X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from omr-d04.mx.aol.com ([205.188.109.201] verified) by logan.com (CommuniGate Pro SMTP 6.0.7) with ESMTPS id 6506195 for flyrotary@lancaironline.net; Tue, 08 Oct 2013 01:26:19 -0400 Received-SPF: pass receiver=logan.com; client-ip=205.188.109.201; envelope-from=Lehanover@aol.com Received: from mtaomg-ma02.r1000.mx.aol.com (mtaomg-ma02.r1000.mx.aol.com [172.29.41.9]) by omr-d04.mx.aol.com (Outbound Mail Relay) with ESMTP id B638570000094 for ; Tue, 8 Oct 2013 01:25:43 -0400 (EDT) Received: from core-moe003b.r1000.mail.aol.com (core-moe003.r1000.mail.aol.com [172.29.188.73]) by mtaomg-ma02.r1000.mx.aol.com (OMAG/Core Interface) with ESMTP id 79B7EE000086 for ; Tue, 8 Oct 2013 01:25:43 -0400 (EDT) From: Lehanover@aol.com Full-name: Lehanover Message-ID: Date: Tue, 8 Oct 2013 01:25:43 -0400 (EDT) Subject: Re: [FlyRotary] Re: Prop and PSRU efficiency To: flyrotary@lancaironline.net MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="part1_ac22d.100f157.3f84f157_boundary" X-Mailer: AOL 9.7 sub 1028 X-Originating-IP: [184.57.118.173] x-aol-global-disposition: G DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=mx.aol.com; s=20121107; t=1381209943; bh=1hVzTz8cmspxgGO7LMB53kbU4+XTAeN/dWiUtsAvbmA=; h=From:To:Subject:Message-ID:Date:MIME-Version:Content-Type; b=FpjrLysPPF1YjM537W8tZQV5cE2kI/8i/IWWXgBIbhXNkHGSKpDIioKB3KPQtF2sU 7TjCslDh7EK/TrMSdRfHvHqy/jP2/V3XX2cXlybdoEYDEPvpIUctXoSu0RspXeAZl/ yemRAmiRUHJdqzIpkSKn0XxAe24K08M7+B0+ktlg= x-aol-sid: 3039ac1d2909525397577582 --part1_ac22d.100f157.3f84f157_boundary Content-Type: text/plain; charset="UTF-8" Content-Transfer-Encoding: quoted-printable Content-Language: en The loss of HP in the reduction unit is not affected by engine accessories,= =20 or tuning errors or anything but the actual torque applied to the input=20 shaft of the reduction unit.=20 There will be losses as reflected in the rise of oil (coolant) =20 temperatures. Prop losses have no effect on reduction unit losses so long = as all=20 available HP is absorbed by the propeller. I suggest that losses are actua= lly=20 minimal. The format is the best choice for the job and the most efficient= =20 system, (planetary). A loss of 10 to 20 HP would require=20 a very large cooler for just the reduction unit. This system is the first = =20 gear reduction from a Ford diesel truck with 450 to 600 foot pounds of=20 torque. It would seem to be under little stress in an aircraft situation w= here =20 a rotary engine is used as the torque output is low compared to piston=20 engines. 160 HP at 5,500 RPM requires 152.78 foot pounds of torque. 160 HP= at =20 6,000 RPM requires 140 foot pounds of torque. =20 Modern automotive applications now use up to 6 reduction speeds to achieve= =20 better fuel economy.So, adding planetary's to improve mileage tells me =20 that planetary's are very efficient indeed.=20 =20 Lynn E. Hanover=20 =20 =20 =20 =20 In a message dated 10/7/2013 3:42:21 P.M. Eastern Daylight Time, =20 bbradburry@bellsouth.net writes: =20 Mark,=20 They would be included if the engine was tested on a dyno, so I consider= =20 them to be part of the engine. But not so the PSRU if measuring from the = =20 flywheel.=20 Ernest,=20 I don=E2=80=99t know what you mean by .98 to .99?? Certainly you don=E2= =80=99t think it=20 would only be a loss of 1 or 2%!?? It would have to be in the range of 10= =20 to 20 HP or even greater. That is 5 to 10% in our HP range. Just the los= s=20 due to prop efficiency is in the range of 30 HP!=20 Bill =20 =20 =20 ____________________________________ =20 From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On=20 Behalf Of Mark Steitle Sent: Monday, October 07, 2013 1:26 PM To: Rotary motors in aircraft Subject: [FlyRotary] Re: Prop and PSRU efficiency =20 =20 Bill,=20 =20 =20 While you're at it don't forget to account for the water pump and=20 alternator(s).=20 =20 =20 Mark S. =20 =20 On Mon, Oct 7, 2013 at 12:18 PM, Bill Bradburry <_bbradburry@bellsouth.net_= =20 (mailto:bbradburry@bellsouth.net) > wrote:=20 I have asked this question a couple of times and no one has hazarded a guess. How much HP is lost from our engines due to the PSRU? I have been interested in determining what the HP output of my engine is and that info would be needed for that estimation. They tell me that most props are about 80-85% efficient, so to calculate= =20 the hp, you take the difference between your climb rate and your glide descent rate at the same airspeed, multiplied by the weight, and then divided by 33000. Wt * V / 33000 =3D HP This would be the prop HP, so to get the prop flange HP, you would divide= =20 by the prop efficiency, between .8 and .85. To get the engine flywheel HP, you would have to add something for the los= s of the PSRU. Is anyone willing to take a shot at that number?? Third or forth chance! :>) Bill B -- Homepage: http://www.flyrotary.com/ Archive and UnSub: =20 http://mail.lancaironline.net:81/lists/flyrotary/List.html --part1_ac22d.100f157.3f84f157_boundary Content-Type: text/html; charset="UTF-8" Content-Transfer-Encoding: quoted-printable Content-Language: en
The loss of HP in the reduction unit is not affected by engine accesso= ries,=20 or tuning errors or anything but the actual torque applied to the input sha= ft of=20 the reduction unit.
There will be losses as reflected in the rise of oil (coolant)=20 temperatures. Prop losses have no effect on reduction unit losses so l= ong=20 as all available HP is absorbed by the propeller. I suggest that losses are= =20 actually minimal. The format is the best choice for the job and the=20 most efficient system,  (planetary). A loss of 10 to 20 HP w= ould=20 require
a very large cooler for just the reduction unit. This system is the fi= rst=20 gear reduction from a Ford diesel truck with 450 to 600 foot pounds of torq= ue.=20 It would seem to be under little stress in an aircraft situation = where=20 a rotary engine is used as the torque output is low compared to piston engi= nes.=20 160 HP at 5,500 RPM requires 152.78 foot pounds of torque. 160 HP= at=20 6,000 RPM requires 140 foot pounds of torque.
 
Modern automotive applications now use up to 6 reduction speeds t= o=20 achieve better fuel economy.So, adding planetary's to improve mileage tells= me=20 that planetary's are very efficient indeed.
 
Lynn E. Hanover 
 
 
 
In a message dated 10/7/2013 3:42:21 P.M. Eastern Daylight Time,=20 bbradburry@bellsouth.net writes:
=

Mark,

 

They would be= =20 included if the engine was tested on a dyno, so I consider them to be par= t of=20 the engine.  But not so the PSRU if measuring from the=20 flywheel.

 

Ernest,

 

I don=E2=80=99= t know what you=20 mean by .98 to .99??  Certainly you don=E2=80=99t think it would onl= y be a loss=20 of 1 or 2%!??  It would have to be in the range of 10 to 20 HP or ev= en=20 greater.  That is 5 to 10% in our HP range.  Just the loss due = to=20 prop efficiency is in the range of 30 HP!

 

Bill=20

 

From: Rotary=20 motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of Mark Steitle
Sent: Monday, October 07, 2013 1:2= 6=20 PM
To: Rotary motors i= n=20 aircraft
Subject: [Fly= Rotary]=20 Re: Prop and PSRU efficiency

 

Bill,

 

While you're at it don't forget to account for = the=20 water pump and alternator(s).

 

Mark S.

 

On Mon, Oct 7, 2013 at 12:18 PM, Bill Bradburry <bbradburry@bellsouth.net>=20 wrote:

I have asked this question a couple of times an= d no=20 one has hazarded a
guess.

How much HP is lost from our engines = due=20 to the PSRU?  I have been
interested in determining what the HP o= utput=20 of my engine is and that info
would be needed for that=20 estimation.

They tell me that most props are about 80-85% efficien= t, so=20 to calculate the
hp, you take the difference between your climb rate a= nd=20 your glide descent
rate at the same airspeed, multiplied by the weight= , and=20 then divided by
33000.

Wt * V / 33000 =3D HP

This would = be the=20 prop HP, so to get the prop flange HP, you would divide by
the prop=20 efficiency, between .8 and .85.

To get the engine flywheel HP, you= =20 would have to add something for the loss
of the PSRU.

Is anyone= =20 willing to take a shot at that number??  Third or forth=20 chance!
:>)

Bill B




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