X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from poplet2.per.eftel.com ([203.24.100.45] verified) by logan.com (CommuniGate Pro SMTP 5.2.14) with ESMTP id 3682568 for flyrotary@lancaironline.net; Mon, 15 Jun 2009 02:20:30 -0400 Received-SPF: none receiver=logan.com; client-ip=203.24.100.45; envelope-from=lendich@aanet.com.au Received: from sv1-1.aanet.com.au (sv1-1.per.aanet.com.au [203.24.100.68]) by poplet2.per.eftel.com (Postfix) with ESMTP id 47392173A4C for ; Mon, 15 Jun 2009 14:19:52 +0800 (WST) Received: from ownerf1fc517b8 (203.171.92.134.static.rev.aanet.com.au [203.171.92.134]) by sv1-1.aanet.com.au (Postfix) with SMTP id 13E5C1153538 for ; Mon, 15 Jun 2009 14:19:49 +0800 (WST) Message-ID: From: "George Lendich" To: "Rotary motors in aircraft" References: Subject: Re: [FlyRotary] Air/Fuel Ratio?? Date: Mon, 15 Jun 2009 16:19:53 +1000 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0033_01C9EDD5.1DA55040" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.5512 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.5579 X-Antivirus: avast! (VPS 0617-3, 04/28/2006), Outbound message X-Antivirus-Status: Clean This is a multi-part message in MIME format. ------=_NextPart_000_0033_01C9EDD5.1DA55040 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Thanks Ed,=20 I will have to work through that a couple of times. George George, not exactly certain what you mean by "calculating Air/Fuel = ratio" . =20 I used the ratio of the mass of the air to the fuel - which basically = relies on the approximation of 0.0765 lbs per cubic foot of air. So you = calculate your air flow in CFM then times 0.0765 to give you the air = mass. The you can take your desired A/F ratio - say 15:1 and divide 15 = into your air mass and that would give you the fuel mass required to = achieve that ratio. Of if you have any two of the three factors you can = find the third. =20 IF you mean from a run time perspective - how do you know you = Air/Fuel ratio - then there are expensive testing instruments and I = believe some fairly accurate air/fuel ratio meters (based on the newer = broad band O2 sensor) But still several hundred dollars when I last = checked. =20 The narrow band O2 sensor is much cheaper and works just fine with 100 = LL for OUR use. I generally get closer to 200 hours using 100LL before = the sensor appears to lose too much sensitivity to continue to be = useful. The common notion that a few seconds running on leaded fuel = will "Kill" an O2 sensor (at least the narrow band ) is simply not true = - at least not for our use. Doing that WILL apparently degrade it for = its intended use in an automobile fuel system (where it needs to help = the engine computer maintain a 14:1 A/F ratio), but if you just want a = general indication of whether you are lean, rich or in the middle, the = cheap O2 sensor works well. =20 =20 You can even find some narrow band units which will read out A/F in = numeric values but if using a narrow band O2 sensor, I question the = accuracy of such units myself. I suppose you could use a microprocessor = and accurate analog/digital converter and if you had the "Z" curve of = your O2 sensor - you might get close. =20 So basically if you know the mass of air and mass of fuel, you have = your Air/Fuel ratio =20 So how to arrive at those two factors, if you know the air pressure = and temperature (at normal atmospheric values) then you essentially = know the air density from which you can calculate air mass and then = using your engine flow rates and with your chosen Air/Fuel ratio - = calculate your fuel flow, etc. But, frequently it's easier to use our = fuel flow (which can be measured fairly accurately) =20 So one r way to approach the problem is as follows: You know the = displacement of your engine and assuming some Ve (volumetric efficiency) = (85% - 110%) you can calculate your air mass flow through the engine for = any rpm. So how to get an approximation of our volumetric efficiency = (at least at WOT), its fairly simple to get close. =20 Note the ambient atmospheric pressure (manifold gauge pressure without = engine running), fire up your engine and when warmed up advance it to = WOT and note the atmospheric pressure inside your intake (i.e. your = manifold pressure). Lets say ambient pressure is 29.92 inches HG and = lets say you are so lucky to read 29.92 " Hg in manifold pressure - then = theoretically your Ve is 100%. But lets say your design is not perfect = (few are) and your read 28.75" then your Ve is 28.75/29.92 =3D 0.9608 = or 96.08 % Ve -not bad, but not perfect. =20 OK calculate your volumetric flow using our old displacement formulas = and as best I recall at 6000 rpm with a 13B at 100% VE =3D 277 CFM. = Since our intake is not perfect we take our Ve of 0.96.08Ve*277 =3D 266 = CFM actually going through your engine. Recalling that a cubic foot of = air approx =3D 0.0765 lbm/Cubic Foot, we have 266 * 0.0765 =3D 20.36 = lbsm of air per minute.=20 =20 Now we don't know our Air/Fuel ratio - but we do know our fuel flow = at that rpm. Lets say its 16 gallon/hour, turning that in to lbm/min we = have 16 / 60 =3D 0.2666 gallon/min and we know mass of gasoline is = approx 6 - 6.25 lbs/gallon. So taking 6.0 lb/gallon we have 0.2666 * 6 =3D 1.6 lbm/min of fuel based on our fuel flow indication. =20 Now taking both the air mass 20.36 lbm/min and the fuel 1.6 lbm/min = and we get 20.36/1.6 =3D 12.725 air/fuel ratio. =20 Which is very close to the common "Best Power" ratio used by many. =20 So don't know if this answered any of your question - but, best I = could come up with {:>) Ed Ed Anderson Rv-6A N494BW Rotary Powered Matthews, NC eanderson@carolina.rr.com http://www.andersonee.com http://www.dmack.net/mazda/index.html http://www.flyrotary.com/ http://members.cox.net/rogersda/rotary/configs.htm#N494BW http://www.rotaryaviation.com/Rotorhead%20Truth.htm -------------------------------------------------------------------------= ----- From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] = On Behalf Of George Lendich Sent: Sunday, June 14, 2009 6:21 PM To: Rotary motors in aircraft Subject: [FlyRotary] Re: Wire separation? =20 Tracy or anyone, Can you tell me what sensors you use to calculate fuel ratios. I notice the mention of a MAP sensor (vacuum type), but that would = give pressure, not density. I notice the absence of an oxygen sensor in the case of 100LL is used = - this is understood. I suppose an oxygen sensor could be used = independently. I assume RPM would be in the calculations for fuel flow, but I wonder = how you actually arrive at air density? perhaps OAT is used. Basically I wondering what the simplest air density sensor arrangement = would be. I thought Oxygen sensor (not for 100LL), with a RPM and = throttle position sensor. George (down under) =20 Old time distributors and carburetors didn't care how badly you = wired the electrical system. Digital systems are not so forgiving of = poor wiring practices. Noisy power or ground wiring is one of the most = common causes of initial builder problems with computerized equipment = (including ECUs). =20 In addition to the wire bundling answers you got, the basic power = wiring is as or even more important. Here is a basic aircraft wiring = scheme that I recommend in latest version of the installation guide. = Not detailed but somewhere to start. Note the independent power feeds = from the battery for the various systems. This is a physical as well as = electrical representation.=20 Tracy -------------------------------------------------------------------------= --- -- Homepage: http://www.flyrotary.com/ Archive and UnSub: = http://mail.lancaironline.net:81/lists/flyrotary/List.html __________ Information from ESET NOD32 Antivirus, version of virus = signature database 3267 (20080714) __________ The message was checked by ESET NOD32 Antivirus. http://www.eset.com ------=_NextPart_000_0033_01C9EDD5.1DA55040 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
 Thanks Ed,
I will have to work through that a = couple of=20 times.
George

George, not = exactly=20 certain what you mean by =93calculating Air/Fuel ratio=94=20 .

 

I used the = ratio of=20 the mass of the air to the fuel =96 which basically relies on the = approximation=20 of 0.0765 lbs per cubic foot of air.  So you calculate your air = flow in=20 CFM then times  0.0765 to give you the air mass.  The you = can take=20 your desired A/F ratio =96 say 15:1  and divide 15 into your air = mass and=20 that would give you the fuel mass required to achieve that ratio. Of = if you=20 have any two of the three factors you can find the=20 third.

 

IF you mean = from a=20 run time perspective  - how do you know you Air/Fuel ratio - =  then=20 there are expensive testing instruments and I believe some fairly = accurate=20 air/fuel ratio meters (based on the newer broad band O2 sensor)  = But=20 still several hundred dollars when I last=20 checked.

 

The narrow = band O2=20 sensor is much cheaper and works just fine with 100 LL for OUR = use.  I=20 generally get closer to 200 hours using 100LL before the sensor = appears to=20 lose too much sensitivity to continue to be useful.  The common = notion=20 that a few seconds running on leaded fuel will =93Kill=94 an O2 sensor (at least the narrow = band ) is=20 simply not true =96 at least not for our use.  Doing that WILL = apparently=20 degrade it for its intended use in an automobile fuel system (where it = needs=20 to help the engine computer maintain a 14:1 A/F ratio), but if you = just want a=20 general indication of whether you are lean, rich or in the middle, the = cheap=20 O2 sensor works well. 

 

You can = even find=20  some narrow band units which will read out A/F in numeric values = but if=20 using a narrow band O2 sensor, I question the accuracy of such units=20 myself.  I suppose you could use a microprocessor and accurate=20 analog/digital converter and if you had the =93Z=94 curve of your O2 = sensor =96 you=20 might get close.

 

So = basically if you=20 know the mass of air and mass of fuel, you have your Air/Fuel=20 ratio

 

So how to = arrive at=20 those two factors, if you know the air pressure and temperature (at = normal=20 atmospheric  values) then you essentially know the air density = from which=20 you can calculate air mass and then using your engine flow rates and = with your=20 chosen Air/Fuel ratio - calculate your fuel flow, etc.  But, = frequently=20 it=92s easier to use our fuel flow (which can be measured fairly=20 accurately)

 

So one r = way to=20 approach the problem is as follows:  You know the displacement of = your=20 engine and assuming some Ve (volumetric efficiency) (85% - 110%) you = can=20 calculate your air mass flow through the engine for any rpm.  So = how to=20 get an approximation of our volumetric efficiency (at least at WOT), = its=20 fairly simple to get close.

 

Note the = ambient=20 atmospheric pressure (manifold gauge pressure without engine running), = fire up=20 your engine and when warmed up advance it to WOT and note the = atmospheric=20 pressure inside your intake (i.e. your manifold pressure).  Lets = say=20 ambient pressure is 29.92 inches HG and lets say you are so lucky to = read=20 29.92 =93 Hg in manifold pressure - then theoretically your Ve is = 100%. =20 But lets say your design is not perfect (few are) and your read = 28.75=94 =20 then your Ve is 28.75/29.92 =3D 0.9608 or 96.08 % Ve =96not bad, but = not=20 perfect.

 

OK calculate your volumetric flow using = our old=20 displacement formulas and as best I recall at 6000 rpm with a 13B at = 100% VE =3D=20 277 CFM.  Since our intake is not perfect we take our Ve of=20  0.96.08Ve*277 =3D 266 CFM actually going through your = engine. =20 Recalling that a cubic foot of air approx =3D 0.0765 lbm/Cubic Foot, = we have 266=20 * 0.0765 =3D 20.36  lbsm of air per minute. =

 

 Now = we don=92t=20 know our Air/Fuel ratio =96 but we do know our fuel flow at that = rpm.  Lets=20 say its 16 gallon/hour, turning that in to lbm/min we have 16 / 60 =3D = 0.2666=20 gallon/min and we know mass of gasoline is approx 6 =96 6.25 = lbs/gallon. =20 So taking 6.0 lb/gallon  we have

0.2666 * 6 = =3D 1.6=20 lbm/min of fuel based on our fuel flow=20 indication.

 

Now taking = both the=20 air mass 20.36 lbm/min and the fuel 1.6 lbm/min and we  get = 20.36/1.6 =3D=20 12.725 air/fuel ratio.

 

Which is = very close=20 to the common =93Best Power=94 ratio used by = many.

 

So don=92t = know if this=20 answered any of your question =96 but, best I could come up with=20 {:>)

Ed

Ed=20 Anderson

Rv-6A = N494BW Rotary=20 Powered

Matthews,=20 NC

eanderson@carolina.rr.com

http://www.andersonee.com

http://www.dmack.net/mazda/index.html

http://www.flyrotary.com/

http://members.cox.net/rogersda/rotary/configs.htm#N494BW

http://www.r= otaryaviation.com/Rotorhead%20Truth.htm

From:=20 Rotary motors in aircraft = [mailto:flyrotary@lancaironline.net] On=20 Behalf Of George Lendich
Sent: Sunday, June 14, 2009 = 6:21=20 PM
To: = Rotary motors in aircraft
Subject: [FlyRotary] Re: Wire=20 separation?

 

Tracy or=20 anyone,

Can you tell me what = sensors you=20 use to calculate fuel ratios.

I notice the mention of = a MAP=20 sensor (vacuum type), but that would give pressure, not=20 density.

I notice the absence of = an oxygen=20 sensor in the case of 100LL is used - this is understood. I suppose an = oxygen=20 sensor could be used independently.

I assume RPM would be in = the=20 calculations for fuel flow, but I wonder how you actually arrive at=20 air density? perhaps OAT is = used.

Basically I wondering = what the=20 simplest air density sensor arrangement would be. I thought Oxygen = sensor (not=20 for 100LL), with a RPM and throttle position=20 sensor.

George (down=20 under)

 

Old time distributors and carburetors = didn=92t care=20 how badly you wired the electrical system.  Digital systems are = not so=20 forgiving of poor wiring practices.  Noisy power or ground = wiring is=20 one of the most common causes of initial builder problems with = computerized=20 equipment (including ECUs). 

In addition to the wire = bundling=20 answers you got, the basic power wiring is as or even more = important. =20 Here is a basic aircraft wiring scheme that I recommend in latest = version of=20 the installation guide.  Not detailed but somewhere to = start. =20 Note the independent power feeds from the battery for the various=20 systems.  This is a physical as well as electrical = representation.=20

Tracy

--
Homepage: =20 http://www.flyrotary.com/
Archive and UnSub:  =20 = http://mail.lancaironline.net:81/lists/flyrotary/List.html



__________ Information from ESET = NOD32=20 Antivirus, version of virus signature database 3267 (20080714)=20 __________

The message was checked by ESET NOD32 = Antivirus.

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