Return-Path: Received: from [24.25.9.102] (HELO ms-smtp-03-eri0.southeast.rr.com) by logan.com (CommuniGate Pro SMTP 4.2) with ESMTP id 375717 for flyrotary@lancaironline.net; Mon, 23 Aug 2004 20:24:38 -0400 Received-SPF: none receiver=logan.com; client-ip=24.25.9.102; envelope-from=eanderson@carolina.rr.com Received: from EDWARD (cpe-069-132-183-211.carolina.rr.com [69.132.183.211]) by ms-smtp-03-eri0.southeast.rr.com (8.12.10/8.12.7) with SMTP id i7O0O5iB022520 for ; Mon, 23 Aug 2004 20:24:06 -0400 (EDT) Message-ID: <001c01c48970$ae5e13c0$2402a8c0@EDWARD> From: "Ed Anderson" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Re: Compressor maps Date: Mon, 23 Aug 2004 20:24:12 -0400 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0019_01C4894F.2715AC50" 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_0019_01C4894F.2715AC50 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable John, Your pressure ratio formula is (Pa + Pb)/Pa. First, I prefer using PSI, = so taking that route first. =20 At sea level we have at MAP 50 =3D (50-30)/30*14.7 =3D 9.8 psi gauge of = boost pressure A check (14.7+9.8)/14.7 =3D 1.66 pressure ratio. Pb is your = differential gauge pressure or Boost pressure Going to 10,000 MSL with the same boost reading we have (10+9.8)/10 =3D = 1.98 pressure ratio due to ambient pressure decrease. I used 10 psi for = atmospheric which is probably too high, but too lazy to look it up. = Stick in the correct figure.. You ambient pressure is an absolute measurement like 29.92 "HG or = 14.7PSI. You boost pressure is differential or the amount above ambient = (whatever ambient is). =20 I think you have mixed MAP measured in inches of mercury and MAP = measured in Lbs/in^2. It helps to always state your units of measure. Using your approach we have=20 MAP =3D 50 at sea level =3D (30+20 )/30 =3D 1.666, so that checks = Boost here in " Hg is 50" - 30" =3D 20" this is your Pb. It can be = stated in either inches Hg, PSI, Pascals, etc. The important thing is = that it is a differential pressure that is Pb =3D PT - Pa. If PT is = your Total absolute pressure (including boost) then your differential = pressure Pb =3D Pt-Pa. In your example at sea level your Pb =3D 50 - 30 = =3D 20" Hg At 10,000 ft altitude if you are going to use "hg then - and if the air = pressure is 10 psi (as we assumed) then the MAP is 10/30*14.7 =3D 20" Hg = for the ambient pressure in inches mercury not 10. Each psi is approx 2 = " Hg. Our Boost pressure remains the same as at sea level or 20 " Hg =3D = 20"/30"*14.7 =3D 9.8 psi so that checks. Using your MAP =3D 40 at 10,000 MSL, we first find our boost Pb =3D = 40"(Pt) - 20"(Pa)(not 10) =3D 20 "hg boost same as we had at sea level Tanking these numbers and putting them in our formula we have (20"Pa{not = 10} + 20"Pb)/20"Pa{not 10) =3D (20+20)/20 =3D 2 or. Close enough to my = 1.98 and far away from your calculate pressure ratio of 4.0 Hope this helped. Just remember you really need to keep track of your = units of measurement - it helps clarify and prevent confusion - I know I = need all the help I can get, so hope this helped you a bit., Summary: =20 You ambient pressure (Pa) is an absolute measurement like 29.92 "HG or = 14.7PSI. You boost pressure(Pb) is differential or the amount above = ambient (whatever ambient is). Your Total pressure (PT) is the sum of = the two. Pressure ratio =3D (Pa+Pb)/Pa Pa =3D Ambient atmospheric pressure (at = whatever altitude). Pb is your differential boost pressure Pb =3D Pt - Pa Where Pt is your total absolute pressure (Pt =3D Pa+Pb) To convert from inches Hg to PSI, I use this forumla Taking an = absolute pressure X in " Hg, we have X/30*14.7psi =3D " Hg. For example to convert 30 " Hg we have 30/30*14.7 =3D 14.7 psi. = Conversely to convert psi to " Hg we use X/14.7*30 =3D " Hg=20 Taking 14.7 psi we have 14.7/14.7*30 =3D 30" Hg. =20 In your case for altitude - if the ambient pressure were 10 psi then to = convert to " Hg, we have 10/14.7*30 =3D 20.48" Hg or 20" Hg rounded = off. Ed Ed Anderson RV-6A N494BW Rotary Powered Matthews, NC ----- Original Message -----=20 From: John Slade=20 To: Rotary motors in aircraft=20 Sent: Monday, August 23, 2004 9:45 AM Subject: [FlyRotary] Re: Compressor maps Let's see if I understand this. To help me read the map, I made a tiny spreadsheet (attached) to show = lbs/min at various rpm=20 This tells me I have 23lb/min at 6000. So... At SL and MAP 50 I have a PR of (30+20)/30 =3D 50/30 =3D 1.66 So on the map I read up from 23 till I hit 1.66 on the y axis and see = that I'm on the bottom of the island.=20 Now go to 10000ft and set the MAP at, say 40. My PS is now (10+30)/10 = =3D 40/10 =3D 4.0 which is off the scale. :( Looking at the (compressor) MAP again I see that my max PR at 23lb is = about 1.8, so working backwards my max MAP at 6000 ft is 18 if I want to = stay on the island. To get into higher PR's I need more rpm. I'd need = 9000+ rpm to reach the top of the island where the available PR is about = 2.5. Doesn't sound right. Where did I go wrong? Comments? -------------------------------------------------------------------------= ----- >> Homepage: http://www.flyrotary.com/ >> Archive: http://lancaironline.net/lists/flyrotary/List.html ------=_NextPart_000_0019_01C4894F.2715AC50 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
John,
 
Your pressure ratio formula  is = (Pa +=20 Pb)/Pa. First, I prefer using PSI, so taking that route=20 first.  
 
 At sea level we have at MAP 50 = =3D=20 (50-30)/30*14.7 =3D 9.8 psi gauge of boost pressure
 
A check (14.7+9.8)/14.7 =3D 1.66 = pressure ratio.=20   Pb is your differential gauge pressure or Boost = pressure
 
Going to 10,000 MSL with the same boost = reading we=20 have (10+9.8)/10 =3D 1.98 pressure ratio due to ambient pressure = decrease.  I=20 used 10 psi for atmospheric which is probably too high, but too lazy to = look it=20 up.  Stick in the correct figure..
 
You ambient pressure is = an absolute=20 measurement like 29.92 "HG or 14.7PSI.  You boost pressure is = differential=20 or the amount above ambient (whatever ambient is).
 
 
 
  I think you=20 have mixed MAP measured in inches of mercury and MAP measured in = Lbs/in^2. =20 It helps to always state your units of measure.
 
 
Using your approach we have =
MAP =3D 50  at sea level =3D = (30+20 )/30 =3D 1.666,=20 so that checks  Boost here in " Hg is 50" - 30" =3D 20" this is = your=20 Pb.  It can be stated in either inches Hg, PSI, Pascals, etc.  = The=20 important thing is that it is a differential pressure that is Pb =3D PT = -=20 Pa.  If PT is your Total absolute pressure (including boost) then = your=20 differential pressure Pb =3D Pt-Pa.  In your example at sea level = your Pb =3D=20 50 - 30 =3D 20" Hg
 
 
At 10,000 ft altitude if you are going = to use "hg=20 then - and if the air pressure is 10 psi (as we assumed) then = the MAP=20 is 10/30*14.7 =3D 20" Hg for the ambient pressure in inches mercury not = 10. =20 Each psi is approx 2 " Hg.
 
Our Boost pressure  remains the = same as at sea=20 level or 20 " Hg =3D 20"/30"*14.7 =3D 9.8 psi so that = checks.
 
Using your MAP =3D 40 at  =  10,000 MSL,=20 we first find our boost  Pb =3D  40"(Pt) - 20"(Pa)(not = 10) =3D 20 "hg boost same as we had at sea = level
 
Tanking these numbers and putting them = in our=20 formula we have (20"Pa{not 10} + 20"Pb)/20"Pa{not 10) =3D (20+20)/20 =3D = 2 or. Close=20 enough to my 1.98 and far away from your calculate pressure ratio of=20 4.0
 
 
  Hope this helped.  Just = remember you=20 really need to keep track of your units of measurement - it helps = clarify and=20 prevent confusion - I know I need all the help I can get, so hope this = helped=20 you a bit.,
 
Summary: 
 
You ambient pressure (Pa) is = an absolute=20 measurement like 29.92 "HG or 14.7PSI.  You boost pressure(Pb) is=20 differential or the amount above ambient (whatever ambient is).  = Your Total=20 pressure (PT) is the sum of the two.
 
 
Pressure ratio =3D (Pa+Pb)/Pa  Pa = =3D Ambient=20 atmospheric pressure (at whatever altitude).  Pb is your = differential boost=20 pressure
 
Pb =3D Pt - Pa  Where Pt is your = total absolute=20 pressure (Pt =3D Pa+Pb)
 
To convert from inches Hg to PSI, I use = this =20 forumla  Taking an absolute pressure X in " Hg,  we have=20     X/30*14.7psi =3D " Hg.
 
For example to convert 30 " Hg we=20 have    30/30*14.7 =3D 14.7 psi.   Conversely = to convert=20 psi to " Hg we use   X/14.7*30 =3D "=20 Hg 
 
Taking 14.7 psi  we have = 14.7/14.7*30 =3D 30"=20 Hg.  
 
 In your case for altitude - if = the ambient=20 pressure were 10 psi then to convert to " Hg, we have   =20 10/14.7*30 =3D 20.48" Hg or 20" Hg = rounded=20 off.
 
Ed
 
 Ed Anderson
RV-6A = N494BW Rotary=20 Powered
Matthews, NC
----- Original Message -----
From:=20 John=20 Slade
Sent: Monday, August 23, 2004 = 9:45=20 AM
Subject: [FlyRotary] Re: = Compressor=20 maps

Let's see if I understand this.
To=20 help me read the map, I made a tiny spreadsheet (attached) to show = lbs/min at=20 various rpm
This=20 tells me I have 23lb/min at 6000. So...
At SL and MAP 50 I have a PR of (30+20)/30 =3D = 50/30 =3D=20 1.66
So on the map I = read up from=20 23 till I hit 1.66 on = the y axis=20 and see that I'm on the bottom of the island.
 
Now=20 go to 10000ft and set the MAP at, say 40. My PS is now (10+30)/10 = =3D 40/10=20 =3D 4.0 which is off the scale. :(
 
Looking at the (compressor) MAP again I see that my max PR at = 23lb is=20 about 1.8, so working backwards my max MAP at 6000 ft is 18 if I want = to stay=20 on the island.  To get into higher PR's I need more rpm. I'd=20 need 9000+ rpm to reach the top of the island where = the available PR=20 is about 2.5.
 
Doesn't sound right. Where did I go = wrong?
 
Comments?
 
 
 
 

>>  Homepage: =20 http://www.flyrotary.com/
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