X-Virus-Scanned: clean according to Sophos on Logan.com X-SpamCatcher-Score: 2 [X] Return-Path: Received: from ms-smtp-02.southeast.rr.com ([24.25.9.101] verified) by logan.com (CommuniGate Pro SMTP 5.1.9) with ESMTP id 2061988 for flyrotary@lancaironline.net; Tue, 22 May 2007 18:32:15 -0400 Received-SPF: pass receiver=logan.com; client-ip=24.25.9.101; 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-02.southeast.rr.com (8.13.6/8.13.6) with SMTP id l4MMVLxd017640 for ; Tue, 22 May 2007 18:31:22 -0400 (EDT) Message-ID: <000701c79cc1$301c2300$2402a8c0@edward2> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Gear box oil temps Date: Tue, 22 May 2007 18:33:14 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0004_01C79C9F.A89CCD10" 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_0004_01C79C9F.A89CCD10 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Other's have done so, George, so it certainly works. In fact, I believe = George Graham did so on his E racer. By the way the numbers of my example are correct, but a 0.5 crept into = the denominator of the equations (7.3 *0.5)*0.4) . The temperature is = correct for 1 gallon/min flow so the 0.5 should be removed - I had stuck = it in there when trying to get a handle on 1/2 gal/min flow but decided = that would lead to a 139F increase which did not seem even in the ball = park. Ed ----- Original Message -----=20 From: George Lendich=20 To: Rotary motors in aircraft=20 Sent: Tuesday, May 22, 2007 6:02 PM Subject: [FlyRotary] Re: Gear box oil temps Ed, Your 249F is convincing me to run a separate oil and cooler. George ( down under) 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_0004_01C79C9F.A89CCD10 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Other's have done so, George, so it certainly = works. =20 In fact, I believe George Graham did so on his E racer.
 
By the way the numbers of my example are = correct, but a=20 0.5 crept into the denominator of the equations (7.3 = *0.5)*0.4) .  The=20 temperature is correct for 1 gallon/min flow so the 0.5 should be = removed - I=20 had stuck it in there when trying to get a handle on 1/2 gal/min flow = but=20 decided that would lead to a 139F increase which did not seem even in = the ball=20 park.
 
Ed
----- Original Message -----
From:=20 George=20 Lendich
Sent: Tuesday, May 22, 2007 = 6:02 PM
Subject: [FlyRotary] Re: Gear = box oil=20 temps

Ed,
Your 249F is convincing me to run a = separate oil=20 and cooler.
George ( down under)
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=20 cruise - say 160 HP-  then the gear box will produce heat with = about 3%=20 of the input  power.    So 3% of 160 HP =3D 4.8 = HP =20 which is mostly converted to heat.
 
I assume the  oil is  already = cooled by the=20 oil cooler before  it is fed to the PSRU.  So the temp of = the oil=20 going into the gear box might be around 180F.  I am not certain = how=20 many GPM of 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=20 =3D 1  gal a min.  Probably not that high but = lower flow=20 just means the temperature calculation will be higher than  in = this=20 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=20 what temperature rise that might cause in the oil.
 
Delta T =3D Heat/(Mass flow *Cp)  or if = we start=20 out 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=20 up,  I would expect the oil temps coming out of the gear box to = be=20 around 249F. 
 
Less efficiency would mean higher = temps.  More=20 oil flow would lower the temp and less would increase it.  Less = power=20 being produced would also decrease the temperatures.  There may = be some=20 heat lost through air flow around the gearbox, but probably offset = by heat=20 from the engine through the spacers and plate.
 
  In my case, I would likely producing = around 80=20 HP at cruise based on fuel burn numbers (8 gph).  My oil temp = at that=20 power is around 170F.
 
So Tf =3D 170F + (102/(7.3*0.4) =3D 170F + = 35F =3D 205=20 F. 
 
So don't know if that tracks or not with = what you guys=20 are seeing, but that's my 0.02 on it.
 
Ed
 
Ed Anderson
Rv-6A N494BW Rotary=20 Powered
Matthews, NC
eanderson@carolina.rr.comhttp:/= /members.cox.net/rogersda/rotary/configs.htm#N494BW
http://www.dmack.net/mazda= /index.html
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