Return-Path: Received: from swan.mail.pas.earthlink.net ([207.217.120.123] verified) by logan.com (CommuniGate Pro SMTP 4.1.8) with ESMTP id 2882022 for flyrotary@lancaironline.net; Mon, 08 Dec 2003 13:14:34 -0500 Received: from h-68-166-182-146.sfldmidn.covad.net ([68.166.182.146] helo=richard) by swan.mail.pas.earthlink.net with smtp (Exim 3.33 #1) id 1ATPtw-0001If-00 for flyrotary@lancaironline.net; Mon, 08 Dec 2003 10:14:32 -0800 Message-ID: <007c01c3bdb7$473757e0$6601a8c0@richard> From: "Eric Ruttan" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Air Density at altitude Date: Mon, 8 Dec 2003 13:15:35 -0500 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0079_01C3BD8D.5DB00140" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2800.1158 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2800.1165 This is a multi-part message in MIME format. ------=_NextPart_000_0079_01C3BD8D.5DB00140 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Ah yes. But I was thinking of a turbo engine making full power up to = 25K. Which of course destroys the symmetry of the math. So I restate. If at X mpg IAS at 2000' you can get rid of Y BTU heat=20 Then can you get rid of Y BTU heat at X mph IAS at 25K' If IAS is a direct measurement of Air Mass then yes, as the Air Mass is = the same. If not then IAS is not a direct measurement of Air Mass. Or my logic is faulty. Thanks for your tolerant indulgence. ----- Original Message -----=20 From: Jim Sower=20 To: Rotary motors in aircraft=20 Sent: Monday, December 08, 2003 11:25 AM Subject: [FlyRotary] Re: Air Density at altitude But of course, with only 47% of air mass flow through the radiator, = you also have only 47% of air mass flow through the engine and 47% as = much heat to reject. So, for an NA engine, there shouldn't be a = significant change in the ability of a cooling system.=20 Or so it seems to me .... Jim S.=20 Ed Anderson wrote:=20 Found a table of air density vs Altitude=20 Sea level Density =3D .00237 Slug/Ft^3=20 Density at 20,000 =3D 0.001267 Slug/Ft^3 or a 47% decrease=20 So taking formula for air mass W =3D p*V*A with p 47% less than at = sea level=20 means you would get 47% less air mass flow (with the same cubic = feet/minute=20 of air volume flow) at 20,000 ft compared to what you would get at = sea level=20 for the same volume flow.=20 While cooler temps would help, it would not compensate for a 45% = less air=20 mass flow.=20 Ed=20 Ed Anderson=20 RV-6A N494BW Rotary Powered=20 Matthews, NC=20 eanderson@carolina.rr.com=20 >> Homepage: http://www.flyrotary.com/=20 >> Archive: http://lancaironline.net/lists/flyrotary/List.html --=20 Jim Sower=20 Crossville, TN; Chapter 5=20 Long-EZ N83RT, Velocity N4095T=20 =20 ------=_NextPart_000_0079_01C3BD8D.5DB00140 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Ah yes.  But I was thinking of a = turbo engine=20 making full power up to 25K.  Which of course destroys the symmetry = of the=20 math.
 
So I restate.
 
If at X mpg IAS at 2000' you can = get rid of Y=20 BTU heat
 
Then can you get rid=20 of Y BTU heat at X mph IAS at 25K'
 
If IAS is a direct measurement of Air = Mass then=20 yes, as the Air Mass is the same.
 
If not then IAS is not a direct = measurement of Air=20 Mass.
 
Or my logic is faulty.
 
Thanks for your tolerant=20 indulgence.
 
----- Original Message -----
From:=20 Jim=20 Sower
Sent: Monday, December 08, 2003 = 11:25=20 AM
Subject: [FlyRotary] Re: Air = Density at=20 altitude

But of course, with only 47% of air mass flow through = the=20 radiator, you also have only 47% of air mass flow through the = engine=20 and 47% as much heat to reject.  So, for an NA engine, there = shouldn't be=20 a significant change in the ability of a cooling system.
Or so it = seems to=20 me .... Jim S.=20

Ed Anderson wrote:=20

Found a table of air density vs Altitude=20

Sea level Density =3D .00237 Slug/Ft^3
Density at 20,000 =3D = 0.001267=20 Slug/Ft^3 or a 47% decrease=20

So taking formula for air mass  W =3D p*V*A with p 47% less = than at=20 sea level
means you would get 47% less air mass flow (with the = same=20 cubic feet/minute
of air volume flow) at 20,000 ft compared to = what you=20 would get at sea level
for the same volume flow.=20

While cooler temps would help, it would not compensate for a 45% = less air=20
mass flow.=20

Ed=20

Ed Anderson
RV-6A N494BW Rotary Powered
Matthews, NC=20
eanderson@carolina.rr.com=20

>>  Homepage:  http://www.flyrotary.com/=20
>>  Archive:   http://lancai= ronline.net/lists/flyrotary/List.html

--
Jim Sower
Crossville, TN; Chapter 5
Long-EZ N83RT, = Velocity=20 N4095T
 

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