP at cruise based on = fuel burn numbers (8 gph). My oil temp at that power is around 170F. So Tf =3D 170F + (102/(7.3*0.4) =3D 170F + 35F =3D 205 F. =20 So don't know if that tracks or not with what you guys are seeing, but = that's my 0.02 on it. Ed Ed Anderson Rv-6A N494BW Rotary Powered Matthews, NC eanderson@carolina.rr.com http://members.cox.net/rogersda/rotary/configs.htm#N494BW http://www.dmack.net/mazda/index.html ------=_NextPart_000_0006_01C79C74.F405AAE0 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable

Figures I found on the web indicates 3% is a = fairly=20 standard figure used for a planetary gearbox efficiency.

So if that figure is close and you are producing = in cruise=20 - say 160 HP- then the gear box will produce heat with about 3% of = the=20 input power. So 3% of 160 HP =3D 4.8 HP = which is=20 mostly converted to heat.

I assume the oil is already cooled = by the oil=20 cooler before it is fed to the PSRU. So the temp of the oil = going=20 into the gear box might be around 180F. I am not certain how many = GPM of=20 oil Tracy's PSRU flows but the old Ross flowed around 1 to 1 1/2=20 pint/minute, but that was apparently rather marginal. So = assume=20 Tracy's flows 8 times that or 8 pints/minute = =3D 1 gal a=20 min. Probably not that high but lower flow just means the = temperature=20 calculation will be higher than in this example.

The Cp of oil is around 0.4 and its mass is = approx 7.3=20 lbs/gallon.

4.8Hp converts to 203 BTU/Min, so we need to = find out what=20 temperature rise that might cause in the oil.

Delta T =3D Heat/(Mass flow *Cp) or if we = start out=20 with 180F oil (Ti) then we need to find the final temperature (Tf) = . =20 So Tf =3D Ti + (Heat/(mass flow *cp)

Tf =3D 180F + (203/((7.3 *.5) *0.4) =3D = 180F=20 + 69F =3D 249F

So using this example and assuming I haven't = screwed up,=20 I would expect the oil temps coming out of the gear box to be = around=20 249F.

Less efficiency would mean higher temps. = More oil=20 flow would lower the temp and less would increase it. Less power = being=20 produced would also decrease the temperatures. There may be some = heat lost=20 through air flow around the gearbox, but probably offset by heat from = the engine=20 through the spacers and plate.

In my case, I would likely producing = around 80 HP=20 at cruise based on fuel burn numbers (8 gph). My oil temp at that = power is=20 around 170F.

So Tf =3D 170F + (102/(7.3*0.4) =3D 170F + 35F = =3D 205 F. =20

So don't know if that tracks or not with what = you guys are=20 seeing, but that's my 0.02 on it.

Ed Anderson
Rv-6A N494BW Rotary = Powered
Matthews,=20 NC
eanderson@carolina.rr.comhttp:/= /members.cox.net/rogersda/rotary/configs.htm#N494BW
http://www.dmack.net/mazda= /index.html

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