X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Wed, 28 Dec 2005 22:50:25 -0500 Message-ID: X-Original-Return-Path: Received: from imo-d03.mx.aol.com ([205.188.157.35] verified) by logan.com (CommuniGate Pro SMTP 5.0.5) with ESMTP id 903997 for lml@lancaironline.net; Wed, 28 Dec 2005 11:45:11 -0500 Received-SPF: pass receiver=logan.com; client-ip=205.188.157.35; envelope-from=REHBINC@aol.com Received: from REHBINC@aol.com by imo-d03.mx.aol.com (mail_out_v38_r6.3.) id q.1c4.374a7cb3 (48600) for ; Wed, 28 Dec 2005 11:44:21 -0500 (EST) From: REHBINC@aol.com X-Original-Message-ID: <1c4.374a7cb3.30e41ae5@aol.com> X-Original-Date: Wed, 28 Dec 2005 11:44:21 EST Subject: Re: [LML] Re: Where has all the power gone? X-Original-To: lml@lancaironline.net MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="-----------------------------1135788261" X-Mailer: 9.0 for Windows sub 5120 X-Spam-Flag: NO -------------------------------1135788261 Content-Type: text/plain; charset="US-ASCII" Content-Transfer-Encoding: 7bit Walter, I hope Santa was good to you. **. As rpm and/or combustion duration approach zero the ideal theta PP approaches TDC. As theta PP drifts away from TDC the engine becomes less efficient.** Unless I'm missing something about your point, I don't think that's quite correct. The optimum thetaPP for maximum mechanical advantage is always about 15-16dATDC. By definition and measurement, approximately half of the charge is burned at the peak pressure. If I've misunderstood, please try again! Thanks. And, Merry Christmas and Happy Holidays to you all. Walter Consider a perfect ottocycle engine with perfect fuel. Mechanically the engine can survive whatever pressure the fuel can generate and the fuel burns infinitely fast. In this perfect world, with no cylinder leakage and no heat transfer, the cylinder pressure at any theta after ignition on the power stroke is always the same, no matter where ignition takes place. If the fuel is ignited at TDC, then the entire charge is combusted at TDC and Theta PP is also at TDC. Now, the piston sees the maximum possible pressure force at every inch of downward travel. Energy is force times distance so this has to extract the most shaft power from the fuel charge. If we ignite the fuel prior to TDC, then theta PP still occurrs at TDC. We still see the same energy extracted from the fuel charge as we did at TDC, however we now have to do additional work on the compression side due to the higher presure of the precombusted fuel. Thus the net shaft power of the engine is reduced. Next, consider the condition where ignition and theta PP are delayed until ATDC. During the period of crank rotation between TDC and ignition the piston is driven with only compression pressure. As this pressure, and the resulting piston force, is much lower than the combustion pressure would have been, less energy is imparted to the piston over this distance and shaft power is reduced accordingly. In the real world, fuels take time to combust. If the engine turns relatively slowly, then the combustion occurs over such a short crank angle that both ignition and theta PP will occur at essentialy the same point and optimum theta PP is still essentially TDC. As the engine turns faster, the fuel burn takes place over a larger crank angle and theta PP is delayed many degrees after ignition. If the crank speed is fast we must start making make compromises. To extract the most energy from the combusted fuel during the power stroke, theta PP should still be at TDC. Unfortunately, nearly all of the fuel would have to be burned BTDC and the addtional compression pressure would absorb substantially more energy on the compression stroke, greatly reducing the net shaft power. Accordingly, ignition is timed to provide for a minimal amount of combustion BTDC to mimimize the compression stroke power requirement and a minimal theta PP to maximize power stroke power. The faster the engine turns, the bigger the compromise. Of course other factors enter into the equation, like heat transfer, bearing friction, valve timing, fuel properties, cylinder temperature and, yes, even ignition energy just to name a few. Consequently, it should be clear that there is no universal value for optimal theta PP or even the ratio of burned to unburned fuel at theta PP. These values change for different engines as well as different speeds on the same engine. Hope this clears up the issue for you. Rob -------------------------------1135788261 Content-Type: text/html; charset="US-ASCII" Content-Transfer-Encoding: quoted-printable
Walter,
 
I hope Santa was good to you.
**. As rpm and/or combustion duration approach= zero the ideal theta PP
approaches TDC. As theta PP drifts away from TD= C the engine becomes
less efficient.**

Unless I'm missing somethi= ng about your point, I don't think that's
quite correct.  The optim= um thetaPP for maximum mechanical advantage is
always about 15-16dATDC.&= nbsp; By definition and measurement, approximately
half of the charge is= burned at the peak pressure.  If I've
misunderstood, please try ag= ain!

Thanks.  And, Merry Christmas and Happy Holidays to you all= .

Walter
Consider a perfect ottocycle engine with perfect fuel. Mechanically the= engine can survive whatever pressure the fuel can generate and the fuel bur= ns infinitely fast. In this perfect world, with no cylinder leakage and no h= eat transfer, the cylinder pressure at any theta after ignition on the=20= power stroke is always the same, no matter where ignition takes place. = If the fuel is ignited at TDC, then the entire charge is combusted at TDC an= d Theta PP is also at TDC. Now, the piston sees the maximum possible pressur= e force at every inch of downward travel. Energy is force times distanc= e so this has to extract the most shaft power from the fuel charge.
 
If we ignite the fuel prior to TDC, then theta PP still occurrs at TDC.= We still see the same energy extracted from the fuel charge as we did=20= at TDC, however we now have to do additional work on the compression side du= e to the higher presure of the precombusted fuel. Thus the net shaft po= wer of the engine is reduced.
 
Next, consider the condition where ignition and theta PP are=20= delayed until ATDC. During the period of crank rotation between TDC and= ignition the piston is driven with only compression pressure. As this=20= pressure, and the resulting piston force, is much lower than the combustion=20= pressure would have been, less energy is imparted to the piston over this di= stance and shaft power is reduced accordingly.
 
In the real world, fuels take time to combust. If the engine turns rela= tively slowly, then the combustion occurs over such a short crank angle that= both ignition and theta PP will occur at essentialy the same point and opti= mum theta PP is still essentially TDC. As the engine turns faster, the=20= fuel burn takes place over a larger crank angle and theta PP is delayed many= degrees after ignition.
 
If the crank speed is fast we must start making make compromises. To ex= tract the most energy from the combusted fuel during the power stroke, theta= PP should still be at TDC. Unfortunately, nearly all of the fuel would have= to be burned BTDC and the addtional compression pressure would absorb=20= substantially more energy on the compression stroke, greatly reducing the ne= t shaft power. Accordingly, ignition is timed to provide for a minimal amoun= t of combustion BTDC to mimimize the compression stroke power requirement&nb= sp;and a minimal theta PP to maximize power stroke power. The faster the eng= ine turns, the bigger the compromise.
 
Of course other factors enter into the equation, like heat transfer, be= aring friction, valve timing, fuel properties, cylinder temperature and, yes= , even ignition energy just to name a few. Consequently, it should be clear=20= that there is no universal value for optimal theta PP or even the ratio= of burned to unburned fuel at theta PP. These values change for different e= ngines as well as different speeds on the same engine.
 
Hope this clears up the issue for you.
 
Rob
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