Return-Path: Received: from [24.25.9.102] (HELO ms-smtp-03-eri0.southeast.rr.com) by logan.com (CommuniGate Pro SMTP 4.2b5) with ESMTP id 148453 for flyrotary@lancaironline.net; Fri, 11 Jun 2004 08:01:26 -0400 Received: from EDWARD (clt25-78-058.carolina.rr.com [24.25.78.58]) by ms-smtp-03-eri0.southeast.rr.com (8.12.10/8.12.7) with SMTP id i5BC0nLr018397 for ; Fri, 11 Jun 2004 08:00:51 -0400 (EDT) Message-ID: <001501c44fab$bfd7cf50$2402a8c0@EDWARD> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Cooling oil Date: Fri, 11 Jun 2004 08:00:54 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0012_01C44F8A.3872B820" 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_01C44F8A.3872B820 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable ----- Original Message -----=20 From: Al Gietzen=20 To: Rotary motors in aircraft=20 Sent: Friday, June 11, 2004 12:46 AM Subject: [FlyRotary] Re: Cooling oil Subject: [FlyRotary] Re: Cooling oil ----- Original Message -----=20 From: "Joseph Berki" To: "Rotary motors in aircraft" Sent: Thursday, June 10, 2004 12:39 PM Subject: [FlyRotary] Re: Cooling oil > It would be interesting to measure flow in both engines. I thought = that > both Lycoming and Mazda engines rejected 2/3 heat load through the > oil that is why I started going down this road. If the engines = generated > the same Hp than the heat load should be similar. > > Joe Berki Joe, both engines may generate the same heat load, but the proportion rejected through the coolant in case of the Mazda is 2/3 of its waste = heat while the oil rejects another 1/3 of the waste heat. Neither engine = rejects anywhere near 2/3 of its waste heat through the oil. Most aircraft engines reject on the order of 300-600 BTU/Min through = the oil, the Mazda at 160HP rejects approx 2446 BTU/Min through the oil. Ed Anderson Ed; That number looked a bit high to me, so I went in to my file to check. = My data shows 28% of the fuel burn energy in the rotary gets converted = to HP, 18% goes to the coolant, and about 7% to the oil. Most of the = rest goes out the exhaust pipe. For 160 HP output, I think that should = be 1725 BTU/Min going to the oil cooler. So about 3 times the = comparable powered Lyc. Double check me on this. Al Thanks Al, I did double check and here are my calcuations - in case I've = missed an error in them somewhere and can't spot it. I use a conservative 25% for HP, 25% for Coolant and 50% for = Exhaust. So I am interested in where you got the 28% for HP. That = would mean less waste heat to the coolers.=20 Here are my calculations 1 Lb of gasoline containts approx 19,000 BTU of energy. For 160HP I = get 1.58 Lb/min of fuel flow. which is approx 1.58*19,000 =3D 30020 BTU/min. Using Hp =3D 25%, = Waste =3D 25%, Exhaust =3D 50% Waste =3D 0.25*30020 =3D 7505 BTU/Min. The oil dumps approx 1/3 of = the waste heat of a rotary, so 0.3333*7505 =3D 2501 BTU/Min at 160HP. So unless my fuel flow is = off I think my initial=20 calculation is approximately correct. I think I used 18,800 BTU in my = orginal calculations which=20 would give a slightly lower value. I have seen the BTU content of a = lbm of gasoline vary between=20 about 18,800 to 20,000 BTU/Lbm. Checking my fuel flow for a 160HP engine.=20 I have air flow =3D 80*6000/1728 =3D 277.77 CFM for a 2 rotor 13B 1 cubic foot of air mass at sea level =3D approx 0.076 lbm. So that = gives .076*277.77 =3D 21.11 lbm/minute of air flow for the 2 rotor at = 6000 rpm at sea level. I used an air/fuel ration for best power of 13.3 (just the value I = happen to have in my calculator). So my fuel needed for that air flow = (21.11 lbm/minute) is 21.11/13.3=3D 1.587 lb/min of fue. So 1.587*19,000 =3D 30157.14 BTU/Min of which 0.25*0.333*30157 =3D = 2510 BTU/Min for the cooler. So unless I've screwed up somewhere in the calculations or unless the = my allocations of the BTU for waste heat rejection are not valid, I = think my calcuation is pretty close. Ed ------=_NextPart_000_0012_01C44F8A.3872B820 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
----- Original Message = -----
From:=20 Al = Gietzen=20
Sent: Friday, June 11, 2004 = 12:46=20 AM
Subject: [FlyRotary] Re: = Cooling=20 oil

 

Subject: [FlyRotary] Re: Cooling = oil

 

 

----- Original Message ----- =

From: "Joseph Berki" <joseph.berki@grc.nasa.gov&g= t;

To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>

Sent: Thursday, June 10, 2004 = 12:39=20 PM

Subject: [FlyRotary] Re: Cooling = oil

 

 

> It would be interesting to measure flow = in both=20 engines.  I thought that

> both Lycoming and Mazda engines = rejected 2/3 heat=20 load through the

> oil  that is why I started going = down this=20 road.  If the engines generated

> the same Hp than the heat load should = be=20 similar.

>

> Joe Berki

 

Joe, both engines may generate the same heat = load, but=20 the proportion

rejected through the coolant in case of the = Mazda is=20 2/3 of its waste heat

while the oil rejects another 1/3 of the = waste=20 heat.  Neither engine rejects

anywhere near 2/3 of its waste heat through = the=20 oil.

 

Most aircraft engines reject on the order of = 300-600=20 BTU/Min through the

oil, the Mazda at 160HP rejects approx 2446 = BTU/Min=20 through the oil.

 

Ed Anderson

 

Ed;

 

That = number looked=20 a bit high to me, so I went in to my file to check.  My data = shows 28% of=20 the fuel burn energy in the rotary gets converted to HP, 18% goes to = the=20 coolant, and about 7% to the oil. Most of the rest goes out the = exhaust pipe.=20 For 160 HP output, I think that should be 1725 BTU/Min going to the = oil=20 cooler.  So about 3 times the comparable powered = Lyc.

 

Double = check me on=20 this.

 

Al

 

Thanks=20 Al,

 

   =20 I did double check and here are my calcuations - in case I've missed = an error=20 in them somewhere and can't spot it.

 

   =20 I use a conservative 25% for HP, 25% for Coolant and 50% for=20 Exhaust.  So I am interested in where you got the 28% for = HP.  That=20 would mean less waste heat to the coolers.

 

Here are = my=20 calculations

 

1 Lb of = gasoline=20 containts approx 19,000 BTU of energy.  For 160HP I get 1.58 = Lb/min of=20 fuel flow.

 

which is = approx=20 1.58*19,000 =3D 30020 BTU/min. Using  Hp =3D 25%, Waste =3D = 25%, Exhaust =3D=20 50%

 

Waste = =3D 0.25*30020=20 =3D 7505 BTU/Min.  The oil dumps approx 1/3 of the waste = heat of a=20 rotary,

 

so = 0.3333*7505=20  =3D 2501 BTU/Min at 160HP.  So unless my fuel flow is off I = think my=20 initial

 

calculation is=20 approximately correct.  I think I used 18,800 BTU in my orginal=20 calculations which

 

would = give a=20 slightly lower value.    I have seen the BTU content of a = lbm of=20 gasoline vary between

 

about = 18,800 to=20 20,000 BTU/Lbm.

 

Checking = my fuel=20 flow for a 160HP engine. 

 

 I = have air=20 flow =3D 80*6000/1728 =3D 277.77 CFM for a 2 rotor = 13B

 

1 cubic = foot of=20 air mass at sea level =3D approx 0.076 lbm. So that gives .076*277.77 = =3D 21.11=20 lbm/minute of air flow for the 2 rotor at 6000 rpm at sea=20 level.

 

I used = an air/fuel=20 ration for best power of 13.3 (just the value I happen to have in my=20 calculator).  So my fuel needed for that air flow (21.11 = lbm/minute) is=20 21.11/13.3=3D 1.587 lb/min of fue.

 

So = 1.587*19,000 =3D=20 30157.14 BTU/Min of which 0.25*0.333*30157 =3D 2510 BTU/Min for the=20 cooler.

 

So = unless I've=20 screwed up somewhere in the calculations or unless the my allocations = of the=20 BTU for waste heat rejection are not valid, I think my calcuation is = pretty=20 close.

 

Ed

 

 

 

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