Return-Path: Sender: (Marvin Kaye) To: lml@lancaironline.net Date: Tue, 25 May 2004 13:20:36 -0400 Message-ID: X-Original-Return-Path: Received: from imo-d06.mx.aol.com ([205.188.157.38] verified) by logan.com (CommuniGate Pro SMTP 4.2b3) with ESMTP id 89226 for lml@lancaironline.net; Tue, 25 May 2004 11:42:27 -0400 Received: from Sky2high@aol.com by imo-d06.mx.aol.com (mail_out_v37_r1.3.) id q.36.59052021 (16781) for ; Tue, 25 May 2004 11:42:22 -0400 (EDT) From: Sky2high@aol.com X-Original-Message-ID: <36.59052021.2de4c35e@aol.com> X-Original-Date: Tue, 25 May 2004 11:42:22 EDT Subject: Ram Air And Injector Consequences X-Original-To: lml@lancaironline.net MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="-----------------------------1085499742" X-Mailer: 9.0 for Windows sub 910 -------------------------------1085499742 Content-Type: text/plain; charset="UTF-8" Content-Transfer-Encoding: quoted-printable Content-Language: en George, Walter, et al,=20 It has been suggested that a 1.5 psi differential between the air supply to= =20 an injector and the deck pressure (manifold) would optimize atomization of=20 the fuel.=20 Verrrrry Interrrrresting!=20 Factoids:=20 Let=E2=80=99s remember that 1.5 psi is approximately equal to 3=E2=80=9D Hg= . At sea level=20 standard conditions, 29.92=E2=80=9D Hg is approximately 14.7 psi (one atmos= phere). =20 Theoretically, at 200 Kts a natural ram air system can raise the pressure=20 about 2=E2=80=9D Hg (1 psi, Lancair 320/360) and at 245 Kts the rise is abo= ut 3=E2=80=9D Hg=20 (1.5 psi, Lancair Legacy).=20 An effective air cooled engine cooling system should see a pressure drop=20 between the plenum and the exit air. Some reasonable estimates are 5 to 7= =20 inches of H2O (.37 to .51 inches of Hg or .18 to .25 psi). While the intake= =20 plenum is not a high pressure chamber (sealed plenums notwithstanding), the= plenum=20 pressure (in Lancairs) is higher than ambient =E2=80=93 I couldn=E2=80=99t=20= find my old=20 measurements =E2=80=93 For now, let=E2=80=99s use .5=E2=80=9D Hg at cruise.= =20 How could a naturally aspirated engine in a commercially built airplane=20 achieve the optimal pressure differential? This assumes no ram air, filte= red=20 induction air and perhaps (as in the case of some Continentals) the air is=20 drawn from the engine cooling plenum thus making no pressure differential a= t WOT.=20 =20 Answer: Reduced throttle operation. For example, a Continental powered=20 Skymaster operating under std. conditions at 3000 feet MSL (approx 26=E2= =80=9D Hg) and=20 2500 RPM with the throttle reduced to 23=E2=80=9D MAP would provide the des= ired=20 pressure differential (1.5 psi) and be running at about 75% power.=20 Those of us flying Lancairs with good cooling plenums and taking advantage=20 of ram air boosts to the induction system whilst also always using WOT are=20= not=20 optimizing fuel atomization at the injector. As a matter of fact, we are=20 pushing back against the air the injector is trying to utilize. Some of us=20 have neutralized this problem by using shrouded injectors to which we have=20 supplied the same type of ram pressure boosted air. This is not air from=20= the=20 post throttle-body induction system since this would steal air from the=20 induction system much like a leak and always operate with no differential.=20 Using my Lancair and the above example, 3000=E2=80=99 MSL, 2500 RPM, 23=E2= =80=9D MAP and=20 about 170 Kt cruise, the pressure differential would probably be closer to=20= 2=20 psi because the injectors are receiving ram pressure boosted air. =20 At 8000=E2=80=99 (21=E2=80=9D Hg), 2500 RPM, WOT, 180 KIAS, 23=E2=80=9D MAP= , the pressure=20 differential is zip because of the WOT =E2=80=93 but it isn=E2=80=99t negat= ive. Note that the =20 engine thinks it=E2=80=99s operating at 6000=E2=80=99 whilst the wings are a= t 8K and I=E2=80=99m still =20 about 73% power.=20 Solutions to get constant 1.5 psi differential:=20 1. Electric air pump with regulator.=20 2. Always fly at less than WOT.=20 3. ?=20 4. Don=E2=80=99t worry about it.=20 5. Get Honda Motorcycle 12 port electronic injectors that don=E2=80= =99t need no=20 stink=E2=80=99n air.=20 6. Get a Turbo-Prop.=20 7. Get a glider.=20 Scott Krueger AKA Grayhawk Sky2high@aol.com II-P N92EX IO320 Aurora, IL (KARR) LML, where ideas collide and you decide! -------------------------------1085499742 Content-Type: text/html; charset="UTF-8" Content-Transfer-Encoding: quoted-printable Content-Language: en

 

George, Walter, et al,

 

It has been suggested tha= t a 1.5=20 psi differential between the air supply to an injector and the deck pressure= =20 (manifold) would optimize atomization of the fuel.

 

Verrrrry Interrrrresting!=

 

Factoids:

Let=E2=80=99s remember th= at 1.5 psi is=20 approximately equal to 3=E2=80=9D Hg. &nbs= p;At=20 sea level standard conditions, 29.92=E2=80=9D Hg is approximately 14.7 psi (= one=20 atmosphere). 

 

Theoretically, at 200 Kts= a=20 natural ram air system can raise the pressure about 2=E2=80=9D Hg (1 psi, La= ncair=20 320/360) and at 245 Kts the rise is about 3=E2=80=9D Hg (1.5 psi, Lancair Le= gacy).

 

An effective air cooled e= ngine=20 cooling system should see a pressure drop between the plenum and the exit=20 air.  Some reasonable estimate= s are=20 5 to 7 inches of H2O (.37 to .51 inches of Hg or .18 to .25 psi).  While the intake plenum is not a h= igh=20 pressure chamber (sealed plenums notwithstanding), the plenum pressure (in=20 Lancairs) is higher than ambient =E2=80=93 I couldn=E2=80=99t find my old me= asurements =E2=80=93 For=20 now, let=E2=80=99s use .5=E2=80=9D Hg at cruise.

 

How could a naturally asp= irated=20 engine in a commercially built airplane achieve the optimal pressure=20 differential?  This assumes no= ram=20 air, filtered induction air and perhaps (as in the case of some Continentals= )=20 the air is drawn from the engine cooling plenum thus making no pressure=20 differential at WOT. 

 

Answer: Reduced throttle=20 operation.  For example, a=20 Continental powered Skymaster operating under std. conditions at 3000 feet M= SL=20 (approx 26=E2=80=9D Hg) and 2500 RPM with the throttle reduced to 23=E2=80= =9D MAP would provide=20 the desired pressure differential (1.5 psi) and be running at about 75%=20 power.

 

Those of us flying Lancai= rs with=20 good cooling plenums and taking advantage of ram air boosts to the induction= =20 system whilst also always using WOT are not optimizing fuel atomization at t= he=20 injector.  As a matter of fact= , we=20 are pushing back against the air the injector is trying to utilize.  Some of us have neutralized this p= roblem=20 by using shrouded injectors to which we have supplied the same type of ram=20 pressure boosted air.  This is= not=20 air from the post throttle-body induction system since this would steal air=20= from=20 the induction system much like a leak and always operate with no=20 differential.

 

Using my Lancair and the=20= above=20 example, 3000=E2=80=99 MSL, 2500 RPM, 23=E2=80=9D MAP and about 170 Kt cruis= e, the pressure=20 differential would probably be closer to 2 psi because the injectors are=20 receiving ram pressure boosted air.     

 

At 8000=E2=80=99 (21=E2= =80=9D Hg), 2500 RPM, WOT,=20 180 KIAS, 23=E2=80=9D MAP, the pressure differential is zip because of the W= OT =E2=80=93 but it=20 isn=E2=80=99t negative.  Note=20= that the=20 engine thinks it=E2=80=99s operating at 6000=E2=80=99 whilst the wings are a= t 8K and I=E2=80=99m still=20 about 73% power.

 

Solutions to get constant= 1.5 psi=20 differential:

 

  1. Electric=20 air pump with regulator.
  2. Always=20 fly at less than WOT.
  3. ?
  4. Don=E2=80=99t=20 worry about it.
  5. Get=20 Honda Motorcycle 12 port electronic injectors that don=E2=80=99t need no s= tink=E2=80=99n=20 air.
  6. Get=20 a Turbo-Prop.
  7. Get=20 a glider.
    8.  
    Scott=20 Krueger AKA Grayhawk
    Sky2high@aol.com
    II-P N92EX IO320 Aurora, IL=20 (KARR)

    LML, where ideas collide and you=20 decide!
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